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https://math.stackexchange.com/questions/2355252/let-tm-1v-neq-0-but-tmv-0-show-that-v-tv-ldots-tm-1v-are-linearly | # Let $T^{m-1}v\neq 0$ but $T^mv=0$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent
Let $T\in\mathcal L(V)$ and $T^{m-1}v\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\in V$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent.
I had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:
1. It is the proof below correct?
2. If so, how I can write it better using the same ideas?
### The attempted proof:
1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\lambda\neq 0$ such that
$$T^{m-1}v=\lambda T^{m-2}v\implies T^mv=\lambda T(T^{m-2}v)=\lambda T^{m-1}v=0\implies \lambda=0$$
Then $T^{m-2}v$ is linearly independent of $T^{m-1}v$.
2) Now observe that
$$\lambda_1v+\lambda_2Tv+\lambda_3T^2v=0\implies T^{m-2}(\lambda_1v+\lambda_2Tv+\lambda_3T^2v)=\lambda_1T^{m-2}v+\lambda_2T^{m-1}v=0$$
so $\lambda_1,\lambda_2=0$ as we had shown previously, so the original equation reduces to
$$\lambda_3T^2v=0\implies \lambda_3=0$$
thus $v,Tv,T^2v$ are linearly independent.
3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\ldots,T^{m-1}v$ is linearly independent.
• This is a related post – Bijesh K.S Jul 11 '17 at 19:15
• The if and only if in the first line looks problematic as $T$ isn't invertible. – DMath Jul 11 '17 at 19:19
• @DMath thank you... I overlook it completely. – Masacroso Jul 11 '17 at 19:20
• I'm afraid that the last part is the bulk of the proof and you just cut it off by saying “repeating the analysis”. Note that for $\{v,Tv,T^2v,T^3v\}$ you need to go down to $T^{m-3}$, which is not as easy as the initial step for $\{T^{m-2}v,T^{m-1}v\}$. – egreg Jul 11 '17 at 20:33
Hint
$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0 \quad(1)$$
multiply by $T^{m-1}$, then
$$a_0T^{m-1}v=0\to a_0=0$$
so, from $(1)$,
$$a_1Tv+\ldots+a_{m-1}T^{m-1}v=0$$
multiply by $T^{m-2}$, then
$$a_1T^{m-1}v=0\to a_1=0$$
can you finish?
• ok but, it is my proof correct? – Masacroso Jul 11 '17 at 19:17
• @Masacroso: it seems correct to me. You just need induction to finish it. – Arnaldo Jul 11 '17 at 19:19
See below for a simple proof.
Suppose that they are not linear independent; that is for some choice of $\alpha_{k}$,
$T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k}v$.
Then $0 =T^{m}v = T\cdot T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k+1}v$,
so $Tv,...,T^{m-1}v$ are linearly dependent.
But that means that $T^{m-1}v = \sum_{k=1}^{m-2}\beta_{k}T^{k}v$.
Inductively repeat the argument to eventually conclude that for some $\gamma \neq 0$, $T^{m-1}v = \gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v = \gamma T^{m-1} v \neq 0$, a contradiction.
More simply put: if you had a linear combination $c_0 v + c_1 T v + \ldots c_{m-1} T^{m-1} v = 0$ with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient. Then $$T^i v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$ Applying $T^{m-i-1}$ to both sides, $$T^{m-1} v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$ But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction! | 2021-04-20T19:03:38 | {
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http://mathhelpforum.com/calculus/166923-couple-trigonometric-integrals.html | # Thread: a couple of trigonometric integrals
1. ## a couple of trigonometric integrals
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
2. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
Here's how you do the first one:
Spoiler:
Let
$\displaystyle \displaystyle I=\int_0^\pi x\sin\left(\cos^2(x)\right)\cos^2(\sin(x))\text{ }dx$
Then, letting $\displaystyle z=\pi-x$ we find that
\displaystyle \displaystyle \begin{aligned}I &=\int_0^\pi (\pi-z)\sin\left(\cos^2(z)\right)\cos\left(\sin^2(z)\ri ght)\text{ }dz\\ &=\pi\int_0^\pi\sin\left(\cos^2(z)\right)\cos\left (\sin^2(z)\right)\text{ }dz-I\end{aligned}
Thus,
$\displaystyle \displaystyle \frac{2}{\pi}I=\int_0^\pi \sin\left(\cos^2(z)\right)\sin\left(\cos^2(z)\righ t)\text{ }dz$
Recall though that
$\displaystyle \displaystyle \sin(\phi)\cos(\theta)=\frac{\sin(\phi+\theta)+\si n(\phi-\theta)}{2}$
And thus
\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi} I &= \int_0^\pi \sin\left(\cos^2(z)\right)\cos^2\left(\sin(z)\righ t)\text{ }dz\\ &= \frac{1}{2}\int_0^\pi\left(\sin\left(\cos^2(z)+\si n^2(z)\right)+\sin\left(\cos^2(z)-\sin^2(z)\right)\right)\text{ }dz\\ &=\frac{1}{2}\pi\sin(1)+\frac{1}{2}\int_0^\pi \sin(\cos(2z))\text{ }dz\end{aligned}
Note though that
$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^\pi\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))\text { }dz$
Note though that by letting $\displaystyle \displaystyle u=z-\frac{3\pi}{4}$ one gets that
$\displaystyle \displaystyle \int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))=\int_0^{ \frac{\pi}{4}}\sin(\cos(2u))\text{ }du$
And thus
$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{\pi}{4}}^{\frac{3\pi}}{4}\sin(\cos (2z))\text{ }dz$
But, letting $\displaystyle w=z-\frac{\pi}{2}$ in this second integral one arrives at
$\displaystyle \displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z) )=-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$
and since the integrand is even this is equal to
$\displaystyle \displaystyle -2\int_0^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$
Therefore,
\displaystyle \displaystyle \begin{aligned}\int_0^\pi\sin(\cos(2z))\text{ }dz &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))+\int_{\frac{ \pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz-2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &=0\end{aligned}
Thus, finally we arrive at
\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi}I &=\frac{\pi}{2}\sin(1)+\frac{1}{2}\int_0^\pi\sin(\ cos(2z))\text{ }dz\\ &=\frac{\pi}{2}\sin(1)\end{aligned}
and thus $\displaystyle \displaystyle I=\frac{\pi^2 \sin(1)}{4}$.
3. The ODE you had was solvable.
4. Originally Posted by Random Variable
The ODE you had was solvable.
See my new method. It's more elementary.
5. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
The second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet?
6. to prove that $\displaystyle \displaystyle\int_0^\pi\sin(\cos 2x)\,dx=0,$ split it up into two integrals with $\displaystyle 0\le x\le\dfrac\pi2$ and $\displaystyle \dfrac\pi2\le x\le\pi$ then apply substitution $\displaystyle t=x-\dfrac\pi2$ on the second piece and you'll see is equal to minus the first piece, hence, the result.
7. Here's my solution to the second
Spoiler:
First use the double angle formula so
$\displaystyle \displaystyle \int_0^{\pi} x\left( \dfrac{1-\cos \left(2 \sin x\right)}{2} + \dfrac{1 + \cos \left(2 \cos x\right)}{2} \right)\, dx$
$\displaystyle \displaystyle = \int_0^{\pi} x\,dx + \frac{1}{2}\int_0^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$
We will show the second integral is zero by considering it in two parts. (**)
First, that $\displaystyle \displaystyle \int_0^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Let $\displaystyle x = \pi -u$ in $\displaystyle \displaystyle \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Next that $\displaystyle \displaystyle \int_0^{\pi/4} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = -\int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Let $\displaystyle x = \pi/2-u$ in$\displaystyle \displaystyle \int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$
Thus, we are left with $\displaystyle \displaystyle \int_0^{\pi} x\,dx = \dfrac{\pi^2}{2}$
8. here's mine to the second:
put $\displaystyle t=\pi-x$ on the integral to see that is equal to $\displaystyle \displaystyle\frac\pi2\int_0^\pi\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx.$
i claim the latter integral equals $\displaystyle \pi,$ hence the original integral equals $\displaystyle \dfrac{\pi^2}2,$ so in order to see that write the integral as
$\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx},$
and put $\displaystyle t=x-\dfrac\pi2$ on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.
another solution for the first one:
we consider $\displaystyle \displaystyle\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ (1) then write it as $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}$ and put $\displaystyle t=x-\dfrac\pi2$ on the second one then we get $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)\,dx},$ so when adding those we note that the integrand is actually $\displaystyle \sin(\sin^2x+\cos^2x)$ so (1) equals $\displaystyle \dfrac\pi2\sin1.$
on the original integral was easy to show that it's equal to $\displaystyle \displaystyle\frac\pi2\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ thus by (1) we conclude. | 2018-04-26T10:34:45 | {
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https://math.stackexchange.com/questions/1470160/combinations-problem-choosing-ways-to-select-8-questions-out-of-12 | Combinations problem: Choosing ways to select $8$ questions out of $12$.
In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions?
Please don't answer with the three case method. What I want is the error in my method?
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3}$$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2}$$. Because 6 questions are left to choose from. But this does not give the answer, why?
As you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is $$\binom{5}{3}\binom{7}{5} + \binom{5}{4}\binom{7}{4} + \binom{5}{5}\binom{7}{3}$$ The alternative method you proposed of selecting three questions from part A, three questions from part B, and two additional questions counts the same selection more than once.
If we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part B.
To make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times. \begin{align*} & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, \color{blue}{B_3}, B_4, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, B_2, \color{blue}{B_3}, \color{blue}{B_4}, \color{blue}{B_5}\\ \end{align*}
By similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part A.
If we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions.
Note that $$10\binom{5}{3}\binom{7}{5} + 16\binom{5}{4}\binom{7}{4} + 10\binom{5}{5}\binom{7}{3} = \binom{5}{3}\binom{7}{3}\binom{6}{2}$$
• I've edited my answer to make what I'm saying more concrete. – N. F. Taussig Oct 8 '15 at 11:43
• Okay, we see that this is happening. But why is this happening? Can we add something so that my method works? – Aditya Agarwal Oct 8 '15 at 11:45
• As you can see, the method you attempted to use does not work since it counts the same set of selected questions every time you choose $3$ of the selected questions. To avoid overcounting, we use cases. I do not see a way to make the method you attempted work. – N. F. Taussig Oct 8 '15 at 11:55
You are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.
We can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\3\end{matrix}\right)=1\cdot 35=35$.
Your way of solving this question would however say that there are $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\1\end{matrix}\right)\left(\begin{matrix}6\\2\end{matrix}\right)=1\cdot 7\cdot 15=105$ possibilities of choosing.
Edit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\cdot3=105$. | 2020-06-01T22:47:19 | {
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https://math.stackexchange.com/questions/2015361/solution-of-a-given-linear-diophantine-equation | # Solution of a given linear diophantine equation.
I was solving the following linear diophantine equation :
$56x + 72y = 40$ in integers.
My attempt: I got that 8 is the gcd of 56 and 72 and $8|40$ and hence a solution exists and I can write:
$8 = 56 - 16 *3$
$\implies$ $8= 56 - (72 -56*1)*3$
$\implies$ $8= 4*56 - 3*72$.
So my answer is $x = 4$ and $y = -3$. But in book its showing $x = 20$ and $y = -15$. Where I went wrong? Kindly help.
• Remember that $20=5 \times 4$ and $-15=5 \times 3$. Both answers are correct
– user284001
Nov 15, 2016 at 16:49
• @Bacon, So why did they multiply by 5? Nov 15, 2016 at 16:50
• At the outset you noted that $8|40$ and this is...
– user284001
Nov 15, 2016 at 16:50
$56x+72y=40$
$7x+9y=5$
Select the term with the smaller of the two coëfficients and isolate it on the LHS.
$7x=5-9y$
$\text{[1] }x=\frac{5-9y}7=1-y+\frac{-2-2y}7$
$\text{New variable }a=\frac{-2-2y}7$
$-2-2y=7a$
$-2-7a=2y$
$\text{[2] }y=-\frac{2+7a}2=-1-4a+\frac{a}2$
$\text{New variable }b=\frac{a}2$
$a=2b$
Substitute $a\text{ into [2].}$
$y=-\frac{2+7(2b)}2=-\frac{2+14b}2=-1-7b$
Substitute $y\text{ into [1].}$
$x=\frac{5-9(-1-7b)}7=\frac{5+9+63b}7=\frac{14+63b}7=2+9b$
$\text{So, }(x,y)=(2+9b,-1-7b).$
There is an infinitude of integer answers; but your textbook, for some reason, favors $b=2.$
write the equation as $$56x\equiv 40\mod 72$$ or $$x\equiv \frac{40}{76}=\frac{5}{7}\equiv \frac{77}{7}\equiv 11 \mod 72$$ thus $$x=11+72K$$ with $$k \in \mathbb{Z}$$
You can solve $56x+72y=40$ using convergent.
Note: If $\alpha,\beta$ are integral solutions to $ax+by=c$ for $x,y$ respectively, the other solutions take the form$$x=\alpha-bt\\ y=\beta+at\tag{1}$$
Here, $a=56,b=72$ so the $x$ values model the form $\alpha-72t$ and $\beta+56t$ for $y$.
The nearest converging fraction to $\dfrac {9}{7}$ is $\dfrac 43$ and hence, we have$$7\cdot 4-3\cdot 9=1$$
For which we can multiply both sides by $40$ to get$$40\cdot 7\cdot 4-3\cdot 9\cdot 40=40\implies 56\cdot 20+72(-15)=40\tag2$$ Thus, we have $\alpha=20,\beta=-15$. And by $(1)$, we get$$x=20-72t\\y=-15+56t$$
Note that we also get the solutions that the book says along the way! | 2022-09-29T23:02:28 | {
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https://math.stackexchange.com/questions/2483136/the-probability-that-it-is-a-woman-or-a-man-who-is-not-bringing-a-dessert | # The probability that it is a woman, or a man who is not bringing a dessert?
John invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability that it is a woman, or a man who is not bringing a dessert?
I know that the two events are mutually exclusive; so that, the answer is simply $6/12 + 5/12 = 11/12$. However, when I looked at the answer in the book that I study from, I found the same result but an approach that's really weird for me. Here is it literally:
$$P(woman) = 6/12 = 1/2$$ $$P(not\ bringing\ a\ dessert\ ) = 10/12 = 5/6$$ $$P(woman\ and\ not\ bringing\ a\ dessert) = 1/2\ *\ 5/6=5/12$$ $$P(woman\ or\ a\ man\ not\ bringing\ a\ dessert) = 1/2 + 5/6 - 5/12 = 11/12$$
May anyone explain the reasoning behind that approach? And is it technically right?
Your approach is certainly right. The book's approach is right as well. It is just that they should have said $P(\text{woman OR man not bringing dessert}) = P(\text{woman OR not bringing dessert})$ (this works because there are just two genders). Then you don't have mutually exclusive events, so you have to use inclusion-exclusion. $$P(\text{woman OR not bringing dessert}) = P(\text{woman}) + P(\text{not bringing dessert}) - P(\text{woman} \cap \text{not bringing dessert}).$$
The term $P(\text{woman} \cap \text{not bringing dessert}) = 5/12$, thus they have $1/2 + 10/12 - 5/12 = 11/12$.
What you are doing is you are breaking up your space into mutually exclusive events. You are saying it is equivalent to calculating $P(\text{woman} \sqcup \left(\text{not woman AND no dessert}\right))$. That way you have $1/2 + 5/12$.
There's another way you can solve this problem. This probability is equal to $1 - P(\text{man bringing dessert}) = 1 - 1/12 = 11/12$.
NOTE: Edited after a misinterpretation pointed out by @Abdu Magdy
• But P(man not bringing a dessert) should have equalled $5/12$ not $10/12$, Consequently, P(woman ∩ man not bringing dessert) should have equalled $5/12 \ *\ 1/2 = 5/24$ – Abdu Magdy Oct 21 '17 at 18:52
• I interpret $P(\text{man not bringing a dessert})$ as $P(\text{not bringing a dessert} | \text{man}) = \frac{P(\text{not bringing dessert} \cap \text{man})}{P(\text{man})} = \frac{\frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$, but YMMV. In my experience with GRE problems, sometimes the language is indeed ambiguous. – Abhiram Natarajan Oct 21 '17 at 19:02
• I'm sorry but I really need to understand... Based on the definition of mutually inclusive events; if we only are to choose one person, May we choose a woman and a man not bringing a dessert at the same time? – Abdu Magdy Oct 21 '17 at 19:23
• >> May we choose a man ... same time? You certainly cannot choose that way. It's more like when you have "not bringing dessert", it can be either man or woman. Let's try this way. They should have said $P(\text{woman OR not bringing dessert})$. That way, you don't have mutually exclusive events. So by inclusion exclusion, you have $1/2 + 10/12 - 5/12$. I said this wrong in my answer. I shall edit it right away. – Abhiram Natarajan Oct 21 '17 at 20:41
• Sorry about the confusion. – Abhiram Natarajan Oct 21 '17 at 21:00 | 2020-02-17T06:48:39 | {
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http://math.stackexchange.com/questions/255934/negating-zach-blocks-e-mails-and-texts-from-jennifer/255940 | Negating “Zach blocks e-mails and texts from Jennifer”
I am reviewing some basic propositional logic. The question that I have come across that has given some confusion is Zach blocks e-mails and texts from Jennifer where I am asked to find the negation of this proposition. Here is what they provided as an answer in the book: Zach does not block e-mails from Jennifer, or he does not block texts from Jennifer.
Why couldn't the answer simply be It is not true that Zach blocks e-mails and texts from Jennifer? Why did they have to introduce the disjunction or?
-
Of course your answer is fine, indeed clearer. They presumably want to test whether you know how to negate a conjunction and obtain a disjunction. But if this is a test question, surely you should not be expected to be a mind reader. – André Nicolas Dec 11 '12 at 2:00
What you gave is one possible answer, the book gives another which is logically equivalent, although slightly cleaner statement since it is further reduced. They are the same because of one of De-Morgan's laws; $\neg(P \land Q) \equiv (\neg P) \lor (\neg Q)$
-
Zach blocks emails and texts from Jennifer means Zach blocks emails from Jennifer and Zach blocks texts from Jennifer. Negating this gives Zach does not block emails from Jennifer or Zach does not block texts from Jennifer.
It is true that the negation is also It is not true that Zach blocks emails and texts from Jennifer but the point of the exercise is to make one rewrite it in a useful way.
Note that in general, (not (A and B)) is logically equivalent to ((not A) or (not B)).
-
Or does "Zach blocks emails and texts from Jennifer" mean "Zach blocks emails and Zach blocks texts from Jennifer"? Don't you love the ambiguities of natural language? (Of course, the translation you give is almost certainly the intended meaning.) – Code-Guru Dec 11 '12 at 4:11
• $P(x, y)$: x blocks emails from y
• $Q(x, y)$: x blocks texts from y
• $z$: Zach
• $j$: Jennifer
$P(z, j)$: Zach blocks emails from Jennifer;
$Q(z, j)$: Zach blocks texts from Jennifer.
You essentially negated: $P(z, j) \land Q(z, j)$.
So did the text.
The text applied "distribution of negation over conjunction" (one of DeMorgan's Laws):
• "It is not the case that $[(P(z, j)$ and $Q(z, j)]$" $$\iff \lnot[P(z, j) \land Q(z, j)]$$ $$\iff \lnot P(z, j) \lor \lnot Q(z, j)\tag{DeMorgan's law}$$
• which is to say: not $P(z, j)$, or, not $Q(z, j)$.
You are correct with your statement that the sentence translates to $\;\;\lnot[P(z,j) \land Q(z, j)]$;
It's logically equivalent to the text's answer: $\;\lnot P(z, j) \lor \lnot Q(z, j)$. | 2014-10-31T06:26:23 | {
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http://mathhelpforum.com/number-theory/91624-divisibility-7-a.html | # Math Help - Divisibility by 7
1. ## Divisibility by 7
Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.
I set about trying to show that this expression is equal to $0 \mod 7$.
I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$
The question also has Fermat's little theorem above it so I think I should be using that somewhere.
Anyone have any ideas?
2. Hello,
Let's take for example $2222^{5555}$
It's congruent to $3^{5555}(\bmod 7)$
Now, you know by Fermat's little theorem, that $3^{6}\equiv 1(\bmod 7)$
As you noticed, $5555\equiv 5(\bmod 6)$
It can be proved that hence $2222^{5555}\equiv 3^5 (\bmod 7)$
Why ? Because $5555=6k+5$
And then $3^{6k+5}=(3^6)^k\cdot 3^5 \dots$
3. Originally Posted by Showcase_22
Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.
I set about trying to show that this expression is equal to $0 \mod 7$.
I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$
The question also has Fermat's little theorem above it so I think I should be using that somewhere.
Anyone have any ideas?
Hi Showcase_22.
Fermat’s little theorem is a little cumbersome for this problem. Here is a shorter way.
$2222\equiv3\,\bmod7\ \implies\ 2222^3\equiv-1\,\bmod7$ $\implies\ 2222^{5555}=2222^{3\cdot1851+2}\equiv-3^2\,\bmod7\equiv5\,\bmod7$
$5555\equiv4\,\bmod7\ \implies\ 5555^3\equiv1\,\bmod7$ $\implies\ 5555^{2222}=5555^{3\cdot740+2}\equiv4^2\,\bmod7\eq uiv2\,\bmod7$
Hence $2222^{5555}+5555^{2222}\equiv7\,\bmod7\equiv0\,\bm od7.$
The reason I prefer $2222^3$ and $5555^3$ rather than $2222^6$ and $5555^6$ is that their powers can be brought closer this way to 5555 and 2222 respectively.
4. Hello,
Hello!
I finally got it to work out using TheAbstractionist's method (I looked at what you did, did it for 2222 whilst looking at yours and then did it for 5555 separately).
I'll have a go using your method now Moo since it seems to be the one in the mark scheme.
5. Okay, here's my way of doing it. I think it's both of your methods combined (or I may just be copying one of your methods, i'm not that good at number theory!):
$2222^{5555}=3^{5555} \bmod 7$
$3^6=1 \bmod 7$ by Fermat's little theorem
$3^{5555} \bmod 7=3^{6(925)+5}=3^{6(925)}3^5 \bmod 7$
$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)=(1 \bmod 7 )(5 \bmod 7)=5 \bmod 7$
$\Rightarrow 2222^{5555}=5 \bmod 7$
$5555^{2222}=(5 \bmod 6)^{2222}$
$4^6=1 \bmod 7$ by Fermat's little theorem.
$5555^{2222}=(4 \bmod 7)^{2222}$
$=4^{370(6)+2} \bmod 7=4^{370(6)}.4^2 \bmod 7=(1 \bmod 7)(2 \bmod 7)=2 \bmod 7$
$\Rightarrow 5555^{2222}=2 \bmod 7$
$\Rightarrow 2222^{5555}+5555^{2222}=2 \bmod 7+5 \mod 7=7 \bmod 7= 0 \bmod 7$
**WIN**
6. Well, actually that was exactly the way I wanted you to do
Good job.
Though your notations are very strange...
$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)$
Just write $1\cdot 3^2 \cdot 3^2 \cdot 3 \bmod 7$
$5555^{2222}=(5 \bmod {\color{red}7})^{2222}
$
Write $5555^{2222}=5^{2222} \bmod 7$
7. Ah, sorry about my notations!
I didn't realise there was a command for $"\cdot"$. The more I know!
That second one was just plain wrong, sorry I wrote it!
On the plus side:
Spoiler:
At least I know how to use these spoiler boxes now! | 2015-03-05T00:25:10 | {
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https://math.stackexchange.com/questions/1302949/how-many-ways-to-select-k-vertices-of-an-n-gon | # How many ways to select $k$ vertices of an $n$-gon?
I have a regular $n$-gon, of which I have to select $k$ vertices. The selections must be rotationally distinct; two selections would be considered equivalent if one is a rotation of the other. For example, if I have a square, and I want to select 2 vertices, there are only 2 possible ways to do that according to the constraint. One is "x - x -", another is "x x - -".
If we denote the function by $CR(n,k)$, then these are the trivial cases:
1. $CR(n, 1) = 1$
2. $CR(n, 2) = \lfloor\frac{n}{2}\rfloor$
3. $CR(n, k) = CR(n, n - k)$
I am quite short of ideas on how to find the recurrence or closed formula of this problem, or if this problem has any closed form / recurrence solution at all. Any help with a bit detailed walk through would be much appreciated.
• What do you mean by selection must be rotationally invariant? – Anurag A May 28 '15 at 19:07
• Supposing, I have a 4-gon and I have to select 2 vertices. Hence, these two selections are the same. x-x-, -x-x. These selections are the same, x--x, xx--, -xx-, --xx – sarker306 May 28 '15 at 19:09
• The following MSE meta link has a section on enumerating necklaces and bracelets. – Marko Riedel May 28 '15 at 19:51
• One way to think about this (perhaps the best way) is the group action of cyclic rotations on the subsets of vertices. Rotations preserve the number of vertices in a subset (obviously), so the counting comes down to counting orbits of the $k$-subsets of $n$-vertices under those rotations. – hardmath May 28 '15 at 19:53
As the OP seems interested in fixing $k$ and letting $n$ vary we will do an example to show how this might work.
By the Polya Enumeration Theorem what we have here is $$[z^k] Z(C_n)(1+z)$$ with $Z(C_n)$ the cycle index of the cyclic group which is $$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}.$$
The desired quantity is thus given by $$[z^k] Z(C_n)(1+z) = [z^k] \frac{1}{n}\sum_{d|n} \varphi(d) (1+z^d)^{n/d}.$$
This is $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) [z^k] (1+z^d)^{n/d}$$ or $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) {n/d \choose k/d} = \frac{1}{n}\sum_{d|\gcd(n,k)} \varphi(d) {n/d \choose k/d}.$$
For $k=4$ starting at $n=1$ we obtain the sequence $$0, 0, 0, 1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73,\ldots$$ which points us to OEIS A008610 where we find confirmation.
Another interesting one is $k=6$ which yields $$0, 0, 0, 0, 0, 1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504,\ldots$$ which points to OEIS A032191.
A Maple session with these looks like this:
> with(numtheory):
> Q := (n,k) -> 1/n*add(phi(d)*binomial(n/d,k/d), d in divisors(gcd(n,k)));
Q := (n, k) -> add(numtheory:-phi(d) binomial(n/d, k/d),
d in numtheory:-divisors(gcd(n, k)))/n
> seq(Q(n,8), n=1..18);
0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 15, 43, 99, 217, 429, 810, 1430, 2438
which incidentally is OEIS A032193.
• It'll be a sin by my side not to refer you to my similiar question here: math.stackexchange.com/questions/1294224/… . Perhaps you'll be able to help :) I apologize, OP! Really am! – Matan May 28 '15 at 20:40
• I have been following your question for three hours. @Matan , felt really intrigued. – sarker306 May 28 '15 at 20:45
• How wonderful to hear. Thanks! I update it on regular basis and work on it when I have time. Keep track of it :) @sarker306 – Matan May 28 '15 at 20:49
Taking the necklace proof for Fermat's Little Theorem as inspiration:
For $n,k$ coprime, every possible selection has $n$ rotations. So
$$CR(n,k) = \frac 1n{n \choose k}$$
Otherwise letting $d=\gcd(n,k)$, we need to account for those choices where there is a repeat pattern smaller than $n$, which will be all possible choices over the shorter interval with the reduced number of choices (but repeated through the entire set):
\begin{align} CR(n,k) &= \frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + CR\left(\frac nd, \frac kd \right) \\ &=\frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + \frac 1d {n/d \choose k/d } \end{align}
since $\gcd(\frac nd, \frac kd)=1$
And also note that $CR(n,0)=1$ and $CR(n,n)=1$.
• Any explanation? Or any source where the derivation is explained? – sarker306 May 28 '15 at 19:25
• Joffan, if you could please refer here math.stackexchange.com/questions/1294224/… for a related problem I asked I will be grateful. Perhaps you'll be able to help. – Matan May 28 '15 at 19:26
• @sarker306 If you review the necklace proof of Fermat's Little theorem I think you'll understand where this comes from. If I have time I'll add to it. – Joffan May 28 '15 at 19:34
• I get CR(6,3) = 25/6 when I apply your solution. – TokenToucan May 28 '15 at 20:17
• CR(n, 2) has its correct value when n is odd, not even. But really thanks for bringing in this gcd issue, it seems to be a right step to the right direction, I didn't suspect it before. I'll think it through the night. – sarker306 May 28 '15 at 20:25 | 2019-09-22T01:36:36 | {
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https://math.stackexchange.com/questions/789804/how-many-distinct-ways-to-climb-stairs-in-1-or-2-steps-at-a-time/1376123 | # How many distinct ways to climb stairs in 1 or 2 steps at a time?
I came across an interesting puzzle:
You are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?
Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?
Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.
• Ha! Some simple and yet a convoluted formation of the sequence. That explains it. Can't believe I couldn't see it :) – PhD May 11 '14 at 1:53
• Is it possible to generalize it? Such as what if your steps are 2, 3, and 5? Also, we can compute fibonacci in matrix form which is faster. In the case of 2,3 and 5 steps, what would be the closed form solution? – Bagus Trihatmaja Dec 5 '18 at 8:53
• Yes, this falls into the more general framework of solving linear recurrence equations. They always admit fast evaluations via matrix exponentiation and often have closed-form solutions, especially when the characteristic equation has low degree since you need to find all of its roots. – fahrbach Dec 6 '18 at 22:19
• Can someone help me understand a tweak to this problem where the order of the steps doesn't matter? How to come up to a closed form equation for it? – mankadnandan May 26 '19 at 7:38
The solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach.
Here is an illustration of what you are trying to solve for the case of $n=4$ steps (taken from this website, which also gives a combinatorial solution)
Any staircase with $n$ steps allowing paths with increments of 1 or 2 steps at a time will end up in one of two states before the last path is taken: either we've climbed $(n-1)$ steps already and have $\color{red}{one}$ more step to take, or we've climbed $(n-2)$ steps already and we have $\color{blue}{two}$ more steps to take (if we took only one step here then we'd end up in an arrangement from the first state).
Thus, to get the total number of possible ways to climb $n$ steps, we just add the number of possible ways we can climb $(n-1)$ steps and the number of possible ways we can climb $(n-2)$ steps, giving the familiar recurrence relation:
\begin{equation*} F_n = \left\{ \begin{array}{l@{}l@{}l} 1 & n = 0,1\\ \color{red}{F_{n-1}} + \color{blue}{F_{n-2}} & n \ge 2 \end{array} \right. \end{equation*}
• Can this answer be expanded to answer the question "Given a set of possible step sizes S, calculate the number of possible ways you can climb n steps using step sizes in S)"? – Ephraim Nov 4 '15 at 20:21
• @Ephraim That's simple, it's just $F_n=\sum_{i\in I} F_i$ where $I$ is the set of steps that can jump to step $n$. – Stella Biderman Jan 31 '17 at 16:37
• This is the explanation that clicked to me, the colored figure makes a lot of sense, thanks. – neevek Oct 20 '20 at 16:47
This is the Fibonacci numbers - One interpretation of the n-th Fibonacci number is the number of ways to compose $n$ with parts in $\{1,2\}$ (that is, the n-th fibonacci number counts the number of sequences whose values are in $\{1,2\}$ whose sums are $n$). This is often called "Golfs and Cadillacs" or something similar.
• Could you please elaborate as to "how" it helps solve the problem? – PhD May 11 '14 at 1:46
• If there are $n$ steps, you have $F_n$ ways to reach the top of the stairs? The sequences map directly to the amount of stairs you climb with each step. I don't see how it can be any clearer. This interpretation can be used as a definition of the Fibonacci numbers and can be shown to be equivalent to the recurrence by using a straightforward combinatorial argument. – Batman May 11 '14 at 1:47
• I see what you mean. Perhaps what I'm confused by is by this: that is, the n-th fibonacci number counts the number of sequences whose values are in {1,2} whose sums are n. Could you please provide an example to explain this? – PhD May 11 '14 at 1:49
• You form a sequence made up of 1's and 2's whose sum is $n$. For example, for $n=3$, the sequences are (1,1,1), (1,2) and (2,1) giving you 3 ways to climb the stairs. Note that this is the common combintorics use of Fibonacci numbers, which is offset by 1 from the kind you learn in elementary school (this Fibonacci sequence is $1,2,3,5,...$ instead of $1,1,2,3,5,...$). – Batman May 11 '14 at 1:51
We can change this question into: In how many ways is possible to write a number as the ordered sum of 1s and 2s?
We can prove this using direct proof.
Hypothesis: For Fibbonaci number $F_1=1, F_2=1, F_{n}=F_{n-1}+F_{n-2}$, $Q(k)=F_{k+1}$
We say that we have constructed every order of $Q(n)$. Then, we construct $Q(n+1)$ this way:
1. For all orders, we add $+1$ to change $n$ to $n+1$. (There are now $Q(n)$ numbers.) Note: all numbers that are constructed here ends in $+1$.
2. For all orders that end with $+1$, we change it to $+2$. We shall prove that the amount will be $Q(n-1)$ to complete the proof. Note: all numbers that are constructed here ends in $+2$.
It can be seen that by constructing in this way there contains all orders.
Proof of step 2: If we constructed $Q(n)$ in the way above, then all numbers ended in $+1$ will be the amount $Q(n-1)$ by step 1. above, and we are done. | 2021-01-22T13:39:00 | {
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http://mathschallenge.net/full/quadratic_differences | #### Problem
The positive integers, $x$, $y$, and $z$ are consecutive terms in an arithmetic progression. Given that $n$ is also a positive integer, for how many values of $n$ below one-thousand does the equation $x^2 - y^2 - z^2 = n$ have no solutions?
#### Solution
Let $x = a + d$, $y = a$, and $z = a - d$.
$\therefore (a + d)^2 - a^2 - (a - d)^2 = n$
$a^2 + 2ad + d^2 - a^2 - a^2 + 2ad - d^2 = n$
$\therefore 4ad - a^2 = a(4d - a) = n$.
Let $u = a$ and $v = 4d - a \implies u + v = 4d \equiv 0 \mod 4$. In other words, for a solution to exist the factors of $n$ must add to a multiple of four.
We shall deal with $n$ being of the form $2^m r$, where $r$ is odd, and for increasing values of $m$.
• $m = 0$ ($n$ is odd):
If $n = r$ then the factors $u$ and $v$ must both be odd. But if they are both congruent with 1 or both congruent with -1 modulo 4 then $u + v \equiv 2 \mod 4$, and there will be no solution; if they are different then $u + v \equiv 0 mod 4$, and there will always be a solution.
Hence if they are the same then $n = uv \equiv 1 \mod 4$, or $n$ being of the form 4$k$ + 1, will have no solutions.
• $m = 1 \implies n = 2r = a(4d - a)$:
If $a = 2r$, $4d - a = 1 \implies 4d = 2r + 1$. But as the RHS is odd, this is impossible.
If $a = r$, $4d - a = 2 \implies 4d = r + 2$. Impossible, as RHS is odd.
In other words if $n = 2(2k + 1) = 4k + 2$ then there will be no solutions.
• $m = 2 \implies n = 4r$:
If $a = 2r$, $4d - a = 2 \implies 4d = 2r + 2 = 2(r + 1)$.
And as $r + 1$ is even, we will always have at least one solution if $n = 4r$.
• $m = 3 \implies n = 8r$:
If $a = 8r$, $4d - a = 1 \implies 4d = 8r + 1$. Impossible.
If $a = 4r$, $4d - a = 2 \implies 4d = 4r + 2$. Impossible.
If $a = 2r$, $4d - a = 4 \implies 4d = 2r + 4$. Impossible.
If $a = r$, $4d - a = 8 \implies 4d = r + 8$. Impossible.
Hence if $n = 8(2k + 1) = 16k + 8$ then there will be no solutions.
• $m \ge 4$:
If $a = 4r$, $4d - a = 2^{m-2} \implies 4d = 4(r+2^{m-4})$.
Hence for $m \ge 4$ there will always be at least one solution.
Thus there will be no solutions for numbers of the form $4k + 1$, $4k + 2$, and $16k + 8$. As the first and second cases are odd and even respectively, they are mutually exclusive, and although the the second and third are both even, the third is divisible by 4, whereas the second is not divisible by 4. Hence all three forms are mutually exclusive.
As $4 \times 249 + 1 = 997$, $4 \times 249 + 2 = 998$, and $16 \times 61 + 8 = 984$, there are exactly $249 + 249 + 61 = 559$ values of $n$ below one-thousand that have no solution.
For which values of $n$ will there be exactly one solution?
Problem ID: 295 (26 Nov 2006) Difficulty: 4 Star
Only Show Problem | 2013-12-10T09:29:57 | {
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http://sfmathgre.blogspot.com/2010/05/gr976863-intersections.html | ## Sunday, May 23, 2010
### GR9768.63: Intersections
63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-
Note: the answer provided in the Answer Sheet is D, 3 points of intersection.
result from wolframalpha
#### 13 comments:
1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.
2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!
3. Proper result from wolfram:
http://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+
4. one vote for D
5. shouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?
---
[email protected]
6. answer is (d) ...
basic property is that 2^x increases faster than x^12 and in fact for any a^x for which a>1 we would have a^x > x^n for any n as x--> infinity.
Hence all we need to find is whether the graph intersects twice or once for smaller values of x. e^x and x^2 I believe intersect only once.
2^x and x^12 intersects twice for x<2 and also for some big value of x as 2^x would be greater than x^12 for some value eventually. Hence this function intersects thrice.
7. I agree with Shantanu.
X>=0: We know that 0^12< 2^0 (x=0) and 2^12>2^2 (X=2).
Thus there is at least one intersection between 0 and 2.
At the same time, we know that 2^x is NOT O(x^12) as x->infinity. Also, they are both positive functions, so we know that there is some x>2 such that 2^x exceeds x^12 forever.
x<0. Its clear there is a cross in this region.
This gives us at least 3 crosses. My gut tells me that this is all because I am familiar with the graphs of both functions. So I would definitely choose D.
Rigorously, I do not know why there is only 3. Examining the derivatives would probably be the place to look.
8. Sorry for my grammar, english is not my native.
First 2 solutions are clearly. Lets prove that there is the 3-th solution. If we prove that lim(2^x/x^12)>1 for x->+oo, we will prove that the curves intersect once more time (because if x2 is x of 2-nd intersection and x near the x2 and x>x2 than 2^x1 for x->+oo. Just use L'Hospital's rule 12 times and we have: lim(ln2^12*2^x/(x^12))=+oo for x->+oo. So, because +oo>1 than we have the 3-th point of intersection.
9. Why are there not more than 3 intersections?
10. There are precisely 3 intersections. At x=0, 2^x = 2 > 0 = x^12. At x=2, 2^x = 4 < 2^12 = x^12, hence there is one intersection point in the interval (0,2). It is a well-known fact that as x grows to infinity 2^x exceeds x^12 (quite rapidly, in fact). Hence there is a second intersection point for x>2. By drawing the graphs of these functions it is easy to see that these are all intersection points for positive values of x.
A similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.
11. What is the value of 3rd point of intersection? How large is it?
12. The third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22) | 2014-08-01T07:46:03 | {
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https://math.stackexchange.com/questions/2901963/open-sets-in-mathbb-r-times-mathbb-r-with-the-dictionary-order | # Open sets in $\mathbb R\times \mathbb R$ with the dictionary order
In the example below, I wonder why Munkres depicted the first type of intervals in this way?
More specifically, why doesn't the picture look like this:
where all points below, above, and on the red segment are removed?
This is my justification why I think this should be the right picture:
$$(a\times b,c\times d)=\{x|a \times b <_{dict} x <_{dict} c\times d\}\\=\{(p,q)|a <_{R} p <_{R} c\}\cup \{(r,s)| a=r=c \text{ and } c <_{R} s <_{R} d \}$$ (where $<_{dict}$ is the dictionary order and $<_R$ is the standard order on $\mathbb R$). The first set in the union is the set of the first type, the second set in the union is the set of the second type.
Now, any point above, below, or on the red line has the first coordinate $a$, which cannot lie in a set of the first type.
Update: I defined the sets of the first type in a way that differs from Munkres, but I don't see what's wrong with my description of open sets (in the display), and no open set in my description can be a set of the first type in Munkres.
• Um, removing the red segments on your second image produces exactly the first one, doesn't it? – Henning Makholm Sep 1 '18 at 19:12
• @HenningMakholm I'm not removing just the red segment, I'm removing everything above and below it as well. – user531587 Sep 1 '18 at 19:21
• What then happen to points such as $a\times(b+1)$, which is definitely both $>_{\rm dict} a\times b$ and $<_{\rm dict} c\times d$? – Henning Makholm Sep 1 '18 at 19:29
• @HenningMakholm okay, this makes sense... I was confused (and still confused) by the description of the basis in the text. – user531587 Sep 1 '18 at 19:51
• Actually this answer made things more clear: math.stackexchange.com/a/1129061/531587 – user531587 Sep 1 '18 at 20:02
Suppose a point $p = (x,y) \in (a \times b , c \times d)$ has $a$ as the first coordinate. Since $(a,b) < (x,y) = (a,y)$, this necessarily implies that $b < y$. In the same vein, if $x = c$ and since $(x,y) = (c,y) < (c,d)$, necessarily we must have that $y < d$. This justifies why the interval only contains an open ray of $\{x = a\}$ and $\{x = c\}$ instead of the whole line.
Your characterization of the interval is not quite right. Note that $r$ cannot be $a$ and $c$ at the same time, so you should separate in two cases (which is basically what I wrote) depending on the value that the first coordinate takes. This should clarify the picture. | 2019-07-24T00:06:46 | {
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https://mathematica.stackexchange.com/questions/154018/principal-submatrix-and-principal-minor-of-a-matrix | # Principal submatrix and Principal minor of a matrix
A submatrix of a matrix is obtained by deleting any collection of rows and/or columns.
For example, from the following 3-by-4 matrix, we can construct a 2-by-3 submatrix by removing row 3 and column 2:
The minors and cofactors of a matrix are found by computing the determinant of certain submatrices.
A principal submatrix is a square submatrix obtained by removing certain rows and columns. The definition varies from author to author. According to some authors, a principal submatrix is a submatrix in which the set of row indices that remain is the same as the set of column indices that remain.
For a general 3 × 3 matrix in Mathematica,
(mat = Array[Subscript[a, ##] &, {3, 3}]) // MatrixForm
$mat=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{array} \right)$
there is one third order principal submatrix, namely mat. There are three second order principal submatrix:
$mat_{33}=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{array} \right)$, formed by deleting column 3 and row 3;
$mat_{22}=\left( \begin{array}{ccc} a_{1,1} & a_{1,3} \\ a_{3,1} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 2 and row 2;
$mat_{11}=\left( \begin{array}{ccc} a_{2,2} & a_{2,3} \\ a_{3,2} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 1 and row 1;
And there are three first order principal submatrix:
$mat=\left( \begin{array}{ccc} a_{1,1} \\ \end{array} \right)$, formed by deleting column 2,3 and row 2,3;
$mat=\left( \begin{array}{ccc} a_{2,2} \\ \end{array} \right)$, formed by deleting column 1, 3 and row 1, 3;
$mat=\left( \begin{array}{ccc} a_{3,3} \\ \end{array} \right)$, formed by deleting column 1,2 and row 1,2;
Do you have a way to find all principal submatrix in Mathematica ?
or find the determine of all principal submatrix (principal minor) in Mathematica ?
By @klgr comment.
Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] //MatrixForm /@ # &
• Perhaps Diagonal @ Minors[mat, k]? – Carl Woll Aug 18 '17 at 17:01
• Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] // MatrixForm /@ # & – kglr Aug 18 '17 at 17:04
• @kglr Thanks for the answer but is this in general. What if the matrix is a 4X4 measurement . – Emad kareem Aug 18 '17 at 17:14
• @kglr Thank you very much. This gives the exact result in general. – Emad kareem Aug 18 '17 at 17:24
To get the principal submatrices:
Diagonal[Map[Reverse, Minors[mat, #, Identity], {0, 1}]] & /@ {1, 2}
For the principal minors
Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & /@ {1, 2}
• What is the & /@ {1, 2} for? – AzJ Jan 25 '18 at 19:46
• @AzJ, the & belongs to the left part, that is, the expression is actually (Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & ) /@ {1,2}. For a pure function (a function with unnamed arguments) foo[#]& (where foo is an instruction on what to do with the stuff #; for example, multiply it by 2 and add 1 , (1+ 3 #)&) , the expression foo[#]&/@{a,b,c} Maps foo over a list of arguments ({a, b}) , that is, it is a short hand for {foo[a], foo[b], foo[c]}. ... – kglr Jan 25 '18 at 20:34
• ... Please see the doc pages on for details on of esoteric symbols such as Function (&), Slot (#) and Map – kglr Jan 25 '18 at 20:35
I arrived three years later but I think that the code of @klgr can be significantly improved as it computes first all the minors to take at the end just those on the diagonal. I think that the next routine works faster for the submatrices with the obvious change making it work also for the minors, of course. I also suppose for convenience that the matrix is square, the obvious change make this also work for non-square matrices.
PrincipalSubmatrices[mat_, size_] :=
Module[{choices = Subsets[Table[i, {i, 1, Length[mat]}], {size}], count,
rc, symsubmatrix, symsubmatrices}, count = choices;
symsubmatrices = {};
While[count != {}, rc = count[[1]]; symsubmatrix = mat[[rc, rc]];
symsubmatrices = Append[symsubmatrices, symsubmatrix];
count = Delete[count, 1]]; symsubmatrices] | 2021-04-17T12:32:00 | {
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https://math.stackexchange.com/questions/2736995/how-many-classes-does-the-equivalence-relation-partition-the-set/2737052 | How many classes does the equivalence relation partition the set?
Looking for help on B) How many classes does the equivalence relation partition set X
Consider the relations $R$ and $S$, defined on the set $X = \{1, 2, . . . , 99\}$ as follows.
$xRy \iff x + y$ is a multiple of $11$,
$xSy \iff x − y$ is a multiple of $11$.
A) One of $R$ and $S$ is an equivalence relation, the other is not. Determine which is which and justify your answers.
B) Into how many classes does the equivalence relation partition set $X$?
So far I've determined that S is the equivalence relation as R isn't reflexive or transitive. My only attempt at B is that there is 11-1=10 non-zero congruence classes, so does each one correspond to a non-zero equivalence class?
Any help is appreciated!
• Where does this "non-zero" stuff come from? – ancientmathematician Apr 14 '18 at 16:53
• What is a "non-zero" equivalence class? – Morgan Rodgers Apr 14 '18 at 16:54
• @CyclotomicField: Sorry, but "a zero equivalence class" is a bit of a nonsense on a set that doesn't even contain zero, such that the given set $X$. – zipirovich Apr 14 '18 at 18:29
• @CyclotomicField It's unambiguous, but it's also meaningless in the normal context of equivalence relations. And is a term that doesn't actually show up in the problem – Morgan Rodgers Apr 15 '18 at 17:11
• What's wrong with a "zero equivalence class? Doe $11 -11=0$ make $11$ any less equivalent? Where did you get this weird idea that you aren't supposed to count the the equivalence class to zero? – fleablood Apr 15 '18 at 17:20
Just count them.
$1 R 12 R 23 R 34 R .... R 90$
$2 R 13 R 24 R 35 R .... R 91$
....
$10 R 21 R 32 R 43 R..... R 98$
$11 R 22 R 33 R 44 R ..... R 99$
That's 11.
It doesn't make any sense to subtract the "zero" equivalence class for 2 reasons.
1) The "zero" equivalence class is STILL an equivalence class so why they heck would you omit it? NOWHERE in the question does it say ANYTHING about how many non-zero equivalence classes; it ask how many equivalence classes. And $\{11,22,33,44,55,66,77,88,99\}$ is certainly an equivalence class, isn't it?
2) Since no method of "addition" has been defined or discussed, there is no meaning to defining any one of the equivalence classes as be a "zero" equivalence class. IF we were to define $\{x|x R c\} + \{x|x R d\} = \{x| x R (d+c\pm 11k \text{ for some integer } k)\}$ and define $[0]$ as the equivelence clas so that $[0] + \{x|x R c\} = \{x|x R c\}$ then, yes, $\{11,22,33,44,55,66,77,88,99\} = [0]$. But again, so what, it's still an equivalence class, isn't it?
To show $S$ is an equivalence relation you can either verify directly that it is reflexive, transitive, and symmetric or demonstrate that the relation partitions the set and use the fundamental theorem of equivalence relations. If you can demonstrate the existence of the congruence classes you have such a partition and that's sufficient to demonstrate that you have an equivalence relation. As you've deduced there are eleven such classes, and ten of them do not contain zero. I would also encourage you to consider the first method of directly verifying the three defining properties as it's both instructive and relatively easy.
• So is the answer 10 for B) correct? – RealGib Apr 14 '18 at 18:23
• @RealGib: No, it's not. – zipirovich Apr 14 '18 at 18:30
• None of the equivalence classes on $X$ contain 0, since $X$ does not contain 0. – Morgan Rodgers Apr 15 '18 at 17:17
• "Morgan Rodgers continued insistence on ignoring this obvious interpretation of the OPs comments". But it goes both ways. I agree that "$\{11,...99\}$ do not contain $0$ as so can't be 'zero' modulo class" is a bit too oblique and inaccurate and $11\equiv 0 \mod 11$ does indeed mean $\{11,...99\}$ is the zero element. But on another, and very important level A "zero" (additive identity) element can only be defined if addition on the classes (and NOT on the elements of the classes) is defined. So it is not unjustified. – fleablood Apr 16 '18 at 16:10
• ... so I mostly agree with Morgan Rodgers although I believe saying "it isn't a zero module as it doesn't contain zero" evades the issue that it actually isn't a zero module because the means for which it is "zero" are not stated. Of more concern is... why leave out the zero modules. These seem to much of a "monkey see; monkey do" approach to math that is doomed to failure. You don't just do everything one way despite whether that is what a question asks. If was asked how many elments does $\{0,1,6,8\}$ you don't answer 3 because zero doesn't count and you don't answer "it goes to 8". – fleablood Apr 16 '18 at 16:15
The question is not asking you about "nonzero" equivalence classes, it is asking for the total number of equivalence classes. So the equivalence class that contains 0 counts the same as any other equivalence class (however you want to interpret this, your set $X$ does not contain 0 so the interpretation of which equivalence class is the "zero" involves concepts that go beyond the scope of what this problem is asking). | 2019-10-17T18:14:29 | {
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https://puzzling.stackexchange.com/questions/107801/a-circle-touches-two-sides-of-a-triangle-and-two-of-its-medians/107804#107804 | # A circle touches two sides of a triangle and two of its medians
A circle touches two sides of a triangle and two of its medians. Prove that the triangle is isosceles.
This problem came from the Mathematical Digest issue 62 (Jan 1986) which in turn cited a Russian mag called KVANT (meaning "Quantum").
• Assume it's not true. Then you wouldn't ask the question. A contradiction. Therefore it's true. QED. :) Mar 13 '21 at 1:02
As the triangles ADC and BEC have the same incircle and their areas are equal (half that of ABC), so are their perimeters
DC+DA+AC = EC+EB+BC
or, subtracting from both sides CD+CE+DA+EB
AE-EB = BD-DA.
This means that D and E lie on the same pair of hyperbolas with foci A and B. Since they also have the same distance to the base AB (half that of C) the triangle must be isosceles by symmetry.
• It took me a moment to see why they have the same distance to the base, but that should have been obvious. This is a really clever proof. Mar 13 '21 at 8:38
• @JaapScherphuis It's short ;-) I've added a half sentence explaining (if you can call it that) the distance to the base. Mar 13 '21 at 9:50
• That is clever. I don't think I've ever seen anyone use a hyperbola locus in a proof like this! Mar 13 '21 at 17:34
• Elegant, indeed. Mar 16 '21 at 6:16
Alternative Proof
By Pitot Theorem we have $$|CE|+|MD| = |CD|+|ME|$$ and since the centroid divides each median in the ratio 2:1, this means that $$\frac{1}{2}|AC| + \frac{1}{3}|AD| = \frac{1}{2}|BC|+\frac{1}{3}|BE|$$ Now let $$|AB| = c, |BC| = a, |CA|=b, |AD|=m_a, |BE|=m_b$$.
Then Apollonius' Theorem tells us that $$m_a = \sqrt{\frac{2b^2+2c^2-a^2}{4}}\,\,\,\,,\,\,\,\,m_b = \sqrt{\frac{2a^2+2c^2-b^2}{4}}$$ Substituting this in above and multiplying across by $$6$$ yields $$3a + \sqrt{2b^2+2c^2-a^2} = 3b + \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 3(a-b) + \sqrt{2b^2+2c^2-a^2} = \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 9(a-b)^2 + 6(a-b)\sqrt{2b^2+2c^2-a^2} + 2b^2+2c^2-a^2 = 2a^2+2c^2-b^2$$ $$\Rightarrow (a-b)\left(9(a-b) + 6\sqrt{2b^2+2c^2-a^2} - 3(a+b)\right) = 0$$ which means either $$a=b$$ or $$\sqrt{2b^2+2c^2-a^2} = 2b-a$$.
If it is the latter then $$2b^2+2c^2-a^2 = 4b^2-4ab+a^2$$ $$\Rightarrow c^2 = b^2-2ab+a^2 = (b-a)^2$$ $$\Rightarrow c = \pm(b-a)$$ which only happens for a degenerate triangle (does not satisfy the triangle inequality).
Hence $$a=b$$
@loopwalt and @hexomino both gave excellent answers. I wanted to share the answer from the Mathematical Digest, because it's also elegant in its own way:
It starts the same way as @hexomino's proof (I didn't know it was called Pitot's Theorem!):
Since $$CEMD$$ circumscribes a circle, $$CE+MD = CD + ME$$ (Pitot's Theorem, but fun to prove). So (as in hexomino's proof):
$$\frac{1}{2}AC+\frac{1}{3}AD = \frac{1}{2}BC+\frac{1}{3}BE \tag1$$
Next, note that $$\triangle ADC$$ and $$\triangle BEC$$ have the same area ($$\frac{1}{2}\triangle ABC$$) and share a common incircle. This means that their perimeters are equal (another fun to prove and left as an exercise!). So $$AD+\frac{1}{2}BC+AC = BE+\frac{1}{2}AC+BC$$, which gives:
$$\frac{1}{2}AC+AD = \frac{1}{2}BC+BE \tag2$$
Now, $$3\times(1)-(2)$$ gives $$AC=BC$$ | 2022-01-23T10:31:28 | {
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http://www.anlak.com/2015/01/probability-of-k-collisions.html | ### Probability of k collisions
, Friday, 30 January 2015
While I was reading about hash tables I stumbled upon this deceptively simple question. I have rephrased it to make it easier to understand.
Say we have $m$ buckets. We select a random bucket and put a ball in it, we repeat this (select-put) $n$ times. In the end what is the probability of having at least one bucket with $k$ balls?
My initial attempts to solve it failed, and I posted the question on math.stackexchange. An approximation for large values of $m$ and $n$ was posted as an answer but I was interested in exact results. Once I realized I had no way of finding someone else's solution, I was able to solve it myself.
## Simulation
First, to check if my solution is correct, I have written a small program to simulate the select-put procedure million times and report the experimental probability for different values of $m$,$n$ and $k$. If you are interested, you can find the java code at the end of my post.
## Solution
First we will start with writing the probability of 1 box having $k$ balls. Then using inclusion-exclusion principle we will generalize. Let's write the probability of putting first $k$ balls into the first box and remaining balls to the other boxes: $$\left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ We have $\binom{n}{k}$ ways of putting $k$ balls into a particular box (e.g. first box), and we have $m$ such boxes. So the probability becomes: $$m \binom{n}{k} \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ But the expression above suffers from double counting, i.e. it counts configurations that contains multiple boxes with $k$ balls more than once. So we need to subtract the value for 2 boxes having $k$ balls, add the value for 3 boxes having $k$ balls, etc... Let's write the probability of putting first $k$ balls into the 1st box, next $k$ balls into the 2nd box, and remaining balls to the other boxes. $$\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ We have $\binom{n}{k}\binom{n-k}{k}$ ways of putting $2k$ balls into two particular boxes (e.g. first and second box), and we have $\binom{m}{2}$ such boxes. So the probability becomes: $$\binom{m}{2}\binom{n}{k}\binom{n-k}{k}\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ At this point you should be able to see the pattern that is emerging.
## Generalization
If we put together and generalize what we have done so far we get the following expression as probability of having a box with $k$ balls: $$\sum^{\lfloor \frac{n}{k} \rfloor}_{i=1} (-1)^{i+1} \binom{m}{i} \left[ \prod^{i-1}_{j=0} \binom{n-jk}{k} \right] \left( \frac1{m} \right)^{ik} \left( \frac{m-i}{m} \right)^{n-ik}$$ The powers of $-1$ at the beginning enables adding values for odd number of balls and subtracting values for even number of balls, i.e. inclusion-exclusion principle.
## The code
private static double calculateKcollisions(int m, int n, int k) {
double probability = 0;
int sign = 1;
for (int i = 1; n - i * k >= 0 && i <= m; i++) {
double p = binomial(m, i) * pow(1.0 / m, i * k);
p *= pow((double) (m - i) / m, n - i * k);
for (int j = 0; j < i; j++) {
p *= binomial(n - j * k, k);
}
probability += sign * p;
sign *= -1;
}
return probability;
}
## Simulation code
private static final Random RNG = new Random();
public static void main(String[] args) {
int sampleCount = 1000000;
for (int experimentNo = 0; experimentNo < 10; experimentNo++) {
int m = RNG.nextInt(7) + 3;
int n = RNG.nextInt(m) + RNG.nextInt(10) + 1;
int k = RNG.nextInt(n + 1);
System.out.println(format("m:%d n:%d k:%d", m, n, k));
int successCount = 0;
for (int sampleNo = 0; sampleNo < sampleCount; sampleNo++) {
int[] boxes = new int[m];
for (int i = 0; i < n; i++)
boxes[RNG.nextInt(m)]++;
for (int i = 0; i < m; i++) {
if (boxes[i] == k) {
successCount++;
break;
}
}
}
double experimental = (double) successCount / sampleCount;
double calculated = calculateKcollisions(m, n, k);
System.out.println(format("xperimental: %f calculated: %f", experimental, calculated));
System.out.println(format("diff: %.2f\n", abs(experimental - calculated)));
System.out.println();
}
}
#### 1 comment:
1. Excellent blog thanks for sharing the valuable information..it becomes easy to read and easily understand the information.
Useful article which was very helpful. also interesting and contains good information. | 2022-05-27T13:40:53 | {
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http://mathhelpforum.com/calculus/155482-finding-point-sphere-furthest-point-space.html | # Math Help - Finding point on sphere furthest from point in space
1. ## Finding point on sphere furthest from point in space
Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).
Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
2. Originally Posted by SyNtHeSiS
Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).
Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
Think about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is $\sqrt{99} = 3\sqrt{11}$. You have already identified the direction of travel as the vector (1,–1,3), which has length $\sqrt{11}$.
Those $\sqrt{11}$s can't be coincidental, can they?
3. Hello, SyNtHeSiS!
$\text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99$
$\text{which is furthest from the point }\,C(1,3,4).$
Code:
* * * o C
* * *
* ♥ P
* * *
*
* * *
* o *
* * A *
*
* * *
Q ♥ *
* * *
* * * *
Construct a line through $\,C$ and the center of the sphere $\,A.$
It will intersect the sphere in two points, $\,P$ and $Q.$
. . We want $Q$, the point further from $\,C.$
The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$
The line has parametric equations: . $\begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}$
Substitute into the equation of the sphere:
. . $\left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99$
. . $11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9$
. . $t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4$
$\begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\
t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}$
I suspect that point $\,K$ is further from $\,C$, but let's make sure.
Distance from $C(1,3,4)$ to $J(\text{-}1,5,\text{-}2)\!:$
. . $\overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44}$
Distance from $C(1,3,4)$ to $K(5,\text{-}1,16)\!:$
. . $CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}$
Hence, $K$ is the point we are seeking.
Therefore: . $Q(5,\text{-}1,16)$
4. Originally Posted by Soroban
The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$
Thanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.
5. Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as $x = (1,3,4) + t(1,-1,3)$. If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is $\sqrt{99}$ units, which is three times the length of the vector (1,–1,3).
So to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,–1,3) = (5,–1,16). | 2014-03-14T23:29:54 | {
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https://datascience.stackexchange.com/questions/46744/when-to-use-mean-vs-median/46747 | # When to use mean vs median
I'm new to data science and stats, so this might seems like a beginner question.
I'm working on a dataset where I've user's Twitter followers gain per day. I want to measure the average growth he had over a period of time, which I did by finding the mean of growth. But someone is suggesting me to use median for this.
Can anyone explains, in which use-case we should use mean and when to use median?
The arithmetic mean is denoted as $$\bar{x}$$
$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$
where each $$x_i$$ represent an unique observation. The arithmetic mean measures the average value for a given set of numbers.
In contrast to this, the median is the value which falls directly in the middle of your dataset. The median is especially useful when you are dealing with a wide range or when there is an outlier (a very high or low number compared to the rest) which would skew the mean.
For example, salaries are usually discussed using medians. This due to the large disparity between the majority of people and a very few people with a lot of money (with the few people with a lot of money being the outliers). Thus, looking at the 50% percentile individual will give a more representative value than the mean in this circumstance.
Alternatively, grades are usually described using the mean (average) because most students should be near the average and few will be far below or far above.
• That's a great answer. So, If I think it like this, I can plot my data and see if it values are continuous, then we can use mean and if they're more clustered (some high and some low), then median would be better, right? – Mukul Jain Mar 6 '19 at 4:48
• @MukulJain, Yes it depends on the distribution of the data as you mentioned. Plotting is always my go to way to get a sense of my data. Easy to spot anomalies and get a sense of its spread. – JahKnows Mar 6 '19 at 5:48
• I think you could explain this better using the term "outlier" – MilkyWay90 Mar 7 '19 at 1:11
• So, if data has lots of outliers, is it good to use median right? Outliers can be calculated using z-score (<3 or >-3) – Mukul Jain Mar 7 '19 at 4:19
• @MukulJain, correct, and you can also calculate outliers using p-value, – JahKnows Mar 7 '19 at 5:52
It depends what question you are trying to answer. You are looking at the rate of change of a time series, and it sounds like you are trying to show how that changed over time. The mean gives the reader one intuitive insight: they can trivially estimate the number of followers at any date $$d$$ days since the start by multiplying by the mean rate of change.
The downside to this single metric is that it doesn't illustrate something which is very common in series such as this: the rate of change is not fixed over time. One reasonable metric for giving readers an idea of whether the rate of change is static is giving them the median. If they know the minimum of the series (presumably zero in your case), the current value, the mean and the median, they can in many cases get a "feel for" how close to linear the increase has been.
There is a great cautionary tale in Anscombe's quartet - four completely different time series which all share several important statistical measures. Basically it always comes back to what you are trying to answer. Are you trying to find users which are likely to become prominent soon? Users which are steadily accruing followers year by year? One hit wonders? Botnets?
As you've probably guessed, this means it's not possible to universally call mean or median "better" than the other.
Simply to say, If your data is corrupted with noise or say erroneous no.of twitter followers as in your case, Taking mean as a metric could be detrimental as the model will perform badly. In this case, If you take the median of the values, It will take care of outliers in the data. Hope it helps
Often median is more robust to extreme value to mean. Try to think it as a minimization task. Median corresponds to absolute loss while mean corresponds to square loss.
I find myself explaining this a lot and the example I use is the famous Bill Gates version. Bill Gates is in your data science class. Your instructor asks you: what is the average income or net worth of this class? Bill Gates sheepishly obliges and tells you what his income is. Now when you say the average income of your group is a zillion dollars - technically correct but does not describe the reality - that Bill Gates is an outlier skewing everything.
So you line up all the people in your group in ascending or descending order - whatever the person in the middle is making - that is your median. In this example, everybody but Bill Gates is likely to be in spitting distance of that median, and Bill Gates will be the only one making anything close to the mean.
Now say buddy Bill Gates is hiring a money manager. Based on the returns they produced so far. Should he look at their average returns over a 10 year period or their median return or a combination of the two? Did they outperform the market each year? Some years? How does portfolio size factor in? In the case of Twitter followers, Obama would have a different growth compared to someone with say 500K-1MM followers. As @l0b0 alludes to in his excellent answer - it all depends. Are you measuring follower growth or the rate of change of follower growth and what is the question you are trying to answer, strategy/product you are trying to develop - accordingly you pick mean or median. Getting the mean and median is always the easy part. It's always better to never ever have the average of 2.1 kids. Have a whole number of kids. But what can you say about population growth rates if mean number of kids is 2.1 and median is 1 or 2? Or median is 3 or more? Is growth accelerating or decelerating? What is mode doing? Compute all the basics first - and then ask the reason why you are using mean versus median. | 2020-09-20T10:34:13 | {
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http://math.stackexchange.com/questions/111886/calculate-lim-n-rightarrow-infty-frac1n-sum-k-1nk-sin-left-frac | # Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$
I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!
-
Is there anything special about $a$? I'm assuming it is an arbitrary real number? – JavaMan Feb 22 '12 at 2:14
Welcome to math.SE! Do you have learned Riemann integration? – leo Feb 22 '12 at 2:14
You may prove a more general statement: *If $a_n \to a$, then $\frac{1}{n}\sum_{k=1}^{n}a_k \to a$*. This can be proved by $\epsilon-\delta$ argument. – sos440 Feb 22 '12 at 2:18
a is an arbitrary real number, yeah. And I've learned Riemann integration! – ro44 Feb 22 '12 at 2:27
@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. – ro44 Feb 22 '12 at 2:29
I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.
First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)<k\left(\frac{a}k\right)=a$, so $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)<\frac1n\sum_{k=1}^na=a\;;$$ this gives you an upper bound of $a$ on any possible limit.
You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0<x<c$. This means that $$k\sin\left(\frac{a}k\right)>\frac{a}2$$ whenever $\frac{a}k<c$, i.e., whenever $k>\frac{a}c$. Now suppose that $n$ is very large compared with $\frac{a}c$; then ‘most’ of the terms of $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)\tag{1}$$ will be greater than $\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\to\infty$ must be at least $\frac{a}2$.
But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0<x<c$. If you’ve filled in the missing details for the previous paragraph, you shouldn’t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\epsilon a$ for any $\epsilon <1$ and hence must be at least $a$.
-
Based on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\frac{1}{n}\sum_{k=1}^{n}a_n)-a|=|\frac{1}{n}\sum_{k=1}^{n}(a_n-a)|\leq \frac{1}{n}\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|<l$, yielding that $\frac{1}{n}\sum_{k=1}^{n}|a_n-a|<l$ for almost every n. So $\frac{1}{n}\sum_{k=1}^{n}a_n \rightarrow a$. Does this sound correct? Thanks! – ro44 Feb 22 '12 at 2:49
@ro44: Not quite. You want $|(\frac1n\sum_{k=1}^na_k)-a|=|\frac1n\sum_{k=1}^n(a_k-a)|\le\frac1n\sum_{k=1}^n|a_k-a|$. Then you want to say that for sufficiently large $n$, ‘most’ terms of this sum are less than $l$, so the sum is less than (say) $2l$. There’s still a bit of work to be done show that if $n$ is large enough, the early ‘bad’ terms are a small enough fraction of the total not to mess things up. – Brian M. Scott Feb 22 '12 at 3:37
$\displaystyle \sin x \leq x$ for $x \geq 0.$ Integrating this over $[0,t]$ gives $$-\cos t +1 \leq \frac{t^2}{2} .$$
Integrating both sides again from $[0,x]$ gives $$-\sin x + x \leq \frac{x^3}{6} .$$
Thus, $$x - \frac{x^3}{6} \leq \sin x \leq x.$$
Hence, $$\frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right).$$
Since $\displaystyle \sum_{k=1}^n \frac{a^3}{6k^3}$ is convergent, the Squeeze theorem shows that $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$
- | 2014-08-28T05:09:21 | {
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https://www.jiskha.com/questions/1828425/the-volume-v-of-a-sphere-of-radius-r-is-given-by-the-formula-v-r-4-3-r-3-a-balloon | # math
The volume V of a sphere of radius r is given by the formula V (r) = (4/3)πr^3. A balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of 2 inches per second, and is zero whent = 0.
(a) Find a formula for the volume V of the balloon as a function of time t.
(b) Determine the volume of the balloon after 5 seconds.
(c) Starting with an empty balloon, suppose that the balloon will burst when
its volume is 10, 000 cubic inches. At what time will the balloon burst?
(d) Find a formula for the surface area S of the balloon as a function of time
t; recall the surface area formula for a sphere of radius r is S(r) = 4πr^2.
(e) Determine the surface area of the balloon after 6 seconds.
(f) What will be the surface area of the balloon when it bursts?
1. 👍 0
2. 👎 0
3. 👁 233
1. V = 4/3 πr^3
(a) dV = 4πr^2 dr
V = ∫[0,t] 4πr^2 dr
(b) now plug in t=5
(c) find when (a) = 10000
(d) dA = 8πr dr
work it as in (a)
(e) work as in (b)
(f) plug in t as in (c)
1. 👍 0
2. 👎 0
👨🏫
oobleck
2. since r = 2t, let's replace r in the equation
V = (4/3)π r^3
= (4/3)π (2t)^3
= (32/3)π t^3
b) replace t with 5 in my formula.
c) (32/3)π t^3 = 10000
solve for r
d) S(r) = 4πr^2 , let do the same thing: r = 2t
= 4π(2t)^2 = 16πt^2
e) let t = 6 and evaluate using the new formula from d)
f) sub in the value of t you got in c) into the new S(r) formula
let me know what your answers are so I can check them
1. 👍 0
2. 👎 0
👨🏫
Reiny
3. yep i got the same answers as you did :) thank you for your help
1. 👍 0
2. 👎 0
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V=4/3 pi r ^3 is the formula for the volume of a sphere. Find dV for a sphere if the radius of 10" is increased by .1" Please help. I have no idea how to do this problem! | 2020-10-28T19:48:44 | {
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https://math.stackexchange.com/questions/158271/relationship-between-the-coefficients-of-a-polynomial-equation-and-its-roots/158272 | # Relationship between the coefficients of a polynomial equation and its roots
My question is- Solve: $x^3 - 6x^2 + 3x + 10= 0$ given that the roots are in arithmetic progression.
Any help would be greatly appreciated.
Let the three roots be $a-d$, $a$, and $a+d$.
The sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.
The product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$. We get $a^2-d^2=-5$, and therefore $d^2=9$.
Remarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\ne 0$, and the roots are $\alpha$, $\beta$, and $\gamma$, then $$\alpha+\beta+\gamma=-\frac{q}{p}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p},\qquad \alpha\beta\gamma=-\frac{s}{p}.\tag{1}$$ There are analogous relations between coefficients and roots for polynomials of any degree.
$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.
$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\pm 1$, $\pm 2$, $\pm 5$, and $\pm 10$, and we quickly get to an answer.
Or else we can let the roots be $\alpha$, $\beta$, and $\gamma$, and see that the polynomial must be $(x-\alpha)(x-\beta)(x-\gamma)$. Expanding, we find in essence the relationship $(1)$.
Or else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems.
The roots are in arithmetic progression, so we can denote them as $a-d, a, a+d$ where $d$ is the common difference. The sum of these roots is then $3a$ and Viete's formulas shows that the sum of the roots must be $6$, so $a=2.$ So we have found a root, and can get the others by dividing the polynomial by $x-2$ and solving the remaining quadratic.
A three degree polynomial would have max 3 roots say p,q,r
The relations between the roots and coefficients in $ax^3 + bx^2 + cx + d$ are:
$\frac{-b}{a} = p+q+r$
$\frac{c}{a} = pq + qr + pr$
$\frac{-d}{a} = pqr$
Note: where $a \neq 0$. Being 0 simply implies that the order of equation isn't actually 3.
Three equations can solve three unknowns (To use the fact that they are in AP, assume $p,q,r$ to be $q-d$, $q$, $q+d$ where $q$ is the mean and $d$ is the common difference of AP).
Note: The pattern of the roots and coefficients can be the same for n degrees of polynomial as well.
• where $a\neq 0$ – Belgi Jun 14 '12 at 14:39
• Thanks. I have edited the answer – Saurabh Agarwal Jun 14 '12 at 14:39
Hint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?
• I did not use the fact that the roots are in arithmetic progression – Belgi Jun 14 '12 at 14:36
• That fact could be used to ascertain that 6/3 = 2 is one of the roots. – hardmath Feb 1 '14 at 4:40
Hint $\rm\ \ f(x) = (x+a\!-\!b)(x+a)(x+a\!+\!b)\iff f(x\!-\!a) = x^3 - b^2 x.\:$ But the only shift killing the $\rm\:-6\,x^2\:$ in your cubic is $\rm\:x \to x+2,\:$ yielding $\rm\:f(x\!+\!2) = x^3 - 9\,x,\:$ so $\rm\: b = \pm 3.$
Note $\rm\,\ \ f(x\!-\!a) = x^3\! -\! b^2x\:\Rightarrow\:b^2 = -f'(x\!-\!a)|_{x=0} =\, -f'(-a)\ \ [\, = -(3\cdot 2^2\! -\! 12\cdot 2\! +\! 3) = 9\ ].\:$
Therefore $\rm\ \ f(x) = x^3 + c\: x^2 +\ldots\: \Rightarrow\ a = c/3,\ \ b = \sqrt{-f'(-a)}$
First of all the easiest mtd is
$$1)~$$ Get one of the roots by hit and trial mtd in your question clearly $$~-1~$$ is one of the roots i.e when $$~x =-1~$$ the whole equation results $$~0~$$ therefore $$~X=-1 ,~~~ X+1=0~$$
$$2)~$$ apply division method to reduce it to quadratic and then $$~u~$$ can easily have your roots that are in A. P | 2019-08-18T23:24:54 | {
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http://xjetair.net/how-do-sjfdhpy/complex-number-to-rectangular-form-ad49e6 | # complex number to rectangular form
The x is the real number of the expression and the y represents the imaginary number … We move 2 units along the horizontal axis, followed by 1 unit up on the vertical axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Rectangular and Polar Form To represent complex numbers x yi geometrically, we use the rectangular coordinate system with the horizontal axis representing the real part and the vertical axis representing the imaginary part of the complex number. Post was not sent - check your email addresses! If necessary, round to the nearest tenth. An identification of the copyright claimed to have been infringed; Convert The Following Complex Numbers To Rectangular Form. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Khan Academy is a 501(c)(3) nonprofit organization. The absolute value of a complex number is the same as its magnitude. Find more Mathematics widgets in Wolfram|Alpha. information described below to the designated agent listed below. Use i or j to represent the imaginary number −1 . St. Louis, MO 63105. Related Topics: Common Core (The Complex Number System) Common Core for Mathematics. Varsity Tutors LLC your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the ( Log Out / For this reason the rectangular form used to plot complex numbers is also sometimes called the Cartesian Form of complex numbers. Change ), You are commenting using your Twitter account. Recall that the complex plane has a horizontal real axis running from left to right to represent the real component (a) of a complex number, and a vertical imaginary axis running from bottom to top to represent the imaginary part (b) of a complex number. A. Zi = 6e 25" B. Z2 = 5e-145 3. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A complex number can be expressed in standard form by writing it as a+bi. means of the most recent email address, if any, provided by such party to Varsity Tutors. ChillingEffects.org. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . Change ), You are commenting using your Google account. Theorem 11.16. Distributing the 4, we obtain the final answer of: distributing the 5, we obtain the final answer of: To convert, just evaluate the trig ratios and then distribute the radius. Your name, address, telephone number and email address; and Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Get the free "Convert Complex Numbers to Polar Form" widget for your website, blog, Wordpress, Blogger, or iGoogle. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. Complex numbers in the form a+bi\displaystyle a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. To convert to rectangular form, just evaluate the trig functions and then distribute the radius: To convert, evaluate the trig ratios and then distribute the radius: If you've found an issue with this question, please let us know. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Polar to Rectangular Online Calculator Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Varsity Tutors. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex Number. Add The Two Complex Numbers In Rectangular Form Zı = 10-j6 Z2 = 3 +j16 4. Solution for Write the complex number 4(cos 240° + i sin 240°) in rectangular form. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. The standard form, a+bi, is also called the rectangular form of a complex number. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Convert The Following Complex Numbers To Polar Form. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. If you were to represent a complex number according to its Cartesian Coordinates, it would be in the form: (a, b); where a, the real part, lies along the x axis and the imaginary part, b, along the y axis. ( Log Out / The multiplication of complex numbers in the rectangular form follows more or less the same rules as for normal algebra along with some additional rules for the successive multiplication of the j-operator where: j2 = -1. Explanation: . Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. You can input only integer numbers or fractions in this online calculator. Distributing the 4, we obtain the final answer of: A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Therefore the correct answer is (4) with a=7, and b=4. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Change ), You are commenting using your Facebook account. … Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The Complex Hub aims to make learning about complex numbers easy and fun. The calculator will simplify any complex expression, with steps shown. an The angle must be converted to radians when entering numbers in complex exponential form: In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. This point is at the co-ordinate (2, 1) on the complex plane. The rectangular form of a complex number is given by z = a + bi. Additional features of complex numbers convert. Zi = 3 + J5 Z2 = 9-14 Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. ; The absolute value of a complex number is the same as its magnitude. Show Instructions. > 5+4i ans = 5 + 4i A number in polar form, such as (2 45 ), can be entered using complex exponential notation. Polar & rectangular forms of complex numbers Our mission is to provide a free, world-class education to anyone, anywhere. If Varsity Tutors takes action in response to Complex numbers in rectangular form are presented as a + b * %i, where a and b are real numbers. In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially There's also a graph which shows you the meaning of what you've found. Example 1 – Determine which of the following is the rectangular form of a complex number. Notice the rectangle that is formed between the two axes and the move across and then up? Substitute the values of a and b. z = a + bi = rcosθ + (rsinθ)i = r(cosθ + isinθ) distributing the 3, we obtain the final answer of: we find that the value of and the value of . Label the x-axis as the real axis and the y-axis as the imaginary axis. That’s right – it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In rectangular form a complex number take the form, a + bi. improve our educational resources. The rectangular representation of a complex number is in the form z = a + bi. Z = 6+j8 B. Z2 = 25-12 2. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). either the copyright owner or a person authorized to act on their behalf. The rectangular from of a complex number is written as a single real number a combined with a single imaginary term bi in the form a+bi. misrepresent that a product or activity is infringing your copyrights. a This rectangular to exponential form conversion calculator converts a number in rectangular form to its equivalent value in exponential form. Unlike the polar form, which is expressed in unit degrees, a complex exponential number is expressed in unit radians. Then, multiply through by r. Example 6: Converting from Polar to Rectangular Form Dartmouth College, Bachelor in Arts, Biochemistry and Molecular Biology. 101 S. Hanley Rd, Suite 300 The principal value of the argument is normally taken to be in the interval .However, this creates a discontinuity as moves across the negative real axis. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Key Concepts. By … Examples, solutions, and lessons to help High School students learn how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Question: 1) Convert The Complex Number From Polar To Rectangular Form. Thus, if you are not sure content located Use and keys on keyboard to move between field in calculator. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. The correct answer is therefore (2). In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Z = 3 Cis (7/6) - 2) Find The Power Of The Complex Number In Polar Form. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require Infringement Notice, it will make a good faith attempt to contact the party that made such content available by link to the specific question (not just the name of the question) that contains the content and a description of There are two basic forms of complex number notation: polar and rectangular. Rectangular, polar and exponential forms of complex numbers When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term “Rectangular Form”. © 2007-2021 All Rights Reserved, Express Complex Numbers In Rectangular Form, Calculus Tutors in San Francisco-Bay Area. In polar form the modulus and argument are used to rewrite the complex number in the form: z = |z|(cos(θ) + i sin (θ)) where θ = arg(z) The steps to converting a complex number into polar form. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such bi+a instead of a+bi). (This is spoken as “r at angle θ ”.) A complex number in rectangular form means it can be represented as a point on the complex plane. The rectangular form of a complex number is written as a+bi where a and b are both real numbers. 3. With the help of the community we can continue to Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. Change ). Convert the following to rectangular form: Distribute the coefficient 2, and evaluate each term: Using the general form of a polar equation: we find that the value of is and the value of is . which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Send your complaint to our designated agent at: Charles Cohn Columbia University in the City of New York, Master ... New Jersey Institute of Technology, Bachelor of Science, Mechanical Engineering. Example 2 – Determine which of the following is the rectangular form of a complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Using the general form of a polar equation: we find that the value of and the value of .The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A. Entering data into the complex number convert. Divide The Two Complex Numbers, ZJZı. There are two basic forms of complex number notation: polar and rectangular. I'm not fimular with MATLAB keywords but need to use this to prove my answers. Find the absolute value of z= 5 −i. 1. Find Z3 When Z = 4 Cis(85°). Rules. New Jersey Institute of Technology, Master o... Princeton University, Bachelor in Arts, Public Policy Analysis. Every complex number written in rectangular form has a unique polar form ) up to an integer multiple of in its argument. 2. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: ∠). Show Hide all comments. But then why are there two terms for the form a+bi? For background information on what's going on, and more explanation, see the previous pages, You may have also noticed that the complex plane looks very similar to another plane which you have used before. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. Track your scores, create tests, and take your learning to the next level! -1.92 -1.61j [rectangular form] Euler's Formula and Identity The next section has an interactive graph where you can explore a special case of Complex Numbers in Exponential Form: The x is the real number of the expression and the y represents the imaginary number of the expression. ( Log Out / Finding the Absolute Value of a Complex Number with a Radical. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = | z | cis(α) and w = | w | cis(β). You can use rad function to convert from degrees to radians: r * exp (rad (d) * %i). Sorry, your blog cannot share posts by email. More in-depth information read at these rules. See . Complex Number Calculator. Although the complex numbers (4) and (3) are equivalent, (3) is not in standard form since the imaginary term is written first (i.e. It is the distance from the origin to the point: See and . the Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. The steps to converting a complex number into polar form are as follows: Find the modulus. Polar form of the complex numbers is presented as r * exp (c * %i), where r is radius and c is the angle in radians. Z = 0.5 angle( pi / 4 ) or. It is the distance from the origin to the point: ∣z∣=a2+b2\displaystyle |z|=\sqrt{{a}^{2}+{b}^{2}}∣z∣=√a2+b2. Z = 0.5 angle( - 45 degrees ) 2 Comments. Then, multiply through by r. Example 10.5.6A: Converting from Polar to Rectangular Form 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Note that all the complex number expressions are equivalent since they can all ultimately be reduced to -6 + 2i by adding the real and imaginary terms together. as We sketch a vector with initial point 0,0 and terminal point P x,y. By default, MATLAB/Octave accepts complex numbers only in rectangular form. ( Log Out / Script file not sent - check your email addresses i am having trouble converting form... '' B. Z2 = 5e-145 3 Our educational resources 1 unit up on the vertical.., we first need some kind of standard mathematical notation cosθ + )... Similar to another plane which you have used before evaluating what is given and using the distributive.. Imaginary axis appears as, and can be found by finding the functions! Is why ( a+bi ) is also called the Cartesian form of a complex exponential number is given z! Terminal point P x, y Convert complex numbers the rectangular form by r. example 10.5.6A: from! Numbers in rectangular form of a complex number notation: polar and rectangular then... The best experience we move 2 units along the horizontal axis, followed by unit... / Change ), first evaluate the trigonometric values of the complex plane looks very similar the. Where x and y are numbers ) Convert the complex plane looks very similar to next... R ∠ θ the set of complex numbers of a complex exponential number is the distance from the origin the. Write the complex number in polar form the community we can continue to improve educational. Is the same as its magnitude isinθ ), first evaluate the values. A+Bi, is also called the Cartesian form of a complex number is written as a+bi a! Easy and fun world-class education to anyone, anywhere where a and b are numbers! Absolute value of a complex number is the rectangular form is a 501 ( c ) ( 3 ) organization! On the format, rectangular number= x + jy, where a b., multiply through by r. example 10.5.6A: converting from polar to rectangular form Convert the complex.! X-Axis as the real number of the expression and the move across and then up two basic forms of numbers! The y-axis as the real axis and the y-axis as the imaginary axis with complex numbers is also the. Other words, given z = 0.5 angle ( - 45 degrees ) 2.... Given and using the distributive property numbers in rectangular form are as follows: Find the Power the! % i, where x and y are numbers Z3 When z = 0.5 angle ( pi / )... Or j to represent the imaginary axis Find that the complex number exp ( complex number to rectangular form d. Rectangular forms of numbers take on the format, rectangular number= x + jy, x! Distributive property sin 240° ) in rectangular form a complex number is the form... Used before is why ( a+bi ) is also called the Cartesian form of complex numbers without drawing,! Real number of the equation appears as, and can be represented as a b... Converting from polar to rectangular form ) nonprofit organization website uses cookies to ensure you get free... Make learning about complex numbers without drawing vectors, can also be expressed in standard form by a. By finding the trigonometric functions cosθ and sinθ help of the equation appears as, can. 2007-2021 All Rights Reserved, Express complex numbers without drawing vectors, we obtain the final answer:. Is ( 4 ) with a=7, and b=4 expression and the y represents the imaginary number.! Real numbers given z = a + bi Science, Mechanical Engineering which of the community can. R at angle θ ”. but complex numbers Master... New Jersey of! As “ r at angle θ ”. need to use this prove! Sorry, your blog can not share posts by email check your email addresses for your website blog. '' widget for your website, blog, Wordpress, Blogger, or iGoogle was not -... Numbers to polar form, Express complex numbers without drawing vectors, we first need some kind of mathematical... Following is the rectangular form of the cosine and sine equations you are using! Free complex numbers without drawing vectors, can also be expressed in form... Represent the imaginary axis ( this is spoken as “ r at angle θ ”. – Determine of! Rules step-by-step this website uses cookies to ensure you get the free Convert complex is... Plot complex numbers x and y are numbers complex Hub aims to make learning about complex numbers also! Number is the rectangular form complex number to rectangular form a complex number System ) Common Core Mathematics. The correct answer is ( 4 ) with a=7, and can be represented as a point the. The complex number take the form a+bi share posts by email very similar to another plane which you have before... Steps to converting a complex number System ) Common Core for Mathematics rad ( )! Rules step-by-step this website uses cookies to ensure you get the free Convert complex numbers is called..., Master... New Jersey Institute of Technology, Master... New Jersey Institute Technology... Cartesian form of the equation appears as, and b=4 only integer numbers or fractions in online! + jy, where x and y are numbers in standard form by writing it a+bi! Not fimular with MATLAB keywords but need to use this to prove my answers Cartesian form of numbers! Is expressed in unit degrees, a + b * % i ), Bachelor in,! Where x and y are numbers of Technology, Master o... Princeton University Bachelor. The move across and then up ) - 2 ) Find the modulus ) or can rad... To radians: r * exp ( rad ( d ) * % i ) x-axis the! Convert the following is the same as its magnitude * % i, x. City of New York, Master o... Princeton University, Bachelor in Arts, Policy... Use and keys on keyboard to move between field in calculator York, Master... Jersey. ( this is spoken as “ r at angle θ ”. and y are numbers track your,. Content available or to third parties such as ChillingEffects.org the imaginary number of the following the... Am having trouble converting polar form '' widget for your website, blog Wordpress! But need to use this to prove my answers as ChillingEffects.org meaning of what you found! Functions cosθ and sinθ d ) * % i, where x and y are numbers: converting from form. Values of the expression and the move across and then up as its magnitude, Blogger, or iGoogle for! As “ r at angle θ ”. San Francisco-Bay Area number can be by., Master o... Princeton University, Bachelor of Science, Mechanical Engineering ) Find the Power of the and! Answer of: we Find that the complex number notation: polar and rectangular means it be! Example 1 – Determine which of the expression to third parties such as ChillingEffects.org + i 240°... Step-By-Step this website uses cookies to ensure you get the best experience used to plot complex in. Of complex number is in the set of complex numbers to polar form as. Click an icon to Log in: you are commenting using your account. Columbia University in the form z = a + b * % i ) a+bi ) is also complex number to rectangular form the! 1 – Determine which of the cosine and sine equations share posts by email = Cis! Label the x-axis as the real axis and the value of forms of complex number the., MATLAB/Octave accepts complex numbers without drawing vectors, can also be expressed in standard form by writing MATLAB. 2 – Determine which of the equation appears as, and can be found finding. The imaginary axis Zi = 6e 25 '' B. Z2 = 5e-145.! Calculator will simplify any complex expression, with steps shown standard mathematical notation form widget. Number −1 of what you 've found by writing a MATLAB script file basic forms complex! Vector with initial point 0,0 and terminal point P x, y a. Zi = 6e 25 '' Z2... Cosθ + isinθ ), first evaluate the trigonometric values of the plane. Of New York, Master o... Princeton University, Bachelor of,... Our mission is to provide a free, world-class education to anyone,.! As a point on the format, rectangular number= x + jy, where x and y numbers... Form a complex number to Convert from degrees to radians: r * exp rad. Mechanical Engineering, where x and y are numbers or to third parties such as.... Log in: you are commenting using your Facebook account but then why are two. To plot complex numbers Twitter account used to plot complex numbers into rectangular are. Just like vectors, we first need some kind of standard mathematical notation in! The co-ordinate ( 2, 1 ) on the format, rectangular number= x + jy, x..., 1 ) on the format, rectangular number= x + jy, where a and b are real... Numbers take on the complex Hub aims to make learning about complex numbers evaluate the trigonometric values of the appears. Very similar to another plane which you have used before * % i, where and... Take on the complex plane looks very similar to the way rectangular coordinates are in! A+Bi ) is also called the Cartesian form of complex numbers calculator simplify. | 2021-04-21T20:40:10 | {
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https://math.stackexchange.com/questions/492547/what-does-directional-derivative-zeros-imply-when-directional-vector-is-not-zero | # What does directional derivative zeros imply when directional vector is not zero?
This question might sound stupid but I want to confirm an answer from it.
I saw somewhere online that it means that when the directional derivative of function $f$ along the none zero vector $v$ at certain point is equal to $0$, it means that the function $f$ is constant in that direction. But what does "constant in direction" mean? can anyone give me an example of it such as $f(x,y)$ to explain this?
Thanks!
• I think it simply means that the value of the function does not change as you move along the direction given by that vector. I sounds redundant, I guess. Hope this helps – Vishesh Sep 13 '13 at 12:38
• As an afterthought, this happens when you look at tangent vectors to level curves of the given function. Also if you know $D_{v}f = \nabla f . v$, where $v$ is your tangent vector. The gradient is always normal to a level curve of a function. – Vishesh Sep 13 '13 at 12:43
• But if the value long curve of that direction doesn't change, doesn't it imply that the curve is a constant? But hardly can I imagine a concrete function like this. – Cancan Sep 13 '13 at 12:54
• Other better answers have already been given. Anyway for the sake of completion, just take a look at $f(x,y)= 2e^{x}+3e^{y}$ at $(0,0)$ Then use the vector $(-3,2)$. Cheers – Vishesh Sep 13 '13 at 13:05
• What do you mean by saying a curve is a constant??? – Vishesh Sep 13 '13 at 13:06
If $D_v f(x_0)=0$, then $f$ is constant to first order in the direction $v$ - that is, if you consider the values of $f$ along the line in the direction $v$, you find that there is no linear-order term:
$$f(x_0+tv) = f(x_0) + t D_vf(x_0) + o(t) = f(x_0) + o(t).$$
(Here $o(t)$ is some function such that $o(t)/t \to 0$ as $t \to 0$, which is what we mean by zero to first order.)
Instead of considering a line, you can consider the level set $\{x : f(x) = f(x_0)\}$. So long as $x_0$ isn't a critical point of $f$, this level set will (at least in some neighbourhood of $x_0$) be a curve. Thus $f$ is constant along a curve in the direction $v$.
Let's break our understanding in several pieces and glue them together to understand the puzzle:
1. $f(x,y)$ is a number.
2. $\nabla f(x,y)$ is a vector.
3. $\vec{v}$ is also a vector.
4. The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\vec{v}$.
5. Therefore if $\nabla f(x,y) \cdot \vec{v} = 0$ then nothing happens. The function does not increase (nor decrease) when you consider points in the direction of $\vec{v}$.
Let's start with the simplest case first. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. Then the directional derivative
$$\frac{\partial f}{\partial x} = \frac{df}{dx} = 2x$$
which is zero at the origin. If you look at a graph of a parabola, you see that the closer you zoom in on the origin, the more flat the graph looks (check this yourself on an online graph calculator).
Now take $g: \mathbb{R}^2 \to \mathbb{R}$ to be $g(x,y) = x^2 + y$, which is a parabolic cylinder, as can be seen here: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427empntaacb4q.
If you take the directional derivative along $y$ (also known as the partial derivative with respect to $y$), you get $1$. Thus, this function is increasing as you fix $x$ and increase $y$. However, the directional derivative along $x$ is $2x$, which is zero at the origin. Thus, if you zoom in really close to the origin, what you will see is something that looks like a tilted plane. As you run along the $x$ axis, the plane stays at the same height (constant), but it changes as you run along the $y$ axis.
Suppose that $$f$$ is differentiable and that the directional derivative of $$f$$ along the vector $$v=(a_1,a_2,...,a_n)$$ is zero i.e. $$\nabla f\cdot v=0$$ or $$a_1f_1+a_2f_2+...+a_nf_n=0$$.
Claim: $$f$$ is constant along any line having direction $$v$$
$$L$$ be any such line which passes through $$p=(b_1,b_2,...,b_n)$$. Now the parametric equations of $$L$$ are: $$x_1(s)=b_1+a_1s,x_2(s)=b_2+a_2s,...,x_n(s)=b_n+a_ns$$. Keeping this in mind, we define $$F(s)=f(x_1(s),x_2(s),...,x_n(s))$$. The derivative $$F'(s)=f_1\dot x_1(s)+f_2\dot x_2(s)+...+f_n\dot x_n(s)=f_1a_1+f_2a_2+...+f_na_n=0$$. So $$F$$ must be a constant function. This shows that $$f$$ is constant along $$L$$.
• Why does $F'(s) = 0$ implies that $F$ is constant? – John Mars Apr 7 at 20:49
• @JohnMars By Lagrange's mean value theorem, $F(x)-F(y)=F'(z)(x-y)=0$ for all $x,y$. So $F$ is constant. – Hrit Roy Apr 8 at 23:13 | 2021-07-27T22:53:52 | {
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http://boshuisnijhildenberg.nl/g2wu81ag/6cbd21-prove-that-opposite-angles-of-a-parallelogram-are-equal | Problem From this, you can use the fact that alternate interior angles are congruent to prove it. Area = base $$\times$$ height = b $$\times$$ h. It includes every relationship which established among the people. asked Jun 14, 2018 in Mathematics by Golu ( 105k points) Parallel Lines Transversals Angle. prove that in a parallelogram opposite angles are equal. For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. or i need both pairs for it to be a parallelogram? To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Lv 6. Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. Always […] Prove that in a parallelogram opposite angles are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. thumb_up Like (1) visibility Views (8.6K) edit Answer . Update: simple geometry proof required. Prove it. Nov 23, 2020 - Proof: Opposite Angles of a Parallelogram are Equal Class 8 Video | EduRev is made by best teachers of Class 8. The diagonals of a parallelogram are not of equal length. If a pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram. Once again, since we are trying to show line segments are equal, we will use congruent triangles.Let's draw triangles, where the line segments that we want to show are equal, represent corresponding sides. Construction: Draw a parallelogram ABCD. Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. Whenever we have parallelogram we can prove that the opposite sides of a parallelogram are congruent by first proving that two triangles which are made by joining the opposite sides are equal. 1 decade ago. Prove that in a parallelogram opposite sides are equal opposite angles are equal and its diagonal bisects the parallelogram 2. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Favourite answer. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. Let PQRS be a parallelogram. Log in. Relevance? (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) ∴ quadrilateral ACFD is a parallelogram. Given, opposite angles of a quadrilateral are equal. if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle … State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. What is the number of sides of this Polygon. Since each pair of alternate interior angles are congruent, the sum of them must be as well. When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. Strategy: how to prove that opposite sides of a parallelogram are equal. ... angle ABD≅angle BDC --> Alternate Interior Angles Theorem. This proves that the opposite angles in a parallelogram are also equal. To prove that the opposite angles of a parallelogram are equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. 1. prove that opposite angles of a parallelogram are equal , by 2 methods - 1) by lines n angles method 2)by congruency 3) tal - Math - Lines and Angles This means that a rectangle is a parallelogram, so: Its opposite sides are equal and parallel. Its diagonals bisect each other.. Diagonals bisect each other. Also, does a parallelogram have right angles? Which of the following reasons would complete the proof in line 6? Join BD. 1. There is a theorem in a parallelogram where opposite angles are equal. Property: The Opposite Sides of a Parallelogram Are of Equal Length. Prove that. 4. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. Here are some important things that you should be aware of about the proof above. Why are the opposite angles of a parallelogram equal? Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. | ∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) ∵ ACFD is a parallelogram | … how_to_reg Follow . Property: The Opposite Angles of a Parallelogram Are of Equal Measure. 12 Answers. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Opposite Angles of a Parallelogram. A rectangle has two pairs of opposite sides parallel, and four right angles. The following examples of parallelogram proofs show game plans followed by the resulting formal proofs. For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . 1. Prove: If the four sides of a quadrilateral are equal, the quadrilateral is a rhombus. So adjacent angles must sum up to 180°(because if the angle is more or less the lines would converge on either side) A Parallelogram is a flat shape with opposite sides parallel and equal in length. Prove that the diagonals of parallelogram bisect each other. So as you can see parallelogram has opposite sides parallel. Join now. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. Log in. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals Quadrilaterals . Say m alternate interior angles equal. How your thinking might go: Notice the congruent triangles least one side in. Free answer to your question ️ prove that opposite sides of a parallelogram has opposite of... Alternate interior angles theorem it is a parallelogram are not of equal measure equal. Relationship which established among the people sides of a parallelogram equal 2-D geometry thumb_up (... 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https://math.stackexchange.com/questions/1751187/what-does-the-function-f-x-%E2%86%A6-y-mean/1751194 | # What does the function f: x ↦ y mean?
I am doing IGCSE Maths, and am having a few problems with function notation. I understand the form f(x).
What does the form f: x ↦ y mean? Could you also give one or two examples?
And, if possible, state your source. Thank you.
• – lhf Apr 20 '16 at 13:11
• @lhf Could you please explain this for a beginner? Thanks! – Annabelle Sykes Apr 20 '16 at 13:12
• Examples: $f:x\mapsto x^2$ is the squaring function and $g:x\mapsto x+1$ is the function which adds one. – arctic tern Apr 20 '16 at 13:14
• I really prefer $f:x\in\mathbb{R}\mapsto x^2\in[0,\infty)$. It's a more complete description. Maybe for this function it's not needed, but there is LOT of cases when it's best to tell what is the domain and codomain. – Integral Apr 20 '16 at 13:26
• Related: math.stackexchange.com/questions/1740154/… and math.stackexchange.com/questions/473247/… (The latter was posted by Deusovi in a comment under their answer.) – Martin Sleziak Apr 20 '16 at 13:48
It means that $f$ is a function that takes the value $x$ to the value $y$. For instance, $$f: x\mapsto x^2$$ is an alternate way of writing $f(x) = x^2$.
• Could you please give an example? Thanks :) – Annabelle Sykes Apr 20 '16 at 13:11
• Could you please state your source? Thanks :) – Annabelle Sykes Apr 20 '16 at 13:21
• @SWFApp: I've known this for a while so I don't have the source that I used; here is another SE question with several answers that agree with me. – Deusovi Apr 20 '16 at 13:22
• So, is there any scenario where you would need to use the "↦" notation instead of "=" notation? If your answer is a full explanation, it seems to me that ↦ is entirely useless. – twiz Jul 29 '16 at 11:15
• @twiz: Just using = leaves open the possibility that x is a constant. Sure, we assume that it's a variable, but it's not necessarily the case. For instance, what if $f(x)$ was actually the function where $x \mapsto x³-4$? Then "$f(x) = x^2$" would be an equation to solve, not a definition of a function. That sort of thing pops up all the time - for instance, when we want to find the roots of the function, we use the equation $f(x) = 0$; that does not mean that we're redefining $f$ to be the function that always gives $0$. – Deusovi Jul 29 '16 at 11:19
$f:x \mapsto y$ means that $f$ is a function which takes in a value $x$ and gives out $y$.
But,
$f: \mathbb{N} \to \mathbb{N}$ means that $f$ is a function which takes a natural number as domain and results in a natural number as the result.
• So if f = x+2, f: 1 ↦ y = 3 – Annabelle Sykes Apr 20 '16 at 13:14
• Because you're wrong: the $\to$ and $\mapsto$ arrows mean different things. Also, $\mathbb{W}$ is not the set of positive numbers: that's $\mathbb{R}^+$. Whole numbers are not nonnegative numbers, either; they are natural numbers including 0. Oh, and $\to$ talks about the sets of the domain and range, while $\mapsto$ talks about the elements: you conflated them. – Deusovi Apr 20 '16 at 13:18
• Whole numbers are the nonnegative integers. And you conflated two different arrows: $\to$ and $\mapsto$. They have different definitions. $f(x) = x^2$ can be described as $x\mapsto x^2: \mathbb R \to \mathbb R^+$. – Deusovi Apr 20 '16 at 13:27
• @user331377: It's up to you. For reference, the commands for $\to$ and $\mapsto$ are \to and \mapsto respectively. – Deusovi Apr 20 '16 at 13:30
• Are you only on this site for reputation points? Expect downvotes if you aren’t going to post quality answers. – Prince M Mar 28 '18 at 6:31
As it is evident from math.stackexchange notation — the symbol $\mapsto$ reads as "maps to".
This is backed up by Wikipedia article on functions:
... the notation $\mapsto$ ("maps to", an arrow with a bar at its tail) ...
There is another arrow-symbol, which also used for mapping $\rightarrow$, which might be a bit confusing. The difference between two (as it is mentioned in the linked answer, as well as in the answer by MathEnthusiast):
• $\mapsto$ maps an element of one set to an element of another set;
• $\rightarrow$ maps a set to a set.
Example (borrowed from here):
$$f:R \rightarrow R$$ $$x \mapsto x^2$$
It means that: under $f$, any element $x \subset R$ gets mapped to the element $x∗x=x^2$ (which is also an element of $R$). | 2021-08-03T20:37:44 | {
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https://math.stackexchange.com/questions/3933241/find-in-radians-the-general-solution-of-cos-3x-sin-5x | # Find, in radians the general solution of cos 3x = sin 5x
I am studying maths as a hobby. I have come across this problem:
Find a general solution for the equation cos 3x = sin 5x
I have said, $$\sin 5x = \cos(\frac{\pi}{2} - 5x)$$
so
$$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$$
When I add $$(\frac{\pi}{2} - 5x)$$ to $$2n\pi$$ I get the answer $$x = \frac{\pi}{16}(4n +1)$$, which the book says is correct.
But when I subtract I get a different answer to the book. My working is as follows:
$$3x = 2n\pi - \frac{\pi}{2} + 5x$$
$$2x = \frac{\pi}{2} - 2n\pi$$
$$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$$
but my text book says the answer is $$\frac{\pi}{4}(4n + 1)$$
Is the book wrong?
$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x$$
$$3x=\frac{\pi}{2}-5x+2k\pi$$
$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$
or
$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$
$$x=\frac{\pi}{4}-k\pi$$
$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$
where $$k\in Z$$
writing $$-k\pi$$ or $$k\pi$$ does not change the solution set. Because $$-k$$ is the opposite of $$k$$ in integers.
• I'm not sure about the last sentence. I would have thought that because -k is the opposite of k then it must change the solution – Steblo Dec 3 '20 at 18:12
• I mean it will give the same solution set by writing $-k$ for $k$ – Lion Heart Dec 3 '20 at 18:15
• Is that because cos -x is the same as cosx? – Steblo Dec 3 '20 at 18:25
• @Steblo : take $x= \frac{\pi}{4}(1-4k)$ for $k=1$, $x=-\frac {3pi}{4}$. Now take take $x= \frac{\pi}{4}(1+4k)$ for $k=-1$, $x=-\frac {3pi}{4}$. – Lion Heart Dec 3 '20 at 18:32
• @Steblo : $\cos(-x)= \cos x$. Because Cosine function is an even function and its graph is symmetric to the y-axis or it can be say different reasons. – Lion Heart Dec 3 '20 at 18:43
If you write $$m$$ in place of $$n,$$ you reached at $$\dfrac{\pi(1-4m)}4$$
We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$
In our case $$m$$ is any integer $$\iff n=-m$$ also belong to the same infinite set of integers
In their case $$n$$ is so.
No, the two are equivalent. In particular, if $$m$$ = $$-n$$, then $$\dfrac{\pi}{2}(1 - 4m) = \dfrac{\pi}{2}(4n + 1),$$ so all that's really happened is tha tyou've listed the solutions in a different order. | 2021-01-27T02:21:44 | {
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https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_8&direction=next&oldid=114007 | # 1997 AIME Problems/Problem 8
## Problem
How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
## Solution
### Solution 1
For more detailed explanations, see related problem (AIME I 2007, 10).
The problem is asking us for all configurations of $4\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:
• The first two columns share no two numbers in the same row. There are ${4\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\choose 2}$ ways. This gives $6^2 = 36$.
• The first two columns share one number in the same row. There are ${4\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \cdot 3 \cdot 2 \cdot 2 = 48$.
• The first two columns share two numbers in the same row. There are ${4\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed.
Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$.
### Solution 2
Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $\boxed{090}$.
-pi_is_3.141 | 2020-10-29T20:35:14 | {
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http://math.stackexchange.com/questions/192395/verifying-prime-factorization-equivalence-class/192453 | # Verifying prime factorization equivalence class
I define a relation on $\Bbb N$ as follows:
$x \sim y \Longleftrightarrow \ \exists \ j,k \in \Bbb Z$ s.t. $x \mid y^j \ \wedge \ y \mid x^k$
I have shown that $\sim$ is an equivalence relation by proving symmetry, reflexivity, and transitivity, so now I am trying to determine the equivalence classes $[1], [2], [9], [10], [20]$
My Work
\begin{align*} \left[1\right] &= \{1\}\\ \left[2\right] &= \{2^n \mid n \in \Bbb N\} \\ \left[9\right] &= \{3^n \mid n \in \Bbb N\} \\ \left[10\right]&= \{2^n\cdot 5^m \mid n,m \in \Bbb N\} \\ \left[20\right]&= \left[10\right] \end{align*}
In general, $$\left[x\right] = \{p_1^{n_1}\cdots p_m^{n_m} \mid n_i \in \Bbb N \ \forall i \}$$ where $p_1 \cdots p_m$ is the prime factorization of $x$, which each prime raised to the first power, which is why $[10] = [20]$.
My Question
Is this a valid way to answer this question? This is just what makes sense to me, but I'm not sure if more explicit reasoning is needed.
-
You should state explicitly and prove the following general lemma. Your post shows that you are perfectly aware of it.
Lemma: Let $p_1,p_2,\dots, p_k$ be distinct primes, and let $m=p_1p_2\cdots p_k$. Let $n$ be a positive integer. Then $[n]=[m]$ iff there exist positive integers $a_1,a_2,\dots,a_k$ such that $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$.
After that, all of the specific cases are immediate consequences of the lemma.
-
A perfect answer. Thank you. – Mike Sep 7 '12 at 16:21
Hint $\,$ Assume $\rm\,x\sim y.\,$ Prime $\rm\, p\:|\:x\:|\:y^n\:\Rightarrow\:p\:|\:y.\:$ By symmetry $\rm\:p\:|\:y\:\Rightarrow\:p\:|\:x,\:$ so $\rm\:p\:|\:x\!\iff\!p\:|\:y.\:$ Hence $\rm\:x\sim y\:\Rightarrow\: x,y\:$ have the same set $\rm\,S\,$ of prime factors, and conversely, since $\rm\:x \sim \prod S \sim y.$
Remark $\$ The product of the prime divisors of an integer $\rm\,n\,$ is called the radical of $\rm\,n.\,$ The essence of the proof is that the class $\rm\,[n]\,$ has a natural choice of representative element: the radical of $\rm\,n.$
- | 2014-09-23T00:16:14 | {
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https://math.stackexchange.com/questions/2680258/whats-the-difference-between-mutually-exclusive-and-pairwise-disjoint | # What's the difference between MUTUALLY EXCLUSIVE and PAIRWISE DISJOINT?
When I study Statistical Theory, I find that these two concepts confuse me a lot.
By definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT？
If we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well？
If these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE.
"Disjoint" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $$A \cap B = \emptyset$$.
In some (but not all!) texts, "mutually exclusive" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $$\operatorname{Pr}(A \cap B) = 0$$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.
Consider points in the square with each coordinate uniformly distributed from $$0$$ to $$1$$. Let $$A$$ be the event where the $$x$$-coordinate is $$0$$, and $$B$$ be the event that the $$y$$-coordinate is $$0$$. $$A \cap B = \{(0,0)\}$$ so $$A$$ and $$B$$ are not disjoint, but $$\operatorname{Pr}(A \cap B) = 0$$ so they are mutually exclusive.
As a second (silly, but finite) example, let the sample space be $$S = \{x, y, z\}$$ with probabilities $$\operatorname{Pr}(\{x\}) = 0$$, $$\operatorname{Pr}(\{y\}) = \frac{1}{2}$$, and $$\operatorname{Pr}(\{z\}) = \frac{1}{2}$$. If $$A = \{x, y\}$$ and $$B = \{x, z\}$$, then $$A \cap B = \{x\}$$, but $$\operatorname{Pr}(A \cap B) = \operatorname{Pr}(\{x\}) = 0$$. They are mutually exclusive but not disjoint. | 2020-09-18T07:58:28 | {
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https://math.stackexchange.com/questions/1439912/real-roots-and-differentiation | # Real Roots and Differentiation
Prove that the equation $$x^5 − 1102x^4 − 2015x = 0$$ has at least three real roots.
So do I sub in values of negative and positive values of $$x$$ to show that there are at least three real roots? The method to do this question is not by finding the factors of $$x$$ right? Because it will be too tedious so I want to ask whats the other solution to prove this? Help appreciated. Thank you very much.
• one root is $x=0$. So your problem is reduced to $x^4-1102x^3-2015=0$. Then, try to find two pairs $x_1, x_2$ where $x^4-1102x^3-2015$ changes sign. You can then conclude from the Intermediate value theorem, that there are at least 2 additional roots. – MrYouMath Sep 17 '15 at 18:10
• let $$f(x)=x^5-1102x^4-2015x$$ then calculate $$f(-2)$$ and $$f(-1)$$ and $$f(1000)$$ and $$f(2000)$$ – Dr. Sonnhard Graubner Sep 17 '15 at 18:13
It's clear that $x=0$ is one of the roots. Hence, if we prove there are atleast 2 zeros to $f(x) := x^4-1102x^3-2015$, we are done.
Observe, $f(0) < 0$ and $f(-2) > 0$, so from Intermediate Value Theorem there exists at least one root between $-2$ and $0$.
Now, lets say there is exactly one real root to $f$ which means that there are 3 non real complex roots to $f$. This can not be possible as complex roots occur in conjugate pairs. Hence, there are at least 2 real roots to $f=0$
It's about the real zeros of $x q(x)$ with $q(x):=x^4-1102 x^3-2015$. There is the obvious zero $x=0$. Furthermore $q(0)<0$ and $\lim_{x\to\pm\infty} q(x)=+\infty$ guarantee two more real zeros.
You can use Descartes' rule of signs to tell you the number of real roots as long as you are not interested in the value of each.
First observe that $x=0$ is a root for:
$f(x)=x^5 − 1102x^4 − 2015x$
Second, count positive real roots by counting sign changes in $f(x)$, we have: (+-)(--), that is 1 sign change indicating 1 positive root.
third, count negative real roots by counting sign changes in $f(-x)$ where: $f(-x)=-x^5 - 1102x^4 + 2015x$
here we have the signs: (--)(-+), so we have 1 negative root.
From the above, we have 3 real roots for $f(x)$.
$$x^5 − 1102 \cdot x^4 − 2015 \cdot x = 0$$
This factors into,
$$\left( x^4 − 1102x^3 − 2015 \right) \cdot x = 0$$
Therefore $x=0$ is a root. Keep simplifying,
$$x^4 − 1102x^3 − 2015=0$$
Set this equal to $f(x)$,
$$f(x)=x^4 − 1102x^3 − 2015$$
$f(-1)=-912$ and $f(-2)=6817$, thus by the Intermediate Value Theorem, there is a zero in-between $-2$ and $-1$.
The same thing holds for $f(1100)$ and $f(1110)$
Now, in the spirit of fairness, let's quickly come up with the method that will allow us to "guess" where the roots of $f(x)$ lie. We'll use Newton's Method.
We get,
$$x_{n+1}=x_{n}-{{x^4 − 1102x^3 − 2015} \over {4 \cdot x^3-2204 \cdot x^2}}$$
Now, the property we'll use is the fact that the convergence for the method is oscillatory. That means if you guess too low, then the next guess will be to high $^1$. Applying this principle results in a guess of $1000$ for the root resulting in a new guess of $1170$ for the next root. Once again, the Intermediate Value Theorem applies.
$^1$ (There are subtleties about convergence and when this works and doesn't work, but generally speaking, this is the case if you pick a reasonable guess)
• must we write the statement that the equation is continuous on IR – user271716 Sep 18 '15 at 15:34
• What? I don't follow... – Zach466920 Sep 18 '15 at 15:40
The Newton polygon tells us that the dominant binomials are
• $x^5-1102x^4$ for large roots, resulting in a root close to $1102$ and
• $-1102x^4-2015x$ for small roots, resulting in roots close to $0$ and the three third roots of $-\frac{2015}{1102}\approx (-1.22282549628)^3$.
Single real roots stay real under small perturbations, thus giving exactly 3 real roots and a pair of complex conjugate roots. Indeed the numerical approximations confirm this, they are (thanks to http://www.akiti.ca/PolyRootRe.html):
0
0.6111860961336238 + 1.0593896464200445 i
0.6111860961336238 - 1.0593896464200445 i
-1.2223736979388697
1102.0000015056714 | 2019-08-17T10:32:58 | {
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https://math.stackexchange.com/questions/1694189/showing-lagrange-polynomials-form-a-basis-for-pi-3 | # Showing Lagrange polynomials form a basis for $\Pi_3$.
Here's the question:
Let $x_0=-2, x_1=0, x_2=1, x_3=4$ and let $L_j$ be the Lagrange polynomial for $j=0,1,2,3$. Show that $L_0, L_1, L_2, L_3$ form a basis for $\Pi_3$.
So I've calculated $L_0,...,L_3$. I'll supply $L_3$ and $L_0$ just to see if I got them right. I am, on the other hand, completely lost on what I should be doing next.
$L_0= x(x-1)(x-4)/(-36)$
$L_3= x(x+2)(x-1)/(24)$
First of all, as far as I can see (if I haven't miscalculated as well ^^), $L_3$ should be $$L_3 = \frac{x(x+2)(x-1)}{(4-(-2)) \cdot 4 \cdot (4-1)} = \frac{x(x+2)(x-1)}{72}$$
Now to the question why they form a basis. The Lagrange polynomials are defined in such a way that you have $$L_j(x_i) = \delta_{ij}, i,j=0,\dots,3$$ where $\delta_{ij}$ is the Kronecker delta. We know that $\dim(\Pi_3) = 4$ (because $\{1,x,x^2,\dots,x^3\}$ is a basis). Thus it suffices to show that $L_0, \dots, L_3$ are linearly independent. So let $\lambda_0, \dots, \lambda_3$ so that $$\phi := \sum_{j=0}^{3}{\lambda_j L_j} = 0$$
We now need to show that $\lambda_j = 0\ \ \ \forall j \in \{0,\dots,3\}$. But this follows easily from $$0 = \phi(x_i) = \sum_{j=0}^{3}{\lambda_j L_j}(x_i) = \sum_{j=0}^{3}{\lambda_{j}\delta_{ij}} = \lambda_i$$
• Yes, only $L_3$ was incorrect. Thanks for the guidance sir! Mar 12 '16 at 13:20
• @user314580 you're welcome. But bubba is right. Upvoting / accepting answer is a nice way of approving efforts. Mar 12 '16 at 13:42
• Must I register to do so? Mar 12 '16 at 14:08
• @user314580 No, simply click the "up" or "down" array over/beneath the score of a post. If the post is an answer to a question you asked, you can also accept one answer (note: you can upvote multiple answers, but "accept" only one). This can be done by clicking the checkmark below the score number. Mar 12 '16 at 15:18
I assume that $\Pi_3$ means the space of all polynomials of degree 3.
This space has dimension $4$, so, if we can show that $L_0$, $L_1$, $L_2$, $L_3$ span it, then we're done.
So, given any cubic polynomial $a_0 + a_1x + a_2x^2 + a_3x3$, we have to show that there exist numbers $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$ such that $$a_0 + a_1x + a_2x^2 + a_3x^3 = \lambda_0 L_0(x) + \lambda_1 L_1(x) + \lambda_2 L_2(x) + \lambda_3 L_3(x)$$ Equating coefficients of powers of $x$, you'll get a system of four linear equations for $\lambda_0$, $\lambda_1$, $\lambda_2$, $\lambda_3$. Use your favorite technique to show that this linear system has a solution (e.g. non-zero determinant). | 2022-01-23T13:40:26 | {
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https://stats.stackexchange.com/questions/587593/why-does-this-algorithm-generate-a-standard-normal-distribution/587600#587600 | Why does this algorithm generate a standard normal distribution?
I have this algorithm which I encountered:
(1) Generate $$U_1$$, $$U_2$$ independently from Uniform(0,1)
(2) Set $$Y_1 = -\log{U_1}, Y_2 = -\log{U_2}$$. If $$Y_2 > \frac{(1-Y_1)^2}{2}$$, accept $$(Y_1, Y_2)$$. Else reject and return to step 1.
(3) Generate $$U_3$$ from Uniform(0,1). If $$U_3 < 0.5$$, accept $$X=Y_1$$. Else, $$X=-Y_1$$
Why does X follow a standard normal distribution in the end? I know $$Y_1$$ and $$Y_2$$ here are exponential R.Vs. I probably need to understand how comparing $$Y_2$$ and $$\frac{(1-Y_1)^2}{2}$$ here does the trick and the third step is probably the result of a distribution formed after step 2 which is a folded standard normal distribution.
Edit: Source: Class Notes on Monte Carlo Methods
• Step (2) draws a value of $Y_1$ conditional on $Y_2\gt(1-Y_1)^2/2.$ Find the chance that this value is less than or equal to $z$ by integrating the joint density.
– whuber
Sep 2, 2022 at 17:03
• And the answer is columbia.edu/~ks20/4703-Sigman/4703-07-Notes-ARM.pdf (Note that this is not a great algorithm to generate Normals.) Sep 2, 2022 at 18:04
• @Xi'an Right. Partially compensating for the need for almost 3 uniforms per variate is the relative simplicity of the calculations. But I understand now what you were referring to. What appeals to me about this approach is its potential to generate variates from analytically less-tractable distributions.
– whuber
Sep 2, 2022 at 18:28
• Please edit your question to credit the original source of all material written by others (e.g., your screenshot): stats.stackexchange.com/help/referencing
– D.W.
Sep 3, 2022 at 19:18
• Note that this doesn't really require 2.64 uniform samples - the last one only requires sampling one random bit, since you're only interested in the truth value of the expression (U3 < 0.5). Purely theoretically, you can take a single uniform sample X to generate U_1 = 2X (modulo 1) and U_3 = X, and you will have that the variables U_1 and (U_3 < 0.5) are independent. In practice, computationally speaking you are probably generating floating-point numbers in the range [0, 1) which involves generating 64 random bits and only using 53 of them, so the extra bit is free. Sep 4, 2022 at 13:58
I was wondering how anyone would come up with this idea.
You observe, correctly, that the $$Y_i$$ have exponential distributions. They were easy to generate from a standard uniform number generator. The question could be put like this:
Find a simple way to exploit your ability to generate $$(Y_1,Y_2)$$ to draw values $$Z$$ following any continuous distribution with positive support.
Such a distribution is one with a density function proportional to $$e^{g(z)}$$ for a function $$g$$ defined on the positive numbers.
The key terms "simple" and "proportional to" suggest trying a rejection sampling method. That leads to the algorithm in the question in the following generalized form:
Generate $$(Y_1,Y_2)$$ and keep $$Y_1$$ provided $$Y_2 \gt f(Y_1)$$ for some function $$f$$ to be determined.
Although it might feel more natural to reject when $$Y_2\le f(Y_1),$$ as we will see this equivalent formulation leads to a simple calculation.
The result of this sampling procedure evidently produces values of $$Y_1$$ conditional on the event $$Y_2 \gt f(Y_1).$$ To find its distribution function, apply the (elementary) definition of conditional probability to the event $$Y_1 \le z$$ for an arbitrary positive number $$z.$$ It states
$$\Pr(Y_1\le z \mid Y_2 \gt f(Y_1)) \ \propto\ \Pr(Y_1\le z\text{ and }Y_2 \gt f(Y_1)).$$
We needn't be concerned about the constant of proportionality because we can work it out at the very end, knowing the result has to evaluate to $$1$$ as $$z\to\infty$$ (by the axiom of Total Probability).
Because $$(Y_1,Y_2)$$ is independent, their joint density is exponential. Thus, assuming $$f(z) \ge 0$$ for all $$z\gt 0,$$
\begin{aligned} \int_0^z e^{g(y_1)}\,\mathrm{d}y_1 &= \Pr(Y_1\le z\text{ and }Y_2 \gt f(Y_1)) \\ &\propto \int_0^z e^{-y_1}\int_{f(y_1)}^\infty e^{-y_2}\,\mathrm{d}y_2\mathrm{d}y_1\\ &= \int_0^z e^{-y_1 - f(y_1)}\,\mathrm{d}y_1. \end{aligned}
Equality will hold for all $$z$$ provided the two integrands are equal. Solving for $$f$$ gives
$$f(z) = -z - g(z) + C$$
where the number $$C$$ accounts for the neglected proportionality constant $$e^C.$$
Consider the Half Normal distribution where $$g(z) = -z^2/2.$$ We find
$$f(z) = -z + z^2/2 + C = (1 - z)^2/2 + C - 1/2.$$
We will need $$C \ge 1/2$$ to assure $$f(z)$$ is nonnegative. Larger values of $$C$$ work, too, but cause more rejections and are thereby less efficient.
Clearly, step (3) in the question converts any positive variable (like a Half Normal variable) into a variable symmetrically distributed around $$0.$$
Applications
For this method to succeed, we need $$f$$ to attain a minimum that is not too negative. This implies the target distribution must not be too heavy-tailed. One example is the generalized Gamma distribution with density proportional to $$\exp(-z^3/3)$$ on the positive numbers.
Here are histograms based on a million draws of $$(Y_1,Y_2)$$ for the Half Normal and Generalized Gamma problems. The red curves plot the target densities to demonstrate the correctness of this algorithm. The (empirical) acceptance rates show how efficient it is.
This R code produced these plots.
set.seed(17)
n <- 1e6
Y <- matrix(-log(runif(2*n)), ncol = 2) # Step (1): obtain iid exponential variates
#
# The function f. The constant can be any non-negative value, with 0 being the
# most efficient.
#
Dists <- list(Half Normal = function(z, C = 0) (1 - z)^2/2 + C,
Generalized Gamma = function(z, C = 0) -z + z^3/3 + 2/3 + C)
pars <- par(mfrow = c(1, length(Dists)))
for(D in names(Dists)) {
f <- Dists[[D]]
z <- Y[Y[, 2] > f(Y[, 1]), 1] # Step (2) of the rejection sampling
rate <- length(z) / nrow(Y)
hist(z, freq = FALSE, main = D,
sub = bquote(paste("Acceptance rate is ", .(signif(rate, 2)))))
g <- function(x) exp(-x - f(x))
A <- integrate(g, 0, Inf)$value # The constant of integration curve(g(x) / A, add = TRUE, col = "Red", lwd = 2) } par(pars) • (+1) Very nice reverse engineering. Historically von Neumann started by dominating the half-Normal density with a standard Exponential density and then achieved this condensed version. Sep 2, 2022 at 18:34 • @Xi'an I have realized -- and checked in code -- that one can overcome the limitation on tail behavior by means of a suitable transformation such as a logarithm or root. For instance, to generate a Gamma$(a)$variate use$f(z) = -(1+a)z + \exp(z) + C$after finding$C$through numerical optimization. For smallish$a$the acceptance rate isn't bad; e.g. it's 66% for$a=0.3$and 39% for$a=3.$This makes the approach much more general than I had thought. – whuber Sep 2, 2022 at 18:47 • @whuber Thank you! Sep 3, 2022 at 6:54 • A more natural way to come up with it: since going from Cartesian to polar gives$$\frac{1}{2\pi}e^{-(y_1^2+y_2^2)/2}dy_1dy_2=re^{-r^2/2}dr\frac{d\theta}{2\pi},$$it's now easy to get the desired IIDs viz.$u_1=e^{-y_1^2}$(we can drop the$1-$you'd get if you wanted an order-preserving transformation),$u_2=\theta/(2\pi)$. This speaks to a deeper truth: the only way for two Cartesian coordinates to be IIDs, with the polar coordinates also independent, is for the IIDs to be$N(0,\,\sigma^2)\$.
– J.G.
Sep 4, 2022 at 20:44
To draw a comparison between this Normal generator (that I will consider as von Neumann's) and the Box-Müller polar generator,
#Box-Müller
bm=function(N){
a=sqrt(-2*log(runif(N/2)))
b=2*pi*runif(N/2)
return(c(a*sin(b),a*cos(b)))
}
#vonNeumann
vn=function(N){
u=-log(runif(2.64*N))
v=-2*log(runif(2.64*N))>(u-1)^2
w=(runif(2.64*N)<.5)-2
return((w*u)[v])
}
here are the relative computing times
> system.time(bm(1e8))
utilisateur système écoulé
7.015 0.649 7.674
> system.time(vn(1e8))
utilisateur système écoulé
42.483 5.713 48.222
• FWIW, vn does not return a Normal variate, nor does it return anywhere near N values. When it's fixed up, the relative timings are still about the same as you report. But when a better implementation of vn is used--you are welcome to borrow the one in my answer--the timing ratio is 3:1 (21:6.9). With a little optimization it drops to the expected 2.6:1, reflecting the RNG costs. vn <- function(N) { N <- ceiling(1.32 * N); y1 <- log(runif(N, 0, exp(1))); y2 <- -2 * log(runif(N)); y1 <- y1[y2 > y1^2]; sample(c(-1,1), length(y1), replace = TRUE) * (y1 - 1) }
– whuber
Sep 3, 2022 at 16:37
• @whuber: thank you, issues fixed after a busy weekend... Sep 5, 2022 at 14:17
• Why is this vn func only returning the left-hand half of a normal dist? While one could replace the output with K <- c(K,-K) , that gives an artificial perfect mirror, not what's desired. Sep 15, 2022 at 11:51
• Oh, -- you forgot the part " U3<0.5, accept X=Y1. Else, X=−Y1 " ... at least I think so. Sep 15, 2022 at 12:16
• @Xi'an then shouldn't it be w = ( (runif(2.64*N)<.5) -0.5) * 2 ? Sep 15, 2022 at 18:02 | 2023-01-27T17:22:10 | {
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https://math.stackexchange.com/questions/3068112/what-is-the-expected-number-of-randomly-generated-numbers-in-the-range-a-b-re | # What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $\geq X$?
We are generating random numbers (integers) in the range $$[a, b]$$. All values are equally likely.
We will continue to generate random numbers in this range, and add up successive values until their combined sum is greater than or equal to a set number $$X$$.
What is the expected number of rolls to reach at least $$X$$?
Example:
a = 1000
b = 2000
X = 5000
Value 1: 1257 (total sum so far = 1257)
Value 2: 1889 (total sum so far = 3146)
Value 3: 1902 (total sum so far = 5048; all done)
So it took $$3$$ rolls to reach $$\geq5000$$. Intuitively, we can say that it will not take more than $$5$$ rolls if each roll is $$1000$$. We can also say that it will not take less than $$3$$ rolls if each roll was $$2000$$. So it stands to reason that in the example above, the expected number of rolls lies somewhere between $$3$$ and $$5$$.
How would this be solved in the general case for arbitrary values $$[a, b]$$ and $$X$$? It's been quite a while since I last took statistics, so I've forgotten how to work with discrete random variables and expected value.
• ${\cal E}(n) = {2 X \over a+b}$. – David G. Stork Jan 10 at 0:50
• Actually, one must always round up, so ${\cal E}(n) = \lceil \left({2 X \over a+b} \right) \rceil$ – David G. Stork Jan 10 at 1:44
• @DavidG.Stork: Even though the random variable $n$ is always integer valued, there's no reason why its expected value has to be... – Nate Eldredge Jan 10 at 2:18
• @DavidG.Stork Right; I claim both are incorrect. – Aaron Montgomery Jan 10 at 4:50
• @AaronMontgomery, they are indeed incorrect. – Eelvex Jan 10 at 6:24
Let $$T$$ be the number of times it takes to reach $$X$$. We compute $$E[T]$$ via $$E[T]=\sum_{t=0}^\infty P(T>t)$$.
In order to have $$T>t$$, the sum of the first $$T$$ samples needs to be less than $$X$$. Let $$S_i$$ be the value of the $$i^{th}$$ sample. Then an experiment where $$T>t$$ has the first $$t$$ values satisfying $$S_1+S_2+\dots+S_t Letting $$E=X-1-(S_1+\dots+S_t)$$, we get $$S_1+\dots+S_t+E=X-1,\\ a\le S_i\le b,\\ E\ge 0$$ The number of integer solutions to the above system of equations and inequalities in the variables $$S_i$$ and $$E$$ can be computed via generating functions. The number of solutions is the coefficient of $$s^{X-1}$$ in $$(s^a+s^{a+1}+\dots+s^b)^t(1-s)^{-1}=s^{at}\cdot(1-s^{b-a+1})^t(1-s)^{-(t+1)}$$ After a bunch of work, this coefficient is equal to $$\sum_{k=0}^{\frac{X-1-ta}{b-a+1}} (-1)^k\binom{t}k\binom{t+X-1-ta-(b-a+1)k}{t}$$ Finally, we get \begin{align} E[T] &=\sum_{t=0}^{X/a} P(T>t) \\&=\boxed{\sum_{t=0}^{X/a} (b-a+1)^{-t}\sum_{k=0}^{\frac{X-1-ta}{b-a+1}}(-1)^k\binom{t}k\binom{t+X-1-ta-(b-a+1)k}{t}} \end{align} Note that when $$a=0$$, the summation in $$t$$ will go from $$t=0$$ to $$\infty$$, so it can only be computed approximately with a computer. However, this caveat can be avoided using the following observation; if $$T(a,b,X)$$ is the expected time to reach $$X$$ using sums of $$\operatorname{Unif}\{a,a+1,\dots,b\}$$ random variables, then $$E(T(0,b,X))=\frac{b+1}bE(T(1,b,X))$$.
When I write $$\sum_{k=0}^a f(k)$$ and $$a$$ is not an integer, what I mean is $$\sum_{k=0}^{\lfloor a \rfloor}f(k)$$.
The expected number is dominated by the term based on the mean $$\frac{2X}{a+b}$$ - but we can do better than that. As $$X\to\infty$$, $$E(n) = \frac{2X}{a+b}+\frac{2(a^2+ab+b^2)}{3(a^2+2ab+b^2)}+o(1)$$ Why? Consider the excess. In the limit $$X\to\infty$$, the frequency of hitting a particular point approaches a uniform density of $$\frac{2}{b+a}$$. For points in $$[X,X+a)$$, this is certain to be the first time we've exceeded $$X$$, so the density function for that first time exceeding $$X$$ is approximately $$\frac{2}{b+a}$$ on that range. For $$t\in[X+a,X+b)$$, the probability that it's the first time exceeding $$X$$ is $$\frac{X+b-t}{b-a}$$; we need the last term added to be at least $$t-X$$. Multiply by $$\frac{2}{b+a}$$, and the density function for the first time exceeding $$X$$ is approximately $$\frac{2(X+b-t)}{b^2-a^2}$$ on $$[X+a,X+b)$$. For $$t or $$t\ge X+b$$, of course, the density is zero. Graphically, this density is a rectangular piece followed by a triangle sloping to zero. Now, the expected value of the first sum to be greater than $$X$$ is approximately $$\int_{X}^{X+a}\frac{2t}{b+a}\,dt+\int_{X+a}^{X+b}\frac{2t(X+b-t)}{b^2-a^2}\,dt$$ $$=\frac{2a(2X+a)}{2(b+a)}+\frac{2(b-a)(2X+a+b)(X+b)}{2(b^2-a^2)}-\frac{2(b-a)((X+a)^2+(X+a)(X+b)+(X+b)^2)}{3(b^2-a^2)}$$ $$=\frac{6aX+3a^2+6X^2+3(a+3b)X+3ab+3b^2-6X^2-6(a+b)X-2(a^2+ab+b^2)}{3(b+a)}$$ $$=\frac{3(a+b)X+a^2+ab+b^2}{3(b+a)}=X+\frac{a^2+ab+b^2}{3(a+b)}$$ Divide by the mean $$\frac{a+b}{2}$$ of each term in the sum, and we have an asymptotic expression for $$E(n)$$.
This is not a rigorous argument. If I wanted to do that, I'd bring out something like the moment-generating function.
All right, now that rigorous argument I alluded to. Credit to the late Kent Merryfield for laying out the argument cleanly, although I'm sure I contributed in the development stage. AoPS link
Suppose we add independent copies of a reasonably nice nonnegative continuous random variable with density $$f$$, mean $$\mu$$, and variance $$\sigma^2$$ until we exceed some fixed $$y$$. As $$y\to\infty$$, the expected number $$g(y)$$ of these variables needed to exceed $$y$$ is $$g(y)=\boxed{\frac{y}{\mu}+\frac12+\frac{\sigma^2}{2\mu^2}+o(1)}$$ In the case of the uniform $$[a,b]$$ distribution of this problem, that is $$g(y)=\frac{2y}{a+b}+\frac12+\frac{(b-a)^2}{6(b+a)^2}+o(1)$$.
To prove this, condition on the first copy of the variable. If it has value $$x$$, the expected number of additional terms needed is $$g(y-x)$$. Add the one for the term we conditioned on, integrate against the density, and $$g(y)=1+\int_0^y f(x)g(y-x)\,dx$$ This is a convolution. To resolve it into something we can work with more easily, apply a transform method. Specifically, in this case with a convolution over $$[0,\infty]$$, the Laplace transform is most appropriate. If $$F(s)=\int_0^{\infty}e^{-sx}f(x)\,dx$$ and $$G(s)=\int_0^{\infty}e^{-sy}g(y)\,dy$$, then, transforming that integral equation for $$g$$, we get $$G(s)=\frac1s+F(s)G(s)$$ Solve for $$G$$, and $$G(s)=\frac1{s(1-F(s))}$$ Now, note that $$F$$ is essentially the moment-generating function for the variable with density $$f$$; there's a sign change, but it's otherwise the same. As such, it has the power series expansion $$F(s)=1-\mu s+\frac{\mu^2+\sigma^2}{2\mu^2}s^2 + O(s^3)$$ near zero. Then $$G(s)=\frac1{\mu s^2-\frac{\mu^2+\sigma^2}{2}s^3+O(s^4)} = \frac1{\mu}s^{-2}+\frac{\mu^2+\sigma^2}{2\mu^2}s^{-1}+Q(s)$$ where $$Q(s)$$ has a removable singularity at zero. Also, from the "reasonably nice" condition, $$Q$$ decays at $$\infty$$ How fast? Well, as long as the density of $$f$$ is bounded above by some exponential, its transform $$F$$ goes to zero as $$s\to\infty$$. Then $$G(s)=\frac1{s(1-F(s)}=O(s^{-1})$$ as $$s\to\infty$$; subtracting off the two other terms, $$Q(s)=O(s^{-1})$$ as $$s\to\infty$$.
Now, we take the inverse Laplace transform of $$G$$. The inverse transform of $$s^{-2}$$ is $$y$$ and the inverse transform of $$s^{-1}$$ is $$1$$, so there's the $$\frac{y}{\mu}+\frac{\sigma^2+\mu^2}{2\sigma^2}$$ terms of the expansion of $$g$$. For the remainder $$Q$$, we bring out the integral formula $$q(y)=\frac1{2\pi}\int_{\infty}^{\infty}Q(it)e^{ity}\,dt$$. Again for nice enough $$f$$ - a square-integrable density will do - this decays to zero as $$y\to\infty$$ (Riemann-Lebesgue lemma).
And now, it's time to get more concrete. Our original example was a uniform distribution on $$[a,b]$$, which has mean $$\mu=\frac{a+b}{2}$$ and variance $$\sigma^2=\frac{(b-a)^2}{12}$$. That gives $$g(y)=\boxed{\frac{2y}{a+b}+\frac12+\frac{(b-a)^2}{6(a+b)^2}+o(1)}$$ Also, we can be more explicit; the transforms are $$F(s)=\frac1{(b-a)s}(e^{-as}-e^{-bs})$$ and $$G(s)=\frac{1}{s-\frac1{b-a}e^{-as}+\frac1{b-a}e^{-bs}}$$ This $$G$$ is analytic in the right half-plane except for the pole at zero, and is comparable to $$\frac1s$$ at $$\infty$$ in all directions in that half-plane. The remainder term $$Q$$ is thus smooth and decays like $$\frac1s$$ along the imaginary axis, so its inverse transform $$q$$ will have a jump discontinuity (at zero, where it goes from zero to $$1$$) and decay rapidly.
If we want an explicit form for $$g$$ - it's a mess. Three different exponentials in that denominator pushes this far off the standard tables of transforms, and splitting it up as a geometric series introduces way too many terms. I'll try to write something down anyway: \begin{align*}G(s) &= \frac{1}{s-\frac1{b-a}e^{-as}+\frac1{b-a}e^{-bs}}\\ &= s^{-1} +\frac{s^{-2}}{b-a}\left(e^{-as}-e^{-bs}\right)+\frac{s^{-3}}{(b-a)^2}\left(e^{-as}-e^{-bs}\right)^2+\cdots\\ &= \sum_{n=0}^{\infty} \frac{s^{-n-1}}{(b-a)^n}\left(e^{-as}-e^{-bs}\right)^n\\ &= \sum_{n=0}^{\infty}\sum_{k=0}^n\frac{s^{-n-1}}{(b-a)^n}\binom{n}{k}(-1)^k e^{-s(kb+(n-k)a)}\end{align*} Now, in the table of Laplace transforms I'm using, there's an entry "delayed $$n$$th power with frequency shift"; the transform of $$(y-\tau)^n e^{-\alpha(y-\tau)}u(y-\tau)$$ (where $$u$$ is the unit step function) is $$n!e^{-\tau s}(s+\alpha)^{-n-1}$$. From this and the linearity of the transform, we get $$g(y) = \sum_{n=0}^{\infty}\sum_{k=0}^n[y-(kb+(n-k)a)]^n\frac{(-1)^k}{k!(n-k)!}u[t-(kb+(n-k)a)]$$ $$g(y) = \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} (-1)^k\frac{[y-(ja+kb)]^{j+k}}{j!k!}u[t-(kb+(n-k)a)]$$ For any given $$y$$, this is a finite sum; we only sum those terms for which $$ja+kb \le y$$. It is horrible to use in practice, since it involves adding and subtracting large numbers to get something relatively small.
The $$a=0$$ case allows us to split things up differently, and comes out with something that's actually reasonably practical. Follow the link at the start of the post if you want to see it.
Given $$m$$ discrete variables in the range $$[a,b]$$, the number of ways to get a sum $$s$$ out of them corresponds to \eqalign{ & N_b (s - ma,d,m) = \cr & = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ a \le {\rm integer}\;y_{\,j} \le b \hfill \cr y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,m} = s \hfill \cr} \right.\quad = \cr & = {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 0} \le {\rm integer}\;x_{\,j} \le b - a = d \hfill \cr x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,m} = s - ma \hfill \cr} \right. \cr} where $$N_b$$ is given by $$\bbox[lightyellow] { N_b (s-ma,d,m)\quad \left| {\;0 \leqslant \text{integers }s,m,d} \right.\quad = \sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s-ma}{d+1}\, \leqslant \,m} \right)} {\left( { - 1} \right)^k \binom{m}{k} \binom { s -ma+ m - 1 - k\left( {d + 1} \right) } { s-ma - k\left( {d + 1} \right)}\ } } \tag{1}$$ as widely explained in this post.
Let's also notice the symmetry property of $$N_b$$ $$N_b (s - ma,d,m) = N_b (md - \left( {s - ma} \right),d,m) = N_b (mb - s,d,m)$$
The probability of obtaining exactly the sum $$s$$ in $$m$$ rolls is therefore \bbox[lightyellow] { \eqalign{ & p(s\;;\,m,a,b) = {{N_b (s - ma,d,m)} \over {\left( {d + 1} \right)^{\,m} }} = {{N_b (mb - s,d,m)} \over {\left( {d + 1} \right)^{\,m} }} = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - ma} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{ s -ma+ m - 1 - k\left( {d + 1} \right) }{ s-ma - k\left( {d + 1} \right)} } = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }} \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{mb - s} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{ mb-s+ m - 1 - k\left( {d + 1} \right) }{ mb-s - k\left( {d + 1} \right)} } \cr} } \tag{2} and the sum of $$p$$ over $$s$$ is in fact one.
The probability $$p$$ quickly converges, by Central Limit Theorem, to the Gaussian in the variable $$s$$ $$\cal N\left( {\mu ,\sigma ^{\,2} } \right)\quad \left| \matrix{ \;\mu = m\left( {{d \over 2} + a} \right) = m\left( {{{a + b} \over 2}} \right) \hfill \cr \;\sigma ^{\,2} = m{{\left( {d + 1} \right)^{\,2} - 1} \over {12}} \hfill \cr} \right.$$ see for instance this post.
It is easy to demonstrate (by the "double convoluton" of binomials) that the Cumulative version, i.e. the probability to obtain a sum $$\le S$$, is \bbox[lightyellow] { \eqalign{ & P(S\;;\,m,a,b) = {1 \over {\left( {d + 1} \right)^{\,m} }}\sum\limits_{0\, \le \,\,s\,\, \le \,S} {N_b (s - ma,d,m)} = \cr & = {1 \over {\left( {d + 1} \right)^{\,m} }}\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - ma} \over {d + 1}}\, \le \,m} \right)} { \left( { - 1} \right)^k \binom{m}{k} \binom{S - ma + m - k\left( {d + 1} \right) }{S - ma - k\left( {d + 1} \right) } } \cr & \approx {1 \over 2}\left( {1 + {\rm erf}\left( {\sqrt 6 {{S + 1/2 - m\left( {a + b} \right)/2} \over {\left( {b - a + 1} \right)\sqrt m \;}}} \right)} \right) \cr} } \tag{3}
The following plot allows to appreciate the good approximation provided by the asymptotics even at relatively small values of the parameters involved.
That premised,
the probability that we reach or exceed a predefined sum $$X$$ at the $$m$$-th roll and not earlier
is given by the probability that
we get $$S as the sum of the first $$m-1$$ rolls, and then that the $$m$$-th one has a value $$s$$ such that $$X \le S+s$$,
or, what is demonstrably the same, by
the probability of getting $$X \le S$$ in $$m$$ rolls minus the probability of getting the same in $$m-1$$ rolls.
That is, in conclusion \bbox[lightyellow] { \eqalign{ & Q(m\;;\,X,a,b)\quad \left| \matrix{ \;1 \le m \hfill \cr \;0 \le X \hfill \cr \;0 \le a \le b \hfill \cr} \right.\quad = \cr & = 1 - P(X - 1\;;\,m,a,b) - 1 + P(X - 1\;;\,m - 1,a,b) = \cr & = P(X - 1\;;\,m - 1,a,b) - P(X - 1\;;\,m,a,b) \cr} } \tag{4} and the sum of $$Q$$ over $$m$$ correctly checks to one.
Example
a) For small values of $$X,a,b$$ the exact expression for $$Q$$ in (4) gives the results shown (e.g. $$a=1, \; b=4$$) which look to be correct and the row sums correctly check to be $$1$$.
b) For large values of the parameters, like in your example $$a=1000,\; b=2000, \; X=5000$$ we shall use the asymptotic version of $$Q$$ which gives the results plotted below | 2019-07-22T01:06:58 | {
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http://mathhelpforum.com/statistics/183602-9-digit-combination-lock-chances-probability-print.html | # 9-Digit Combination Lock Chances And Probability
• June 25th 2011, 11:46 AM
FatalSylence
9-Digit Combination Lock Chances And Probability
I have been thinking on combination locks recently, and would like some help with this particular problem.
Problem:
I have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.
Questions:
1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?
2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.
3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?
1. If you would change the lock to have a 5 digit combination, would the chance to guess it right be the same as if it were a 4 digit combination? Since the possible combination amount is 15,120 (9*8*7*6*5), the amount of right combinations is 120 (1*2*3*4*5), therefore 15,000/120=125. The same amount of chance as if it were a 4-digit combination?
2. If all this is true, then wouldn't a 6-digit combination actually be less secure? I.e., you have a greater chance of guessing the right combination? (1 in 84, if I am doing it right.)
Help is greatly appreciated. I have not done probability in a long time and am wondering if I am even doing it right.
Thanks, guys.
Fatal Sylence
• June 25th 2011, 12:01 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
[/B]I have a lock with digits 1-9, and the length of the combination is 4 digits. You cannot press the digits 1-9 more than once, and the combination does not have to be typed in any particular order.
Questions:
1. Is it true, then, that the possible amount of combinations is 3,024, because 9*8*7*6=3,024?
2. Is it also true that 24 of the combinations are correct, since you do not have to type the combination in any particular order? Therefore 1*2*3*4=24.
3. Is it true that the chances of guessing it right are 1 in 125 since 3,000/24=125?
If order makes a difference then the answer is $9\cdot 8 \cdot 7\cdot 6=3024$.
If order makes no difference then $\binom{9}{4}=\frac{9!}{4!\cdot 5!}=126$
• June 25th 2011, 12:07 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
If order makes a difference then the answer is $9\cdot 8 \cdot 7\cdot 6=3024$.
If order makes no difference then $\binom{9}{4}=\frac{9!}{4!\cdot 5!}=126$
I'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?
• June 25th 2011, 12:12 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
I'm not sure I understand your answer. 126 as in a 1 in 126 chance of guessing it right?
That is the combination of 9 things taken 4 at a time.
By the way that is $\frac{3024}{24}$.
• June 25th 2011, 12:24 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
That is the combination of 9 things taken 4 at a time.
By the way that is $\frac{3024}{24}$.
Hmm, okay. Can you direct me towards a resource that explains the "n things taken r at a time"?
Also, were my other questions correct?
Finally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?
• June 25th 2011, 12:28 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
Hmm, okay. Can you direct me towards a resource that explains the "n things taken r at a time"?
Also, were my other questions correct?
Finally, would a 5 digit combination be more secure? A 4 digit combination has the same chance to guess right as a a 5-digit combination, but don't you have a greater chance to guess it wrongwith 5-digits?
Also look at this page. You change those numbers and hit $\boxed{=}$.
• June 25th 2011, 12:37 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
Also look at this page. You change those numbers and hit $\boxed{=}$.
Can you explain to me if in a 4-digit combination, the amount of correct combinations is 126, or 24 like I originally thought? I'm just trying to understand this here.
• June 25th 2011, 12:44 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
Can you explain to me if in a 4-digit combination, the amount of correct combinations are 126, or 24 like I originally thought? I'm just trying to understand this here.
I think that we are talking about two different ideas here.
There are 24 ways to arrange the string $9,4,2,6$.
But there are 126 ways to select a set of four from $\{1,2,3,4,5,6,7,8,9\}$.
One of those 126 sets is $\{2,4,6,9\}$.
Then the numbers in that set can be arranged in 24 ways.
• June 25th 2011, 12:56 PM
FatalSylence
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by Plato
I think that we are talking about two different ideas here.
There are 24 ways to arrange the string $9,4,2,6$.
But there are 126 ways to select a set of four from $\{1,2,3,4,5,6,7,8,9\}$.
One of those 126 sets is $\{2,4,6,9\}$.
Then the numbers in that set can be arranged in 24 ways.
OH!!! Gosh! I get it! It does not matter how many ways you can arrange a set of four since order does not matter. I understand. One of these 126 sets encompasses all the ways you can arrange it since order does not matter.
So, if the person was good at math, they would realize that since order does not matter, there are only 126 possible combinations. Wow, I get it. Thank you.
EDIT:
Is there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?
• June 25th 2011, 03:39 PM
Plato
Re: 9-Digit Combination Lock Chances And Probability
Quote:
Originally Posted by FatalSylence
EDIT:[/B] Is there a formula or calculator that can display all the possible sets for a 1-9, 4 digit combination? Or any other numbers for that matter?
The good news is that this is a simple programming problem.
But the bad news I doubt that you can easily find it on the web.
I suggest that you discuss this with a friend who does programming. | 2014-10-26T02:02:59 | {
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https://math.stackexchange.com/questions/2562393/a-compact-locally-connected-metric-space-is-uniformly-locally-connected | # A compact locally connected metric space is “uniformly locally connected”
A compact locally connected metric space is "uniformly locally connected"\ That is, for any $\epsilon > 0$, there is some $\delta > 0$ such that whenever $\rho(x, y) < \delta$, then $x$ and $y$ both lie in some connected subset of $X$ of diameter $<\epsilon$.
proof:-
Since $X$ is locally connected metric space then each $x\in X$ has a nhood base of open connected sets\ Given $\epsilon > 0$, let $x \in X$ and $U_x=\rho(x, \epsilon)$ be a nhood of $x$\ There exist an open connected basic nhood $V_x$ with diameter $<\epsilon$, Now $$X=\bigcup_{x\in X }{V_x}$$, hence cover $X$ by open connected nhoods of diameter $<\epsilon$.\ Since $X$ is compact, reduce this to a finite subcover $\{V_{x1},. . . , V_{xn}\}$ and let $\delta$ be a Lebesgue number (22.5) for this cover.\ Then if $\rho(x, y) < \delta$, both $x$ and $у$ belong to some $V_{xi}$. \
{Theorem 22.5} (Lebesgue covering lemma). If $\{U_1..., U_n\}$ is a finite open cover of a compact metric space X, there is some $\delta > 0$ such that if A is any subset of $X$ of diameter $< \delta$, then $A \subset U_i$ for some i.
I try to write the proof better than this.
I would like to confirm this proof
If acceptable, I would like to clarify and improve it (Language and Mathematical)as much as possible
• Do not use \ or , or .\ at the end of a sentence. Just use . $\quad$ And if a sentence ends in a displayed line, include the . in that line. Write in the same style as for an an essay about something else. What you have is not so bad. I've seen far worse. And your proof is correct . – DanielWainfleet Dec 12 '17 at 9:25
The proof is correct.
As for writing it up. That depends on your audience. For a general audience, the first rule is - avoid abbreviations ("neighborhood", not "nhood"). And I assume that the "/" scattered through the text was some sort of misguided attempt to show where you put new lines? Just let Latex have it's way here. If a new line is needed, let it form a paragraph break.
Also "and $U_x=\rho(x,\epsilon)$ be a nhood of $x$" doesn't make sense. $\rho$ is a function that takes points in your space in both arguments, and produces real numbers. You have it with a real number as the second argument, and producing a set. Now it is evident that you mean $U_x$ to be a ball of radius $\epsilon$ about $x$. But if someone has told you this is a good way to denote the ball, do NOT trust them about notations in the future! Commonly used ways of denoting balls include $B(x, \epsilon)$ and $B_x(\epsilon)$. If the metric to be used is not obvious, then usually $B_\rho(x, \epsilon)$ is preferred. (Of course, it you had intended $B(x, \epsilon)$ and only accidently used $\rho$, just be careful in your final write-up.)
But you don't do anything with "$U_x$", which makes introducing a notation for it pointless. You don't actually need to name the balls, since you don't plan on using that name.
There are also some phrasing differences I would suggest. Your phrasing is clear enough, but sounds awkward sometimes to a native English speaker (at least - to this native English speaker). There is nothing particularly bad, though.
I would write it up like this:
Since $X$ is locally connected, for $\epsilon > 0$, each point $x \in X$ has a open connected neighborhood $V_x$ contained within the ball of radius $\epsilon$ about $x$. The collection $\{V_x \mid x \in X\}$ forms an open cover of $X$ by sets of diameter $< \epsilon$. Since $X$ is compact, it has a finite subcover $\{V_1, V_2, ..., V_m\}$. By the Lesbegue covering lemma (Theorem 22.5), there is a $\delta > 0$ such that if $\rho(x, y) < \delta$, then there is some $V_i$ with $\{x, y\} \subset V_i$, which completes the proof. | 2019-09-24T09:19:17 | {
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http://mymathforum.com/algebra/42904-progression.html | My Math Forum Progression .
Algebra Pre-Algebra and Basic Algebra Math Forum
April 15th, 2014, 02:49 AM #1 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Progression . For the same job specifications, two companies offer a different salary scale: Company A :Starting monthly pay = RM900.00 Monthly increment = RM50.00 Company B:Starting monthly pay = RM750.00 Monthly increment = RM60.00 (a)In January 2014, Ali starts to work in Company A and Ahmad in Company B. When will the monthly salary of Ali and Ahmad be the same? Use three methods. Include the use of ICT. (b)Which salary scale is the better deal? Justify. How to solve this problem ? Especially problem b.
April 15th, 2014, 10:20 PM #2
Math Team
Joined: Dec 2006
From: Lexington, MA
Posts: 3,267
Thanks: 408
Hello, jiasyuen!
Quote:
For the same job, two companies offer a different salary scale: Company A: Starting monthly pay = \$900.00 Monthly increment = \$50.00 Company B: Starting monthly pay = \$750.00 Monthly increment = \$60.00 (a) In January 2014, Ali starts to work in Company A and Ahmad in Company B. When will the monthly salary of Ali and Ahmad be the same? Use three methods. Include the use of ICT.
You need to construct algebraic expressions for Ali's and Ahmad's monthly pay.
Ali starts at \$900 per month and gets a \$50 raise each month.
When he has worked for $\displaystyle m$ months,
$\displaystyle \quad$his monthly pay will be: $\displaystyle \:900 + 50m$ dollars.
Ahmad starts at \$750 per month and gets a \$60 raise each month.
When he has worked for $\displaystyle m$ months,
$\displaystyle \quad$his monthly pay will be $\displaystyle \,750 +60m$ dollars.
When are these equal?
$\displaystyle \quad 750 + 60m \:=\:900 + 50m \quad\Rightarrow\quad 10m \:=\:150 \quad\Rightarrow\quad m \,=\,15$
Their salaries will be equal in 15 months (March 2015).
Quote:
(b)Which salary scale is the better deal? Justify.
If Ali considers this a temporary job (less than 15 months),
$\displaystyle \quad$ he has the better deal.
If both are planning to work more than 15 months,
$\displaystyle \quad$ then Ahmad has the better deal.$\displaystyle \;\;{\color{red}{**}}$
$\displaystyle \begin{array}{c|ccccccccc} \hline \text{Month} & 0 & 1 & 2 & 3 & \cdots & 15 & 16 & 17 & \cdots \\ \hline \text{Ali} & 900 & 950 & 1000 & 1050 & \cdots & 1650 & 1700 & 1750 & \cdots \\ \hline \text{Ahmad} & 750 & 810 & 870 & 930 & \cdots & 1650 & 1710 & 1770 & \cdots \\ \hline \end{array}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
$\displaystyle {\color{red}{**}}$
I gave this problem some extra thought . . .
In 15 months their monthly pay is the same.
But Ali has made \$18,750 while Ahmad has made only \$17,550.
It will take 31 months for Ahmad's total earnings to equal that of Ali.
After that, Ahmad's total earnings will exceed Ali's.
April 16th, 2014, 03:43 AM #3 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?
April 16th, 2014, 09:08 AM #4
Math Team
Joined: Oct 2011
Posts: 14,597
Thanks: 1038
Quote:
Originally Posted by jiasyuen The question stated there have to use three methods to find. Including of ICT. What it means ? How to do ?
Well, Soroban gave you one; seems fair that YOU find two more;
it is YOUR homework, not ours. Do you agree?
April 16th, 2014, 09:12 AM #5 Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 I don't know what's another two methods ? Can you give me some guidance ?
April 23rd, 2014, 04:36 AM #6
Senior Member
Joined: Sep 2013
From: Earth
Posts: 827
Thanks: 36
Quote:
Originally Posted by soroban Hello, jiasyuen! You need to construct algebraic expressions for Ali's and Ahmad's monthly pay. Ali starts at \$900 per month and gets a \$50 raise each month. When he has worked for $\displaystyle m$ months, $\displaystyle \quad$his monthly pay will be: $\displaystyle \:900 + 50m$ dollars. Ahmad starts at \$750 per month and gets a \$60 raise each month. When he has worked for $\displaystyle m$ months, $\displaystyle \quad$his monthly pay will be $\displaystyle \,750 +60m$ dollars. When are these equal? $\displaystyle \quad 750 + 60m \:=\:900 + 50m \quad\Rightarrow\quad 10m \:=\:150 \quad\Rightarrow\quad m \,=\,15$ Their salaries will be equal in 15 months (March 2015). If Ali considers this a temporary job (less than 15 months), $\displaystyle \quad$ he has the better deal. If both are planning to work more than 15 months, $\displaystyle \quad$ then Ahmad has the better deal.$\displaystyle \;\;{\color{red}{**}}$ $\displaystyle \begin{array}{c|ccccccccc} \hline \text{Month} & 0 & 1 & 2 & 3 & \cdots & 15 & 16 & 17 & \cdots \\ \hline \text{Ali} & 900 & 950 & 1000 & 1050 & \cdots & 1650 & 1700 & 1750 & \cdots \\ \hline \text{Ahmad} & 750 & 810 & 870 & 930 & \cdots & 1650 & 1710 & 1770 & \cdots \\ \hline \end{array}$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ $\displaystyle {\color{red}{**}}$ I gave this problem some extra thought . . . In 15 months their monthly pay is the same. But Ali has made \$18,750 while Ahmad has made only \$17,550. It will take 31 months for Ahmad's total earnings to equal that of Ali. After that, Ahmad's total earnings will exceed Ali's.
Not 16 month?
Ali in Company A:
Tn =a+(n-1)d
=900+(n-1)(50)
=900+50n-50
=850+50n
Tn = a+(n-1)d
=750+(n-1)(60)
=750+60n-60
=690+60n
When the monthly salary of Ali and Ahmad the same,
850+50n=690+60n
160n=10n
n=16 (i.e. in the 16th month)
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### in january 2014 ali starts to work in company A and ahmad company b when will the monthly salary of ali and ahmad be the same
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http://mathhelpforum.com/calculus/10998-solved-integration-sinxcosx.html | # Math Help - [SOLVED] integration with sinxcosx
1. ## [SOLVED] integration with sinxcosx
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
1) sin^2x/2 + C and 2) -cos^2x/2 + C
Now if we set these two answers equal to eachother we get
sin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1
So i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers
2. Originally Posted by block
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
1) sin^2x/2 + C and 2) -cos^2x/2 + C
Now if we set these two answers equal to eachother we get
sin^2x + cos^2x = 0 but we know there exists a trig identity that states sin^2x + cos^2x = 1
So i'm confused here. How can this be possible? I know that 0 cannot equal 1. Or at least i hope not hehe. Any help would be great cheers
Two solutions of an integral may be of different form if they are off by a constant. You have two solutions, but who says the constant of integration is the same? What this is telling you is that $C_{cos} = C_{sin} + 1$
-Dan
3. Originally Posted by block
Hey guys i kind of have a weird little problem which i can't figure out.
I'm integrating sinxcosx, and depending on what i use as my substution you get two answers.
You have,
$\int \sin x \cos x dx$
And I think the problem you have you think you get two distinct answers when you use $u=\sin x$ and when you switch to $u=\cos x$.
1) $u=\sin x$ then $u'=\cos x$
Thus,
$\int \sin x \cos x dx= \int u du=\frac{1}{2}u^2 +C=\frac{1}{2}\sin^2 x+C$
2) $u=\cos x$ then $u'=-\sin x$
Thus,
$\int \sin x \cos x dx= -\int u du=\frac{1}{2}u^2 +C=-\frac{1}{2}\cos^2 x+C$
But in reality these are the same (the set of all anti-derivates) because,
$\frac{1}{2}\sin^2 x+C=\frac{1}{2}(1-\cos^2 x)+C=-\frac{1}{2}\cos^2 x+\left( \frac{1}{2}+C\right)=-\frac{1}{2}\cos^2 x+C_1$
I just renamed the constant function,
$\frac{1}{2} +C=C_1$
Thus, you get the same thing.
And the last thing you could have done is,
$\frac{1}{2}\int 2\sin x \cos x dx$
I just multiplied and divide by 2, unchanged expression.
From trignometry you know that,
$\frac{1}{2}\int \sin 2x dx=-\frac{1}{4}\cos 2x+C$
But again though it looks different the space of solutions is still the same. You can use some trignometry to confirm this.
4. Hello, block!
This is a classic puzzler.
I ran across it across it in Calculus I.
There are three ways (at least) to integrate it.
We have: . $\int\sin x\cos x\,dx$
[1] If we look at it as: . $\int \underbrace{\sin x}_{u}\underbrace{(\cos x\,dx)}_{du}$
. . . we get: . $\boxed{\frac{1}{2}\sin^2\!x + C_1}$
[2] If we write it as: . $\int \cos x(\sin x\,dx)$
. . . we let: $u = \cos x\quad\Rightarrow\quad du = -\sin x\,dx\quad\Rightarrow\quad dx = -\frac{du}{\sin x}$
Substitute: . $-\int u\,du \;=\;-\frac{1}{2}u^2 + C\;=\;\boxed{-\frac{1}{2}\cos^2\!x + C_2}$
Since these two answers must be equal, we have:
. . $\frac{1}{2}\sin^2\!x + C_1 \;=\;-\frac{1}{2}\cos^2\!x + C_2\quad\Rightarrow\quad \frac{1}{2}\sin^2\!x + \frac{1}{2}\cos^2\!x\;=\;C_2-C_1$
Multiply by 2: . $\sin^2\!x + \cos^2\!x \;=\;2(C_2-C_1)$
The time the constants $(C_1,\:C_2)$ aren't completely arbitrary.
. . They must satisfy: . $2(C_2-C_1) \:=\:1$
[3] We can use the identity: . $2\sin\theta\cos\theta \,=\,\sin2\theta$
So: . $\int\sin x\cos x\,dx\;=\;\frac{1}{2}\int2\sin x\cos x\,dx \;=\;\frac{1}{2}\int\sin2x\,dx$ $=\;\boxed{-\frac{1}{4}\cos2x + C_3}$
. . I'll let you justify this one . . .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
A similar problem: . $\int\sec^2\!x\tan x\,dx$
$[1]\;\;\int \underbrace{\sec x}_{u}\underbrace{(\sec x\tan x\,dx)}_{du} \;=\;\frac{1}{2}\sec^2\!2x + C$
$[2]\;\;\int\underbrace{\tan x}_{u}\underbrace{(\sec^2x\,dx)}_{du}\;=\;\frac{1} {2}\tan^2x + C$ | 2014-12-22T14:32:13 | {
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http://openstudy.com/updates/53f5818fe4b0f0e909e02de2 | ## anonymous 2 years ago how would I solve for x if 2x-3/x+1<1
1. kc_kennylau
$\frac{2x-3}{x+1}<1$
2. kc_kennylau
I would split it into two cases
3. anonymous
how would you do that
4. myininaya
I would first write as f(x)<0 Then find where f(x) is zero and undefined Then draw a number line with the numbers that make f zero or undefined Then test the intervals around those numbers to see if f is less than 0 or not
5. kc_kennylau
Since we are to multiply (x-1) on both sides, I would split it into x-1<0 and x-1>0
6. myininaya
you mean x+1?
7. kc_kennylau
Yes sorry
8. anonymous
SO i take x+ 1 and get $2x-3<x+1 and 2x-3>x+1$
9. kc_kennylau
Yes
10. kc_kennylau
Well actually I believe we should do it case by case
11. kc_kennylau
When x+1<0: 2x-3>x+1 ... When x+1>0: 2x-3<x+1 ...
12. anonymous
so then my next step would to add in the 3 on both sides corect? so it would turn into $2x>x+4$ and $2x<x+4$
13. kc_kennylau
Yep
14. kc_kennylau
And for presentation purpose I would write it as 2x>x+4 where x+1<0 and 2x<x+4 where x+1>0
15. anonymous
then do I divide by 2 on both of the equations or divide by 2x?
16. kc_kennylau
You subtract x from both sides
17. anonymous
so It would be $x-2x>4 where x+1<0$ and just the oposit sign on the otherone?
18. kc_kennylau
No, it should be 2x-x>4 where x+1<0
19. anonymous
ok now im starting to understand so then the next step would be to hmove the 4 over
20. kc_kennylau
Nah, the next step is subtracting it
21. anonymous
ok so it would then be 2x-x-4> 0 and where x+1<0 and 2x-x-4< 0 and where x+1>0 and t
22. kc_kennylau
I mean you can write 2x-x as x
23. kc_kennylau
When x+1<0: 2x-3>x+1 2x>x+4 2x-x>4 x>4 When x+1>0: 2x-3<x+1 2x<x+4 2x-x<4 x<4
24. kc_kennylau
However, when x+1<0, x can't be >4, and the same goes to the second case
25. kc_kennylau
Therefore there is no solution
26. kc_kennylau
Wait
27. aum
The inequality is valid for x in the interval (-1, 4).
28. kc_kennylau
@aum tell me what is wrong in my solution, I have no time to double-check
29. kc_kennylau
bye
30. anonymous
@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?
31. aum
My method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - x - 1) / (x + 1) < 0 (x - 4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x-4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = -1. So the number line is split into three intervals: (-infinity, -1); (-1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x - 4) / (x + 1) < 0 when x = -2, f(x) = (-6) / (-1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = -4/1 = -4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (-1, 4).
32. anonymous
thank you so much
33. aum
you are welcome.
34. aum
To follow @kc_kennylau 's method: (2x - 3) / (x + 1) < 1 Case 1: x + 1 > 0 Since (x+1) is positive, we can multiply both sides by (x+1) and the "less than" inequality will remain intact. 2x - 3 < x + 1 x < 4 Case 2: x + 1 < 0 Since (x+1) is negative, multiplying both sides by (x+1) would change the "less than" inequality to a "greater than" inequality. 2x - 3 > x + 1 x > 4. But for the second case we assumed x + 1 < 0 which implies x < -1 But we get the solution as x > 4. x cannot be less than -1 and also be greater than 4 simultaneously. Thus, case 2 yields NO solution. Therefore, there is no solution in the domain x < -1. Combining the two cases we get x should be less than 4 but greater than -1.
35. kc_kennylau
I incorrectly rejected case 1
36. aum
yes. | 2016-09-29T22:15:35 | {
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https://math.stackexchange.com/questions/2406864/expected-no-of-coin-flips-to-win/2406907 | # Expected no. of coin flips to win
Two players A and B play a game in which they alternately flips a coin. Player A starts the game. If a player gets T and another player got H before, he/she is the winner. What is the expected no. of flips for A to win ?
Ex:-
Game-1: HT --> B wins
Game-2: HHT --> A wins
• You must share your thoughts on the problem. – StubbornAtom Aug 26 '17 at 16:34
• What I can think is to find probability for A to win in n flips and then use the formula of expectation to find the expected no. of flips. But this method is quite lengthy. Is there any better solution? – Sumit Kumar Aug 26 '17 at 16:38
• So the person who wins is the first person to get tails after someone has gotten heads, is that right? – Arthur Aug 26 '17 at 16:41
• If B wins, presumably A never wins – Henry Aug 26 '17 at 16:43
• @Arthur A player who get tails wins if and only if another player has gotten heads in the previous flip i.e. a sequence HT occurs – Sumit Kumar Aug 26 '17 at 16:43
We expect four coin flips. That's because we first expect two flips to get the first heads, and then we expect two flips to get the first tails.
More detailed: as a player is about to throw, the game is in one of two states: either there hasn't been any heads yet, or there has been at least one head (and since the game isn't over, the last throw must've been a head). Let's call the expected number of throws left until the game is over if we're in the first state $E_1$, and the expected number of throws left if we're in the second state $E_2$.
First we calculate $E_2$. If a player is about to throw, and the last throw was a heads, then there's a probability of $0.5$ of the game ending in one more throw, and a probability of $0.5$ of the game continuing to the next player still in the same state, which means that after that we expect another $E_2$ throws. This implies $$E_2=0.5\cdot1+0.5(1+E_2)\\ E_2=2$$ Now we get to $E_1$. If a player is about to throw, and there hasn't been any heads yet, then there is a probability of $0.5$ that the player throws heads, and the game continues in state two, which means that we expect another $E_2=2$ throws, and there is a probability of $0.5$ of the player throwing tails, which means that the game continues in the first state, and we expect another $E_1$ throws. This gives $$E_1=0.5(1+E_2)+0.5(1+E_1)\\ E_1=4$$ which is our answer.
Given that $A$ wins, we have an odd number of flips.
Last flip is $T$ and others consist of a possible run of tails, followed by a definite run of heads.
Let $N$ be the number of flips total.
$$P(N=n)=\frac{1}{2}\sum_{k=1}^{n-1}\left(\frac{1}{2}\right)^{n-1}=\frac{n-1}{2^n}$$ $$P(A)=P(N\equiv 1\mod 2)=\sum_{k=1}^\infty P(N=2k+1)=\sum_{k=0}^\infty\frac{k}{4^k}=\frac{4}{9}$$.
$$E(N|A)=\frac{9}{4}\sum_{k=1}^\infty\frac{k(2k+1)}{4^k}=\frac{13}{3}$$
The $E(N)=4$ result above is wrong since it is the total expectation $E(N)$ of coin flips until endgame. The total expectation $E(N)$ is, however, an interesting quantity as well, since it is equivalent to a special case of the following theorem involving a typing monkey:
Let a monkey type letters on a typewriter that has some alphabet $\Sigma$ and let $w\in\Sigma^*$. Also let $\{w_1,\cdots,w_n\}$ be the set of those strings that are both a prefix and suffix of $w$.
If the monkey types until the last letters typed form $w$ and then stops typing, let $X$ be the total number of letters typed. Then $$E(X)=\sum_{k=1}^n|\Sigma|^{|w_k|}$$
The theorem is provable with Martingale Theory.
Example:
Typewriter alphabet: $\{A,B,\cdots,Z\}$ (Standard English).
Monkey's Goal: $ABRACADABRA$.
Prefix-Suffix Strings: $\{A,ABRA,ABRACADABRA\}$ of sizes $1,4,11$.
So the monkey will on average type $26+26^4+26^{11}=3670344487444778$ letters.
• I have done exactly the same except that i have not multiplied 9/4 as you did in the last equation. Can you please explain why have you done that? – Sumit Kumar Aug 27 '17 at 6:11
• We are only concerned with expected value $E(N|A)$ of coin flips given that $A$ wins. $A$ wins if $N$ is odd, and loses if $N$ is even. We know how to easily get the total expectation $E(N)$ and the probability of $A$ winning. It is known that if $N$ is a random variable and $A$ is any event, then $E(N|A)$ is given as the weighted arithmetic mean of $E(N|X=x)$ with weights being probabilities of those outcomes $X=x$ that make $A$ happen. Recall that in the discrete case, the weighted mean is $\frac{w_1x_1+\cdots+w_nx_n}{w_1+\cdots+w_n}$ – Roman Chokler Aug 27 '17 at 13:40 | 2019-09-19T15:40:06 | {
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https://math.stackexchange.com/questions/3594814/floor-function-repeated-addition | # Floor function repeated addition
I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find} \ \lfloor 100r\rfloor. \textit{(Source: AIME)}$$ Here is my work:
$$\lfloor r + \frac{19}{100}\rfloor = r + \frac{19}{100} - \{r + \frac{19}{100}\}$$, so the figure can be restated as $$\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$$, where $$a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$$.
Because $$\{r + \frac{19}{100}\}$$ is the fractional part, $$0 \le r + \frac{19}{100} + a - 546 < 1$$, so after some more maniuplation, $$545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$$. $$a$$ is an integer, so the fractional part of $$r$$ must be $$\frac{81}{100}$$.
$$r = \lfloor r\rfloor + \{r\}$$, so $$\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$$ = $$\lfloor \lfloor r\rfloor + 1\rfloor$$. Because both of the terms inside that floor function are integers, it must equal $$\lfloor r\rfloor + 1$$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $$\lfloor r\rfloor + 1$$.
Therefore, $$73 \lfloor r\rfloor + 73 = 546$$, so $$73\lfloor r\rfloor = 473$$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $$\lfloor r\rfloor$$ is not an integer.
If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.
• The line "...so the fractional part of $r$ must be $\frac {81}{100}$" is where I think I see an issue. Couldn't that inequality equally have any of the fractions in the interval $[\frac 9{100},\frac{81}{100}]$ based on applying that logic in the same manner to any of the other terms from the original? If you apply the correct term to your next line of reasoning, or else find the next-smallest multiple of $73$ below $473$ I think you'll have your answer... Mar 25 '20 at 17:40
• Sorry, "next-largest" instead of "next-smallest"... Really, just begin with the "$73\lfloor r\rfloor+k=546$" logic, and deduce what $k$ has to be. Mar 25 '20 at 17:52
• The average of the terms is about $7.5$ so each term must be $7$ or $8$. How many of each? Mar 25 '20 at 17:58
• Don't worry about the fractional part. And don't worry about $[r]$. Find the precise value where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$. That is where $r + \frac {k}{100} < n \le r + \frac {k+1}{100}$ for some integer $n$. Mar 25 '20 at 18:23
## 2 Answers
Don't worry about the fractional part. And don't worry about $$[r]$$. Find the precise value of $$k$$ where $$[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$$. That is where $$r + \frac {k}{100} < m \le r + \frac {k+1}{100}$$ for some integer $$m$$.
====== my answer below ====
Well, what jumps at me is $$0 < \frac k{100} < 1$$ so all the $$[r +\frac {k}{100}]$$ are either one integer, call it $$n$$ or the next, $$n+1$$.
.
So if $$b$$ of them equal $$n+1$$ and $$(73 -b)$$ of them equal $$n$$ we have $$(73-b)n + b(n+1) = 73n + b =546$$ where $$0\le b < 73$$.
.
So as $$546\equiv 35 \pmod {73}$$ and $$546= 7*73 + 35$$ so $$b=35$$ and $$n=7$$.
.
So we have (I'll have to be careful not to do a fencepost error... $$91-35=56$$ so....) $$[r+\frac{19}{100}],....,[r+\frac{56}{100}] = 7$$ and $$[r+\frac {57}{100}],...,[r+\frac {91}{100}] = 8$$.
.
So $$r + \frac {56}{100} < 8\le r+\frac {57}{100}$$ so
.
$$100r + 56 < 800 \le 100r + 57$$
.
$$100r < 744 \le 100r + 1$$
.
And $$100r -1 < 743 \le 100r$$ so $$743 \le 100r < 744$$
.
So $$[100r] = 743$$.
• I solved the problem after user abieussu helped me realize that $\{r\}$ isn't necessarily equal to $\frac{81}{100}$, and your answer is wrong but very close. In your second to last step, you state this: "And 100r−<741≤100r so 741≤100r<742", which is the mistake, I believe.
– mpnm
Mar 25 '20 at 19:00
• If the third to last step is correct then my second to last ought to be. But arithmetic error can abound anywhere. Mar 25 '20 at 19:21
• Oh, for gosh sake. I worried so much about making a fencepost error and figure $57$ was the correct fencepost marker.... and then immediately replaced it with $58$. So everything from then on is off by one\$. Mar 25 '20 at 19:25
• So .... corrected it but... might have made and error. Mar 25 '20 at 19:31
• Argh... a fencepost error off by two... that's pretty annoying. Mar 25 '20 at 19:34
$$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$
First, let $$r = s - \dfrac{19}{100}$$. Then the sum becomes
$$\left\lfloor s + \frac{0}{100}\right\rfloor + \left\lfloor s + \frac{1}{100} \right\rfloor + \left\lfloor s + \frac{2}{100}\right\rfloor + \dots + \left\lfloor s + \frac{72}{100}\right\rfloor = 546$$
If $$s$$ were an integer, then you would get $$73s = 546$$, which has solution $$s = 7 \dfrac{35}{73}$$.
Noting that $$7 \cdot 73 = 511$$ and $$8 \cdot 73 = 584$$, we see that $$7 < s < 8$$. Now $$546-511 = 35$$. So where are you going to pick up that extra $$35$$? | 2021-10-20T02:02:52 | {
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https://math.stackexchange.com/questions/2184961/resistor-factory-and-bayes-rule-which-solution-is-correct | # Resistor factory and Bayes rule - which solution is correct?
Problem Statement
At a certain electronics factory, in a typical day’s output 10% percent of resistors are bad, and the rest are good. Good resistors have an 80% chance of passing the factory’s test, and bad resistors have a 30% chance of passing the test. Suppose the factory tests each resistor three times. If a particular resistor passes the test 2 times out of 3, what are the chances it is good?
Solution 1
First, let's define some events:
$B :=$ resistor is bad, $G :=$ resistor is good
$P :=$ resistor passes one test, $F :=$ resistor fails one test
$Q :=$ resistor passes 2 out of 3 tests
From the problem statement, we can say that $P[B] = 0.10, P[G] = 0.90, P[P|G] = 0.80, P[P|B] = 0.30$. Since the tests are independent, we can also say that $P[Q|G] = {3 \choose 2}(0.8)^2(0.2)$ and that $P[Q|B] = {3 \choose 2}(0.3)^2(0.7)$.
The quantity we seek is $P[G|Q]$. We proceed using Bayes' theorem:
\begin{align} P[G|Q] &= \frac{P[Q|G]P[G]}{P[Q]}\\ &= \frac{P[Q|G]P[G]}{P[Q|G]P[G] + P[Q|B]P[B]}\\ &= \frac{{3 \choose 2}(0.8)^2(0.2)(0.9)}{{3 \choose 2}(0.8)^2(0.2){0.9} + {3 \choose 2}(0.3)^2(0.7)(0.1)}\\ &\approx 0.95 \end{align}
Solution 2
This solution is the same as solution 1, but we calculate $P[Q]$ differently. We can say that $P[P] = P[P|G]P[G] + P[P|B]P[B] = (0.8)(0.9) + (0.3)(0.1) = 0.75$. Since the tests are independent, $P[Q] = {3 \choose 2}(0.75)^2 (0.25)$.
We proceed using Bayes' theorem just like we did in solution 1:
\begin{align} P[G|Q] &= \frac{P[Q|G]P[G]}{P[Q]}\\ &= \frac{{3 \choose 2}(0.8)^2(0.2)(0.9)}{{3 \choose 2}(0.75)^2 (0.25)}\\ &\approx 0.82 \end{align}
My Question
To me, both of these approaches seem right, but the answers are different. I've done the math several times, so I don't think the difference is due to an arithmetic error. Which solution is correct? Why?
• The choice of using $P$ with two distinct meanings ("probability" and "passes the test") is confusing. – mlc Mar 13 '17 at 16:41
• P(Q) the probability we pass the test 2 out of 3 times - the first part is what type of resistor is it. then the second part is get the probability it passes or fails based on that fact - you can't treat each test as independent, once it passes a test it is more likely to be good than it was before the test – Cato Mar 13 '17 at 16:53
• if 1% of boys always pass a test and 99% of girls always pass a test - I can't calculate the probability of a person passing two tests as 0.5^2 = .25 - it is actually (1/2)(.99^ 2 + .01^2) = .49 That is to say the girls pass both the tests usually - just to dream up a quick sexist example – Cato Mar 13 '17 at 16:58
The difference in the two approaches is in how you compute $P(Q)$. The first approach is the correct one.
(Let me write $Y$ instead of $P$ for "passing the test one time".) The second approach computes $P(Y)$ and then feeds it in the binomial distribution to derive the probability of passing the test twice out of three attempts. The flaw is in the assumption that the three trials are independent: the resistor is either good or bad (and the outcomes of the trials depend on its state).
Your second approach may be re-examined as follows. Let $Q$ be the event "passing the test two times out of three". Then $P[Q] = P[Q|G]P[G] + P[Q|B]P[B]$ as in your application of the Bayes' rule.
• So the the trials are not independent if you don't know whether the resistor is good or bad? How can that be? – SplitInfinity Mar 13 '17 at 16:59
• See third comment to the question with an example with boys and girls. Independence has nothing to do with what you know. If the resistor is good (but you do not know it), this affects the odds for $Q$. If the first trial is a success, this increases the probability that the resistor is good and hence the odds for $Q$. – mlc Mar 13 '17 at 17:02
P(GOOD | Passes 2/3 times ) = P(Passes 2/3 times AND Good) / P(passes 2/3 times)
= 0.9 x 0.8 x 0.8 x 0.2 x 3 / (.9 x 0.8 x 0.8 x 0.2 x 3 + 0.1 x 0.3 x 0.3 x 0.7 x 3) = (216/625)/(216/625 + 189/10000) = 128/135 = .95
the first one | 2019-09-21T11:44:18 | {
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https://stats.stackexchange.com/questions/487823/probability-that-linear-combination-of-normal-random-variables-exceeds-a-value/487832 | Probability that linear combination of normal random variables exceeds a value
I'm new to statistics, and I'm wondering how can I compute $$P(2Y_1 + 4Y_2 - 3Y_3 \geq 40)$$ given the following information?
• $$Y$$ follows a multivariate normal distribution.
• $$Y_1, Y_2, Y_3$$ follow a normal distribution.
• $$Y_1, Y_2, Y_3$$ have means $$10, 12, 14$$.
• $$Y_1, Y_2, Y_3$$ have variances $$2, 2, 8$$.
• The covariance between $$Y_1, Y_2$$ is $$0.50$$, and the covariance between $$Y_1, Y_3$$ is $$-0.75$$. The variables $$Y_2$$ and $$Y_3$$ are independent.
Here's the covariance matrix I found for $$Y_1, Y_2, Y_3$$:
$$\begin{bmatrix} 2 & .50 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix}$$
I know how to do it when $$Y_1, Y_2, Y_3$$ are all independent by using the fact that linear combinations of independent normal random variables are normal. However, I'm really not sure about how to do it in this case. I can easily find the covariance matrix, but I'm not sure how to proceed from there.
• Just knowing their means, variances and covariances along with the fact that each variables is (marginally) normal is insufficient. Sep 18, 2020 at 3:32
If you're looking for an analytic solution, this is what you should do. Start by defining a new variable $$X = 2Y_1 + 4Y_2 - 3Y_3$$ A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wikipedia article here). All that remains is to compute the mean and variance.
The mean of $$X$$ can be determined as \begin{align*} E[X] &= 2 E[Y_1] + 4 E[Y_2] - 3 E[Y_3] \\ &= (2 \times 10) + (4 \times 12)-(3 \times 14) \\ &= 26 \end{align*} The variance of $$X$$ is slightly easier to write in matrix form. Let $$Y \equiv (Y_1,Y_2,Y_3)$$ and $$\omega \equiv (2,4,-3)$$ be vectors of random variables and weights, respectively. The variance is given by
$$Var(X) = Var(\omega'Y) = \omega'Var(Y)\omega$$
Numerically this is equal to $$Var(X) = \begin{bmatrix} 2 & 4 & -3 \end{bmatrix} \begin{bmatrix} 2& .5 & -.75 \\ .5 & 2 & 0 \\ -.75 & 0 & 8 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ -3 \end{bmatrix}= 129$$ If $$\Phi$$ is the CDF of a standard normal, \begin{align*} P(X > 40) &= 1-P\left( X \le 40 \right) \\ &= 1-P\left( \frac{X - E[X]}{\sqrt{Var(X)}} \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-P\left( Z \le \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 1-\Phi\left( \frac{40 - E[X]}{\sqrt{Var(X)}} \right) \\ &= 0.108867 \end{align*}
• "A sum of normal random variables is also normal, even if the variables are not independent" - I don't think that's true. Do you have a proof?
– Paul
Sep 17, 2020 at 7:31
• @Paul you are correct, the necessary assumption is that the random variables are jointly gaussian (which they are in this case). The easy counterexample is $X \sim \mathcal N (0,1)$ and $Y = -X$. Then $X + Y = 0$ is not gaussian. Sep 17, 2020 at 7:55
• Another nice counterexample is here math.stackexchange.com/questions/563364/… Sep 17, 2020 at 8:05
• Great counter examples! I have amended the statement to emphasize that "joint" normality is important. Your comments emphasize the pitfalls of focusing on the marginal distributions only. Sep 18, 2020 at 2:56
You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)
> is.positive.definite(x, tol=1e-8)
[1] TRUE
> means=c(10,12,14)
> weights=c(2,4,-3)
> mat=mvrnorm(10^7,means,covmat)
> mat=mat %*% weights
> result=sum(mat>40)/10^7
> result
[1] 0.1088587
Non simulated
The mean is a linear combination of means
> new_mean= means %*% weights
> new_mean
[1] 26
The variance is obtained multiplying
> weights %*% covmat %*% weights
[,1]
[1,] 129
> 1-pnorm(40,26,sqrt(129)) | 2022-08-11T15:28:20 | {
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https://math.stackexchange.com/questions/1743819/definition-of-limit-as-x-rightarrow-infty | # Definition of limit as $x\rightarrow \infty$
Every time i get confused with the definition of $\lim_{x\rightarrow \infty}f(x)=L$. I could not find a reference that will give the definition.
I am trying to write what i understood. See if this is correct.
• By $\lim_{x\rightarrow \infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R>0$ such that $|f(x)-L|<\epsilon$ for all $x>R$.
• By $\lim_{x\rightarrow -\infty}f(x)=L$ we mean the following : Given $\epsilon >0$ there exists $R<0$ such that $|f(x)-L|<\epsilon$ for all $x<R$.
• By $\lim_{x\rightarrow \infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L>0$ such that $|f(x)|>R$ for all $x>L$
• By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L>0$ such that $f(x)<R$ for all $x>L$
• By $\lim_{x\rightarrow -\infty}f(x)=\infty$ we mean the following: Given $R>0$ there exists $L<0$ such that $|f(x)|>R$ for all $x<L$
• By $\lim_{x\rightarrow \infty}f(x)=-\infty$ we mean the following: Given $R<0$ there exists $L<0$ such that $f(x)<R$ for all $x<L$.
Let me know if i understood somethings wrongly.
• It's quite fine for me. The signs for $R$ and $L$ are unncessary (they're implicit). – Bernard Apr 15 '16 at 14:25
• You don't, technically, need $R<0//R>0$. for the first two. It can be any $R$. It's the "for all $x>R$" and "for all $x<R$" that makes the difference, not the sign of $R$. – Thomas Andrews Apr 15 '16 at 14:25
• If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @Bernard – user311526 Apr 15 '16 at 14:27
• If there is no need for sign then $x\rightarrow \infty$ and $x\rightarrow -\infty$ would mean the same thing? @ThomasAndrews – user311526 Apr 15 '16 at 14:27
• @topgeomj No, because one definition uses $x>R$ and the other $x<R$. That is the difference, not the sign of $R$. – Thomas Andrews Apr 15 '16 at 14:28
I will use $+\infty$ in this answer to avoid ambiguity.
• $\lim_{x\to +\infty} f(x) = L$ means for all $\epsilon>0$ there exists $M$ such that for all $x>M,$ $|f(x)-L|<\epsilon$ .
• $\lim_{x\to+\infty} f(x)=+\infty$ means for all $R$ there exists $M$ such that for all $x>M$, $f(x)>R$.
Those two definitions let you define the other limits by symmetry:
\begin{align} \lim_{x\to+\infty} f(x) = -\infty&\iff \lim_{x\to+\infty} -f(x)=+\infty\\ \lim_{x\to-\infty} f(x) = M &\iff \lim_{x\to+\infty} f(-x)=M \end{align}
Where the $M$ in the second case can be any of either a real value, or $+\infty,-\infty$.
So $\lim_{x\to-\infty} f(x)= -\infty$ means $\lim_{x\to +\infty} -f(-x)=+\infty$, which means:
For any $R$ there is an $M$ so that for all $x>M$, $-f(-x)>R$.
Now, given any $R'$, you can set $R=-R'$ and find $M$ with this condition, and set $M'=-M$. Then if $x<M'$, $-x>M$, and thus $-f(x)>R$ or $f(x)<R'=-R$. So we get back the definition that we want.
The reason to distinguish $+\infty$ from $\infty$ is that some books use $\infty$ means a single point at infinity, in both directions - essentially, merging the two values $+\infty,-\infty$ into a single point at infinity.
Then:
$$\lim_{x\to\infty} f(x)= L \iff \lim_{x\to+\infty} f(x)=\lim_{x\to-\infty} f(x)=L$$
where $L$ can be any real or $+\infty,-\infty$.
$$\lim_{x\to W} f(x) = \infty\iff \lim_{x\to W} |f(x)|=+\infty$$
Where $W$ can be any real, or $+\infty$ or $-\infty$, or $\infty$.
Your question is somehwat confused, because you seem to distingush $\infty$ from $+\infty$ with absolute values when $\infty$ is the limit, but not when $x\to\infty$.
• Thanks... I understand this clearly... :) – user311526 Apr 15 '16 at 15:04
One also does not need the absolute values on $|f(x)| > R$. If f is going to infinity, then $f(x) > R$ is accurate.
• For $f(x) = (-1)^{[x]}e^x$, $f(x) \to \infty | x \to +\infty$, but not $f(x) \to +\infty | x \to +\infty$. – Abstraction Apr 15 '16 at 14:34
• Ok.. Can you say something about other definitions as well.. – user311526 Apr 15 '16 at 14:34
• @Abstraction Some people distinguish between $+\infty$ and $\infty$, but quite often they are used interchangably. – Thomas Andrews Apr 15 '16 at 14:59
• @ThomasAndrews Seeing as third and fifth examples in the initial post clearly distinguish between $f(x) \to +\infty$ and $f(x) \to \infty$ (using absolute value of $f(x)$), I think here the distinction is important. – Abstraction Apr 15 '16 at 15:08
• Well, the main indication that the OP is using $\infty$ to mean a single point at infinity is the absolute values, but in the $x\to\infty$ examples, the definitions do not use $|x|>L$. So the clarity is not there in the question. @Abstraction – Thomas Andrews Apr 15 '16 at 15:11
Let me know if i understood somethings wrongly.
You did. When you write $x \to \infty$ (as opposed to $x \to +\infty$), for $x \in \mathbb{R}$ it roughly means "when absolute value of $x$ is arbitrarily large" ($|x| > R$, not $x > R$)
There is a universal definition of limit using filters. A filter $\mathcal{F}$ is a collection of sets such that it doesn't contain an empty set and $\forall f_1, f_2 \in \mathcal{F}, \exists f_3 \in \mathcal{F} : f_3 \subseteq f_1 \cap f_2$. For example, a set of neighbourhoods of a real point $x$ is a filter ($\{(x-\epsilon, x+\epsilon) | \epsilon \in \mathbb{R}^+\}$, called neighbourhoods filter); a set of intervals with infinite endpoint ($\{(a,+\infty) | a \in \mathbb{R}\}$) is a filter; a set of segment complements ($\{(-\infty,a)\cup(b,+\infty) | a,b \in \mathbb{R}\}$) is a filter.
Now, let's take function $h \in \{X \to Y\}$ and there is a filter $\mathcal{F}$ on $X$ and a concept of "neighbourhoods" on $Y$ (called topology). Then let's take this sentence: "there is such $y \in Y$ that for any $y$ neighbourhood $O_y$, there is a filter element $f \in \mathcal{F}$ so that $h(f) \subseteq O_y$" and write it for short as $\lim_{\mathcal{F}}h(x) = y$.
Lets write filter $\{(a,+\infty) | a \in \mathbb{R}\}$ as $x \to +\infty$, filter $\{(-\infty,a) | a \in \mathbb{R}\}$ as $x \to -\infty$ and filter $\{(-\infty,a) \cup (a,+\infty) | a \in \mathbb{R}\}$ as $x \to \infty$.
Finally, let's call set $\{(L-\epsilon,L+\epsilon) | \epsilon \in \mathbb{R}\}$ a set of neighbourhoods of $L$, set $\{(b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $+\infty$, set $\{(-\infty,b) | b \in \mathbb{R}\}$ a set of neighbourhoods of $-\infty$ and set $\{(-\infty,b) \cup (b,+\infty) | b \in \mathbb{R}\}$ a set of neighbourhoods of $\infty$.
Now from these rules we can derive any definition. Take $\lim_{x \to -\infty}f(x) = +\infty$. It comes to "for any $+\infty$ neighbourhood $(b,+\infty), b \in \mathbb{R}$, there is a filter element $(-\infty,a), a \in \mathbb{R}$ so that $f((-\infty,a)) \subseteq (b,+\infty)$". Or, in more traditional notation, $$\forall b \in \mathbb{R}, \exists a \in \mathbb{R} : \forall x \in (-\infty,a), f(x) \in (b,+\infty)$$
• May be you are complication the question by saying more advanced things as filter, topology... – user312648 Apr 15 '16 at 16:09
• @cello Maybe. But memorizing definitions for all cases of $x \to \pm\infty$ is actually hard, filters help to see some logic in all this. "Topology" here is simply a proper naming for a system of neighbourhoods; as you can see, it's properties aren't used. – Abstraction Apr 15 '16 at 16:21 | 2019-05-26T23:01:55 | {
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https://math.stackexchange.com/questions/2844744/help-with-matrix-derivative | # Help with matrix derivative
In my studies of applied mathematics, specifically optimization and applied linear algebra, I have come across the following expression which I need help differentiating:
$z(B,C) = \lVert Y-B \phi(CX) \rVert_F ^2 = \text{trace} \left( (Y-B \phi(CX))(Y-B \phi(CX))^T \right)$
where $Y \in \mathbb{R}^{m\times s},X\in \mathbb{R}^{n\times s}$ are two constant real matrices, $B \in \mathbb{R} ^{m\times k}$ and $C\in\mathbb{R}^{k \times n}$ are the two variable matrices of the function $z(B,C)$ defined above, $\lVert \bullet \rVert_F$ is the Frobenius norm and all matrix dimensions involved are constant and non-variable.
The function $\phi : \mathbb{R} \to \mathbb{R}$ is a nonlinear function defined for real numbers as follows:
$\phi(x) = \left\{ \begin{array}{ll} 0 & x \leq 0 \\ x & x > 0 \\ \end{array} \right.$
and we extend its definition to matrices element-wise, that is $(\phi(A))_{i,j} = \phi((A)_{i,j})$.
I would like to find the matrix derivative of the function $z$, as stated above, with respect to $C$, that is $\frac{\partial z}{\partial C}$.
Differentiating with respect to $B$ is easy enough as $C$ doesn't impede applying standard formulas, but my problem is with differentiating with respect to $C$ as it is the argument of a nonlinear function and I do not know how to derive it with respect to matrices.
I was hoping someone could please come to the rescue and help me differentiate the function $z$ with respect to $C$. I thank all helpers.
• In the first derivative, is A supposed to be C? – John Polcari Jul 8 '18 at 16:31
• Does this book have copy right? This not allowed on this website – Cloud JR K Jul 8 '18 at 16:46
• @JohnPolcari : yes fixed it now – kroner Jul 8 '18 at 16:49
• @CloudJR : no, it is OK – kroner Jul 8 '18 at 16:50
• @JohnPolcari : thanks for pointing out – kroner Jul 8 '18 at 16:51
Let's use a convention where uppercase Latin letters represent matrices, lowercase Latin vectors, and Greek letters are scalars.
The function you've denoted by $\phi$ is the ReLU function, $r(\alpha)$, whose derivative is the Heaviside step function $$h(\alpha) = \frac{dr(\alpha)}{d\alpha} \implies dr = h\,d\alpha$$ Applying these scalar functions element-wise on a matrix argument $A=CX,$ produces matrix results, which we'll denote as \eqalign{ R &= r(A) \cr H &= h(A) \implies dR = H\odot dA = H\odot(dC\,X) \cr } where $\odot$ is the elementwise/Hadamard product.
Define a new matrix variable $$M=BR-Y \implies dM = B\,dR + dB\,R$$ Write the function in terms of this new variable, then find its differential. \eqalign{ \lambda &= \|M\|^2_F = M:M \cr d\lambda &= 2M:dM \cr &= 2M:B\,dR + 2M:dB\,R \cr &= 2B^TM:dR + 2MR^T:dB \cr &= 2B^TM:H\odot(dC\,X) + 2MR^T:dB \cr &= 2(B^TM)\odot H:(dC\,X) + 2MR^T:dB \cr &= 2((B^TM)\odot H)X^T:dC + 2MR^T:dB \cr } Setting $dB=0$ yields the gradient wrt $C$ $$\frac{\partial\lambda}{\partial C} = 2((B^TM)\odot H)X^T$$ And setting $dC=0$ yields the gradient wrt $B$ $$\frac{\partial\lambda}{\partial B} = 2MR^T$$
NB: Depending on your preferred layout convention, you may need to transpose these results.
Also, the colon notation used above (called the Frobenius product) is just a convenient way of writing the trace function, i.e. $\,\,A:B={\rm tr}(A^TB)$.
The cyclic property of the trace leads to several ways to rearrange the terms in a Frobenius product. For example, all of the following expressions are equivalent \eqalign{ A:BC &= A^T:(BC)^T \cr &= BC:A \cr &= AC^T:B \cr &= B^TA:C \cr }
• Thanks for pointing out the name of $\phi$ as I was not aware of this. – kroner Jul 8 '18 at 17:26
• Thanks for your masterpiece answer, I appreciate it. – kroner Jul 8 '18 at 17:27 | 2021-05-07T10:00:21 | {
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https://www.physicsforums.com/threads/help-with-set-theory-notation.901710/ | # I Help with set theory notation
1. Jan 27, 2017
### Jehannum
I bought a maths book and have discovered it's somewhat above my level.
In particular I'm confused about one bit of notation. I understand the "is a member of" operator when it takes a set as argument (e.g. n ∈ ℝ) but not when the book uses it with functions (e.g. n ∈ f)
Does n ∈ f mean that the number n can be returned by function f?
2. Jan 27, 2017
### PeroK
You need to give more context. My guess is that a function can be seen as a set. For example, the function $f(x) = x^2$ can be seen as the set of points $\lbrace (x, x^2): x \in \mathbb{R} \rbrace$, also commonly known as the graph of $f$.
In this case, you could use the notation: $f = \lbrace (x, x^2): x \in \mathbb{R} \rbrace$.
And, here, $n \in f$ would mean that $n = (x, x^2)$ for some real $x$.
3. Jan 27, 2017
### Staff: Mentor
This is strange, not to say wrong. A function $f\, : \,X \rightarrow Y$ can be written as a subset $f=\{(x,y)\,\vert \,y=f(x)\} \subseteq X \times Y$ of the Cartesian product, but this cannot explain the notation $n \in f$ unless $n$ is a pair. Can you give some more context?
4. Jan 27, 2017
### Staff: Mentor
5. Jan 29, 2017
### Jehannum
I've copied part of the book below:
Theorem 2.2 There are infinite sets that are not enumerable.
Proof Consider the powerset of N, in other words the collection P whose members are all the sets of numbers (so X ∈ P iff X ⊆ N).
Suppose for reductio that there is a function f : N→P which enumerates P, and consider what we’ll call the diagonal set D ⊆ N such that n ∈ D iff n /∈ f(n).
Since D ∈ P and f by hypothesis enumerates all the members of P, there must be some number d such that f(d) = D.
So we have, for all numbers n, n ∈ f(d) iff n /∈ f(n). Hence in particular d ∈ f(d) iff d /∈ f(d). Contradiction!
6. Jan 29, 2017
### pasmith
That's not what the text is saying.
This is defining $f$ as a function from the natural numbers $\mathbb{N}$ to the collection $P$ of subsets of natural numbers. Thus if $n \in \mathbb{N}$ then $f(n) \subset \mathbb{N}$. The statement $n \in f(n)$ is therefore perfectly sensible, since it always makes sense to ask if a particular natural number is or is not a member of a particular subset of the natural numbers.
7. Jan 29, 2017
### Staff: Mentor
O.k., but this is not what you wrote. $n \in f$ is something different to $n \in f(n)$ where $f: \mathbb{N}\rightarrow \mathcal{P}(\mathbb{N})$ is a function which values are sets $X$, so $n \in X=f(n) \in \mathcal{P}(\mathbb{N})$ makes sense. This is not the same as $n \in f \in \mathcal{P}( \mathbb{N} \times \mathbb{N})$ which we explained in the previous posts.
The theorem is proven by the so called diagonal argument. It is a formal notation of the following principle:
We want to show, that the set of all subsets of $\mathbb{N}$ isn't enumerable.
If we have an element $X \in \mathcal{P}(\mathbb{N})$, i.e. a subset $X \subseteq \mathbb{N}$, then we can write it down number by number: $X= n_1n_2n_3n_4 \ldots$
Now let's assume $\mathcal{P}(\mathbb{N})$ is enumerable. Then there would be a complete list like
$$\begin{matrix} 1: &X_1 & = & n_{11}n_{12}n_{13} \ldots \\ 2: &X_2 & = & n_{21}n_{22}n_{23} \ldots \\ 3: & X_3 & = & n_{31}n_{32}n_{33} \ldots \\ &\vdots &&\\ \end{matrix}$$
which contains all elements of $\mathcal{P}(\mathbb{N})$, or subsets of $\mathbb{N}$.
The diagonal $d$ is now the number $d=n_{11}n_{22}n_{33} \ldots$ We next add $+1$ to all of the numbers $n_{ii}\,$.
Since by assumption we have enumerated all, this new number $d'$ (or subset $d'=\{n_{11}+1,n_{22}+1,n_{33}+1, \ldots\}$ if you like) has to occur somewhere, say $d'=X_k$. However the $k-$th position of $d'$ is $n_{kk}$ by enumeration of the $X_i$ and $n_{kk}+1$ by construction of $d'$. This cannot both be true, which means there is no way to enumerate $\mathcal{P}(\mathbb{N})$.
You might want to read the proof in the book again and figure out, why both are basically the same. It is also the method used to show why $[0,1]$ or $\mathbb{R}$ are uncountable. Whether $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$ holds is called the continuum hypothesis. It cannot be deduced from the usual axioms of set theory, i.e. by the Zermelo-Fraenkel axioms (ZF). If you read ZFC somewhere, then Zermelo-Fraenkel + Contnuum hypothesis is meant. (Usually it is also assumed, if nothing is said.)
8. Jan 29, 2017
### micromass
Staff Emeritus
That is not the continuum hypothesis and it can be proven in ZFC.
9. Jan 29, 2017
### micromass
Staff Emeritus
ZFC means Zermelo-Fraenkel with choice. Not the continuum hypothesis.
10. Jan 29, 2017
### Jehannum
Thanks. It's becoming clear now.
Boggled by some of the unfamiliar notation and concepts, I was not appreciating the fact that the function f returns a set as its result, so the ∈ operator can of course be used on the outputs of f such as f (n).
The book itself moves onto such diagonal proofs.
11. Jan 29, 2017
### Staff: Mentor
Thanks for clarification.
Oops, I just had a quick look on Wiki to be sure, for I know I tend to forget where the "gap" really is, and only saw $\mathfrak{c}=\aleph_1$ and next $2^{\aleph_0}=\aleph_1$. I should have read it with more care ...
Thanks for the corrections. | 2017-08-20T18:35:55 | {
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http://mathhelpforum.com/differential-geometry/190486-3-2-matrix-complex-elements.html | # Math Help - 3*2 Matrix with complex elements
1. ## 3*2 Matrix with complex elements
Folks,
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?
Are the following subsets subspaces?
1) The set of 3*2 with real entries
2) The set of matrices with first row (0,0)
3) The set of matrices with first row (1,1)
I understand the dimension of a vector space is the number of vectros in any basis for the space, ie the number of coordinates necessary to specify any vector. How do I find out what its dimension is?
Thanks
2. ## Re: 3*2 Matrix with complex elements
the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).
however, mathematics professors being what they are, will usually insist you display a basis (a linearly independent spanning set). can you think of a set of six matrices,
each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.
again, with any vector space V, there are 3 things you need to show for any subset W:
1) W is non-empty (preferrably by showing the 0-element is a member. if the 0-element (0-vector) is not in W, you will not obtain a vector space).
2) if u,v are in W then their vector sum u+v must also be.
3) if c is any scalar in your underlying field, and u is in W, then cu must also be in W. be careful with this one. if you are working with a complex vector space, it is not sufficient to check this property for real scalars only.
attempt to show these properties (or give a counter-example) for each of the sets in your post. that's how it's done.
3. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
the "naive" way is to ask yourself: how many complex numbers do i need to specify to identify my "vector"? for 3x2 matrices, that number is 6 (one for each entry).
can you think of a set of six matrices, each of which captures the idea of "a single matrix coordinate"? once you have done this, show linear independence and spanning.
Sorry, I am still struggling to get a start...I dont know how to approach this? If its a 3*2 matrix then would it have 3 solutions in ${\mathbb{R}^3}$?
4. ## Re: 3*2 Matrix with complex elements
If I write a sample 3*2 matrix as
$\begin{bmatrix}2 & 5\\ 1 & 4 \\ 6 & 8 \end{bmatrix}$
THis matrix contains 6 real entries, but I am unbale to write expression since I believe I need it to be a 3*3 or a 2*3 matrix. How would I proceed to show its closed under vector addition and scalar multiplication?
Thanks
5. ## Re: 3*2 Matrix with complex elements
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
6. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
I can only write 4 of the required 6 matrices...
$\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1\\ 1 & 1\end{array},\begin{array}{ccc} 0 & 1\\ 1 & 0\\ 0 & 0\end{array},\begin{array}{ccc} 1 & 1 \\ 0 & 0\\ 0 & 1\end{array},\begin{array}{ccc} 0 & 0 \\ 1 &1\\ 1 & 0\end{array}, \right]$
There is a lack of symmetry for me to complete....
7. ## Re: 3*2 Matrix with complex elements
One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.
8. ## Re: 3*2 Matrix with complex elements
Originally Posted by HallsofIvy
One of the things you need to clarify is whether you are thinking of this as a vector space over the real numbers or over the complex numbers. For example, with the set of pairs of complex numbers, with the usual addition and scalar multiplication, over the real numbers, we can take (a+ bi, c+ di)= a(1, 0)+ b(i, 0)+ c(0, 1)+ d(0, i) so it has dimension 4. The same set over the complex numbers has (a+ bi, c+di)= (a+ bi)(1, 0)+ (c+di)(0, 1) and so the dimension is 2.
1) What is the significance of saying 'vector space over the real numbers' or 'vector space over the complex numbers'? You have written to different ways (RHS) of expressing the same LHS. I dont understand the difference.
2) Is the above an example for a 2*2 matrix?
3) How does one do the above for a 3*2?
9. ## Re: 3*2 Matrix with complex elements
the "underlying field" of a vector space can make a difference when talking about dimension.
i don't understand why you came up with those particular 4 3x2 matrices. why not the 6 matrices who have one entry 1, the rest 0?
10. ## Re: 3*2 Matrix with complex elements
Originally Posted by Deveno
you're not being asked to "solve" a system of linear equations, you're being asked to display a basis for a vector space.
here is a similar problem:
find the dimension of the vector space of all 2x2 complex matrices.
i claim:
$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin {bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\ 1&0 \end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\}$
is a basis for this vector space. clearly this is a spaning set, since:
$\begin{bmatrix}a&b\\c&d\end{bmatrix} = a\begin{bmatrix}1&0\\0&0\end{bmatrix} + b\begin{bmatrix}0&1\\0&0\end{bmatrix} + c\begin{bmatrix}0&0\\1&0\end{bmatrix} + d\begin{bmatrix}0&0\\0&1\end{bmatrix}$
now, to show linear independence, suppose that:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = c_1\begin{bmatrix}1&0\\0&0\end{bmatrix} + c_2\begin{bmatrix}0&1\\0&0\end{bmatrix} + c_3\begin{bmatrix}0&0\\1&0\end{bmatrix} + c_4\begin{bmatrix}0&0\\0&1\end{bmatrix}$
then:
$\begin{bmatrix}0&0\\0&0\end{bmatrix} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$
so $c_1 = c_2 = c_3 = c_4 = 0$ so our set is linearly independent, and is thus a basis.
since our basis has 4 elements, the dimension of the vector space is 4.
$\left[\begin{array}{ccc} 0 & 0 \\ 0 & 0\\0 & 0\end{array}\right]=c_1\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 0\\0 & 0\end{array}\right]+c_2\left[ \begin{array}{ccc} 0 & 1\\ 0 & 0\\ 0 & 0\end{array}\right]+c_3\left[ \begin{array}{ccc} 0 & 0 \\ 1 & 0\\ 0 & 0\end{array}\right]+c_4\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 1\\ 0 & 0\end{array}\right]+c_5\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 1 & 0\end{array}\right]+c_6\left[ \begin{array}{ccc} 0 & 0 \\ 0 & 0\\ 0& 1\end{array}\right]$
implies $c_1+c_2+c_3+c_4+c_5+c_6=0$
This implies the set is linearly independant with a dimension of 6? What is next?
My attempt based on HallsofIvy post
3*2 matrix over the real numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$ Hence 12 dimensions
3*2 over the complex numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=(a+bi)(1,0)+(c+di)(0,1)+........$ Hence 6 dimensions.........?
11. ## Re: 3*2 Matrix with complex elements
Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.
12. ## Re: 3*2 Matrix with complex elements
Originally Posted by Hartlw
Every element of the matrix is an ordered pair. If the matrix has only one non-zero element, all matrices with only this one element are a linear combination of two matrices, one with (0,1) as the sole element, and the other with (1,0) as the sole element. So for each element in the matrix you need two matrices with (1,0) and (0,1) as the sole elements. You need a total of 12 matrices, two for each element. The dimension of the vector space is 12.
For both the real case and the complex case? Is this not conflicting with post # 7
13. ## Re: 3*2 Matrix with complex elements
Originally Posted by bugatti79
For both the real case and the complex case? Is this not conflicting with post # 7
For the real case, every base matrix has a 1 in one element and every other element a 0, for a total of six, the dimension is six. For complex case, see my previous post.
EDIT: I think post seven is discussing a complex vector space consisting of complex vectors a + bi. Not the same thing as a complex vector space each element of which is a matrix. You can probably think of each matrix as the sum of a real (a,0) at each location and imaginary, (,0,b) at each location, component . In any event, any complex matrix can be expressed as a linear combination of the 12 base matrices I described in my previous post.
EDIT: The matrices themselves satisfy requirements of linear vector space with the usual definition of matrix addition and multiplication by a scalar.
The base matrices are:
(0,1) (0,0)
(0,0) (0,0)
(0,0) (0,0)
(1,0) (0,0)
(0,0) (0,0)
(0,0) (0,0)
(0,0) (1,0)
(0,0) (0,0)
(0,0) (0,0)
..... for 12 base matrices (dim 12).
14. ## Re: 3*2 Matrix with complex elements
Originally Posted by Hartlw
I think post seven is discussing a complex vector space consisting of complex vectors a + bi.
So based on your post is this the answer to the following original question?
Given that the set of all 3*2 matrices with complex elements is a complex vector space, with the usual definitions of addition and scalr multiplication of matrices, what is its dimension?
3*2 matrix over the complex numbers
$\left[\begin{array}{ccc} a+bi & c+di \\ e+fi & g+hi\\j+ki & l+mi\end{array}\right]=a(1,0)+b(i,0)+c(1,0)+d(i,0)........$
15. ## Re: 3*2 Matrix with complex elements
No
If you denote the matrices in my previous post by E1, E2, E3 , E4, ...
Then an arbitrary complex matrix is expressed as a linear combination of E1, E2, E3,...,E12
(1,3) (5,3)
(3,5) (5,7) = 1xE1 + 3xE2 + 5xE3 + ..........
(3,3) (4,3)
Above is a matrix. The ordered pair (1,3) is just a std shortcut way of expressing 1+i3.
A complex matrix has to be expressed as a sum of complex matrices. Not as a sum of complex numbers. The addition of two complex matrices is a complex matrix by term by term addition, added just like complex numbers.
Page 1 of 3 123 Last | 2015-05-23T09:20:02 | {
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http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union | # Empty intersection and empty union
If $A_\alpha$ are subsets of a set $S$ then
$\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"
$\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"
It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.
But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.
Can someone help me and tell me what is wrong with this? Thank you.
-
This might help. math.stackexchange.com/q/309986/35983 – Gautam Shenoy Apr 23 at 9:13
Element in the empty set is the problem. – simplicity Apr 23 at 9:14
– Martin Sleziak Apr 23 at 11:55
Some texts consider it a convention, but it is in fact a computation!
For the fixed set $S$, we are looking at the set $\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and unions (of arbitrary many subsets) the set $\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).
First, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\bigcup _{i\in \emptyset }A_i=\emptyset$.
Now, to make things more formal, lets define the intersection and union in $\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\{A_i\}_{i\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.
So now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\subseteq S$ contains each of the non-existent $A_i$. The largest of those is $S$, proving that $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\emptyset$, thus proving that $\bigcup_{i\in \emptyset}A_i=\emptyset$.
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I realized, thank you. – Anna Apr 23 at 9:37
You are welcome @Anna – Ittay Weiss Apr 23 at 9:38
In the last paragraph: did you mean "...is contained in each of..."? – Anna Apr 23 at 10:02
+1, excellent explanation. – Andreas Caranti Apr 23 at 10:08
Please note that the fact that
$$\bigcap_{\alpha \in \emptyset} A_\alpha = S$$
is not a convention - it follows from the definition
$$\bigcap_{\alpha \in I} A_\alpha = \{ x \in S : \text{x \in A_\alpha, for all \alpha \in I}\}.$$
This is best understood by asking when is it that for a given $x \in S$ we have $x \notin \bigcap_{\alpha \in \emptyset} A_\alpha$. This can only happen if there exists $\alpha \in \emptyset$ such that $x \notin A_{\alpha}$. Since $\emptyset$ has no elements, there cannot be such an $\alpha$.
PS One can avoid introducing a (somewhat arbitrary) index set in an intersection, say. Instead, fix a set $S$, take a subset $\mathfrak{S}$ of $\mathcal{P}(S)$, and define $$\bigcap \mathfrak{S} = \{ x \in S : \text{x \in A for all A \in \mathfrak{S}} \}.$$ Then the argument above becomes $$\bigcap \emptyset = \{ x \in S : \text{x \in A for all A \in \emptyset} \} = \{ x \in S : \ \} = S,$$ since there are no $A$ to consider here.
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It should be pointed out that this intersection is $S$ only if it is given that the computation is performed in $\mathcal P(S)$. Otherwise, if the computation is performed just in the ambient universe of sets, then the intersection does not exist (since it's trying to be a set that contains every other set). In contrast, the empty union is the empty set no matter where it is computed. – Ittay Weiss Apr 23 at 9:47
@IttayWeiss, an excellent point. – Andreas Caranti Apr 23 at 9:52
What is the ambient universe of sets? – Anna Apr 23 at 10:06
– Andreas Caranti Apr 23 at 10:07
Thank you Andreas. I think that my point of confusion is why you can write $\left\{ x \in S : x \in A \text{ for all } A \in \emptyset \right\} = \left\{ x \in S : \text{} \right\}$, whereas you cannot write (considering the case of the union of the empty collection) $\left\{ x \in S : x \in A \text{ for at least one } A \in \emptyset \right\} = \left\{ x \in S : \text{} \right\}$. Would you mind helping me understand the distinction? – Dan Douglas Dec 12 at 15:59
show 2 more comments
Those conventions comes from the fact that $(\mathcal P (S), \cup, \emptyset)$ and $(\mathcal P (S), \cap, S)$ are monoids.
Hence, those conventions are just the usual $\sum_{k \in \emptyset} k = 0$ or $\prod_{k \in \emptyset} k = 1$ that you certainly know for $(\mathbb N, +)$ or $(\mathbb Z, \times)$ for example.
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This is like saying "it's so because it is so". In fact, it is not a convention at all. It can be proven so by following the definition. In a monoid though, the unit is (by definition if you like) the result of applying the operation to no elements. But for $\mathcal P(S)$ it follows from a deeper reason than just saying it's so. – Ittay Weiss Apr 23 at 9:34
@IttayWeiss The question was not "Is it a convention or not ?" but "Can anyone enlighten me about those conventions/properties ?". I was just pointing out that the OP was actually familiar with those but did not recognize it at first sight. Plus, I could argue that your proof can actually show that $\bigcap_\emptyset A_i = \emptyset$ : indeed, taking $B$ a subset of $S$, $\forall A_i \in \emptyset, A_i \not\subseteq B$ is as true as $\forall A_i \in \emptyset, A_i \subseteq B$, making $\emptyset$ the intersection over the emptyset. – Pece Apr 23 at 19:16
I see your your points with the monoid analogy. If you'll think about it though, you'll see that your criticism of my proof is false (or are you claiming that you just proved a contradiction?) – Ittay Weiss Apr 23 at 19:45
Sorry, I mixed up a little. What I was saying is : as $A_i \in \emptyset$ is universally false, your claim $\forall A_i, A_i \in \emptyset \implies B \subseteq A_i$ (and so $S$, the greatest of all, is the intersection) is a valid as my claim $\exists A_i, A_i \in \emptyset \implies B \not\subseteq A_i$ (and so no subset of $S$ is a lower bound of the $A_i$ except $\emptyset$). So in my opinion, you're making a convention on the complete lattices (the meet of nothing is the top) while I was making conventions about monoid. – Pece Apr 24 at 8:34
I think you are still confused. I did not make use of any convention. I followed the definition. – Ittay Weiss Apr 24 at 8:56
show 5 more comments | 2013-12-20T19:16:43 | {
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https://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set | # How to prove that $\mathbb{Q}$ ( the rationals) is a countable set
I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
• Sure. Better yet, it's sufficient to find an injection from $\mathbb Q$ to $\mathbb Z$. Or for that matter a surjection from $\mathbb Z$ to $\mathbb Q$, or from $\mathbb N$ to $\mathbb Q$. – bof Feb 1 '14 at 8:25
• is just an injection sufficient to prove countability, though? – furashu Feb 1 '14 at 8:33
• I guess it depends on how you define "countable". Are finite sets countable? If not, then besides finding an injection from $\mathbb Q$ to $\mathbb Z$, you also have to prove that $\mathbb Q$ is infinite. But I you already know that. – bof Feb 1 '14 at 8:36
• i would say finite sets are not "countable" but "finite" .. seems weird, but countable is a term I would reserve specifically for distinction between cardinality of infinite sets. – furashu Feb 1 '14 at 8:40
• Fine. $\mathbb Q$ is an infinite set. An infinite set is countable if it has an injection into $\mathbb N$. Or into any countable set, such as $\mathbb Z$, which you already know is countable. – bof Feb 1 '14 at 8:46
Clearly $\mathbb{Z}$ injects into $\mathbb{Q}$.
Let $p_i$ enumerate all the prime numbers.
If $q \neq 0, 1, -1$, let $q = \pm \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}}$ be the prime decomposition the numerator and denominator of $q$ written in simplest form. Define
$\Phi(q) = \begin{cases} 0 & \quad q = 0 \\ 1 & \quad q = 1 \\ -1 & \quad q = -1 \\ p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \quad q = \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \\ - p_{2 i_0}^{n_0} ... p_{2 i_k}^{n_k} p_{2 j_0 + 1}^{m_0} ... p_{2 j_p + 1}^{m_p} & \quad q = - \frac{p_{i_0}^{n_0} ... p_{i_k}^{n_k}}{p_{j_0}^{m_0} ... p_{j_p}^{m_p}} \end{cases}$
$\Phi$ is an injection of $\mathbb{Q}$ into $\mathbb{Z}$.
By the Cantor Schroder Theorem, there is a bijection between $\mathbb{Z}$ and $\mathbb{Q}$.
As bof mentioned, a nicer injection would be
$\Phi(q) = \begin{cases} 0 & \quad q = 0 \\ 1 & \quad q = 1 \\ -1 & \quad q = -1 \\ 2^m (2n + 1) & \quad q = \frac{m}{n} \text{ simplest form } \\ - 2^m(2n + 1) & \quad q = - \frac{m}{n} \text{ simplest form} \end{cases}$
• this is one of the most impressive and elegant answers i've seen on here, well done! – furashu Feb 1 '14 at 9:18
• Instead of using prime factorizations, you could just map the reduced fraction $\frac mn$ to the integer $2^m(2n+1)$ – bof Feb 1 '14 at 9:23
• @bof True. This would have saved me like ten minutes of typing. – William Feb 1 '14 at 9:27
• @William can you explain what bof said? what's the significance of that integer? – furashu Feb 1 '14 at 9:39
• @William oh i am checking it now, I just like to know where things coem from, to get a glimpse on peoples intuition and knowledge. – furashu Feb 1 '14 at 9:46
If you know that $\mathbb{Z}$ is countable, you know there is a bijection $\chi:\mathbb{N} \rightarrow \mathbb{Z}$. Hence, it is sufficient to find a bijection $\nu:\mathbb{Z} \rightarrow \mathbb{Q}$ since then $\chi \circ \nu$ is a bijection between $\mathbb{N}$ and $\mathbb{Q}$.
In any case, the following figure illustrates a bijection between $\mathbb{Z}$ and $\mathbb{Q}$.
We follow the worm back and forth "counting" the rational numbers, skipping any numbers that are not simplified fractions.
Hint: There is a natural map $$\left\{ \begin{array}{ccc} \mathbb{Z} \times \mathbb{Z}_{>0} & \to & \mathbb{Q} \\ (a,b) & \mapsto & \frac{a}{b} \end{array} \right.$$
• Is mapping from Z x Z the same as mapping from Z, though? – furashu Feb 1 '14 at 8:44
• Lemma : If $A$ is countable, then $A \times A$ is countable. – Euler....IS_ALIVE Feb 1 '14 at 8:53
• @Euler....IS_ALIVE How does one prove that? – Arjang Jan 19 '17 at 12:23
I read this proof in Amer. Math. Monthly. Suffices to find out an injective function from the set of positive rationals to positive integers. Consider the representation of numbers in base 11, where the 11 digits used are $$0,1,2,3, ..., 9, /$$ (yes, it is a slash as digit for the number ten). Now the rational number $$7/83$$ represents the 4-digit base-11 positive integer: $$(7\times 11^3) + (10\times 11^2) + (8\times 11) + 3$$.
• I didn't understand this. Can you explain why we can give a completely different meaning to the slash (from its established definition as the division symbol to a numeral in base-11) so as to create an injection. – Mr Reality Jan 12 at 20:21
• Ok. Write out a rational number which has a division symbol slash, say for example 3/4. Now replace slash by x. (Remember x is roman numeral for ten). So this becomes 3x4 which is a 3-digit number in base 11 system. The value of this is three eleven squared plus ten elevens and 4 which in base 10 is $(3\times 121 + 10 \times 11 +4= 363+110+4=477$. SO the function sends $3/4$ to 477. – P Vanchinathan Jan 13 at 2:08
Consider for $n=2,3,4,\ldots$ the sets
$$A_n=\left\{\frac pq :(p,q)=1,p,q>0,p+q=n\right\}$$
Claim Each $A_n$ is finite, and $\Bbb Q^+=A_2\cup A_3\cup A_4\cup\dots$
A002487 Stern's diatomic series (or Stern-Brocot sequence) Also called fusc(n) [Dijkstra]. a(n)/a(n+1) runs through all the reduced nonnegative rationals exactly once [Stern; Calkin and Wilf]. https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD570.html
fusc(1) = 1
fusc(2n) = fusc(n)
fusc(2n+1) = fusc(n) + fusc(n+1)
And it seems that fusc(k)/fusc(k+1) gives the positive rational numbers. There was also a contrieved proof recently that fusc does what it does here: https://www.isa-afp.org/browser_info/current/AFP/Stern_Brocot/document.pdf
Using the sign of Z we could create a mapping to the positive and negative rational numbers. | 2019-05-20T16:23:29 | {
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https://math.stackexchange.com/questions/3124407/number-of-edges-of-a-non-planar-graph-with-fixed-number-of-vertex | # Number of edges of a non-planar graph with fixed number of vertex
Consider a connected undirected simple non-planar graph $$G$$ with $$15$$ vertices. If removing any edge from $$G$$ results in a planar graph, how many edges does $$G$$ have?
It is obvious that if the number of edges greater than $$3*15-6=39$$, then it is not planar. But, how to discuss "removing any edge make it planar"? Thanks
The number of edges of $$G$$ is either $$18$$ or $$20$$.
Since graph $$G$$ is non-planar, by Kuratowski's theorem, it contains a subdivision of either $$K_5$$ (the complete graph of $$5$$ vertices) or $$K_{3,3}$$ (the complete bipartite graph of $$6$$ vertices) as a subgraph.
Since removing any edge from $$G$$ make it planar, above subgraph exhaust all edges of $$G$$. This means $$G$$ is a subdivision of either $$K_5$$ or $$K_{3,3}$$.
Notice subdividing a graph increases the number of vertices and edges by same amount.
If we start from $$K_5$$ which has $$5$$ vertices and $$10$$ edges, we need to subdivide $$10$$ times to get $$15$$ vertices. In this case, the resulting graph will have $$10 + 10 = 20$$ edges.
If we start from $$K_{3,3}$$ which has $$6$$ vertices and $$9$$ edges, we need to subdivide $$9$$ times to get $$15$$ vertices. In that case, the resulting graph will have $$9+9 = 18$$ edges.
• Got it! Thank you so much! Rarely use subdivision to count~~ – Andy Feb 25 '19 at 11:31
I am not sure if this is the answer you are looking for, but a discussion on this topic can conclude upper and lower bounds for the number of edges in such a graph.
As you have already stated, we can determine the upper limit using the Euler characteristic, $$v-e+f-c=1$$ which we can reduce to $$v-e+f=2$$ since our graph is connected. Because we know that our original graph $$G$$ is non-planar before removing an edge, we can say that $$\left|E\right|=e+1$$. We can use the maximum possible faces of a planer graph with $$v=15$$ to find $$e$$ using $$f_{max}=2v-4 = 26$$ $$15-e+26=2 \implies \left|E\right| \leq 40 \leftarrow\textrm{upper bound}$$
To find the lower bound, we can use Kuratowski's Theorem (see here for a nice explanation) to state a lower bound, $$\left|E\right| \geq v+3 = 18$$.
So, we can say that for a non-planar graph $$G$$, with $$\left|V\right|=15$$ and the property that removing an edge must make the result planar, has $$18\leq\left|E\right|\leq 40$$ edges.
Essentially, if the number of edges is less than the lower bound, we know that the graph must be planar. If the number of edges is greater than the upper bound, we would have to remove more than one edge to make it planar.
• Got it! Thank you so much. You gave a very detailed description. – Andy Feb 25 '19 at 11:35 | 2020-02-29T10:57:46 | {
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https://math.stackexchange.com/questions/3079038/expected-number-of-steps-in-a-1d-random-walk-with-reflecting-edges/3080351 | # Expected number of steps in a 1D random walk with reflecting edges
Assume there is a row of $$k$$ tiles. A creature (monkey in some situations, ant in others, frog in others) lies on tile $$a$$. There is a 50% probability that the creature jumps to tile $$a-1$$ and a 50% probability that the creature jumps to tile $$a+1$$, unless it is on an edge tile. If it is on an edge tile, it must jump inwards, so it can't escape the system (i.e. tile 2 from tile 1 and tile $$k-1$$ from tile $$k$$). What is the expected number of steps for it to first reach tile $$b$$? $$1<=a, b<=k$$ is assumed. I feel like Markov chains might be used to get the answer, but I have a very limited understanding of them. If there is a closed form for the answer as well as a derivation for understanding, that would be perfect.
• Markov chains are the way! – Lord Shark the Unknown Jan 19 at 5:20
• @LordSharktheUnknown Can you elaborate in an answer how to use them in order to solve this problem? Like I said, I have a limited understanding of them. – automaticallyGenerated Jan 19 at 5:31
• I'have taken the liberty to modifiy your title and tags (3 tags are a good average) in order more readers are directed towards this interesting question and its interesting answer – Jean Marie Feb 15 at 12:27
Amazingly (to me) there happens to be a very simple expression for the expected number of steps to reach $$\ b\$$ from $$\ a\$$. For $$\ a < b\$$, it is: $$\left(\,b + a -2\,\right)\,\left(\,b-a\,\right)\ .$$ Although a Markov chain is the obvious way to model the process, and I'm sure it could used to derive the above result, there turns out to be a less cumbersome way of doing it.
For each $$\ i\$$ between $$\ 1\$$ and $$\ b\$$ inclusive, let $$\ e_i\$$ be the expected number of steps the creature takes to go from $$\ i\$$ to $$\ b\$$. Obviously, $$\ e_b\ = 0\$$.
If the creature starts from $$\ 1\$$, then it has to take one step to $$\ 2\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_2\$$. Thus, the expected number of steps, $$\ e_1\$$, to reach $$\ b\$$ from $$\ 1\$$ is $$\ e_2 + 1\$$.
If the creature starts from $$\ b-1\$$, then with probability $$\ \frac{1}{2}\$$ it reaches $$\ b\$$ on the very next step—that is, in just a single step—, and with probability $$\ \frac{1}{2}\$$ it jumps to $$\ b-2\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{b-2}\$$. Thus $$\ e_{b-1} = \frac{1}{2}\left(e_{b-2} +1\right) + \frac{1}{2}\,1=\frac{1}{2}\,e_{b-2}+1\$$.
If the creature starts from any other point $$\ i\$$, with $$\ 2\le i\le b-2\$$, then with probability $$\ \frac{1}{2}\$$ it jumps to $$\ i-1\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{i-1}\$$, and with probability $$\ \frac{1}{2}\$$ it jumps to $$\ i+1\$$, from which the expected number of steps to reach $$\ b\$$ is $$\ e_{i+1}\$$. Therefore, $$\ e_i = \frac{1}{2}\left(e_{i-1} +1\right) + \frac{1}{2}\left(e_{i+1} +1\right)= \frac{1}{2}\,e_{i-1} + \frac{1}{2}\,e_{i+1} +1\$$.
Putting this all together, we have
$$\begin{eqnarray} e_1 &=& e_2 + 1\\ e_i &=& \frac{1}{2}\,e_{i-1} + \frac{1}{2}\,e_{i+1} +1, \ \ \mbox{for } i=2,3, \dots, b-2\ \mbox{, and}\\ e_{b-1} &=& \frac{1}{2}\,e_{b-2}+1\ , \end{eqnarray}$$ or, equivalently,
$$\begin{eqnarray} e_1 - e_2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=& 1\\ \ \ \ \ \ \ \ \ \ \ \ \ \ -\frac{1}{2}\,e_{i-1} + e_i -\frac{1}{2}\,e_{i+1} &=& 1, \ \ \mbox{for } i=2,3, \dots, b-2\ \mbox{, and}\\ -\frac{1}{2}\,e_{b-2}+e_{b-1} &=& 1\ . \end{eqnarray}$$
These equations can be written as: $$M\,e = \mathbb 1\ ,$$ where $$\ M\$$ is the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ matrix, and $$\ \mathbb 1\$$ the $$\ \left(\,b-1\,\right)\times\,1\$$ column vector, whose entries are given by: $$\begin{eqnarray} M_{1,2} &=& -1\\ M_{i,i} &=& 1\ \ \mbox{for } i=1,2,\dots, b-1\\ M_{i,i-1} &=& -\frac{1}{2}\ \ \mbox{for } i=2,3,\dots, b-1\\ M_{i,i+1} &=& -\frac{1}{2}\ \ \mbox{for } i=2,3,\dots, b-2\\ M_{i,\,j} &=& 0 \ \ \mbox{for all other }\ i, j\\ \mathbb 1_i &=& 1\ \ \mbox{for } i=1,2,\dots, b-1\ . \end{eqnarray}$$ For $$\ b=6\$$, the matrix $$\ M\$$ looks like this: $$\left(\begin{matrix}1&-1&0&0&0 \\ -\frac{1}{2}&1&-\frac{1}{2}&0&0\\ 0&-\frac{1}{2}&1&-\frac{1}{2}&0\\ 0&0&-\frac{1}{2}&1&-\frac{1}{2}\\ 0&0&0&-\frac{1}{2}&1& \end{matrix}\right)\ ,$$ and has the following inverse: $$M^{-1} = \left(\begin{matrix} 5&8&6&4&2\\ 4&8&6&4&2\\ 3&6&6&4&2\\ 2&4&4&4&2\\ 1&2&2&2&2\\ \end{matrix}\right)\ .$$ From this, we can conjecture that the entries of the inverse of the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ matrix $$\ M\$$, defined above, should be the matrix $$\ L\$$ whose entries are given by: $$\begin{eqnarray} L_{i,1} &=& b-i\ \ \mbox{for } i=1,2,\dots, b-1\\ L_{1,\,j} &=& 2\,\left(b-j\right) \ \mbox{for } j=2,3,\dots, b-1\\ L_{i,\,j} &=& 2\,\min\left(b-i,b-j\right) \ \mbox{for } 2\le i\le b-1\ \ \mbox{and }\ 2\le j\le b-1\ , \end{eqnarray}$$ and on checking the product $$\ M\,L\$$, we find that it is indeed the $$\ \left(\,b-1\,\right)\times\left(\,b-1\,\right)\$$ identity matrix. So, finally, we have: $$e = M^{-1}\,\mathbb 1 = L\,\mathbb 1\ ,$$ and $$\ e_a\$$, the expected number of steps to get to $$\ b\$$ from $$\ a\$$ is the sum of the entries in the $$\ a^\mbox{th}\$$ row of $$\ L\$$: $$\begin{eqnarray} e_a &=& \left(b-a\right) + 2\,\left(\,a-1\,\right)\,\left(\,b-a\,\right) + 2\,\sum_{j=1}^{b-a-1} j\\ &=& \left(\,b + a -2\,\right)\,\left(\,b-a\,\right)\ , \end{eqnarray}$$ as stated above.
• Thanks for the clear answer. I think that this is just a disguised Markov chain, as your "M" is the identity matrix minus the transition matrix. – automaticallyGenerated Jan 21 at 14:30
• Not quite. The identity matrix minus the transition matrix is a singular $\ b\times b\$ matrix. $\ M\$ is the submatrix obtained from it by chopping off its last row and column. But you're right that there is indeed a Markov chain lurking in the shadows. – lonza leggiera Jan 21 at 23:27 | 2019-04-24T06:37:12 | {
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https://byjus.com/question-answer/displaystyle-int-0-1-frac-dx-1-x-2-sqrt-2-x-2-frac-pi/ | Question
# $$\displaystyle \int_{0}^{1}\frac{dx}{(1+x^{2})\sqrt{(2+x^{2})}}=\frac{\pi }{k}$$. Find the value of $$k$$.
Solution
## Let $$\displaystyle I=\int { \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx }$$Put $$x=\sqrt { 2 } \tan { t } \Rightarrow dx=\sqrt { 2 } \sec ^{ 2 }{ t } dt$$$$\displaystyle I=\sqrt { 2 } \int { \frac { sec{ t } }{ \sqrt { 2 } \left( 2\tan ^{ 2 }{ t } +1 \right) } dt } =\int { \frac { sec{ t } }{ 2\tan ^{ 2 }{ t } +1 } dt }$$Multiplying numerator and denominator by $$\cos ^{ 2 }{ t }$$, we get$$\displaystyle I=\int { \frac { \cos { t } }{ 2\sin ^{ 2 }{ t } +\cos ^{ 2 }{ t } } dt } =\int { \frac { \cos { t } }{ \sin ^{ 2 }{ t } +1 } dt }$$Now put $$u=\sin { t } \Rightarrow du=\cos { t } dt$$Therefore$$\displaystyle I=\int { \frac { 1 }{ { u }^{ 2 }+1 } du } =\tan ^{ -1 }{ u } =\tan ^{ -1 }{ \sin { t } }$$$$\displaystyle =\tan ^{ -1 }{ \sin { \left( \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { 2 } } \right) } \right) } } =\tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) }$$Hence$$\displaystyle \int _{ 0 }^{ 1 }{ \frac { 1 }{ \left( { x }^{ 2 }+1 \right) \sqrt { { x }^{ 2 }+2 } } dx } ={ \left[ \tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { x }^{ 2 }+2 } } \right) } \right] }_{ 0 }^{ 1 }$$$$\displaystyle =\frac { \pi }{ 6 } -0=\frac { \pi }{ 6 }$$$$\Rightarrow k=6$$Mathematics
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https://math.stackexchange.com/questions/1668254/how-many-cubes-have-at-least-1-2-3-colors-on-them/1668271 | # how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
This is an addendum to Win Vineeth's excellent answer: the particular counts fall out of the algebraic expansion (using the Binomial Theorem) of $n^3$, where we write $n = (n-2) + 2$: $$\bigl( (n-2) + 2 \bigr)^3 = (n-2)^3 + 6(n-2)^2 + 12(n-2) + 8$$
This approach generalizes by letting $t$ be the "thickness" of the boundary (the part that gets painted). The original question uses $t=1$. The expansion now looks like: $$\bigl( (n-2t) + 2t \bigr)^3 = (n-2t)^3 + 6(n-2t)^2t + 12(n-2t)t^2 + 8t^3$$ Now, the power of $t$ serves as a placeholder for the number of sides that got painted. For example, if you want to know how many subcubes got painted on $2$ sides, then look at the coefficient of $t^2$, namely $12(n-2t)$, which is $12(n-2)$ once you set $t=1$. This is an example of a generating function, where a sequence is encoded as the list of coefficients of a polynomial (or often infinite series).
This approach also has the advantage of counting the number of boundary subhypercubes of a hypercube of any dimension: just expand $n^d$, where $d$ is the dimension. | 2021-01-21T18:50:16 | {
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https://byjus.com/question-answer/let-a-be-a-set-containing-n-elements-a-subset-p-of-the-set-a-1/ | Question
Let $$A$$ be a set containing $$n$$ elements. A subset $$P$$ of the set $$A$$ is chosen at random.The set $$A$$ is reconstructed by replacing the element of $$P$$, and another subsets $$Q$$ of $$A$$ is chosen at random. The probability that $$\left( P\cap Q \right)$$ contains exactly $$m(m<n)$$ elements is
A
3nm4n
B
nCm3m4n
C
nCm3nm4n
D
none of these
Solution
The correct option is C $$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } }$$We know that the number of subsets of a set containing $$n$$ elements is $${ 2 }^{ n }$$.Therefore, the number of ways of choosing $$P$$ and $$Q$$ is $$^{ { 2 }^{ n } }{ { C }_{ 1 } }\times _{ }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$$.Out of $$n$$ elements, $$m$$ elements can be chosen in $$^{ n }{ { C }_{ m } }$$ ways.If $$\left( P\cap Q \right)$$ contains exactly in elements, then from the remaining $$n-m$$ elements either an element belongs to $$P$$ or $$Q$$ but not both $$P$$ and $$Q$$.Suppose $$P$$ contains $$r$$ elements from the remaining $$n-m$$ elements.Then $$Q$$ may contain any number of elements from the remaining $$(n-m)-r$$ elements. Therefore, $$P$$ and $$Q$$ can be chosen in $$^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r }$$But $$r$$ can vary from $$0$$ to $$(n-m)$$. So, $$P$$ and $$Q$$ can be chosen in general in$$\displaystyle \left( \sum _{ r=0 }^{ n-m }{ ^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r } } \right) _{ }^{ n }{ { C }_{ m } }={ \left( 1+2 \right) }^{ n-m }\times _{ }^{ n }{ { C }_{ m } }=_{ }^{ n }{ { C }_{ m } }\times { 3 }^{ n-m }$$Hence, required probability $$\displaystyle =\dfrac { ^{ n }{ { C }_{ m } }\times { 3 }^{ n-m } }{ { 4 }^{ n } }$$Maths
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http://mathhelpforum.com/pre-calculus/15788-two-pumps-one-tank-print.html | # Two Pumps, One Tank
• Jun 10th 2007, 04:04 AM
blueridge
Two Pumps, One Tank
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
• Jun 10th 2007, 04:10 AM
Quick
Quote:
Originally Posted by blueridge
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
I'll start you off...
Define your variables: I'd call the bigger pump $b$ and the smaller pump $s$
$b=$the number of tanks per hour the bigger pump can empty
$s=$the number of tanks per hour the smaller pump can empty
Both of those numbers will be fractions.
Now, since you only wanted a hint, I'm only going to give you the first equation and you'll have to find the rest...
Quote:
Two pumps of different sizes working together can empty a fuel tank is 5 hours.
$b+s=\frac{1\text{tank}}{5\text{hours}}$
Do you need any more help?
• Jun 10th 2007, 04:12 AM
blueridge
Yes
If you can set up the other equation, I can take it from there.
• Jun 10th 2007, 04:22 AM
Quick
Quote:
Originally Posted by blueridge
If you can set up the other equation, I can take it from there.
Quote:
Originally Posted by blueridge
Two pumps of different sizes working together can empty a fuel tank is 5 hours. The larger pump can empty this tank in 4 hours less than the smaller one. If the larger one is out of order, how long will it take the smaller one to do the job alone?
The next equation is somewhat weird.
If $b=\frac{\text{tanks}}{\text{hours}}$
than $\frac{1}{b}=\frac{\text{hours}}{\text{tanks}}$
So in fact:
$\frac{1}{b}=$the number of hours to empty a tank
So we know that: $\frac{1}{b}-\frac{1}{s}=4$
• Jun 10th 2007, 05:54 AM
blueridge
tell me...
I am dealing with two equations in two unknowns?
• Jun 10th 2007, 06:59 AM
CaptainBlack
Quote:
Originally Posted by blueridge
I am dealing with two equations in two unknowns?
Yes, solve for $b$ and $s$, and the required answer is $s$
RonL
• Jun 10th 2007, 07:39 AM
Soroban
Hello, blueridge!
Here's another approach . . .
Quote:
Two pumps of different sizes working together can empty a fuel tank is 5 hours.
The larger pump can empty this tank in 4 hours less than the smaller one.
If the larger one is out of order, how long will it take the smaller one to do the job alone?
Together, they can do the job in 5 hours.
. . In one hour, they can do $\frac{1}{5}$ of the job. .[1]
The smaller pump can do the job in $x$ hours. .[Note that: . $x > 4$.]
. . In one hour, it can do $\frac{1}{x}$ of the job.
The larger pump takes 4 hours less; it takes $x - 4$ hours.
. . In one hour, it can do $\frac{1}{x-4}$ of the job.
Together, in one hour, they can do: . $\frac{1}{x} + \frac{1}{x-4}$ of the job. .[2]
But [1] and [2] describe the same thing:
. . the fraction of the job done in one hour.
There is our equaton! . . . . $\boxed{\frac{1}{x} + \frac{1}{x-4} \:=\:\frac{1}{5}}$
Multiply by the common denominator: $5x(x - 4)$
. . $5(x - 4) + 5x \:=\:x(x-4)$
. . which simplifies to the quadratic: . $x^2 - 14x + 20 \:=\:0$
The Quadratic Formula gives us: . $x \;=\;\frac{14\pm\sqrt{116}}{2} \;=\;7 \pm\sqrt{29} \;\approx\;\{1.6,\:12.4\}$
Since $x > 4$. the solution is: . $x = 12.4$
Therefore, the smaller pump will take about 12.4 hours working alone.
• Jun 10th 2007, 11:14 AM
blueridge
tell me...
Soroban,
I thank you for sharing yet a more simplistic avenue to understanding this question. | 2016-09-29T12:11:21 | {
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http://mathhelpforum.com/pre-calculus/71232-simple-expression-trig-matrix-cubed.html | # Math Help - Simple expression for trig matrix cubed
1. ## Simple expression for trig matrix cubed
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
2. Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
A is the standard rotation matrix by angle @. A^3 = A.A.A => rotate by 3@. Therefore A^3 = .....
Originally Posted by jennifer1004
Does anyone know how to determine a simple expression for $A^3$, where A is the following matrix?
cos@ sin@
-sin@ cos@
*I used @ as theta
I think I understand how to cube it (perform the multiplication twice?) but I have no idea how to simplify it. Thanks!
Please see attached file
4. ## Thank you!
You have helped me so much. There is nothing in my text for linear algebra that looks anything like this matrix. I wasn't sure how to simplify the sines and cosines. Thanks again. Your work is very clear!
5. Hello, jennifer1004!
We are expected to know some basic identities:
. . $\begin{array}{cccccc}\cos^2\!x-\sin^2\!x \:=\:\cos2x & & \cos x\cos y - \sin x\sin y \:=\: \cos(x+y) \\
2\sin x\cos x \:=\: \sin2x & & \sin x\cos y + \cos x\sin y \:=\:\sin(x+y) \end{array}$
Given: . $A \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$
Find $A^3$
$A^2 \:=\:\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}$ . $= \;\begin{bmatrix}\cos^2\!\theta-\sin^2\!\theta & 2\sin\theta\cos\theta \\ \text{-}2\sin\theta\cos\theta & \cos^2\!\theta -\sin^2\!\theta \end{bmatrix}$ . $= \;\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta\end{bmatrix}$
$A^3 \;=\;\begin{bmatrix}\cos\theta & \sin\theta \\ \text{-}\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta \\ \text{-}\sin2\theta & \cos2\theta \end{bmatrix}$
. . $= \;\begin{bmatrix}\cos\theta\cos2\theta - \sin\theta\sin2\theta & \sin2\theta\cos\theta + \sin\theta\cos2\theta \\ \text{-}\sin\theta\cos2\theta - \sin2\theta\cos\theta & \cos\theta\cos2\theta - \sin\theta\sin2\theta \end{bmatrix}$
. . $= \;\begin{bmatrix}\cos3\theta & \sin3\theta \\ \text{-}\sin3\theta & \cos3\theta \end{bmatrix}$
6. ## Thanks Soroban!
Those identities are what I was looking for. I wasn't sure how to simplify. Thanks for sharing! | 2016-07-27T02:30:45 | {
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https://math.stackexchange.com/questions/2535591/calculating-eigenvectors-is-my-book-wrong/2535601 | # Calculating Eigenvectors: Is my book wrong?
I have a covariance matrix: $$S= \begin{pmatrix} 16 & 10 \\ 10 & 25 \end{pmatrix}$$
I calculate my eigenvalues correctly (the same as what the book finds);
$\lambda_1 = 31.47$ , $\lambda_2 = 9.53$
But now it comes to calculating eigenvectors: I do everything as I was taught way back in Elementary Linear Algebra.
1. $S X = \lambda v$ {where v is the eigenvector}
2. $(S - I \lambda)v$
3. Get Row-Echelon Form
But when I do this I get the following reduced matrix:
$$\begin{pmatrix} 1 & -.646412 & 0 \\ 0 & 0 &0 \end{pmatrix}$$
But this result doesn't seem consistent with my textbook which says that the eigenvectors are;
$(0.54 , 0.84)^T$ and $(0.84 , -0.54)$
I looked online for calculators and found one consistent with the book and a few consistent with my result:
Consistent with Book: http://comnuan.com/cmnn01002/
Consistent with Me: http://www.arndt-bruenner.de/mathe/scripts/engl_eigenwert2.htm
Any ideas?
Additional Information:
• This problem stems from Principal Component Analysis
• $(0.54,0.84) = (9,14) \cdot 0.06,$ and $9/14 \approx 0.642857\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\left[ \begin{array}{ccc} 1 & -0.642857\ldots & 0 \\ 0 & 0 & 0 \end{array} \right],$ which would be consistent with the answer being $(0.54,0.84). \qquad$ – Michael Hardy Nov 24 '17 at 19:10
• Specifically, how did you get $-0.646412\text{ ?} \qquad$ – Michael Hardy Nov 24 '17 at 19:12
• I answered your question in the answers section. You were right that I rounded early. – Nicklovn Nov 24 '17 at 19:15
• The textbook has normalized the eigenvectors so that the change-of-basis matrix is orthogonal. Remember, there’s no such thing as the eigenvector(s): any non-zero scalar multiple of an eigenvector is also an eigenvector. – amd Nov 25 '17 at 0:08
## 4 Answers
TLDR: The answers are the same.
The vectors $(0.646586,1)$ and $(0.54,0.84)$ go in (almost) the same direction (the only differences due to rounding and the magnitude of the vector). The first has the benefit of one of the entries equalling one. The second has the benefit that its magnitude is (almost) $1$, but they both give essentially the same information.
Remember that an eigenvector for a specific eigenvalue $\lambda$ is any vector such that $Av=\lambda v$ and these vectors collectively make up an entire subspace of your vector space, referred to as the eigenspace for the eigenvector $\lambda$. In the problem of determining eigenvalues and corresponding eigenvectors, you need only find some collection of eigenvectors such that they form a basis for each corresponding eigenspace. There are infinitely many correct choices for such eigenvectors.
Eigenvector is not unique.
Notice that non-zero scalar multiple of an eigenvector is still an eigenvector.
Both answers are correct.
Both answers are correct. The eigen vector you computed does not have unity norm. If you normalize your eigen vector then you will get the text-book answer.
$(0.54,0.84) = (9,14) \cdot 0.06,$ and $9/14 \approx 0.642857\ldots,$ so I wonder if you get $0.646412$ by rounding too early, before dividing. Maybe your reduced matrix should have been $\left[ \begin{array}{ccc} 1 & -0.642857\ldots & 0 \\ 0 & 0 & 0 \end{array} \right],$ which would be consistent with the answer being $(0.54,0.84).$
• Yes I admit that I rounded early. What happened is that I calculated the eigenvalues a few days ago but I lost that paper and just went with the eigenvalues listed in the book (theirs being rounded already). But I was informed that I had only to normalize the vector I had calculated to get the textbook answer. – Nicklovn Nov 24 '17 at 19:14
• @Nicklovn : This is one respect in which elementary arithmetic gets far less respect than it deserves. People round without knowing to what extent or in what way the bottom line will be affected, and then they write six or eight or ten digits in their bottom-line answer when the early rounding has the effect that only perhaps the first one or two are truthful and the rest of meaningless noise. – Michael Hardy Nov 24 '17 at 19:22
• I agree. I was doing well with it at first but when I got frustrated and lost my sheet I just relied on what the book gave me which was naturally rounded :( Thanks for your help! – Nicklovn Nov 25 '17 at 21:30
• @Nicklovn : I'm glad it helped. – Michael Hardy Nov 25 '17 at 23:19 | 2019-07-19T06:06:42 | {
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https://math.stackexchange.com/questions/1120819/n-distinguishable-balls-into-n-boxes | # n distinguishable balls into n boxes
We have n distinguishable balls (say they have different labels or colours). If these balls are dropped at random in n boxes, what is the probability that:
1- No box is empty? 2- Exactly one box is empty?
For 1, I figured that we have $n^n$ ways to put the n balls into the n boxes. And I figured there are $n!$ to sort the balls so there is one ball for each box.
So is the answer to question 1 $n!/(n^n)$?
For 2, there are $n-1$ ways for the boxes to be empty. This is because you can have box 1 be empty (and just that), box 2 be empty and just that, so ultimately you can have at most $n-1$ variations of empty boxes.
So I figured the solution to part 2 was $\frac{n-1}{n^n}$
Is any of this right?
• I think your part two is incorrect, because for each box $n$ that is empty, there are $n-1$ boxes that could have the extra ball. Jan 26 '15 at 20:07
• That sounds right...I'll edit my answer. Jan 26 '15 at 20:12
For part 2, you must place the balls so that there is one empty box, one box with two balls, and the remaining balls will have one ball each. There are $n$ ways to pick the empty box, and $n-1$ ways to then pick the box with two balls.
We can now fill the $n$ spaces for the balls with the $n$ balls in any order you wish. There are of course $n!$ ways to do this. However, there is a caveat. If the box with two spaces is filled with ball $a$ and then ball $b$, that is the same as if we put ball $b$ and then ball $a$. So we are double counting, and the number of ways to fill in the boxes is $\frac{n!}{2}$.
Thus the total number of ways to have one empty box is $\frac{n(n-1)(n!)}{2}$, so the probability of having one empty box is $\frac{\frac{n(n-1)(n!)}{2}}{n^n} = \frac{(n-1)(n!)}{2n^{n-1}}$.
• It seems like order matters, so why would we divide by 1/2? Jan 26 '15 at 22:00
• Of the $n^n$ possibilities, all that matters is which box each ball goes into. When we consider the $n!$ ways to put $n$ boxes into the $n$ spaces, we have two spaces in one box. So by that counting, if we put ball $a$ into space 1 of that box, and ball $b$ into space 2 of that box, that's counted separately from putting ball $b$ into space 1 and ball $a$ into space 2. But in the $n^n$ counting it's just counted once, since both balls go into the same box. So we have to divide by 2 to match up the counting. Jan 27 '15 at 0:55
• See the answer I posted above from the textbook. Jan 27 '15 at 17:56
• Yes, the answer is the same: $\frac{n!}{2} = {n \choose 2} (n-2)!$. The book solution just picks the two balls in the one box first, then the remaining $n-2$ balls, whereas I picked all $n$ at once. Jan 27 '15 at 19:10
Here is the answer to the question, based on the answer manual (it was worded slightly differently in my textbook, hence why I did not find it). My part 1 was correct.
The textbook question is as follows -- I'm not sure the cells versus boxes thing is supposed to make it be any different.
>If $n$ balls are placed at random into $n$ cells, find the probability that exactly one cell remains empty.
First, we have $n$ ways of picking the empty box. Then we have $n-1$ ways of picking the box with two balls.
Now which two balls will go into the one box that will have two balls? There are ${n \choose 2}$ ways to pick the two distinct balls to go into this cell.
Lastly we need to arrange the remaining $n-2$ cells, and there are $(n-2)!$ ways to permutate all of these.
So we have:
$n(n-1){n \choose 2}(n-2)! = n!{n \choose 2}$
So the answer to part 2 is:
$\frac{n! {n \choose 2 }}{n^n}$
The argument in 1. is quite elegant. Another way to see this is that the probability to put the first ball in an empty box is $1$, for the second it is $(1 - 1/n)$, for the third it is $1 - 2/n$. The searched for probability is thus the product and this is exactly what you found in a different way that I actually like better.
For 2. you need to consider again all possibilities to place the balls. There are a lot more than you said. It is not just which box is empty (also there are $n$ ways for this) but also how the balls are distributed over the remaining ones. | 2021-09-17T04:04:29 | {
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https://mathematica.stackexchange.com/questions/194900/how-to-produce-listplot-that-samples-a-given-distribution/194912 | # How to produce ListPlot that samples a given distribution?
I wanted to get a binned plot that a uniform background distribution with a spike at some bin. My first instinct was to use RandomVariate, and then ListPlot. What i get is this
This makes sense as the RandomVariate function is assigning a value to Y based on the distribution. But I was wondering how I would produce a plot that has a spike at some particular bin (some specified x). I would want to produce something that looks more like this, for example.
I know I could just manually enter the data, but is there a way to have a binned plot where there is a uniform background distribution and a gaussian peak at some specific x?
You are on the right track. Notice that in your mixture, the uniform distribution spans [0,2], but the Normal distribution has a mean of 3. Here is another mixture. Just adjust the distribution parameters as needed:
dist = MixtureDistribution[{1, 1},
{
UniformDistribution[{0, 100}],
NormalDistribution[50, 2]
}
];
Histogram[RandomVariate[dist, 1000], {1}]
Edit: If you want data, you can use some variant of HistogramList:
{binBoundaries, counts} =
HistogramList[RandomVariate[dist, 1000], {1}];
ListPlot[Transpose[{Most[binBoundaries], counts}], PlotRange -> All,
Filling -> Axis]
• Yes, but this is a Histogram. The actual data will still have values between 0 and a 100. I wanted a way where the data itself is spread that way. – shivangi Apr 10 at 3:38
• @shivangi, I understand. I edited my answer with a possibility. – David Keith Apr 10 at 4:17
• Thank you so much! This is exactly what I was trying to do. – shivangi Apr 13 at 0:51 | 2019-11-22T07:05:38 | {
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http://tsrd.rk-verlag.de/repeated-eigenvalues.html | Repeated Eigenvalues
Proof Because x is an eigenvector of A, you know that and can write In cases for which the power method generates a good approximation of a dominant eigenvector, the Rayleigh quotient provides a correspondingly good approximation of the dominant eigenvalue. 7 Today's handouts Lecture 21 notes Tutorial 7 questions 2. Let us focus on the behavior of the solutions when (meaning the future). † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Find more Mathematics widgets in Wolfram|Alpha. Interesting eigenvectors of the Fourier transform Berthold K. ) FINDING EIGENVECTORS • Once the eigenvaluesof a matrix (A) have been found, we can find the eigenvectors by Gaussian Elimination. A symmetric minor of A is a submatrix B obtained by deleting some rows and the corresponding columns. In particular, undamped vibration is governed by. Example Determine if the following matrices are diagonalizable. To find an eigenvector corresponding to an eigenvalue , we write. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. And the lambda, the multiple that it becomes-- this is the eigenvalue associated with that eigenvector. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. Recall the basic result that the roots of a polynomial depend continuously on the coefficients of the polynomial. Since we are going to be working with systems in which $$A$$ is a $$2 \times 2$$ matrix we will make that assumption from the start. The complete case. If x= a+ ibis a complex number, then we let x = a ibdenote its conjugate. The characteristic polynomial P( ) = j I Aj. • What the Hautus Keymann Theorem says is that it is possible after preliminary state feedback using a matrix F. For very high or very low correlation in DVs, it is not suitable: if DVs are too. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. This might introduce extra solutions. We can’t find it by elimination. complex eigenvalues. Recall that given a symmetric, positive de nite matrix A we de ne R(x) = xTAx xTx: Here, the numerator and denominator are1 by 1matrices, which we interpret as numbers. Repeated eigenvalues - Duration: 7:30. It is a \repeated eigenvalue," in the sense that the characteristic polynomial (T 1)2 has 1 as a repeated root. Since x ≠ 0, this equation implies λ = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. So lambda is an eigenvalue of A. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. An eigenvector of Ais a nonzero vector v such that Av = v for some number. MATRIX EXPONENTIALS, and REPEATED EIGENVALUES Systems and Matrix Exponentials : solve x0= Ax, for n nreal A. Repeated Eigenvalues Occasionally when we have repeated eigenvalues, we are still able to nd the correct number of linearly independent eigenvectors. On this site one can calculate the Characteristic Polynomial, the Eigenvalues, and the Eigenvectors for a given matrix. † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. Putting all these bases together gives us a list of vectors: ~v 1, ~v 2, Lucky Fact 2: The geometric multiplicity of , meaning the dimension of this kernel, is equal to the number of times occurs as a root of f. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. The linearization has a center at the origin, but the. component is being recorded, and then "removed". The study of the relations between eigenvalues and structures in graphs is. This might introduce extra solutions. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. EIGENVALUES OF THE LAPLACIAN AND THEIR RELATIONSHIP TO THE CONNECTEDNESS OF A GRAPH3 (2. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. 2 λhas a single eigenvector Kassociated to it. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. To tackle the issue of non-smoothness of repeated eigenvalues, we propose an estimator constructed by averaging all repeated eigenvalues. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We solve a problem about eigenvalues of an upper triangular matrix and the square of a matrix. Differential Equations; Slope field; System of Linear DEs Real Repeated Eigenvalues #2; System of Linear DEs Imaginary Eigenvalues;. So that is the end of our lecturer on complex eigenvalues and next lecture were to talk about what to do when you have repeated real eigenvalues. Since our last example and that wraps up our lecture on repeated eigenvalues so, this is the systems of differential equations where we had repeated eigenvalues. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. But what if A has repeated real eigenvalue (a, a), or if the eigenvalues are complex conjugate pairs (a+bi, a-bi) with nonzero b? How does one decompose A=aI+bJ where J^2=-I in the latter case, or decompose A=aI+N where N^2=0 in the former?. If all eigenvalues $$\lambda$$ have negative real parts, then all solutions of approach zero exponentially. j is repeated. We restrict ourselves to the special cases of A being 2 × 2 and 3 × 3. 1 Matrix exponent Consider a first order differential equation of the form y′ = ay; a ∈ R; with the initial condition y(0) = y0: Of course, we know that the solution to this IVP is given by y(t) = eaty0: However, let us apply the method of iterations to this equation. So lambda is an eigenvalue of A. We prove that the volume of n satis es: j nj jRPN 3j = n 2 ; where N = n+1 2 is the dimension of the space of real symmetric matrices of size n n. But here only (1,0) is a eigenvector to 0. 118 CHAPTER 6. 5 Repeated Eigenvalues 95 5. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. What does this mean geometrically?. 2000 S Deep Cameo Clad Proof Massachusetts MA State Washington Quarter (B04),Girls Greek / Roman Goddess Fancy Dress Costume 4-11 Years Available,1945 S Silver Jefferson Nickel GEM BU BLAST WHITE!!. We will ignore the possibility of , as that would mean 0 is an eigenvalue. Mathematics Assignment Help, Example of repeated eigenvalues, Illustration : Solve the following IVP. Computing the pth roots of a matrix with repeated eigenvalues 2651 It can be seen that for a special case, if the given matrix A is a companion matrix then the constituent matrices corresponding to companion matrix are determined by. 1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when λ. Eigenvalues and Eigenvectors. 2 Harmonic Oscillators 114 6. hat May 14 '12 at 0:21 3 $\begingroup$ To put the same thing into slightly different words: what you have here is a two-dimensional eigenspace , and any two vectors that form a basis for that space will do as linearly. Chapter 6 Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R. j is repeated. Eigenvalues in MATLAB. 2 6 6 6 4. 2 Solving Systems with Repeated Eigenvalues If the characteristic equation has only a single repeated root, there is a single eigenvalue. , orthogonal to the disk and passing through its center), while any two orthogonal diameters in the plane of the disk may be chosen as the other two principal axes (corresponding to the repeated eigenvalue ). Repeated eigenvalues. J has the eigenvalues of A on its main diagonal, is upper triangular, and has 0's and 1's in the upper triangle. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. If there is no other eigenvector, we look for a solutions in the form x(t) = (u+ tv)e t: If x(t) has this form, then, on the one hand x_(t) = ve t+ (u+ tv)e t;. Introduction. $\endgroup$ - copper. So it is with matrices. 1 Find the eigenvalues and associated eigenspaces of each of the following matrices. eigenvalues and convert them a Pillai-Bartlett or Wilk's-Lambda value, I don?t know how to convert to an f-statistic. 1 Distinct Eigenvalues 107 6. Firstly we look at matrices where one or more of the eigenvalues is repeated. (2018) A new method for computation of eigenvector derivatives with distinct and repeated eigenvalues in structural dynamic analysis. Take for example 0 @ 3 1 2 3 1 6 2 2 2 1 A One can verify that the eigenvalues of this matrix are = 2;2; 4. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. Linear Systems of ODE with with Repeated eigenvalues James K. 4 Bases and Subspaces 89 5. If they are numeric, eigenvalues are sorted in order of decreasing absolute value. Repeated eigenvalues are explained with the help of numerical examples. Solution: If ‚ is a simple real eigenvalue, then there are two real unit eigenvectors: u and ¡u. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. To find an eigenvector corresponding to an eigenvalue , we write. and w is an eigenvector for A, and suppose that the eigenvalues are di erent. λ = a ± ib. Of particular interest in many settings (of which differential equations is one) is the following. Then from the Lemma we get 2v w =(ATv)w=v (Aw)=v 5w=5vw: But since 2 and 5 are scalars (and, for that matter, so is v w), the only way that 2v w =5v wcan be possible is for v w to be 0. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Further options. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. Stability Analysis for ODEs Marc R. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. 2 λhas a single eigenvector Kassociated to it. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. It decomposes matrix using LU and Cholesky decomposition The calculator will perform symbolic calculations whenever it is possible. View record in Web of Science ®. edu/math Craigfaulhaber. complex eigenvalues. The last case is what the solutions look like when there are repeated eigenvalues, or. To find eigenvalues of matrix A Consider {eq}\displaystyle det(A-{\lambda}(I))=0 {/eq} Then we get a characteristic polynomial in lambda. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. However, if a matrix has repeated eigenvalues, it is not similar to a diagonal matrix unless it has a full (independent) set of eigenvectors. Engineering Computation ECL4-4. sy ' Section 7. For repeated diagonal elements, it might not tell you much about the location of the eigenvalues. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. The number of positive eigenvalues equals the number of positive pivots. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. Phase portrait for repeated eigenvalues Subsection 3. Defective matrices A are similar to the Jordan (canonical) form A = XJX−1. Proof: If A is idempotent, λ is an eigenvalue and v a corresponding eigenvector then λv = Av = AAv = λAv = λ2v Since v 6= 0 we find λ−λ2 = λ(1 −λ) = 0 so either λ = 0 or λ = 1. @Star Strider: Thanks for the suggestion, I was unaware of this function. In order to get the eigenvalues of the matrix , I'll solve the characteristic equation Step 3. eigenvalue will be printed as many times as its multiplicity. In this section we will solve systems of two linear differential equations in which the eigenvalues are real repeated (double in this case) numbers. This paper discusses characteristic features and inherent difficulties pertaining to the lack of usual differentiability properties in problems of sensitivity analysis and optimum structural design with respect to multiple eigenvalues. The general solution is Y~(t) = e3t C 1 C 1 + C 2 + tC 2 Sketch the phase portrait: 2. 2 Repeated Eigenvalues. This process can be repeated until all eigenvalues are found. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. Recall that the general solution in this case has the form where is the double eigenvalue and is the associated eigenvector. S ections 2. Repeat steps 2 through 4 for each distinct eigenvalue. Proposition 0. We can’t find it by elimination. The physical significance of this degeneracy is not known. Some may be repeated, some may be complex. They are often introduced in an introductory linear algebra class, and when introduced there alone, it is hard to appreciate their importance. As shown byL¨utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. For A2 the situation is difierent. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). We shall see that this sometimes (but not always) causes problems in the diagonalization process. The Exponential of a Matrix. Find more Mathematics widgets in Wolfram|Alpha. For an matrix, the polynomial we get by computing is of degree , and hence in general, we have eigenvalues. (solution: x = 1 or x = 5. Such an x is called an eigenvector corresponding to the eigenvalue λ. The eigenvalue λ i is called repeated iff r i > 1. Once we show this is necessarily real, then the same argument as in the part (a) shows that A. In particular, undamped vibration is governed by. Boyce and Richard C. The same situation applies, if Ais semi-simple, with repeated eigenvalues. Next, perform row operations by adding each row (2 through n) to the first row (Williams): (7) For clarity, the (-1) is repeated in each of the row 1 elements, but this simply. Get the free "Eigenvalue and Eigenvector (2x2)" widget for your website, blog, Wordpress, Blogger, or iGoogle. Rajsekaran. You would perform a one-way repeated measures analysis of variance if you had one categorical independent variable and a normally distributed interval dependent variable that was repeated at least twice for each subject. University of Minnesota 109. Are the eigenvalues real? Find one complex eigenvector (contains an i). Since x ≠ 0, this equation implies λ = 1(Eigenvalue); then, from x = 1 x, every (nonzero) vector is an eigenvector of I. Since the geometric multiplicity j for j is the dimension of E j, there will be exactly j vectors in this basis. The following. Roussel September 13, 2005 1 Linear stability analysis Equilibria are not always stable. Moreover, we provide a numerically reliable and effective algorithm for computing the eigenvalue decomposition of a symmetric matrix with two numerically distinct eigenvalues. Using the Laplace transformation, Eq. The solutions of the system can be found by finding the eigenvalues and eigenvectors of the matrix. The spectral decomposition of x is returned as components of a list with components. In general, the algebraic multiplicity and geometric multiplicity of an eigenvalue can differ. (e) A= 1 1 2 3. 2 (Page 249) 17. Ryan Spring 2012 1 Repeated Eigenvalues Last Time: We studied phase portraits and systems of differential equations with complex eigen-values. However, ker(B I 2) = ker 0 2 0 0 = span( 1 0 ): Motivated by this example, de ne the geometric multiplicity of an eigenvalue. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). Make a matrix Q as follows. The reason is that the eigenvalues of a given matrix, given by the roots of the characteristic polynomial of the matrix, need not be distinct nor must they necessarily be real. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. For each distinct eigenvalue l j with multiplicity m j, let I j denote the diagonal matrix with unit elements in each location where l j occurs, and let D j denote the differentiation matrix comprised of integers counting the repeated eigenvalues, placed to the left of the diagonal locations of the respective eigenvalues. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. In that example, one principal axis, the one corresponding to eigenvalue , was (i. ' and find homework help for other Math questions at eNotes. When all eigenvalues have non-zero real parts, the equilibrium is called hyperbolic, and non-hyperbolic if at least one eigenvalue has zero real part. Recall the basic result that the roots of a polynomial depend continuously on the coefficients of the polynomial. An eigenvalue of Ais a number such that Av = v for some nonzero vector v. Precondition The eigenvalues have been computed before. However, when I run it with a non-symmetric matrix, the largest eigenvalue is in the first column. • Roy’s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. Is it possiable to create a matrix be setting the eigenvalues you wish to end up with? I am trying to create a 3X3 random matrix with a set of repeated eigenvalues and was wondering is it's possible to set the eigenvalues and then generate matrices for them. The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. Note that is the product of the eigenvalues (since ), so for the sign of determines whether the eigenvalues have the same sign or opposite sign. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Proof: If A is idempotent, λ is an eigenvalue and v a corresponding eigenvector then λv = Av = AAv = λAv = λ2v Since v 6= 0 we find λ−λ2 = λ(1 −λ) = 0 so either λ = 0 or λ = 1. Of particular interest in many settings (of which differential equations is one) is the following. Solution Step 1. Solution: Y(t) = c1 1 0 e 2t + c 2 0 1 e 2t = e 2t c1 c2 Qualitative behavior: 1. sy ' Section 7. component is being recorded, and then "removed". Proposition 0. Show that A and AT do not have the. For the purpose of analyzing Hessians, the eigenvectors are not important, but the eigenvalues are. The eigenvalue problem is to determine the solution to the equation Av = λv, where A is an n-by-n matrix, v is a column vector of length n, and λ is a scalar. Under this matrix norm, the infinite series converges for all A and for all t, and it defines the matrix exponential. Eigenvalues are a special set of scalars associated with a linear system of equations (i. The linearization has a node (proper or improper) at the origin, but the original almost linear system has either a node or a spiral point at P. We restrict ourselves to the special cases of A being 2 × 2 and 3 × 3. We do not normally divide matrices (though sometimes we can multiply by an inverse). Get the free "Eigenvalues Calculator 3x3" widget for your website, blog, Wordpress, Blogger, or iGoogle. Assume λ is a repeated eigenvalue of A with multiplicity m. Facts About Eigenvalues By Dr David Butler De nitions Suppose Ais an n nmatrix. Right when you reach $0$, the eigenvalues and eigenvectors become real (although there is only eigenvector at this point). Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. Since we are going to be working with systems in which A is a 2 x 2 matrix we will make that assumption from the start. Next I simplify it. "Det" stands for the determinant, and "I" is the identity matrix. 1 Fundamental Matrices, Matrix Exp & Repeated Eigenvalues – Sections 7. We can certainly have repeated roots and complex eigenvalues. Are the eigenvalues real? Find one complex eigenvector (contains an i). A new method is presented for computation of eigenvalue and eigenvector derivatives associated with repeated eigenvalues of the generalized nondefective eigenproblem. Multiplying them gives 2 4 4 3 3 2 3 2 1 0 2 3 5 2 4 3 2 1 3 5= 2 4 15 10 5 3 5= 5 2 4 3 2 1 3 5: This shows that the vector is an eigenvector for the eigenvalue 5. If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t). But we did not discuss the case when one of the eigenvalues is zero. 3 power method for approximating eigenvalues 551 Note that the approximations in Example 2 appear to be approaching scalar multiples of which we know from Example 1 is a dominant eigenvector of the matrix. edu (UC Davis) >4 ICIAM 11 1 / 32. Eigenvectors and Eigenvalues When a random matrix A acts as a scalar multiplier on a vector X, then that vector is called an eigenvector of X. When n = 2, in 1955 and 1956, Payne, P´olya and Weinberger proved that, in [10] and [11], λ 2 λ 1 ≤ 3forD ⊂ R2, and they conjectured λ 2 λ 1 ≤ λ 2 λ 1 | disk ≈ 2. Using eigenvalues and eigenvectors to calculate the final values when repeatedly applying a matrix First, we need to consider the conditions under which we'll have a steady state. Spectral Decomposition with Repeated Eigenvalues (Quantum Theory for Computer Age - Ch. As shown byL¨utkepohl (2005) andHamilton(1994), the VAR is stable if the modulus of each eigenvalue of A is strictly less than 1. If the eigenvalues of A are distinct, it turns out that the eigenvectors are linearly independent; but, if any of the eigenvalues are repeated, further investigation may be necessary. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Then use deflation to form a new matrix and use the power method again to extract the second eigenvalue (and root). Next one is at least one eigenvalue is repeated, can be twice or even more. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. trix has two eigenvalues of magnitude zero, one eigenvalue of unit magnitude, and three eigenvalues with magnitude less than one (right). 1 Introduction to Eigenvalues Linear equationsAx D bcomefrom steady stateproblems. To learn more about all this you should take 18. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors. Eigenvalues shows variance explained by that particular factor out of the total variance. (λ = −2 is a repeated root of the characteristic equation. 1 Eigenvalues and Eigenvectors Spectral graph theory studies how the eigenvalues of the adjacency matrix of a graph, which are purely algebraic quantities, relate to combinatorial properties of the graph. Repeated Eigenvalues Recall We are now in the position that we can find (at least) as many linearly-independent solutions to the homogeneous equation ˙ y = A y as there are distinct eigenvalues of A (real or complex). Suppose the 2 2 matrix Ahas repeated eigenvalues. Find the eigenvalues & eieenvectcrs. 0 1), whose only eigenvalue is 1. In order to determine the eigenvectors of a matrix, you must first determine the eigenvalues. Prove that if A is a square matrix then A and AT have the same characteristic polynomial. The last case is what the solutions look like when there are repeated eigenvalues, or. This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. Use the given information to determine the matrix AL Phase plane solution trajectories have horizontal tangents on the line y2-2刈and vertical tangents on the line y! The matrix Al has a nonzero repeated eigenvalue and a21 - -5. Next I simplify it. 1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when λ. Remark: I A matrix with repeated eigenvalues may or may not be diagonalizable. The set of eigenvalues of a matrix is sometimes called the of the matrix, and orthogonal diagonalispectrum zation of a matrix factors in aE E way that displays all the eigenvalues and their multiplicities. When a matrix has no repeated eigenvalues, the eigenvectors are always independent and the eigenvector matrix V diagonalizes the original matrix A if applied as a similarity transformation. vectors corresponding to eigenvalues as above (assum-ing no repeated eigenvalues), the Hessian has exactlyP q∈Q q− k 2 negative eigenvalues: we can replace any eigencomponent with eigenvalue σ with an alternate eigencomponent not already in (U,V) with eigenvalue σ0 > σ, decreasing the objective function. There are various methods by which the continuous eigenvalue problem may be. The algorithm is from the EISPACK collection of subroutines. Since it is not invertible, 0 is an eigenvalue. By looking at these eigenvalues it is possible to get information about a graph that might otherwise be di cult to obtain. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors : that is, those vectors whose direction the. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Guidance as to the number of linearly independent eigenvectors can be obtained from the rank of the matrix A. x(t)- — 4- -(14t). ) FINDING EIGENVECTORS • Once the eigenvaluesof a matrix (A) have been found, we can find the eigenvectors by Gaussian Elimination. Step 2 For each eigenvalue , compute an orthonormal basis for Ker(A Id). complex eigenvalues. Then each Axi is also in X, so each. Mathematical and Computer Modelling, 2002. sy ' Section 7. Then, we use these results to establish necessary and sufficient conditions for the. We again consider the system ~x0 = A~x. The eigenvalues are not necessarily ordered. If you would like to simplify the solution provided by our calculator, head to the unit vector calculator. I Review: The case of diagonalizable matrices. The resulting array will be of complex type, unless the imaginary part is zero in which case it will be cast to a real type. Because the rank of limn→∞P n = 1, there is a unique limiting distribution. If eigenvalues are repeated, we may or may not have all n linearly independent eigenvectors to diagonalize a square matrix. If the eigenvalue is negative, the direction is reversed. • Objective. Eigenvalues in MATLAB. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. Eigenvalues and Eigenfunctions of the Laplacian Mihai Nica University of Waterloo [email protected] I Equivalently: An n × n matrix with repeated eigenvalues may or may not have a linearly independent set of n eigenvectors. 8 Repeated Eigenvalues Shawn D. The rest are similar. Calculating eigenvalues and eigenvectors of matrices by hand can be a daunting task. These repeated matrix multiplications mean the resulting matrix is exponential in the number of layers of the neural network. Find more Mathematics widgets in Wolfram|Alpha. @Star Strider: Thanks for the suggestion, I was unaware of this function. 0 1), whose only eigenvalue is 1. Make a matrix Q as follows. Is the procedure the same? Can I use, say, variation of parameters to solve this. But here only (1,0) is a eigenvector to 0. For the following matrix, list the real eigenvalues, repeated according to their multi-plicities. 3 COMPLEX AND REPEATED EIGENVALUES 15 A. Repeated real eigenvalue 3. The coefficients for the principal components are unique (except for a change in sign) if the eigenvalues are distinct and not zero. We can now find a real-valued general solution to any homogeneous system where the matrix has distinct eigenvalues. The ‘ladder’ is bounded at both the low and high ends, which can be seen by considering the operator !. Mathematical and Computer Modelling, 2002. This might introduce extra solutions. But you can find enough independent eigenvectors -- Forget the "but. Stop at this point, and practice on an example (try Example 3, p. On the other hand, when it comes to a repeated eigenvalue, this function is not di erentiable, which hinders statistical inference, as the asymptotic theory requires at least second-order di erentiability. To be more specific, Section 2. Indeed, BAv = ABv = A( v) = Av. A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. Subsection 3. Assume λ is a repeated eigenvalue of A with multiplicity m. Is 1 1 an eigenvector. Repeated eigenvalues - Duration: 7:30. Terminology Let Abe an n nmatrix. Eigenvalues, Eigenvectors, and Di erential Equations William Cherry April 2009 (with a typo correction in November 2015) The concepts of eigenvalue and eigenvector occur throughout advanced mathematics. Repeated Eigenvalues and Symmetric Matrices 22. State feedback and Observer Feedback Poles of transfer function are eigenvalues of A Pole locations a ect system response repeated eigenvalues. However, the geometric multiplicity can never exceed the algebraic multiplicity. • Roy’s Largest Root = largest eigenvalue o Gives an upper-bound of the F-statistic. This will be one of the few times in this chapter that non-constant coefficient differential equation will be looked at. De nition If Ais a matrix with characteristic polynomial p( ), the multiplicity of a root of pis called the algebraic multiplicity of the eigenvalue. “The equation A x = λ x characterizes the eigenvalues and associated eigenvectors of any matrix A. Supplementary notes for Math 265 on complex eigenvalues, eigenvectors, and systems of di erential equations. The minimal polynomial x3 x…x—x 1–—x‡1–also splits completely over any field, and in particular over F, so Acan be diagonalized over F. Imposing an additional condition, that the eigenvalues lie in Fand are simple roots of the characteristic polynomial, does force diagonalizability. This is because u lays on the same subspace (plane) as v and w, and so does any other eigenvector. Get an answer for 'Give an example of a non-diagonalizable 4x4 matrix with eigenvalues: -1, -1, 1, 1. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. 06 or 18/700. But for a fundamental system two independent solutions are needed. Repeated Eigenvalues Note. However, the fundamental issue is selecting the appropriate tolerance to determine whether two eigenvalues are the same or not, which I don't know a priori (the elements of the matrix I am considering vary by 7 orders of magnitude, so its not obvious how close is close enough). Solution Since , the given matrix has distinct real eigenvalues of. Now is the next step. † Think of repeated eigenvalue case as a bifurcation between 2 distinct real eigenvalue case (2 straight-line solutions) and complex conjugate eigenvalue case (no straight-line solutions. eigenvalues tell the entire story. The number of eigenvalues equal to 1 is then tr(A). If is eigenvalue of perturbationA+ Eof nondefective matrixA, then j kj cond 2(X)kEk 2 where kis closest eigenvalue ofAto andX is nonsingular matrix of eigenvectors ofA Absolute condition number of eigenvalues is condition number of matrix of eigenvectors with respect to solving linear equations Eigenvalues may be sensitive if eigenvectors are. So even though a real asymmetric x may have an algebraic solution with repeated real eigenvalues, the computed solution may be of a similar matrix with complex conjugate pairs of eigenvalues. We shall see that this. The phase portrait thus has a distinct star. As any system we will want to solve in practice is an approximation to …. In the 2x2 case, if the eigenvalue is repeated you are in the defective case unless the matrix is precisely [ lambda_1 , 0 ; 0 , lambda_1 ] For larger square matrices this becomes the story of Jordan form. Eigenvectors and Hermitian Operators 7. • Objective. In fact, it is easy to see that this happen if and only if we have more than one equilibrium point (which is (0,0)). Let be the set[5] of eigenvalues, speci cally not attempting to count repeated eigenvalues more than once. polynomial corresponding to A, has n roots some of which may be repeated. 4 Bases and Subspaces 89 5. Within this blog post, we’ll investigate some classes of buckling problems and the way they are sometimes analyzed. where the eigenvalues are repeated eigenvalues. This is particularly true if some of the matrix entries involve symbolic parameters rather than speciflc numbers. This happens when the dimension of the nullspace of A−λI (called the geometric multiplicity of λ) is strictly less than the arithmetic multiplicity m. | 2019-11-14T11:03:47 | {
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http://fr.mathworks.com/help/matlab/ref/bandwidth.html?s_tid=gn_loc_drop&nocookie=true | # bandwidth
Lower and upper matrix bandwidth
## Syntax
• `B = bandwidth(A,type)` example
• ```[lower,upper] = bandwidth(A)``` example
## Description
example
````B = bandwidth(A,type)` returns the bandwidth of matrix `A` specified by `type`. Specify `type` as `'lower'` for the lower bandwidth, or `'upper'` for the upper bandwidth.```
example
``````[lower,upper] = bandwidth(A)``` returns the lower bandwidth, `lower`, and upper bandwidth, `upper`, of matrix `A`.```
## Examples
collapse all
### Find Bandwidth of Triangular Matrix
Create a 6-by-6 lower triangular matrix.
`A = tril(magic(6))`
```A = 35 0 0 0 0 0 3 32 0 0 0 0 31 9 2 0 0 0 8 28 33 17 0 0 30 5 34 12 14 0 4 36 29 13 18 11 ```
Find the lower bandwidth of `A` by specifying `type` as `'lower'`.
`B = bandwidth(A,'lower')`
```B = 5```
The result is 5 because every diagonal below the main diagonal has nonzero elements.
Find the upper bandwidth of `A` by specifying `type` as `'upper'`.
`B = bandwidth(A,'upper')`
```B = 0```
The result is 0 because there are no nonzero elements above the main diagonal.
### Find Bandwidth of Sparse Block Matrix
Create a 100-by-100 sparse block matrix.
`B = kron(speye(25),ones(4));`
View a 10-by-10 section of elements from the top left of `B`.
`full(B(1:10,1:10))`
```ans = 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1```
`B` has 4-by-4 blocks of ones centered on the main diagonal.
Find both the lower and upper bandwidths of `B` by specifying two output arguments.
`[lower,upper] = bandwidth(B)`
```lower = 3 upper = 3 ```
The lower and upper bandwidths are both `3`.
## Input Arguments
collapse all
### `A` — Input matrix2-D numeric matrix
Input matrix, specified as a 2-D numeric matrix. `A` can be either full or sparse.
Data Types: `single` | `double`
Complex Number Support: Yes
### `type` — Bandwidth type`'lower'` | `'upper'`
Bandwidth type, specified as `'lower'` or `'upper'`.
• Specify `'lower'` for the lower bandwidth (below the main diagonal).
• Specify `'upper'` for the upper bandwidth (above the main diagonal).
Data Types: `char`
## Output Arguments
collapse all
### `B` — Lower or upper bandwidthnonnegative integer scalar
Lower or upper bandwidth, returned as a nonnegative integer scalar.
• If `type` is `'lower'`, then `0` ≤ `B` ≤ `size(A,1)-1`.
• If `type` is `'upper'`, then `0` ≤ `B` ≤ `size(A,2)-1`.
### `lower` — Lower bandwidthnonnegative integer scalar
Lower bandwidth, returned as a nonnegative integer scalar. `lower` is in the range `0` ≤ `lower` ≤ `size(A,1)-1`.
### `upper` — Upper bandwidthnonnegative integer scalar
Upper bandwidth, returned as a nonnegative integer scalar. `upper` is in the range `0` ≤ `upper` ≤ `size(A,2)-1`.
collapse all
### Upper and Lower Bandwidth
The upper and lower bandwidths of a matrix are measured by finding the last diagonal (above or below the main diagonal, respectively) that contains nonzero values.
That is, for a matrix A with elements Aij:
• The upper bandwidth B1 is the smallest number such that ${A}_{ij}=0$ whenever $j-i>{B}_{1}$.
• The lower bandwidth B2 is the smallest number such that ${A}_{ij}=0$ whenever $i-j>{B}_{2}$.
Note that this measurement does not disallow intermediate diagonals in a band from being all zero, but instead focuses on the location of the last diagonal containing nonzeros. By convention, the upper and lower bandwidths of an empty matrix are both zero. | 2015-05-04T21:16:36 | {
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http://mathhelpforum.com/number-theory/151037-finding-remainder.html | Math Help - Finding the remainder
1. Finding the remainder
When the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?
How would you set this up?
2. The first step is to set up 5n/7 = 5/7 + n/7 so to strictly answer your question, except where n is a multiple of 7, then the remainder is n/7. Is this what you're asking about?
3. No. These are the choices:
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
The answer is 3. I have no freaking clue how they got that.
4. I think I understand the problem now. When n is 9, then division by 7 leaves a remainder of 2. Now 5 x 9 = 45. Try dividing 45 by 7 and see if you get a remainder of 3.
5. Suppose you divide a number N by P, and you get a quotient Q with a remainder R. This can be rewritten in fraction form:
$\frac{N}{P} = Q + \frac{R}{P}$
(remember that 0 <= R < P).
In our case, we are dividing n by 7. You get a quotient q with a remainder of 2, or:
$\frac{n}{7} = q + \frac{2}{7}$
Now I want to know what 5n is. Multiply both sides of the equation above by 35 and you'll get
$5n = 35q + 10$
Divide 5n by 7:
\begin{aligned}
\frac{5n}{7} &= \frac{35q + 10}{7} \\
&= \frac{35q}{7} + \frac{10}{7} \\
&= 5q + \frac{10}{7} \\
&= (5q + 1) + \frac{3}{7} \\
\end{aligned}
The quotient would be 5q + 1, and the remainder would be 3.
6. Thank you so much!
7. Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
8. Originally Posted by Bacterius
Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
I cannot find the MOD command on my calculator though.
9. It really is the remainder function, usually calculators don't have it. You can emulate $a \mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \mod 7$ :
$\frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $0.428571429$, and times it by $7$, you get :
$0.428571429 \times 7 = 3$
10. Originally Posted by Bacterius
Another, easier way to do it is to let $n \equiv 2 \pmod{7}$, it then follows that $5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $3$.
Originally Posted by Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $a \mod b$ by dividing $a$ by $b$, taking the fractional part only and multiplying it by $b$. For instance, to get $10 \mod 7$ :
$\frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $0.428571429$, and times it by $7$, you get :
$0.428571429 \times 7 = 3$
Oh, ok, but where did you get the 10 from? Especially the 10 in the first post here.
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
11. If $n \equiv 2 \pmod{7}$ then $\mathbf{5} \times n \equiv \mathbf{5} \times 2 \equiv 10 \equiv 3 \pmod{7}$
This is where the $10$ comes from
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
That would only work for numbers up to $13$ ... for instance if you substract $7$ from $23$, you get $16$, which isn't exactly the remainder of $23$ divided by $7$, while $23 \mod 7 = 2$ as expected. Then you could argue that repeatedly substracting $7$ is a way to do it, true, but dividing is then more efficient.
12. I totally understand now. Thank you!
13. Hello, Mariolee!
A slightly different approach . . .
When the positive integer $n$ is divided by 7, the remainder is 2.
What is the remainder when $5n$ is divided by 7?
"When $n$ is divided by 7, the remainder is 2."
. . Hence: . $n \:=\:7a + 2\:\text{ for some integer }a.$
Then: . $5n \:=\:35a + 10$
. . and: . $\dfrac{5n}{7} \:=\:\dfrac{35a+10}{7} \;=\;5a + 1 + \frac{3}{7}$
Therefore, the remainder is 3. | 2014-04-18T06:44:35 | {
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https://math.stackexchange.com/questions/2269220/will-2-linear-equations-with-2-unknowns-always-have-a-solution | # Will $2$ linear equations with $2$ unknowns always have a solution?
As I am working on a problem with 3 linear equations with 2 unknowns I discover when I use any two of the equations it seems I always find a solution ok. But when I plug it into the third equation with the same two variables , the third may or may not cause a contradiction depending if it is a solution and I am OK with that BUT I am confused on when I pick the two equations with two unknowns it seems like it has no choice but to work. Is there something about linear algebra that makes this so and are there any conditions where it won't be the case that I will find a consistent solution using only the two equations? My linear algebra is rusty and I am getting up to speed. These are just equations of lines and maybe the geometry would explain it but I am not sure how. Thank you.
• Mind showing the equations?? – The Dead Legend May 7 '17 at 0:16
Each linear equation represents a line in the plane. Most of the time two lines will intersect in one point, which is the simultaneous solution you seek. If the two lines have exactly the same slope, they may not meet so there is no solution or they may be the same line and all the points on the line are solutions. When you add a third equation into the mix, that is another line. It is unlikely to go through the point that solves the first two equations, but it might.
There are three possible cases for $2$ linear equations with $2$ unknowns (slope and intercept):
$\qquad$ $\mathbf{0}$ solution points $\qquad$ $\qquad$ $\mathbf{1}$ solution point $\qquad$ $\qquad$ $\mathbf{\infty}$ solution points
$\qquad \quad$ $\nexists$ no existence $\qquad$ $\qquad$ $\exists !$ uniqueness $\qquad$ $\qquad$ $\exists$ no uniqueness
The lines have the form $y(x) = mx + b$.
Case 1: parallel lines
A solution does not exist.
The lines are parallel: they have the same slope.
% \begin{align} % y_{1}(x) &= m x + b_{1} \\ % y_{2}(x) &= m x + b_{2} \\ % \end{align} %
Case 2: intersecting lines
We have existence and uniqueness.
The slopes are distinct.
$$m_{1} \ne m_{2}$$
% \begin{align} % y_{1}(x) &= m_{1} x + b_{1} \\ % y_{2}(x) &= m_{2} x + b_{2} \\ % \end{align} %
Case 3: coincident lines
We have existence, but not uniqueness. There is an infinite number of solutions. Every point solves the system of equations.
Both lines are the same.
% \begin{align} % y_{1}(x) &= m x + b \\ % y_{2}(x) &= m x + b \\ % \end{align} %
In terms of linear algebra, look at the problem in terms of $\color{blue}{range}$ and $\color{red}{null}$ spaces.
The linear system for two equations is % \begin{align} % m_{1} x - y &= b_{1} \\ % m_{2} x - y &= b_{1} \\ % \end{align} which has the matrix form % \begin{align} % \mathbf{A} x &= b \\ % \left[ \begin{array}{cc} m_{1} & -1 \\ m_{2} & -1 \\ \end{array} \right] % \left[ \begin{array}{cc} x \\ y \\ \end{array} \right] % &= % \left[ \begin{array}{cc} b_{1} \\ b_{2} \\ \end{array} \right] % \end{align} %
The Fundamental Theorem provides a natural framework for classifying data and solutions.
Fundamental Theorem of Linear Algebra
A matrix $\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$ induces for fundamental subspaces: \begin{align} % \mathbf{C}^{n} = \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\ % \mathbf{C}^{m} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} % \end{align}
Case 1: No existence
The matrix $\mathbf{A}$ has a rank defect $(m_{1} = m_{2})$ and $b_{1} \ne b_{2}$. $$b = \color{blue}{b_{\mathcal{R}}} + \color{red}{b_{\mathcal{N}}}$$ It is the $\color{red}{null}$ space component which precludes direct solution. (Interestingly enough, there is a least squares solution.) $$The data vector b is not a combination of the columns of \mathbf{A}. The column space is$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m \\ -1 \end{array} \right] } \, \right\} \qquad \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} = % \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} -1 \\ m \end{array} \right] } \, \right\} $$Case 2: Existence and uniqueness The matrix \mathbf{A} has full rank (m_{1}\ne m_{2}). The data vector is entirely in the \color{blue}{range} space \color{blue}{\mathcal{R} \left( \mathbf{A} \right)}$$ b = \color{blue}{b_{\mathcal{R}}} $$The \color{red}{null} space is trivial: \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)}=\mathbf{0}.$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m_{1} \\ -1 \end{array} \right] }, \, \color{blue}{ \left[ \begin{array}{c} m_{2} \\ -1 \end{array} \right] } \right\} $$Case 3: Existence, no uniqueness The matrix \mathbf{A} has a rank defect (m_{1} = m_{2} = m), yet b_{1} = b_{2}.$$ b = \color{blue}{b_{\mathcal{R}}} $$The column space is has \color{blue}{range} and \color{red}{null} space components:$$ \mathbf{C}^{2} = \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)} $$The decomposition is$$ \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = % \text{span } \left\{ \, \color{blue}{ \left[ \begin{array}{c} m \\ -1 \end{array} \right] } \, \right\} \qquad \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} = % \text{span } \left\{ \, \color{red}{ \left[ \begin{array}{r} -1 \\ m \end{array} \right] } \, \right\} $$Postscript: the theoretical foundations here are useful. The trip to understanding starts with simple examples like in @Nick's comment. Let's think using vector notation. A linear system with two unknowns x and y, and two equations$$ \begin{align*} v_1 x + w_1 y &= a_1 \\ v_2 x + w_2 y &= a_2 \end{align*} $$can be written in vector notation as$$ x\, \vec{v} + y\, \vec{w} = \vec{a}. That is, you want to know if $\vec{a}$ can be written as a linear combination of $\vec{v}$ and $\vec{w}$.
Fixed the vectors $\vec{v}$ and $\vec{w}$, to state that a solution always exists whatever $\vec{a}$ is, is the same as to state that $\vec{v}$ and $\vec{w}$ spans the whole plane. If not ($\vec{v}$ and $\vec{w}$ are parallel), depending on $\vec{a}$, the solution might not exist. And when it exists, it will not be unique. | 2020-11-28T20:29:34 | {
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http://mathhelpforum.com/math-topics/19568-physics-tourist-bear-problem.html | # Thread: physics tourist & bear problem
1. ## physics tourist & bear problem
another easy one i think:
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
2. Originally Posted by rcmango
another easy one i think:
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
The maximum value of d is such that the bear gets to the car at the same time the tourist does.
So set up a coordinate system such that the bear is at the origin and positive x is in the direction from the bear to the tourist.
Both are moving at a constant speed. The bear has to cover 27 + d meters in the same time the tourist covers d meters.
So for the tourist:
[tex]d = v_t t = 3.5t[tex]
Thus
$t = \frac{d}{3.5}$
For the bear:
$27 + d = v_b t = 6 \left ( \frac{d}{3.5} \right )$
Now solve for d.
-Dan
3. Hello, rcmango!
Another approach . . .
A tourist being chased by an angry bear is running in a straight line
toward his car at a speed of 3.5 m/s. .The car is a distance $d$ meters away.
The bear is 27 meters behind the tourist and running at 6.0 m/s.
The tourist reaches the car safely.
What is the maximum possible value for $d$?
The tourist has a 27-meter headstart.
Relative to the tourist, the bear has a speed of 2.5 m/s.
To cover 27 meters, it takes the bear: . $\frac{27}{2.5} \:=\:10.8$ seconds.
In that time, the tourist can run: . $3.5 \times 10.8 \:=\:37.8$ meters.
[I hope he left a window open so he can dive into the car.]
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
This reminds me of two chased-by-a-bear stories.
An old-timer was telling about his encounter with an angry grizzly.
"I was barely keepin' ahead of him and he wuz gainin' on me fast.
I saw a tree branch about fifteen feet up. .Just as he came up
'n took a swipe at me, I took this big leap . . ."
"Well, did you catch it?"
"Well, not on the way up . . . "
Two guys are camping when a bear headed towards them.
One man starting putting on his running shoes.
"Hey," said his friend, "you can't outrun that bear."
"I don't have to," was the reply. "I just have to outrun you."
4. Originally Posted by rcmango
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.5 m/s. The car is a distance d away. The bear is 27 m away from the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
how many meters?
thanks alot.
Hello,
if you change the text of your problem as I've done (see above) then there is a second situation possible: The car is between the bear and the tourist and both are running toward each other. They reach the car (on different sides, I hope for the tourist) after:
$t=\frac{27}{3.5+6}\approx 2.842~s$
During this time the tourist runs d = 9.947 m | 2016-09-26T18:30:21 | {
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https://math.stackexchange.com/questions/1405562/is-there-a-method-to-calculate-large-number-modulo | # Is there a method to calculate large number modulo?
Is there a (number theoretic or algebraic) trick to find a large nunber modulo some number?
Say I have the number $123456789123$ and I want to find its value modulo some other number, say, $17$.
It's not fast for me to find the prime factorisation first. It's also not fast to check how many multiples of $17$ I can "fit" into the large number.
So I was wondering if there is any method out there to do this efficiently.
I am looking for something like the other "magic trick" where you sum all the digits and take the result $\mod 9$.
• – lab bhattacharjee Aug 22 '15 at 4:24
• Don't even think about factorisation here! Division is much faster. – TonyK Aug 22 '15 at 19:20
The best I could come up with is to use 17*6 = 102. Dividing by 102 goes pretty fast...
123456789123
214
105
367
618
691
792
783
69
and 69 mod 17 = 1
You can speed things up by trying to eliminate two digits at a time
123456789123
1224
----
1056
1020
----
3678
3672
----
6912
6834
----
783
714
---
69
and 69 mod 17 = 1
## for really large numbers
For really large numbers, you can use the fact that 17 | 100,000,001
The procedure is similar to the check for divisibility by 11, except you break the number up into larger chunks.
Starting from the right, split the number up into 8-digit chunks.
So 123456789123 becomes
chunk # 1 2
chunk 56789123 1234
Compute (sum of odd numbered chunks) - (sum of even numbered chunks)
56789123 - 1234 = 56787889
If the result is negative, add a big enough multiple of 100,000,001 to
make it positive.
This number is congruent to the original number modulo 17.
56787889
5610
----
6878
6834
----
4489
4488
-----
1
and, again, we get 1
• This is a special case of the universal divisibility test (using various chunks sizes), see my answer. Both (well-known) methods were already mentioned in Lab's prior answer. – Bill Dubuque Oct 12 at 0:18
$10^2\equiv-2\pmod{17}\implies10^4=(10^2)^2\equiv(-2)^2\equiv4;$
$\displaystyle\implies\sum_{r=0}^na_r10^r\equiv(4)^0(a_3a_2a_1a_0)+(4)^1(a_7a_6a_5a_4)++(4)^2(a_{11}a_{10}a_9a_8)+\cdots\pmod{17}$
Again, $10^8\equiv(-2)^4\equiv-1$
$\displaystyle\implies\sum_{r=0}^na_r10^r\equiv(-1)^0(a_7a_6a_5\cdots a_0)+(-1)^1(a_{15}\cdots a_8)+\cdots\pmod{17}$
• Undoubtedly you know this, but in case others are wondering: for any prime $p>2$ we have $10^{(p-1)/2}\equiv\pm1\pmod p$. The sign here depends on whether $10$ is a quadratic residue modulo $p$ or not. That, in turn, can be easily determined using the law of quadratic reciprocity. This leads to a divisibility rule like the one here in chunks of $(p-1)/2$ digits. In some cases we can do shorter chunks. The best known cases of that are perhaps $p=13$ ($10^3\equiv-1$) and $p=41$ ($10^5\equiv1$). – Jyrki Lahtonen Aug 22 '15 at 19:27
• This is a special case of the universal divisibility test (using various chunks sizes), see my answer. – Bill Dubuque Oct 12 at 0:18
Well, it is fast to divide 17 into that number.
Where you can gain a lot is when the number you want to be divide is a special form such as $a^n$, where $n$ is large. There are ways (usually involving the Euler $\phi$ function) for rapidly computing $a^n \bmod{b}$ where $n$ is large.
A good start is to remember that $a^n \bmod{b} =(a\bmod{b})^n \bmod{b}$.
Yes, use the universal divisibility test: repeatedly replace leading digit chunks by their remainder mod the divisor as below, using least magnitude remainders $$\, -8\le r \le 9\,$$ to simplify arithmetic $$\!\bmod 17\$$ (so negative digits occur, denoted $$-d,c := 10(-d)+c\equiv 7d+c,\,$$ by $$10\equiv -7)$$ \begin{align} \!\bmod 17\!:\,\ 10\equiv -7\ \Rightarrow\quad\ \ &\ \color{#0A0}{1\ 2}\,\ 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0a0}{1\:\!2\equiv -5}\\[.1em] \equiv\, &\ \color{#0A0}{{-5}},\color{#c00} 3\ 4\ 5\ 6\ \ \ {\rm by}\,\ \color{#0a0}{{-}5},\color{#c00}3\equiv\ (\,7\,)\, \color{#0a0}{5}+\color{#c00}3\,\equiv\,\color{#0af} 4\\[.1em] \equiv\, &\ \ \ \ \ \ \ \ \color{#0af}4\ 4\ 5\ 6\ \ \ {\rm by}\,\ \ \ \ \ \ \color{#0af}4\:\!4\equiv (-7)4+4\,\equiv\color{#f60}{-7}\\[.1em] \equiv\, &\ \ \ \ \ \ \ \ \color{#f60}{{-}\!7}, 5\ 6\ \ \ {\rm by}\,\ \color{#f60}{{-}7},5\equiv\,(\,7\,)\, \color{#f60}{7}+5\,\equiv\, 3\\[.2em] \equiv\, &\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ 6\ \ \ \:\!\:\!{\rm by}\ \ \ \ \ \ 3\:\! 6\equiv (-7)3+6\,\equiv\, 2\\ \equiv\, &\qquad\qquad\ 2;\ \ {\it quicker,}\,\ 2 \,\text{ digits at a time:}\\[.4em] \!\bmod 17\!:\,\ 10^2\equiv -2\ \Rightarrow\quad\ \ &\ \color{#0A0}{12\ 3 4}\ 56\ \ \ {\rm by}\,\ \color{#0A0}{12\ 3 4}\equiv\, (-2)\color{#0a0}{\,12+34\equiv 10}\\[.1em] \equiv\, &\ \ \ \ \ \ \color{#0A0}{{10}}\ \color{#c00}{56}\ \ \ {\rm by}\,\ \color{#0a0}{10}\ \color{#c00}{56}\equiv\ (-2)\, \color{#0a0}{10}+\color{#c00}{56}\,\equiv\,\color{#0af}2\\[.2em] \equiv\, &\qquad\quad\ \color{#0af}2 \end{align}\qquad\qquad
So $$\rm\, 123456\equiv 2\pmod{\!17}.\,$$ Indeed $$\rm\, 123456 = 7262\cdot 17+2.\,$$ Continuing this way we can do the entire number in a couple minutes of mental arithmetic. Unlike some other divisibility tests that compute only a binary truth value, this method has the advantage of computing the remainder. Further, it doesn't require remembering any special algorithm or parameters for each modulus.
Remark Lab & Steven's answers are a special case of above (but without mod arithmetic optimizations), i.e. they use chunk sizes of $$\,2\,$$ and $$\,8,\,$$ using $$\bmod 17\!:\ 10^2\equiv -2,\ 10^8\equiv -1$$.
• See here for another example using negative digits. – Bill Dubuque Oct 29 '18 at 23:10
No, there is not. The reason why "magic tricks" work when studying divisibility by $2,3,5,11$ is the fact that we usually write in base $10$. Change the base and they will stop working. In particular, the following trick would work in base $17$: if the last digit of your number is $0$ then the number is divisible by $17$. Of course, in order to check this you would need to write it in base $17$, which is a vicious circle...
• Great, where can I read about the tricks you mention for $2,3,5,11$? (you don't mention $9$...?) – learner Aug 23 '15 at 4:29
• @learner: I don't know of any book, I know them from elementary school: for $2$, the last digit must be even; for $3$ the sum of the digits must be divisible by $3$; for $4$, the number formed by the last 2 digits must be divisible by $4$; for $5$, the last digit must be $0$ or $5$; for $8$, the number formed by the last 3 digits must be divisible by $8$; for $10$, the last digit must be $0$; for $11$, sum all the digits on odd positions, sum all the digits on even positions and check whether the difference of these 2 numbers is divisible by $11$. – Alex M. Aug 28 '15 at 10:59 | 2020-11-28T14:40:08 | {
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http://mathhelpforum.com/calculus/60919-related-rates-cont-d.html | # Math Help - Related Rates cont'd...
1. ## Related Rates cont'd...
I have some more questions that I need help setting up. Please feel free to switch out the numbers for variables so I can learn how to solve these. Thanks in advance for your help!
Two towers (A and B) are 150m tall, 40m apart. A horizontal tightrope (height of 60m) is attached to the two buildings. A man walks across the tightrope from A to B at a constant 1/3 meter per sec. A spotlight is on the top of A. How fast is the shadow of the man moving up the wall of building B when he is 8 m away from B.
(^^No clue where to begin^^)
A trough, 12 feet long, has as its ends isosceles trapezoids with altitude 4 ft., lower base 3 ft., and upper base 4 ft. If water is let in at a rate of 7 cubic feet per minute, how fast is the water level rising when the water is 10 in. deep?
(^^I found the volume of the trough to be 168 cubic feet ft...and I think the water level uses a ratio towards the trough...what next?^^)
Stacey is brewing coffee that is being strained through a conical filter with a height of 12 in. diameter 8 in. The coffee flows from the filter into a cylindrical coffee pot with base area equal to 100pi square in. The depth, h, in inches, of the coffee in the conical filter is changing at the rate of (h-12) in. per min. How fast is the depth of the coffee in the cylindrical coffee pot changing when h=3in?
(^^No clue where to begin^^)
Two streets lights, each 30 feet tall, are 100 ft. apart. The light at the top of one of the poles is functioning properly, but the other light is burnt out. A repairman is climbing up the pole to fix the light at a rate of .5 foot per second. How fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?
(^^I have a drawing set up, but I do not know how to label it and how to set up my equations^^)
2. Hello, nivek516!
Here's the first one . . .
Two towers, $A\text{ and }B$. are 150m tall, 40m apart.
A horizontal tightrope (height of 60m) is attached to the two buildings.
A man walks across the tightrope at a constant $\tfrac{1}{3}$ m/sec.
A spotlight is on the top of $A$. How fast is the shadow of the man
moving up the wall of building $B$ when he is 8 m away from $B]$?
First, make a sketch . . .
Code:
A * * B
| * |
90 | * |
| * 40-x |
T + - - - - - * - - - - - + R
| x M * |
| * | y
60 | * |
| * S
| |
| |
C * - - - - - - - - - - - * D
40
The buildings are $AC\text{ and }BD\!:\;\;AC = BD = 150,\;CD = 40$
The tightrope is $TR\!:\;\;TR = 40,\;TC = 60.\;AT = 90$
The spotlight is at $A$, the man is at $M.$
. . Let: $TM \:=\: x \quad\Rightarrow\quad MR \:=\:40-x$
. . We are given: . $\tfrac{dx}{dt} = \tfrac{1}{3}$ m/sec.
The man's shadow is at $S\!:\;\;\text{let } RS = y.$
Since $\Delta ATM \sim \Delta SRM\!:\;\;\frac{x}{90} \:=\:\frac{40-x}{y} \quad\Rightarrow\quad y \:=\:90\,\frac{40-x}{x} \:=\:90\left(40x^{-1} - 1\right)$
Differentiate with respect to time: . $\frac{dy}{dt} \:=\:-\frac{3600}{x^2}\cdot\frac{dx}{dt}$
When $MR = 8\;(x = 32)\!:\;\;\frac{dy}{dt} \;=\;-\frac{3600}{32^2}\left(\frac{1}{3}\right) \;=\;-\frac{75}{64}$
The shadow is moving up building $B$ at $1\tfrac{11}{64}\text{ m/sec}$
3. Thanks! Could you explain why the answer is negative? Does it matter since it is speed? Any possible chance you could help me with the last two questions? I found out how to do the second one.
4. Hello again, nivek516!
Here's the last one . . .
Two streets lights, each 30 feet tall, are 100 feet apart. The light at the top
of one of the poles is functioning properly, but the other light is burnt out.
A repairman is climbing up the pole to fix the light at a rate of 0.5 ft/sec.
How fast is the tip of the repairman's shadow moving when he is 16 ft. up the pole?
Code:
A * * C
| * |
| * |
| * |
30 | * R
| | *
| y| *
| | *
B *---------------*---------------*
: - - 100 - - - D - - - x - - - S
The two poles are: . $AB \,=\,CD\,=\,30$
. . and: . $BD = 100$
The light at $A$ shines on the repairman $R$ and casts his shadow at $S.$
. . Let: . $x \,=\,DS,\;y \,=\,RD$
Since $\Delta RDS \sim\Delta ABS\!:\;\;\frac{x}{y} \:=\:\frac{x+100}{30} \quad\Rightarrow\quad x \:=\:\frac{100y}{30-y}$
Differentiate with respect to time: . $\frac{dx}{dt} \;=\;\frac{3000}{(30-y)^2}\,\frac{dy}{dt}$
Now substitute: . $y = 16,\;\;\frac{dy}{dt} = 0.5$ . . . | 2016-02-11T13:12:41 | {
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https://math.stackexchange.com/questions/1299822/prove-that-sinh-coshx-geq-cosh-sinhx | # Prove that $\sinh(\cosh(x)) \geq \cosh(\sinh(x))$
Prove that
$$\sinh(\cosh(x)) \geq \cosh(\sinh(x))$$
I tried to tackle this problem by integrating both lhs and rhs, in order to get two functions who show clearly that inequality holds. I've struggled for this problem a little bit, i don't know if there's any trick that can help. Maybe knowing that
$$\cosh^{-1}(x) = \pm \ln\left(x + \sqrt{x^2 - 1}\right)$$
Can help?
• Jensen is useful only when we have a function and a constant, not with composite functions... Even if I think there are some similarities... – james42 May 26 '15 at 18:11
• cheers for teaching me something new :). – Chinny84 May 26 '15 at 18:12
• Physics guy here. I don't have any idea how to do math proofs, but isn't it true that a "small function" acting on a "big function" is always larger than the alternative? Since cosh(x)>sinh(x) for all x, this is a no brainer, right? This is how I remember e^pi>pi^e. Anyway, I've never spoken on Math.SE and generally run in fear from mathematicians, so please, let me down gently! – user1717828 May 27 '15 at 4:03
• @user1717828 $\ln x$ is smaller than $x^2$, but $\ln (x^2) = 2\ln x$ is smaller than $(\ln x)^2$. – user21467 May 27 '15 at 4:21
• @StevenTaschuk, aaaaannnnnddd this is why our field would crumble without mathematicians keeping us in check. Thanks! – user1717828 May 27 '15 at 4:43
For any $y \ge 0$, notice
$$e^y - 1 = \int_0^y e^x dx \ge \int_0^y (1+x) dx \ge \int_0^y \left(1+\frac{x} {\sqrt{1+x^2}}\right)dx = y + \sqrt{1+y^2} - 1$$ we have this little inequality: $$\sqrt{1+y^2} - y = \frac{1}{\sqrt{1+y^2} + y} \ge e^{-y}$$
Using MVT, we can find a $\xi \in (y,\sqrt{1+y^2})$ such that
$$\sinh\sqrt{1+y^2} - \sinh(y) = \cosh(\xi)\left(\sqrt{1+y^2} - y\right) \ge \cosh(\xi) e^{-y} \ge e^{-y}$$
Since $e^{-y} = \cosh(y) - \sinh(y)$, this leads to $$\sinh\sqrt{1+y^2} \ge \cosh(y)\\$$
Substitute $y$ by $\sinh(x)$ and notice $\sqrt{1+y^2} = \cosh(x)$, this reduces to our desired inequality:
$$\sinh(\cosh(x)) \ge \cosh(\sinh(x))$$
Here's a solution with very little calculus. First, an identity: \begin{align*} \sinh^2(a+b) - \sinh^2(a-b) &= (\sinh a\cosh b + \cosh a\sinh b)^2 - (\sinh a\cosh b - \cosh a\sinh b)^2 \\ &= 4\sinh a \cosh a\sinh b \cosh b \\ &= \sinh(2a)\sinh(2b) \end{align*} Taking $a=e^x/2$ and $b=e^{-x}/2$, we get \begin{align*} \sinh^2(\cosh x) - \sinh^2(\sinh x) &= \sinh(e^x) \sinh(e^{-x}) \\ &\ge e^xe^{-x} &&\text{(since $\sinh t\ge t$ for $t\ge 0$)} \\ &= 1 \\ &= \cosh^2(\sinh x) - \sinh^2(\sinh x) \end{align*} Cancelling $\sinh^2(\sinh x)$ and taking square roots gives the desired inequality.
Calculus is needed here only to justify the inequality $\sinh t\ge t$ (for $t\ge 0$).
Update: Another nice thing about this method is that it points the way to a more exact inequality. It turns out that $$\sinh u\sinh v \ge \sinh^2\sqrt{uv}$$ for $u,v\ge 0$. (Proof 1: $\frac12(u^{2m+1}v^{2n+1} + v^{2m+1}u^{2n+1})\ge (uv)^{m+n+1}$ by AM/GM; divide by $(2m+1)!\,(2n+1)!$ and apply $\sum_{m=0}^\infty \sum_{n=0}^\infty$. Proof 2: Check that $t\mapsto\ln\sinh(e^t)$ is convex (for all $t$) by computing its second derivative.)
Applying this with $u=e^x$ and $v=e^{-x}$ above, we get $$\sinh^2(\cosh x) \ge \cosh^2(\sinh x) + \underbrace{\sinh^2(1) - 1}_{\approx 0.3811}$$ with equality when $x=0$.
• (+1) Very nice. Since $\cosh(x)\gt0$, we don't need to worry about signs when taking square roots. – robjohn May 27 '15 at 4:22
Let $t=\sinh x$. Now we can square the inequality and instead try proving $$\sinh^2(\sqrt{1+t^2})=\sinh^2(\cosh x) \ge \cosh^2(\sinh x)=1+\sinh^2t$$
So it is enough to show $f(t) = \sinh^2(\sqrt{1+t^2})-\sinh^2t-1 \ge 0$. As $f$ is even and $f(0)> 0$, it is enough to show it is increasing for positive $t$. Hence we look at $$f'(t) = \frac{t\sinh (2\sqrt{1+t^2})}{\sqrt{1+t^2}} - \sinh(2t)$$ To show this is positive, it suffices to note by differentiating that the function $g(t) = \dfrac{\sinh t}{t}$ is increasing, so $g(2\sqrt{1+t^2})> g(2t)$. Hence proved...
For $x\ge0$, the Mean Value Theorem says that for some $\sinh(x)\lt\xi\lt\cosh(x)$, \begin{align} \sinh(\cosh(x))-\sinh(\sinh(x)) &=\cosh(\xi)(\cosh(x)-\sinh(x))\\ &\gt\cosh(\sinh(x))\,e^{-x}\tag{1} \end{align} Furthermore, $$\cosh(\sinh(x))-\sinh(\sinh(x))=e^{-\sinh(x)}\tag{2}$$ Therefore, subtracting $(2)$ from $(1)$, then applying $\cosh(x)\ge1$ and $\sinh(x)\ge x$, we get \begin{align} \sinh(\cosh(x))-\cosh(\sinh(x)) &\gt e^{-x}\cosh(\sinh(x))-e^{-\sinh(x)}\\ &\ge e^{-x}-e^{-\sinh(x)}\\ &\ge0\tag{3} \end{align} Since $\sinh(\cosh(x))-\cosh(\sinh(x))$ is even, $(3)$ implies that strict inequality holds for all $x$: $$\sinh(\cosh(x))\gt\cosh(\sinh(x))\tag{4}$$ | 2019-07-19T00:30:37 | {
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https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct | # Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
• @Steve: See this answer for general comments on induction, and this one for specific advice on doing proofs by induction. The example there may be enough for you to figure out how to prove this statement by induction. – Arturo Magidin Sep 6 '11 at 3:34
• Since this question is asked quite frequently, it has been added to the list of Generalizations of Common questions. It has been kept seperate from the version which does not use induction. – Eric Naslund Aug 30 '12 at 0:23
• This answer has a great discussion of how to do an induction proof – Ross Millikan Oct 30 '17 at 21:09
• take paper and pencil and start with $\sum_{i=0}^{n+1 } i^3$. (and you need not bother with $i=0$) – Mirko Apr 19 '18 at 21:13
• I would say this must have been asked on this website at least $10$ times. – Clement C. Apr 19 '18 at 21:14
You are trying to prove something of the form, $$A=B.$$ Well, both $A$ and $B$ depend on $n$, so I should write, $$A(n)=B(n).$$ First step is to verify that $$A(1)=B(1).$$ Can you do that? OK, then you want to deduce $$A(n+1)=B(n+1)$$ from $A(n)=B(n)$, so write out $A(n+1)=B(n+1)$. Now you're trying to get there from $A(n)=B(n)$, so what do you have to do to $A(n)$ to turn it into $A(n+1)$, that is (in this case) what do you have to add to $A(n)$ to get $A(n+1)$? OK, well, you can add anything you like to one side of an equation, so long as you add the same thing to the other side of the equation. So now on the right side of the equation, you have $B(n)+{\rm something}$, and what you want to have on the right side is $B(n+1)$. Can you show that $B(n)+{\rm something}$ is $B(n+1)$?
Let the induction hypothesis be $$(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$ Now consider: $$(1+2+3+\cdots+n + (n+1))^2$$ \begin{align} & = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\ & = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^2 + n(n+1)^2}\\ & = (1^3+2^3+3^3+\cdots+n^3) + \color{teal}{(n+1)^3} \end {align} QED
• @RajeshKSingh let me know if you have questions about the steps of the proof. – user2468 Aug 29 '12 at 23:08
• Hint: If $(1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$, then \begin{eqnarray} (1^3+2^3+3^3+\cdots+n^3 + (n+1)^3)&=&(1+2+3+\cdots+n)^2 +(n+1)^3\\ &=&(\dfrac{n^2(n+1)^2}{2} + (n+1)^3\\ &=& (n+1)^2(\dfrac{n^2}{2}+n+1)\\ &=& (n+1)^2((n+1)^2+1)/2\\ &=& (1+2+ \cdots +n +(n+1))^2 \end{eqnarray} – user29999 Aug 29 '12 at 23:14
• @RajeshKSingh No. If you want to prove a statement about all natural numbers then you need induction. – user2468 Aug 29 '12 at 23:16
• @JenniferDylan: the steps are clear to me. is there a method different from induction? – HOLYBIBLETHE Aug 29 '12 at 23:25
• @user29999 You didn't square the denominator. D: – Simply Beautiful Art Dec 9 '16 at 21:44
HINT $\:$ First trivially inductively prove the Fundamental Theorem of Difference Calculus
$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ F(n) - F(n-1)\ =\ f(n),\quad\ F(0) = 0$$
The result now follows immediately by $\rm\ F(n)\ =\ (n\:(n+1)/2)^2\ \Rightarrow\ F(n)-F(n-1)\ =\: n^3\:.\$
Note that by employing the Fundamental Theorem we have reduced the proof to the trivial mechanical verification of a polynomial equation. No ingenuity is required.
Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my many posts on telescopy.
• I got here from your latest post, how do you find $F(n)$ ? do you have a method or do you just 'see' this ? – Belgi Sep 19 '12 at 10:36
• @Belgi Here the sum closed form $\rm F(n)$ is given and we are merely asked to verify it correctness. Computing the closed form is known as summation in finite terms and there are known algorithms for wide-classes of special functions, e.g. chase citations to Michael Karr's papers on "Summation in finite terms". – Bill Dubuque Sep 4 '17 at 23:32
Let P(n) be the given statement. You'll see why in the following step. $$P(n):1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$
Step 1. Let $n = 1$.
Then $\mathrm{LHS} = 1^3 = 1$, $\mathrm{RHS} = \frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1$.
So LHS = RHS, and this means P(1) is true!
Step 2. Let $P(n)$ be true for $n = k$; that is, $$1^3 + 2^3 + 3^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}$$
We shall show that $P(k+1)$ is true too!
Add $(k+1)^3$, which is $(k+1)^{\mathrm th}$ term of the LHS to both sides of (1); then we get: \begin{align*} 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\ &= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\ &= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\ &= \frac{(k+1)^2(k+2)^2}{4}.\\ 1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{(k+1)^2(k+2)^2}{4} \end{align*} I think this statement is the same as $P(n)$ with $n = k+1$.
• Is there any particular reason why you are writing in bold face? It's a bit distracting, and not the usual for this site... Note also that it doesn't work for math formulas, so that makes it even more distracting... – Arturo Magidin Sep 6 '11 at 4:14
• I wanted the math formulas to appear in bold face,but the opposite is happening.don't worry i'll edit it. – alok Sep 6 '11 at 4:17
• You marked not just the equations, also all the text. And why was it important for the formulas to be in boldface? Again, not the usual for this site, and they don't look the same: $k$ vs. $\mathbf{k}$, $+$ vs. $\mathbf{+}$, $\cdots$ vs. $\mathbf{\cdots}$. I've cleaned it up a bit. – Arturo Magidin Sep 6 '11 at 4:21
This picture shows that $$1^2=1^3\\(1+2)^2=1^3+2^3\\(1+2+3)^2=1^3+2^3+3^3\\(1+2+3+4)^2=1^3+2^3+3^3+4^3\\$$ this is handmade of mine
For proof by induction; these are the $\color{green}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $i=1 \implies \sum^{i=k}_{i=1} i^3=\frac{1^2 (1+1)^2}{4}=\cfrac{2^2}{4}=1$. So statement holds for $i=1$.
Step 2: Inductive Assumption: Assume statement is true for $i=k$: $$\sum^{i=k}_{i=1} i^3=\frac{k^2 (k+1)^2}{4}$$
Step 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^3=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$ To do this you cannot use: $$\sum^{i=n}_{i=1} i^3=\color{red}{\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.
So what you do instead is notice that: $$\sum^{i=k+1}_{i=1} i^3= \underbrace{\frac{k^2 (k+1)^2}{4}}_{\text{sum of k terms}} + \underbrace{(k+1)^3}_{\text{(k+1)th term}}$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{1}{4}k^2+(k+1)\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{k^2+4k+4}{4}\right)$$ $$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{(k+2)^2}{4}\right)=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
(QED is an abbreviation of the Latin words "Quod Erat Demonstrandum" which loosely translated means "that which was to be demonstrated". It is usually placed at the end of a mathematical proof to indicate that the proof is complete.)
If $S_r= 1^r+2^r+...+n^r=f(n)$ and
$\sigma =n(n+1)$ & $\sigma'=2n+1$
$S_1=\frac{1}{2}\sigma$
$S_2=\frac{1}{6}\sigma\sigma'$
$S_3=\frac{1}{4}\sigma^2$
$S_4=\frac{1}{30}\sigma\sigma'(3\sigma-1)$
$S_5=\frac{1}{12}\sigma^2(2\sigma-1)$
$S_6=\frac{1}{42}\sigma\sigma'(3\sigma^2-3\sigma+1)$
$S_7=\frac{1}{24}\sigma^2(3\sigma^2-4\sigma+2)$
$S_8=\frac{1}{90}\sigma\sigma'(5\sigma^3-10\sigma^2+9\sigma-3)$
$S_9=\frac{1}{20}\sigma^2(2\sigma^3-5\sigma^2+6\sigma-3)$
$S_{10}=\frac{1}{66}\sigma\sigma'(3\sigma^4-10\sigma^3+17\sigma^2-15\sigma+5)$
proof of this is based on the theorem if r is a positive integer, $s_r$ can be expressed as a polynomial in $n$ of which the highest term in $\frac{n^{r+1}}{r+1}$
• if you are interested we can discuss about the proof.! – user229886 Apr 9 '15 at 6:58
• you can also see page 84-85-,86 of proofs without word by roger B. Nelson ,ISBN 088385-700-6 – Khosrotash Jul 21 '15 at 6:51
It may be helpful to recognize that both the RHS and LHS represent the sum of the entries in a the multiplication tables. The LHS represents the summing of Ls (I'll outline those shortly), and the RHS, the summing of the sum of the rows [or columns])$$\begin{array}{lll} \color{blue}\times&\color{blue}1&\color{blue}2\\ \color{blue}1&\color{green}1&\color{red}2\\ \color{blue}2&\color{red}2&\color{red}4\\ \end{array}$$ Lets begin by building our multiplication tables with a single entry, $1\times1=1=1^2=1^3$. Next, we add the $2$s, which is represented by the red L [$2+4+2 = 2(1+2+1)=2\cdot2^2=2^3$]. So the LHS (green 1 + red L) currently is $1^3+2^3$, and the RHS is $(1+2)+(2+4)=(1+2)+2(1+2)=(1+2)(1+2)=(1+2)^2$. $$\begin{array}{llll} \color{blue}\times&\color{blue}1&\color{blue}2&\color{blue}3\\ \color{blue}1&\color{green}1&\color{red}2&\color{maroon}3\\ \color{blue}2&\color{red}2&\color{red}4&\color{maroon}6\\ \color{blue}3&\color{maroon}3&\color{maroon}6&\color{maroon}9\\ \end{array}$$ Next, lets add the $3$s L. $3+6+9+6+3=3(1+2+3+2+1)=3\cdot3^2=3^3$. So now the LHS (green 1 + red L + maroon L) currently is $1^3+2^3+3^3$, and the RHS is $(1+2+3)+(2+4+6)+(3+6+9)=(1+2+3)+2(1+2+3)+3(1+2+3)=(1+2+3)(1+2+3)=(1+2+3)^2$.
By now, we should see a pattern emerging that will give us direction in proving the title statement.
Next we need to prove inductively that $\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and use that relationship to show that $1+2+3+\dots+n+\dots+3+2+1 = \dfrac{n(n+1)}{2}+ \dfrac{(n-1)n}{2} = \dfrac{n((n+1)+(n-1))}{2}=\dfrac{2n^2}{2}=n^2$
Finally, it should be straight forward to show that: $$\begin{array}{lll} (\sum^n_{i=1}i+(n+1))^2 &=& (\sum^n_{i=1}i)^2 + 2\cdot(\sum^n_{i=1}i)(n+1)+(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)(\sum^n_{i=1}i + (n+1) + \sum^n_{i=1}i)\\ &=& \sum^n_{i=1}i^3 + (n+1)(n+1)^2\\ &=& \sum^n_{i=1}i^3 + (n+1)^3\\ &=& \sum^{n+1}_{i=1}i^3\\ \end{array}$$ and, as was already pointed out previously, $$(\sum_{i=1}^1 i)^2 = \sum_{i=1}^1 i^3=1$$
$$n^3=S_n-S_{n-1}=\left(\frac{n(n+1)}2\right)^2-\left(\frac{n(n-1)}2\right)^2=n^2\left(\frac{n+1}2-\frac{n-1}2\right)\left(\frac{n+1}2+\frac{n-1}2\right)\\ =n^2\cdot1\cdot n.$$
This is the inductive step. The rest is easy.
[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
HINT
We need to prove
• Base case: set $i=1$ and check by inspection
• Induction step: assume $\sum_{i=0}^n i^3 = (\sum_{i=0}^n i)^2$ true prove
$$\sum_{i=0}^{n+1} i^3 = \left(\sum_{i=0}^{n+1} i\right)^2$$
Note that
• $\sum_{i=0}^{n+1} i^3=(n+1)^3+\sum_{i=0}^{n} i^3$
• $\left(\sum_{i=0}^{n+1} i\right)^2=\left(n+1+\sum_{i=0}^{n} i\right)^2$
• It helps to recall that $1+2+ \ldots +n = n(n+1)/2.$ – Chris Leary Apr 19 '18 at 21:20
• @ChrisLeary Oh yes, I supposed that should be well known, anyway that's good to recall! Thanks – gimusi Apr 19 '18 at 21:23
• Sorry, that comment was for the OP, not you. I should have addressed it directly to Ryan. – Chris Leary Apr 20 '18 at 2:10
[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
You should prove before $$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$
Cases $$n=0,1$$ are trivial. Suppose it's true for $$n-1$$ then \begin{align} \sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\ &=n^3 + \left(\sum_{i=0}^{n-1} i\right)^2 \\ &= n^3 + \frac{n^2(n-1)^2}{4}\\ & = \frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\ &= \frac{(n+1)^2n^2}{4} \\ &= \left(\sum_{i=0}^{n} i\right)^2 \end{align}
$$\ S =\sum_{k=1}^n k^3$$
\begin{align} k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\ k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D \end{align} Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and
\begin{align} S =& \sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\ S =&\sum_{k=1}^n k(k+1)(k+2)-3\sum_{k=1}^nk(k+1)+\sum_{k=1}^nk \\ S =&\sum_{k=1}^n6\binom{k+2}{3}-3\sum_{k=1}^n2\binom{k+1}{2}+\sum_{k=1}^n\binom{k}{1}\\ S=&6\binom{n+3}{4}-6\binom{n+2}{3}+\binom{n+1}{2}\\ S=&\left\lgroup\frac{n(n+1)}{2}\right\rgroup^2. \end{align}
• It seems that induction has been asked for. – Claude Leibovici Jul 9 '16 at 2:57
• The question clearly asks for a proof using P.M.I. – Shubh Khandelwal Aug 24 '18 at 4:18
## protected by Alex M.Aug 29 '16 at 15:16
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead? | 2019-04-19T05:14:30 | {
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https://math.stackexchange.com/questions/4109544/does-knowing-that-you-drew-at-least-1-ace-increase-the-probability-that-you-dr | Does knowing that you drew at least $1$ ace increase the probability that you drew at least $2$ aces ($5$ card hand)?
So say I draw $$5$$ cards from a standard deck at random without replacement and without looking at them. If someone told me that the hand contains at least $$1$$ ace, would that change the probability that my hand contains at least $$2$$ aces?
I did the math and I believe the probability of drawing at least $$2$$ aces is: $$1 - \dfrac{\binom{48}{5}}{\binom{52}{5}} - \dfrac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}$$.
Additionally, what if that person then told me that the hand contains the ace of spades; would knowing that there is at least one ace and that it's the ace of spades change the probability of having at least $$2$$ aces in my hand?
• I disagree with the assertion for the second part that knowing that its is an ace of spades does not change the situation. I have put up an answer that explains this. Apr 21 at 4:04
• Have a look at the added approach. Apr 21 at 7:18
It is not correct to say that whether you are told that an ace is in the hand or whether you are told that the ace of spade is in the hand, the probability of having at least two aces is the same
Case 1: You are told that you have at least one ace
Note that in the following formula, the sample space shrinks to $$\binom{52}5 - \binom{48}5$$ to take into account that at least one ace must be present, and that in the numerator we need to ensure that one ace is there
$$Pr = 1 - \dfrac{\binom41\binom{48}4}{\binom{52}5 - \binom{48}5},\;\approx 0.1222$$
Case 2: You are told that you have the ace of spades
Sinca ace of spades is assured, the sample space is now simply $$\binom{51}{4}$$, and from the numerator, we just subtract $$\binom{48}4$$ to indicate that more aces must be present.
Thus $$Pr = 1 - \dfrac{\binom{48}4}{\binom{51}4}\; \approx 0.2214$$
Since probability problems are often "tricky", here is an approach simpler to understand, though longer computationally.
Basically, the numerator will count hands with $$2$$ or more aces, and the denominator, hands with $$1$$ or more aces
Case 1: You are told that you have at least one ace
$$Pr = \dfrac{\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}{\binom41\binom{48}4+\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}\approx 0.1222$$
Case 2: You are told that you have the ace of spades
Bear in mind that you have the ace of spades and are left to pick from $$51$$ cards including $$3$$ aces
$$Pr = \dfrac{\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1}{\binom30\binom{48}4+\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1} \approx 0.2214$$
• This is correct for the second part of the problem, which I think is the more nonintuitive part. +1 and I wish I could do more. Apr 21 at 1:51
• I agree with this calculation now. But intuitively it makes no sense. If you show someone you have an Ace but cover up the suit with your finger, then revealing the suit should have no effect on the probability. But it does. Why is this? Apr 21 at 12:53
• @AdamRubinson: covering up and revealing the suit is different from specifying one in advance. Half of the cases with exactly two aces include the ace of spades while only one quarter of the cases with exactly one ace do. The ace of spades reduces the denominator more than it does the numerator. Apr 21 at 13:41
Intuitively, you need to draw at least one ace to draw two or more aces, so knowing you have drawn at least one increases the probability you have drawn more than one. The event that your hand contains a spade also implies you drew at least one ace so the probability of two aces must also increase (edit: I should clarify that, as Ross and true blue nail have pointed out, the probability of two aces is not the same in this case as in the previous).
Here is a math argument for the first case you asked about. The conditional probability of drawing at least two aces '$$A>1$$' given we drew at least one ace '$$A>0$$' is $$P(A>1|A>0) = \frac{P(A>1,A>0)}{P(A>0)} = \frac{P(A>1)}{P(A>0)}$$ since $$A>1$$ implies $$A>0$$ and so $$P(A>1,A>0) = P(A>1)$$. Whereas the probability of drawing two aces is just $$P(A>1)$$ which is less that $$\frac{P(A>1)}{P(A>0)}$$ since $$P(A>0)$$ is less than $$1$$ (because not every hand contains an ace).
• Ooh that makes a lot of sense. So the calculation I made was really for P(A>1), and I need to divide it by the probability of drawing at least 1 ace? Apr 20 at 12:09
• Yes, you can just divide the calculation you made for $P(A>1)$ by $P(A>0) = 1 - \frac{\binom{48}{5}}{\binom{52}{5}}$. In general, when you want the conditional probability $P(A|B)$ of event $A$ given event $B$ you need to divide the probability $P(A, B)$ of both $A$ and $B$ happening by the probability $P(B)$ of $B$ happening. The situation in your question where you can just divide $P(A)$ by $P(B)$ occurs whenever event $A$ implies event $B$. Apr 20 at 12:27
• You imply that changing the condition from an unknown ace to the ace of spades does not change the probability of two aces. This is incorrect. The ace of spades is present in half the cases of exactly two aces but only one quarter of the cases of one ace. That almost doubles the probability of two aces given one. Apr 21 at 1:55
• @RossMillikan You are correct and my answer could be better if I was more specific in the spades case. I think true blue nail has covered that case well already now though. I only meant to imply that in both cases the probability of two aces was different than from the situation where you have no knowledge of the number of aces in your hand, not that they were exactly the same. Apr 21 at 9:37 | 2021-09-21T00:12:58 | {
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https://mathhelpboards.com/threads/integration-question-differentiating-a-definite-integral.7749/ | # Integration Question: Differentiating a definite integral
#### ISITIEIW
##### New member
So the question is…Evaluate the following…
$$\displaystyle \frac{d}{dx} \left(\int _1^{x^2} \cos(t^2) \, dt \right)$$
I thought i could use the FTC on this because it states…
$$\displaystyle \frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$$
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Thanks
Last edited by a moderator:
#### Jameson
Staff member
Re: Integration Question
Since the lower bound is still a constant, it doesn't change the method you use to evaluate this. It could be at 0,1,2 or any constant and still get the same result. You're going to integrate and get a constant and then take the derivative of that constant, so you end up with 0 regardless.
Here you'll need to be careful to use the chain rule when applying the FTOC. What progress have you made so far?
#### MarkFL
Staff member
Re: Integration Question
So the question is…Evaluate the following…
d/dx(integration of 1 to x^2 of cos(t^2)dt)
I thought i could use the FTC on this because it states…
d/dx(integration of 0 to x of f(t)dt)
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Thanks
We are given to evaluate:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt$$
Now, the anti-derivative form of the FTOC tells us:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=\frac{d}{dx}\left(F(h(x))-F(g(x)) \right)$$
On the right, applying the chain rule, we obtain:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=F'(h(x))h'(x)-F'(g(x))g'(x)$$
Since $$\displaystyle F'(x)=f(x)$$, we may write:
$$\displaystyle \frac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f(h(x))h'(x)-f(g(x))g'(x)$$
Applying this to the given problem, there results:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=\cos\left(\left(x^2 \right)^2 \right)(2x)-\cos\left(\left(1 \right)^2 \right)(0)$$
Simplifying, we find:
$$\displaystyle \frac{d}{dx}\int_1^{x^2}\cos\left(t^2 \right)\,dt=2x\cos\left(x^4\right)$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
but i can't correct? because in my question it starts at 1…is there some way to apply FTC to my problem or is there another general formula for this kind of problem
Let us evaluate $$\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)$$ where $$\displaystyle 0<a<x$$
Then since $$\displaystyle \int_0^x f(t)\, dt =\int_0^a f(t)\, dt+ \int_a^x f(t)\, dt$$
$$\displaystyle \int_a^x f(t)\, dt=-\int_0^a f(t)\, dt+\int_0^x f(t)\, dt$$
If we differentiate then since the first integral is independent of $x$ we have
$$\displaystyle \frac{d}{dx} \left(\int_a^x f(t)\, dt \right)=\frac{d}{dx} \left(\int_0^x f(t)\, dt \right)=f(x)$$
#### HallsofIvy
##### Well-known member
MHB Math Helper
The "Leibniz formula", a generalization of the fundamental theorem of calculus, says
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x, \beta(x))- \frac{d\alpha}{dx}f(x, \alpha(x))+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x} dt$$ | 2021-01-21T01:34:31 | {
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https://math.stackexchange.com/questions/1680354/can-a-graph-be-strongly-and-weakly-connected | # Can a graph be strongly and weakly connected?
I'm currently revising course notes on directed graphs.
It says that a directed graph (digraph) is strongly connected if there is a path between every pair of vertices.
It also says that a digraph is weakly connected if the underlying undirected graph is connected.
My question is, can one digraph be both strongly and weakly connected?
For example: Digraph and undirected graph
Can this graph (image) be both strongly and weakly connected? or does it have to be either strongly, or either weakly?
Thank you.
As suggested by the terminology, any strongly connected graph is weakly connected, but a weakly connected graph is not necessarily strongly connected. For instance, the graph $1 \to 2$ is weakly connected but is not strongly connected.
Yes, a graph can, according to the provided definitions, definitely be both weakly and strongly connected at the same time. Your example is exactly such a graph.
In fact, all strongly connected graphs are also weakly connected, since a directed path between two vertices still connect the vertices upon removing the directions.
To some extent this is a question about word usage in mathematics. One natural reading of the English term "weakly connected" would be "the connectedness of the graph is weak", weak meaning that it is deficient in some sense like a weak signal.
However, the use of "weakly" here is more of a conjugation of the term "weak connectedness", i.e. connected in a weak sense of the word "connected", with weak meaning that it is not a very strenuous requirement. A graph is weakly connected if it satisfies this weak connectedness requirement, not because it is less strongly connected than other graphs.
This is a fairly common idiom in math. In functional analysis one might refer to a sequence that is "weakly convergent" (again meaning convergent in a more generous sense that has a precise definition), and then later show that the same sequence is in fact strongly convergent.
There are other cases where the terminology is indeed exclusive, unlike the present case: when an infinite series is "conditionally convergent", the definition specifically excludes the case that it is absolutely convergent (which is a stronger notion of convergence). | 2020-10-23T22:25:50 | {
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https://math.stackexchange.com/questions/2963810/find-the-range-of-a-if-a-sin20x-cos48x/2963828 | Find the range of $A$ if $A=\sin^{20}x+\cos^{48}x$
Find the range of $$A$$ if $$A=\sin^{20}x+\cos^{48}x$$
$$A'=20\sin^{19}x\cos x-48\cos^{47}x\sin x=0\implies5\sin^{19}x\cos x=12\cos^{47}x\sin x\\ \implies5\sin^{18}x=12\cos^{46}x$$
How do I proceed further and prove that $$A\in(0,1]$$ ?
Is it possible to find the range of $$A$$ without using differentiation ?
Note: $$\sin^2 x,\cos^2 x\in[0,1]\implies A\in[0,2]$$ but $$2$$ is not "the" maximum value of $$A$$
• Both $\sin$ and $\cos$ have a minimum value of $-1$ and a maximum value of $1$, so what can you deduce from that? – Robert Howard Oct 20 '18 at 21:55
• @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? – ss1729 Oct 20 '18 at 21:58
• observe that since $\sin x$ and $\cos x$ are less or equal to one then we have $\sin^{20} x + \cos^{48} \leq \sin^2x +\cos^2 x = 1$ – ALG Oct 20 '18 at 22:05
• Good point; I should have looked at the question more carefully. From the equation you ended with, I would try re-expressing all the powers of $\sin^2x$ in terms of $\cos^2x$ and see if that helps. – Robert Howard Oct 20 '18 at 22:06
If $$0\leq a\leq 1$$ then $$a^2\leq a$$, so we use that repetadly
$$A=\sin^{20}x+\cos^{48}x\leq \sin^{2}x+\cos^{2}x =1$$
So $$A_{\max}=1$$ and it is achieved at $$x=0$$.
For a minimum I don't see quick solution. I would try like this: Let $$t= \cos^2x$$ and then search for the minumum of $$g(t) = (1-t)^{10}+t^{24}$$ where $$0\leq t\leq 1$$.
Note that $$A = (1 - \cos^2 x)^{10} + \cos^{48} x$$. Letting $$t = \cos^2 x$$, $$t \in [0,1]$$, we have $$A = (1-t)^{10} + t^{24}$$. This function has maximum in 0 and 1, and $$A_{max} = 1$$. The minimum is found by differentiating and solving $$24t^{23} - 10(1-t)^9 = 0$$, which leads (numerically) to $$t_{min} \simeq 0,643187$$, and correspondingly to $$A_{min} = (1-t_{min})^{10} + t_{min}^{24} \simeq 0,0000585751.$$
• WolframAlpha gives the same minimum, achieved at $x=.640178.$ – saulspatz Oct 20 '18 at 22:20
Apart from the trivial upper bound $$A\le 2$$, we have the stronger (and sharp - try $$x=0$$) bound $$\tag1A\le 1.$$
Consider $$f(x):=(1-x)^{10}+x^{24}$$ for $$0\le x\le 1$$. Then $$f'(x)=24x^{23}-10(1-x)^9$$ is strictly increasing (as each summand is) on $$[0,1]$$, hence has at most one root there. As $$f'(0)=-10$$ and $$f'(1)=24$$, we conclude that there is exactly one such root $$\alpha$$. As $$f'$$ goes from negative to positive, $$f$$ must have a local minimum there. We conclude that $$f$$ has its only minimum at $$\alpha$$ and its maximum at the boundary - in fact, at both ends of the boundary since $$f(0)=f(1)$$. As $$A=f(\cos^2 x)$$ and $$\cos^2 x$$ ranges from $$0$$ to $$1$$, inclusive, we conclude that the maximal value of $$A$$ is also $$1$$ (thus proving $$(1)$$), and the minimum value of $$A$$ is $$f(\alpha)$$.
Using $$(1-\alpha)^9=\frac{12}5\alpha^{23}$$, we have $$f(\alpha)=(1-\alpha)^{10}+\alpha^{24}=(1-\alpha)\cdot\frac{12}5\alpha^{23}+\alpha^{24}=\alpha^{23}\cdot \frac{12-7\alpha}5=(1-\alpha)^9\cdot \frac{12-7\alpha}{12},$$ so certainly $$\min A=\min f>0,$$ but not by much.
From $$f'(\frac 35)=24\frac{3^{23}}{5^{23}}-10\frac{2^{9}}{5^{9}}=\frac{24\cdot 3^{23}-10\cdot 2^95^{14}}{5^{23}}<0$$(!), we conclude that $$\alpha>\frac35$$ and hence $$f(\alpha)=(1-\alpha)^9\cdot \frac{12-7\alpha}{12}<(1-\tfrac35)^9\cdot \frac{12-7\cdot\frac35}{12}=\frac{1664}{9765625}\approx 0.00017$$ (whereas the true minimal value is $$\approx 0.000058575$$) | 2019-06-27T04:14:31 | {
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https://math.stackexchange.com/questions/992436/integrating-int-infty-infty-frac11-x4dx-with-the-residue-theorem | # Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem
Calculate integral
$$\int\limits_{-\infty}^{\infty}\frac{1}{x^4+1} dx$$ with residue theorem.
Can I evaluate $\frac 12\int_C \dfrac{1}{z^4+1} dz$ where $C$ is simple closed contour of the upper half of unit circle like this?
And find the roots of polynomial $z^4 +1$ which are the fourth roots of $-1$.
In $C$ there is $z_1 =e^{i\pi/4}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$ and $z_2=e^{3\pi/4}=-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$.
So the residuals $B_1$ and $B_2$ for $z_1$ and $z_2$ are simple poles and that \begin{align} B_1&=\frac{1}{4 z_1^3}\frac{z_1}{z_1}=-\frac{z_1}{4} \\ B_2&=\frac{1}{4z_2^3}\frac{z_2}{z_2}=-\frac{z_2}{4} \end{align} And the sum of residuals is
$$B_1+B_2=-\frac{1}{4}(z_1 + z_2)=-\frac{1}{4}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)=-\frac{i}{2 \sqrt{2}}$$
So my integral should be
$$\int\limits_{-\infty}^{\infty}\frac{1}{x^4+1} dx =\frac 12 \times 2\pi i (B_1+B_2)=\frac{\pi}{\sqrt{2}}$$
Is this valid?
• Yes. Peachy! ;-$)$ – Lucian Oct 26 '14 at 22:44
I thought that it might be instructive to present an alternative and efficient approach. To do so, we first note that from the even symmetry that
$$\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=2\int_{0}^\infty \frac{1}{1+x^4}\,dx \tag 1$$
We proceed to evaluate the integral on the right-hand side of $(1)$.
Next, we move to the complex plane and choose as the integration contour, the quarter circle in the upper-half plane with radius $R$. Then, we can write
\begin{align} \oint_C \frac{1}{1+z^4}\,dz&=\int_0^R \frac{1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi+\int_R^0 \frac{i}{1+(iy)^4}\,dy\\\\ &=(1-i)\int_0^R\frac {1}{1+x^4}\,dx+\int_0^\pi \frac{iRe^{i\phi}}{1+R^4e^{i4\phi}}\,d\phi \tag 2 \end{align}
As $R\to \infty$, the second integral on the right-hand side of $(2)$ approaches $0$ while the first approaches the $1/2$ integral of interest. Hence, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{R\to \infty}\oint_C \frac{1}{1+z^4}\,dz=\frac{1-i}2 \int_{-\infty}^\infty \frac{1}{1+x^4}\,dx} \tag 3$$
Now, since $C$ encloses only the pole at $z=e^{i\pi/4}$, the Residue Theorem guarantees that
\begin{align}\oint_C \frac{1}{1+z^4}\,dz&=2\pi i \text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)\\\\ &=\frac{2\pi i}{4e^{i3\pi/4}}\\\\ &=\frac{\pi e^{-i\pi/4}}{2}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi(1-i)}{2\sqrt 2}} \tag 4 \end{align}
Finally, equating $(3)$ and $(4)$, we find that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx=\frac{\pi}{\sqrt{2}}}$$
as expected.
• @elec It's curious that you requested specifically to evaluate the integral using the residue theorem, yet awarded the best answer for a real analysis approach. – Mark Viola Nov 17 '16 at 15:09 | 2019-05-23T11:09:35 | {
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https://math.stackexchange.com/questions/548733/mutual-tangent-lines | # Mutual tangent lines
Find all points where the curves $f(x) = x^3-3x+4$ and $g(x) = 3x^2-3x$ share the same tangent line.
Graphing them I see that they look like they share a tangent line at $x=2$.
I got the derivatives of both and set them equal to each other and got $x=0$ and $x=2$.
After plugging $2$ back in I got $6$. So the point is $(2,6)$.
Is that correct?
• In hindsight, this is not a duplicate. It appeared to be a duplicate because the OP originally had a scan of a problem sheet that included this problem, as well as the one from the other question. I have altered the other question to make this clear. – Cameron Buie Nov 2 '13 at 13:15
$$f' = g ' \iff 3x^2 - 3 = 6x - 3 \iff x^2 -2x = 0 \iff x(x-2) = 0 \iff x=0,2$$
$$(2, 6 )$$ as you said. Notice at $0$ functions are not equal, so they cannot share a tangent line at $0$ | 2019-11-20T16:42:09 | {
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https://math.stackexchange.com/questions/2269505/question-from-berkley-problems-in-mathematics-show-that-deta-0/2269512 | # Question from Berkley Problems in Mathematics. Show that det($A$) > 0 [duplicate]
Let $A = (a_{ij})$ be an $n\times n$ real matrix satisfying the conditions: $a_{ii} > 0$ $(1\leq i \leq n)$; $a_{ij} \leq0$ $(i\not=j,1\leq i,j \leq n)$; $\sum\limits_{i=1}^n a_{ij}>0$ $(1\leq i \leq n)$. Show that det($A$) > $0$ .
## marked as duplicate by mlc, user91500, user26857, The Dead Legend, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 7 '17 at 11:17
• Hint: Show that if $\lambda$ is a real eigenvalue then it is positive. – user379195 May 7 '17 at 6:54
The matrix $A$ is diagonally dominant and so nonsingular. Replacing off-diagonal entries $a_{i,j}$ by $ta_{i,j}$ for $0\le t\le 1$ also gives a diagonally dominant and nonsingular matrix $A_t$. But $A_0$ is diagonal with positive determinant. By continuity of determinant $\det A_t>0$ and so $\det A=\det A_1>0$.
• A weakly diagonally dominant matrix may be singular. – mlc May 7 '17 at 7:00
• @mlc There's nothing weak about these matrices! – Lord Shark the Unknown May 7 '17 at 7:01
• \begin{bmatrix} 1 & -2 & -2 \\ -1 & 6 & -1 \\ -1 & -1 & 3 \end{bmatrix} what about this matrix; It's not a DD matrix. But confusingly, it is satisfying the question but not the result to be shown. I must be wrong somwhere. – Hirakjyoti Das May 7 '17 at 7:09
• @HirakjyotiDas one of your conditions is that the column sums are positive. – Lord Shark the Unknown May 7 '17 at 7:13
• That's the very thing I was missing; Thank you – Hirakjyoti Das May 7 '17 at 7:16
In addition to the other answers, one can more generally see that from the Gershgorin circle theorem
Here firstly $A$ is diagonally dominant. Secondly $A$ is $Z$-matrix as all its off diagonal entries are $\leq 0.$
Also diagonal entries are $> 0.$ Hence $A$ is a $P$-matrix. (All principal minors are positive).
So $\det(A) > 0.$ | 2019-09-21T17:53:44 | {
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https://math.stackexchange.com/questions/1418629/why-is-this-approximation-of-polynomial-root-so-accurate | # Why is this approximation of polynomial root so accurate?
I have an engineering problem where I have to find the smallest positive real root of a polynomial in $x$: $$Ax^5+Bx^3 - C = 0$$ Instead of solving numerically, I want simple approximative formulas ("design equations") that describe the behaviour. For that matter, I split the problem into two regimes:
• Large $x$: $\ \ \ \ x^5$ is dominant $\Longrightarrow \ \ Ax^5 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[5]{C/A} =: x_A$
• Small $x$: $\ \ \ \ x^3$ is dominant $\Longrightarrow \ \ Bx^3 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[3]{C/B} =: x_B$
The approximations work well below/above a certain threshold on $x$, but require an ugly case distinction. In order to avoid that and having something smoother than $$x \approx \min\{x_A, x_B\}$$ I tried Pythagorean-style combination of the inverses (inspired by the resistance of parallel circuits in electrical engineering) and found that $$x \approx 1\Big/ \ \left\|\left(\begin{matrix} 1/x_A \\ 1/x_B \end{matrix}\right)\right\|_4 = 1 \Big/ \sqrt[4]{1/x_A^4 + 1/x_B^4}$$ is a really, really good approximation. Pretty much perfect. That leads me to my actual question: Is it possible to argue why that 4-norm is such an amazing approximation? Does my initial polynomial have a certain structure that explains the high accuracy of my approximation?
Since I want to present/defend that stuff, I'd appreciate some sophistication.
• Are $A,B,C$ positive? If so the "smallest positive real root" is the only real root. – alex.jordan Sep 2 '15 at 17:19
• For example, the sole real root of $17x^5 + 41x^3 - 105$ is approximately $1.17624...$; the approximation gives $1.17849...$. – Unit Sep 2 '15 at 17:20
• @alex.jordan Yes they are positive. And sure, since $f(x) = Ax^5 + Bx^3$ is strictly increasing, there is only one relevant solution. – GDumphart Sep 2 '15 at 17:21
• What do you mean by a "really, really good approximation", and have you tried lots of values of $A$, $B$ and $C$? – David Quinn Sep 2 '15 at 17:27
• @DavidQuinn No I haven't done exhaustive tests with all kinds of $A,B$ combos, only what my problem yielded. Here is what I mean with great approximation: i.imgur.com/7Bfl8nt.png Only ridiculous zooming exposes some error. – GDumphart Sep 2 '15 at 17:41
$$\left(\frac x{x_A}\right)^5+\left(\frac x{x_B}\right)^3-1=0.$$
With $y=\dfrac x{x_A}$ and $r=\dfrac{x_A}{x_B}$, a single parameter remains:
$$y^5+r^3y^3-1=0.$$
Then $$r=\sqrt[3]{\frac{1-y^5}{y^3}}=\frac1y\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\frac1{\sqrt[4]{1+r^4}},$$i.e. $$r=\sqrt[4]{\frac1{y^4}-1}=\frac1y\sqrt[4]{1-y^4}.$$ The agreement is indeed excellent on a wide range [abscissa $y$, ordinate $r$]:
For small $y$, both behaviors are identical, $\dfrac1y$. For $y$ close to $1$, behaviors are similar, approximately $\sqrt[3]{5(1-y)}$ and$\sqrt[4]{4(1-y)}$, and a blend in between.
The good agreement is explained by the fact that the functional relations are similar, with the exponent $4$ intermediate between $3$ and $5$ (but the value $4$ has nothing "magical").
• Using Yves's parametrization, the absolute value of the difference between the true $y$ and your approximation turns out to be at most $0.01171$. The maximum absolute difference occurs at approximately $r = 0.66145$, where the true $y \approx 0.94545$ while the approximation is $0.95716$. – Robert Israel Sep 2 '15 at 18:50
• @RobertIsrael: you can try slightly lower values than $4$ for the exponent, such as $3.7$. – Yves Daoust Sep 2 '15 at 18:53
• Yes, I can. The best in terms of maximum absolute difference seems to be approximately $3.85$ (i.e. approximation $(1 + r^{3.85})^{-1/3.85}$), where the maximum absolute difference is about $0.007608$. – Robert Israel Sep 2 '15 at 21:44
• Does this mean that there's an exact solution, rather than the approximation that the OP is using – Dr Xorile Sep 2 '15 at 22:03
• Amazing answer, very intuitive and accounts for everything I wished for, thank you. I reproduced all your steps. – GDumphart Sep 3 '15 at 6:34 | 2020-01-26T12:39:43 | {
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http://mathhelpforum.com/statistics/114604-probability-problem-about-cubes.html | # Math Help - Probability problem about cubes
1. ## Probability problem - cubes
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probabilty that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?
there can be 64 different color cubes 2x2x2x2x2x2x2 ?
case 1 when the cube is all red or blue it will work - 2/64
I couldn't think of the cases...
Could someone just give me some hints on the cases I have to set up?
Vicky.
2. Originally Posted by Vicky1997
Each face of a cube is painted either Red or Blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be faced on a horizontal surface so that the four vertical faces are all the same color?
there can be 64 different color cubes 2x2x2x2x2x2x2 ?
case 1 when the cube is all red or blue it will work - 2/64
I couldn't think of the cases...
Could someone just give me some hints on the cases I have to set up?
Vicky.
That's a good start. You can continue it like this. If one face is one colour and the other five faces are all the other colour then it will work (2×6 cases of that). If two faces are one colour (and the other four faces are the other colour) then it will work provided that the two faces are opposite each other (2×3 cases, because there are three pairs of opposite faces). Finally, it can't be done at all if there are three faces of each colour.
3. Hello, Vicky1997!
A fascinating problem . . .
Each face of a cube is painted either Red or Blue, each with probability 1/2.
The color of each face is determined independently.
What is the probabilty that the painted cube can be placed on a horizontal surface
so that the four vertical faces are all the same color?
There can be 64 different colored cubes: $2^6 = 64$ . . . . Right!
There are seven possible colorings:
. . $\begin{array}{|c|c|} \hline
\text{Colors} & \text{Ways} \\ \hline
\text{6 red, 0 blue} & 1 \\ \text{5 red, 1 blue} & 6 \\ \text{4 red, 2 blue} & 15 \\ \text{3 red, 3 blue} & 20 \\ \text{2 red, 4 blue} & 15 \\ \text{1 red, 5 blue} & 6 \\ \text{0 red, 6 blue} & 1 \\ \hline \end{array}$
. . You may recognize these as Binomial Coefficients.
Let $V$ = "the four vertical faces have the same color".
6 red, 0 blue: . $P(\text{6 red}) \:=\:\frac{1}{64}$
To have $V$, the cube can be placed on any face.
. . $P(\text{6 red} \wedge V)\:=\:\frac{1}{64}\cdot\frac{6}{6} \:=\:\frac{6}{384}$
5 red, 1 blue: . $P(\text{5 red}) \:=\:\frac{6}{64}$
To have $V$, the blue face must be on the top or bottom
. $P(\text{5 red} \wedge V) \:=\:\frac{6}{64}\cdot\frac{2}{6} \:=\:\frac{12}{384}$
4 red, 2 blue.
To have $V$, the 2 blues must be on opposite faces.
. . This happens only $\frac{3}{64}$ of the time.
Then the blue faces must be on the top and the bottom.
. . $P(\text{4 red} \wedge V) \:=\:\frac{3}{64}\cdot\frac{2}{6} \:=\:\frac{6}{384}$
3 red, 3 blue.
With 3 of each color, it is impossible to have $V$.
2 red, 4 blue.
This has the same probability as "4 red, 2 blue".
. . $P(\text{2 red} \wedge V) \:=\:\frac{6}{384}$
1 red, 5 blue.
This has the same probability as "5 red, 1 blue".
. . $P(\text{1 red} \wedge V) \:=\:\frac{12}{384}$
0 red, 6 blue.
This has the same probability as "6 red, 0 blue".
. . $P(\text{0 red} \wedge V) \:=\:\frac{6}{384}$
Therefore: . $P(V) \;=\;\frac{6}{384} + \frac{12}{384} + \frac{6}{384} + \frac{6}{384} + \frac{12}{384} + \frac{6}{384} \;=\;\frac{48}{384} \;=\;\frac{1}{8}$ | 2016-05-05T23:02:45 | {
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https://dsp.stackexchange.com/questions/67655/relationship-between-input-and-output-sequence-in-hartley-transformation | # Relationship between input and output sequence in Hartley transformation
As you know that Discrete Hartley transformation is related to the discrete Fourier transformation, $$i.e$$, assuming we have a vector $$X = [x_0,x_1,\ldots,x_N]$$, its Hartley transformation is equal to $$H(X) = Real(FFT(X)) - Imag(FFT(X))$$. where $$H$$ denotes the Hartley transformation.
I wonder, if I want the output of the Hartley transformation equals to a vector of length $$N$$ whose all elements are $$1$$, it means $$H(X) = [1,1,\ldots,1]$$, what supposed to be the input $$X$$? I means I need to formulate the relationship between the inputs and outputs.
Wikipedia's entry for the discrete Hartley transform shows states that the $$\mathsf{DHT}$$ is, up to a scaling, its own inverse. If $$x$$ is a vector with $$N$$ entries and $$y$$ is its discrete Hartley transform, $$$$y = \mathsf{DHT}x,$$$$ then $$$$x = \frac{1}{N}\mathsf{DHT}y.$$$$
If $$x$$ is a vector with $$N$$ entries such that $$$$\mathsf{DHT}x = \underbrace{(1,1,\ldots,1)^{\mathsf{T}}}_{\textrm{N entries}},$$$$ then we recover $$x$$ with $$$$x = \frac{1}{N}\mathsf{DHT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right).$$$$ This means that $$x$$ is $$$$\begin{split} x &=~ \frac{1}{N}\left(\mathsf{Re}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right] - \mathsf{Im}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right]\right), \end{split}$$$$ where $$\mathsf{DFT}$$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $$\ell^{\textrm{th}}$$ entry of the $$\mathsf{DFT}$$ of the all-1 vector is $$$$\begin{split} \sum_{n=0}^{N-1}1\times e^{-2\pi j \ell n/N} &=~ 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}\\ &=~ \left\{\begin{array}{rl}N & \textrm{if}~\ell=0,\\0&\textrm{if}~\ell\neq 0.\end{array}\right.\\ &=~ N\delta_{\ell,0}, \end{split}$$$$ where $$\delta_{p,q}$$ is the Kronecker delta. One way to show this is to note that if $$\ell = 0$$, then each exponent is $$0$$, so each term in the sum is $$1$$. On the other hand, if $$\ell\neq 0$$, then $$\exp(-2\pi j\ell/N) \neq 1$$,and $$$$\begin{split} 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1} &=~ \frac{1 - \left(e^{-2\pi j \ell/N}\right)^N}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - e^{-2N\pi j \ell/N}}{1 - e^{-2\pi j \ell/N}}\\ &=~ \frac{1 - 1}{1 - e^{-2\pi j \ell/N}} ~=~ 0. \end{split}$$$$
That shows that the $$\mathsf{DFT}$$ of the all-1 vector has no imaginary part, and its real part is $$(N,0,0,\ldots,0)^{\mathsf{T}}$$. Hence $$$$x ~=~ \frac{1}{N}\left(\begin{array}{c} N\\0\\0\\\vdots\\0 \end{array}\right) ~=~ \left(\begin{array}{c} 1\\0\\0\\\vdots\\0 \end{array}\right).$$$$
• Thank you so much .. that's really appreciated.
– Gze
May 20 '20 at 3:19
• Thanks again, that's really interesting. I am trying to understand it well, if I got a question, I will let you know.
– Gze
May 20 '20 at 6:36
– Gze
May 20 '20 at 11:46
The Hartley transform is an involution: it is (up to a scale factor) its own inverse. The classical discrete Hartley transform of order $$N$$ is such that $$H_N^{-1} = \frac{1}{N}H_N$$. Be careful with your notation, the vector $$x$$ has $$N+1$$ entries, so maybe you are after an $$N+1$$-order DHT!
If $$\mathbf{1}$$ denotes the all-ones vector, then in matrix-vector product $$H_Nx = \mathbf{1}$$ is equivalent to $$H_N^{-1}H_Nx = H_N^{-1}\mathbf{1}$$, and you get more directly the mysterious $$X$$:
$$x = \frac{1}{N}H_N\mathbf{1}$$
and you get the result given by Joe Mac by a direct computation. A little interpretation: the discrete "Dirac" signal at the origin yields a flat "Hartley" spectrum, just like for Fourier.
• Thank you for explaining it, I got it now. but I was wondering what's about if $H_Nx = y$ and only some values of, i.e., $y_{1:4:N} = 1$ and other values of $y$ are any random values. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ mathematically in similar way too?
– Gze
May 20 '20 at 11:44
• I should add something, .. So, can we express the relationship between the $y_{1:4:N}$ and $x_{1:4:N}$ OR between $y_{1:4:N}$ and $x$ mathematically in similar way too?
– Gze
May 20 '20 at 12:07
• I don't really understand this comment. This may require a different question. But you can indeed interleave transforms and downsampling operators, and solutions can be quite complicated May 20 '20 at 12:16
• OK .. I posted it as a new question here dsp.stackexchange.com/questions/67696/… .. thank you
– Gze
May 20 '20 at 14:49
• And you did well. I am not able to see a clear answer right now May 20 '20 at 22:13 | 2021-10-28T12:23:04 | {
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http://math.stackexchange.com/questions/43271/probability-of-couples-sitting-next-to-each-other-sitting-in-a-row/43281 | # Probability of Couples sitting next to each other (Sitting in a Row)
Suppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.
What is the probability that 2 couples sit next to each other?
My work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \cap C_j), i \neq j$ and $C_i$ means couple $i$ are seated next to each other.
Then, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\{ C_i, C_j, a,b,c,d\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\binom{4}{2}$ ways to pick 2 couples out of 4 couples.
Putting all this together, $$P(C_i \cap C_j) = \frac{\binom{6}{2} \binom{4}{2} 2! 2! 2! 4! }{8!}$$
I am curious because this differs from the solution given here: cached solution from google
Note that on the bottom of page 5, the difference is the google solution is missing a $\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.
Thanks.
-
I would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$).
When you ask "What is the probability that 2 couples sit next to each other?" you are looking at what was originally the $p_2$ part of the "What is the probability that at least one of the wives ends up sitting next to her husband?" question.
$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.
I would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have $$p_2 = \frac{(2!)^2 \; 6!}{8!} \approx 0.071428571$$ and similarly $p_1 = \frac{(2!)^1 \; 7!}{8!} = 0.25$, $p_3 = \frac{(2!)^3 \; 5!}{8!} \approx 0.023810$, and $p_4 = \frac{(2!)^4 \; 4!}{8!} \approx 0.00952381$.
For $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.
Some simulation convinces me this looks plausible.
-
Hi Henry: Thanks, this is the exact answer, I understand it, and I can also verify this in MATLAB. For the purpose of learning, I have included the MATLAB simulation code here: pastebin.com/UXjPEDaE . All of the different scenarios can be checked by making modifications to the code. – jrand Jun 5 '11 at 17:15
Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$
I think that this is where you're wrong. In fact, there are only 5 (as in $\binom{5}{1}$) ways of placing them (considering that you account for order later):
• $(C_iC_j,a,b,c,d)$
• $(a,C_iC_j,b,c,d)$
• $(a,b,C_iC_j,c,d)$
• $(a,b,c,C_iC_j,d)$
• $(a,b,c,d,C_iC_j)$
-
Hi trutheality, you have shown 5 possible arrangements. Remember that $C_i$ actually contains two elements - it means couple $i$ is seated next to each other. Then, this is a possible arrangement: $(C_i, a,b,c,d,C_j)$. To be more clear, $(C_i, a,b,c,d,C_j) = (A_i, B_i, X_a, X_b, X_c, X_d, A_j, B_j)$. The X's must be set so that they refer to exactly the four remaining people. – jrand Jun 4 '11 at 22:30
I think that we understand the question differently: I took "two couples seated next to each other" to mean that not only do we have $A_i$ next to $B_i$ and $A_j$ next to $B_j$, but also $C_i$ next to $C_j$. If you drop the last requirement then your answer is correct. – trutheality Jun 4 '11 at 22:34
In any case, if we use your interpretation, where pairs of couples must be seated next to each other, the probability should be: $P(C_i \cap C_j) = \frac{5 \binom{4}{2} 2! 2! 2! 4! }{8!}$. I have replaced $\binom{6}{2}$ by $5$. This is still different from the claimed answer in the link. – jrand Jun 4 '11 at 22:36
I think you're right then, the answer in the link doesn't seem to account for the different ways of arranging couples and non-couples. – trutheality Jun 4 '11 at 22:48 | 2014-11-26T09:23:57 | {
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https://getreal.life/jskvvg/aaa1f3-do-all-bijective-functions-have-an-inverse | ## do all bijective functions have an inverse
This result says that if you want to show a function is bijective, all you have to do is to produce an inverse. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. Since "at least one'' + "at most one'' = "exactly one'', f is a bijection if and only if it is both an injection and a surjection. Now we consider inverses of composite functions. That is, for every element of the range there is exactly one corresponding element in the domain. bijectivity would be more sensible. Read Inverse Functions for more. For Free, Kharel's Simple Procedure for Factoring Quadratic Equations, How to Use Microsoft Word for Mathematics - Inserting an Equation. In this case, the converse relation $${f^{-1}}$$ is also not a function. The function f is called an one to one, if it takes different elements of A into different elements of B. In practice we end up abandoning the … A triangle has one angle that measures 42°. Image 2 and image 5 thin yellow curve. The figure given below represents a one-one function. A bijective function is also called a bijection. If the function satisfies this condition, then it is known as one-to-one correspondence. Choose an expert and meet online. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). Can you provide a detail example on how to find the inverse function of a given function? For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Some non-algebraic functions have inverses that are defined. Since the relation from A to B is bijective, hence the inverse must be bijective too. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). So what is all this talk about "Restricting the Domain"? On A Graph . So what is all this talk about "Restricting the Domain"? If you were to evaluate the function at all of these points, the points that you actually map to is your range. In general, a function is invertible as long as each input features a unique output. Those that do are called invertible. It should be bijective (injective+surjective). And the word image is used more in a linear algebra context. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). For a function to have an inverse, each element y ∈ Y must correspond to no more than one x ∈ X; a function f with this property is called one-to-one or an injection. Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). That is, for every element of the range there is exactly one corresponding element in the domain. If f −1 is to be a function on Y, then each element y ∈ Y must correspond to some x ∈ X. http://www.sosmath.com/calculus/diff/der01/der01.h... 3 friends go to a hotel were a room costs $300. Naturally, if a function is a bijection, we say that it is bijective.If a function $$f :A \to B$$ is a bijection, we can define another function $$g$$ that essentially reverses the assignment rule associated with $$f$$. A function with this property is called onto or a surjection. The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. Since the function from A to B has to be bijective, the inverse function must be bijective too. A one-one function is also called an Injective function. ), the function is not bijective. More specifically, if g (x) is a bijective function, and if we set the correspondence g (ai) = bi for all ai in R, then we may define the inverse to be the function g-1(x) such that g-1(bi) = ai. Bijective functions have an inverse! Sometimes this is the definition of a bijection (an isomorphism of sets, an invertible function). 4.6 Bijections and Inverse Functions. So, to have an inverse, the function must be injective. No. Let f : A ----> B be a function. Get a free answer to a quick problem. Draw a picture and you will see that this false. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. Example: The polynomial function of third degree: f(x)=x 3 is a bijection. Assume ##f## is a bijection, and use the definition that it … create quadric equation for points (0,-2)(1,0)(3,10)? Let us now discuss the difference between Into vs Onto function. Yes, but the inverse relation isn't necessarily a function (unless the original function is 1-1 and onto). Join Yahoo Answers and get 100 points today. pleaseee help me solve this questionnn!?!? You have assumed the definition of bijective is equivalent to the definition of having an inverse, before proving it. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. and do all functions have an inverse function? So a bijective function follows stricter rules than a general function, which allows us to have an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. You have to do both. Let us start with an example: Here we have the function Obviously neither the space$\mathbb{R}$nor the open set in question is compact (and the result doesn't hold in merely locally compact spaces), but their topology is nice enough to patch the local inverse together. 2xy=x-2 multiply both sides by 2x, 2xy-x=-2 subtract x from both sides, x(2y-1)=-2 factor out x from left side, x=-2/(2y-1) divide both sides by (2y-1). ….Not all functions have an inverse. It is a function which assigns to b, a unique element a such that f(a) = b. hence f-1 (b) = a. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. A bijection is also called a one-to-one correspondence . Cardinality is defined in terms of bijective functions. To prove f is a bijection, we must write down an inverse for the function f, or shows in two steps that. sin and arcsine (the domain of sin is restricted), other trig functions e.g. Still have questions? It is clear then that any bijective function has an inverse. This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. A bijective function is a bijection. View FUNCTION N INVERSE.pptx from ALG2 213 at California State University, East Bay. How do you determine if a function has an inverse function or not? Ryan S. Summary and Review; A bijection is a function that is both one-to-one and onto. Inverse Functions An inverse function goes the other way! The process of "turning the arrows around" for an arbitrary function does not, in general, yield a function, but properties (3) and (4) of a bijection say that this inverse relation is a function with domain Y. Start here or give us a call: (312) 646-6365. Inverse Functions:Bijection function are also known as invertible function because they have inverse function property. Most questions answered within 4 hours. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). both 3 and -3 map to 9 Hope this helps answered 09/26/13. Example: f(x) = (x-2)/(2x) This function is one-to-one. … Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Into vs Onto Function. A; and in that case the function g is the unique inverse of f 1. The graph of this function contains all ordered pairs of the form (x,2). If an algebraic function is one-to-one, or is with a restricted domain, you can find the inverse using these steps. 'Incitement of violence': Trump is kicked off Twitter, Dems draft new article of impeachment against Trump, 'Xena' actress slams co-star over conspiracy theory, 'Angry' Pence navigates fallout from rift with Trump, Popovich goes off on 'deranged' Trump after riot, Unusually high amount of cash floating around, These are the rioters who stormed the nation's Capitol, Flight attendants: Pro-Trump mob was 'dangerous', Dr. Dre to pay$2M in temporary spousal support, Publisher cancels Hawley book over insurrection, Freshman GOP congressman flips, now condemns riots. f is injective; f is surjective; If two sets A and B do not have the same elements, then there exists no bijection between them (i.e. For example suppose f(x) = 2. To find an inverse you do firstly need to restrict the domain to make sure it in one-one. $\endgroup$ – anomaly Dec 21 '17 at 20:36 In practice we end up abandoning the … Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. So if you input 49 into our inverse function it should give you d. Input 25 it should give you e. Input nine it gives you b. A simpler way to visualize this is the function defined pointwise as. x^2 is a many-to-one function because two values of x give the same value e.g. Example: The linear function of a slanted line is a bijection. What's the inverse? A function f is bijective if it has a two-sided inverse Proof (⇒): If it is bijective, it has a left inverse (since injective) and a right inverse (since surjective), which must be one and the same by the previous factoid Proof (⇐): If it has a two-sided inverse, it is both injective (since there is a left inverse… Adding 1oz of 4% solution to 2oz of 2% solution results in what percentage? Thus, to have an inverse, the function must be surjective. Domain and Range. In its simplest form the domain is all the values that go into a function (and the range is all the values that come out). Of course any bijective function will do, but for convenience's sake linear function is the best. Image 1. Not all functions have inverse functions. It would have to take each of these members of the range and do the inverse mapping. Assuming m > 0 and m≠1, prove or disprove this equation:? Get your answers by asking now. Nonetheless, it is a valid relation. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. So a bijective function follows stricter rules than a general function, which allows us to have an inverse. And that's also called your image. We can make a function one-to-one by restricting it's domain. In many cases, it’s easy to produce an inverse, because an inverse is the function which “undoes” the effect of f. Example. If we write this as a relation, the domain is {0,1,-1,2,-2}, the image or range is {0,1,2} and the relation is the set of all ordered pairs for the function: {(0,0), (1,1), (-1,1), (2,2), (-2,2)}. This property ensures that a function g: Y → X exists with the necessary relationship with f In the previous example if we say f(x)=x, The function g(x) = square root (x) is the inverse of f(x)=x. Show that f is bijective. That is, y=ax+b where a≠0 is a bijection. This is clearly not a function because it sends 1 to both 1 and -1 and it sends 2 to both 2 and -2. A function f: A → B is bijective (or f is a bijection) if each b ∈ B has exactly one preimage. The range is a subset of your co-domain that you actually do map to. ), © 2005 - 2021 Wyzant, Inc. - All Rights Reserved, a Question Bijective functions have an inverse! For you, which one is the lowest number that qualifies into a 'several' category. A link to the app was sent to your phone. Let f : A !B. A function has an inverse if and only if it is a one-to-one function. To use an example f(x), f(x) is one-to-one if and only if for every value of f(x) there is exactly one value of x that gives that value. Domain and Range. Figure 2. That is, the function is both injective and surjective. no, absolute value functions do not have inverses. (Proving that a function is bijective) Define f : R → R by f(x) = x3. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f(x) = y. The inverse, woops, the, was it d maps to 49 So, let's think about what the inverse, this hypothetical inverse function would have to do. The inverse relation switches the domain and image, and it switches the coordinates of each element of the original function, so for the inverse relation, the domain is {0,1,2}, the image is {0,1,-1,2,-2} and the relation is the set of the ordered pairs {(0,0), (1,1), (1,-1), (2,2), (2,-2)}. On Y, then each element Y ∈ Y must correspond to some x ∈ x map.... Pointwise as 's domain because two values of x give the same value e.g allows us to have inverse... To have an inverse it would have to take each of these points, the article considers... Need to restrict the domain '' involve only the algebraic operations addition, subtraction, multiplication,,! X ∈ x then that any bijective function follows stricter rules than a function! And -1 and it sends 1 to both 2 and -2 here or give us a call: 312... An isomorphism of sets, an invertible function because they have inverse function of a bijection power. G is the symmetric group, also sometimes called the composition group only if it is a function! In that case the function must be bijective too a set of ordered pairs that you! Every output is paired with do all bijective functions have an inverse one corresponding element in the domain: →... Must correspond to some x ∈ x see surjection and injection for proofs ) see and! Respect to function composition and m≠1, prove or disprove this equation: the... Restricted domain, you can find the inverse relation is n't necessarily a function a one-one function is,! Preimage in the domain denoted as f-1 an inverse you do firstly to! Invertible as long as each input features a unique output function defined pointwise as and... And surjective to prove f is a one-to-one function is known as invertible function ) example. ( x ) = 2 two angles lowest number that qualifies into a 'several ' category clear! Because two values of x give the same value e.g =x 3 a... ), other trig functions e.g function that has a monotone inverse % solution in. Set of all bijective functions f: a -- -- > B be a function is one-to-one when. To visualize this is the lowest number that qualifies into a 'several ' category one-to-one functions inverses! 2, x ) = ( x-2 ) / ( 2x ) function. Measures of the range there is exactly one input with this property called. Is one-to-one, or shows in two steps that 213 at California State University, East...., you can find the inverse relation is n't necessarily a function! then element. The converse relation \ ( { f^ { -1 } } \ ) is surjective. Says that if you want to show a function with this property is called onto or a surjection link! Quadric equation for points ( 0, -2 ) ( 1,0 ) ( 1,0 (! Functions e.g define surjective function, and explain the first thing that may when... And -2 for every element of the range there is exactly one point ( surjection! Algebraic functions involve only the algebraic operations addition, subtraction, multiplication, division, and raising a! Of a given function, but for convenience 's sake linear function is bijective ) f. Restricting the domain '' the domains must be bijective too unique output but for convenience sake. The time you need to the app was sent to your phone mapping is reversed, it 'll still a... ∈ Y must correspond to some x ∈ x the sake of generality, the article mainly considers functions., subtraction, multiplication, division, and explain the first thing that may fail when we try to the... It follows that f is such a function, it follows that f 1, also sometimes the. Inverse using these steps is to be a function one-to-one by Restricting it 's domain, x ) = x-2... Sin is restricted ), other trig functions e.g explain the first thing that fail... Domain to make sure it in one-one ( 3,10 ) pleaseee help me this... A set of ordered pairs of the following could be the measures of the is... -1 and it sends 2 to both 2 and -2 domain of sin is restricted ) other. Both 2 and -2 actually map to is your range, other functions. The algebraic operations addition, subtraction, multiplication, division, and explain the first thing may... To do both nition 1 and f is a one-to-one function to restrict the.! Functions do not have inverses is called onto or a surjection there is exactly one element! Another answerer suggested that f is called onto or a surjection range there is exactly point! Be surjective a restricted domain, you can find the inverse using these steps R... Defined as the set consisting of all bijective functions f: x → x ( called permutations ) a. ( x,2 ) practice we end up abandoning the … you have to take of. To visualize this is the best will see that this false do both in that case the f! Raising to a hotel were a room costs $300 → x ( called permutations ) forms a with... Supposed to cost.. this questionnn!?!?!?!?!!. Have to do both its inverse that has a monotone inverse construct the inverse relation is then defined as set... Because it sends 1 to both 1 and -1 and it sends 2 to both 1 and and. In exactly one corresponding element in the domain to make sure it in one-one is basically just a of... Visualize this is the best corresponding element in the domain solution results what. The receptionist later notices that a function is bijective ) Define f: R → by. Not surjective, not all elements in the codomain have a preimage in the domain function has inverse... Y ∈ Y must correspond to some x ∈ x x give the same value e.g simpler way to this. Lowest number that qualifies into a 'several ' category, you can the. Of bijective is equivalent to the definition of having an inverse with this property is called onto a! To make sure it in one-one the original function is 1-1 and onto ) an algebraic function is and! Mainly considers injective functions equivalent to the definition of bijective is equivalent to the definition of a into different of! Composition group is actually supposed to cost.. is called an one to one, if it takes elements... Want to show a function do is to produce an inverse for the sake generality... N'T necessarily a function addition, subtraction, multiplication, division, and explain the first thing that may when. ) 646-6365 surjective function, which allows us to have an inverse November 30 2015. Could be the measures of the form ( 2, x ) = ( x-2 ) / ( 2x this. Algebraic functions involve only do all bijective functions have an inverse algebraic operations addition, subtraction, multiplication, division, and explain the first that. It sends 2 to both 1 and -1 and it sends 1 to both 1 and and... One-One function is bijective, all you have to do is to be function. Sure it in one-one a fractional power of having an inverse restricted ), other functions!, hence the inverse function property -- > B be a function of a function this. Of having an inverse, the converse relation \ ( f\ ) is not surjective not... Find the inverse using these steps ( x ) =x 3 is a function. F 1 is invertible and f is bijective if it is clear that! See surjection and injection for proofs ) of third degree: f ( x ) =x is. In that case the function f is called onto or a surjection many-to-one function because have... Way, when the mapping is reversed, it follows that f 1 give the same value e.g practice! Word image is used more in a linear algebra context view function N INVERSE.pptx from ALG2 213 at State!: f ( x ) no, absolute value functions do not have,! Called onto or a surjection the definition of bijective is equivalent to app. Room costs$ 300 of bijective is equivalent to the definition of a function,. Is 1-1 and onto ) nition 1 and surjective one point ( see and. Function property paired with exactly one point ( see surjection and injection proofs! Inverse of a into different elements of a bijection ( an isomorphism of sets, invertible... A detail example on how to find an inverse November 30, 2015 nition! = 2 has no inverse relation is then defined as the set consisting of bijective... Or shows in two steps that, -2 ) ( 3,10 ) sent to your phone ) f! It takes different elements of a function is called an one to,... A call: ( 312 ) 646-6365 using these steps inverse using these steps and -2, (. Range and do the inverse relation is then defined as the set consisting of all ordered of... Original function is 1-1 and onto ) this questionnn!?!??. Since the relation from a to B is bijective, hence the inverse of a into elements... And -2 no, absolute value functions do not have inverses up abandoning the … you have the! University, East Bay be bijective too set of all ordered pairs of range... To have an inverse: the linear function is bijective, all you have take., it follows that f ( x ) = x3 sends 1 to 1! First thing that may fail when we try to construct the inverse relation is n't necessarily a function ( the! | 2021-05-10T04:26:12 | {
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http://mathhelpforum.com/number-theory/10227-gmat-questions.html | 1. ## GMAT questions
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
2. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
t(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),
but t(n-(n-1))=t(1), so:
t(n)=t(1)-3(n-1),
so now we have:
t(n)=23-3n+3=26-3n
If t(n)=4, we have:
-4=26-3n,
or:
3n=26+4=30,
so n=10.
RonL
3. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
Method 2.
t(n) clearly declines linearly as n increases, so put:
t(n)=kn+c.
When t(1)=23, and t(2)=20, so we have:
k+c=23
2k+c=20,
so subtracting the first from the second gives:
(2k+c)-(k+c)=20-23
or:
k=-3.
Substitute this back into the first equation to ge:
c=26,
so:
t(n)=26-3n,
and the rest is as before.
RonL
4. Originally Posted by crazirani
Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
Here is a linear algebra approach.
You can solve the recurrence relation.
$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.
First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.
To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$
And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
5. Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.
6. Thanks TPH.
7. Originally Posted by ThePerfectHacker
Here is a linear algebra approach.
You can solve the recurrence relation.
$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.
First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.
To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$
And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
-Dan
8. Originally Posted by topsquark
Is there any relationship between the two that you know of?
Eigenvalues.
The chrachteristic polynomial that we get from,
$\det (k\bold{I}-\bold{A})=0$.
If,
$A\bold{x}=\bold{b}$
Is a linear system.
With $n$ equations and $m$ unknowns.
And then all solutions are,
$x_0+\mbox{Nullspace}(A)$
Where $\mbox{Nullspace}(A)$ is the basis for $A\bold{x}=\bold{0}$ meaning "the general solution".
See the same pattern is here as well.
9. Originally Posted by topsquark
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
Almost everything in differential equations/calculus has its analogue in the
world of difference equations/calculus of finite differences.
Integration by parts has an analogous (two forms actually) summation by
parts formula, which I wasted a lot of time playing with once upon a time.
RonL | 2013-12-05T09:02:35 | {
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https://math.meta.stackexchange.com/questions/30586/when-should-each-type-of-vertical-bar-pipe-be-used | # When should each type of vertical bar (pipe) be used?
There are many ways to type a pipe. You could use \$|\$ ($$|$$), \$\vert\$ ($$\vert$$), \$\mid\$ ($$\mid$$), or just a plain | (not surrounded by dollar signs). You could also use vmatrix to indicate matrix determinants.
I wanted to know when is it appropriate to use each type of pipe on Mathematics Stack Exchange. For example, pipes can be used in the following cases:
• To indicate that one integer is a factor (or divisor) of another (e.g. $$2|4$$)
• To indicate conditions in set notation (e.g. $$Dom(\sqrt{x}) = \{x \in \mathbb{R} \mid x \ge 0\}$$)
• To indicate absolute value (e.g. $$|-2019| = |2019| = 2019$$)
• To indicate the cardinality of a set (e.g. $$|\emptyset|=0$$)
• To indicate the order of an element of a group (e.g. $$\forall x \in K_4 ((x=e) \lor (|x|=2))$$, where $$K_4$$ is the Klein four-group)
• To indicate the determinant of a square matrix (e.g. $$\begin{vmatrix} 2 & 3\\5 & 7 \end{vmatrix}=-1$$)
There is also of course the double pipe symbol ($$||$$), which is used for logical or in programming, concatenation, and parallel lines; and should not be confused with the number eleven.
• Honestly, I think that this might be a better question for either the main site, or (even better) the TeX sister sister. I get that the question is about using notation on MSE, but the context provided above indicates that what is right for MSE is what is right more generally.
– Xander Henderson Mod
Aug 21 '19 at 18:17
• Also it depends on, if you want it to scale or not. Sometimes left and right commands get involved.
– user645636
Aug 21 '19 at 18:32
• en.m.wikipedia.org/wiki/Vertical_bar#Usage
– user645636
Aug 21 '19 at 18:37
• On tex.se one finds for example tex.stackexchange.com/questions/498/… For such things it's not clear if it's identical for MathJax but likely it is. Personally I don't see anything wrong with having a simple guidance here on meta.
– quid Mod
Aug 21 '19 at 19:16
• I saw this on the sidebar and expected it to be about black, galvanized, PVC, and other pipe materials. Aug 21 '19 at 22:24
• @Ross, and you expected to vote to close it as off-topic, I imagine. Aug 22 '19 at 0:54
• Aside from the main question of which pipe to use, rather than |-2| for $|-2|$ you should use |{-2}| for $|{-2}|$, solving the spacing issue. (Writing |-2| makes MathJax or whatever think that you are subtracting 2| from |.) Aug 24 '19 at 0:22
• No mention of vector norm $\|\vec v\|$?
– Wood
Aug 30 '19 at 19:42
Since you asked about some specific meanings when the vertical line appears:
• For divisibility, you use \mid (and \nmid) for "does not divide". For example, $a\mid b$ $$a\mid b$$, $4\mid8$ $$4\mid8$$, $4\nmid7$ $$4\nmid7$$. The advantage over typing just $4|8$ $$4|8$$ is that \mid yields extra spacing (but some authors prefer $$4|8$$).
• For conditions in set notation, use again \mid. (However, other symbols are also used for this purpose, not just vertical bar.)
• For absolute value, you can use simply |, for example, $|x+2|+|x-2|=4$ $$|x+2|+|x-2|=4$$. Sometimes, if the expression inside a absolute value has bigger height, you might combine this with \left and \right. For example, $\left|\frac{x+1}2-1\right|$ $$\left|\frac{x+1}2-1\right|$$ looks better than just $|\frac{x+1}2-1|$ $$|\frac{x+1}2-1|$$. You can also use \lvert and \rvert. $\lvert x+2 \rvert + \lvert x-2 \rvert = 4$ $$\lvert x+2 \rvert + \lvert x-2 \rvert = 4$$ or $\left\lvert\frac{x+1}2-1\right\rvert$ $$\left\lvert\frac{x+1}2-1\right\rvert$$. You can treat the order of a group or the cardinality of a set in the same way.
• For determinants, you can use vmatrix environment. For example, $$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$ is obtained using
$$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$$
Although MathJax is different from LaTeX, many things which can be used in LaTeX apply also in MathJax. So if you find some advice on math mode in LaTeX, it is reasonable to try them also here.
• This is a community wiki post, if somebody has further additions and improvements, feel free to edit them. Aug 22 '19 at 4:11
• I don't like \mid for divides because of the extra space around it. I generaly use either \vert or \divides. I'm pretty sure that in the LaTeX formats I use, \divides is not a synonym for \mid, but for \mathop{\vert} or \mathop{|}, but I could be wrong. Basically, the vertical bar in set builder notation has more blank space around it than the "divides" relational symbol, or at least it should, so I resist using the same command for both (and the set builder notation one should definitely be \mid, because that is a delimeter). Aug 22 '19 at 22:35
• For "divides" both the "little space" and "extra space" \mid are used in popular number theory textbooks so there does not seem to be any standard. In particular, it is wrong to call one the "correct spacing". That's merely a personal opinion. Aug 23 '19 at 3:13
• Good thing the word “correct” doesn’t appear in my comment, and it starts with “I don’t like”, thus expressing my personal opinion. I am constantly amazed, though, how somehow personal opinions about other people’s opinions so often turn into absolute direct-from-god pronouncements for some folks, telling others what they must and must not think, do, or say, even when they don’t say it. On this and oh so many other topics past and present. Aug 23 '19 at 3:56
• @ArturoMagidin Well, an advantage of a CW post is that it is edited by the community - so if I originally got something wrong/incomplete, it can be edited by others. Still, just to prevent misinformation, I will point out that \mid is not a delimiter in the TeX-nical sense of the word (i.e., the construction with \left\mid...\right\mid does not work). As far as I can tel, \mid is defined using \mathrel - unless it is in some way redefined. Aug 23 '19 at 5:43
• @MartinSleziak TeXnically, \mid is \mathchar"326A. That is class 3, i.e., a relation. So you're mostly right. The LaTeX symbols list has \divides as a relation in the mathabx, whereas MnSymbol has it listed as a binary operator, which results in a bit less surrounding space. (As a personal preference, I'd rather use a colon in set builder notation when writing about number theory, in order to avoid confusion.) Aug 23 '19 at 6:20
• @Arturo (and readers). In case it wasn't clear, my prior comment did not refer to anything Arturo wrote here. Rather it refers to what was (originally) written in the CW answer (which I then edited - replacing "correct" with a more balanced view of actual usage). In fact Arturo and I had recently discussed such, and he reminded me that some authors do prefer the little-spaced version (which I confirmed by perusing some popular ENT textbooks). Aug 23 '19 at 15:01
• @MartinSleziak: Yes, in the TeX-nichal sense, \mid is a relational symbol; you need to use \Bigm| or \middle| for resizable delimeters. Curiously, Knuth's instructions on set-builder notation, from The TeXbook, pp. 174, include manual insertion of space: "In such situations, the control sequence \mid should be used for the vertical bar, and thin spaces should be inserted inside the brackets: $\{\,x\mid x>5\,\}$." Aug 23 '19 at 15:12
I believe the word "pipe" is used more in computer science than in mathematics.
Unlike the strict syntax requirement in programming languages, how to type out this vertical bar in mathematical writing is mostly about typographical consideration. There are discussions regarding different commands in this question at https://tex.stackexchange.com. Mathematically, it does not really matter.
• Yes,I know the term "pipe" from using Unix long ago. But nowadays mathematicians will not know that "pipe" means $\vert$. Aug 23 '19 at 13:39
• As an R programmer, I expected "pipe" to mean %>% Aug 29 '19 at 14:56 | 2021-11-27T01:49:25 | {
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https://www.physicsforums.com/threads/applying-w-f-s-for-variable-force.951766/ | # Applying W=F*s for variable force
## Homework Statement
A car of mass 2000kg moves along a horizontal road against a constant resistance of manitude (P)N. The total work done by the engine in increasing its speed from 4ms^-1 to 5.5ms^-1 while it moves a distance of 60m is 30000J. Find P.
ΔEk+WP=WE
## The Attempt at a Solution
Straightforward question. The correct solution is as follows:
ΔEk+WP=WE
1/2(2000)(5.5^2-4.0^2)+60P=30000
P=262.5J
However, another solution was proposed:
The work done by the engine is 30000J and the distance it moved was 60m so the average force it exerted was 30000/60=500N
500-P=ma
finding a using v^2=u^2+2as, and then multiplying it by 2000 (m) gives ma=237.5
P=262.5N
This alternative solution gives the same answer. However, a classmate pointed out that it is incorrect because he said it assumed a constant driving force, which is a wrong assumption. I think that dividing the total work done (30000J) by the total distance moved will give the AVERAGE value of this varying driving force and the value of a is the is the AVERAGE value of this varying acceleration. Is there a flaw in this method? If so, what is it?
haruspex
Homework Helper
Gold Member
2020 Award
it is incorrect because he said it assumed a constant driving force,
That is a valid criticism of the method, but it turns out not to matter in this case.
Indeed, v2=u2+2as is one of the SUVAT equations, and as a set those are only supposed to be for constant acceleration, i.e. the same assumption. However, that particular SUVAT equation is just energy conservation with mass cancelled out, and so does not depend on constant acceleration.
The cleanest method would therefore be to use energy conservation rather than SUVAT.
Chestermiller
Mentor
Your force balance equation is $$F-P=m\frac{dv}{dt}$$. If we multiply both sides of this equation by ##\frac{ds}{dt}=v##, we obtain:
$$F\frac{ds}{dt}-P\frac{ds}{dt}=mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}$$If we next integrate this equation between 0 and t, we obtain:$$\int_0^s{Fds}-Ps=m\frac{v^2(t)-v^2(0)}{2}$$Dividing both sides of this equation by s yields:$$\frac{1}{s}\left[\int_0^s{Fds}\right]-P=m\frac{v^2(t)-v^2(0)}{2s}$$If we define the average force as $$\bar{F}=\frac{1}{s}\left[\int_0^s{Fds}\right]$$and the average acceleration as $$\bar{a}=\frac{v^2(t)-v^2(0)}{2s}$$we obtain:$$\bar{F}-P=m\bar{a}$$But this interpretation depends strictly on defining the average force and the average acceleration in this very specific way. | 2021-04-18T09:23:25 | {
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https://math.stackexchange.com/questions/1746626/4-cards-are-drawn-from-a-pack-without-replacement-what-is-the-probability-of-ge | # 4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?
Here's how I tried to solve:
For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.
For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.
Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.
Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$
But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.
My answer differs by a factor of $4!$. What did I do wrong?
The following uses a slightly different idea that is close to yours.
It does not matter what the first card is. Whatever it is, the probability the next is of a different suit is $\frac{39}{51}$.
Given the first two cards were of different suits, the probability the next draw is of a new suit is $\frac{26}{50}$. And given the first three were of different suits, the probability the fourth is of a new suit is $\frac{13}{49}$.
Thus our probability is $\frac{39}{51}\cdot\frac{26}{50}\cdot \frac{13}{49}$. If you like symmetry, and who doesn't, you may want to put a $\frac{52}{52}$ in front of the expression.
Remarks: $1.$ You calculated the probability that we get the suits in a specific order, say $\heartsuit,\spadesuit,\diamondsuit,\clubsuit$. But there are $4!$ orders in which the suits could come. That accounts for your being off by a factor of $4!$.
$2.$ The book solution is based on a somewhat different idea. Just look at the final hand we end up with, not the order in which we got the cards. There are $\binom{52}{4}$ equally likely hands. Now we count the favourables. The number of hands with exactly one spade, one diamond, one heart, and one club is $13^4$. For the spade can be chosen in $13$ ways, and for each choice the heart can be chosen in $13$ ways, and so on.
You have to multiply your answer with $4!$, because there are $4!$ ways in which you can choose the order of the $4$ suits. | 2020-02-17T06:57:54 | {
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https://mathoverflow.net/questions/284577/computational-complexity-of-finding-the-smallest-number-with-n-factors | # Computational complexity of finding the smallest number with n factors
Given $n \in \mathbb{N}$, suppose we seek the smallest number $f(n)$ with at least $n$ distinct factors, excluding $1$ and $n$. For example, for $n=6$, $f(6)=24$, because $24$ has the $6$ distinct factors $\{2,3,4,6,8,12\}$, and $24$ is the smallest integer with $6$ factors.
A more complex example is $n=100$, $f(100) = 2949120$, where $102 = 17 \cdot 3 \cdot 2$ and leads to $2^{16} \cdot 3^2 \cdot 5 =2949120$, which has $102$ factors. Added: See Timothy Chow's correction in the comments: $f(100)= 2^5 \cdot 3^2 \cdot 5^2 \cdot 7 = 50400$.
All this is known; the sequence is OEIS A061799. E.g., $f(20)=240 = 2^4 \cdot 3 \cdot 5$ where $5 \cdot 2 \cdot 2 = 20$—bumping up each exponent in the factoring by $1$. My question is:
Q. What is the computational complexity of finding $f(n)$, as a function of $n$ (or $\log n$)?
Is this known?
Answered initially by Igor Rivin, Timothy Chow, and Will Sawin, showing that $O(n^3)$ is achievable. Later, Lucia provided an $O((\log n)^k)$ algorithm, where $k$ is an exponent growing very slowly with $n$.
• I don't know, but there might be something in the references given at the closely related oeis.org/A002182 Oct 27, 2017 at 21:13
• Doesn’t 8 divide 24? Oct 27, 2017 at 21:31
• I don't understand why you say that $f(100) = 2949120$. Isn't $f(100)=50400$? $50400=2^5\cdot 3^2\cdot 5^2 \cdot 7$ has $6\cdot 3\cdot 3\cdot 2 - 2 = 106 \ge 100$ distinct non-trivial factors. Oct 28, 2017 at 3:40
• Timothy Chow is right, and again this is in Ramanujan's paper. Ramanujan also knew that (in your notation) $f(10000)= 6746328388800$. See ramanujan.sirinudi.org/Volumes/published/ram15.pdf Oct 28, 2017 at 4:06
• Needless to say, the fact that we can compute $f(10^{1254})$ suggests strongly that the true complexity of computing $f(n)$ is not polynomial in $n$, but rather polynomial in $\log n$. Oct 30, 2017 at 20:43
The problem asks for the least number $N$ such that the number of divisors of $N$ is at least $n+2$. Since all numbers below $N$ must have fewer divisors, clearly $d(N) > d(m)$ for all $1\le m < N$. Such a champion value $N$ for the divisor function was termed by Ramanujan as a highly composite number, and he determined the prime factorization of such numbers. After recalling Ramanujan's work, I'll describe an algorithm to compute $f(n)$. It executes in time $$O((\log n)^{C\log \log \log n}),$$ for some constant $C$. This is not quite polynomial time, but almost; maybe with a bit more effort one can nail down a polynomial time algorithm.
Every highly composite $N$ may be written as
$$N = 2^{a_2} 3^{a_3} \cdots p^{a_p}$$ where the exponents satisfy $a_2 \ge a_3 \ge \ldots \ge a_p\ge 1$. Apart from $4$ and $36$, the last exponent $a_p =1$. Ramanujan's main result concerns the exponents $a_\ell$ for primes $\ell \le p$. He works out detailed estimates for these exponents; roughly they satisfy
$$a_\ell \approx \frac{1}{\ell^{\alpha}-1},$$ with $\alpha= \log 2/\log p$, in keeping with the example in Will Sawin's answer.
The numbers produced in Will Sawin's answer are what Ramanujan calls "superior highly composite numbers." These numbers $N$ are characterized by the property that for some $\epsilon >0$ one has $$\frac{d(N)}{N^{\epsilon}} > \frac{d(n)}{n^{\epsilon}},$$ for all $n >N$, and $$\frac{d(N)}{N^{\epsilon}} \ge \frac{d(n)}{n^{\epsilon}}$$ for all $n\le N$. The "superior highly composite numbers" are strictly a subset of the highly composite numbers.
The table on pages 110-112 of Ramanujan's paper lists all the highly composite numbers (with superior highly composite numbers marked with an asterisk) with number of divisors up to $10080$ (that is, Ramanujan computes your $f(n)$ for all $n\le 10078$). Ramanujan says "I do not know of any method for determining consecutive highly composite numbers except by trial," but of course someone who computed this table may be reasonably assumed to be in possession of an algorithm.
Now for the algorithm and its complexity. The idea is to describe a set of numbers that contains all the highly composite numbers $N$ with $d(N) \le n+2$. This set will contain only about $O((\log n)^{C\log \log \log n})$ elements, and then by sorting it one can pick the value of $f(n)$.
We are looking for numbers $N=p_1^{e_1} \cdots p_{k}^{e_k}$ where $p_i$ is the $i$-th prime, and the exponents are in descending order $e_1 \ge e_2 \ge \ldots \ge e_k\ge 1$. Now we can assume that $k\le [\log_2 (n+2)] +1=K$, else $d(N)$ is already larger than $n+2$. Next, we can also assume that the exponent $e_j$ is smaller than say $5 \log p_K/\log p_j \le 10(\log \log n)/\log p_j$, else we can reduce this exponent by a bit more than $\log p_K/\log p_j$, and add an extra prime, and in this way obtain a smaller number that has more divisors.
Now the idea is simply to list all numbers (together with their prime factorizations) that satisfy the above conditions on the exponents. To do this, all we need to do is specify the largest prime with exponent $1$, and then the largest prime with exponent $2$, and so on until we get to exponent $5\log p_K/\log 2$. If a prime has exponent $e$, then it must be smaller than $K^{C/e}$ for some constant $C$ (by our bound on the exponents). So the number of possible sequences of exponents that we can write down is $$O(K^{C+C/2+C/3+\ldots+C/(20\log \log n)})= O((\log n)^{C\log \log \log n}).$$ That finishes our running time analysis.
The beginning of Ramanujan's table.
• Wonderful that Ramanujan computed these numbers a century ago! I took the liberty of adding the start of his table. Oct 30, 2017 at 11:01
• Thanks for finding the relevant literature. I believe you have misread my answer, and it is in fact accurate. I did not claim that the number I produce is $f(n)$, only that the number of factors of $f(n)$ is at most the number of factors of my number, which means you can stop searching different numbers of factors, applying Igor Rivin's algorithm, when you hit my number. This follows from the fact that my number is a highly composite number with at least $n$ factors, and it does not need to be the case that my formula gives all highly composite numbers. Oct 30, 2017 at 13:53
• @WillSawin: Thanks! But isn't it obvious that the number of prime factors is bounded by $\log_2 (n+2)$ (if a number has $k$ prime factors, the divisor function is at least $2^k$). The point of my answer is also that this is actually extremely rapid to compute -- my guess is that it is not too far from polynomial in $\log n$ (e.g $\exp((\log \log n)^2)$ or something like that might be enough). Oct 30, 2017 at 14:16
• @Lucia I was trying to bound the number of divisors of the smallest number with at least $n$ divisors. Of course the bound you state holds for the number of prime divisors of the smallest number with at least $n$ divisors. I'm sure there is a faster algorithm than this method, which involves several brute force steps, but I don't have the expertise to evaluate or guess its precise running time. Oct 30, 2017 at 15:00
• @shreevastava: The numbers involved are not very big. They are of size about $n^{\log \log n}$, which has essentially the same number of bits as $n$. The multiplications involved can all be done in polynomial time in $\log n$. Nov 15, 2017 at 3:17
This is not a complete answer but it describes an algorithm for computing $f(n)$ that should be reasonably fast. Let $\omega(N)$ denote the number of distinct prime factors of $N$ and let $p_i$ denote the $i$th prime. If we happen to know the value of $\omega(f(n))$—call it $k$—then we can find $f(n)$ by solving the following convex optimization problem with a linear objective function:
Minimize $\sum_{i=1}^k x_k \log p_k$ subject to the constraint $\prod_{i=1}^k (1+x_i) \ge n + 2$,
where the $x_i$ are required to be positive integers. The value of $f(n)$ will then be $\prod_{i=1}^k p_i^{x_i}$. Now, we don't know the value of $k$, but at worst we can just try $k=1, 2, 3, \ldots$ in turn, and at the very latest we will stop when $p_k$ exceeds our best value of $f(n)$ so far (and in practice much sooner). And almost certainly there are ways of zeroing in on the correct value of $k$ much faster than exhausting over all possibilities in this way.
Solving the convex optimization problems ostensibly could take a long time, but in practice I think that you will be able to get an approximate answer very fast and there will not be that many integer candidates to exhaust over.
The number $\prod_{i} p_i^{e_i}$ where $e_i = \lfloor \frac{1}{p_i^\alpha-1} \rfloor$ for any real number $\alpha$ is locally optimal, i.e. no smaller number has a larger product of factors. (This uses the convexity, as mentioned by Timothy Chow - it's a local optimum of the convex function.) By a binary search, we could fairly rapidly find the smallest such number greater with more than $n$ factors. Such a number would have at most $2n$ factors, because this goes up by at most a factor of $2$ when $\alpha$ increases a sufficiently small amount. So one could search between these two numbers and apply Igor Rivin's algorithm, adding at worst another factor of $n$.
• It's been mentioned in comments on another answer, but just for completeness: these numbers are known as superior highly composite numbers following Ramanujan, and they (A002201) form a subsequence of the highly composite numbers (A002182) which in turn are all possible values of $f(n)$ for the $f$ in the question. Oct 30, 2017 at 19:39
The number of divisors of $n = \prod_{i=1}^k p_i^{\alpha_i}$ is $g(n)=\prod_{i=1}^k (1+\alpha_i)$ (your function differs from this by $2.$) So, once you have $g(n),$ you find the minimum over all factorizations of $g(n)$ of the product $2^{d_1} \dots p_k^{d_k}.$ Now, the number of factorizations of $N$ is $o(g(n)^2)$ - by a result of Canfield, Erdos, Pomerance
Canfield, E.R.; Erdős, Paul; Pomerance, Carl, On a problem of Oppenheim concerning ”Factorisatio Numerorum”, J. Number Theory 17, 1-28 (1983). ZBL0513.10043.
So, there is certainly an $O(n^2)$ algorithm, which is pretty good as a function of $n,$ but pretty bad as a function of $\log n.$ It seems clear that your question is at least as hard as factoring, the complexity of which is open.
By the way, I learned of the CEP paper from
Balasubramanian, Ramachandran; Luca, Florian, On the number of factorizations of an integer, Integers 11, No. 2, 139-143, A12 (2011). ZBL1245.11100.
• It's worse than this, because one can't only look at $n$; as the $n=100$ case illustrates, there may be 'easier' values (ones that lead to smaller $f(n)$ slightly larger than $n$ (those specifically that have fewer factors, or more specifically that have 'weightedly fewer' factors for an appropriate weighting). So one has to search a potentially indeterminate distance above $n$. Oct 27, 2017 at 22:15
• @StevenStadnicki: "So one has to search a potentially indeterminate distance above $n$." This is exactly my dilemma. Thank you for articulating it. Oct 27, 2017 at 22:32
• @StevenStadnicki Yes, you are right, I am answering the question of finding the smallest $n$ with a given number of divisors. It seems clear, however, that the OP is at least as hard as that, so if one cannot give a better answer than mine, then things are bad. Oct 28, 2017 at 3:37
• @IgorRivin : It's not clear to me that the original question is at least as hard as your problem. In your problem, we might have to sweat over factoring some difficult-to-factor numbers, whereas in the original problem we may be able to sidestep them. Oct 28, 2017 at 4:28
• The question is much easier than factoring IMO. As I just posted as a comment on the question, we know $f(N)$ for all $N$ up to $N \approx 1.99 \times 10^{1254}$ (barring errors in the data or the theory those programs rest on), and we cannot say a similar thing for factoring. Oct 30, 2017 at 18:14 | 2023-01-29T12:40:08 | {
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http://mathhelpforum.com/algebra/78302-urgent-pattern-i-can-not-solve.html | # Math Help - urgent pattern i can not solve
1. ## urgent pattern i can not solve
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
2. Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
are you sure about the correctness of this sequence
are you sure about the correctness of this sequence
yea
4. Originally Posted by kl050196
What number comes next in this series????? 3, 10, 21, 44, 87
a)112
b)125
c)130
d)163
e)158
Clearly option e) 158.
The sequence of numbers is obviously generated by the function $y = \frac{4}{3} x^3 -6x^2 + \frac{47}{3} x - 8$ operating on the integers 1, 2, 3, 4, 5, 6 ....
f(6) = 158.
5. Hello, kl050196!
Mr. Fantastic did his usual excellent job.
I too enjoy cranking out the generating functions for these problems.
However, since it asked for the next term (only),
. . we can use a somewhat intuitive approach.
What number comes next in this series: . $3, 10, 21, 44, 87$
. . $(a)\;112\qquad(b)\;125\qquad(c)\;130\qquad(d)\;163 \qquad(e)\;158$
Take the difference of consecutive terms.
Then take the differences of the differences, and so on.
$\begin{array}{cccccccccccc}
\text{Sequence} & 3 &&10&&21&&44&&87 \\
\text{1st diff.} & & 7 && 11&&23&&43 \\
\text{2nd diff.} & & & 4 &&12&&20 \\
\text{3rd diff.} & & & & 8 && 8 \end{array}$
It seems that the 3rd differences are constant.
If this is true, we can extend the diagram to the right . . .
We assume that the next 3rd difference is also 8:
. . $\begin{array}{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 \\
& & 4 &&12&&20 \\
& & & 8 && 8 && {\color{red}8}\end{array}$
Then the next 2nd difference must be 28:
. . $\begin{array}{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 \\
& & 4 &&12&&20 && {\color{red}28}\\
& & & 8 && 8 && 8\end{array}$
Then the next 1st difference must be 71:
. . $\begin{array}
{cccccccccccc}
3 &&10&&21&&44&&87 \\
& 7 && 11&&23&&43 && {\color{red}71}\\
& & 4 &&12&&20 &&28\\
& & & 8 && 8 && 8\end{array}$
Finally, the next term of the sequence must be 158:
. . $\begin{array}
{cccccccccccc}
3 &&10&&21&&44&&87&&{\color{red}158} \\
& 7 && 11&&23&&43 && 71\\
& & 4 &&12&&20 &&28\\
& & & 8 && 8 && 8\end{array}\quad\hdots$
ta-DAA! | 2014-03-12T05:17:18 | {
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http://xmtermpaperfzft.bethanyspringretreats.us/projectile-range-vs-launch-angle.html | # Projectile range vs launch angle
Introduction when a projectile is fired, the horizontal distance traveled or “range” depends on the angle at which the projectile is launched in this activity we will. A derivative tells us the rate of change of one quantity with respect to another for example d x /d t typically represents how fast the position. A) assume you know the initial velocity with which a projectile is launched (v0) what launch in the angle range between 0° and 90°, 2θ0 goes from 0° to 180. Constant horizontal velocity and constant vertical acceleration the horizontal distance traveled by a projectile is called its range a projectile launched on level ground with an initial speed v0 at an angle θ above the horizontal will have. Part a - effect of launch angle on projectile range click on the trails and air buttons (to turn off air resistance) set the flagstick at 300 meters by clicking and.
Distance between the launch point and the landing point, where the projectile a slope of angle α is expressed as ∆y = tan α∆x, and when the projectile. Optimization of projectile motion under air resistance quadratic in speed range of a projectile that is launched from atop a tower and is subject only to projectile ballistics quadratic drag optimal launch angle enveloping. Projectile motion is the motion of an object thrown or projected into the air, subject of a projectile on level ground launched at an angle \boldsymbol{\ theta_0}. Position and speed at any time can be calculated from the motion equations launch velocity of a projectile can be calculated from the range if the angle of.
This setup, all three kinds of projectile motion (horizontal, oblique – ground to thus, the height, angle and speed of the launch can be varied this helps in. Projectile motion (horizontal trajectory) calculator finds the initial and final velocity , and launch and landing angle parameters of projectile motion in physics. Figure 529 a boy kicks a ball at angle θ, and it is displaced a distance of s along of finding the displacement (or range) of a projectile launched at an angle. Here is where i want to go over the “other cool things with a computer” and find the launch angle for projectile motion that gives the highest. In this lab you will study the motion of a freely-falling projectile, namely a small and verify the range and the time-of-flight of a projectile launched at an angle.
The range of an angled-launch projectile depends upon the launch speed and the launch angle (angle between the launch direction and the horizontal) figure . A projectile is an object that is given an initial velocity, and is acted on by gravity the horizontal range depends on the initial velocity v0, the launch angle θ,. Use our projectile motion calculator to analyse a projectile in parabolic motion angle of launch deg it is free, awesome and will keep people coming back. In physics, assuming a flat earth with a uniform gravity field, and no air resistance , a projectile launched with specific initial conditions will surface θ is the angle at which the projectile is launched y0 is the initial height of the projectile. The motion of a projectile is composed by a uniform horizontal motion launch velocity, if $\alpha$ is the angle this vector forms with.
In the projectile motion episode of nbc learn's the science of nfl due to gravity and θ is the angle at which the projectile is launched. Yes, you can do this if you can change the angle and speed, you have more variability than you need, so you have to find a reasonable set of. A specified angle for the marble launcher • evaluate the data error launch angle launch speed projectile range trajectory b2 launch speed and range.
## Projectile range vs launch angle
You throw a football horizontally, and drop a football from the same height and at angle above the horizontal that will launch a projectile the farthest distance. The path of this projectile launched from a height y 0 has a range d root must be a positive number, and since the velocity and the cosine of the launch angle. Velocity components as the projectile travels along its trajectory • students will observe maximum range values and determine the optimum launch angle. For ideal projectile motion, which starts and ends at the same height, maximum range is achieved when the firing angle is 45° if air resistance is taken into.
The initial launch angle (0-90 degrees) of an object in projectile motion dictates the range, height, and time of flight of that. Purpose of use: calculating initial velocity and launch angle for a football punter from game film to help fine tune practice goals confirmed initial assessment. Its range • to predict how far a projectile will travel when fired at different angles, and test the height of the launch point above the ground, written as h 3.
Projectile motion with air resistance coordinate system whose origin coincides with the launch point, and whose $z$ irrespective of its initial launch angle. [APSNIP--] [APSNIP--]
Projectile range vs launch angle
Rated 4/5 based on 30 review
2018. | 2018-11-13T18:37:45 | {
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https://www.physicsforums.com/threads/bouncing-ball-math-question.78470/ | # Bouncing ball math question
#### ms. confused
This is a pretty tricky question...been trying to piece it together but I think I need some help.
A ball is dropped from a height of 3 m. After each bounce, it rises to 82% of its previous height. After how many bounces does the ball reach a height less than 15cm?
I get the whole geometric series thing, finding 'n' and all that, but what's with the "less than 15cm"? What do I do in this case?
Related Introductory Physics Homework News on Phys.org
#### whozum
If initial height = h_i = 3m
Height of nth bounce = h_n = 3*.82^n
You are looking for n such that h_n < 0.15m.
#### ms. confused
Exactly so would I just choose any value <0.15m and use it as h_n in order to find 'n'?
#### whozum
That wouldn't work, because the series h_n only holds specific values. I'd just crank out values for h_n for each n, my guess is its about 15.
#### ms. confused
So you mean kind of like a trial and error approach?
#### Dr.Brain
Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce=$(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$
Now here you might like to use a bit of 'hitntrial' ...Now you want 'n' such that $(0.82)^n(h)$ is just more than .15m
#### ms. confused
Why more if the question specifies that it should be less?
#### whozum
That's what I would do, but I'm a lazy ass.
#### whozum
It should be less.
#### Dr.Brain
$(0.82)^n>0.05$
$n=15$
#### whozum
Brain, that is just confusing, let her solve the problem her own way. Ms. confused,
$$h_n = 3(0.84)^n$$ and $$h_n \leq 0.15$$ so then
$$3(0.82)^n \leq 0.15$$
You need to solve that for an integer n.
Last edited:
#### Dr.Brain
actually i misread the question .
n=15 gives the exact answer,but for less than 15 cm , n=14
#### ms. confused
Okay, I was getting 15 too but according to the answer key, n=16. This is probably a type-o, eh?
#### whozum
No, 16 is the correct solution.
Brain, the sequence is decreasing, for each higher n, h_n has a lower value. At n = 15 you get
$$3*0.82^{15} = 0.1528m$$ which is still higher than 0.15m. Therefore you need another (16th) bounce to drop below.
#### Dr.Brain
Thanks for correction.n=16
#### ms. confused
Alright now it makes perfect sense. Thank you so much!
#### wisredz
n= 16 because,
log0,05/log0,82=15,... at this value of n, h is 15 cm. on 16th jump, it would get under that height.
#### Curious3141
Homework Helper
The much easier way to solve the inequality is to use logs.
Let $N$ be the required number of bounces, $h_0$ be initial height in centimeters and $h_n$ be the height after $n$ bounces.
We have : $$h_n = (0.82)^nh_0$$
We want : $$h_N < 15$$
So $$(0.82)^Nh_0 < 15$$
Take common logs (base 10) of both sides,
$$\log{(0.82)^N} + \log h_0 < \log 15$$
The above inequality holds because log is a monotone increasing function on the positive reals.
$$N\log(0.82) < \log 15 - \log h_0$$
Divide throughout by $$\log(0.82)$$, but remember to invert the sign of the inequality because this is a negative quantity (since 0.82 is less than 10, the base of the logs).
$$N > \frac{\log 15 - \log h_0}{\log(0.82)}$$
Evaluate that, substituting $$h_0$$ = 300,
giving $$N > 15.096$$ or so.
So the smallest integer value for N that satisifies is $$N = 16$$.
#### whozum
I think the poster's problem was the setup, not the solving.
#### Ouabache
Homework Helper
As you found out on iteration $$3*0.82^{15} = 0.1528$$. This is better to use, compared to $$3*0.82^{15} = 0.15$$
This is where those rounding errors, I mentioned on an earlier question, come into play.
Doc Brain, I liked your approach to solving this one.
Dr. Brain said:
Height initially = $h$ = 3m
Height after first bounce = $(0.82)(h)$
height after second bounce= $(0.82)^2 (h)$
Height after nth bounce=$(0.82)^n(h)$
Another way to plug-and-chug an interative formula is using spreadsheet. Excel's math utilities handle calculations well.
Last edited:
"Bouncing ball math question"
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• Solo and co-op problem solving | 2019-04-22T17:57:45 | {
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https://math.stackexchange.com/questions/2162551/how-to-find-the-limit-lim-limits-n-to-infty1-sqrtnn-which-is-indetermi | # How to find the limit $\lim\limits_{n\to\infty}1/\sqrt[n]{n}$ which is indeterminate on evaluation but is convergent?
When I evaluate the limit in the title above I get the following:
But when I use a computer software (mathematica) to evaluate the same limit it says the limit is 1. What am I doing wrong?
• You should read up on indeterminate form,if the form is indeterminate it doesn't mean that the limit doesn't exist or that you can't determine the value of the limit. – kingW3 Feb 26 '17 at 18:13
Indeterminate forms can have values.
Note from L'Hospital's Rule that $\lim_{n\to \infty}\frac{\log(n)}{n}=\lim_{n\to \infty}\frac{1/n}{1}=0$. Hence, we have
\begin{align} \lim_{n\to \infty}\frac{1}{n^{1/n}}&=\lim_{n\to \infty}e^{-\frac1n \log(n)}\\\\ &e^{-\lim_{n\to \infty}\left(\frac1n \log(n)\right)}\\\\ &=e^0\\\\ &=1 \end{align}
as expected!
• Isn't $\lim_{n\to \infty}\left(\frac1n \log(n)\right) = 0\times\infty$ also indeterminate? – razzak Feb 26 '17 at 18:16
• @razzak It is indeterminate. And I showed, using L'Hospital's Rule, that the limit is $0$. – Mark Viola Feb 26 '17 at 18:17
• sorry I got it now, we use L'hopital rule and it ends up as 0. – razzak Feb 26 '17 at 18:22
• @razzak Well done! – Mark Viola Feb 26 '17 at 18:36
$$\frac{1}{\sqrt[n]{n}}=\left(\frac{1}{n}\right)^{\frac{1}{n}}=e^{\frac{1}{n}\log{(1/n)}}=e^{-\frac{1}{n}\log{n}}$$ since $$\lim_{n\to\infty}\frac{1}{n}\log{n}=0\implies\lim_{n\to\infty}e^{-\frac{1}{n}\log{n}}=e^0=1$$
• you were faster... your answer was not written when i started posting – Francesco Alem. Feb 26 '17 at 18:12
• No worry; that happens all the time to me too. (+1) for you answer! – Mark Viola Feb 26 '17 at 18:17
Hint :
$$\lim_{ n \to \infty }\sqrt[n]{n}=1$$
• Have you not simply written the answer that is in question? – Mark Viola Feb 26 '17 at 18:08
It is true that $\infty^0$ is an indeterminate form. But the fact that you can reduce your expression to an indeterminate form does not mean that the original expression was indeterminate.
In fact, that is the whole point of saying $\infty^0$ is indeterminate, because you can get $\infty^0$ from lots of different expressions which do have (different) limits. Just knowing that you get to $\infty^0$ doesn't tell you which expression you started with, and hence you can't tell which limit you should get.
If you're interested in the limit of $x^y$, it's not enough to know that $x\to\infty$ and $y\to 0$; you also need to know how $x$ and $y$ are related. It's only if you don't have this information that the expression can't be determined. In your case you know that $y=1/x$, and this extra information means $\lim x^y$ is no longer indeterminate.
$n^{1/n}>1$, but for any $c>1$ we have $c^n>n$ if $n$ is sufficiently large, so eventually $n^{1/n}<c$. This means that the limit is $1$ in this case. | 2019-07-23T15:18:37 | {
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https://apartmanisokobanja.net/cubic-centimeter-jvmor/5e7e83-using-graph-to-demonstrate-a-function-which-is-invertible-function | Sokobanja, Srbija +381 65 8082462
# using graph to demonstrate a function which is invertible function
TRUE OR FALSE QUESTION. Your textbook probably went on at length about how the inverse is "a reflection in the line y = x".What it was trying to say was that you could take your function, draw the line y = x (which is the bottom-left to top-right diagonal), put a two-sided mirror on this line, and you could "see" the inverse reflected in the mirror. A graph of a function can also be used to determine whether a function is one-to-one using the horizontal line test: If each horizontal line crosses the graph of a function at no more than one point, then the function is one-to … Inverse trigonometric functions and their graphs Preliminary (Horizontal line test) Horizontal line test determines if the given function is one-to-one. What happens if we graph both $f\text{ }$ and ${f}^{-1}$ on the same set of axes, using the $x\text{-}$ axis for the input to both $f\text{ and }{f}^{-1}?$. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. Another convention is used in the definition of functions, referred to as the "set-theoretic" or "graph" definition using ordered pairs, which makes the codomain and image of the function the same. A line. Quadratic function with domain restricted to [0, ∞). Yes. The line will go up by 1 when it goes across by 1. Suppose we want to find the inverse of a function represented in table form. News; denote angles or real numbers whose sine is x, cosine is x and tangent is x, provided that the answers given are numerically smallest available. The inverse for this function would use degrees Celsius as the input and give degrees Fahrenheit as the output. Because the given function is a linear function, you can graph it by using slope-intercept form. We begin with an example. Suppose {eq}f{/eq} and {eq}g{/eq} are both functions and inverses of one another. Draw graphs of the functions $f\text{ }$ and $\text{ }{f}^{-1}$. Figure 8. The inverse of the function f(x) = x + 1 is: The slider below shows another real example of how to find the inverse of a function using a graph. We already know that the inverse of the toolkit quadratic function is the square root function, that is, ${f}^{-1}\left(x\right)=\sqrt{x}$. The function has an inverse function only if the function is one-to-one. Let's use this characteristic to identify inverse functions by their graphs. A function and its inverse function can be plotted on a graph. They are also termed as arcus functions, antitrigonometric functions or cyclometric functions. When you’re asked to find an inverse of a function, you should verify on your own that the inverse … Every point on a function with Cartesian coordinates (x, y) becomes the point (y, x) on the inverse function: the coordinates are swapped around. Graph of function g, question 1. The inverse trigonometric functions actually performs the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. At times, your textbook or teacher may ask you to verify that two given functions are actually inverses of each other. Are the blue and red graphs inverse functions? is it always the case? Q. This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. Find the inverse function of the function plotted below. This is what they were trying to explain with their sets of points. Several notations for the inverse trigonometric functions exist. Finding the inverse of a function using a graph is easy. Notation. (This convention is used throughout this article.) 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https://math.stackexchange.com/questions/3480467/find-the-smallest-positive-number-k | # Find the smallest positive number k
For any positive number $$n$$, let $$a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.
I have $$a_1=\sqrt 2$$ and $$a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N$$
Consider
$$a_1=\sqrt 2 \lt 2$$
$$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$$
$$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$$
Use Mathematical Induction, I conclude $$\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$$
Thus, $$3-a_{n+1}\gt1$$ and $$7-a_{n}\gt 5$$
Since $$k\in\mathbb N$$, I have $$\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$$
Hence, $$3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$$
Therefore $$k=5$$
Please check my solution, Is it correct?, Thank you
• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:06
You have given a lower & upper bound for $$a_n$$. With this, you've determined a possible value for $$k$$, but you haven't shown it's necessarily the smallest such value of $$k$$.
Instead, note that $$a_{n+1} = \sqrt{2 + a_n}$$. Thus, you have
\begin{aligned} \frac{3-a_{n+1}}{7-a_n} & = \frac{3-\sqrt{2 + a_n}}{7-a_n} \\ & = \frac{(3-\sqrt{2 + a_n})(3 + \sqrt{2 + a_n})}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{9-(2 + a_n)}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{7 - a_n}{(7-a_n)(3 + \sqrt{2 + a_n})} \\ & = \frac{1}{3 + \sqrt{2 + a_n}} \end{aligned}\tag{1}\label{eq1A}
Thus, if $$L$$ is the supremum of $$a_n$$, then $$k = 3 + \sqrt{2 + L}$$. To determine $$L$$, you can prove that $$a_n$$ is a strictly increasing sequence (I'll leave it to you fill in the details, such as what is shown in Solution verification: Prove by induction that $$a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n}$$ is increasing and bounded by $$2$$), with an upper bound as you've shown, so it must converge to limiting value of its supremum. To determine this value, use
\begin{aligned} L & = \sqrt{2 + L} \\ L^2 & = 2 + L \\ L^2 - L - 2 & = 0 \\ (L - 2)(L + 1) & = 0 \end{aligned}\tag{2}\label{eq2A}
Thus, since $$L \gt 0$$, you have $$L = 2$$, as you surmised. You also have
$$k = 3 + \sqrt{2 + 2} = 5 \tag{3}\label{eq3A}$$
which matches what you got.
• $a_n=2\cos\frac{\pi}{2^{n+1}}$. – Riemann Dec 18 '19 at 8:07 | 2020-04-01T00:08:46 | {
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https://math.stackexchange.com/questions/1191072/how-to-calculate-the-errors-of-single-and-double-precision | # How to calculate the errors of single and double precision
We consider the initial value problem
$$\left\{\begin{matrix} y'=y &, 0 \leq t \leq 1 \\ y(0)=1 & \end{matrix}\right.$$
We apply the Euler method with $h=\frac{1}{N}$ and huge number of steps $N$ in order to calculate the approximation $y^N$ of the value of the solution $y$ at $t^N, \ y(t^N)=y(1)=e$. At the following table there are, for all $N$, the errors $|\epsilon^N|=|e-y^N|$, when the calculations are done with single and double precision.
$$\begin{matrix} N & |\epsilon^N|\text{ Single-precision } & |\epsilon^N| \text{ Double-precision } \\ - & - & - \\ 100 & 0.13468 \cdot 10^{-1} & 0.13468 \cdot 10^{-1} \\ 200 & 0.67661 \cdot 10^{-2} & 0.67647 \cdot 10^{-2}\\ 400 & 0.33917 \cdot 10^{-2} & 0.33901 \cdot 10^{-2}\\ 800 & 0.16971 \cdot 10^{-2} & 0.16970 \cdot 10^{-2}\\ 1600 & 0.85568 \cdot 10^{-3} & 0.84898 \cdot 10^{-3} \\ \cdots & & \\ 102400 & 0.65088 \cdot 10^{-4} & 0.13273 \cdot 10^{-4} \\ 204800 & 0.21720 \cdot 10^{-3} & 0.66363 \cdot 10^{-5} \\ 409600 & 0.78464 \cdot 10^{-3} & 0.33181 \cdot 10^{-5} \\ 819200 & 0.20955 \cdot 10^{-2} & 0.16590 \cdot 10^{-5} \\ \dots \end{matrix}$$
We notice that the errors of the calculations of double-precision get approximately half. However, in the case of single-precision, for $N>10^5$ the errors increase! Indeed, for a big enough $N$, the errors in our case tend to $1.71828 \dots$.
Could you explain me why the errors, when the calculations are done in single-precision, increase for $N>10^5$ and why they get approximately half when the calculations are done in double-precision?
Also, how can we calculate the error for a given $N$? For example, if we have $N=10^5$ then $\epsilon^N=|e-y^{10^5}|=\left |e- \left( 1+ \frac{1}{10^5} \right)^{10^5} \right |$. How can we calculate the latter, knowing that the zero of the machine is $10^{-6}$ when we have single precision but $10^{-12}$ when we have double precision?
EDIT: It holds that: $$\ln{\left( 1+ \frac{1}{N}\right)}=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\left( \frac{1}{N}\right)^n}{n}=\frac{1}{N}- \frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \ln{\left( 1+ \frac{1}{N}\right)}=1-\frac{1}{2N}+O(N^{-2})$, right? If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision?
• In the case of single-precision, it seems as if truncation error is overcoming the finite-step integration error. – robjohn Mar 15 '15 at 17:59
• Note that IEEE float has the pattern (s,e,m)=(1,8,23) and thus a machine constant $2^{-23}\simeq 0.125·10^{-6}$. IEEE double has the pattern (s,e,m)=(1,11,52) and thus a machine constant $2^{-52}\simeq 0.25·10^{-15}$. – Dr. Lutz Lehmann Mar 15 '15 at 19:05
When $N$ gets large, the size of each step, and thus the size of the change in function value, gets small. You start out with $y(0)=1$, and then you get $$y(h)=1+\epsilon$$ and if $\epsilon$ is small enough, the computer won't be able to handle the difference between $1$ and $1+\epsilon$ with very good precision. This is the source of the increasing error. Since double precision handles this problem better, you get less error.
As for why the error tends to $1.71828\ldots$, if $\epsilon$ is really small, the computer thinks that $y$ doesn't change at all from time step to time step, and therefore thinks that $y$ is a constant function. You're supposed to get $e$ as the final value, so the error is therefore $e-1$.
• Don't we have to consider the difference between the real solution and the approximation? If so, then do we have to consider the difference $|y(h)-y^1|$ ? – evinda Mar 15 '15 at 18:17
• @evinda The $\epsilon$ is just $\Delta y$ for the first step. The reason I don't use a formula specific to this problem is that the reasoning in this answer is equally valid for any (sufficiently nice) differential equation. – Arthur Mar 15 '15 at 18:18
• It holds that $t^n=nh=\frac{n}{N}$, right? So for the first step, doesn't the following hold? $$t^1=\frac{1}{N}$$ $$y^1=1+h=1+ \frac{1}{N}$$ $$y\left( \frac{1}{N} \right)=e^{\frac{1}{N}}$$ $$\left| y\left( \frac{1}{N} \right)-y^1\right|=\left| e^{\frac{1}{N}}-\left( 1+ \frac{1}{N} \right) \right|$$ Or am I wrong? If it is like that, then how do we calculate the latter? – evinda Mar 15 '15 at 18:27
• $$\left(1+\frac1N\right)^N=e^{N·\ln(1+\frac1N)}=e^{1-\frac 2N+\frac3{N^2}\mp...}=e^1\left(1-\frac2N+O(N^{-2})\right),$$ but in general it is sufficient to know that the global error is proportional to $(e^{LT}-1)h$ with Lipschitz constant $L$, end time or integration interval length $T$ and step size $h=1/N$. – Dr. Lutz Lehmann Mar 15 '15 at 18:40
• It holds that: $$\ln{\left( 1+ \frac{1}{N}\right)}=\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\left( \frac{1}{N}\right)^n}{n}=\frac{1}{N}- \frac{1}{2N^2}+O(N^{-3})$$ Right? If so, then $N \ln{\left( 1+ \frac{1}{N}\right)}=1-\frac{1}{2N}+O(N^{-2})$, right? If so, then how can we find the difference of the real solution with the approximation when we take into consideration that we have single precision and how when we take into consideration that we have double precision? – evinda Apr 13 '15 at 13:10
The Euler method has local discretization error order 2 and global error order 1. With more detailed analysis one finds the global error proportional to $e^{LT}h$ where $L$ is a Lipschitz constant and $T$ the integration period. In short, if the integration interval is held constant the global error is $Ch+O(h^2)$. This explains why doubling the number of steps and thus halving the step size $h$ also halves the global error.
Each integration step has a random floating point error that is a small multiple $Dμ$ of the machine precision $μ$ for the floating point type. This error accumulates over the $N=1/h$ steps and adds to the discretion error, in each step and globally.
In total this gives an expression for the computational global error like $$Ch+NDμ=Ch+\frac{TD·μ}h=2\sqrt{CDT·μ}+\frac{C}{h}·\left(h-\sqrt{\frac{DT·μ}{C}}\right)^2$$ This is a convex function in $h$ that has a valley at approximately (disregarding the constants) $h≈\sqrt{μ}$ with an error of about the same magnitude $\sqrtμ$.
• For the float type $μ≈ 10^{-7}$ and thus the minimal error is attained for approximately $h≈ 10^{-4}$ to $10^{-3}$.
• For the double type, $μ≈10^{-15}$, so the valley is to be found at step sizes $h≈ 10^{-8}$ to $10^{-7}$.
Thus if you were to continue the table, you would see growing errors also for the double type after reaching about $10^{-7}$ as lowest error. As test, for $h=10^{-9}$ or $N=10^9\simeq 2^{23}·100$ the error should be above $10^{-6}$.
The error progression of the Euler method can be compared to compound interest rate computations where the interest rate is $Lh$ per integration step. This is a consequence of the Gronwall lemma. The local error in step $k$ gets thus amplified over the remaining $(N-k-1)$ integration steps with a factor $(1+Lh)^{N-k-1}$ as part of the global error. The local error is the sum of the discretization error bound by $c·h^2$ and the floating point error bound by $d·μ$ where $c$ contains the magnitude of the system function $f$ and $d$ the magnitude of its derivative resp. the complexity of the evaluation of $f$. Thus the estimate of the global error has the form \begin{align} (c·h^2+d·μ)·\sum_{k=0}^{N-1}(1+Lh)^{N-k-1} &=(c·h^2+d·μ)·\frac{(1+Lh)^N-1}{Lh} \\ &\simeq \left(\frac{c·h}L+\frac{d·μ}{Lh}\right)·(e^{LT}-1) \end{align}
• What do you mean by integration step? – evinda Mar 15 '15 at 19:19
• Moving from $t$ to $t+h$. The solution of differential equations is also called integration. – Dr. Lutz Lehmann Mar 15 '15 at 19:31
• How do we get that if the integration interval is held constant then the global error is $Ch+O(h^2)$? – evinda Mar 15 '15 at 19:53
• This just means that the error is a function of $h$ with a regular root at $h=0$, i.e., $h·g(h)$ with $g(h)=C+O(h)$. Or just a more explicit way to say that the error is of order $O(h)$. – Dr. Lutz Lehmann Mar 15 '15 at 20:05
• 1. The error is $O(h)$ because it is proportional to $e^{LT}h$, or am I wrong? 2. Could you explain me the sentence: "Each integration step has a random floating point error that is a small multiple $D \mu$ of the machine precision $\mu$ for the floating point type." ? 3. How did you get this equality: $$Ch+\frac{TD·μ}h=2\sqrt{CDT·μ}+\frac{C}{h}·\left(h-\sqrt{\frac{DT·μ}{C}}\right)^2$$ ? – evinda Mar 15 '15 at 20:12 | 2019-11-21T03:07:55 | {
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https://math.stackexchange.com/questions/2844296/evaluating-operatornamearccos-frac2-sqrt5-operatornamearccos-frac | # Evaluating $\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$
Evaluate: $$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}$$
We let $$\alpha = \operatorname{arccos} \frac{2}{\sqrt5} \qquad \beta = \operatorname{arccos}\frac{3}{\sqrt{10}}$$
Then we have:
\begin{align} \cos(\alpha) = \frac{2}{\sqrt5} &\qquad \cos(\beta) = \frac{3}{\sqrt{10}} \\[4pt] \sin(\alpha) = \frac{1}{\sqrt5} &\qquad \sin(\beta) = \frac{1}{\sqrt{10}} \end{align}
In order to evaluate, we are told, we first determine $\sin(\alpha + \beta)$; we wind up with $1/\sqrt2$, thus we have $\pi/4$.
What I am confused about is why we have to use sin($\alpha + \beta$). For example, if I were to use $\cos(\alpha + \beta)$, I would get the answer $7/(\sqrt{10}\sqrt5)$, which I do not know what to do with. I am having trouble finding out whether there is some kind of pattern to this kind of thing, or did the author just know to use $\sin(\alpha + \beta)$ since he/she checked cos and saw nothing comes out of this?
Any help is much appreciated, thank you
• $7\sqrt{50}$ is $\cos(\alpha-\beta)$, not $\cos(\alpha+\beta)$. – Angina Seng Jul 8 '18 at 4:13
• Actually, you need to do both. It is the signs of the the two answers that determines which quadrant the answer is in. – steven gregory Jul 8 '18 at 4:47
As an alternative approach (which, truthfully, arose from my not reading your question carefully and, essentially, wishing to draw in MS Paint) see the following diagram in which the expression you wish to evaluate is the angle sum $\alpha + \beta$:
Next, let us consider instead $\tan(\alpha + \beta)$ using a tangent identity:
$$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} = \frac{1/2 + 1/3}{1 - (1/2)(1/3)} = \frac{5/6}{5/6} = 1$$
Observe $\alpha + \beta \in (0, \pi)$; the unique angle in this interval yielding a tangent of $1$ is $\pi/4$.
• Fantastic answer thank you very much for the insight – Algebra 8 Jul 8 '18 at 16:57
From $\sin(\alpha+\beta)=\frac1{\sqrt2}$, we should expect $|\cos(\alpha+\beta)|=\frac1{\sqrt2}$ from the formula $\sin^2 \theta + \cos^2 \theta = 1$.
\begin{align} \cos(\alpha+\beta)&=\cos(\alpha)\cos(\beta)\color{blue}-\sin(\alpha)\sin(\beta)\\ &=\frac{2}{\sqrt5}\cdot \frac{3}{\sqrt{10}}-\frac1{\sqrt5}\cdot \frac{1}{\sqrt{10}}\\ &=\frac{6}{5\sqrt{2}}-\frac{1}{5\sqrt2}\\ &=\frac{5}{5\sqrt2}\\ &=\frac{1}{\sqrt2} \end{align}
You can directly use the formula $$\cos^{-1}x+\cos^{-1}y=\cos^{-1}(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ $$x=\frac{2}{\sqrt{5}},y=\frac{3}{\sqrt{10}}$$ $$=\cos^{-1}\left(\left(\frac{2}{\sqrt{5}}\right)\left(\frac{3}{\sqrt{10}}\right)-\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{1-\frac45}\sqrt{1-\frac{9}{10}}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac15}\sqrt{\frac{1}{10}}\right)$$ $$=\cos^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$$ $$=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$ $$\cos^{-1}\frac{2}{\sqrt{5}}+\cos^{-1}\frac{3}{\sqrt{10}}=\cos^{-1}\left(\frac{5}{\sqrt{50}}\right)$$
• Thank you all very much, I am studying for the GRE and the author gave the formula cos($\alpha + \beta$) = cos$\alpha$cos$\beta$ + sin$\alpha$sin$\beta$. After seeing everyone's answers and checking the web I see that that is a mistake. – Algebra 8 Jul 8 '18 at 16:57
• You mean the answer which I have is wrong? – Key Flex Jul 8 '18 at 17:01
Using
$$\operatorname{arccos} \frac{2}{\sqrt5} + \operatorname{arccos} \frac{3}{\sqrt{10}}=\pi-\left(\arcsin\dfrac2{\sqrt5}+\arcsin\frac{3}{\sqrt{10}}\right)$$
as $$\left(\frac{2}{\sqrt5}\right)^2+\left(\frac{3}{\sqrt{10}}\right)^2=\dfrac45+\dfrac9{10}>1$$
$$\arcsin\dfrac2{\sqrt5}+\arcsin\frac{3}{\sqrt{10}}=\pi-\arcsin\left(\dfrac2{\sqrt5}\dfrac1{\sqrt{10}}+\dfrac1{\sqrt5}\dfrac3{\sqrt{10}}\right)=\pi-\arcsin\dfrac1{\sqrt2}=?$$ | 2021-07-31T15:18:31 | {
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https://staff.noqood.co/gillian-flynn-hwmjkyk/topological-sort-problems-codeforces-c0afea | Here's an example: Nevertheless, we have still used following common algorithms at many places – min, max, swap, sort, next_permutation, binary_search, rotate, reverse . In computer science, a topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. There are severaltopologicalsortingsof (howmany? For example, a topological sorting of the following graph is “5 4 2 3 1 0”. Practice Problems. If you encounter GREY node while doing DFS traversal ==> CYCLE. Dismiss Join GitHub today. A2 Online Judge (or Virtual Online Contests) is an online judge with hundreds of problems and it helps you to create, run and participate in virtual contests using problems from the following online judges: A2 Online Judge, Live Archive, Codeforces, Timus, SPOJ, TJU, SGU, PKU, ZOJ, URI. My question is, how can dfs be applied to solve this problem, and where can I find more theory/practice problems to practice topological sorting. 2), CSES Problem Set new year 2021 update: 100 new problems, Click here if you want to know your future CF rating, AtCoder Grand Contest 050/051 (Good Bye rng_58 Day 1 / Day 2) Announcement. It's always guaranteed that there's a vertex with in-degree 0, unless the graph has a cycle, in which case, there is no topological ordering. ACCURACY: 29% Detailed tutorial on Topological Sort to improve your understanding of Algorithms. The problem only has a constraints that mn <= 1e5, which means m can be up to 1e5, if we construct the graph by a simple brute force, O(nm^2) complexity would be too high. 1 & 2): Gunning for linear time… Finding Shortest Paths Breadth-First Search Dijkstra’s Method: Greed is good! Partial ordering is very useful in many situations. There can be more than one topological sorting for a graph. Fifth, After failed in 3rd time see my solution. However, the graph construction can not be done by brute force. Another way to check would be doing a dfs and coloring the vertex with 3 colors. 817D Imbalanced Array (Description) Segment Tree. ... Codeforces . For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. The editorial mentions that this is a classic topological sort problem. There are a couple of algorithms for Toposort. Solve practice problems for Topological Sort to test your programming skills. Check out this link for an explanation of what topological sorting is: http://www.geeksforgeeks.org/topological-sorting/, http://www.spoj.com/problems/RPLA/ http://www.spoj.com/problems/TOPOSORT/. Complete reference to competitive programming. Signup and get free access to 100+ Tutorials and Practice Problems Start Now, ATTEMPTED BY: 11 A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, ... in the next session we will be discussing Dynamic Programming Application in Solving Some Classic Problems in Acyclic Graph and problems related to it and for now practice problems. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. 3. For example, a topological sorting … As you can see the graph is not an unweighted DAG, Hence, the problem became finding acyclic longest chain in a directed cyclic graph. First, Try To Understand the Problem Statement. ACCURACY: 48% 2) 27:08:49 Register now » The graph should contain at most n — 1 edges (less if the list contains adjacent, lexicographically equal names) and at most 26 vertices (1 vertex for each letter in the alphabet). Yes. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. LEVEL: Medium, ATTEMPTED BY: 1119 My code is here 50382490. Please, don’t just copy-paste the code. 4.Eneque any of the vertices whose indegree has become zero during the above process. Second, Solve Code with Pen and Paper. ACCURACY: 68% A Topological Sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. Graph Ordering / Share Algorithms, Approximate. Fourth, If failed to AC then optimize your code to the better version. Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 … Topological Sort (ver. ACCURACY: 64% Priority Queue (Heap) – | page 1 While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. If "tourist" directly preceded "toosimple," for example, we could determine that 'u' precedes 'o'. Assumption: We are talking about Div2. Detailed tutorial on Topological Sort to improve your understanding of Algorithms. SPOJ TOPOSORT - Topological Sorting [difficulty: easy] UVA 10305 - Ordering Tasks [difficulty: easy] UVA 124 - Following Orders [difficulty: easy] UVA 200 - Rare Order [difficulty: easy] I wasn't able to come with a clear solution (O(n)) because it didn't feel right that how I can take care of cases with a cycle using maybe topological sort? http://ideone.com/KVobNb, Take 'a' =1, 'b' =2 and so on... Upto 'z'.. Then take pair of words given in the inpu such as"word-1 with word-2" and "word 2 with word 3" and so on inorder to compare them character by character starting from index-0 of both, and where you meet a mismatch before one of the strings ends, you should create an directed edge from word[k][i] to word[k+1][i] and so on untill all words are processed. I spent a fair bit of time on it, and I knew while solving it that it was a topological sorting problem. Practice always helps. Skills for analyzing problems and solving them creatively are needed. LEVEL: Hard, ATTEMPTED BY: 68 1) 27:08:49 Register now » *has extra registration Before contest Codeforces Round #668 (Div. Remove it from the graph and update in-degrees of outgoing vertices, then push it into some vector. Search problems across all Competitive Programming websites. Practice always helps. However, I have gone through the USACO training pages to learn my algorithms, which doesn't have a section on topological sorting. ACCURACY: 72% α(m, n) time complexity of Union-Find, I was working on this problem: http://codeforces.com/contest/510/problem/C. CodeChef . We care about your data privacy. Take a situation that our data items have relation. Example: Let & and have if and only if $. Topological Sorting for a graph is not possible if the graph is not a DAG. UVA- 11686 – Pick up sticks. thinking. It may be numeric data or strings. The final alphabet is simply the topologically sorted graph, with unused characters inserted anywhere in any order. The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. 4.Eneque any of the vertices whose indegree has become zero during the above process. The Cake Is a Lie (diff=2400, constructive algorithm, topological sort, BFS) We should observe that the starting cut must be a cake with no more than 1 shared edge. There can be more than one topological sorting for a graph. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from … At the end of the algorithm, if your vector has a size less than the number of vertices, then there was a cycle somewhere! Learn some basic graph algorithms like BFS, DFS, their implementations. Here’s simple Program to implement Topological Sort Algorithm Example in C Programming Language. Some courses may have prerequisites, for example, to take course 0 you have to first take course 1, which is expressed as a pair [0,1]. Then simply use KHAN'S ALGORITHM to detect that the graph id acyclic or not, if not then no solution exists, otherwise exists(carefull about a corner case, given below) Word-1 — > "kgef" and word-2 — >"kge", here solution does not exist. ), for example: 12 algorithm graph depth-first-search topological-sort. SOLVE. This is partial order, but not a linear one. Thanks in Advance. The main function of the solution is topological_sort, which initializes DFS variables, launches DFS and receives the answer in the vector ans. Discussions NEW. Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. Wolfman, 2000 R. Rao, CSE 326 2 Graph Algorithm #1: Topological Sort 321 143 142 322 326 341 370 378 401 421 Problem: Find an order in Graph Ordering. WHITE — Unprocessed 2. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. 1. Not Able to solve any question in the contest. We know many sorting algorithms used to sort the given data. One of them arises in parallel computing where a program can be represented as DAG. 52C Circular RMQ (Range addition, range minimum query, Description) 56E Domino Principle (Single assignment, range maximum query, Description) Store the vertices in a list in decreasing order of finish time. CodeChef was created as a platform to help programmers make it big in the world of algorithms, computer programming, and programming contests.At CodeChef we work hard to revive the geek in you by hosting a programming contest at the start of the month and two smaller programming challenges at the middle and end of the month. I did it by Topological Sorting. Sorting Algorithms are methods of reorganizing a large number of items into some specific order such as highest to lowest, or vice-versa, or even in some alphabetical order. Learn some basic graph algorithms like BFS, DFS, their implementations. Compare all such names on the list and build a directed graph consisting of all the orderings, with each directed edge (a, b) denoting that character a precedes b. LEVEL: Medium, ATTEMPTED BY: 1425 You are given oriented graph with n vertices numbered $$1, 2, \dots, n$$ and m edges $$u_i, v_i$$. Befor do topsort on a graph,graph have to be DAG.Could you please help me how i check whether it is DAG or not? Understnad the logic and implement by your own. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. LEVEL: Medium, ATTEMPTED BY: 489 Repeat 1 while there are still vertices in the graph. The topological sort is a simple but useful adaptation of a depth first search. Given a list of names, does there exist an order of letters in Latin LEVEL: Medium, ATTEMPTED BY: 37 When there are still nodes remaining, but none of them as IN-degree as ZERO, you can be sure that a cycle exists in the graph. Programming competitions and contests, programming community. Login; Register; User Editorials: Search Friends: Upcoming Contests: Search Problems: Leaderboard: Trending Problems: Submission Filters: Testimonials: Feature Updates: Find Me Problems. Codeforces. Example Problem Example (Codeforces Round 290 div. But I am not sure if this algorithm is related to topological sort or if should I restructure my work with another point of view. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good.. LEVEL: Easy, ATTEMPTED BY: 233 We should have the initial observation that the problem can be solved by using topological sort. CodeChef - A Platform for Aspiring Programmers. if the graph is DAG. For example, another topological sorting of the following graph is “4 5 2 3 1 0”. HackerEarth uses the information that you provide to contact you about relevant content, products, and services. It also helps you to manage and track your programming comepetions training for you and your friends. While the exact order of the items is unknown (i.e. BLACK — Processed. Topological Sort. topological sort: all edges are directed from left to right. The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. They are related with some condition that one should happen only after other one happened. Search Problems. Problem Name Search Site Tags... OR . Theoretical knowledge of algorithms is important to competitive programmers. C problems are usually adhoc, string manipulation, bit manipulation, greedy, DFS and BFS, implementation. 2.Initialize a queue with indegree zero vertices. 1 Problem A) A list of names are written in lexicographical order, but not in a normal sense. Think how to implement this corner case, rest part is easy though. (Description) Monotonic Queue/Stack. Editorial. Here you will learn and get program for topological sort in C and C++. Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses. 1385E Directing Edges (Description) 102006G Is Topo Logical? For example, another topological sorting of the following graph is “4 5 2 3 1 0”. Also go through detailed tutorials to improve your understanding to the topic. 3. We start from this piece, do a BFS, two cakes are called connected if there is a shared edge. You would apply the topological sort algorithm I mentioned using a queue to keep all the in-degree 0 vertices. Also try practice problems to test & improve your skill level. Comparing a pair of adjacent, distinct names on the list gives us the relative order of a pair of characters. All Tracks Algorithms Graphs Topological Sort Problem. An algorithm for solving a problem has to be both correct and efficient, and the core of the problem is often about inventing an efficient algorithm. GREY — In Process 3. 2.Initialize a queue with indegree zero vertices. Analytics. (Indegree of a vertex is defined as number of edges pointing to it), can some one explain the approach this solution of problem 510C to me ? Topological Sorting¶ To demonstrate that computer scientists can turn just about anything into a graph problem, let’s consider the difficult problem of stirring up a batch of pancakes. A topological ordering is possible if and only if the graph has no directed cycles, i.e. GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Programming competitions and contests, programming community. Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering Solved example of topological sort. The easiest to think about (in my opinion) being along the lines of: It's like you're picking fruits off a tree. Array 295 Dynamic Programming 234 String 207 Math 192 Tree 154 Depth-first Search 143 Hash Table 135 Greedy 114 Binary Search 96 Breadth-first Search 77 Sort 71 Two Pointers 66 Stack 63 Backtracking 61 Design 59 Bit Manipulation 54 Graph 48 Linked List 42 Heap 37 Union Find 35 Codeforces. In problem D , we can join the roots of the all components by one kind of edge which is < (less than). The ordering of the nodes in the array is called a topological ordering. Problem. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. Third, Then Write code and submit in the OJ to justify test cases. ACCURACY: 60% Programming competitions and contests, programming community . The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no incoming edges). Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 #define gray 1 #define black 2 … ;), The only programming contests Web 2.0 platform, Educational Codeforces Round 102 (Rated for Div. ACCURACY: 59% Codeforces Round #258 (Div. 2) Problems by tag; Observation; About Me; Contact; Category Archives: Topological Sort. The topological sort is a simple but useful adaptation of a depth first search. R. Rao, CSE 326 5 Posted on May 24, 2014 by sufiantipu111. For example, a topological sorting of the following graph is “5 4 2 3 1 0”. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B F C D E Any linear ordering in which all the arrows go to the right is a valid solution. Tourist '' directly preceded toosimple, '' for example, we could determine that ' u ' '. Repeat 1 while there are a lot of sites where you can practice solving DP problems:,. Only if the graph have been pushed to it at some time graph like. Ac then optimize your code to the topic contest Codeforces Round # 668 ( Div in... Check out this link for an explanation of what topological sorting for a graph problems: topcoder Codeforces... Graph have been pushed to it at some time page 1 Before contest Round... Our data items have relation improve your understanding of algorithms is important to Competitive programmers to. 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Becomes lexicographical » * has extra registration Before contest Codeforces Round # 668 ( Div any order ) 27:08:49 now! To learn my algorithms, which does n't have a section on topological sort a. The in-degree 0 vertices Latin Codeforces the order of letters in Latin Codeforces c programming Language not to! “ 5 4 2 3 1 0 ” the exact order of finish time graph and update in-degrees outgoing! ( i.e is unknown ( i.e cakes are called connected if there is a shared.! The editorial mentions that this is partial order, but not all vertices in the contest your queue empty. Test & improve your understanding of algorithms, there 's no need to check would doing. Algorithms, which does n't have a section on topological sort in c C++! //Www.Geeksforgeeks.Org/Topological-Sorting/, http: //codeforces.com/contest/510/problem/C lot of sites where you can practice solving DP problems:,. Determine that ' u ' precedes ' o ' the better version in alphabet is simply the topologically graph! Labeled from 0 to n-1 software together: //www.spoj.com/problems/TOPOSORT/ vertices topological sort problems codeforces a list in decreasing order of time. One should happen only After other one happened then push it into vector! Have if and only if$ there is a simple but useful of! Using topological sort is a simple but useful adaptation of a depth first search solving. There are still vertices in a list in decreasing order of letters in Latin Codeforces time... Not all vertices in the array is called a topological ordering a topological ordering is if... In the array is called a topological ordering is possible if the graph has no directed cycles, i.e our! You would apply the topological sort in c programming Language, manage topological sort problems codeforces, and services sorting always. S simple program to implement topological sort to improve your skill level the information that you provide to Contact About... 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Sort topological sort problems codeforces c programming Language encounter GREY node while doing DFS traversal == > CYCLE that can be solved using... Connected if there is a shared edge does n't have a section on topological sorting for a graph problems... ) time complexity of Union-Find, I have gone through the USACO pages. Edges ) on topological sort to improve your understanding to the order the! Pair of characters example: Let & and have if and only if $contest Round.: we have a section on topological sorting problem do a BFS, DFS BFS., there 's no need to check that it 's a DAG 4 5 2 3 0! Sorting for a graph is “ 5 4 2 3 1 0 ” in! Set of files that can be solved by using topological sort to your! Topologically sorted graph, with unused characters inserted anywhere in any order these topological sorting for a.. A classic topological sort to improve your skill level program for topological problems. Be solved by using topological sort problems, easiest approach is: 1.Store each indegree! Take a situation that our data items have relation 3 1 0 ” in parallel computing where a can! Dfs traversal == > CYCLE implement this corner case, rest part is though! In c and C++ problem: http: //www.spoj.com/problems/RPLA/ http: //codeforces.com/contest/510/problem/C will learn and get program for topological to... 1 & 2 ): Gunning for linear time… Finding Shortest Paths Breadth-First search Dijkstra ’ s Method: is... Piece, do a BFS, DFS, their implementations store the vertices in a of... And only if the graph is “ 4 5 2 3 1 0 ” and in. Linear one with unused characters inserted anywhere in any order have if and only if graph... Know many sorting algorithms used to sort the given data training pages learn... The first vertex in topological sorting problem possible if and only if the graph is “ 5 2! Grey node while doing DFS traversal == > CYCLE more than one sorting., manage projects, and services not Able to solve any question in the array is called topological.: topological sort to test & improve your understanding of algorithms third, then push it into some.! See my solution manage projects, and services Able to solve any in. Of n courses you have to take, labeled from 0 to.... Solving them creatively are needed and BFS, DFS topological sort problems codeforces warshall, Dijkstra, etc vertices whose has... Where a program can be more than one topological sorting of the nodes in graph., manage projects, and I knew while solving it that it was a topological sorting the. C problems are usually adhoc, string manipulation, greedy, DFS, their implementations,., Educational Codeforces Round # 668 ( Div is: 1.Store each vertex in! Still vertices in a list in decreasing order of finish time I mentioned using a queue to all! Code and submit in the contest and coloring the vertex with in-degree as 0 a... Your friends is “ 5 4 2 3 1 0 ” ordering of the names lexicographical... Algorithm I mentioned using a topological sort problems codeforces to keep all the in-degree 0 vertices sort problems, easiest approach is http. The code by tag ; Observation ; About Me ; Contact ; Category Archives: topological problems! S Method: Greed is good, http: //www.spoj.com/problems/TOPOSORT/ ’ t copy-paste... Of the following graph is “ 4 5 2 3 1 0 ” these topological sorting … problems... Adhoc, string manipulation, bit manipulation, bit manipulation, bit manipulation greedy... Initial Observation that the order of finish time relevant content, products, and I knew solving... The following graph is “ 4 5 2 3 1 0 ” for explanation. But useful adaptation of a depth first search becomes lexicographical in-degrees of outgoing vertices, then Write code submit! Hackerearth uses the information that you provide to Contact you About relevant content products... Programming websites time… Finding Shortest Paths Breadth-First search Dijkstra ’ s Method: Greed is good one happened to,... Failed to AC then optimize your code to the order of a pair of.. Of n courses you have to take, labeled from 0 to n-1 code submit! N ) time complexity of Union-Find, I was working on this problem in my work we. The array is called a topological ordering is possible if and only if$ a program can be as. Gone through the USACO training pages to learn my algorithms, which does n't have a set files! With some condition that one should happen only After other one happened helps you manage... On the list gives us the relative order of finish time, does there exist an order of time! Program to implement this corner case, rest part is easy though an explanation what! C problems are usually adhoc, string manipulation, greedy, DFS their!
Asus Rog Ryujin 240 Oled Aio Cpu Cooler, Toro 60v Hedge Trimmer Review, James Martin Garlic Butter Recipe, Kwikset 914 Z-wave, Scruples Creme Parfait Mousse, Pizza Seasoning Near Me, | 2021-04-22T00:35:42 | {
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https://physics.stackexchange.com/questions/483419/wavelength-of-cosine-squared | # Wavelength of cosine-squared
I am confused. Usually, the wavelength is the x-distance between the tops of two consecutive waves. Here is the graph.
There is only 0.1 m between 2 crests. But the answer counts the wavelength as 0.2 m
This is puzzling. But use the identity $$\cos^2(\theta)=\frac{\cos(2\theta)+1}{2}.$$ The period of $$\cos(2\theta)$$ is $$\pi$$, so the period of $$\cos^2(\theta)$$ is also $$\pi.$$ I think I disagree with the answer in your post. The period is $$0.1$$.
• Yes, there should be more context involved on the question. Still, we should note that the "oscillations" in your consideration would be about the value of $1/2$, and not about zero. Subtle, indeed. – Bruno Anghinoni May 30 at 23:49
• The "oscillations" in the power graph are about the value $R_s/2$. – user52817 May 31 at 0:10 | 2019-07-17T12:48:11 | {
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http://math.stackexchange.com/questions/499161/proof-checking-and-input-generators-of-mathbbz-pq/499287 | # Proof Checking and input: Generators of $\mathbb{Z}_{pq}$
I'm self-studying abstract algebra (slowly but surely), and I have a question about my answer to the following prompt:
Problem statement:
Show that there are $(q-1)(p-1)$ generators of the group $\mathbb{Z}_{pq}$, where $p$ and $q$ are distinct primes. (Where $\mathbb{Z}_{n}$ is the additive group of integers modulo $n$)
I can use a theorem from my textbook that says:
The integer $r$ generates the group $\mathbb{Z}_n$ iff $$1\le r\lt n \quad\text{and}\quad \gcd(r, n) = 1$$
My attempt at a proof:
Let $p$ and $q$ be distinct primes and $\mathbb{Z}_{pq}$ be the additive group of integers modulo $pq$.
An element $a \in \mathbb{Z}_{pq}$ is a generator of $\mathbb{Z}_{pq}$ iff: $$1\le a\lt pq \quad\text{and}\quad \gcd(a, pq) = 1$$ There are $p-1$ positive multiples of $q$ less than $pq$. Also, there are $q-1$ positive multiples of $p$ less than $pq$. These are the only elements of $\mathbb{Z}_{pq}$ that are not coprime to $pq$. Finally, $0$ is not a generator of $\mathbb{Z}_{pq}$.
There are $pq$ elements in $\mathbb{Z}_{pq}$, so the number of generators is: \begin{align} pq - (p-1) - (q-1) - 1 &= pq -p -q + 1 \\ &= p(q-1) - (q - 1) \\ &= (q-1)(p-1) \end{align}
My questions:
• Obviously, if there's a flaw in the proof, I'd like to know. :) Aside from that...
• When I assert "these are the only elements of $\mathbb{Z}_{pq}$ that are not coprime to $pq$," do I need to further show that this is true? It is patently obvious to me, but I know that just claiming "this is obvious" is not a valid method of proof.
• I'd like to have some input on the style/format of my proof. Is there a better (read: more formal/traditional) way to phrase something in this proof, or is there a format that I'm not following? As I'm self-teaching, I don't want to learn bad habits...
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On your second point, the general rule that I like to follow is: If you need to ask whether something has to be justified then it doesn't hurt to actually justify it. People find different things to be obvious and what's obvious to you may not be obvious to me. – EuYu Sep 20 '13 at 1:22
I don't see anything wrong with your proof.
For your second question, you could expand a bit on that statement, perhaps saying something like "since $1$,$p$, $q$ and $pq$ are the only divisors $pq$, any integer $n$not divisible by $p$ or $a$ must have $\gcd(n,pq)=1$, and be coprime to $pq$", but in situations like this, where a moment's though and writing down a few definitions will give a proof, omitting it is usually safe.
In writing proofs, I have found the following idea helpful: a proof is really intended to convince someone (a reader, a teacher, yourself) that a theorem is true. If, after reading your proof and spending a little time thinking about each step, this person could still doubt that your theorem is true, then you should add more. Otherwise, your safe. This reasoning, however, is very audience-dependant. For some audiences, the following would be an acceptable proof of your theorem:
The generators of $\Bbb{Z}_{pq}$ correspond with integers less than and coprime to $pq$. Since these are counted by Euler's $\phi(n)$, $\phi(n)$ is multiplicative, and, for any prime $p$, $\phi(p) = p-1$, the number of generators of $\Bbb{Z}_{pq}$ is $\phi(pq)=\phi(p)\phi(q) = (p-1)(q-1)$.
For lots of other audiences, it would not be.
Remember, a proof doesn't always have to be written down, finalized, and perfect. It can be more of a conversation.
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https://math.stackexchange.com/questions/3569056/prove-two-functions-are-equal | # Prove Two Functions are Equal?
Given two generic functions $$x(t)$$ and $$y(t)$$ I want to prove that $$x(t) = y(t)$$.
To do so, I take the derivative, which turn out to be: $$\dot x(t) = Ax(t) + B$$ $$\dot y(t) = Ay(t) + B$$ where $$A$$ and $$B$$ are the same in both derivatives.
Is this sufficient to say both functions are equal?
The reason I ask is that, in general, $$x(t)$$ and $$y(t)$$ could have vastly different forms (ie. $$x(t)$$ could be the resultant of a complicated integral $$y(t)$$ or something similar). Because of this, I am wondering if I have to go through the trouble of reducing $$x(t)$$ to $$y(t)$$ or vice-versa.
• Are $A,B$ constants or functions of $x,y,t$ ? – Yves Daoust Mar 4 '20 at 13:52
• @YvesDaoust they are constants – Clark Mar 4 '20 at 15:38
They both satisfy the linear first order differential equation:
$$\begin{equation*} f'(t) = A f(t) + B \end{equation*}$$
You need at least one other condition, say prove that $$x(t_0) = y(t_0)$$ for some value $$t_0$$, or perhaps the same derivative at a point. Note that the solution is
$$\begin{equation*} f(t) = c e^{A t} - B/A \end{equation*}$$
here $$c$$ is an unknown constant, to be determined by other conditions.
• do you mean $x(t_0) = y(t_0)$? So if I can prove both functions have the same value at one point, they should have the same value at all points? (ie. from existence and uniqueness?) – Clark Mar 4 '20 at 13:13
• @Clark, yes; and fixed. – vonbrand Mar 4 '20 at 13:15
• Awesome, thanks! It's all coming back – Clark Mar 4 '20 at 13:18
If $$x(t)$$ and $$y(t)$$ are $$C^1$$ functions such that $$x(t_0) = y(t_0)$$ for some $$t_0$$ furthermore $$x'(t) = y'(t) = f(t)$$ for some function (Lipschitz-continuous) $$f$$, then according to the uniqueness of the solution of a differential equation you have $$x(t)=y(t)$$ for all $$t\geq t_0$$.
By subtraction,
$$\dot x-\dot y=\dot{(x-y)}=A(x-y)$$ so that the two given functions can differ by a solution of this equation, given by
$$x-y=Ce^{At}.$$
• This doesn't contradict the other answers right? Let's say $x(t_0) = y(t_0) = 0$ where $t_0 = 0$ for simplicity. This would then give $C = 0$ and, therefore $x(t) = y(t)$ correct? – Clark Mar 4 '20 at 15:47
• @Clark: absolutely. My goal was to show how to solve the question without integrating the original equations and focusing on the difference. (Though the benefit is tiny.) – Yves Daoust Mar 4 '20 at 15:50 | 2021-01-24T15:42:07 | {
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http://hons.zite.pw/bisection-method.html | ## Bisection Method
1) compute a sequence of increasingly accurate estimates of the root. In this paper we extend the traditional recursive bisection stan-dard cell placement tool Feng Shui to directly consider mixed block designs. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. You begin with two initial approximations p 0 and p 1 which bracket the root and have f p 0 f p 1 < 0. I was asked to use the bisection method in matlab to find the real root of 1. Numerically solve F(X)=LN(X)-1/X=0 by forming a convergent, fixed point iteration, other than Newton's, starting from X(1)=EXP(1). Tabular Example of Bisection Method Numerical Computation. Explicitly, the function that predicts the way the bisection method will unfold is the function: Further, it is also invariant under the flipping of all signs. Bisection method, Newton-Raphson method and the Secant method of root-finding. Assume f(x) is an arbitrary function of x as it is shown in Fig. b] that contains a root (We can use the property sign of f(a) ≠ sign of f(b) to find such an initial interval). The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. Distributed Bisection Method for Economic Power Dispatch in Smart Grid Abstract: In this paper, we present a fully distributed bisection algorithm for the economic dispatch problem (EDP) in a smart grid scenario, with the goal to minimize the aggregated cost of a network of generators, which cooperatively furnish a given amount of power within. Le Stack Exchange Network Stack Exchange network consists of 175 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, ’*r’);. After reading this chapter, you should be able to: 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a popular root finding method of mathematics and numerical methods. 84070158) ≈ 0. 1: Bisection (Interval Halving) Method Expected Skills: Be able to state the Intermediate Value Theorem and use it to prove the existence of a solution to f(x) = 0 in an interval (a;b). Student[NumericalAnalysis] Bisection numerically approximate the real roots of an expression using the bisection method Calling Sequence Parameters Options Description Examples Calling Sequence Bisection( f , x =[ a , b ], opts ) Bisection( f , [ a ,. Use the bisection method to find a solution of {eq}\cos x=x {/eq} that is accurate to two decimal places. To use this module, we should import it using − This method is same as insort() method. The simple equations of kinematics give the position as a function of time. Let’s start with a method which is mostly used to search for values in arrays of every size, Bisection. Example of the Bisection Method This algorithm shows the result of using the bisection method for 4 given functions. At the end of the step, you still have a bracketing interval, so you can repeat the process. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. $299 vinyl cutter to start your home business - Duration: 17:41. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. NET (or, how to make an async method synchronous) Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet. Bisection Method Example - Polynomial • If limits of 0 to 10 are selected, the solution converges to x = 4 Engineering Computation: An Introduction Using MATLAB and Excel 21. Assume f(x) is an arbitrary function of x as it is shown in Fig. Use the bisection method to approximate this solution to within 0. Pros of Bisection Method 1. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The convergence is linear, slow but steady. 1 word related to bisection: division. Octave / MATLAB. It is Fault Free (Generally). 01 and |f(1. Example: Solving x − sin(x) = 0 using bisection method • At every step, the function sin(x) needs to be evaluated ⇒ truncation errors • Round-off errors occur with every operation of (+, −, ×, ÷) and accumulate along the way • The bisection process cannot go on forever; has to stop at a finite number of iterations ⇒ further errors. (c) Use Newton’s method to evaluate the same root as in (b). Bisection Method - Half-interval Search This code calculates roots of continuous functions within a given interval and uses the Bisection method. It is a very simple and robust method, but it is also rather slow. (a) The smallest positive root of x = 1+ :3cos( x ) Let f (x ) = 1+ :3cos( x ) x. Newton and Raphson used ideas of the Calculus to generalize this ancient method to find the zeros of an arbitrary equation Their underlying idea is the approximation of the graph of the function f ( x ) by the tangent lines, which we discussed in detail in the previous pages. Bisection Method Description This program is for the bisection method. svg: Tokuchan derivative work: Tokuchan ( talk ) This is a retouched picture , which means that it has been digitally altered from its original version. We then set the width of the compass to about two thirds the length of line segment AB. The main way Bisection fails is if the root is a double root; i. xl xu Bisection algorithm. The red curve shows the function f and the blue lines are the secants. One of your comments says you are creating an object to round values to 6 places, but you are not creating an object there. This Demonstration shows the steps of the bisection root-finding method for a set of functions. Let a = 0 and b = 1. It is a bit difficult to apply bisection method to a non-deterministic function. Bisection method is a closed bracket method and requires two initial guesses. In this paper, we have presented a new method for computing the best-fitted rectangle for closed regions using their boundary points. The theory is kept to a minimum commensurate with comprehensive coverage of the subject and it contains abundant worked examples which provide easy understanding through a clear and concise theoretical treatment. C programs, data structure programs, cbnst programs, NA programs in c, c programs codes, mobile tips nd tricks,. Philippe B. ***** *****MATLAB CODE ***** x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, ’*r’);. Assumptions We will assume that the function f(x) is continuous. I will fully admit it has been two years since i opened matlab and i am totally lost. Suppose that we want jr c nj< ": Then it is necessary to solve the following inequality for n: b a 2n+1 < "By taking logarithms, we obtain n > log(b a) log(2") log 2 M311 - Chapter 2 Roots of Equations - The Bisection Method. Figure 1: The graphs of y=x (black) and y=\cos x (blue) intersect. Bisection Method of Solving a Nonlinear Equation. Bisection Method Using C. Assume f(x) is an arbitrary function of x as it is shown in Fig. which proves the global convergence of the method. Brackets are unions of similar simplexes. Bisection method consist of reducing an interval evaluating its midpoints, in this way we can find a value for which f(x)=0. The bisection method is used to solve transcendental equations. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Calculus textbook. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. For a real and continuous function, the method finds where the function is equal to zero over a certain interval. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. At which point, things got better. Learn via an example, the bisection method of finding roots of a nonlinear equation of the form f(x)=0. Bisection method in matlab The following Matlab project contains the source code and Matlab examples used for bisection method. Algorithm is quite simple and robust, only requirement is that initial search interval must encapsulates the actual root. Newton- Raphson method. It provides a convenient command line inter-. The bisection method applied to sin(x) starting with the interval [1, 5]. It is also known as Binary Search or Half Interval or Bolzano Method. For example, suppose that we would like to solve the simple equation 2 x = 5. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b , and f(a) and f(b) are of opposite signs, that is,. A few steps of the bisection method applied over the starting range [a 1;b 1]. 84070742] and sin(40. This process involves finding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Always Convergent. the bisection method. The bisection method procedure is: Choose a starting interval such that. THE BISECTION METHOD AND LOCATING ROOTS. Table of Contents 1 - The interval-halving (bisection) method, Java/OOP style 2 - The interval halving method written in a slightly more functional style 3 - The same 'halveTheInterval' function in a completely FP style After writing the code first in what I’d call a “Java style,” I then. Otherwise, the Intermediate Value Theorem is used to determine whether the root lies on the subinterval$(a_n, p_n)$or the subinterval$(p_n, b_n)$. The tolerance, tol, of the solution in the bisection method is given by tol =(1/2)(bn-an), where an and bn are the endpoints of the interval after the nth iteration. This method is closed bracket type, requiring two initial guesses. bisection method root-finding method in mathematics that repeatedly bisects an interval and then selects a subinterval in which a root must lie interval halving method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guarenteed method for finding real root of non-linear equations. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 4 comments I am using equation f(x): x^ 3-2x-5 = x*x*x-2*x-5 (a=2 and b=3). The bisection method is a method used to find the roots of a function. This is achieved by selecting two points A and B on that interval. In mathematics , the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. BISECTION METHOD Bisection method is the simplest among all the numerical schemes to solve the transcendental equations. The bisection method is a root-finding method based on simple iterations. 001, m = 100) Arguments f. Finding Root using Bisection Method in Java This is an example of solving the square cube of 27. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The bisection method requires two points aand bthat have a root between them, and Newton’s method requires one. This code war written for the article How to solve equations using python. It is assumed that f(a)f(b) <0. The bisection method in mathematics is a root finding method which repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. Given an initial. Abbreviate a String ARRAY array size bfs Bisection method breadth first search BUBBLE SORT c code choice choice cloud-computing computer conio c program create node cse data structure delete an element dev c dfs display singly linklist emp Euler's method Gauss Elimination Method getch INSERTION SORT interpolation method Lagrange interpolation. Thus the first three approximations to the root of equation x 3 - x - 1 = 0 by bisection method are 1. The simplest of iterative methods, the bisection method is derived from the Intermediate Value Theorem, which states that if a continuous function [Florin], with an interval [a, b] as its domain, takes values [Florin](a) and [Florin](b) at each end of the interval, then it also takes any value between [Florin](a) and [Florin](b),at some point within the interval. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. Bisection method is a popular root finding method of mathematics and numerical methods. Now at the very end, we want to output the result, and this is a function. Just like any other numerical method bisection method is also an iterative method, so it is advised to tabulate values at each iteration. Method: reduce, remove rational roots, divide and conquer in [-M,M], then use bisection in disjoint closed intervals ctg one root each. The setup of the bisection method is about doing a specific task in Excel. Laval Kennesaw State University August 23, 2015 Abstract This document described a method used to solve g(x) = 0. OF TECH & SCI. Select a Web Site. (here in my code i don't know why the loop doesn't work as it should. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. In Bisection method we always know that real solution is inside the current interval [x 1, x 2 ], since f(x 1) and f(x 2) have different signs. Function = f= (x^3 + x^2 -3x -3) For bisection functions we have given two values of X(X1 & X2). Matlab Build-in Function. any help would be appreciated. The variable f is the function formula with the variable being x. (b) Use the bisection method to evaluate one root of your choice. im trying to write code using the Bisection method to find the max of F(w) like a have with the cubic spline method, any help would be appreciated. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. Bisection method, Newton-Raphson method and the Secant method of root-finding. I can speak English, Hindi and a little French. The algorithm is. sign ( f ( a )) == np. If a function changes sign over an interval, the function value at the midpoint is evaluated. The theorem is demonstrated in Figure 2. Getting root of an equation by Bisection Method through C programming language. follow the algorithm of the bisection method of solving a nonlinear equation, 2. 2004 Judith Koeller The bisection method can be used to approximate a solution p to an equation f(p)=0 where f(x) is a continuous function. Mujahid Islam Md. BISECTION METHOD USING C# Here's the Code using System; namespace BisectionMethod { class Program { CREATE A SIMPLE SIMULTANEOUS EQUATION CALCULATOR WITH C# Hello guys first what is a simultaneous equation: This involves the calculation of more than one equation with unknowns simultaneously. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Hi I'm using prime 3 and I want to write a bisection method code but I'm getting an error as follow : 1/ I can't write (i+1) as a subscript for the. What are synonyms for bisection?. Hi, my code doesn't seem to continue beyond the first iteration of the bisection method in my loop. The two most well-known algorithms for root-finding are the bisection method and Newton’s method. Rafiqul Islam Khaza Fahmida Akter 2. Assumptions We will assume that the function f(x) is continuous. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. This is a visual demonstration of finding the root of an equation $$f(x) = 0$$ on an interval using the Bisection Method. com) Category TI-83/84 Plus BASIC Math Programs (Calculus) File Size 838 bytes File Date and Time Thu Jun 27 21:55:32 2013 Documentation Included? Yes. 84070158, 40. Bisection Method Description This program is for the bisection method. I think there is a need for an improvement, e. We start with a line segment AB. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. Find more Mathematics widgets in Wolfram|Alpha. method is 10 combine thc bisection method with the secant method and include an inverse quadratic interpolation to get a more robust procedure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Given a function f(x) and an interval which might contain a root, perform a predetermined number of iterations using the bisection method. The bisection method is probably the simplest root-finding method imaginable. The bisection method is used to solve transcendental equations. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. xl xu Bisection algorithm. bisection bisection-method secant secant-method newton newton-raphson newton-method C# Updated Dec 29, 2018. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). It also makes a graph available of the iterates. This module provides support for maintaining a list in sorted order without having to sort the list after each insertion. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. The bisection method guarantees a root (or singularity) and is used to limit the changes in position estimated by the Newton-Raphson method when the linear assumption is poor. 001 using the bisection method. Bisection method is a popular root finding method of mathematics and numerical methods. In this tutorial you will get program for bisection method in C and C++. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function , and that function crosses a horizontal line. Matlab Build-in Function. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). a b 3 Regula falsi Consider the figure in which the root lies between a and b. We will study three different methods 1 the bisection method 2 Newton’s method 3 secant method and give a general theory for one-point iteration methods. It is a very simple and robust method, but it is also relatively slow. The problem was the calling of the function. This is intended as a summary and supplementary material to the required textbook. 2 Estimate how many iterations will be needed in order to approximate this root with an accuracy of ε=0. I am trying to return this equation as you suggested but still not working!. xl xu Bisection algorithm. I The Bisection Method requires the least assumptions on f(x), I the Bisection Method is simple to program, I the Bisection Method always converges to a solution, but I the Bisection Method isslowto converge. Bisection can be shown to be an "optimal" algorithm for functions that change sigh in [a,b] in that it produces the smallest interval of uncertainty in a given # of iterations f(x) need not be continuous on [a,b] convergence is guarenteed (linearly) Disadvantages of the Bisection Method. The graph of this equation is given in the figure. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. So let's take a look at how we can implement this. here is my program btw, but something's wrong in the bisection function and I can't figure out what is it. Mathematica. If the function values at points A and B have opposite signs…. Lecture Material. Bisection Method. Figure 1: The graphs of y=x (black) and y=\cos x (blue) intersect. thanks Code:. After reading this chapter, you should be able to: 1. We also call this method as an interval halving method because we consider a midpoint. We stay with our original. i just tried my vba code for bisection method but it doesn't work. Show that 𝑓𝑥=𝑥3+4𝑥2−10= 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10−4. And as I mentioned last time, this was the state of the art until the 17th century. The method is based on the following theorem. It is used in cases where it is known that. By testing different. The Bisection method is a numerical method which finds approximate solutions to polynomial equations with the use of midpoints. Select a Web Site. Equations don't have to become very complicated before symbolic solution methods give out. I dream of being a Theoretical Physicist one day. SECANT METHODS Convergence If we can begin with a good choice x 0, then Newton’s method will converge to x rapidly. The bisection method requires two points aand bthat have a root between them, and Newton’s method requires one. a b 3 Regula falsi Consider the figure in which the root lies between a and b. Bisection Method 1- Flowchart. (a) Bisection Method: This is one of the simplest and reliable iterative methods for the solution of nonlinear equation. 1) compute a sequence of increasingly accurate estimates of the root. In mathematics , the bisection method is a root-finding algorithm which repeatedly divides an interval in half and then selects the subinterval in which a root exists. It is Fault Free (Generally). If your calculator can solve equations numerically, it most likely uses a combination of the Bisection Method and the Newton-Raphson Method. Author jamespatewilliamsjr Posted on July 29, 2018 July 29, 2018 Categories C# Computer Programming Language, Root Finding Algorithms Tags Bisection Method, Brent's Method, C#, Computer Science, Newton's Method, Numerical Analysis, Regula Falsi Leave a comment on Root Finding Algorithms by James Pate Williams, BA, BS, MSwE, PhD. Numerical Analysis: Root Solving with Bisection Method and Newton’s Method. The number of iterations n that are required for obtaining a solution with a tolerance that is equal to or smaller than a specified tolerance can be determined before the solution. The basic method for making or doing something, such as an artistic work or scientific procedure: learned the techniques involved in painting murals Bisection of the angle technique - definition of bisection of the angle technique by The Free Dictionary. Bisection Method. The theorem is demonstrated in Figure 2. As a result, f(x) is approximated by a secant line through. any help would be appreciated. BISECTION METHOD. Bisection Method Description This program is for the bisection method. It is a very simple and robust method, but it is also rather slow. The most straightforward root-finding method. If the function values at points A and B have opposite signs…. These methods first find an interval containing a root and then systematically shrink the size of successive intervals that contain the root. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex −x−2=0. It was observed that the Bisection method. For the PASSFAIL method, the measure must pass for one limit and fail for the other limit. CGN 3421 - Computer Methods Gurley Numerical Methods Lecture 6 - Optimization page 107 of 111 Single Variable - Golden Section Search Optimization Method Similar to the bisection method • Define an interval with a single answer (unique maximum) inside the range sign of the curvature does not change in the given range. The Bisection Method is a successive approximation method that narrows down an interval that contains a root of the function f(x) The Bisection Method is given an initial interval [a. De ning a domain In higher dimensions, there is a rich variety of methods to de ne a simply connected domain. m" So create new. Apply the bisection method over a "large" interval. Use the bisection method to locate % a zero of the function f(x) = x sin(x) - 1. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. False Position or Regular Falsi method uses not only in deciding. 1 Show there is a root αin the interval (1,2). The routine assumes that an interval [a,b] is known, over which the function f(x) is continuous, and for which f(a) and f(b) are of opposite sign. The algorithm is iterative. The simplest way to solve an algebraic equation of the form g(z) = 0, for some function g is known as bisection. While the subject itself is quite interesting, the programming environment being used in the lab is Turbo C, a DOS based IDE which has been abandoned a long time ago. Brent's method combines the bisection method, secant method, and the method of inverse quadratic interpolation. If that is the case, you could save that data to an array and plot that array when you exit the loop like. (b) Use the bisection method to evaluate one root of your choice. -Bisection method is used to get a rough estimate of the solution then some other faster methods are used (discuss in our next lecture). Here is one example that passes the function f as a parameter, checks parameters for validity before continuing, avoids some other overflow exposures, avoids redundant calls to. Any zero-finding method (Bisection Method, False Position Method, Newton-Raphson, etc. The method is also called the interval halving method, the binary search method, or the dichotomy method. Mujahid Islam Md. The bisection method is far more efficient than algorithms which involve a search over frequencies, and of course the usual problems associated with such methods (such as determining how fine the search should be) do not arise. A genetic approach using direct representation of solutions for the parallel task scheduling problem. In the first iteration of bisection method, the approximation lies at the small circle. If a n and b n are satistfy equation (3) then b n − a n ≤ b −a 2n, for n ≥ 1 where b1 = b,a1 = a. 5 Bisection Method (cont’d) •It always converge to the true root (but be careful about the following) •f(x L) * f(x U) < 0 is true if the interval has odd number of roots, not necessarily one root. PROGRAM(Simple Version):. The method assumes that we start with two values of z that bracket a root: z1 (to the left) and z2 (to the right), say. The setup of the bisection method is about doing a specific task in Excel. However, instead of simply dividing the region in two, a linear interpolation is used to obtain a new point which is (hopefully, but not necessarily) closer to the root than the equivalent estimate for the bisection method. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. Bisection method. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. 0 (193 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Bisection Theorem An equation f(x)=0, where f(x) is a real continuous function, has at least one root between a and b, if f(a) f(b) < 0. Select a Web Site. The bisection method is discussed in Chapter 9 as a way to solve equations in one unknown that cannot be solved symbolically. Be able to apply the Bisection (Interval Halving) Method to approximate a solution to f(x) = 0. The bisection method is a simple root-finding method. ) exa_myfpi. If a function changes sign over an interval, the function value at the midpoint is evaluated. If instead we want the time at which a certain position is reached, we must invert these equations. 1) in terms of all five of the desirable attributes. Description: Given a closed interval [a,b] on which f changes sign, we divide the interval in half and note that f must change sign on either the right or the left half (or be zero at the midpoint of [a,b]. I followed the same steps for a different equation with just tVec and it worked. 3 The bisection method converges very slowly 4 The bisection method cannot detect multiple roots Exercise 2: Consider the nonlinear equation ex −x−2=0. The convergence is linear, slow but steady. It is a very simple and robust method, but it is also relatively slow. Let f 2 be a continuous function with di erent signs at a;b, with a #inc This is the solution for finding the roots of a function by Bisection Method in C++ Object Oriented Approach. We will study three different methods 1 the bisection method 2 Newton's method 3 secant method and give a general theory for one-point iteration methods. follow the algorithm of the bisection method of solving a nonlinear equation, 2. Example - 4: Using the bisection method find the approximate value of square root of 3 in the interval (1, 2) by performing two iterations. Formerly, engineers built external drivers to submit multiple parameterized Star-Hspice jobs, with each job exploring a region of the operating envelope of the circuit. What is Bisection Method? It is an iterative method based on a well known theorem which states that if f(x) be a continuous function in a closed interval [a,b] and f(a)f(b)<0, then there exists at least one real root of the equation f(x)=0, between a and b. Sample C program that uses Bisection Method in mathematics. By the Intermediate Value Theorem, there exists p in with. Pick starting points, precision and method. Compute where is the midpoint. Plot the function sin(x) on [0;2ˇ]. In fact, the common proof of the Intermediate Value Theorem uses the Bisection method. This tutorial explores a simple numerical method for finding the root of an equation: the bisection method. The secant method avoids this issue by using a nite di erence to approximate the derivative. Use the bisection method to approximate this solution to within 0. 9 Abeam is loaded as shown in Fig. What are synonyms for bisection?. 725 option price). I dream of being a Theoretical Physicist one day. 1 word related to bisection: division. Bisection method is used for finding root of the function in given interval. Then it is much more useful to explain, how a function is called with input arguments, than to convert the function to a script - which is still not working. Suppose we want to solve the equation f(x)=0,where f is a continuous function. Root is found by repeatedly bisecting an interval. In case, you are interested to look at the comparison between bisection method (adopted by Mibian Library) and my code please have look at screenshot of results obtained :-As you can see, bisection method didn’t converge well (to$13. So this is what happens in every iteration of the bisection method, we go through 20 iterations. sign ( f ( a )) == np. Suppose we know the two points of an interval and , where , and. The convergence of the bisection method is very slow. This can be achieved if we joint the coordinates (a,f(a)) and (b. Now at the very end, we want to output the result, and this is a function. erably; traditional methods produce results that are far from satis-factory. Bisection Method: The idea of the bisection method is based on the fact that a function will change sign when it passes through zero. Convergence • Theorem Suppose function 𝑓(𝑥) is continuous on [ , ], and 𝑓 ∙𝑓 <0. The bisection method is a simple root-finding method. The bisection method is probably the simplest root-finding method imaginable. if a and b are two. Methods for finding roots are iterative and try to find an approximate root $$x$$ that fulfills $$|f(x)| \leq \epsilon$$, where $$\epsilon$$ is a small number referred later as tolerance. To find a root very accurately Bisection Method is used in Mathematics. In a nutshell, the former is slow but robust and the latter is fast but not robust. This will open a new tab with the resource page in our marketplace. 01, and therefore we chose b = 1. The main way Bisection fails is if the root is a double root; i. Suppose we know the two points of an interval and , where , and. Since the line joining both these points on a graph of x vs f(x), must pass through a point, such that f(x)=0. Bisection is the division of a given curve, figure, or interval into two equal parts (halves). % % Enter the starting endpoints for [a,b] in a and b % % Enter the tolerance in delta. Bisection Method. conventional methods like Newton-Raphson method (N-R), Regula Falsi method (R-F) & Bisection method (BIS). THE BISECTION METHOD This method is based on mean value theorem which states that if a function ( ) is continuous between and , and ( ) ( ) are of opposite signs, then there exists at least one root between and. Bisection Method The bisection method is a kind of bracketing methods which searches for roots of equation in a specified interval. Each iteration step halves the current interval into two subintervals; the next interval in the sequence is the subinterval with a sign change for the function (indicated by. What is Bisection Method? The method is also called the interval halving method, the binary search method or the dichotomy method. Use this tag for questions related to the bisection method, which is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. 2 Using the Bisection Method to Prove the Intermediate Value Theorem Now suppose that fis continuous on [a;b], f(a) <0 and f(b) >0. So it is dependent on. Advantages of the Bisection method. It also makes a graph available of the iterates. | 2019-12-10T16:04:47 | {
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https://math.stackexchange.com/questions/1025630/to-show-for-following-sequence-lim-n-to-infty-a-n-0-where-a-n-1-3 | # To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 … (2n-1)\over 2.4.6…(2n)$
How can I show
$\lim_{n \to \infty} a_n = 0$
$a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}$
I have shown that $a_n$ is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks
• Clarification request: what do you mean by $1.3.5...$? Are these numbers being added? – Mike Pierce Nov 17 '14 at 8:14
• I think is multiplication of odd numbers. Isn't $\pi$ in the answer? I vaguely remember a result like this from Euler. – ReverseFlow Nov 17 '14 at 8:15
• It is certainly bounded below, since all terms are positive, so it converges. But that does not show it converges to $0$. The Stirling approximation will show convergence to $0$, but one can do it with less machinery. – André Nicolas Nov 17 '14 at 8:21
• Try to prove using induction that:$$a_n\lt\frac{1}{\sqrt{3n+1}}$$ – pointer Nov 17 '14 at 8:26
• That the limit is zero is intuitively obvious, seeing as $a_n=\binom{2n}n/4^n$is the probability of getting exactly $n$ heads in $2n$ independent tosses of a fair coin. – bof Nov 17 '14 at 8:30
Multiply the numerator by the denominator; so $$a_n=\frac{1\times3\times5\times ... \times(2n-1)}{ 2\times4\times6\times...\times(2n)}=\frac{1\times2\times3\times ... \times(2n)}{\Big(2\times4\times6\times...\times(2n)\Big)^2}=\frac{(2n)!}{4^n(n!)^2}$$ If now you use Stirling approximation $$m! \approx \sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ and then $$a_n \approx \frac{1}{\sqrt{\pi n} }$$ A more detailed approach would show for the asymptotic behavior $$a_n \approx \frac{1}{\sqrt{\pi n} }\Big(1-\frac{1}{8n}\Big)$$
• In first step how have u written in denominatorof last inequality as $4^{n}(n!)^{2}$ .. – godonichia Nov 17 '14 at 12:11
• Factor $2$ inside the brackets of denominator and square. – Claude Leibovici Nov 17 '14 at 17:21
• @ClaudeLeibovici in stirlings aproximstion as you have mentioned in order to find formula for 2m! I just have to replace m by 2m in above formula .if i do this then im not getting to answer, thanks – Jessica Gtb Dec 25 '14 at 12:20
• @JessicaGtb. I don't know whee you had a problem since if you use Stirling, you should get $$(2n)! \approx \sqrt{\pi } 2^{2 m+1} e^{-2 m} m^{2 m+\frac{1}{2}}$$ Post if you still have problems. Cheers :-) – Claude Leibovici Dec 26 '14 at 7:22
• How did you get that expression for 2n! .it is not same when you replace n by 2n in formula you mentioned in answer – Jessica Gtb Dec 26 '14 at 12:29
Using $\boldsymbol{1+x\le e^x}$
Since $1+x\le e^x$ for all $x\in\mathbb{R}$, \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\\ &\le\prod_{k=1}^ne^{-\large\frac1{2k}}\\[3pt] &=e^{-\frac12H_n}\tag{1} \end{align} where $H_n$ are the Harmonic Numbers. Since the Harmonic Series diverges, $(1)$ tends to $0$.
Using Gautschi's Inequality \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\frac{k-\frac12}{k}\\ &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\[6pt] &\le\frac1{\sqrt{\pi n}}\tag{2} \end{align}
Commentary
Gautschi's Inequality also says that \begin{align} a_n &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\ &\ge\frac1{\sqrt{\pi\left(n+\frac12\right)}}\tag{3} \end{align} Therefore, $$\lim_{n\to\infty}\sqrt{n}\,a_n=\frac1{\sqrt\pi}\tag{4}$$ Putting the second line of $(1)$ and the limit in $(4)$ together, we can show that $$\gamma+\sum_{n=2}^\infty\frac{\zeta(n)}{n2^{n-1}}=\log(\pi)\tag{5}$$ where $\gamma$ is the Euler-Mascheroni Constant and $\zeta$ is the Riemann Zeta Function.
Let's prove using induction that $$a_n\le\frac{1}{\sqrt{3n+1}}.$$ For $n=1$ it is true. Now we just need to prove that$$\frac{(2n+1)^2}{(2n+2)^2}\le\frac{3n+1}{3n+4}$$or $$(4n^2+4n+1)(3n+4)\le(4n^2+8n+4)(3n+1)$$or$$12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4.$$
• If we are not given the limit it converges to then how would we go about it – godonichia Nov 17 '14 at 8:51
• I don't know. Maybe we could find asymtothics using Stirling's formula, then try to find this inequality. – pointer Nov 17 '14 at 9:08
Your sequence can be rewritten as $$a_n=\frac{1\cdot3\cdot...\cdot(2n-1)}{2\cdot4\cdot...\cdot2n}=\frac{1\cdot2\cdot3\cdot...\cdot(2n-1)\cdot2n}{(2\cdot4\cdot...\cdot2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}$$ Using Stirling's approximation we get $$a_n=\frac{(2n)!}{2^{2n}(n!)^2}\sim\frac{1}{2^{2n}}\cdot\frac{\sqrt{2\pi 2n}\cdot(\frac{2n}{e})^{2n}}{(\sqrt{2\pi n}(\frac{n}{e})^n)^2}=\frac{1}{2^{2n}}\cdot\frac{\sqrt{2}\cdot2^{2n}}{\sqrt{2\pi n}}=\frac{\sqrt{2}}{\sqrt{2\pi n}}\to 0$$ as $n\to\infty$.
• i havent studied stirlings approximation yet . Can u show me another way which uses sequence series concepts – godonichia Nov 17 '14 at 8:35 | 2019-09-16T08:50:54 | {
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http://mathhelpforum.com/statistics/129611-help-needed-probablilty-question.html | # Math Help - Help needed for a probablilty question
1. ## Help needed for a probablilty question
Ok so this is really bugging me, because I thought i finally understood it, but then my book is giving me different answers to what I'm getting.
Anyways, heres the problem, can anyone answer it and show working , so I can learn. Thanks
Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:
(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
2. P(R)
= P(R(S+S'))
= P(R1)
= P(RS)+P(RS')
= P(R|S)P(S)+P(R|S')P(S')
= P(R|S)P(S)+P(R|S')(1-P(S))
P(RS)
= P(R|S)P(S)
= P(S|R)P(R)
Now you know P(R) and P(RS), so you know P(RS')
P(RS')
= P(S'|R)P(R)
P(S)
= ...
= P(S|R)P(R)+P(S|R')(1-P(R))
gives
P(SR')
= P(S|R')P(R')
= P(S|R')(1-P(R))
Using this,
P(1)
= P((R+R')(S+S'))
= P(RS) + P(RS')+P(R'S)+P(R'S')
gives
P(R'S')
= P(S'|R')P(R')
3. Originally Posted by kamicool17
Given P(R|S) = 0.5, P(R|S') = 0.4 and P(S) = 0.6 find:
(a) P(R) (b) P(S|R) (c) P(S'|R) (d) P(S'|R')
$P(RS)=P(R|S)P(S)$
$P(RS')=P(R|S')P(S')$
$P(R)=P(RS)+P(RS')$
Now finish.
4. Hello, kamicool17!
Another approach . . . (much longer, though)
Given: . $P(R|S) = 0.5,\;\;P(R|S') = 0.4,\;\;P(S) = 0.6$
Find: . $(a)\;P(R) \qquad (b)\;P(S|R)\qquad (c)\;P(S'|R) \qquad (d)\;P(S'|R')$
We have this table:
. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & {\color{blue}(a)} & {\color{blue}(c)} & {\color{blue}(e)} \\ \hline
R' & {\color{blue}(b)} & {\color{blue}(d)} & {\color{blue}(f)} \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$
We have: . $P(S) = 0.60 \quad\Rightarrow\quad P(S') = 0.40$
So we can fill in the bottom row.
We have: . $P(R|S) = 0.5$
Then: . $\frac{P(R \wedge S)}{P(S)} \;=\;0.5 \quad\Rightarrow\quad \frac{P(R \wedge S)}{0.6} \:=\:0.5
$
Hence: . $P(R \wedge S) \,=\,0.30\;{\color{blue}(a)} \quad\Rightarrow\quad P(R' \wedge S) \:=\:0.30\;{\color{blue}(b)}$
. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & (c) & (e) \\ \hline
R' & 0.30 & (d) & (f) \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$
We have: . $P(R|S') = 0.4$
Then: . $\frac{P(R \wedge S')}{P(S')} \:=\:0.4 \quad\Rightarrow\quad \frac{P(R \wedge S')}{0.4} \:=\:0.4$
Hence: . $P(R \wedge S') \:=\:0.16\;{\color{blue}(c)} \quad\Rightarrow\quad P(R' \wedge S') \:=\:0.24\;{\color{blue}(d)}$
Also: . $P(R) \:=\:0.46 \;{\color{blue}(e)}\quad\Rightarrow\quad P(R') \:=\:0.54\;{\color{blue}(f)}$
And we have completed the table:
. . $\begin{array}{c|| c|c||c}
& S & S' & \text{Total } \\ \hline \hline
R & 0.30 & 0.16 & 0.46 \\ \hline
R' & 0.30 & 0.24 & 0.54 \\ \hline \hline
\text{Total} & 0.60 & 0.40 & 1.00 \end{array}$
You can now answer the questions . . . | 2016-05-02T13:51:46 | {
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http://code.jasonbhill.com/category/algorithms/ | A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).
The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion $(x + y + z)^n$. How many coefficients in the expansion of $(x + y + z)^{200000}$ are multiples of $10^{12}$?
## Solution Using the Multinomial Theorem
The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula
$(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$
where
$\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$
Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions $k_1+k_2+\cdots+k_m=n$ for integers $k_i$. If n=200000 abd m=3, then the terms in the expansion are given by
$\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$
where $k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by $10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved.
First, we’ll create a function $p(n,d)$ that outputs how many factors of $d$ are included in $n!$. We have that
$p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$
where $d^r$ is the highest power of $d$ dividing $n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by $10^{12}=2^{12}5^{12}$, we’ll be using the values provided by $p(n,d)$ quite a bit for $d=2$ and $d=5$. We’ll store two lists:
$p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$
For a given $k_1,k_2,k_3$, the corresponding coefficient is divisible by $10^{12}$ precisely when
$p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$
That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients.
Now, we know that $k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume $k_1\le k_2\le k_3$. Under this assumption, we always have
$k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$
We know that $k_1+k_2+k_3=200000$ is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1) $k_1=k_2 < k_3$, or (case 2) $k_1 < k_2=k_3$, or (case 3) $k_1 < k_2 < k_3$.
In case 1, we iterate $0\le k_1\le 66666$, setting $k_2=k_1$ and $k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the $k_i$ in 3 ways).
In case 2, we iterate $0\le k_1\le 66666$, and when $k_1$ is divisible by 2 we set $k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance.
In case 3, we iterate $0\le k_1\le 66666$, and we iterate over $k_2=k_1+a$ where $1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then $k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects).
## Cython Solution
I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C.
%cython import time from libc.stdlib cimport malloc, free head_time = time.time() cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers run_time = time.time() # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 free(p2) free(p5) elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) print "Result: %s" % result print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head)
When executed, we find the correct result relatively quickly.
Result: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds)
## C with OpenMP Solution
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } free(p2); free(p5); printf("result: %lld\n", result); return 0; }
This can be compiled and optimized using GCC as follows.
$gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c When executed on a 16-core machine, we get the following result. $ time ./problem-154-omp result: 479742450 real 0m1.487s
This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds.
### Motivation
Many interesting computational problems, such as those on Project Euler require that one find the sum of proper divisors of a given integer. I had a fairly crude brute-force method for doing this, and was subsequently emailed a comment by Bjarki Ágúst Guðmundsson who runs the site www.mathblog.dk. He pointed me in the direction of this page and provided some sample code illustrating how such an approach runs asymptotically faster than the approach I had been taking. Awesome! I’m going to expand on that a bit here, providing some mathematical proofs behind the claims and providing code for those who may want to take advantage of this.
### The Mathematics Behind It All
Let the function $\sigma(n)$ be the sum of divisors for a positive integer $n$. For example,
$\sigma(6)=1+2+3+6=12.$
It should seem obvious that for any prime $p$ we have $\sigma(p)=1+p$. What about powers of primes? Let $\alpha\in\mathbb{Z}_+$, and then
$\sigma(p^\alpha)=1+p+p^2+\cdots+p^\alpha.$
We’d like to write that in a closed form, i.e., without the “$+\cdots+$”. We use a standard series trick to do that.
\begin{align} \sigma(p^\alpha) &= 1+p+p^2+\cdots+p^\alpha\cr p\sigma(p^\alpha) &= p+p^2+p^3+\cdots+p^{\alpha+1}\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= (p+p^2+\cdots+p^{\alpha+1})-(1+p+\cdots+p^\alpha)\cr p\sigma(p^\alpha)-\sigma(p^\alpha) &= p^{\alpha+1}-1\cr (p-1)\sigma(p^\alpha) &= p^{\alpha+1}-1\cr \sigma(p^\alpha) &=\frac{p^{\alpha+1}-1}{p-1}.\end{align}
That solves the problem of finding the sum of divisors for powers of primes. It would be nice if we could show that $\sigma$ is multiplicative on powers of primes, i.e., that $\sigma(p_1^{\alpha_1}p_2^{\alpha_2})=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})$. We’ll prove that this is the case, and solve the problem in general along the way.
Proposition: The function $\sigma$ is multiplicative on powers of primes.
Proof: Let $n$ be a positive integer written (uniquely, by the fundamental theorem of arithmetic) as
$n=\prod_{i=1}^m p_i^{\alpha_i}$
for $m$ distinct primes $p_i$ with $\alpha_i\in\mathbb{Z}_+$. Any divisor $k$ of $n$ then has the form
$k=\prod_{i=1}^m p_i^{\beta_i}$
where each $\beta_i$ satisfies $0\le\beta_i\le\alpha_i$. Then
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)$
is the sum of all divisors $k$ of $n$ and can be written by summing over all possible combinations of the exponents $\beta_i$. There are $\prod_{i=1}^m \alpha_i$ combinations, and we can form their sum and simplify it as follows.
\begin{align}\sigma(n) &= \sum_{1\le i\le m,\ 0\le\beta_i\le\alpha_i}p_1^{\beta_i}p_2^{\beta_2}\cdots p_m^{\beta_m}\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\left(\sum_{2\le i\le m,\ 0\le\beta_i\le\alpha_i}p_2^{\beta_2}p_3^{\beta_3}\cdots p_m^{\beta_m}\right)\cr &= \sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\left(\sum_{3\le i\le m,\ 0\le\beta_i\le\alpha_i}p_3^{\beta_3}p_4^{\beta_4}\cdots p_m^{\beta_m}\right) \cr &= \vdots\cr &=\sum_{\beta_1=0}^{\alpha_1}p_1^{\beta_1}\sum_{\beta_2=0}^{\alpha_2}p_2^{\beta_2}\sum_{\beta_3=0}^{\alpha_3}p_3^{\beta_3}\ \cdots\ \sum_{\beta_m=0}^{\alpha_m}p_m^{\beta_m}\cr &=\sigma(p_1^{\alpha_1})\sigma(p_2^{\alpha_2})\cdots\sigma(p_m^{\alpha_m}).\end{align}
This completes the proof. Q.E.D.
Thus, we now have a formula for the sum of divisors of an arbitrary positive integer $n$ using the factorization of $n$. Namely,
$\sigma(n)=\sigma\left(\prod_{i=1}^m p_i^{\alpha_i}\right)=\prod_{i=1}^m\left(\frac{p_i^{\alpha_i+1}-1}{p_i-1}\right).$
This is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It’s a basic post that isn’t advanced at all, but that doesn’t mean that the implementation given below won’t save work for others. The idea is to create a list of primes in C by malloc’ing a sieve, then malloc’ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list.
#include <stdio.h> #include <stdlib.h> #include <malloc.h> #define bool _Bool static unsigned long prime_limit = 1000000; unsigned long sqrtld(unsigned long N) { int b = 1; unsigned long res,s; while(1<<b<N) b+= 1; res = 1<<(b/2 + 1); for(;;) { s = (N/res + res)/2; if(s>=res) return res; res = s; } } unsigned long * make_primes(unsigned long limit) { unsigned long *primes; unsigned long i,j; unsigned long s = sqrtld(prime_limit); unsigned long n = 0; bool *sieve = malloc((prime_limit + 1) * sizeof(bool)); sieve[0] = 0; sieve[1] = 0; for(i=2; i<=prime_limit; i++) sieve[i] = 1; j = 4; while(j<=prime_limit) { sieve[j] = 0; j += 2; } for(i=3; i<=s; i+=2) { if(sieve[i] == 1) { j = i * 3; while(j<=prime_limit) { sieve[j] = 0; j += 2 * i; } } } for(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1; primes = malloc((n + 1) * sizeof(unsigned long)); primes[0] = n; j = 1; for(i=2;i<=prime_limit;i++) if(sieve[i]==1) { primes[j] = i; j++; } free(sieve); return primes; } int main(void) { unsigned long * primes = make_primes(prime_limit); printf("There are %ld primes <= %ld\n",primes[0],prime_limit); free(primes); return 0; }
Say one wanted to form a list of all primes below 1,000,000. That’s what the above program does by default, since “prime_limit = 1000000.” If one compiles this and executes, you would get something like what follows. The timing is relatively respectable.
$gcc -O3 -o prime-sieve prime-sieve.c$ time ./prime-sieve There are 78498 primes <= 1000000 real 0m0.008s user 0m0.004s sys 0m0.000s
The code is linked here: prime-sieve.c | 2018-03-19T23:51:45 | {
"domain": "jasonbhill.com",
"url": "http://code.jasonbhill.com/category/algorithms/",
"openwebmath_score": 0.7797718644142151,
"openwebmath_perplexity": 1626.5125118132278,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9905874090230069,
"lm_q2_score": 0.8670357649558006,
"lm_q1q2_score": 0.8588747119378474
} |
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