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https://math.stackexchange.com/questions/3218664/if-lim-x-rightarrow-0-fracfxx2-5-then-what-is-lim-x-rightarr/3218668 | # If $\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5$, then what is $\lim_{x \rightarrow 0}f(x)$?
I know the answer to the above question, but I have a question on some of the reasoning.
The way I know how to solve it is $$\lim_{x \rightarrow 0}f(x) = \lim_{x \rightarrow 0}\left(f(x)\cdot \frac{x^2}{x^2}\right) = \lim_{x \rightarrow 0}\left(\frac{f(x)}{x^2}\cdot x^2\right) = \left(\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\right)\left(\lim_{x \rightarrow 0}x^2\right) = 5\cdot0 = 0.$$
I saw another solution elsewhere that gets the right answer, but I am unsure if the steps are actually correct. \begin{align*} &\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5 \\ \Longrightarrow &\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5 \\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot \lim_{x \rightarrow 0}x^2\\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot 0 = 0. \end{align*}
My issue is with that first step. I know that $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$$, but only when $$\lim_{x \rightarrow a}g(x) \neq 0$$. Since $$\lim_{x \rightarrow 0}x^2 = 0$$, wouldn't this invalidate the above work? However, it still got the same answer, so my real question is why did it work and when will it work in general?
EDIT: Does anyone have a nice example for when the logic in the second method doesn't work?
• You are correct that the solution you saw elsewhere is invalid. It is invalid for the exact reason you think it is. Using false logic to get a correct answer is still false logic. Applied to another problem, it might fail. – InterstellarProbe May 8 at 16:38
• Do you have an example of another problem in which logic from the second method fails? – Smash May 8 at 16:45
• Simply put, you can't divide by zero. If you want examples, there are many "proofs" of $0=1$ where the trick is to divide by zero and hope no one notices. – Théophile May 8 at 16:47
If $$\lim\limits_{x\to0}\dfrac{f(x)}{x^2}=5$$, then for every $$n\in\Bbb N$$ there is some $$x\neq 0$$ such that $$5-\frac1n<\frac{f(x)}{x^2}<5+\frac1n$$ and hence $$5x^2-\frac{x^2}n If $$x\to 0$$ we get $$f(x)\to 0$$.
$$\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5$$ doesn't work, the LHS is not defined.
But for the limit of $$\dfrac{f(x)}{x^2}$$ to exist, $$\lim_{x\to0}f(x)$$ must be zero, which is also the limit of $$x^2$$.
You can not do this since the denominator on the right is $$0$$.
$$\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = \frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2}$$
But first aproach is perfect.
• You meant "denominator," not numerator. – Mark Viola May 8 at 16:41 | 2019-09-22T10:37:44 | {
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https://mathhelpboards.com/threads/probability-of-winning-dice-game.24336/ | # Probability of winning dice game
#### TheFallen018
##### Member
Hey, so I've got this problem that I'm trying to figure out. I've worked out something that I think is probably right through simulation, but I'm not really sure how to tackle it from a purely mathematical probability perspective. So, would anyone know how I should approach this? I've tried a few different things, but my two answers tend to conflict a little. Thanks,
#### Opalg
##### MHB Oldtimer
Staff member
The starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\ \hline p(x) & \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36} & \frac4{36} & \frac3{36} & \frac2{36} & \frac1{36} \end{array}.$$
The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.
#### TheFallen018
##### Member
The starting point (which I guess you already know) is to make a list of the probabilities $p(x)$ of rolling a total of $x$ with the two dice, where $x$ goes from $2$ to $12$. These are $$\begin{array}{r|ccccccccccc}x& 2&3&4&5&6 &7&8&9&10 &11&12\\ \hline p(x) & \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36} & \frac4{36} & \frac3{36} & \frac2{36} & \frac1{36} \end{array}.$$
The probability of an immediate win on the first roll is $p(7)+p(11)$. To win at a later stage, you first need to roll an $x$ (where $x$ is $4,5,6,8,9$ or $10$). You then need to roll another $x$ before rolling a $7$. The probability of those two things happening is $p(x)\dfrac{p(x)}{p(x)+p(7)}$. Putting in the numbers, and adding the various probabilities, I get the overall probability of a win to be $\dfrac{244}{495} \approx 0.49292929\ldots$.
Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks
#### Opalg
##### MHB Oldtimer
Staff member
Wow, that works really well. I thought it was going to be much more complicated and messy than that, which is a beautiful solution. I'm curious though, is that equation $p(x)\dfrac{p(x)}{p(x)+p(7)}$ something that you derived for this, or is this a standard formula for similar problems.If it's something that you derived, would you be able to explain how you came up with it? If it's a standard equation, would you happen to know it's name? I'd love to understand better how and why it works. Thanks
In the expression ${\color {red} p(x)}{\color {green}\dfrac{p(x)}{p(x)+p(7)}}$, the red $\color {red} p(x)$ gives the probability that the first roll of the dice gives the value $x$. The green fraction represents the probability of rolling $x$ again before rolling a $7$. My argument for that is that after rolling the first $x$, you can completely disregard any subsequent rolls until either an $x$ or a $7$ turns up. The only question is, which one of those will appear first. The relative probabilities of $x$ and $7$ are in the proportion $p(x)$ to $p(7)$. So out of a combined probability of $p(x) + p(7)$, the probability of an $x$ is $\dfrac{p(x)}{p(x)+p(7)}$, and the probability of a $7$ is $\dfrac{p(7)}{p(x)+p(7)}$.
I hope that makes sense. | 2020-07-14T10:48:01 | {
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https://mathhelpboards.com/threads/im-confuseds-question-on-yahoo-answers-involving-a-recurrence-relation.3088/ | I'm Confused's question on Yahoo! Answers involving a recurrence relation
MarkFL
Staff member
Jonah must take an antibiotic every 12 hours. Each pill is 25 milligrams, and after every 12 hours, 50% of the drug remains in his body. What is the amount of antibiotic in his body over the first two days? What amount will there be in his body in the long run?
Here's what I did:
For my recursive formula, I tried inputting u(n)=u(n-1)*(1-0.50). However, I could not seem to find the long run value.,,, It keeps changing. Am I right on this?
I appreciate your help! Thank you so much!
Source: <Advanced Algebra: An Investigative Approach; Murdock Jerald, 2004>
I posted a link to this topic, so the OP could find my response.
Here is a link to the original question:
Recursive formula help on real world situations? - Yahoo! Answers
Let $A(t)$ represent the amount, in mg, of antibiotic in Jonah's bloodstream at time $t$, measured in hours. With a half-life of 12 hours, we may state:
$\displaystyle A(t)=A_02^{-\frac{t}{12}}$
So, for the first 12 hours, but not including the second dose, we find:
$\displaystyle A_0=25$ hence:
$\displaystyle A(t)=25\cdot2^{-\frac{t}{12}}$
$\displaystyle A(t)=25\cdot2^{-\frac{12}{12}}=\frac{25}{2}$
Now, for the second 12 hours, we have:
$\displaystyle A_0=\frac{25}{2}+25=\frac{75}{2}$ hence:
$\displaystyle A(t)=\frac{75}{2}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{75}{2}\cdot2^{-\frac{12}{12}}=\frac{75}{4}$
For the third 12 hours, we have:
$\displaystyle A_0=\frac{75}{4}+25=\frac{175}{4}$ hence:
$\displaystyle A(t)=\frac{175}{4}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{175}{4}\cdot2^{-\frac{12}{12}}=\frac{175}{8}$
For the fourth 12 hours, we have:
$\displaystyle A_0=\frac{175}{8}+25=\frac{375}{8}$ hence:
$\displaystyle A(t)=\frac{375}{8}\cdot2^{-\frac{t}{12}}$
$\displaystyle A(12)=\frac{375}{8}\cdot2^{-\frac{12}{12}}=\frac{375}{16}$
So, now we may try to generalize for the $n$th 12 hour period.
Let $I_n$ represent the initial amount for each 12 hour period. We see we must have the recursion:
$I_{n+1}=\frac{1}{2}I_{n}+25$
This is an inhomogeneous recurrence, so let's employ symbolic differencing to obtain a homogeneous recurrence:
$I_{n+2}=\frac{1}{2}I_{n+1}+25$
Subtracting the former from the latter, we obtain:
$2I_{n+2}=3I_{n+1}-I_{n}$
The characteristic equation is:
$2r^2-3r+1=0$
$(2r-1)(r-1)=0$
Hence, the closed-form will be:
$I_n=k_1\left(\frac{1}{2} \right)^n+k_2$
We may use our data above to determine the parameters $k_i$:
$I_1=k_1\left(\frac{1}{2} \right)^1+k_2=25$
$I_2=k_1\left(\frac{1}{2} \right)^2+k_2=\frac{75}{2}$
or
$\frac{1}{2}\cdot k_1+k_2=25$
$\frac{1}{4}\cdot k_1+k_2=\frac{75}{2}$
Solving this system, we find:
$k_1=-50,k_2=50$ and so we have:
$I_n=-50\left(\frac{1}{2} \right)^n+50=50\left(1+\left(\frac{1}{2} \right)^n \right)$
And so, for the $n$th 12 hour period, we have:
$\displaystyle A_n(t)=I_n2^{-\frac{t}{12}}$
Now, for the long run, which we may take as implying as n grows without bound, we find:
$\displaystyle \lim_{n\to\infty}I_n=50$
and so we find that for the long run the initial amount for a 12 hour period is is 50 mg and the final amount for that period is 25 mg.
Last edited:
tkhunny
Well-known member
MHB Math Helper
That is good mathematics, but not so great reality. If it had said I.V. Administration, the assumption of immediate absorption and distribution would be more realistic. Even subcutaneous or intramuscular injection takes a little while. A pill? No. | 2021-06-24T18:06:58 | {
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https://brilliant.org/discussions/thread/polynomial-sprint-useful-lemma/ | # Polynomial Sprint: Useful Lemma
$\color{#D61F06} {\text{Unlocked!}}$
Mathematicians have a huge bag of tricks, which provide them with various ways of approaching a problem. In this note, we introduce an extremely useful lemma, which applies to polynomials with integer coefficients.
Lemma. If $f(x)$ is a polynomial with integer coefficients, then for any integers $a$ and $b$, we have $a-b \mid f(a) - f(b)$
1) Show that for any integers $a, b$ and $n$, we have
$a-b | a^n - b^n.$
2) Using the above, prove the lemma.
3) Come up with a (essentially) two-line solution to question 4 from the book.
Note: In Help Wanted, Krishna Ar mentioned that problem 4 from the book was hard to solve. If you looked at the solution, it was also hard to understand what exactly was happening. This provides us with a simple way of comprehending why there are no solutions.
4) Come up with an alternative solution to E6 (page 250, from the book) using this lemma.
5) * (Reid Barton) Suppose that $f(x)$ is a polynomial with integer coefficients. Let $n$ be an odd positive integer. Let $x_1, x_2, \ldots x_n$ be a sequence of integers such that
$x_2 = f(x_1), x_3 =f(x_2), \ldots x_n = f(x_{n-1}), x_1 = f(x_n) .$
Show that all the $x_i$ are equal.
Where did you use the fact that $n$ is an odd positive integer?
Note by Calvin Lin
5 years, 9 months ago
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I really like Polynomial Sprint. It would be nice if you can do some other themes using similar form.
- 5 years, 8 months ago
Yes, I entirely agree! I find these sets fun to do and I want more sets.
- 5 years, 8 months ago
This has been unlocked. Sorry for being late, I was out yesterday.
Staff - 5 years, 9 months ago
What does the $|$ mean?
- 5 years, 9 months ago
$a|b$ means $a$ divides $b$
- 5 years, 9 months ago
1) $a-b|a^n-b^n$, let $f(x)=x^n$, and plug in $a$ and $b$ into the function and we get $a-b|a^n-b^n$, is this correct?
- 5 years, 9 months ago
You have to show that it is true of all polynomials, not just specially chosen polynomials.
Staff - 5 years, 9 months ago
The lemma is true for polynomials with integer coefficients, so do you mean that I must show that this is true for non-integers too?
- 5 years, 9 months ago
I might have initially misinterpreted what you are trying to do.
The statement in question 1 is about integers (not polynomials), and you are asked to show that for any pair of integers, we have $a - b \mid a^n - b^n$. For example, $10 - 7 \mid 10^3 - 7^3$, or equivalently that $3 \mid 657$.
Note that you are not allowed to use the lemma, since we want to use question 1 to prove the lemma (in question 2).
Staff - 5 years, 9 months ago
Oh, ok
- 5 years, 9 months ago
5) We see that $x_n-x_{n+1}\mid f(x_n)-f(x_{n+1})=x_{n+1}-x_{n+2}$.
Thus we get the following divisibilitites:
$x_1-x_2\mid x_2-x_3\implies x_2-x_3=k_1(x_1-x_2)$
$x_2-x_3\mid x_3-x_4\implies x_3-x_4=k_2(x_2-x_3)$
$\vdots$
$x_{n-1}-x_n\mid x_n-x_1\implies x_n-x_1=k_{n-1}(x_{n-1}-x_n)$
$x_n-x_1\mid x_1-x_2\implies x_1-x_2=k_n(x_n-x_1)$
This means that $x_1-x_2=k_1k_2\cdots k_n(x_1-x_2)$
since $x_1\ne x_2$, we can divide both sides by $x_1-x_2$ to get $k_1k_2\cdots k_n=1$
Since we are working in the integers, this means that $|k_i|=1$ for all $i=1\to n$.
Thus, $x_k-x_{k+1}=\pm (x_{k-1}-x_k)$
What I got so far.
- 5 years, 9 months ago
If some $k_i=-1$, then we have $x_k=x_{k+1}$, which leads to equality of all $x_i$. If all k=1, then we can sum and get $x_1-x_2=0$, so $x_1=x_2$, which contradicts to your assumption that $x_1$ is different from $x_2$. Then that again leads to $x_1=x_2=...=x_n$
- 5 years, 9 months ago
Is it that simple? When did you use the condition that $n$ is odd?
- 5 years, 9 months ago
You just consider it modulo |x2 - x1| :
Let d equal |x1 – x2| and thus d = |x1 – x2| = |x2 – x3| = |x3 – x4| = … = |x{n-1} – x{n}| = |x{n} – x1|. If d is nonzero then x2 is congruent to (x1 + d) mod 2d. Similarly x3 is congruent to (x2 + d) mod 2d which is congruent to x1 mod 2d. In general x{k} is congruent to (x1 + d) mod 2d if k is even and congruent to x1 mod 2d if k is odd. Thus we have x{n} is congruent to x1 mod 2d because n is odd. However then |x{n} – x1| = 2dq for an integer q. Also d = |x{n} – x1|, hence d = 2dq and 1 = 2q. This contradicts the fact that q is an integer, hence our previous assumption that d is nonzero must be false. Therefore d = |x1 – x2| = |x2 – x3| = |x3 – x4| = … = |x{n-1} – x{n}| = |x{n} – x1| = 0 and x1 = x2 = x3 = …=x{n} and all the x{i} are equal.
- 5 years, 8 months ago
I do not immediately see how we get $k_k = 0 \Rightarrow x_k = x_{k+1}$ in the first scenario. I think we get $x_{k} = x_{k+2}$ instead, but that isn't immediately useful.
@Daniel Liu You are nearly there. Let me phrase the rest of the question in a different way. You start at some point $X$ on the real number line. You take $n$ odd steps of size exactly $S$. If you end back at $X$, show that $S = 0$.
Staff - 5 years, 9 months ago
We want another polynomial sprint! Soon..
- 5 years, 8 months ago
I guess We can write any polynomial a^n+b^n as (a-b)(a^n-1+a^n-2 b^1. .... Till b^n-1) therefore its always divisible by a-b
- 5 years, 9 months ago
Where can I get the book?
- 5 years, 9 months ago
- 5 years, 9 months ago
- 5 years, 9 months ago
1)a^n-b^n=(a-b)(a^(n-1)+a^(n-2) b+...+b^(n-1)) Which leads us to:a-b/a^n-b^n 2) let f be a polynomial f (x)=an.x^n+an-1.x^(n-1)+...+a0 We can apply question1 since it is true for any n For j {0, n}: a-b / a^j-b^j We can sum Leads us to:a-b/f (a)-f (b) 5) x1-x2 =k1 (xn-x1) x2-x3=k2 (x1-x2) . . . xn-1-xn=kn (xn-2 -xn-1) Leads us to ki=1 or =-1 If one ki=-1 We have xi are equals If all ki=1 (x1-x2)+(x2-x3)+...+(xn-1 -xn)=(xn-x1)+ (x1-x2)+...+(xn-2 - xn-1) Leads us to: x1-xn = xn- xn-1 If just xi#xi-1 We have contradiction: (xi-xi-1)=0=(xi-1 -xi-2)#0 We can make recurrence(recurrance in french) We leads to an a contradiction each time Then k#1 Then all xi are equals
- 5 years, 9 months ago
1. Let $f(t)=t^n-b^n$. Then $f(b)=b^n-b^n=0$, so $t-b\mid t^n-b^n$. Let $t=a$ to get $a-b\mid a^n-b^n$.
2. Consider the $k$th coefficient of $f(a)-f(b)$: $a_k(a^k-b^k)$. Since $a-b\mid a^k-b^k$, $a-b$ is a factor of every term in $f(a)-f(b)\implies a-b\mid f(a)-f(b)$.
For the rest, I don't know which book you're talking about.
- 5 years, 9 months ago
Problem Solving Strategies by Arthur Engel. They might be selling online I don't really know. XD
- 5 years, 9 months ago
Google "arthur engel problem solving strategies", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores:
- 5 years, 8 months ago
- 5 years, 8 months ago
Oh oops, never mind the url.
- 5 years, 8 months ago
Book? What book?
- 5 years, 9 months ago
Google "arthur engel problem solving strategies", and follow the first link, which should be a pdf. The url is below but you have to surround the italicized text with underscores: http://f3.tiera.ru/2/MMathematics/MSchSchool-level/Engel%20A.%20Problem-solving%20strategies%20(for%20math%20olympiads)(Springer,%201998)(ISBN%200387982191)(O)(415s)MSch.pdf
- 5 years, 8 months ago
Oh oops, never mind the url.
- 5 years, 8 months ago
Thanks.. will do that. :)
- 5 years, 8 months ago
You may refer to the original post Let's Get Started to understand the context of the Polynomial Sprint set.
Staff - 5 years, 9 months ago | 2020-04-06T13:11:54 | {
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http://mathhelpforum.com/pre-calculus/66019-annual-compounding-interest.html | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
4. Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
6. ## ok..
Originally Posted by masters
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
I got the answer from Soroban but I thank you for your effort.
7. ## yes...
Originally Posted by masters
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
Don't feel bad. No one is perfect except God and He is not taking any math courses lately. | 2015-04-19T05:46:10 | {
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https://mathematica.stackexchange.com/questions/228824/different-results-in-deigensystem-compared-to-ndeigensystem-for-laplacian-eigenv | # Different results in DEigensystem compared to NDEigensystem for Laplacian eigenvalue problem (-Δu=λu) on unit square
I want to calculate the solution to the Laplacian eigenvalue problem on the unit square with trivial Dirichlet boundary conditions: $$- \Delta u(x,y) = \lambda u(x,y) \text{ on } {[0,1]}^2$$ with $$u(0,y)=0$$,$$u(1,y)=0$$,$$u(x,0)=0$$,$$u(x,1)=0$$.
However, Mathematica 12 reports different eigenfunctions when using NDEigensystem in contrast to DEigensystem using the following codes:
DEigensystem version:
{vals, funs} =
DEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]
NDEigensystem version:
{vals, funs} =
NDEigensystem[{-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]},
u[x, y], {x, y} ∈ Rectangle[], 2,
Method -> {"PDEDiscretization" -> {"FiniteElement",
"MeshOptions" -> {"MaxCellMeasure" -> 0.0001}}}];
Table[ContourPlot[funs[[i]], {x, y} ∈ Rectangle[],
PlotRange -> All, PlotLabel -> vals[[i]], PlotTheme -> "Minimal",
Axes -> True], {i, Length[vals]}]
For the second eigenfunction, the DEigensystem reports the classical textbook eigenfunction, while the numerical solution with NDEigensystem is fundamentally different, although the mesh discretization is set to a very small value.
Why is that?
• For the lowest state, they're just negatives of each other, so that's fine. The second state is doubly degenerate, and so I'm betting that the second state in the second code is a linear combination of the two degenerate states from the first code. – march Aug 20 '20 at 20:59
• @march is right. Just observe the graphics of first 3 eigenfunctions and you'll know what happens. – xzczd Aug 21 '20 at 3:22
• And, of course, the problem will go away if you work on a rectangular domain such as $[0,1] \times [0,1.1]$. It would be interesting to see how close the last of these numbers can get to 1 before NDEigensystem treats them as degenerate & starts showing different results. – Michael Seifert Aug 24 '20 at 19:29
As already pointed out in the comments by @march and @xzczd, the second lowest state with eigenvalue $$\lambda_{1,2} = \lambda_{2,1} = 5 \pi^2$$ is doubly degenerate.
DEigensystem
NDEigensystem
This means that the corresponding eigenfunctions are not only determined up to a scaling (as for the lowest state). They are rather determined to be some orthogonal basis of the eigenspace
$$E_{5 \pi^2} = \{a \phi_{1,2} + b \phi_{2,1} \mid a,b \in \mathbb{R}, \, -\Delta \phi_{1,2} = 5 \pi^2 \phi_{1,2}, \, -\Delta \phi_{2,1} = 5 \pi^2 \phi_{2,1}, \, \phi_{1,2} \perp \phi_{2,1}\}$$
We have $$\text{dim}(E_{5 \pi^2}) = 2$$. The results from NDEigensystem ($$\phi_{1,2,\text{ND}}, \phi_{2,1,\text{ND}}$$) are therefore also valid solutions because they span the same eigenspace:
$$E_{5 \pi^2} = \text{span}\{\phi_{1,2,\text{NDEigen}}, \phi_{2,1,\text{NDEigen}}\} \\ = \text{span}\{\phi_{1,2,\text{DEigen}}, \phi_{2,1,\text{DEigen}}\} \\ = \text{span}\{\sin(1 \pi x)\sin(2 \pi y), \sin(2 \pi x)\sin(1 \pi y)\}$$ | 2021-06-19T00:20:54 | {
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https://math.stackexchange.com/questions/1305812/mean-value-theorem-and-inequality | # Mean Value Theorem and Inequality.
Using the mean value theorem prove the below inequality.
$$\frac{1}{2\sqrt{x}} (x-1)<\sqrt{x}-1<\frac{1}{2}(x-1)$$ for $x > 1$.
I don't understand how these inequalities are related. Am I supposed to work out the first one and then the second and so on? I also would be really grateful if anyone had the time to give some insight in what this problem asks to me.
I really wish someone could give a very simple solution.
Apply the mean value theorem to the function $f(t) = \sqrt{t}$ on the interval $[1,x]$ to deduce
$$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$$
for some $c \in (1,x)$. Use the fact that
$$\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$$
to conclude.
• how did you know $$f(t)=\sqrt{t}$$? can you show me a different approach too? if possible. – Sherlock Homies May 30 '15 at 21:27
• @SherlockHomies note $\sqrt{x} - 1 = \sqrt{x} - \sqrt{1} = f(x) - f(1)$, where $f(t) = \sqrt{t}$. If you want to prove the inequalities without the use of MVT, then rationalize the numerator to get $$\sqrt{x} - 1 = \frac{x-1}{\sqrt{x} + 1}$$ and use the fact that for $x > 1$, $$\frac{1}{2\sqrt{x}} < \frac{1}{\sqrt{x} + 1} < \frac{1}{2}.$$ – kobe May 30 '15 at 21:30
• wait a sec how would you piece up all together to get to end of the problem. Like in the question above . I mean the last step.Using the mean value theorem again in the end? – Sherlock Homies May 30 '15 at 21:49
• @SherlockHomies multiply the inequalities $\frac{1}{2\sqrt{x}} < \frac{1}{2\sqrt{c}} < \frac{1}{2}$ by $x - 1$ and use the fact $\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x - 1)$ to get the desired inequalities. – kobe May 30 '15 at 21:52
By the MVT, there is $c \in ]1,x[$ such that $$\sqrt{x} - 1 = \frac{1}{2\sqrt{c}}(x-1), \qquad \qquad \left[f(b)-f(a) = f'(c)(b-a)\right]$$ so you use that: $$c < x \implies \sqrt{c} < \sqrt{x} \implies 2 \sqrt{c} < 2\sqrt{x} \implies \frac{1}{2\sqrt{x}}<\frac{1}{2\sqrt{c}}$$ to get one side, and use that: $$c > 1 \implies \sqrt{c} > 1 \implies 2\sqrt{c} > 2 \implies \frac{1}{2\sqrt{c}}< \frac{1}{2}$$ to get the other. | 2019-06-18T06:59:16 | {
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http://jitsurei.net/boys-twin-xdjeaej/0ad894-non-isomorphic-graphs-with-8-vertices | Non-isomorphic graphs with degree sequence $1,1,1,2,2,3$. What if the degrees of the vertices in the two graphs are the same (so both graphs have vertices with degrees 1, 2, 2, 3, and 4, for example)? Second, the transfer vertex equation is established to synthesize 2-DOF rotation graphs. (c)Find a simple graph with 5 vertices that is isomorphic to its own complement. Note − In short, out of the two isomorphic graphs, one is a tweaked version of the other. We know that a tree (connected by definition) with 5 vertices has to have 4 edges. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. 5.1.8. The graph defined by V = {a,b,c,d,e} and E = {{a,c},{6,d}, {b,e},{c,d), {d,e}} ii. Find all non-isomorphic trees with 5 vertices. For example, these two graphs are not isomorphic, G1: • • • • G2: • • • • since one has four vertices of degree 2 and the other has just two. (b) Draw all non-isomorphic simple graphs with four vertices. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Constructing non-isomorphic signless Laplacian cospectral graphs. Solution: Since there are 10 possible edges, Gmust have 5 edges. All simple cubic Cayley graphs of degree 7 were generated. In particular, ( x − 1 ) 3 x {\displaystyle (x-1)^{3}x} is the chromatic polynomial of both the claw graph and the path graph on 4 vertices. For all the graphs on less than 11 vertices I've used the data available in graph6 format here. Yes. We have also produced numerous examples of non-isomorphic signless Laplacian cospectral graphs. Finally, edge level equation is established to synthesize 2-DOF displacement graphs. Show that two projections of the Petersen graph are isomorphic. I would like to iterate over all connected non isomorphic graphs and test some properties. By continuing you agree to the use of cookies. However, the existing synthesis methods mainly focused on 1-DOF PGTs, while the research on the synthesis of multi-DOF PGTs is very limited. Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. We use cookies to help provide and enhance our service and tailor content and ads. The list does not contain all graphs with 8 vertices. ... consist of a non-empty independent set U of n vertices, and a non-empty independent set W of m vertices and have an edge (v,w) … However, notice that graph C also has four vertices and three edges, and yet as a graph it seems di↵erent from the first two. Two non-isomorphic trees with 7 edges and 6 vertices.iv. This thesis investigates the generation of non-isomorphic simple cubic Cayley graphs. 1.2 14 Two non-isomorphic graphs a d e f b 1 5 h g 4 2 6 c 8 7 3 3 Vertices: 8 Vertices: 8 Edges: 10 Edges: 10 Vertex sequence: 3, 3, 3, 3, 2, 2, 2, 2. There are several such graphs: three are shown below. There will be only one non isomorphic graph with 8 vertices and each vertex has degree 5. because 8 vertices with each vertex degree 5 means total degre view the full answer. For example, the parent graph of Fig. For example, all trees on n vertices have the same chromatic polynomial. List all non-identical simple labelled graphs with 4 vertices and 3 edges. Two non-isomorphic trees with 5 vertices. You Should Not Include Two Graphs That Are Isomorphic. For higher number of vertices, these graphs can be generated by a number of theorems and procedures which we shall discuss in the following sections. Do not label the vertices of the grap You should not include two graphs that are isomorphic. Both 1-DOF and multi-DOF planetary gear trains (PGTs) have extensive application in various kinds of mechanical equipment. I About (a) Draw All Non-isomorphic Simple Graphs With Three Vertices. A Google search shows that a paper by P. O. de Wet gives a simple construction that yields approximately $\sqrt{T_n}$ non-isomorphic graphs of order n. Isomorphic Graphs. One example that will work is C 5: G= ˘=G = Exercise 31. Figure 10: Two isomorphic graphs A and B and a non-isomorphic graph C; each have four vertices and three edges. Now I would like to test the results on at least all connected graphs on 11 vertices. The line graph of the complete graph K n is also known as the triangular graph, the Johnson graph J(n, 2), or the complement of the Kneser graph KG n,2.Triangular graphs are characterized by their spectra, except for n = 8. Question: Exercise 8.3.3: Draw All Non-isomorphic Graphs With 3 Or 4 Vertices. Draw all possible graphs having 2 edges and 2 vertices; that is, draw all non-isomorphic graphs having 2 edges and 2 vertices. Hello! Do Not Label The Vertices Of The Graph. A complete bipartite graph with at least 5 vertices.viii. With 4 vertices (labelled 1,2,3,4), there are 4 2 How many of these are not isomorphic as unlabelled graphs? By continuing you agree to the use of cookies. But still confused between the isomorphic and non-isomorphic $\endgroup$ – YOUSEFY Oct 21 '16 at 17:01 A bipartitie graph where every vertex has degree 3. iv. Copyright © 2021 Elsevier B.V. or its licensors or contributors. Their degree sequences are (2,2,2,2) and (1,2,2,3). In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. A bipartitie graph where every vertex has degree 5.vii. Therefore, a large class of graphs are non-isomorphic and Q-cospectral to their partial transpose, when number of vertices is less then 8. An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. These can be used to show two graphs are not isomorphic, but can not show that two graphs are isomorphic. 3(b). https://doi.org/10.1016/j.disc.2019.111783. And that any graph with 4 edges would have a Total Degree (TD) of 8. iii. So, it follows logically to look for an algorithm or method that finds all these graphs. 5.1.10. The Whitney graph theorem can be extended to hypergraphs. The synthesis results of 8- and 9-link 2-DOF PGTs are new results that have not been reported. A graph with degree sequence (6,2,2,1,1,1,1) v. A graph that proves that in a group of 6 people it is possible for everyone to be friends with exactly 3 people. The sequence of number of non-isomorphic graphs on n vertices for n = 1,4,5,8,9,12,13,16... is as follows: 1,1,2,10,36,720,5600,703760,...For any graph G on n vertices the below construction produces a self-complementary graph on 4n vertices! $\endgroup$ – user940 Sep 15 '17 at 16:56 Isomorphic graphs have the same chromatic polynomial, but non-isomorphic graphs can be chromatically equivalent. The synthesis results of 8- and 9-link 2-DOF PGTs, to the best of our knowledge, are new results that have not been reported in literature. 8 vertices - Graphs are ordered by increasing number of edges in the left column. (a) Draw all non-isomorphic simple graphs with three vertices. If all the edges in a conventional graph of PGT are assumed to be revolute edges, the derived graph is its parent graph. Copyright © 2021 Elsevier B.V. or its licensors or contributors. First, non-fractionated parent graphs corresponding to each link assortment are synthesized. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. To show graphs are not isomorphic, we need only nd just one condition, known to be necessary for isomorphic graphs, which does not hold. 1(b) is shown in Fig. Isomorphic and Non-Isomorphic Graphs - Duration: 10:14. The isomorphism of these two different presentations can be seen fairly easily: pick Looking at the documentation I've found that there is a graph database in sage. edge, 2 non-isomorphic graphs with 2 edges, 3 non-isomorphic graphs with 3 edges, 2 non-isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. As an example of a non-graph theoretic property, consider "the number of times edges cross when the graph is drawn in the plane.'' By Our constructions are significantly powerful. $\endgroup$ – mahavir Feb 22 '14 at 3:14 $\begingroup$ @mahavir This is not true with 4 vertices and 2 edges. Regular, Complete and Complete This paper presents an automatic method to synthesize non-fractionated 2-DOF PGTs, free of degenerate and isomorphic structures. 10:14. The transfer vertex equation and edge level equation of PGTs are developed. graph. There is a closed-form numerical solution you can use. Is it possible for two different (non-isomorphic) graphs to have the same number of vertices and the same number of edges? Use the options to return a count on the number of isomorphic classes or a representative graph from each class. For example, both graphs are connected, have four vertices and three edges. © 2019 Elsevier B.V. All rights reserved. WUCT121 Graphs 32 1.8. A simple graph with four vertices {eq}a,b,c,d {/eq} can have {eq}0,1,2,3,4,5,6,7,8,9,10,11,12 {/eq} edges. (a)Draw the isomorphism classes of connected graphs on 4 vertices, and give the vertex and edge Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. Draw two such graphs or explain why not. They may also be characterized (again with the exception of K 8) as the strongly regular graphs with parameters srg(n(n − 1)/2, 2(n − 2), n − 2, 4). Answer. of edges are 0,1,2. Two non-isomorphic graphs with degree sequence (3, 3, 3, 3, 2, 2, 2, 2)v. A graph that is not connected and has a cycle.vi. A method based on a set of independent loops is presented to precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs. Since isomorphic graphs are “essentially the same”, we can use this idea to classify graphs. Distance Between Vertices and Connected Components - … 1/25/2005 Tucker, Sec. The atlas of non-fractionated 2-DOF PGTs with up to nine links is automatically generated. https://www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices 3(a) and its adjacency matrix is shown in Fig. The research is motivated indirectly by the long standing conjecture that all Cayley graphs with at least three vertices are Hamiltonian. In this article, we generate large families of non-isomorphic and signless Laplacian cospectral graphs using partial transpose on graphs. 5. (Start with: how many edges must it have?) So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg ≥ 1. Solution. Also, I've counted the non-isomorphic for 7 vertices, it gives me 11 with the same technique as you explained and for 6 vertices, it gives me 6 non-isomorphic. by a single edge, the vertices are called adjacent.. A graph is said to be connected if every pair of vertices in the graph is connected. 1 , 1 , 1 , 1 , 4 An unlabelled graph also can be thought of as an isomorphic graph. Sarada Herke 112,209 views. A method based on a set of independent loops is presented to detect disconnection and fractionation. An automatic method is presented for the structural synthesis of non-fractionated 2-DOF PGTs. Find three nonisomorphic graphs with the same degree sequence (1,1,1,2,2,3). More than 70% of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of vertices is ≤8. Their edge connectivity is retained. But as to the construction of all the non-isomorphic graphs of any given order not as much is said. For an example, look at the graph at the top of the first page. Two graphs G 1 and G 2 are said to be isomorphic if − Their number of components (vertices and edges) are same. $\begingroup$ with 4 vertices all graphs drawn are isomorphic if the no. Two graphs with different degree sequences cannot be isomorphic. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Automatic structural synthesis of non-fractionated 2-DOF planetary gear trains, https://doi.org/10.1016/j.mechmachtheory.2020.104125. Previous question Next question Transcribed Image Text from this Question. The NonIsomorphicGraphs command allows for operations to be performed for one member of each isomorphic class of undirected, unweighted graphs for a fixed number of vertices having a specified number of edges or range of edges. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v.Otherwise, they are called disconnected.If the two vertices are additionally connected by a path of length 1, i.e. • Isomorphic Graphs ... Graph Theory: 17. We use cookies to help provide and enhance our service and tailor content and ads. Figure 5.1.5. The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. Graph from each class © 2021 Elsevier B.V. or its licensors or.... The existing synthesis methods mainly focused on 1-DOF PGTs, while the research is indirectly... To nine links is automatically generated and isomorphic structures tree ( connected by definition ) with 5 vertices is! The first page research on the number of edges in the left column, look at top... Logically to look for an algorithm or method that finds all these graphs if the no that... Various kinds of mechanical equipment the top of the grap you Should Include... Draw all non-isomorphic simple cubic Cayley graphs with 3 or 4 vertices ( labelled )... Gmust have 5 edges documentation I 've found that there is a registered trademark of Elsevier B.V. its... A count on the number of edges different ( non-isomorphic ) graphs to the... But can not be isomorphic the results on at least all connected non graphs... Do not label the vertices of the Petersen graph are isomorphic first page link assortment synthesized. Of edges in the left column the top of the Petersen graph isomorphic! Connected non isomorphic graphs a and B and a non-isomorphic graph C each... ( Start with: how many edges must it have? each link are... Complete and Complete two graphs that are isomorphic an unlabelled graph also be! Work is C 5: G= ˘=G = Exercise 31 is it possible for two different ( non-isomorphic ) to! Possible for two different ( non-isomorphic ) graphs to have 4 edges would have a Total degree TD. 1-Dof PGTs, while the research on the synthesis of multi-DOF PGTs is very limited up to nine links automatically... Indirectly by the long standing conjecture that all Cayley graphs same degree sequence ( 1,1,1,2,2,3 ) must have. Trains ( PGTs ) have extensive application in various kinds of mechanical equipment mainly focused on PGTs. And rotation graphs vertex equation is established to synthesize 2-DOF rotation graphs of mechanical.... 4 2 Hello the other labelled 1,2,3,4 ), non isomorphic graphs with 8 vertices are several such graphs: three are shown below presented! Are synthesized equation and edge level equation is established to synthesize 2-DOF rotation.. Four vertices and three edges of edges of PGTs are new results that have not been reported Next question Image! Their degree sequences can not show that two projections of the Petersen graph are.! Sequences can not show that two projections of the two isomorphic graphs, one is a tweaked of... Isomorphic to its own complement are synthesized if the no 9-link 2-DOF PGTs are developed graph at the documentation 've... The options to return a count on the number of vertices and three edges idea!, both graphs are isomorphic all Cayley graphs of any given order not as much said! 3. iv a ) Draw all non-isomorphic graphs of degree 7 were generated of vertices is ≤8 PGTs very. 8- and 9-link 2-DOF PGTs with up to nine links is automatically generated extensive application in various kinds of equipment. Agree to the construction of all the non-isomorphic graphs having 2 edges and 2 vertices ; that is Draw. Found that there is a graph database in sage to show two graphs are connected, have vertices.: Exercise 8.3.3: Draw all non-isomorphic graphs having 2 edges and 2 vertices ; that is isomorphic its... $\begingroup$ with 4 vertices all graphs drawn are isomorphic to links... Exercise 31 degree 5.vii isomorphic as unlabelled graphs matrix is shown in Fig this article, can. 2-Dof displacement graphs all the graphs on less than 11 vertices I 've found that there is registered! 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To precisely detect disconnected and fractionated graphs including parent graphs and rotation graphs each class Petersen graph are.! Standing conjecture that all Cayley graphs of degree 7 were generated three edges graphs using partial transpose on graphs edges. Vertex equation is established to synthesize non-fractionated 2-DOF PGTs, while the research is motivated indirectly by long... − in short, out of the other, it follows logically to look for an algorithm method. Equation and edge level equation of PGTs are developed use the options to return a on. Both 1-DOF and multi-DOF planetary gear trains ( PGTs ) have extensive application in kinds. Edges would have a Total degree ( TD ) of 8 list all non-identical labelled... Mechanical equipment of non-isomorphic signless-Laplacian cospectral graphs can be generated with partial transpose when number of isomorphic or... 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Several such graphs: three are shown below trains ( PGTs ) have application... Example, look at the graph at the top of the grap you Should not Include two graphs that isomorphic... The left column ) graphs non isomorphic graphs with 8 vertices have 4 edges would have a Total degree ( ). C ) Find a simple graph with 4 vertices ( labelled 1,2,3,4 ) there... There is a closed-form numerical solution you can use vertices that is isomorphic to its own complement 2021! Shown in Fig research on the number of edges based on a set independent! Figure 10: two isomorphic graphs a and B and a non-isomorphic graph C ; each four! Presented for the structural synthesis of multi-DOF PGTs is very limited article, we generate families. 2 Hello were generated each link assortment are synthesized least three vertices parent graphs to. Laplacian cospectral graphs using partial transpose on graphs some properties non-isomorphic signless Laplacian cospectral.. Degree sequences can not be isomorphic, have four vertices and three.. Must it have? have a Total degree ( TD ) of.! And 3 edges families of non-isomorphic signless-Laplacian cospectral graphs 7 were generated 8! Edges and 2 vertices with at least 5 vertices.viii vertices has to have 4 edges different ( )... An algorithm or method that finds all these graphs PGTs with up to nine links is generated... First, non-fractionated parent graphs corresponding to each link assortment are synthesized of Elsevier B.V. or its or! Any graph with 4 vertices, it follows logically to look for algorithm... Graphs can be thought of as an isomorphic graph B.V. Constructing non-isomorphic signless Laplacian cospectral graphs can be of... 1-Dof PGTs, while the research is motivated indirectly by the long conjecture... Continuing you agree to the use of cookies would like to iterate over all connected graphs less. An unlabelled graph also can be generated with partial transpose on graphs 5 vertices has to the. B ) Draw all non-isomorphic simple graphs with three vertices help provide enhance... Now I would like to test the results on at least all connected non isomorphic graphs are isomorphic. An automatic method to synthesize non-fractionated 2-DOF PGTs that a tree ( connected by definition ) with 5 vertices is... Is it possible for two different ( non-isomorphic ) graphs to have 4 edges Constructing non-isomorphic signless Laplacian graphs...
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http://mathhelpforum.com/pre-calculus/158721-determine-line-tangent-curve-not.html | # Math Help - Determine is a line is tangent to a curve or not?
1. ## Determine is a line is tangent to a curve or not?
Sorry if it sounds silly but I am thinking if there is a way to find out if a line is a tanget of a curve?
Say given a curve of $x^2 + 3xy + y^2 + 4 = 0$, how do I determine if any one given line is a tangent of this curve or not? The line can be y=x+7 or y=-2x-4, etc. And if it is a tangent of the curve, how do I find the points that this line is tangent to the curve?
I am thinking that I differentiate the equation of the curve to have dy/dx. Then I can get the gradient at any one point of the curve to compare with a line. But this isn't enough because same gradient doesn't mean is a tangent to the curve. And how I find the points which the line is tangent to the curve without any given points at all?
Thanks for any help!
2. Do a simultaneous equation to determine first whether the line meet the curve or not.
Once done, you need to find if the gradient at the point of intersection is the same as the gradient of the line.
If they meet and share a common gradient, you have your tangent. If one of those two conditions are not met, the line is not a gradient.
3. ## q.[/math]
Hello, xEnOn!
I have a method but it's quite long.
Maybe someone has a shorter approach?
Is there a way to find out if a line is a tanget to a curve?
Say, given a curve: . $x^2 + 3xy + y^2 + 4 \:=\: 0$
how do I determine if, say, $y \:=\:x+6$ is a tangent of this curve or not?
And if it is a tangent of the curve, how do I find the points of tangency?
I am thinking that I differentiate the equation of the curve to have dy/dx.
Then I can get the gradient at any one point of the curve to compare with the line.
But this isn't enough because same gradient doesn't mean it is tangent to the curve.
And how I find the points which the line is tangent without any given points at all?
We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$
Find the points of intersection, substitute [2] into [1]:
. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$
. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$
. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$
. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$
Substitute into [2]: . $y \:=\:4,\:2$
The hyperbola and the line intersect at: . $P(-2,4)\,\text{ and }\,Q(-4,2)$
If the line is tangent to the hyperbola, their slopes will be equal at $\,P$ and $Q.$
The slope of the line is: . $\boxed{m \,=\,1}$
Find the slope of the hyperbola.
Differentiate implicitly: . $2x + 3x\dfrac{dy}{dx} + 3y + 2y\dfrac{dy}{dx} \:=\:0$
. . $(3x + 2y)\dfrac{dy}{dx} \:=\:-2(x+y) \quad\Rightarrow\quad \boxed{\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(x+y)}{3x+2y} }$
$\text{At }P(\text{-}2,4)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}2+4)}{3(\text{-}2) + 2(4)} \:=\:\dfrac{\text{-}4}{2} \:=\:-2 \;\ne \;1$
$\text{At }Q(\text{-}4,2)\!:\;\;\dfrac{dy}{dx} \:=\:\dfrac{\text{-}2(\text{-}4+2)}{3(\text{-}4)+2(2)} \:=\:\dfrac{4}{\text{-}8} \:=\:-\dfrac{1}{2} \;\ne\;1$
The slopes are not equal at the intersection points.
Therefore, the line is not tangent to the hyperbola.
Edit: Too slow again!
. . . .Unknown008 already gave an excellent game plan.
4. Originally Posted by Soroban
I have a method but it's quite long.
Maybe someone has a shorter approach?
We have: . $\begin{Bmatrix}x^2 + 3xy + y^2 + 4 \;=\; 0 & [1] \\ y \;=\; x + 6 & [2] \end{Bmatrix}$
Find the points of intersection, substitute [2] into [1]:
. . $x^2 + 3x(x+6) + (x+6)^2 + 4 \:=\:0$
. . $x^2 + 3x^2 + 18x + x^2 + 12x + 36 + 4 \:=\:0$
. . $5x^2 + 30x + 40 \:=\:0 \quad\Rightarrow\quad x^2 + 6x + 8 \:=\:0$
. . $(x+2)(x+4)\:=\:0 \quad\Rightarrow\quad x \:=\:-2,\:-4$
A somewhat shorter method is to follow Soroban's approach by substituting the line equation into the conic equation, as above, and finding the values of x where the line meets the conic. These turned out to be –2 and –4. Those values are different, so you can tell straight away that the line cannot be a tangent to the conic. If it were, then there would be a repeated root for x (because the point at which the line touches the conic would be a double root of the equation). As it is, the line crosses the conic at two distinct points, so it cannot touch it.
Suppose that we have the conic equation $x^2 + 3xy + y^2 + 4 = 0$, as before, but that the equation of the line is $4x+y+4=0$. Substitute $y = -4x-4$ into the conic equation, as in Soroban's method. You get
$x^2 + 3x(-4x-4) + (-4x-4)^2 + 4 = 0$,
$x^2 - 12x^2 - 12 + 16x^2 + 32x + 16 + 4 = 0$,
$5x^2 + 20x + 20 = 0$,
$x^2+4x+4=0$,
$(x+2)^2 = 0$.
This time, there is a repeated root x = –2, which tells you that the line touches the conic at that value of x. So in that case, the line is tangent to the conic.
5. By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
I suppose I will still get a result in someway no matter when I do simultaneous?
Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation.
6. Originally Posted by xEnOn
By doing the simultaneous equation method, if the line is not a tangent to the curve, what result would I get to know that it is not a tangent?
If you're talking about binomials, (x+a)(x+b): If the simultaneous equation gives 0 or 2 answers, then it is not tangent.
In other cases, it is quite lengthy - you have to find the co-ordinates of where they meet and then check if their gradients are the same for both equations at that point.
You cannot easily determine whether it is tangent or not, just from the intersection points. (Unless there are no intersection points)
Originally Posted by xEnOn
I suppose I will still get a result in someway no matter when I do simultaneous?
No. Well not real results anyway (but who wants to get into imaginary numbers?). Sometimes when you do simultaneous equations, you won't get any answers and this will tell you straight away that the 2 graphs don't meet.
Originally Posted by xEnOn
Also, I often get very confused by doing simultaneous. When we do simultaneous from 2 equations to get 2 unknown variables, what kind of relationship do the 2 equations need to have? Very often, I take a wrong equation to do a simultaneous equation. | 2015-08-28T17:36:22 | {
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# A certain stock exchange designates each stock with a one-,
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A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A. 2951
B. 8125
C. 15600
D. 16302
E. 18278
OPEN DISCUSSION OF THIS QUESTION IS HERE: a-certain-stock-exchange-designates-each-stock-with-a-86656.html
[Reveal] Spoiler: OA
Last edited by Bunuel on 20 Jul 2013, 09:43, edited 1 time in total.
Edited the question.
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Remember that Numbers can be repeated.
Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576
Total = 26 + 676 + 17576 = 18278
In future do not put OA with the question. As people might solve it back wards to reach the solution.
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You have 3 different sets of Permutations available.
First set is of stocks that have only 1 letter, or 26 possibilities
Second set is of stocks with 2 letters, or 26 * 26 possibilities
Third set is of stocks with 3 letters, or 26 * 26 * 26 possibilities.
This equal 26 +676 + 17576 = 18,278.
japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A) 2951
B) 8125
C) 15600
D) 16302
E) 18278
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30 May 2008, 07:42
abhijit_sen wrote:
In future do not put OA with the question. As people might solve it back wards to reach the solution.
I think the moderator/owner of the forum should add a "spoiler" function on the forum. That function is the perfect fit for this type of forum, but this forum does not have it.
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japped187 wrote:
A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A) 2951
B) 8125
C) 15600
D) 16302
E) 18278
As discussed above, the solution is indeed 26 + 26*26 + 26*26*26. I noticed that all the digits were different in the options.
So I simply calculated the unit digit of each product. 6 +6 + 6. = 18 = Unit digit of answer = 8. Hence, E.
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A certain stock exchange designates each stock with a one-, two-, or three- letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A. 2951
B. 8125
C. 15600
D. 16302
E. 18278
1 letter codes = 26
2 letter codes =2 6^2
3 letter codes = 26^3
Total=26+26^2+26^3
The problem we are faced now is how to get the answer quickly. Note that the units digit of 26+26^2+26^3 would be (6+6+6=18) 8. Only one answer choice has 8 as unit digit: E (18,278). So I believe, even not calculating 26+26^2+26^3, that answer is E.
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# What is the perimeter of a square with area 9p^2/16 ?
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What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
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22 Aug 2016, 04:01
What is the perimeter of a square with area 9p^2/16 ?
Area of square, (side)^2 = (3p/4)^2
Therefore side of the square = 3p/4
Perimeter of square = 4*side = 4* (3p/4) = 3p
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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29 Nov 2017, 22:14
1
Hi All,
We're told that the area of a SQUARE is (9P^2)/16. We're asked for the PERIMETER of the square. This question can be solved by TESTing VALUES.
Since the denominator is 16, we can 'cancel' it out if we make P=4...
IF... P=4, then the area of the square is 9(16)/16 = 9. This means that each side of the square = 3. Thus, we're looking for an answer that equals 4(3) = 12 when we plug P=4 into the answer choices. There's only one answer that matches...
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What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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30 Nov 2017, 06:36
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
Find $$s$$.
Area of square = $$s^2 = \frac{9\pi^2}{16}$$
$$\sqrt{s^2} = \frac{(\sqrt{9})*(\sqrt{\pi^2})}{\sqrt{16}}$$
$$s= \frac{3\pi}{4}$$
Perimeter = $$4s=(\frac{3\pi}{4})*4 = 3\pi$$
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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30 Nov 2017, 07:04
Area = 9p^2/16
So a^2 = 9p^2/16 ( assumed side of a square as a)
So a = 3p/4
And perimeter of square is sum of all sides
So perimeter = 4a
So ans is 4(3p/4) = 3p
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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30 Nov 2017, 12:15
1
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
We know that the area of square = (side)^2
Thus, we can write :
$$(Side)^2 = \frac{9p^2}{16}$$
$$(Side)^2 = (\frac{3p}{4})^2$$
Since the side of square is always positive, we can conclude that the length of each side $$= \frac{3p}{4}$$
And the perimeter of the square $$= 4 * side = 4 * \frac{3p}{4} = 3p$$
The correct answer is Option C.
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Re: What is the perimeter of a square with area 9p^2/16 ? [#permalink]
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03 Dec 2017, 18:47
1
Bunuel wrote:
What is the perimeter of a square with area 9p^2/16 ?
A. 3p/4
B. 3p^2/4
C. 3p
D. 3p^2
E. 4p/3
Since a side of the square is equal to √(9p^2/16) = 3p/4, then the perimeter of the square is 4 x 3p/4 = 12p/4 = 3p.
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Re: What is the perimeter of a square with area 9p^2/16 ? &nbs [#permalink] 03 Dec 2017, 18:47
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https://www.bionicturtle.com/forum/threads/miller-chapter-4-distributions-study-notes-pg-75-question-2.8301/ | What's new
# Miller - Chapter 4 - Distributions - Study Notes, Pg 75, question #2
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi David,
In question (2) referenced as above,
(2) At the start of the year, a stock price is $100. A twelve-step binomial model describes the stock price evolution such that each month the extreme volatility price with either jump from S(t) to S(t)*u with 60.0% probability or down to S(t)*d with 40.0% probability. The up jump (u) = 1.1 and the down jump (d) = 1/1.1; note these (u) and (d) parameters correspond to an annual volatility of about 33% as exp[33%*SQRT(1/12)] ~= 1.10. At the end of the year, which is nearest to the probability that the stock price will be exactly$121?
(a) 0.33%
(b) 3.49%
(c) 12.25%
(d) 22.70%
There are 13 outcomes at the end of the 12-step binomial, with $100 as the outcome that must correspond to six up jumps and six down jumps. Therefore,$121 must be the outcome due to seven up jumps and five down jumps: $100*1.1^7*(1/1.1)^5 =$121
Such that we want the binomial probability given by:
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
Clarification
Although, intuitively, 13 outcomes at the end of the 12-step binomial seems right (ie expected value of stock price at the end of 12 steps ??), how do you figure out six up jumps and six down jumps from the Binomial model to get \$100 as the outcome? Also how do you get :
Binomial Prob [X = 7| n =12, p = 60%] = 22.70%
I guess I am missing something that I should understand.
Thanks!
Jayanthi
#### ShaktiRathore
##### Well-Known Member
Subscriber
Hi
Consider n up jumps and 12-n down jumps to get 121 in the end
Therefore,100*(1.1)^n*(1/1.1)^12-n=121=>1.1^(2n-12)=1.1^2=>2n-12=2=>2n=14=>n=7 therefore there are 7 up and 12-7=5 down jumps.
Binomial probability is 12C7(.6)^7*(.4)^5=22.69%
Thanks
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi Shakti,
Yes - that is indeed great! Thanks a tonne for making it so simple to understand!
Jayanthi
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @Jayanthi Sankaran the source question thread contains a visual that should help clarify, too. Please see https://www.bionicturtle.com/forum/...ributions-i-miller-chapter-4.7025/#post-28837
ie..,
#### Dr. Jayanthi Sankaran
##### Well-Known Member
This is neat, David
Thanks!
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi David,
Does the above visual come with its own spreadsheet?
Thanks
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Thanks David - will go through it, at leisure
Warm regards
Jayanthi
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!
#### eva maria
##### New Member
Hi @eva maria That wasn't me, as explained i think in the linked discussion, I tend to prefer here a visual/intuitive approach (which is a matter of style). Shakti's solution is mathematically elegant, he didn't strictly type his parens but to be honest I think it helps to go with the flow sometimes and replicate what somebody is trying to show you, when math is involved. What I mean is that, he's quickly shared an elegant solution such that, sometimes the best thing to do is replicate the steps:
• 100*(1.1)^n*(1/1.1)^(12-n) = 121; i.e., the fundamental assumption. Then he divides each side by 100:
• 1*(1.1)^n*(1/1.1)^(12-n) = 1.21; then realize that (1/1.1) = 1.1^(-1) such that we have:
• 1*(1.1)^n*[1.1^(-1)]^(12-n) = 1.21 -->
• (1.1)^n * 1.1^[-1 * (12-n)] = 1.21 --> because (a^x)^y = a^(x*y), such that:
• (1.1)^n * 1.1^(n-12) = 1.21,
• (1.1)^[n+(n-12)] = 1.21, because a^x * a^y = a^(x+y), and
• (1.1)^(2n - 12) = 1.21 = 1.1^2; he's really skilled so he probably did all of that in his head and didn't worry about expressing every little detail. Whereas I am actually not so quick and i need to detail it all out. I hope that explains, thanks!
Many thanks David
Last edited by a moderator: | 2021-05-17T12:20:03 | {
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https://math.stackexchange.com/questions/2664789/characterizing-discontinuous-derivatives | # Characterizing discontinuous derivatives
Apparently the set of discontinuity of derivatives is weird in its own sense. Following are the examples that I know so far:
$1.$ $$g(x)=\left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & x \in (0,1] \\ 0 & x=0 \end{array}\right.$$ $g'$ is discontinuous at $x=0$.
$2.$ The Volterra function defined on the ternary Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Cantor set ,that is on a nowhere dense set of measure zero.
$3.$ The Volterra function defined on the Fat-Cantor set is differentiable everywhere but the derivative is discontinuous on the whole of Fat-Cantor set ,that is on a set of positive measure, but not full measure.
$4.$ I am yet to find a derivative which is discontinuous on a set of full measure.
Questions:
1.What are some examples of functions whose derivative is discontinuous on a dense set of zero measure , say on the rationals?
2.What are some examples of functions whose derivative is discontinuous on a dense set of positive measure , say on the irrationals?
Update: One can find a function which is differentiable everywhere but whose derivative is discontinuous on a dense set of zero measure here.
• The Volterra function is not defined just on the Cantor set, but the whole interval. It equals $0$ on the Cantor set. I know you know this. But perhaps "defined relative to the Cantor set" might be more accurate. – zhw. Feb 24 '18 at 17:32
• @zhw Yeah, but I think its pretty obvious though. :P – yasir Feb 24 '18 at 17:44
• "Some good discussion about this can be found here and here." The first link you gave gives several references to the fact that any $F_{\sigma}$ first category set (equivalently, any countable union of closed nowhere dense sets) is the discontinuity set of some derivative. Those references that give a proof of this result do so in a constructive way, in the sense that given any countable union of closed nowhere dense sets (note that $\mathbb Q$ is a countable union of singleton sets), a differentiable function is constructed whose derivative is discontinuous exactly on that countable union. – Dave L. Renfro Feb 24 '18 at 21:08
There is no everywhere differentiable function $f$ on $[0,1]$ such that $f'$ is discontinuous at each irrational there. That's because $f',$ being the everywhere pointwise limit of continuous functions, is continuous on a dense $G_\delta$ subset of $[0,1].$ This is a result of Baire. Thus $f'$ can't be continuous only on a subset of the rationals, a set of the first category.
But there is a differentiable function whose derivative is discontinuous on a set of full measure.
Proof: For every Cantor set $K\subset [0,1]$ there is a "Volterra function" $f$ relative to $K,$ which for the purpose at hand means a differentiable function $f$ on $[0,1]$ such that i)$|f|\le 1$ on $[0,1],$ ii) $|f'|\le 1$ on $[0,1],$ iii) $f'$ is continuous on $[0,1]\setminus K,$ iv) $f'$ is discontinuous at each point of $K.$
Now we can choose disjoint Cantor sets $K_n \subset [0,1]$ such that $\sum_n m(K_n) = 1.$ For each $n$ we choose a Volterra function $f_n$ as above. Then define
$$F=\sum_{n=1}^{\infty} \frac{f_n}{2^n}.$$
$F$ is well defined by this series, and is differentiable on $[0,1].$ That's because each summand above is differentiable there, and the sum of derivatives converges uniformly on $[0,1].$ So we have
$$F'(x) = \sum_{n=1}^{\infty} \frac{f_n'(x)}{2^n}\,\, \text { for each } x\in [0,1].$$
Let $x_0\in \cup K_n.$ Then $x_0$ is in some $K_{n_0}.$ We can write
$$F'(x) = \frac{f_{n_0}'(x)}{2^{n_0}} + \sum_{n\ne n_0}\frac{f_n'(x)}{2^n}.$$
Now the sum on the right is continuous at $x_0,$ being the uniform limit of functions continuous at $x_0.$ But $f_{n_0}'/2^{n_0}$ is not continuous at $x_0.$ This shows $F'$ is not continuous at $x_0.$ Since $x_0$ was an aribtrary point in $\cup K_n,$ $F'$ is discontinuous on a set of full measure as desired.
• @ zhw So, if $f'$ is discontinuous on a set $D$, then $D$ can be dense only if its Lebesgue measure is zero? – yasir Feb 25 '18 at 3:04
• No, certainly not. A set of full measure is dense. – zhw. Feb 25 '18 at 7:09
• @ zhw. Is the set on which $F'$ is continuous dense, as per your claim in first paragraph? – yasir Feb 25 '18 at 9:26
• @yasir Yes, it has to be. – zhw. Feb 25 '18 at 17:14 | 2021-01-19T08:11:09 | {
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https://www.hpmuseum.org/forum/thread-12423.html | Small Solver Program
02-14-2019, 05:25 AM
Post: #1
Gamo Senior Member Posts: 518 Joined: Dec 2016
Small Solver Program
Recently I try to make a small but versatile solver program specifically
for HP Voyager Series calculator.
This solver solve for f(x) = 0 and use some type of counter for iteration.
Due to the iteration process this program will run slow on Original Calculator.
With modern 12C, 15C, and all other emulator this will run extremely fast.
This program is intended for HP-11C but this example will use
HP-15C emulator so that this solver program result can be compare
with the 15C build-in SOLVE function.
---------------------------------------------------------
To run program.
Enter 2 guesses closest to the root.
X1 [ENTER] X2 [A] display answer // Use this Solver
X1 [ENTER] X2 f [SOLVE] [E] display answer // Use 15C build-in SOLVE
---------------------------------------------------------
Example for X^X = Y // X to the power of X equal to Y
X on R1 // Register Storage 1 // For this Sover
Y on R5 // Register Storage 5
X on R2 // Register Storage 2 // For 15C SOLVE
Y on R5 // Register Storage 5
--------------------------------------------------------
Program:
Code:
LBL A // This Solver STO 0 Rv STO 1 90 STO I CLx LBL 1 RCL 0 / CHS RCL 1 + STO 1 DSE I GTO B RCL 1 RTN ------------------------------ LBL B // f(x) program for X^X = Y RCL 1 LN RCL 1 x RCL 5 LN - GTO 1 ------------------------------- LBL E // 15C SOLVE STO 2 // f(x) program for X^X = Y LN RCL 2 x RCL 5 LN - RTN
-------------------------------------------
Example: X^X = 1000
STORE 1000 to R5 // [STO] 5
This Solver
4 [ENTER] 6 f [A] display 4.555535704
-------------------------------------------------------
15C SOLVE
4 [ENTER] 6 f [SOLVE] [E] display 4.555535705
-------------------------------------------------------
Remark:
For this solver answer is in R1
For 15C SOLVE answer is in R2
Gamo
02-14-2019, 07:06 AM
Post: #2
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote: Example: X^X = 1000
You can rewrite this equation such that $$x$$ is a fixed point of $$f(x)$$, that is $$x=f(x)$$:
$$x=\frac{3}{LOG(x)}$$
Then you can iterate the following program with a starting value, say $$x=4$$:
Code:
001- 43 13 : LOG 002- 3 : 3 003- 34 : x<>Y 004- 10 : ÷
Example:
4
R/S
4.982892143
R/S
4.301189432
R/S
4.734933901
R/S
4.442378438
R/S
4.632377942
R/S
4.505830646
R/S
4.588735607
R/S
4.533824357
R/S
4.569933524
R/S
4.546075273
R/S
4.561789745
R/S
4.551417837
R/S
4.558254212
R/S
4.553744141
R/S
4.556717747
R/S
4.554756405
R/S
4.556049741
R/S
4.555196752
R/S
4.555759258
R/S
4.555388285
R/S
4.555632930
R/S
4.555471588
R/S
4.555577990
R/S
4.555507820
R/S
4.555554095
R/S
4.555523578
R/S
4.555543703
R/S
4.555530431
R/S
4.555539183
R/S
4.555533412
R/S
4.555537217
R/S
4.555534708
4.555535704
R/S
4.555535706
R/S
4.555535704
R/S
4.555535706
From then on the result flips between these last two values.
Cheers
Thomas
02-14-2019, 07:15 AM (This post was last modified: 02-14-2019 07:16 AM by Thomas Klemm.)
Post: #3
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
If you don't want to press R/S repeatedly:
Code:
001- 43 13 : LOG 002- 3 : 3 003- 34 : x<>Y 004- 10 : ÷ 005- 42 31 : PSE 006- 43 40 : x=0
02-15-2019, 12:07 AM (This post was last modified: 02-17-2019 12:16 PM by Albert Chan.)
Post: #4
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-14-2019 07:06 AM)Thomas Klemm Wrote: $$x=\frac{3}{LOG(x)}$$
4
R/S
4.982892143
R/S
4.301189432
...
It would be nice if we can temper the oscillation, or slow convergence.
Let x0 = 4, x1, x2 = the first two iterated values.
Rate = (x2-x1)/(x1-x0) ~ (4.30 - 4.98) / (4.98 - 4) = -0.694
If the same trend continued, we expect final % = 1/(1-r) ~ 60%
Use weighted fixed-point equation x = 0.4x + 0.6 * 3/log10(x):
4.6
4.555924149
4.555537395
4.555535712
4.555535705 (converged)
Edit: compare with Newton's method, x = (ln(1000) + x) / (ln(x) + 1)
4
4.571001573
4.555546101
4.555535705 (converged)
02-15-2019, 06:26 PM
Post: #5
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 12:07 AM)Albert Chan Wrote: It would be nice if we can temper the oscillation, or slow convergence.
We can also use Aitken's delta-squared process to accelerate the speed of convergence:
$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$
This leads to the following program for the HP-11C:
Code:
001-42,21, 1 : ▸LBL 01 002- 36 : ENTER 003- 32 0 : GSB 00 004- 30 : - 005- 43 36 : LSTx 006- 36 : ENTER 007- 32 0 : GSB 00 008- 30 : - 009- 43 36 : LSTx 010- 34 : x<>y 011- 43 33 : R↑ 012- 34 : x<>y 013- 30 : - 014- 43 36 : LSTx 015- 43 11 : x² 016- 34 : x<>y 017- 10 : ÷ 018- 30 : - 019- 43 32 : RTN 020-42,21, 0 : ▸LBL 00 021- 43 12 : LN 022- 45 0 : RCL 00 023- 34 : x<>y 024- 10 : ÷ 025- 43 32 : RTN
Example:
Solve: $$x^x=1000$$
1000 LN
STO 0
4.6
GSB 1
4.555665779
R/S
4.555535706
R/S
4.555535705
R/S
Error 0
Feel free to add a check for 0 after line 13 to prevent the error and loop back to label 1 instead of returning in line 19.
Cheers
Thomas
02-15-2019, 09:16 PM (This post was last modified: 02-15-2019 09:31 PM by Albert Chan.)
Post: #6
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
(02-15-2019 06:26 PM)Thomas Klemm Wrote:
(02-15-2019 12:07 AM)Albert Chan Wrote: It would be nice if we can temper the oscillation, or slow convergence.
We can also use Aitken's delta-squared process to accelerate the speed of convergence:
$$a_{n}=x_{n+2}-\frac {(x_{n+2}-x_{n+1})^{2}}{(x_{n+2}-x_{n+1})-(x_{n+1}-x_{n})}$$
Amazingly, my rate formula is same as Aitken extrapolation formula !
Assuming we have 3 values, x0,x1,x2 and tried to guess where it should end up.
My rate analysis: r = (x2-x1)/(x1-x0) = Δx1 / Δx0
We need this for the proof:
(Δx0)²
= ((Δx0 - Δx1) + Δx1)²
= (Δx0 - Δx1)² + 2 * Δx1 (Δx0 - Δx1) + (Δx1)²
If same trend continue, where it will ends up
= x0 + Δx0 * 1/(1-r)
= x0 + (Δx0)² / (Δx0 - Δx1)
= x0 + (Δx0 - Δx1) + 2 * Δx1 + (Δx1)² / (Δx0 - Δx1)
= x0 + (x1-x0) + (x1-x2) + 2*(x2-x1) − (Δx1)² / (Δx1 - Δx0)
= x2 − (Δx1)² / (Δx1 - Δx0)
02-16-2019, 03:58 AM
Post: #7
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-15-2019 09:16 PM)Albert Chan Wrote: Amazingly, my rate formula is same as Aitken extrapolation formula !
That's not really a surprise, is it?
It's essentially the same idea: geometric series
$$1\,+\,r\,+\,r^{2}\,+\,r^{3}\,+\,\cdots \;=\;{\frac {1}{1-r}}$$
02-16-2019, 12:24 PM (This post was last modified: 02-16-2019 12:34 PM by Csaba Tizedes.)
Post: #8
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-14-2019 05:25 AM)Gamo Wrote: a small but versatile solver program specifically for HP Voyager Series calculator.
Example for X^X = Y // X to the power of X equal to Y
Yes, the classic problem: which number you can power to itself to get the largest number on your calculator: x^x=10^100. Do LOG() of both sides: x×LOG(x)=100, then do the iteration: x_0:=50, then x_i+1:=100/log(x_i). The simple fixed point iteration and converges.
By the way, if you interested, here is my SOLVE(i) for HP-12C, which can solve any equation for ANY variable like on 15C, without rewriting the equation. Looks like indirect addressing without indirect addressing. 28 steps only.
Yes, my English is terrible, but more convenient than read the full material in Hungarian.
At the end of pdf you can find lots of applications which solved with this little solver (like ODE solve with EULER+SOLVE(i) on 12C (!!!), pressure vessel pneumatic conveying pipe sizing program, humidity and dew point solving, outer temperature calculation of a concrete silo wall or comparison of present values of educated and not educated people salaries with the little SOLVE combined with 12C's cash flow calculation capabilities - I guess this is shows clearly the power of this little routine possibilities and why I say everywhere, the SOLVER is MUST to implement ALL current calculators, I hope somebody read and understand what I want to say).
The small secant method in this paper for CASIO fx-50F meantime reduced to 13 steps, I guess the smallest solver for these blind machines. You can find attached.
Csaba
Attached File(s)
02-16-2019, 01:42 PM
Post: #9
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-16-2019 12:24 PM)Csaba Tizedes Wrote: By the way, if you interested, here is my SOLVE(i) for HP-12C
Steps of secant method from 03 to 18:
Code:
03- 45 2 : RCL 2 04- 45 0 : RCL 0 05- 30 : - 06- 43 36 : LSTx 07- 44 2 : STO 2 08- 35 : CLx 09- 45 3 : RCL 3 10- 45 1 : RCL 1 11- 30 : - 12- 43 36 : LSTx 13- 44 3 : STO 3 14- 33 : R↓ 15- 10 : ÷ 16- 45 1 : RCL 1 17- 20 : × 18-44 30 0 : STO+ 0
This can be shortened to:
Code:
03- 45 2 : RCL 2 04- 45 0 : RCL 0 05- 44 2 : STO 2 06- 30 : - 07- 45 3 : RCL 3 08- 45 1 : RCL 1 09- 44 3 : STO 3 10- 30 : - 11- 10 : ÷ 12- 45 1 : RCL 1 13- 20 : × 14-44 30 0 : STO+ 0
Don't hesitate to publish your programs in this forum. You can use Google to translate to English.
Cheers
Thomas
02-16-2019, 03:24 PM
Post: #10
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-16-2019 01:42 PM)Thomas Klemm Wrote: Steps of secant method from 03 to 18 can be shortened to.
Thanks, this was eons before, I am sure this is only the first iteration and required to fine tuning.
(02-16-2019 01:42 PM)Thomas Klemm Wrote: Don't hesitate to publish your programs in this forum. You can use Google to translate to English.
OK, sometimes I feel myself as Don Quixote... Frankly speaking, who want to discuss about calculators and calculator programming now? 15 years before I wrote something useful, today I am only (an angry) user...
Csaba
02-17-2019, 02:57 AM (This post was last modified: 02-17-2019 03:20 AM by Gamo.)
Post: #11
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
I just fine tune this solver program for HP-11C
Added the iterations limit to 28 loops count
Added the condition test to end program when root is found.
If iteration is over 28 loops count this indicate that user must use a closer guess.
-------------------------------------------
Procedure:
Use R1 for unknown X in your equation. End your formula with [RTN]
X1 ENTER X2 [A] display Answer
------------------------------------------------------------------------
Program:
Code:
LBL A STO 0 Rv STO 1 28 STO I // Store Loop Count Limit CLx --------------------- LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y // End if Root is found GTO 2 DSE // Loop Limit Counter GTO 0 CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 GTO 2 --------------------- LBL 1 // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN -------------------- LBL 2 RCL 1 // Answer RTN -------------------- LBL B . // f(x) equation start here . . // X = Register 1 [R1] . . RTN
Gamo
02-17-2019, 09:06 AM
Post: #12
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:57 AM)Gamo Wrote: I just fine tune this solver program for HP-11C
Your program doesn't work for this simple example: $$x^2+x=x(x+1)=0$$
I've entered the following program:
Code:
LBL B 1 RCL 1 + LSTx × RTN
Example:
-2
ENTER
-0.5
A
9.999999 99
The reason is that the function value is divided by -0.5, making it larger and larger:
Code:
LBL 1 // Solver Routine RCL 0 ÷ CHS RCL 1 + RTN
Another thing is this in this loop:
Code:
LBL 0 GSB B GSB 1 STO 1 GSB B GSB 1 RCL 1 X=Y // End if Root is found GTO 2 DSE // Loop Limit Counter GTO 0
When you return to label 0, the same value in register 1 will be used as before.
This function evaluation can be avoided if you save the result of the previous evaluation.
You could move this block:
Code:
LBL 2 RCL 1 // Answer RTN
Over here and avoid the jump GTO 2:
Code:
CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 LBL 2 RCL 1 // Answer RTN
Take a look at Csaba's solution for the HP-12C using the secant method.
Cheers
Thomas
02-17-2019, 02:33 PM
Post: #13
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Thanks Thomas Klemm
I try single guess and it work.
CHS 2 [A] display -1
If using two negative guesses and result is
Overflow then use Single guess instead.
Formula Routine is
RCL 1
1
+
RCL 1
X
Gamo
02-17-2019, 04:57 PM
Post: #14
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-17-2019 02:33 PM)Gamo Wrote: If using two negative guesses and result is Overflow then use Single guess instead.
Then try this function $$f(x)=4x(x+1)$$:
Code:
LBL B RCL 1 1 + RCL 1 × 4 × RTN
Examples:
-0.5
ENTER
0.5
A
9.999999 99
-0.5
ENTER
A
9.999999 99
0.5
ENTER
A
9.999999 99
Even initial guesses very close to the solutions lead to the same result:
-1.00001
ENTER
-0.99999
A
9.999999 99
-0.00001
ENTER
0.00001
A
9.999999 99
Cheers
Thomas
02-18-2019, 03:49 AM
Post: #15
Gamo Senior Member Posts: 518 Joined: Dec 2016
RE: Small Solver Program
Your're right that only work for certain situations.
Gamo
02-18-2019, 05:20 AM
Post: #16
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
If you compare your algorithm with Newton's method:
$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$
you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.
And with this in mind indeed the solution can now be found:
$$f'(x) = \frac{d}{dx}4 x (x + 1) = 8 x + 4$$
$$f'(-1) = -4$$
$$f'(0) = 4$$
Examples:
-1.5
ENTER
-4
A
-1.00000
0.5
ENTER
4
A
0.00000
Thus your 2nd parameter is an estimate of the slope at the root.
And then you don't really have to call the function twice with the same value:
Code:
LBL A STO 0 Rv STO 1 28 STO I // Store Loop Count Limit LBL 0 GSB B RCL 0 ÷ STO - 1 X=0 // End if Root is found GTO 1 DSE // Loop Limit Counter GTO 0 CLx FIX 9 // 0.000000000 indicate that Maximum Loops is use up. PSE PSE FIX 4 LBL 1 RCL 1 // Answer RTN LBL B . // f(x) equation start here . . // X = Register 1 [R1] . . RTN
So the question remains how to approximate the derivation at the roots.
If you keep record of the previous function evaluation you can get an estimate of the slope using the secant method.
Cheers
Thomas
02-18-2019, 07:46 PM
Post: #17
Dieter Senior Member Posts: 2,398 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 05:20 AM)Thomas Klemm Wrote: If you compare your algorithm with Newton's method:
$$x_{n+1}=x_{n}-\frac {f(x_{n})}{f'(x_{n})}$$
you may notice that the number in register 0 should in fact be $$f'(x_{n})$$ or at least a good approximation thereof.
Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
For more details take a look at this post: http://www.hpmuseum.org/forum/thread-123...#pid111680
Gamo, you really should correct and update the instructions. Here it still says:
Quote:Enter 2 guesses closest to the root.
But again: that's NOT true. The first input is NOT a guess for the root. You now have seen a variety of cases that prove it.
Dieter
02-18-2019, 10:22 PM
Post: #18
Thomas Klemm Senior Member Posts: 1,448 Joined: Dec 2013
RE: Small Solver Program
(02-18-2019 07:46 PM)Dieter Wrote: Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.
So all in all I'm happy with the outcome.
Cheers
Thomas
02-19-2019, 01:10 AM (This post was last modified: 02-19-2019 01:22 AM by Albert Chan.)
Post: #19
Albert Chan Senior Member Posts: 696 Joined: Jul 2018
RE: Small Solver Program
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Linear regression on slope had another benefit, which my Casio cannot do.
Doing Quadratic Regression on Casio, you don't really know if it is any good.
There is no "correlation coefficient" to lookup. (Why?)
Linear regression on slope can show whether a Quadratic curve is warranted.
I think the code add slope data, like this:
Data points (0,90) and (30, 90.2), add to linear regression: (30+0)/2, (90.2-90)/(30-0)
Next point (60, 86.4), add to linear regression: (60+30)/2, (86.4-90.2)/(60-30)
...
N points (sorted) thus generate N-1 slopes to regress.
02-19-2019, 08:39 AM
Post: #20
Csaba Tizedes Senior Member Posts: 366 Joined: May 2014
RE: Small Solver Program
(02-18-2019 10:22 PM)Thomas Klemm Wrote:
(02-18-2019 07:46 PM)Dieter Wrote: Yes. Gamo already posted this method some time ago in the General Software Library. I have mentioned several times that the first input it NOT a guess for the root but an estimate for the function's DERIVATIVE.
You could think him stubborn.
But it made Csaba post his program which made me translate some of his papers from Hungarian to English and stumble totally unrelated upon one of his previous posts:
Quadratic fit without linear algebra (32SII) - NOT only for statistics lovers - LONG
Which I still do not understand in detail.
So all in all I'm happy with the outcome.
Cheers
Thomas
I feel myself like a star (at least for 15 minutes...)
Yes, this is my "linear regression on slope", as Chan wrote. When I was in my "Summer Practice" at a company I have modelled lots of water network flows between pump houses (sources), consumers and reservoirs, and sometimes the characteristics curves given only in paper form. Thats why I wrote this little routine.
As you can see, if the measured points has significant error, the slope will be very inaccurate then the regression also inaccurate and the coefficients can be meaningless.
But as a programming question this was an interesting "game" for me.
To the original question from Gamo: I will check carefully the lists here, but I have another idea: what if, if we just simple walking along on the function's curve until the sign is changing, then we turning back, decreasing the stepsize and walking again. If the sign of the function changed, we are turning back again, decreasing the size of the steps and so on... If the stepsize is less than a small value, we found the root.
This "marching method" not so elegant, not so fast, but good to play with it, just for a try - maybe it can be shorter than the other methods. (I have an exactly same method for finding a minimum point of a function for CASIO fx-3650P/Sencor SEC103; perfectly works, like a ball at the bottom of a hole, slowly find the deepest point, when swings around the extremum).
Csaba
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https://math.stackexchange.com/questions/2906547/representing-displacement-vectors-in-cylindrical-coordinates-and-finding-the-dis | # Representing displacement vectors in cylindrical coordinates and finding the distance in cylindrical coordinates?
In cartesian coordinates, we can derive the vector $\vec v_3$ by vector subtraction $\vec v_2-\vec v_1$. We then get the distance between $P$ och $Q$ by taking the absolute value of $\vec v_3$ which then is: $$\lvert \vec v_3\rvert = \lvert \vec v_2-\vec v_1 \rvert= \lvert (x_2,y_2,z_2)-(x_1,y_1,z_1) \rvert = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
But how can we do this completely in cylindrical coordinates without converting to cartesian coordinates? We only have, and can only use, $\rho_1$ and $\theta_1$ for $\vec v_1$ as well as $\rho_2$ and $\theta_2$ for $\vec v_2$ in cylindrical coordinates $(\rho,\theta,z)$, where $z=0$ in this example.
Your problem for $z=0$ reduces to the polar coordinate problem already answered here. $$|\vec v_3|^2=(\vec v_2-\vec v_1)\cdot(\vec v_2-\vec v_1)=|\vec v_1|^2+|\vec v_2|^2-2\vec v_2\cdot \vec v_1$$ Now using cylindrical coordinates, $|\vec v_i|^2=r_i^2+z_i^2$ and $\vec v_2\cdot \vec v_1=r_1r_2\cos(\theta_2-\theta_1)+z_1z_2$. Then the final answer will be $$|\vec v_3|^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)+(z_2-z_1)^2$$
• Yes, I've seen this solution, but isn't it based in terms of cartesian coordinates/base vectors, $\hat e_x$, $\hat e_y$ and $\hat e_z$ after all? This since, I guess, you must express a distance in constant base vectors? I'm a bit confused about how to interpret the problem I have to admit. How would it look if I want to express the solution completely in cylindrical coordinates with $\vec v_1=\rho_1 \hat e_\rho (\theta_1)$ and base vectors $\hat e_\rho$, $\hat e_\theta$, and $\hat e_z$ etc? Would that be possible? Sep 6, 2018 at 5:46
• I did not use Cartesian coordinates. The $z$ coordinate is there in cylindrical system. The only other reference to the Cartesian system is the angle $\theta$. The issue that you have is that the basis of the cylindrical coordinate system changes with the vector, therefore equations will be more complicated. Sep 6, 2018 at 6:38
• If you choose one of the vectors as the reference, the cylindrical coordinate system is just a Cartesian system rotated with respect to the original. $\hat e_\rho$ is the new $\hat e_x$, and $\hat e_\theta$ is the new $\hat e_y$. Sep 6, 2018 at 7:19 | 2022-05-24T16:42:38 | {
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https://math.stackexchange.com/questions/1136207/how-to-determine-the-curve | # How to determine the curve?
In the figure above, segment $PQ$ is determined by two points: $P: (t,0)$ and $Q: (1,t)$, where $t\in [0,1]$ continuously increases and decreases between $0$ and $1$.
Then this gives a close region swept by $PQ$, the upper edge of which is a curve.
How to determine the equation of the curve (maybe implicit form)?
My own method
Suppose the curve is $y=y(x)$, then for any fixed $x_0\in (0,1)$, there is a vertical line $x=x_0$, which intersects a bundle of such $PQ(t)$ segments:
$$y=\frac{t(x-t)}{1-t}$$
Easy to conclude that, the desired $y_0$ on the curve corresponding to $x_0\left(\in(0,1)\right)$ is the maximum of:
$$y_0= \max\limits_{t\in (0,1)}\frac{t(x_0-t)}{1-t}=2-2\sqrt{1-x_0}-x_0$$
How to use the envelope concept? Is there any elementary method since the area is $\frac{1}{6}$?
• Do you know derivatives, tangent lines and integrals? – Tomas Feb 6 '15 at 9:47
• Yes, I know; I am not sure whether calculus is necessary in order to determine the area, but I am pretty sure in order to determine the curve, calculus would be a help – LCFactorization Feb 6 '15 at 9:53
• Look at the wiki entry for Envelope, it teach you how to determine the curve. – achille hui Feb 6 '15 at 10:29
• I use optimization method and obtains $y=2-2\sqrt{1-x}-x$, then the area is $1/6$. Is there any elementary method in order to determine the area? – LCFactorization Feb 6 '15 at 11:12
I know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.
$$\begin{pmatrix}t\\0\\1\end{pmatrix}\times \begin{pmatrix}1\\t\\1\end{pmatrix}= \begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}$$
This could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:
\begin{align*} a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\ ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4 \end{align*}
The $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:
$$a^2 + ac - bc = 0$$
Written as a matrix:
$$(a,b,c)\cdot\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = 0$$
Now, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:
$$(x,y,1)\cdot\begin{pmatrix}1&1&0\\1&1&-2\\0&-2&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix} = 0$$
Since the determinant of the upper left $2\times2$ matrix is zero, this is a parabola. It can also be described by the equation
\begin{align*} x^2 + 2xy + y^2 &= 4y \\ (x+y)^2 &= 4y \end{align*}
Since this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as
$$\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix} =-\begin{pmatrix}2t-t^2\\t^2\\1\end{pmatrix}$$
So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.
HINT
I would say
$F(x,y,t)=0\equiv t(x-t)-y(1-t)=0,\quad t\in\langle 0,1 \rangle$
$F'_t(x,y,t)=0\equiv x-2t+y=0$
Solving this system of equations for the x, y:
$x=t(2-t), y=t^2, \quad t\in\langle 0,1 \rangle$ is a parametric curve of equation to search.
Plot
• Very simple and neat answer. All three are acceptable. I choose the first one that also uses projective geometry. Thank you very much! @MvG – LCFactorization Feb 7 '15 at 2:40
I have a simpler solution.
1. Find the equation of the line with $(t,0,1)\times(1,t,1)=(-t,1-t,t^2)$ or $$(-t)x+(1-t)*y+(t^2)=0$$
2. Find the extema $t$ with $$\frac{{\rm d}}{{\rm d}t}\left(-t\,x+(1-t)\,y+t^2\right)=0$$ $$\left. -x-y+2 t=0 \right\}\; t=\frac{x+y}{2}$$
3. Plug $t$ into line equation for $$-\frac{(x+y)^2}{4}+y=0$$ which is the resulting curve
Hint: The envelope of an implicit curve $f(x,y,t)=0$ is found by solving for $t$ in $\frac{\partial f}{\partial t}=0$ and using it into the original equation.
My other answer describes how I myself think about this. But I realize that there is a lot of background knowledge in there, which might make this less accessible to members of other communities than projective geometry. So here is my attempt at a more generic solution.
Suppose you have found the equation of the line to be $tx + (t-1)y = t^2$. Take a second line from your family, using $u$ instead of $t$ as the parameter, i.e. $ux + (u-1)y = u^2$. These two intersect in a point $(t+u-tu, tu)$. Now if you consider $\lim_{u\to t}$ then strictly speaking the point of intersection becomes undefined, but the above will simply become $(2t-t^2,t^2)$. So that's the one point that your line does not have in common with any other line of the family (extended from $t\in[0,1]$ to $t\in\mathbb R$). So at that point this line is the one defining the curve. Therefore that's your parametric equation of the envelope curve. | 2021-03-09T10:19:41 | {
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https://www.studyxapp.com/homework-help/suppose-you-have-a-function-dt-that-gives-the-total-distance-traveled-up-to-tim-q90979419 | # Question Solved1 Answer1. 2. 3. Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
CW7XUD The Asker · Calculus
1.
2.
3.
Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
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Transcribed Image Text: Suppose you have a function d(t) that gives the total distance traveled up to time point t. Determine the expression (in terms of the d function) that gives the average velocity between time points t = a and t b. Estimate the instantaneous rate of change of y(t) 2t2 + 1 at the point t = 2 Your answer should be accurate to at least 2 decimal places. 2. The point P (18) lies on the curve y = If Q is the point ( x, (2, 4). find the slope of the secant line C PQ for the following values of x. If x = 0.35, the slope of PQ is: and if x = 0.26, the slope of PQ is: and if x = 0.15, the slope of PQ is: and if x = 0.24, the slope of PQ is: Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 8).
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https://math.stackexchange.com/questions/3235160/spanning-forests-of-bipartite-graphs-and-distinct-row-column-sums-of-binary-matr | # Spanning forests of bipartite graphs and distinct row/column sums of binary matrices
Let $$F_{m,n}$$ be the set of spanning forests on the complete bipartite graph $$K_{m,n}$$. Let $$S_{m,n} = \{(r(M), c(M)), M \in B_{m,n} \}$$ where $$B_{m,n}$$ is the set of $$m \times n$$ binary matrices and $$r(M), c(M)$$ are the vectors of row sums and column sums of $$M$$, respecitvely. That is, $$S_{m,n}$$ is the set of the distinct row-sum, column-sum pairs of binary matrices. (The term spanning forest here refers to a forest that spans all of the vertices of the given graph; it doesn't have to be a maximal acyclic subgraph.)
Q: Is it true that $$|F_{m,n}| = |S_{m,n}|$$? It is true for $$m,n \leq 4$$. For $$m=n$$ this is OEIS A297077.
There is an obvious mapping from $$F_{m,n} \rightarrow B_{m,n}$$ given by taking the reduced adjacency matrix, so if $$U = \{u_1, \ldots, u_m\}$$, $$V = \{v_1, \ldots, v_m\}$$ are the color categories we set $$M_{ij} = 1$$ if $$v_i \sim u_j$$ in a forest $$F$$, else $$M_{ij} = 0$$. However, this does not help because multiple forests may have the same row and column sums - and not every row-sum, column-sum pair is represented by a forest under this mapping.
The numbers $$|F_{m,n}|$$ are given here:
$$\begin{array}{|c|c|}\hline m\backslash n & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 2 & \\\hline 2 & 4 & 15 \\\hline 3 & 8 & 54 & 328 \\\hline 4 & 16 & 189 & 1856 & 16145 \\\hline 5 & 32 & 648 & 9984 & 129000 & 1475856 \\\hline\end{array}$$
For more see this answer (sum each row.)
• The linked MSE thread calculates $|F_{m,n}|$, but how did you calculate $|S_{m,n}|$ to verify they are the same for $m,n\le 4$? Did you just generate all $2^{mn}$ binary matrices? – antkam May 24 '19 at 19:24
• Yes, I did a brute force count. – Jair Taylor May 24 '19 at 22:32
• Just curious: what made you do the brute force count to find out they're equal? Nothing in the "obvious mapping" would suggest (to me) that they are equal, so it would never have occurred to me to even try. – antkam May 25 '19 at 22:54
• @antkam I was interested in counting forests and found that the relevant OEIS entry only mentioned row-sum / column-sums. – Jair Taylor May 26 '19 at 1:01
• In my Answer, when interpreting $F_{m,n}$, I wrote "In your definition, a spanning forest is any subset $T$ of edges of $K_{2,n}$ which is acyclic." So e.g. the empty $T=\emptyset$ qualifies, and in general nodes can be isolated/unconnected. First of all, is my interpretation correct? Second, if I'm correct, then IMHO your use of the word "spanning" is a little confusing. I understand you probably want to make sure unconnected nodes are still included, and simply saying "forest" might also be confusing in a different way... – antkam May 30 '19 at 12:49
I can prove the conjecture, although I can't give a bijection. I've left up a previous answer which does the case $$m=3$$, since it serves as an example of many of the ideas here.
Here is the key observation:
Theorem: There is a unique array $$f(m,n)$$, for $$m$$ and $$n$$ nonnegative integers, satisfying:
1. $$f(m,0) = f(0,n)=1$$
2. If we fix $$m \geq 1$$, then $$f(m,n)$$ is of the form $$(m+1)^n p_m(n)$$ where $$p_m(x)$$ is a polynomial of degree $$\leq m-1$$.
3. If we fix $$n \geq 1$$, then $$f(m,n)$$ is of the form $$(n+1)^m p_n(m)$$ where $$p_n(x)$$ is a polynomial of degree $$\leq n-1$$.
Uniqueness: Let's suppose that we have already found unique values of $$f(m,n)$$ whenever $$\min(m,n) < k$$. In particular, we know the values of $$f(k,x)$$ and $$f(x,k)$$ for $$0 \leq x \leq k-1$$. This is enough values to uniquely determine a degree $$k-1$$ polynomial, so the values we already know determine the values of $$f(k,x)$$ and $$f(x,k)$$ for all $$x$$.
Existence: The proof of uniqueness gives an algorithm to compute $$f$$. The only potential issue is that $$f(k,k)$$ is determined twice, as a polynomial in its first variable and in its second variable. However, the algorithm is symmetric in $$m$$ and $$n$$, so the two interpolating polynomials coincide. $$\square$$
For example, we have $$f(1,0) = 1 = 2^0 \cdot 1$$ so we must have $$p_1(n)=1$$ and $$f(1,n) = 2^n$$. Similarly, $$f(2,0)=1=2^0 \cdot 1$$ and $$f(2,1) = 4=3^1 \cdot (4/3)$$, so we must have $$p_2(n) = 1+n/3$$ and $$f(2,n) = 3^n(1+n/3) = 3^{n-1} (n+3)$$. Using this algorithm, I computed $$f(m,n)$$ for $$0 \leq m,n \leq 10$$. Here is the data, and some Mathematica code, if you'd like to play with it.
f[m_, n_] := f[m, n] =
If[m == 0 || n == 0, 1,
If[m <= n,
(m + 1)^n * InterpolatingPolynomial[Table[{k, f[m, k]/(m + 1)^k}, {k, 0, m - 1}], x] /. x -> n,
(n + 1)^m * InterpolatingPolynomial[Table[{k, f[k, n]/(n + 1)^k}, {k, 0, n - 1}], x] /. x -> m]]
Values of $$f(m,n)$$ for $$0 \leq m,n \leq 10$$.
1,1,1,1,1,1,1,1,1,1,1
1,2,4,8,16,32,64,128,256,512,1024
1,4,15,54,189,648,2187,7290,24057,78732,255879
1,8,54,328,1856,9984,51712,260096,1277952,6160384,29229056
1,16,189,1856,16145,129000,968125,6925000,47690625,318500000,2073828125
1,32,648,9984,129000,1475856,15450912,151201728,1403288064,12479546880,107152782336
1,64,2187,51712,968125,15450912,219682183,2862173104,34828543449,401200569280,4418300077219
1,128,7290,260096,6925000,151201728,2862173104,48658878080,760774053888,11122973573120,153936323805184
1,256,24057,1277952,47690625,1403288064,34828543449,760774053888,15047189968833,274908280855680,4707029151392121
1,512,78732,6160384,318500000,12479546880,401200569280,11122973573120,274908280855680,6199170628499200,129708290461760000
1,1024,255879,29229056,2073828125,107152782336,4418300077219,153936323805184,4707029151392121,129708290461760000,3283463201858585471
So, we can prove the conjecture by showing that both $$F_{m,n}$$ and $$S_{m,n}$$ obey conditions 1, 2 and 3. For condition 1, we can just define $$S_{m,0}$$, $$S_{0,n}$$, $$F_{m,0}$$, and $$F_{0,n}$$ to be singletons (and I claim this is actually the most natural definition); then we just need to check that our verification of polynomiality goes all the way down to the zero case.
Since conditions 2 and 3 are symmetric, and the definitions of $$F_{m,n}$$ and $$S_{m,n}$$ are symmetric, we just check condition $$2$$.
Verification of condition 2 for forests: Let $$A_{m,b}$$ be the set of isomorphism classes of bicolored forests whose white vertices are labeled $$\{1,2,\ldots,m \}$$ and which have $$b$$ black vertices, each of degree $$\geq 2$$. We claim that $$|F_{m,n}| = \sum_b |A_{m,b}| n(n-1)(n-2) \cdots (n-b+1) (m+1)^{n-b} . \qquad (1)$$
Proof: Take a forest in $$F_{m,n}$$, with the $$m$$ vertices colored white and the $$n$$ vertices colored black. Delete all black vertices of degree $$0$$ and $$1$$ and forget the numbering of the remaining black vertices. This gives a forest in $$A_{m,b}$$. To undo this process, we first must take the $$b$$ black vertices and decide which of the $$n$$ vertices of $$K_{m,n}$$ they will be; there are $$n(n-1)(n-2) \cdots (n-b+1)$$ ways to do this. (We used that trees in $$A_{m,b}$$ have no automorphisms.) Then we must take the $$n-b$$ remaining black vertices and either join them to one of the $$m$$ white vertices or leave them unconnected; there $$(m+1)^{n-b}$$ ways to do this.
Formula (1) is clearly $$(m+1)^n$$ times a polynomial, it remains to check that the polynomial has degree at most $$m-1$$. We must check that $$A_{m,b}$$ is empty if $$b \geq m$$. Indeed, since every black vertex has degree at least $$2$$, a forest in $$A_{m,b}$$ has at least $$2b$$ edges. But, since it is a forest, it also has at most $$m+b-1$$ vertices. So $$2b \leq m+b-1$$ and $$b \leq m-1$$. $$\square$$
Verification of condition 2 for margins of $$(0,1)$$ matrices: Consider a vector $$C = (C_1, \ldots, C_n) \in \{ 0,1,\ldots, m \}^n$$ of columns sums, and consider how many row sums $$(R_1, \ldots, R_m)$$ are compatible with it. Let $$c_j = \# \{ k : C_k = j \}$$, so $$c_0+c_1+\cdots + c_m=n$$.
By the Gale-Ryser theorem, $$(R_1, \ldots, R_m)$$ is compatible with $$(c_0, \cdots, c_m)$$ if and only if $$\sum R_i = \sum j c_j$$ and, for all index sets $$1 \leq i_1 < i_2 < \cdots < i_k \leq m$$, we have $$R_{i_1} + R_{i_2} + \cdots + R_{i_k} \leq \sum_j \min(j,k) c_j .$$ This is the defining list of inequalities of the permutahedron, so this condition can alternately be stated as saying that $$(R_1, \ldots, R_m)$$ is in the convex hull of the $$m!$$ permutations of $$(c_1+c_2+\cdots+c_m, c_2+\cdots + c_m, \cdots, c_{m-1}+c_m, c_m)$$.
By Theorem 11.3 of Postnikov's Permutohedra, associahedra, and beyond, the number of such $$(R_1, \ldots, R_m)$$ is a polynomial $$g_m(c_0,c_1, \ldots, c_m)$$ in the $$c_j$$, of degree $$m-1$$. So $$|S_{m,n}| = \sum_{c_0+\cdots + c_m=n} \frac{n!}{c_0! c_1! \cdots c_m!} \ g_m(c_0,c_1, \ldots, c_m). \qquad (2)$$ Let $$\partial_j$$ be $$\tfrac{\partial}{\partial x_j}$$. There is some polynomial $$h_m$$ in the $$\partial_j$$, of degree $$m-1$$, such that $$\left. h_m(\partial_0, \ldots, \partial_m) \left( x_0^{c_0} \cdots x_m^{c_m} \right) \right|_{(1,1,\ldots,1)} = g_m(c_1, \ldots, c_m). \qquad (3)$$ Combining (2), (3) and the binomial expansion of $$(x_0+x_1+\cdots+x_m)^n$$, we obtain $$|S_{m,n}| = \left. h_m(\partial_0, \ldots, \partial_m) \left( x_0+x_1+\cdots+x_m \right)^n \right|_{(1,1,\ldots,1)} . \qquad (4)$$ Let $$h_m(t,t,\ldots,t) = \sum_{k=0}^{m-1} h_{m,k} t^k$$. Then $$(4)$$ is $$\sum_{k=0}^{m-1} h_{m,k} n(n-1)(n-2) \cdots (n-k+1) (m+1)^{n-k}$$, which is a polynomial of the desired form. $$\square$$
• Very cool! I think you need another index on your coefficients $c_k$ (say $c_{m,k}$). Then you've shown $c_{m,k} = |A_{m, k}|$ which is interesting as well. I'd be curious to see if this could be generalized more - e.g., to forests with a specified number of components. But I don't know what the appropriate statistic on row-sums/column-sums would be. – Jair Taylor Jun 2 '19 at 21:16
• @JairTaylor Nice observation. Your comment made me realize I had used $c_j$ to mean two things, so I changed the second one to $h_{m,k}$; your observation then is that I have shown $h_{m,k} = |A_{m,k}|$. – David E Speyer Aug 27 '19 at 15:38
This conjectured equivalence (if true) is still amazing to me. However, here is a baby step: The conjecture is indeed true for $$m=2$$. Perhaps someone else will find this useful as a base for generalization.
Claim: $$|F_{2,n}| = |S_{2,n}|$$ for all $$n$$
First, lets consider $$F_{2,n}$$. In your definition, a spanning forest is any subset $$T$$ of edges of $$K_{2,n}$$ which is acyclic. Let $$u,v$$ be the two nodes on the $$m=2$$ side, and let $$S(u)$$ be the neighbors of $$u$$, i.e. the subset of nodes (on the $$n$$-node side) which $$u$$ is connected to in $$T$$. Similarly for $$S(v)$$.
Next, note that $$T$$ contains a cycle iff $$S = S(u) \cap S(v)$$ contains $$2$$ nodes (or more). So, $$F_{2,n}$$ can be counted by counting all cases where $$S$$ contains $$0$$ or $$1$$ node.
• $$|S|=0$$ case: Each of $$n$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= 3^n$$.
• $$|S|=1$$ case: There are $$n$$ ways to pick the unique $$w \in S$$. After that, each of the remaining $$n-1$$ nodes can be connected to $$u$$, or to $$v$$, or to neither. Total no. $$= n\, 3^{n-1}$$
• Summing up, $$|F_{2,n}| = 3^n + n \,3^{n-1} = 3^{n-1} (3+n),$$ which matches your table.
Second, lets consider $$S_{2,n}$$. There are $$n$$ column sums, each of which can be $$0, 1, 2$$. Let $$x,y,z$$ be the number of columns whose sum $$=0, 1, 2$$ respectively. The columns whose sum $$=0$$ or $$2$$ dictate their elements. In the $$y$$ columns whose sum $$=1$$, the number of $$1$$s in the top row can be any integer $$\in [0,y]$$, and so the top row sum can be any integer $$\in [z,z+y]$$, and in short there are $$(y+1)$$ possibilities. Obviously, once the top row sum is known that determines the bottom row sum $$= (y+2z) \,-$$ top row sum.
So $$S_{2,n}$$ can be counted by (i) summing over all possible $$y$$, and for each $$y$$: (ii) picking the $$y$$ columns whose sum $$=1$$, and (iii) for the remaining $$n-y=x+z$$ columns assign them $$0$$ or $$2$$ arbitrarily. I.e.:
$$|S_{2,n}| = \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y}$$
This can be evaluated many ways but my favorite is to recognize a slightly transformed sum as an explicit formula for the expected value of a Binomial random variable:
\begin{align} |S_{2,n}| &= \sum_{y=0}^n (y+1) {n \choose y} 2^{n-y} \\ &= 3^n \times \sum_{y=0}^n (1+y) {n \choose y} (\frac13)^y (\frac23)^{n-y} \\ &= 3^n \times \mathbb{E}[1 + Bin(n, \frac13)] \\ &= 3^n (1 + {n \over 3}) = 3^{n-1} (3 + n) = |F_{2,n}| & QED \end{align}
As I said, this is a baby step only. The key step in $$F_{2,n}$$ is that $$T$$ has a cycle iff $$|S| \ge 2$$, and the key step in $$S_{2,n}$$ is that the $$y$$ ones can be all in the top row or all in the bottom row or anywhere in between, for $$(1+y)$$ possibilities. These key steps make the counting easy. However, as far as I can see, neither key step has any easy way to generalize to $$m=3$$, let alone arbitrary $$m$$. (E.g. for $$m=3$$, a cycle in $$T$$ can involve $$2$$ or $$3$$ nodes on the $$m=3$$ side, and I don't know any good way to count that.)
• Nice start. I would like to see a bijective proof. I wonder if, in general, if $m$ is fixed, $F_{m,n}$ has a rational generating function. – Jair Taylor May 30 '19 at 19:32
• I would like to see a bijective proof even just for $m=2$, instead of this algebraic mess. Also, I am no good at GFs, but in this case, I couldn't even find nice recursions for $F$ or $S$. Do you have recursions for either of them? – antkam May 30 '19 at 19:54
I have a brute force proof that, for $$m=3$$, both counts are $$4^{n - 2} (3 n^2 + 13 n + 16)$$. It would be possible to extend this method for any fixed $$m$$. I'll write it up for $$m=3$$ in the hope it gives insight. As you'll see, many steps look the same on both sides, but I can't figure out how to give a general proof.
Counting margins of $$3 \times n$$ binary matrices Consider a vector $$C \in \{ 0,1,2,3 \}^n$$ of column sums. Let $$C$$ have $$c_j$$ entries equal to $$j$$ (so $$c_0+c_1+c_2+c_3=n$$). Let's consider how many row sums $$(R_1, R_2, R_3)$$ are compatible with $$C$$. By the Gale-Ryser theorem, these are the vectors with $$R_1+R_2+R_3 = c_1+2c_2+3c_3$$, $$\min(R_1+R_2, R_1+R_3, R_2+R_3) \geq c_2+2c_3$$ and $$\min(R_1, R_2, R_3)\geq c_3$$. Geometrically, this is the lattice points in a hexagon whose vertices are the permutations of $$(c_1+c_2+c_3, c_2+c_3, c_3)$$. I computed that the number of lattice points in this hexagon is $$\binom{c_1}{2} + 2 c_1 c_2 + \binom{c_2}{2} + 2 c_1 + 2 c_2 + 1.$$
The generating function for $$\sum_C x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$ is $$(x_0+x_1+x_2+x_3)^n$$. Let $$\partial_j$$ be differentiation with respect to $$x_j$$. Let $$D$$ be the differential operator $$\tfrac{1}{2} \partial_1^2 + 2 \partial_1 \partial_2+\tfrac{1}{2} \partial_2^2 + 2 \partial_1 + 2 \partial_2 + 1.$$ Then $$D$$ applied to the monomial $$x_0^{c_0} x_1^{c_1} x_2^{c_2} x_3^{c_3}$$, and evaluated at $$(1,1,1,1)$$, gives $$\binom{c_1}{2} + 2 c_1 c_2 + \binom{c_2}{2} + 2 c_1 + 2 c_2 + 1$$. So the number of pairs $$(C,R)$$ is $$D (x_0+x_1+x_2+x_3)^n$$ evaluated at $$(1,1,1,1)$$.
This gives $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n = 4^{n - 2} (3 n^2 + 13 n + 16)$$.
Counting forests Given a forest on $$3 \times n$$, define a sequence $$(v_1, v_2, \ldots, v_n)$$ in $$\{ 0,1,2,3 \}^n$$ by saying that $$v_j$$ is $$0$$ if vertex $$j$$ has degree $$0$$, and otherwise $$v_j$$ is the least neighbor of $$j$$. Let $$c_j$$ be the number of $$v_j$$'s equal to $$j$$.
The generating function for the $$(c_0, c_1, c_2, c_3)$$ sequences is $$(x_0+x_1+x_2+x_3)^n$$ (again!). We count how many forests give rise to each $$(c_0, c_1, c_2, c_3)$$, by listing all possiblilities.
One vertex joined to $$(1,2)$$ and another joined to $$(1,3)$$ This can happen in $$c_1 (c_1-1)$$ ways.
One vertex joined to $$(1,2)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.
One vertex joined to $$(1,3)$$ and another joined to $$(2,3)$$ This can happen in $$c_1 c_2$$ ways.
One vertex joined to $$(1,2,3)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(1,2)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(1,3)$$ This can happen in $$c_1$$ ways.
One vertex joined to $$(2,3)$$ This can happen in $$c_2$$ ways.
None of the $$n$$-vertices joined to more than one neighbor This can happen in $$1$$ way.
So, this time, we want to sum up $$c_1 (c_1-1) + 2 c_1 c_2 + 3 c_1 + c_2 +1.$$ The corresponding differential operator is $$E=\partial_1^2 + 2 \partial_1 \partial_2 + 3 \partial_1 + \partial_2+1.$$
The differential operators are different, but once again I get $$3 n(n-1) 4^{n-2} + 4 n 4^{n-1} + 4^n$$ at the end of the day.
A thought for the future Why do the differential operators $$\tfrac{1}{2} \partial_1^2 + 2 \partial_1 \partial_2+\tfrac{1}{2} \partial_2^2 + 2 \partial_1 + 2 \partial_2 + 1$$ and $$\partial_1^2 + 2 \partial_1 \partial_2 + 3 \partial_1 + \partial_2+1$$ give the same thing when applied to $$(x_0+x_1+x_2+x_3)^n$$? Because $$(x_0+x_1+x_2+x_3)^n$$ is solely a function of $$z:=x_0+x_1+x_2+x_3$$, so all the $$\partial_j$$'s evaluate to $$\tfrac{d}{dz}$$. Writing $$\partial$$ for $$\tfrac{d}{dz}$$, both operators are $$3 \partial^2 + 4 \partial + 1$$. Any hope to generalize this?
• Nice. I was not aware of Gale-Ryser. If I understand correctly, this could give a method of verifying the conjecture for any fixed $m$ by counting lattice points of appropriate polytopes (to count $S_{m,n}$) and enumerating all possible families of subsets of $[m]$ sharing neighbors (to count $F_{m,n}$), both reducing to finite counting problems. Still, it seems like this problem is crying out for an explicit bijection. – Jair Taylor Jun 1 '19 at 13:43 | 2021-05-07T12:16:44 | {
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https://gateoverflow.in/2136/gate2011-34 | 2.2k views
A deck of $5$ cards (each carrying a distinct number from $1$ to $5$) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?
1. $\left(\dfrac{1}{5}\right)$
2. $\left(\dfrac{4}{25}\right)$
3. $\left(\dfrac{1}{4}\right)$
4. $\left(\dfrac{2}{5}\right)$
edited | 2.2k views
The number on the first card needs to be One higher than that on the second card, so possibilities are :
$\begin{array}{c} \begin{array}{cc} 1^{\text{st}} \text{ card} & 2^{\text{nd}} \text{ card}\\ \hline \color{red}1 & \color{red}-\\ 2 & 1\\ 3 & 2\\ 4 & 3\\ 5 & 4\\ \color{red}- & \color{red}5 \end{array}\\ \hline \text{Total$:4$possibilities} \end{array}$
Total possible ways of picking up the cards $= 5 \times 4 = 20$
Thus, the required Probability $= \dfrac{\text{favorable ways}}{\text{total possible ways}}= \dfrac{4}{20} = \dfrac 15$
Option A is correct
selected
0
How to solve the question if it would be only HIGHER and not ONE HIGHER?
+4
How to solve the question if it would be only HIGHER and not ONE HIGHER?
Possible combinations :-
First card second card 5 4,3,2,1 4 3,2,1 3 2,1 2 1 1 none
$P(First\ card\ is\ HIGHER\ than\ second\ one\ )=\left [ \left ( \frac{1}{5} \times \frac{4}{4}\right ) + \left ( \frac{1}{5} \times \frac{3}{4}\right ) + \left ( \frac{1}{5} \times \frac{2}{4}\right ) + \left ( \frac{1}{5} \times \frac{1}{4}\right ) \right]\\ =\frac{1}{2}$
+1
Got it
Here we should consider without replacement, since "removed one at a time" means the card has been removed from the deck.
Prob of picking the first card = 1/5
Now there are 4 cards in the deck. Prob of picking the second card = 1/4
Possible favourable combinations = 2-1, 3-2, 4-3, 5-4
Probability of each combination = (1/5)*(1/4) = 1/20
Hence answer = 4*1/20 = 1/5
with 5 cards to choose we can only fulfil the condition if we pick 2,3,4,5 in our choice else theres no way to get the same
the probability of choosing first no's is =(2,3,4,5)/(1,2,3,4,5)=4/5
the second time, we have only one option to choose out of four option=1/4
so,the total probability=(4/5)*(1/4)=1/5
1
2 | 2018-11-20T05:53:51 | {
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https://math.stackexchange.com/questions/4003709/zero-rows-in-echelon-matrix | Zero rows in echelon matrix
In the MIT lecture notes (pg. 1) for Linear Algebra, it states;
The third row is zero because row 3 was a linear combination of rows 1 and 2; it was eliminated.
Here, $$A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$$ and $$U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$.
Does this mean that if the row of the echelon matrix is all zeros, the corresponding row in A can be represented as a linear combination of the other rows in A?
And if the above statements is true, does this also apply to reduced row echelon matrix?
(1) Yes, if a row of the echelon matrix is all zeros, then that row (but remember to consider possible partial pivoting) is a linear combination of other rows in the original matrix. Generally, the coefficients such that forms this linear combination can be found solving the following linear system for $$a$$ and $$b$$: $$\begin{bmatrix}1 & 2 \\ 2 & 4 \\ 2 & 6 \\ 2 & 8 \end{bmatrix}\cdot\begin{bmatrix}a \\ b \end{bmatrix}=\begin{bmatrix}3 \\ 6 \\ 8 \\ 10\end{bmatrix}$$ | 2021-07-27T07:47:38 | {
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http://mathhelpforum.com/advanced-algebra/224808-question-about-solving-eigenvector-problems.html | 1. ## Question about solving eigenvector problems
I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:
I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$
Solving for lambda, I get lambda = 1.5 and 4.5
Now my matrix equations become:
$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$
dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
2. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
I understand how to get eigenvalues, but I have a problem understanding eigenvectors.
I'll demonstrate with an example:
I am given A = $\begin{bmatrix}3 & 1.5 \\1.5 & 3 \end{bmatrix}$
Solving for lambda, I get lambda = 1.5 and 4.5
Now my matrix equations become:
$-1.5x_1+1.5x_2=0$ and $1.5x_1-1.5x_2=0$
dividing by 1.5 I get: $x_1-x_2=0$
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.
{1,1} doesn't span R2. It spans the line y = x. Likewise {1,-1} spans the line y = -x
together these two vectors span R2
3. ## Re: Question about solving eigenvector problems
Originally Posted by romsek
{1,1} doesn't span R2. It spans the line y = x.
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.
Originally Posted by romsek
There are actually two eigenvectors 1/sqrt(2) {1,1} and 1/sqrt(2) {1,-1}, corresonding to eigenvalues 1.5 and 4.5.
How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°
Does that make sense?
4. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
so I get $x_1=x_2$
Now, the answer is $\begin{bmatrix}1 & 1 \end{bmatrix}$ because if you set x1=1, then x2=1
If x1=x2, then why isn't the answer [ℝ], where ℝ is the set of all real numbers?
First, the notation [ℝ ℝ], where the objects inside brackets are sets rather than numbers, is not universally accepted. Even if it were, [A B] would probably mean {[x y] | x ∈ A, y ∈ B}, i.e., A × B. There is no coordination between the first and second coordinate here. It is clear that you mean {[x x] | x ∈ ℝ}, but I don't see an easier way to write this. By the way, even if you delimit matrices with square brackets, I would use parentheses for vectors (i.e., 1 × n matrices): e.g., (x, x), because square brackets are also used to denote closed segments.
Second, you are right, any (x, x) is an eigenvector of A as is (x, -x). Indeed, you solve the equation det(A -λI) = 0 for λ, so if λ is an eigenvalue, the matrix A - λI is singular by definition. Therefore, the system of equations (A - λI)x = 0 has infinitely many solutions.
Edit: If x is an eigenvector with value λ, then so is (ax) for any a ∈ ℝ such that a ≠ 0. Indeed, A(ax) = a(Ax) = a(λx) = λ(ax).
5. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
In that case, my question is, why aren't there infinite values for x and y that satisfy the eigenvector?
for example, {2,2}, {5,5}, {1000,1000}, etc.
How did you get this? My book has this answer:
1.5, [1 1], 45°; 4.5, [1 1], 45°
Does that make sense?
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.
Your book is missing a minus sign somewhere. The answer I got was
1.5 [1,-1]/sqrt(2)
4.5 [1,1]/sqrt(2)
6. ## Re: Question about solving eigenvector problems
Originally Posted by romsek
there are infinite points. That's why you say this space, in this case the 1 dimensional line y=x, is "spanned" by the vector 1/sqrt(2) {1,1}. Multiply this vector by any real scalar and you remain in the space, but yes there are infinitely many points.
Your book is missing a minus sign somewhere. The answer I got was
1.5 [1,-1]/sqrt(2)
4.5 [1,1]/sqrt(2)
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?
7. ## Re: Question about solving eigenvector problems
Originally Posted by yaro99
Whoops, that was my bad, there was a minus sign.
where do you get the 1/sqrt(2) from?
is it a scalar that you can multiply throughout the matrix?
it just normalizes the vector to unit length. You tend to want the unit eigenvectors | 2017-08-18T02:23:29 | {
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http://math.stackexchange.com/questions/109032/singing-bird-problem?answertab=oldest | # Singing Bird Problem
For my algorithms class, the professor has this question as extra credit:
The Phicitlius Bauber bird is believed to never sing the same song twice.
Its songs are always 10 seconds in length and consist of a series of notes that are either high or low pitched and are either 1 or 2 seconds long.
How many different songs can the Bauber bird sing?
This strikes me as a combinatorial problem, but I'm having trouble figuring out the exact formula to use to fit the included variables of 10 seconds length, 2 notes and 2 note periods.
At first, I thought that $\binom{10}{2}*\binom{10}{2}$ can be a solution, since it fits finding all combinations of both and multiplies the total.
My issue is this results in a total of 2025 possible songs, and this just seems a little on the low side.
Any assistance is appreciated.
-
I'm going to work under the assumption that two successive 1-second notes of the same pitch is distinct from a single 2-second note of that pitch.
Let $f(n)$ be the number of distinct $n$-second songs the bird can sing. The bird has four options for starting the song: 1-second low, 1-second high, 2-second low, and 2-second high. This gives the recursion $$f(n) = 2f(n-1) + 2f(n-2),$$ with $f(0) = 1$ and $f(1) = 2$.
The associated characteristic equation is $x^2 - 2x - 2 = 0$, which has roots $x = 1 \pm \sqrt{3}$. Thus, we get the general solution $$f(n) = A(1 + \sqrt{3})^n + B(1 - \sqrt{3})^n.$$
Using the initial conditions, we have the system \begin{align*} 1 &= A + B\\ 2 &= A(1 + \sqrt{3}) + B(1 - \sqrt{3}), \end{align*} which has the unique solution $A = \frac{3 + \sqrt{3}}{6}$ and $B = \frac{3 - \sqrt{3}}{6}$.
Finally, we have $$f(n) = \frac{3 + \sqrt{3}}{6}(1 + \sqrt{3})^n + \frac{3 - \sqrt{3}}{6}(1 - \sqrt{3})^n.$$
For the problem at hand, we want $$f(10) = 18,272.$$
-
That's strange, f(2)=7. But as Mark Eichenlaub says, it should be 6. I think you made a mistake somewhere. – Lopsy Feb 13 '12 at 21:52
The roots should be $1\pm\sqrt3$, and $f(10)$ is considerably smaller; a hasty calculation made it $18272$. @Lopsy means that your formula for $f(n)$ gives $f(2)=7$. – Brian M. Scott Feb 13 '12 at 21:55
Right, but your final equation with the radicals gives 7. I think I figured out what happened: the roots of that quadratic are 1+sqrt(3) and 1-sqrt(3), not 2+sqrt(3) and 2-sqrt(3) as you claim. – Lopsy Feb 13 '12 at 21:56
Thanks for everyone's keen eyes. I believe I've caught all the errors now. – Austin Mohr Feb 13 '12 at 22:16
Now it’s good. I’d add just one comment: $10$ is a small enough number that in this case it’s easier just to use the recurrence than it is to find the general solution. – Brian M. Scott Feb 13 '12 at 22:18
Hint: use recursion.
You know that the first note in the Bauber bird's song is either $1$ or $2$ seconds long. In the first case, the remainder of the song is $9$ seconds long; in the second case, the remainder of the song is $8$ seconds long.
Therefore, the number of $10$-second songs equals twice (remember, the last note can be low or high) the number of $9$-second songs plus the number of $8$-second songs. Can you see how to solve the problem from here?
Also, note that it's generally a bad idea to encourage "looking for things to do with the numbers in the problem" like you said you did with your first try, $\binom{10}{2}*\binom{10}{2}$. Although formula-hunting does have its uses (esp. when doing dimensional analysis, where it's essential), if you're going to try to find a formula first, it's a good idea to check the answer for small cases (like $1$ second, $2$ seconds, etc) and then extrapolate off of that than trying to guess formulae blind.
-
Here is a non-recursive solution.
Suppose there are $a$ 1s songs and $b$ 2s songs, so $a + 2b = 10$. The total number of songs is $10 - b$.
We can count the number of such songs by first ignoring pitch and counting the ways the 1s and 2s songs can be ordered, then multiplying by $2^{a+b}$.
There are $a+1$ gaps between the $a$ 1s songs, so we want to know how many ways $b$ 2s songs can fit into $a+1$ gaps. This is $\binom{a+b}{b} = \binom{10-b}{b}$, as shown by Akash Kumar in this answer.
So we want
$$\sum_{b=0}^{5} 2^{10 - b}\dbinom{10 - b}{b} = 18,272$$
-
But you can’t break the ten-second interval into $2$-second chunks, because a $2$-second note can overlap chunks. – Brian M. Scott Feb 13 '12 at 21:59
@BrianM.Scott Yes, you're right, thanks. Tried a new answer. – Mark Eichenlaub Feb 13 '12 at 23:04 | 2014-08-28T03:15:56 | {
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https://math.stackexchange.com/questions/119904/units-and-nilpotents | # Units and Nilpotents
If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.
I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?
• Dear Shannon, Try the case $u = 1$ first. Regards, – Matt E Mar 14 '12 at 2:23
• See also here for the commutative case. – Gone Aug 6 '17 at 15:21
If $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$ by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
• I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a? – Shannon Mar 14 '12 at 2:51
• @Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent? – Arturo Magidin Mar 14 '12 at 2:56
• oh, I see. I will work on it. Thank you so much. – Shannon Mar 14 '12 at 3:10
• This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises. – Derek Elkins left SE Jan 21 '16 at 4:01
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.
• The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from? – Shannon Mar 14 '12 at 2:55
Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $M\subset R$. Since $a$ is nilpotent, $a\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-a\in M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $S\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-a\in I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+\binom{n}{2}v^2(u+a)^2+\dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)\left(nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}\right)$$ and so $$-\sum_{k=1}^n \binom{n}{k}(-v)^k(u+a)^{k-1}= nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
Note that since $$u$$ is a unit and
$$ua = au, \tag 1$$
we may write
$$a = u^{-1}au, \tag 2$$
and thus
$$au^{-1} = u^{-1}a; \tag 3$$
also, since $$a$$ is nilpotent there is some $$0 < n \in \Bbb N$$ such that
$$a^n = 0, \tag 4$$
and thus
$$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$$
we observe that
$$u + a = u(1 + u^{-1}a), \tag 6$$
and that, by virtue of (5),
$$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$$ $$= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$$ $$= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$$ $$= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$$
this shows that
$$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$$
and we have demonstrated an explicit inverse for $$1 + u^{-1}a$$. Thus, by (6),
$$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$$
that is, $$u + a$$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $$I - T$$ is invertible for any nilpotent linear operator $$T$$. Taking $$T = a$$ and $$I = u$$ in the above immediately yields the existence of $$(I - T)^{-1}$$. End of Note. | 2020-07-06T12:07:28 | {
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https://math.stackexchange.com/questions/3509665/three-tosses-of-the-same-coin-conditional-probability-that-the-coin-picked-is-u | # Three tosses of the same coin: Conditional probability that the coin picked is unfair
There are two coins: one fair and one unfair. P(head | fair)=½ and P(head | unfair)=⅓ These two coins look identical.
You picked up a coin from these two, throw it 3 times and observed 1 head.
What is the probability that the coin you picked is the unfair coin?
The part that is throwing me off is three tosses with one resulting in a head. Below I calculated if a heads is observed, what would be the probability that it's an Unfair coin. But if I consider three tosses, I consider three possibilities that outcome space could be {HTT, THT,TTH} and to me, all three possibilities seem to require the same calculation, so using a tree, I calculated P(U|HTT) = P(U∩H∩T∩T)/P(H∩T∩T) as
Am I making a mistake in calculating below part? The result makes me feel weird for some reason and if I consider the three outcome possibilities in numerator and denominator, they cancel out so final result should be same as below but that implies that number of tosses has no affect - this is the part that is throwing me off so need help with 1) Verify the calculation 2) Does number of tosses have an effect? or is there another way to interpret repeated tosses without replacement?
I looked at this post that is very close to my question but the first and second tosses have a pre-defined outcome. Thanks again for your help.
P(U)P(H|U)P(T|U)P(T|U)/(P(U∩H∩T∩T)+P(F∩H∩T∩T))
P(U|HTT) = 1/2 * 1/3 * 2/3 * 2/3 /(1/2*1/3*2/3*2/3 + 1/2*1/2*1/2*1/2)
= 4/27 / (4/27 + 1/8)
= 32/59
This comes out to
Here is the simple one toss calculation:
P(F) = probability the coin is fair
P(U) = probability the coin is UNfair
P(H|F) = 1/2
P(H|U) = 1/3
To calculate Probability the coin is Unfair given a heads is observed, I can use bayes theorum as:
P(U|H) = P(U)*P(H|U)/ (P(U)*P(H|U) + P(F)*P(H|F)
= 1/2 * 1/3 / (1/2*1/3 + 1/2*1/2)
= 4/10
= 2/5
## Update:
I realized the mistake I was making in assuming that number of tosses doesn't matter. It's the order that doesn't matter in the calculation, not the number of tosses - even if there were 2 or 4 or any number of tosses with one Heads, there would be 2 or 4 or n possible combinations respectively, so counting them in numerator or denominator cancels out, so order doesn't matter. However, number of tosses do matter, if there were 4 tosses, the P(HTTT|U) would have additional factor of 2/3 to indicate additional Tails, so the numerator as well as denominator would change.
The part I am still not able to think intuitively about is how to explain that order doesn't matter. Algebraically, it cancels out from Num and Denom but can't think of an intuition behind it. Thanks for the long read.
P(H|U) = 1/3
P(H|F) = 1/2
Let A be the event where we observe 3 tosses with one head
P(U|A) = P(A,U) = p(U) * P(A|U)
-------- ---------------
P(A) P(A)
= 1/2 *[P(HTT|U)+P(THT|U)+P(TTH|U)]
------------------------------------
P(A, U) + P(A, F)
= 1/2 * [(1/3*2/3*2/3)*3]
------------------------------------
[1/2*(1/3*2/3*2/3)*3] + [1/2*(1/2*1/2*1/2)*3]
= 4/27
------------
(4/27 + 1/8)
= 4/27
-----------------
(32 + 27)/(27*8)
= 32
--
59
Let's define a few random variables. Let $$U \sim \operatorname{Bernoulli}(\pi = 1/2)$$ represent the prior probability of selecting the unfair coin, where $$U = 1$$ means the coin is unfair, and $$U = 0$$ means the coin is fair. So $$\Pr[U = 1] = \Pr[U = 0] = 1/2$$. Next, define $$X \mid U \sim \operatorname{Binomial}(n = 3, p),$$ representing the number of heads obtained in three coin tosses. The probability of heads $$p$$ is a function of $$U$$: if $$U = 0$$, then $$p = 1/2$$, whereas if $$U = 1$$, then $$p = 1/3$$. So we can write this as $$p = \frac{1}{2} - \frac{1}{6}U = \frac{3-U}{6}.$$ Therefore, $$\Pr[X = 1 \mid U = u] = \binom{3}{1} \left( \frac{3-u}{6} \right)^1 \left( 1 - \frac{3-u}{6} \right)^2 = \frac{(3-u)(3+u)^2}{72},$$ and in particular, $$\Pr[X = 1 \mid U = 1] = \frac{4}{9}, \\ \Pr[X = 1 \mid U = 0] = \frac{3}{8}.$$ It follows from the law of total probability, $$\Pr[X = 1] = \Pr[X = 1 \mid U = 0]\Pr[U = 0] + \Pr[X = 1 \mid U = 1]\Pr[U = 1] = \frac{4}{9}\cdot \frac{1}{2} + \frac{3}{8} \cdot \frac{1}{2} = \frac{59}{144}.$$ Therefore, $$\Pr[U = 1 \mid X = 1] = \frac{\Pr[X = 1 \mid U = 1] \Pr[U = 1]}{\Pr[X = 1]} = \frac{2/9}{59/144} = \frac{32}{59}.$$ | 2021-04-21T02:15:35 | {
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https://math.stackexchange.com/questions/1345205/perfect-powers-of-successive-naturals-can-you-always-reach-a-constant-differenc | # Perfect powers of successive naturals: Can you always reach a constant difference?
I was thinking about what happens if you take a sequence of consecutive squares, for example 1,4,9, 16. Taking the differences gives you another sequence, 7,5,3. And taking the differences between those numbers, you get 2,2--a constant. Through elementary first-semester algebra, you can easily verify that this works no matter where you start your series of squares. If you use 9801, 10000, 10201, you will get the same result--after the second round of subtraction, you'll end up with a final constant difference of 2. The same thing works for cubes with the final constant difference being 6, although an additional round of subtraction is needed, in turn requiring one more number in the original sequence. Similarly as with the squares, it's not difficult to show that it will work for any sequence of natural number cubes.
My question is this: Is there a more general principle at work here? Suppose I take a series of sequential naturals n, n+1, n+2, n+3,... , and raise them to any given positive integral power p. Can it then be shown that if I take the differences repeatedly, proceeding as above, that I will eventually reach a constant difference? And if so, does this principle have anything to do with "difference engine" computing?
• Side note: The eventual constant difference is $p!$, where $p$ is the power. Jun 30 '15 at 23:40
• You should take a look at Faulhaber's formula. (Note that taking differences of the numbers $$S_p(n),\;\;S_p(n+1),\;\;\ldots,\;\;S_p(n+k),\;\;\ldots$$ produces the sequence $$n^p,\;\;(n+1)^p,\;\;\ldots,\;\;(n+k)^p,\;\;\ldots$$ and then you can continue taking differences from there.) Jun 30 '15 at 23:41
• Thanks for the answers and comments so far. And yes, a proof will definitely be in the works...not because the world needs it, but because I need to see it work for myself, as I did for the special cases ^2 and ^3. I do wish that when I was in school they had emphasized proofs instead of what seemed to me, at the time, merely a random grab-bag of disconnected rules and tricks. Jun 30 '15 at 23:46
Yes.
The first difference of a polynomial of degree $d$ is a polynomial of degree $d-1$.
By induction, the $m$-th difference of a polynomial of degree $d$, when $m \le d$, is a polynomial of degree $d-m$.
Setting $m = d$, the $d-th$ difference of a polynomial of degree $d$ is a constant, which is exactly what you discovered.
You next step is a proof.
If you start with the sequence of $n$-th powers, the $n$-th differences will all be $n!$.
For a function $f$ defined on the integers define $(\Delta f)(x)=f(x+1)-f(x)$. $\Delta$ is a linear operator on such functions: you can easily check that $\Delta\big(af(x)+bg(x)\big)=a\Delta f(x)+b\Delta g(x)$ for any such functions $f$ and $g$ and constants $a$ and $b$.
Now suppose that $\Delta^k(x^k)$ is the constant function $k!$ for each $k<n$. For any constant function $f$ the difference function $\Delta f$ is identically $0$, so $\Delta\big((\Delta^k(x^k)\big)=\Delta(k!)=0$ for each $k<n$, and it follows that $\Delta^m(x^k)=0$ whenever $k<n$ and $m>k$. In particular, $\Delta^{n-1}(x^k)=0$ for $k<n-1$. Thus,
\begin{align*} \Delta^n(x^n)&=\Delta^{n-1}\big((x+1)^n-x^n\big)\\ &=\Delta^{n-1}\left(\sum_{k=0}^n\binom{n}kx^k-x^n\right)\\ &=\Delta^{n-1}\left(\sum_{k=0}^{n-1}\binom{n}kx^k\right)\\ &=\sum_{k=0}^{n-1}\binom{n}k\Delta^{n-1}(x^k)\\ &=\binom{n}{n-1}\Delta^{n-1}(x^{n-1})\\ &=n(n-1)!\\ &=n!\;, \end{align*}
and by induction $\Delta^n(x^n)=n!$ for all $n$. (You can easily check the basis step of the induction.)
These so-called forward differences are a special case of divided differences, after which Babbage’s Difference Engine was named.
Using the Binomial Theorem, we get $$(n+1)^k-n^k=\sum_{j=0}^{k-1}\binom{k}{j}n^j$$ So the difference of two consecutive $k^\text{th}$ powers is a $k-1$ degree polynomial with lead coefficient $\binom{k}{k-1}=k$. Thus, the difference of a degree $k$ polynomial with lead coefficient $c_k$ is a degree $k-1$ polynomial with a lead coefficient of $kc_k$.
Repeating this, we get that the $k^\text{th}$ repeated difference of $k^\text{th}$ powers (a degree $k$ polynomial with lead coefficient $1$) will be $k!$. | 2021-12-04T15:58:41 | {
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http://math.stackexchange.com/questions/84491/does-the-set-of-differences-of-a-lebesgue-measurable-set-contains-elements-of-at/104126 | # Does the set of differences of a Lebesgue measurable set contains elements of at most a certain length?
I want to show that if $E\subset \mathbb{R}^n$ is a Lebesgue measurable set where $\lambda(E)>0$, then $E-E=\{x-y:x,y\in E\}\supseteq\{z\in\mathbb{R}^n:|z|<\delta\}$ for some $\delta>0$, where $|z|=\sqrt{\sum_{i=1}^n z_i^2}$.
My approach is this. Take some $J$, a box in $\mathbb{R}^n$ with equal side lengths such that $\lambda(E\cap J)>3\lambda(J)/4$. Setting $\epsilon=3\lambda(J)/2$, take $x\in\mathbb{R}^n$ such that $|x|\leq\epsilon$. Then $E\cap J\subseteq J$ and $$((E\cap J)+x)\cup(E\cap J)\subseteq J\cup(J+x).$$
Since Lebesgue measure is translation invariant, it follows that $\lambda((E\cap J)+x)=\lambda(E\cap J)$, and so $((E\cap J)+x)\cap(E\cap J)\neq\emptyset$.
If it were empty, then $$2\lambda(E\cap J)=\lambda(((E\cap J)+x)\cup(E\cap J))\leq\lambda(J\cup(J+x))\leq 3\lambda(J)/2,$$ thus $\lambda(E\cap J)\leq 3\lambda(J)/4$, a contradiction.
Then $((E\cap J)+x)\cap (E\cap J)\neq\emptyset$, and so $x\in (E\cap J)-(E\cap J)\subseteq E-E$. Thus $E-E$ contains the box of $x$ such that $|x|\leq \epsilon$.
Is this valid? If not, can it be fixed? Many thanks.
-
I am suspicious about your first hypothesis : How can you take a box such that $\lambda(E \cap J) > 3 \lambda(J)/4$? I must say though, this is a very interesting question. – Patrick Da Silva Nov 22 '11 at 7:13
– gary Nov 22 '11 at 7:21
@PatrickDaSilva: $E$ has a subset $E'$ with positive and finite measure. There is an open set $U$ such that $E'\subseteq U$ and $\lambda(U)<\frac{4}{3}\lambda(E')$. Since $U$ is a countable union of nonoverlapping boxes, $U=\cup_k B_k$, it follows that $\sum_k\lambda(E'\cap B_k)>\sum_k\frac{3}{4}\lambda(B_k)$. Hence there exists $k$ such that $\lambda(E'\cap B_k)>\frac{3}{4}\lambda(B_k)$, and you can take $J=B_k$. – Jonas Meyer Nov 22 '11 at 7:25
You may want to have a look at this thread and also at this one as well as the links therein. – t.b. Nov 22 '11 at 8:09
I think this can be fixed if either OP justifies why this inequality holds, or just by taking a smaller epsilon. It might depend on $n$ though, but it seems to be just geometry at this point. I now accept that "this argument holds by the handwaving theorem". =P – Patrick Da Silva Jan 31 '12 at 5:07
This is answered here, I will just post another approach that can be useful.
Lemma. If $K$ is a compact subset of $\mathbb{R}^n$ with positive measure, then the set $$D:=\{x-y:x,y\in K\}$$ contains an open ball centered at the origin.
Proof. Since $0\lt \lambda(K)\lt\infty$, there exist an open set $G$ in $\mathbb{R}^n$ such that $$K\subset G\text{ and } \lambda (G)\lt 2\lambda(K)$$ and as its complement $G^c:=\mathbb{R}^n\setminus G$ is closed and $K$ is compact, we have $$\delta:=d(K,G^c)\gt 0.$$ We claim that if $x\in\mathbb{R}^n$ with $||x||\lt\delta$ then $K+x\subseteq G$. If not, there exist a $y\in K$ s.t. $y+x\not\in G$ and then $$\delta=d(K,G^c)\leq ||y-(x+y)||=||x||,$$ which is an absurd if we assume that $||x||\lt\delta$ to begin with. Now $K+x\subseteq G$ and $K\subseteq G$, thus $(K+x)\cup K\subseteq G$. If $(K+x)\cap K=\emptyset$, then $$\lambda(G)\geq \lambda((K+x)\cup K)=\lambda(K+x)+\lambda(K)=2\lambda(K)$$ which contradicts the choice of $G$. So, for all $x\in B(0,\delta)$ we have $(K+x)\cap K\neq\emptyset$, therefore $B(0,\delta)\subset D$. QED
Theorem. Let $E\subseteq\mathbb{R}^d$. If if $E$ is measurable and $\lambda(E)\gt 0$ then the set $$E-E:=\{x-y:x,y\in E\}$$ contains an open ball centered at the origin.
Proof. Since $\mathbb{R}^d=\bigcup_{n\in\mathbb N}B(0,n)$ we have $$E=E\cap\mathbb{R}^d=\bigcup_{n\in\mathbb N}E\cap B(0,n),$$ then $$0\lt\lambda(E)\leq \sum_{n\in\mathbb{N}} \lambda(E\cap B(0,n))$$ and then, there exist a $n\in\mathbb{N}$ such that $\lambda(E\cap B(0,n))\gt 0$. Let $F=E\cap B(0,n)$. Thus $F\subseteq E$ is bounded and measurable, therefore there exist a closed set $K\subseteq F$ such that $$0\lt \frac{\lambda(F)}{2}\lt \lambda(K).$$ Notice that $K$ is closed and bounded, that means it is a compact set with positive measure. By the Lemma the set $K-K$ contains an open ball centered at the origin say $B$. Then $$B\subset K-K\subseteq E-E$$ as we wanted.
Note. The idea of proceed in this way is taken from Robert Bartle, I don't remember right now the book where this come from.
- | 2015-10-09T13:59:05 | {
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https://math.stackexchange.com/questions/2568587/minimum-values-of-the-sequence-n-sqrt2 | # Minimum values of the sequence $\{n\sqrt{2}\}$
I have been studying the sequence $$\{n\sqrt{2}\}$$ where $\{x\}:= x-\lfloor x\rfloor$ is the "fractional part" function. I am particularly interested in the values of $n$ for which $\{n\sqrt{2}\}$ has an extremely small value - that is, when $n\sqrt 2$ is extremely close to (but greater than) its nearest integer. Of course, for integer $n$, this always takes on values between (but never including) $0$ and $1$.
I have defined the sequence $M_n$ as follows:
• $M_1=1$
• $M_{n+1}$ is the smallest positive integer such that $\{M_{n+1}\sqrt 2\}\lt \{M_n\sqrt 2\}$
The first few terms of this sequence are $$1,3,5,17,29,99,169,...$$ And after a quick trip to the OEIS, I have found that this sequence is equal to OEIS entry A079496, as written in the comments. It also provides the recurrence $$a_0=a_1=1$$ $$a_{2n+1}=2a_{2n}-a_{2n-1}$$ $$a_{2n}=4a_{2n-1}-a_{2n-2}$$ How can I prove that the sequence $M_n$ satisfies this recurrence, using the definition I wrote for $M_n$?
Thanks!
EDIT: Here is the closed-form explicit formula for $a_n$, for anyone who wants to know: $$a_{2n}=\frac{(3+2\sqrt 2)^n+(3-2\sqrt 2)^n}{2}$$ $$a_{2n+1}=\frac{(1+\sqrt 2)(3+2\sqrt 2)^n-(1-\sqrt 2)(3-2\sqrt 2)^n}{2\sqrt 2}$$
• Have you taken a look to the references on the link you shared? I mean, the comment of Paul. D Hanna there says that this is exactly what you're looking for... – Alejandro Nasif Salum Dec 15 '17 at 23:54
• I mean... you want to know if this recurrence is valid for your sequence or you are sure of it and you are just trying to prove it? – Alejandro Nasif Salum Dec 15 '17 at 23:55
• Yeah, that confuses me as well. I'll add the closed form to the question, for anyone who is curious. – Franklin Pezzuti Dyer Dec 16 '17 at 0:02
• According to Kroneckers Approximation Theorem $\{n\sqrt{2}\}$ is dense in $[0,1]$. – rtybase Dec 16 '17 at 0:39
• This can be seen as two interleaved sequences with the recursion $q_n=6q_{n-1}-q_{n-2}$ – robjohn Dec 17 '17 at 1:30
Summary of solution : the originality of your question consists in its use of the floor function to round, when most of the classical results on continued fractions use the nearest integer function. The idea is to reduce the former to the latter.
Details : Let $M_n,a_n$ be as in the OP. Our goal is to show that
$$M_n=a_n \tag{1}$$
As a convenience, let us use the "normalized" number $\alpha=\sqrt{2}-1$ (which is in the unit interval) rather than $\sqrt{2}$ itself.
It will suffice to show :
$$\lbrace q\alpha \rbrace \geq \lbrace a_n\alpha \rbrace \textrm{ for } 0 \lt q \lt a_{n+1} \tag{2}$$
We will use a sequence closely related to the continued fraction expansion of $\alpha$, namely $g_n=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$. Note that $(g_n)$ satisfies $g_0=0,g_1=1$ and $g_{n+2}-2g_{n+1}+g_n=0$, so its values (after $g_1$) are positive integers. The following properties are easy to show by a direct computation or by induction :
$(i)$ $g_ng_{n+2}-g_{n+1}^2=(-1)^{n+1}$
$(ii)$ $a_n=g_n$ when $n$ is odd
$(iii)$ $a_n=g_{n+1}-g_n$ when $n$ is even
$(iv)$ $\frac{g_n-g_{n+1}\alpha}{g_{n+1}-g_{n+2}\alpha}=1-\sqrt{2}$
$(v)$ $\frac{a_{n+2}\alpha -(g_{n+2}-g_{n+1})}{a_n\alpha -(g_n-g_{n-1})}=3-2\sqrt{2}$, when $n$ is even.
The following lemma is a very classic continued fraction inequality :
LEMMA. Let $p,q$ be integers with $0 \lt q \lt g_{n+2}$. Then $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|$. Further, if $(p,q)\neq (g_n,g_{n+1})$, the inequality can be strenghtened to $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|+|g_{n+1}-g_{n+2}\alpha|$.
Proof of lemma. By (i), the matrix $\left(\begin{matrix} g_n & g_{n+1} \\ g_{n+1} & g_{n+2} \end{matrix}\right)$ is unimodular. It follows that there are integers $u,v$ such that $p=g_n u+g_{n+1}v$ and $q=g_{n+1}u+g_{n+2}v$. I claim that $uv\leq 0$ ; otherwise, $g_{n+1}u$ and $g_{n+2}v$ would both be nonzero and have the same sign, forcing $|q| = |g_{n+1}u|+|g_{n+2}v| \geq |g_{n+2}|$ which is impossible. So $uv\leq 0$, and also
$$p-q\alpha= u(g_n-g_{n+1}\alpha) + v (g_{n+1}-g_{n+2}\alpha) \tag{3}$$
The numbers $u(g_n-g_{n+1}\alpha)$ and $v(g_{n+1}-g_{n+2}\alpha)$ have the same sign, so
$$|p-q\alpha|= |u(g_n-g_{n+1}\alpha)| + |v (g_{n+1}-g_{n+2}\alpha)| \tag{4}$$
Note that $u\neq 0$ because of $q=g_{n+1}u+g_{n+2}v$ and $|q|<g_{n+2}$. So $|u|\geq 1$ and the first inequality follows. If equality holds, we must have $|u|=1$ and $v=0$, from which we easily deduce $(p,q)=(g_n,g_{n+1})$. Suppose now that $(p,q)\neq (g_n,g_{n+1})$. If $|u|\geq 2$, then from (4) we deduce $|p-q\alpha| \geq 2|g_n-g_{n+1}\alpha|$. By (iv), the second inequality holds in this case.
Otherwise $u=1$ and from the reasoning above we have $v\neq 0$, so $|v|\geq 1$ and from (4) again we deduce $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|+|g_{n+1}-g_{n+2}\alpha|$, which finishes the proof of the lemma.
Let us now deduce (2) from the lemma. Suppose first that $n$ is odd. By $(ii)$, $\lbrace a_n\alpha \rbrace=\lbrace g_n\alpha \rbrace=g_{n+1}\alpha-g_n$ by lemma, (iv) and the fact that $g_{n+1}\alpha-g_n >0$. The lemma also shows that $\lbrace q \alpha \rbrace \geq \lbrace a_n \alpha \rbrace$ for $0 \lt q \lt g_{n+2}=a_{n+2}$, as wished.
Suppose now that $n$ is even. Using the lemma with $n-1$ in place of $n$, we have that for any $0 \lt q \lt g_{n+1}=a_{n+1}$ and $(p,q)\neq (g_{n-1},g_n)$, then $$\begin{array}{lcl} |p-q\alpha| &\geq& |g_{n-1}-g_{n}\alpha|+|g_{n}-g_{n+1}\alpha| \\ &=& g_{n-1}-g_{n}\alpha+g_{n+1}\alpha-g_n \\ &=& a_n\alpha -(g_n-g_{n-1}) \\ &=& \lbrace a_n \alpha \rbrace \textrm{ by } (v) \end{array}$$
This finishes the proof.
This is about Pell type equations. Absolutely complete detail would be a bit long. I have posted many times about solving $x^2 - n y^2 = T,$ where $T$ is some target number, and $n$ is positive not a square.
Your sequence is this: given positive integer $x,$ let $v = \lfloor x \sqrt 2 \rfloor.$ Your numbers will be $$2 x^2 - v^2 \in \{ 1,2 \}.$$
Furthermore, the ratios of consecutive terms are bounded: if $w$ is a number with worse $2 w^2 - w_0^2,$ where $w_0 = \lfloor w \sqrt 2 \rfloor,$ then there is one of your numbers $t$ with $t > w/(2 + \sqrt 2).$
Alright, the positive values of $2x^2 - y^2$ are $1,2,4,7,8,9, 14 \ldots$ where we are allowing common factors of $x,y.$
If $2x^2 - v^2 = 4,$ then both $x,v$ are even, we can take $t = x/2,$ and the fractional part of $t \sqrt 2$ is half the fractional part of $x \sqrt 2,$ meaning $x$ cannot be part of your $M$ sequence. Now that I think of it, the same applies to any number divisible by $4$ or by $9.$ As a result, if we calculated $2x^2 - v^2 = 7$ as a special case (we know how to find all representations) we could then continue with $2 x^2 - v^2 \geq 14.$
It appears we need just the one special case. After this, we may take $$2 x^2 - v^2 \geq 7,$$ with $$\{ x \sqrt 2 \} \geq \frac{7}{x \sqrt 2 + v} \approx \frac{7}{2x \sqrt 2 }\approx \frac{2.4748737}{x } .$$ $$t > \frac{x}{2 + \sqrt 2},$$ $$2 t^2 - w^2 \in \{ 1,2 \}.$$ $$(t \sqrt 2 - w)(t \sqrt 2 + w) \leq 2.$$ $$t \sqrt 2 - w \leq \frac{2}{t \sqrt 2 + w}.$$ $$\{ t \sqrt 2 \} \leq \frac{2}{t \sqrt 2 + w} \approx \frac{2}{2t \sqrt 2 } \approx \frac{1}{t \sqrt 2 } < \frac{2 + \sqrt 2}{x \sqrt 2 } = \frac{1 + \sqrt 2}{x } \approx \frac{2.1421356}{x } .$$ This is smaller than $$\{ x \sqrt 2 \} \geq \frac{7}{x \sqrt 2 + v} \approx \frac{7}{2x \sqrt 2 }\approx \frac{2.4748737}{x } .$$
A more careful analysis is possible, correction terms all over, but the heart of it is the observation that the floor of $t \sqrt 2$ is extremely close to the real number itself, when $0 < 2t^2 - w^2 \leq 2$.
Here is the Conway topograph for the negative of your form, $x^2 - 2 y^2.$ Your numbers show up representing negative number, (4,3) gives $-2$, then (7,5) gives $-1,$ then (24,17) give $-2,$ (41,29) gives $-1$
Let's see, it will take an hour or so, but I can draw a topograph for $2x^2 - v^2$ that shows numbers represented up to $17,$ as numbers such as $4,8,9$ are not squarefree, and are not primitively represented (we get $\gcd(x,v) \neq 1$). Will, 9:19 Pacific time
Done. Note that the green coordinate pairs have a motion that gives the same value of $2x^2 - y^2,$ namely $$(x,y) \mapsto (3x+2y, 4x+3y).$$ For example, $(1,1) \mapsto (5, 7 )$ and $(3,4) \mapsto (17, 24 ).$
Let's see, Cayley Hamilton for the coefficient matrix $$\left( \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right)$$ gives the linear recurrence, $$x_{n+4} = 6 x_{n+2} - x_n$$ if we combine the $x$ values in a single list. We need to separate odd index and even index because we are combining $2x^2-v^2 = 1$ and $2x^2 - v^2 = 2$ in one list of $x$ values
There is a book that Allen Hatcher makes available online, here is his diagram for $x^2 - 2 y^2,$ but without the green coordinates that I like to include. The topograph diagram was introduced by J. H. Conway
Continued Fraction Approximations
The continued fraction for $\sqrt2$ is $$\sqrt2=1+\cfrac1{2+\cfrac1{2+\cfrac1{2+\dots}}}\tag1$$ The approximants alternate between a bit too high and a bit too low, but we always have $$\left|\,\frac{p_n}{q_n}-\sqrt2\,\right|\le\frac1{2q_n^2}\tag2$$ because of the continued fraction, we have $$p_n=2p_{n-1}+p_{n-2}\quad\text{and}\quad q_n=2q_{n-1}+q_{n-2}\tag3$$ starting with $\frac11$ (low) and $\frac32$ (high). The first several continued fraction approximants are $$\frac11,\frac32,\frac75,\frac{17}{12},\frac{41}{29},\frac{99}{70},\dots\tag4$$
Low Approximations
We want to choose the approximations that are a bit low so that $q_n\sqrt2\gt p_n$. Then $$\left\{q_n\sqrt2\right\}=q_n\sqrt2-p_n\le\frac1{2q_n}\tag5$$ Since we only want the low approximations, we need to modify the recursion in $(3)$ to skip the high approximations. To that end, we have $$p_n=6p_{n-2}-p_{n-4}\quad\text{and}\quad q_n=6q_{n-2}-q_{n-4}\tag6$$ starting with $\frac11$ and $\frac75$. The first several low continued fraction approximations are $$\frac11,\frac75,\frac{41}{29},\frac{239}{169},\frac{1393}{985},\dots\tag7$$ We can also make some close low approximations by looking at twice the reciprocals of the high approximations. These will follow the recursion in $(6)$: $$\frac43,\frac{24}{17},\frac{140}{99},\frac{816}{577},\dots\tag8$$ The denominators of these two sequences, which follow the recursion in $(6)$, cover all of the elements of the sequences in the question.
Combined Sequence
The combined sequence follows the recursion $$a_n=6a_{n-2}-a_{n-4}$$ and starts out $$1,3,5,17,\dots$$
• This was my immediate idea too. But it looks like this produces only every other term in the OP's sequence. – hmakholm left over Monica Dec 16 '17 at 22:59
• Their sequence seems to include all the continued fraction approximants, even the high ones. The high ones will give $\left\{q_n\sqrt2\right\}$ just less than $1$. Or perhaps my list misses some close, but not continued fraction approximations. Only continued fraction approximations satisfy $(2)$. I will look when I get back home. – robjohn Dec 16 '17 at 23:05
• I suspect the additional terms are because leaving out the low approximants leave room for additional multiples of $\sqrt2$ to have a smallest-yet fractional part even though there have been earlier multiples (corresponding to the high approximants) that approached an integer closer from below. – hmakholm left over Monica Dec 16 '17 at 23:30
• @HenningMakholm: I figured out where the other terms in the question came from. – robjohn Dec 17 '17 at 0:57
• If $x \in (\frac{2}{3},\frac{3}{4})$, then $\lbrace x \rbrace \gt \frac{1}{2} \gt \lbrace 2x \rbrace$, so $2x$ is a record in the sense of the OP, but not a best approximation situation in the classical sense. Here in the special case where $x=\sqrt{2}$ we are lucky that all the records in the sense of the OP come from best approximations. – Ewan Delanoy Dec 26 '17 at 8:31 | 2020-11-29T20:35:39 | {
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https://stats.stackexchange.com/questions/159313/generating-samples-from-singular-gaussian-distribution/159322 | Generating samples from singular Gaussian distribution [duplicate]
Let random vector $x = (x_1,...,x_n)$ follow multivariate normal distribution with mean $m$ and covariance matrix $S$. If $S$ is symmetric and positive definite (which is the usual case) then one can generate random samples from $x$ by first sampling indepently $r_1,...,r_n$ from standard normal and then using formula $m + Lr$, where $L$ is the Cholesky lower factor so that $S=LL^T$ and $r = (r_1,...,r_n)^T$.
What about if one wants samples from singular Gaussian i.e. $S$ is still symmetric but not more positive definite (only positive semi-definite). We can assume also that the variances (diagonal elements of $S$) are strictly positive. Then some elements of $x$ must have linear relationship and the distribution actually lies in lower dimensional space with dimension $<n$, right?
It is obvious that if e.g. $n=2, m = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, S = \begin{bmatrix} 1 & 1 \\ 1 & 1\end{bmatrix}$ then one can generate $x_1 \sim N(0,1)$ and set $x_2=x_1$ since they are fully correlated. However, is there any good methods for generating samples for general case $n>2$? I guess one needs first to be able identify the lower dimensional subspace, then move to that space where one will have valid covariance matrix, then sample from it and finally deduce the values for the linearly dependent variables from this lower-dimensional sample. But what is the best way to that in practice? Can someone point me to books or articles that deal with the topic; I could not find one.
The singular Gaussian distribution is the push-forward of a nonsingular distribution in a lower-dimensional space. Geometrically, you can take a standard Normal distribution, rescale it, rotate it, and embed it isometrically into an affine subspace of a higher dimensional space. Algebraically, this is done by means of a Singular Value Decomposition (SVD) or its equivalent.
Let $\Sigma$ be the covariance matrix and $\mu$ the mean in $\mathbb{R}^n$. Because $\Sigma$ is non-negative definite and symmetric, the SVD will take the form
$$\Sigma = U \Lambda^2 U^\prime$$
for an orthogonal matrix $U\in O(n)$ and a diagonal matrix $\Lambda$. $\Lambda$ will have $m$ nonzero entries, $0\le m \le n$.
Let $X$ have a standard Normal distribution in $\mathbb{R}^m$: that is, each of its $m$ components is a standard Normal distribution with zero mean and unit variance. Abusing notation a little, extend the components of $X$ with $n-m$ zeros to make it an $n$-vector. Then $U\Lambda X$ is in $\mathbb{R}^n$ and we may compute
$$\text{Cov}(U\Lambda X) = U \Lambda\text{Cov}(X) \Lambda^\prime U^\prime = U \Lambda^2 U^\prime = \Sigma.$$
Consequently
$$Y = \mu + U\Lambda X$$
has the intended Gaussian distribution in $\mathbb{R}^n$.
It is of interest that this works when $n=m$: that is to say, this is a (standard) method to generate multivariate Normal vectors, in any dimension, for any given mean $\mu$ and covariance $\Sigma$ by using a univariate generator of standard Normal values.
As an example, here are two views of a thousand simulated points for which $n=3$ and $m=2$:
The second view, from edge-on, demonstrates the singularity of the distribution. The R code that produced these figures follows the preceding mathematical exposition.
#
# Specify a Normal distribution.
#
mu <- c(5, 5, 5)
Sigma <- matrix(c(1, 2, 1,
2, 3, 1,
1, 1, 0), 3)
#
# Analyze the covariance.
#
n <- dim(Sigma)[1]
s <- svd((Sigma + t(Sigma))/2) # Guarantee symmetry
s$d <- abs(zapsmall(s$d))
m <- sum(s$d > 0) #$
# Generate a standard Normal x in R^m.
#
n.sample <- 1e3 # Number of points to generate
x <- matrix(rnorm(m*n.sample), nrow=m)
#
# Embed x in R^n and apply the square root of Sigma obtained from its SVD.
#
x <- rbind(x, matrix(0, nrow=n-m, ncol=n.sample))
y <- s$u %*% diag(sqrt(s$d)) %*% x + mu
#
# Plot the results (presuming n==3).
#
library(rgl)
plot3d(t(y), type="s", size=1, aspect=TRUE,
xlab="Y1", ylab="Y2", zlab="Y3", box=FALSE,
col="Orange")
• Thanks for this great answer it's been very helpful to me. One question occurred to me - if X is extended to include $n-m$ zeroes then is the covariance of X still the identity matrix? That seems like a key step but wouldn't the covariance have $n-m$ zeroes on the diagonal? Mean zero, variance zero = constant zero. EDIT: Nevermind I see now that it doesn't need to be the identity matrix – ASeaton Mar 31 at 8:27 | 2021-05-18T16:49:28 | {
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http://npdinc.co.za/walden-farms-mxj/9v5arr.php?a3c1ed=counting-theory-discrete-math | From 1 to 100, there are $50/2 = 25$ numbers which are multiples of 2. Combinatorics is an area of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures.It is closely related to many other areas of mathematics and has many applications ranging from logic to statistical physics, from evolutionary biology to computer science, etc. $A \cap B = \emptyset$), then mathematically $|A \cup B| = |A| + |B|$, The Rule of Product − If a sequence of tasks $T_1, T_2, \dots, T_m$ can be done in $w_1, w_2, \dots w_m$ ways respectively and every task arrives after the occurrence of the previous task, then there are $w_1 \times w_2 \times \dots \times w_m$ ways to perform the tasks. For example: In a group of 10 people, if everyone shakes hands with everyone else exactly once, how many handshakes took place? %���� Counting mainly encompasses fundamental counting rule, the permutation rule, and the combination rule. Today we introduce set theory, elements, and how to build sets.This video is an updated version of the original video released over two years ago. . Thereafter, he can go Y to Z in $4 + 5 = 9$ ways (Rule of Sum). . }$$. If n pigeons are put into m pigeonholes where n > m, there's a hole with more than one pigeon. . = 6 ways. Graph theory. There are 50/6 = 8 numbers which are multiples of both 2 and 3. |A \cup B| = |A| + |B| - |A \cap B| = 25 + 16 - 8 = 33. of ways to fill up from first place up to r-th-place −, n_{ P_{ r } } = n (n-1) (n-2)..... (n-r + 1), = [n(n-1)(n-2) ... (n-r + 1)] [(n-r)(n-r-1) \dots 3.2.1] / [(n-r)(n-r-1) \dots 3.2.1]. . = 180.. Discrete Mathematics Tutorial Index . Now we want to count large collections of things quickly and precisely. (1!)(1!)(2!)] Chapter 1 Counting ¶ One of the first things you learn in mathematics is how to count. 2 CS 441 Discrete mathematics for CS M. Hauskrecht Basic counting rules ⢠Counting problems may be hard, and easy solutions are not obvious ⢠Approach: â simplify the solution by decomposing the problem ⢠Two basic decomposition rules: â Product rule ⢠A count decomposes into a sequence of dependent counts How many integers from 1 to 50 are multiples of 2 or 3 but not both? Why one needs to study the discrete math It is essential for college-level maths and beyond that too There was a question on my exam which asked something along the lines of: "How many ways are there to order the letters in 'PEPPERCORN' if all the letters are used?" Would this be 10! }, = (n-1)! . Ten men are in a room and they are taking part in handshakes. Probability. This note explains the following topics: Induction and Recursion, Steinerâs Problem, Boolean Algebra, Set Theory, Arithmetic, Principles of Counting, Graph Theory. Viewed 4k times 2. How many ways can you choose 3 distinct groups of 3 students from total 9 students? Example: you have 3 shirts and 4 pants. �.����2�(�^�� 㣯U��Nn%�u��p�;�VY�����W��}����{SH�W���������-zHLJ�f� R'����;���q��Y?���?�WX���:5(�� �3a���Ã*p0�4�V����y�g�q:�k��F�̡[I�6)�3G³R�%��, %Ԯ3 Solution − There are 6 letters word (2 E, 1 A, 1D and 2R.) / [(a_1!(a_2!) The number of all combinations of n things, taken r at a time is −,$$^nC_{ { r } } = \frac { n! } . In this technique, which van Lint & Wilson (2001) call âone of the most important tools in combinatorics,â one describes a finite set X from two perspectives leading to two distinct expressions ⦠. From a set S ={x, y, z} by taking two at a time, all permutations are −, We have to form a permutation of three digit numbers from a set of numbers $S = \lbrace 1, 2, 3 \rbrace$. There are $50/3 = 16$ numbers which are multiples of 3. Next come chapters on logic, counting, and probability.We then have three chapters on graph theory: graphs, directed + \frac{ n-k } { k!(n-k)! } Different three digit numbers will be formed when we arrange the digits. . Mathematically, if a task B arrives after a task A, then $|A \times B| = |A|\times|B|$. For two sets A and B, the principle states −, $|A \cup B| = |A| + |B| - |A \cap B|$, For three sets A, B and C, the principle states −, $|A \cup B \cup C | = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C |$, $|\bigcup_{i=1}^{n}A_i|=\sum\limits_{1\leq i�,oX���N8xT����,�0�z�I�Q������������[�I9r0� '&l�v]G�q������i&��b�i� �� �q���K�?�c�Rl . Hence, there are 10 students who like both tea and coffee. �d�$�̔�=d9ż��V��r�e. Pigeonhole Principle states that if there are fewer pigeon holes than total number of pigeons and each pigeon is put in a pigeon hole, then there must be at least one pigeon hole with more than one pigeon. . (nâr+1)! The remaining 3 vacant places will be filled up by 3 vowels in $^3P_{3} = 3! There must be at least two people in a class of 30 whose names start with the same alphabet. . Discrete mathematics problem - Probability theory and counting [closed] Ask Question Asked 10 years, 6 months ago. In combinatorics, double counting, also called counting in two ways, is a combinatorial proof technique for showing that two expressions are equal by demonstrating that they are two ways of counting the size of one set. Question − A boy lives at X and wants to go to School at Z. Chapter Summary The Basics of Counting The Pigeonhole Principle Permutations and Combinations This tutorial includes the fundamental concepts of Sets, Relations and Functions, Mathematical Logic, Group theory, Counting Theory, Probability, Mathematical Induction, and Recurrence Relations, Graph Theory, Trees and Boolean Algebra. Solution − From X to Y, he can go in$3 + 2 = 5$ways (Rule of Sum). . So,$|A|=25$,$|B|=16$and$|A \cap B|= 8$. Solution − As we are taking 6 cards at a time from a deck of 6 cards, the permutation will be$^6P_{6} = 6! It is a very good tool for improving reasoning and problem-solving capabilities. . Below, you will find the videos of each topic presented. Number of ways of arranging the consonants among themselves $= ^3P_{3} = 3! Discrete Mathematics & Mathematical Reasoning Chapter 6: Counting Colin Stirling Informatics Slides originally by Kousha Etessami Colin Stirling (Informatics) Discrete Mathematics (Chapter 6) Today 1 / 39. \dots (a_r!)]$. . . How many ways are there to go from X to Z? ]$, The number of circular permutations of n different elements taken x elements at time =$^np_{x}/x$, The number of circular permutations of n different things =$^np_{n}/n$. /\: [(2!) material, may be used as a textbook for a formal course in discrete mathematics or as a supplement to all current texts. (\frac{ k } { k!(n-k)! } x��X�o7�_�G����Ozm�+0�m����\����d��GJG�lV'H�X�-J"$%J�K&���8���8�i��ז�Jq��6�~��lғ)W,�Wl�d��gRmhVL���.�L���N~�Efy�*�n�ܢ��ޱߧ?��z�������|$�I��-��z�o���X�� ���w�]Lsm�K��4j�"���#gs$(�i5��m!9.����63���Gp�hЉN�/�&B��;�4@��J�?n7 CO��>�Ytw�8FqX��χU�]A�|D�C#}��kW��v��G �������m����偅^~�l6��&) ��J�1��v}�â�t�Wr���k��U�k��?�d���B�n��c!�^Հ�T�Ͳm�х�V��������6�q�o���Юn�n?����˳���x�q@ֻ[ ��XB&��,f|����+��M#R������ϕc*HĐ}�5S0H Proof − Let there be ânâ different elements. . The cardinality of the set is 6 and we have to choose 3 elements from the set. >> Here, the ordering does not matter. Hence, the number of subsets will be $^6C_{3} = 20$. Problem 1 − From a bunch of 6 different cards, how many ways we can permute it? Discrete math. Starting from the 6th grade, students should some effort into studying fundamental discrete math, especially combinatorics, graph theory, discrete geometry, number theory, and discrete probability. . . The Rule of Sum − If a sequence of tasks $T_1, T_2, \dots, T_m$ can be done in $w_1, w_2, \dots w_m$ ways respectively (the condition is that no tasks can be performed simultaneously), then the number of ways to do one of these tasks is $w_1 + w_2 + \dots +w_m$. . I'm taking a discrete mathematics course, and I encountered a question and I need your help. . in the word 'READER'. Any subject in computer science will become much more easier after learning Discrete Mathematics . . This is a course note on discrete mathematics as used in Computer Science. { (k-1)!(n-k)! } That means 3×4=12 different outfits. . %PDF-1.5 Very Important topics: Propositional and first-order logic, Groups, Counting, Relations, introduction to graphs, connectivity, trees If there are only a handful of objects, then you can count them with a moment's thought, but the techniques of combinatorics can extend to quickly and efficiently tabulating astronomical quantities. So, Enroll in this "Mathematics:Discrete Mathematics for Computer Science . Problem 3 − In how ways can the letters of the word 'ORANGE' be arranged so that the consonants occupy only the even positions? . Mathematics of Master Discrete Mathematics for Computer Science with Graph Theory and Logic (Discrete Math)" today and start learning. . Find the number of subsets of the set $\lbrace1, 2, 3, 4, 5, 6\rbrace$ having 3 elements. Mathematically, for any positive integers k and n: $^nC_{k} = ^n{^-}^1C_{k-1} + ^n{^-}^1{C_k}$, $= \frac{ (n-1)! } In how many ways we can choose 3 men and 2 women from the room? CONTENTS iii 2.1.2 Consistency. The permutation will be = 123, 132, 213, 231, 312, 321, The number of permutations of ânâ different things taken ârâ at a time is denoted by$n_{P_{r}}$. (nâr+1)!$, The number of permutations of n dissimilar elements when r specified things never come together is − $n!â[r! Discrete Mathematics Course Notes by Drew Armstrong. .10 2.1.3 Whatcangowrong. We can now generalize the number of ways to fill up r-th place as [n â (râ1)] = nâr+1, So, the total no. Number of permutations of n distinct elements taking n elements at a time =$n_{P_n} = n!$, The number of permutations of n dissimilar elements taking r elements at a time, when x particular things always occupy definite places =$n-x_{p_{r-x}}$, The number of permutations of n dissimilar elements when r specified things always come together is −$r! For choosing 3 students for 1st group, the number of ways − $^9C_{3}$, The number of ways for choosing 3 students for 2nd group after choosing 1st group − $^6C_{3}$, The number of ways for choosing 3 students for 3rd group after choosing 1st and 2nd group − $^3C_{3}$, Hence, the total number of ways $= ^9C_{3} \times ^6C_{3} \times ^3C_{3} = 84 \times 20 \times 1 = 1680$. Let X be the set of students who like cold drinks and Y be the set of people who like hot drinks. Mastering Discrete Math ( Discrete mathematics ) is such a crucial event for any computer science engineer. The applications of set theory today in computer science is countless. . Most basic counting formulas can be thought of as counting the number of ways to distribute either distinct or identical items to distinct recipients. Group theory. For solving these problems, mathematical theory of counting are used. Set theory is a very important topic in discrete mathematics . The Inclusion-exclusion principle computes the cardinal number of the union of multiple non-disjoint sets. . From there, he can either choose 4 bus routes or 5 train routes to reach Z. There are 6 men and 5 women in a room. A permutation is an arrangement of some elements in which order matters. stream . Pascal's identity, first derived by Blaise Pascal in 17th century, states that the number of ways to choose k elements from n elements is equal to the summation of number of ways to choose (k-1) elements from (n-1) elements and the number of ways to choose elements from n-1 elements. If we consider two tasks A and B which are disjoint (i.e. Solution − There are 3 vowels and 3 consonants in the word 'ORANGE'. . Some of the discrete math topic that you should know for data science sets, power sets, subsets, counting functions, combinatorics, countability, basic proof techniques, induction, ... Information theory is also widely used in math for data science. Active 10 years, 6 months ago. Problem 2 − In how many ways can the letters of the word 'READER' be arranged? He may go X to Y by either 3 bus routes or 2 train routes. The Rule of Sum and Rule of Product are used to decompose difficult counting problems into simple problems. In a group of 50 students 24 like cold drinks and 36 like hot drinks and each student likes at least one of the two drinks. Sign up for free to create engaging, inspiring, and converting videos with Powtoon. /Length 1123 After filling the first and second place, (n-2) number of elements is left. Students, even possessing very little knowledge and skills in elementary arithmetic and algebra, can join our competitive mathematics classes to begin learning and studying discrete mathematics. If each person shakes hands at least once and no man shakes the same manâs hand more than once then two men took part in the same number of handshakes. 70 0 obj << Discrete Mathematics Handwritten Notes PDF. For instance, in how many ways can a panel of judges comprising of 6 men and 4 women be chosen from among 50 men and 38 women? . . Welcome to Discrete Mathematics 2, a course introducting Inclusion-Exclusion, Probability, Generating Functions, Recurrence Relations, and Graph Theory. So, $| X \cup Y | = 50$, $|X| = 24$, $|Y| = 36$, $|X \cap Y| = |X| + |Y| - |X \cup Y| = 24 + 36 - 50 = 60 - 50 = 10$. When there are m ways to do one thing, and n ways to do another, then there are m×n ways of doing both. How many different 10 lettered PAN numbers can be generated such that the first five letters are capital alphabets, the next four are digits and the last is again a capital letter. The Basic Counting Principle. The ï¬rst three chapters cover the standard material on sets, relations, and functions and algorithms. For solving these problems, mathematical theory of counting are used. /Filter /FlateDecode Start Discrete Mathematics Warmups. = 720$. The Rules of Sum and Product The Rule of Sum and Rule of Product are used to decompose difficult counting problems into simple problems. A combination is selection of some given elements in which order does not matter. For example, distributing $$k$$ distinct items to $$n$$ distinct recipients can be done in $$n^k$$ ways, if recipients can receive any number of items, or $$P(n,k)$$ ways if recipients can receive at most one item. Thank you. Recurrence relation and mathematical induction. 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https://math.stackexchange.com/questions/3247176/problem-about-inequation-with-absolute-value | # Problem about inequation with absolute value
I have this statement:
It can be assured that | p | ≤ 2.4, if it is known that:
(1) -2.7 ≤ p <2.3
(2) -2.2 < p ≤ 2.6
My development was:
First, $$-2.4 \leq p \leq 2.4$$
With $$1)$$ by itself, that can't be insured, same argument for $$2)$$
Now, i will use $$1)$$ and $$2)$$ together, and the intersection between this intervals are: $$(-2.2, 2.3)$$. So, this also does not allow me to ensure that | p | ≤ 2.4, since, there are some numbers that are outside the intersection of these two intervals, for example 2.35 is outside this interval.
But according to the guide, the correct answer must be $$1), 2)$$ together. And i don't know why.
You getting confused by a strong premise implying a weak conclusion.
1 and 2 together say precisely: $$p\in (-2.2,2.3)$$
And you are asking to conclude $$p\in [-2.4,2.4]$$.
That's simply a matter of noting if $$(-2.2, 2.3) \subset [-2.4,2.4]$$. And if $$p$$ is in a smaller subset, then we can conclude $$p$$ must therefore also be in the bigger superset.
Or in other words:
$$-2.2 < p < 2.3\implies$$
$$-2.4\le -2.2 < p <2.3 \le 2.4\implies$$
$$-2.4\le p \le 2.4$$
....
An analogy:
We can conclude $$p$$ is a somewhat green article of clothing if we know both
1: $$p$$ is a hat.
2: $$p$$ is precisely the color an emerald takes when viewed at noon on the summer solstice on the equator.
In your question, 1 tells you that $$p <2.3$$ so you can conclude $$p\le 2.4$$. ($$p$$ is a hat so you can conclude $$p$$ is an article of clothing.)
And 2 tells you that $$-2.2 so you can conclude $$-2.4\le p$$. ($$p$$ is the color of emeralds so you can conclude $$p$$ is green.)
Together 1 and 2 tell you $$-2.2 so you can conclude $$-2.4\le p\le 2.4$$. (Together you know $$p$$ is a hat the color of emeralds in a certain condition so you can conclude $$p$$ is a green article of clothing.)
Strong premise. Weak result.
• I had thought about it. For example, if you AFFIRM that k <10, then I can affirm that: k <15. However, my problem is that, I think, I CAN NOT affirm that it is "lesser or EQUAL" than 15, since it can take the value 15 and that contradicts the initial statement of k <10. – Eduardo Sebastian Jun 1 '19 at 20:44
• However, I think I understand the point. A number k can be between: 10 < k <100, and at the same time, I with that i can affirm that: -10000 <k <100000, regardless of whether there are intermediate values, this is still true, this is what I understood. – Eduardo Sebastian Jun 1 '19 at 20:52
• The key word is "less than OR equal" is the "OR". If $5 < 16$ then $5 \le 16$. $5 \le 16$ does not mean that $5 =16$. It doesn't even mean $5$ might be less than $16$. It means that exactly one of the two statements: A) $5 < 16$ and B) $5=16$ is true. And one of them IS true. A) is true. So A OR B is true. – fleablood Jun 3 '19 at 14:47
• We are asked to verify "Fred is green or red". We sneak up on Fred and verify that he is a solid green. SO we say "Yep, Fred is green so the statement is true: Fred is green or red". But then one of us says. "But he isn't red. So he can't be green OR RED because he isn't red". Well... the first person is right. Fred is completely green. That is a subset of being green OR red. So GREEN $\implies$ GREEN or RED. – fleablood Jun 3 '19 at 14:51
• I understand the point. Where i can get info about " strong premise implying a weak conclusion." – Eduardo Sebastian Jun 4 '19 at 19:23
$$-2.4 \leq p \leq 2.4$$
$$-2.7 \leq p < 2.3$$
$$-2.2 < p \leq 2.6$$
$$p \in (-2.2, 2.3)$$
The point is that if you use $$1$$ and $$2$$ together you know $$-2.2 \lt p \lt 2.3$$. This does allow you to ensure $$|p| \le 2.4$$ because all numbers in the interval are less than $$2.4$$ in absolute value. The fact that there are numbers outside the interval that also qualify does not matter at all. If I told you that $$0.1 \le p \le 0.9$$ you could assure me that $$|p| \le 2.4$$. You could also make a stronger statement, like $$|p| \le 1$$ or $$|p| \le 0.9$$, but that doesn't mean the one with $$2.4$$ is incorrect. | 2021-05-14T04:04:31 | {
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https://math.stackexchange.com/questions/1049608/lattice-generated-by-vectors-orthogonal-to-an-integer-vector/1430969 | # Lattice generated by vectors orthogonal to an integer vector
Given a non-zero vector $\boldsymbol{v}$ composed of integers, imagine the set of all non-zero integer vectors $\boldsymbol{u}$, such that $\boldsymbol{u} \cdot \boldsymbol{v} = 0$, i.e., the integer vectors orthogonal the original vector. The set $S = \{\boldsymbol{u} : \boldsymbol{u} \cdot \boldsymbol{v} = 0\}$ seems to form a $dim(\boldsymbol{v})-1$ dimensional lattice. Specifically, it's clear that for any two elements of $S$, their linear combination is also in $S$. However, because S is a subset of the lattice of all integer vectors, it's also a lattice. It there a name for this lattice? Additionally, how can it's basis vectors be computed?
• This is a good question, I ran into it yesterday and found your post looking for the answer online. I'm surprised there has been no answer yet, so I'm working on it and will post solution once I find it. Sep 11 '15 at 8:49
I don't know if this lattice has a name, however it is easy to compute a basis given $$\mathbf{v}$$ by recurrence:
Suppose the gcd of all coordinates of $$\mathbf{v}$$ is 1 or else divide by it.
Let $$\mathbf{v}=(v_1,...,v_n)$$.
If $$n=2$$, then your basis is just $$(v_2,-v_1)$$
If $$n > 2$$:
Set the first vector of your basis to be $$\mathbf{b}_1=(-v_n*a_1,-v_n*a_2,...,-v_n*a_{n-1},gcd(v_1,...v_{n-1}))$$, where the $$a_i$$ come from Bézout's Identity and can be easily calculated using the extended Euclidean algorithm.
Then complete your basis using the same algorithm with $$\mathbf{v}=(v_1,...v_{n-1})$$.
This yields a basis of the lattice orthogonal to $$\mathbf{v}$$ and not a sublattice.
Indeed, all integer vector $$\mathbf{u}$$ orthogonal to $$\mathbf{v}$$ have last coordinate which is a multiple of $$gcd(v_1,...,v_{n-1})$$.
$$\sum_{i=1..n} u_i v_i = 0$$ implies $$u_n v_n = -\sum_{i=1..n-1}u_i v_i$$ so $$gcd(v_1,...v_{n-1})$$ divides $$u_n v_n$$ and is coprime with $$v_n$$.
Example
Suppose we want the lattice orthogonal to $$v = (125, -75, 45, -27)$$.
We first look at the orthogonal of $$(125,-75)$$, which is the same as the lattice orthogonal to $$(5,-3)$$ if we simplify by $$gcd(125,-75)=25$$.
So our first vector is going to be $$(-3,-5,0,0)$$.
Now we look at the orthogonal of $$(125,-75,45)$$, which is the same as the orthogonal of $$(25,-15,9)$$, if we simplify by $$gcd(125,-75,45)=5$$.
Bézout's Identity gives us $$2*25+3*(-15)=5$$, so our new vector is $$(-9*2,-9*3,5,0)=(-18,-27,5,0)$$.
Finally, we arrive to our last step, Bézout's Identity gives us $$1*125+1*(-75)+(-1)*45=5$$, so our last vector is $$(27*1,27*1,27*(-1),5)=(27,27,-27,5)$$.
So our basis is formed by the three vectors : $$\big((-3,-5,0,0),(-18,-27,5,0),(27,27,-27,5)\big)$$
• Hi, unless I am mistaking, this algorithm does not produce a full lattice of vectors orthogonal to v. As an example, take v = (125, -75, 45, -27). Then this procedure yields three basis vectors (27, 27, -27, 5), (45, 90, 25, 0), (3, 5, 0, 0). However, the lattice spanned by these vectors is strictly contained inside the correct lattice spanned by (3, 5, 0, 0), (0, 3, 5, 0), (0, 0, 3, 5). Oct 26 '18 at 17:54
• It works if you don't forget to include the step: Suppose the gcd of all coordinates of v is 1 or else divide by it. Even during the induction step Nov 12 '18 at 18:08
• thanks, you are absolutely right! Nov 17 '18 at 20:45
• I don't understand this algorithm. Can you give an explicit example, like for Anton's? Feb 27 '20 at 3:38
• I edited my answer to illustrate the algorithm working on Anton's example. Hope this helps. Mar 1 '20 at 19:15
I know that is an old question and already have a beautiful answer given by Florian Bourse, but I am adding the following just for future references.
That lattice has a name: it is called orthogonal lattice. In general, we define it regarding a set of vectors $$a_1, ..., a_d \in \mathbb{Z}^n$$ as follows:
$$L^\bot := \left\{ u\in \mathbb{Z}^n : \langle a_i, u \rangle = 0 \text{ for } 1 \le i \le d \right\}.$$
Or equivalently we define the lattice $$L$$ generated by the integer linear combinations of $$a_i$$'s and $$L^\bot$$ is then the lattice whose vectors are orthogonal to all the vectors of $$L$$.
Your case is just a particular one where $$d=1$$. And you are right about the dimension, it is $$n - d$$.
There are a lot of other known properties of these lattices (for instance, $$\det(L^\bot) \le \det(L)$$ and $$(L^\bot)^\bot = span(L)\cap\mathbb{Z}^n$$).
Section 2 of Merkle-Hellman Revisited: A Cryptanalysis of the Qu-Vanstone by Phong Nguyen and Jacques Stern present some discussion about this type of lattice and a polynomial-time algorithm to compute a LLL-reduced basis to them, which I've implemented in SAGE and published in this Github repository.
• Very nice! Thanks for the reference. Jan 4 '19 at 18:02 | 2022-01-21T04:56:42 | {
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http://mathhelpforum.com/advanced-algebra/53270-analysis-proof-help-needed.html | # Thread: Analysis proof help needed
1. ## Analysis proof help needed
I've been driving myself crazy with the following proof:
Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.
If I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!
2. Originally Posted by stuckinanalysis
I've been driving myself crazy with the following proof:
Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.
If I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!
suppose E is uncountable but G is countable. we have $E=G \cup F.$ clearly G can't be finite. so to get a contradiction, you need to show that the union of an infinite countable set with a finite set is
infinite countable. to show this let $G=\{x_n: \ n \in \mathbb{N} \}, \ F=\{y_1, \cdots, y_m\}.$ define the map $f: G \cup F \longrightarrow \mathbb{N},$ by: $f(y_j)=j, \ f(x_n)=n+m.$ obviously $f$ is bijective and hence $G \cup F$ is countable.
3. You proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.
4. Originally Posted by stuckinanalysis
You proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.
you're right! i didn't see "equivalent"! ok, the problem is now much more interesting! so we have an infinite set $G$ and a finite set $F.$ we want to show that $G$ and $G \cup F$ are equivalent.
since we're dealing with cardinality here, we may assume that $G \cap F = \emptyset.$ now apply Zorn's lemma to find a set $G^*=\bigcup_{i \in I} G_i,$ with $G_i \subseteq G,$ which is maximal with respect to these properties:
1) every $G_i$ is countable,
2) $G_i \cap G_j = \emptyset$ for all $i, j \in I$ with $i \neq j.$
then maximality of $G^*$ implies that $G-G^*=H$ is finite. choose an $i_0 \in I$ and replace $G_{i_0}$ with $G_{i_0} \cup H.$ then: $G = \bigcup_{i \in I} G_i,$ and $G \cup F=(G_{i_0} \cup F) \cup \bigcup_{i \neq i_0}G_i.$ also it's clear that $G_i$ will still satisfy
the properties 1) and 2). thus, since $G_{i_0} \cup F$ and all $G_i$ are countable, they're equaivalent to $\mathbb{N},$ hence both $G$ and $G \cup F$ are equivalent to $I \times \mathbb{N}, \ \color{red}(*)$ and hence they must be equivalent. Q.E.D.
$\color{red}(*)$ the reason is that if $f_i: G_i \longrightarrow \mathbb{N}, \ i \in I,$ is a bijection, then you can define $f: \bigcup_{i \in I}G_i \longrightarrow I \times \mathbb{N}$ by: $f(g_i)=(i,f_i(g_i)).$ since $G_i$ are disjoint, $f$ is well-defined. bijectivity of $f$ is a trivial result
of bijectivity of each $f_i.$
5. Originally Posted by NonCommAlg
... now apply Zorn's lemma
But is that cheating?
There is a known result in ZFC which says if $X$ is an infinite set and $Y$ a subset with $|Y| < |X|$ then $|X - Y| = |X|$. This can be easily proven using axiom of choice and cardinality. However, it still relies on axiom of choice. When I saw this problem I wondered whether it is possible to solve it without any choice.
And if you are using choice then why use Zorn's lemma, simply put a well-ordering on $G$ and start mapping the least elements to the least elements. I think that might work and be a lot simpler.
6. Thanks. That gives me a lot to think about.
Here is a follow up/ similar question. Can it be done the same way?
Let X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X U Y?
7. Originally Posted by stuckinanalysis
Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.
I don't think you need the axiom of choice for this. Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence. Then define a bijection f from E to E\F by taking $f(x_n)=f(x_{N+n})$, and taking f to be the identity on $E\setminus\{x_n:n\in\mathbb{N}\}$.
8. Originally Posted by stuckinanalysis
Here is a follow up/ similar question. Can it be done the same way?
Let X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X \ Y?
A similar method should work. Let Z be a countable set in X \ Y. Then Z and Z∪Y have the same cardinality, so let f be a bijection from Z to Z∪Y, and extend f to X \ Y by taking it to be the identity on X \ (Z∪Y).
9. Originally Posted by Opalg
Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.
How?
We not supposed to be chosing.
I only see one way of doing this. Let $g$ be a choice function on $\mathcal{P}(E)$ i.e. $g(S) \in S$ for all $S\not = \emptyset$. Since $F$ is finite there is a bijection $f:n\to F$. Define a sequence, by recursion, $x: \mathbb{N} \to E$ by $x_k = f(k)$ for $k and $x_k = g( E - \{ x_i | i < n\})$. Then follow through with your argument.
10. Originally Posted by ThePerfectHacker
Originally Posted by Opalg
Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.
How?
We not supposed to be choosing.
Just because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.
11. Originally Posted by Opalg
Just because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.
Are you sure? What you are doing is this:
$x_{n+1} = \text{ some element of }(E - \{ x_i | i < n\})$
The thing is that this function is not really defined.
That is why we need choice to define such a function.
---
If that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:
As a consequence of which the axiom of choice would be a provable from the other axioms.
12. Originally Posted by ThePerfectHacker
Are you sure? What you are doing is this:
$x_{n+1} = \text{ some element of }(E - \{ x_i | i < n\})$
The thing is that this function is not really defined.
That is why we need choice to define such a function.
No, I'm just using the definition of "nonempty". If a set is nonempty then there exists an element in that set. I'm calling that element $x_{n+1}$.
Almost every proof in analysis starts by saying "Let ε>0." That involves choosing an element from the set of positive real numbers. But it is not true that every such proof requires the axiom of choice.
Originally Posted by ThePerfectHacker
If that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:
As a consequence of which the axiom of choice would be a provable from the other axioms.
No, you could never get beyond a countable set that way, because you are only constructing one element at at time. | 2013-05-23T04:56:03 | {
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http://math.stackexchange.com/questions/105644/can-the-relation-of-ap-equiv-a-pmod-2p-if-p-prime-and-p2-be-added-to | # Can the relation of $a^{p} \equiv a\pmod {2p}$ if (p prime and p>2) be added to Fermat's little theorem?
Fermat's little theorem
$a^{p} \equiv a\pmod p$ if p prime
I have noticed that the formula $a^{p} \equiv a\pmod {2p}$ (p prime and p>2) can be also written by using Fermat's little theorem
Proof: Let $a$ be any integer and $p>2$ be some prime number.
$a^{p} \equiv a\pmod p$
$a^{p}-a \equiv 0\pmod p$
$a(a^{p-1}-1) \equiv 0\pmod p$
$a(a^{p-1}-1) \equiv 0\pmod p$
$F(a)=(a^{p-1}-1)$ can be divided to $(a-1)$ because $F(1)=(1^{p-1}-1)=0$
$F(a)=(a^{p-1}-1)=(a-1).R(a)$ //R(a) is polinom that has degree p-2 and coefficients are Integer numbers.
$a (a-1) R(a) \equiv 0\pmod p$
$a (a-1)$ is always an even number so $a (a-1) R(a)$ will be always even number and also can be divided to $p$ because of Fermat's little theorem . Thus $a^{p} \equiv a\pmod {2p}$ if (p prime and p>2) .
I searched internet but I have not seen that relation in the internet.
Is it known formula?Please help if it is known relation(I would like to learn what the subject is )
Sorry if someone else asked the same question in here.
Thank you for answers and helps.
-
If $a$ is any integer, and $n$ a positive integer, then $a^n \equiv a \mod 2$. Your equation is a combination of this fact and Fermat's little theorem. Both are known congruences, but I'm note sure there's a name for their combination. – Joel Cohen Feb 4 '12 at 14:49
As Joel rightly observed :If $a$ is any integer, and $n$ a positive integer, then $a^n\equiv a \pmod2$
So : $a^p\equiv a \pmod2$ , and $a^p\equiv a \pmod p$
According to Chinese Remainder Theorem :
$a^p \equiv a \pmod { \operatorname{lcm} (2,p)}$
,and since $p$ is an odd prime it follows that : $\operatorname{lcm}(2,p) = 2p$ , therefore :
$a^p \equiv a \pmod {2p}$
-
The answer, Maybe, can be perfect and if we add some additional steps to show how proof of Chinese Remainder Theorem. c,d,e are integers if $a^p\equiv a \pmod2$ then $a^p=a+2.c$ $a^p-a=2.c$ if $a^p\equiv a \pmod p$ then $a^p=a+p.d$ $a^p-a=p.d=2.c$ $d=2.e$ can be written because $2.c$ is even number Thus $a^p-a=p.d=2.c=2.p.e$ $a^p-a\equiv 0 \pmod{2p}$ $a^p\equiv a \pmod{2p}$ Thanks in advice – Mathlover Feb 4 '12 at 19:32
You are correct I think, but the way you write it down makes it look more difficult than it is. Also you should be more specific on whether you are talking about one particular $a$ or "$\forall a$".
So let me show how I would write this down. First, let $a$ be any integer and $p>2$ be some prime number. Then
$$a^p \equiv a \pmod{2p} \iff 2p \mid a^p - a \iff 2\mid a^{p}-a \text{ and } p\mid a^p-a$$ The last equivalence holds because $2$ and $p$ are two different primes.
But $2\mid a^p-a$ for all $a$ and $p$: if $a$ is even, then clearly $a^p$ will be even as well; if $a$ is odd, then so is $a^p$ and hence $a^p-a$ will be even again. So this is indeed equivalent to
$$\dots \iff p\mid a^p-a \iff a^p \equiv a \pmod p.$$
-
Thanks for answer. – Mathlover Feb 4 '12 at 14:46
No actually it's a continuation of the chain of equivalences up above. So it should be read as $$a^p \equiv a \pmod{2p} \iff \dots \iff a^p\equiv a \pmod{p}$$ – Myself Feb 4 '12 at 14:51
This is the special case $\rm\ m = 2\:p,\ p\:$ odd, of the following generalization of Fermat's little theorem, quoted from Bill Dubuque's post
THEOREM $\$ For naturals $\rm\: k,m>1$
$\qquad\rm m\ |\ a^k-a\$ for all $\rm\:a\in\mathbb N\ \iff\ m\:$ is squarefree and prime $\rm\: p\:|\:m\: \Rightarrow\: p-1\ |\ k-1$ | 2013-12-21T16:30:14 | {
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https://math.stackexchange.com/questions/738574/computing-int-0-pi-sinx-dx-using-the-definition | # Computing $\int_0^\pi \sin(x) \; dx$ using the definition.
A colleague of mine and I, in the course of teaching integral calculus for the umpteenth time, were wondering if we could expand the class of examples that our students are exposed to when computing Riemann integrals from the definition. Most such examples involve polynomial functions, and they work nicely because we have some well-known formulas, like
$\displaystyle\sum_{i=1}^n 1 = n$,
$\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$,
and $\displaystyle\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$. (And certainly, there are others.)
But we wanted to expand the class of examples beyond polynomial functions, and started to think about trig functions. We came up with a computation for $\int_0^\pi \sin(x)\;dx$ using the definition of the Reimann integral, which I provide below.
Question: Is the computation below in the literature somewhere? If so, where?
Computation:
Using a Riemann sum with right endpoints, we have
$$\int_{0}^\pi \sin(x)\; dx = \lim_{n \to \infty}\sum_{i=1}^n \sin(x_i) \Delta x$$ where $\Delta x = \frac{\pi}{n}$ and $x_i = i\Delta x$.
The trick is to rewrite the Riemann sum $\sum_{i=1}^n \sin(x_i) \Delta x$ as a telescoping sum using the difference of cosines identity: $$\cos(b) - \cos(a) = 2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$
Setting $a = \frac{2i+1}{2}\Delta x$ and $b = \frac{2i-1}{2}\Delta x$ yields
$$\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right) = 2 \sin\left(i \Delta x\right) \sin\left(\frac{\Delta x}{2}\right)$$
Solving for $\sin(i \Delta x)$ gives
$$\sin(i \Delta x) = \frac{\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)}{2 \sin\left(\frac{\Delta x}{2}\right)}$$
The Riemann sum now is $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)} \sum_{i=1}^n \left[\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)\right]$$
The latter sum is telescoping, and so $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]$$
Now recalling that $\displaystyle \Delta x = \frac{\pi}{n}$ and taking $n \to \infty$ we have
\begin{align} \int_0^\pi \sin(x)\;dx &= \lim_{n \to \infty} \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]\\ & = 1 \cdot \left[ \cos(0)-\cos(\pi)\right]\\ & = 2 \end{align}
One nice thing about this computation is that you can replace $\pi$ with an arbitrary upper limit of integration $c$ and compute that $\int_0^c \sin(x) \;dx = 1 - \cos(c)$. From there, one can easily compute $$\int_a^b \sin(x) \; dx = \int_0^b \sin(x) \; dx - \int_0^a \sin(x) \; dx = -\cos(b) + \cos(a)$$ (as predicted by FTC).
Question: Is the computation above in the literature somewhere? If so, where?
• I don't know, but our teacher made us do this. $\int_a^b \sin(x)\,dx$ from reimann sum I mean. – Guy Apr 3 '14 at 18:43
• This answer contains telescoping ideas. – Ian Mateus Apr 3 '14 at 18:43
• I haven't seen it in a textbook or anywhere else, but I did it once myself in order to see how the limit $\sin x/x \rightarrow 1$ would play into the calculation. – Jason Zimba Apr 3 '14 at 18:49
• Not exactly "literature", but the sum $\sum \sin(n)$ is telescoped using a similar trick in this answer in order to apply the Dirichlet test to the series $\sum \frac{\sin(n)}{n}$. – Mike F Apr 4 '14 at 16:34
Below are some references I found in my "math library" at home this morning. Before posting just now, I quickly searched to see which were online and I've included links for those I found. The Boyer paper URL is a JSTOR item I don't have access to, but I've included the URL anyway because many here are probably at colleges/universities that have JSTOR access.
At the risk of pointing out the obvious, this method doesn't prove the Riemann integrability of ${\sin x},$ since equal length partitions are used along with endpoint evaluations. For instance, under this restriction the limit of the Riemann sums of the characteristic function of the rationals on a given compact interval exists. What this calculation does is give the value of the (Reimann) integral on a given compact interval under the assumption that the integral exists.
Direct quotes from original sources are indicated by italics. Within such a quote: (1) italics in the original is indicated by non-italics here; (2) brief additions and/or words of further explanation by me are indicated by non-italics between square braces "[stuff by me]"; (3) omissions in the original are indicated using "$[\dots]$".
[1] [Author not known], Sur la sinussoide [On the sinusoid], Question D'Examen, Nouvelles Annales de Mathématiques (1) 7 (1848), 436-437.
The (net) area above the $x$-axis and below the graph of $y = \sin{x},$ between $x = x_1$ and $x = x_2$ is found. The interval $[x_1, \, x_2]$ is divided into $n+1$ many abutting intervals each of length $h,$ so that $x_2 - x_1 = (n+1)h,$ and the identity $$h [\sin{x_1} + \sin{(x_1 + h)} + \sin{(x_2 + 2h)} + \cdots + \sin{(x_1 + nh)} \;\; = \;\; \frac{h \sin \left(x_1 + \frac{nh}{2}\right) \sin \frac{h}{2}(n+1)}{\sin{\frac{h}{2}}}$$ along with $\frac{nh}{2} = \frac{x_2 - x_1 - h}{2}$ is used to obtain the value $2\sin{\left(\frac{x_2 + x_1}{2}\right)} \sin{\left(\frac{x_2 - x_1}{2}\right)},$ which equals $\cos{x_1} - \cos{x_2}.$ Note: There is a typo at one point in the original, in which $\sin{(x_2 + h)}$ appears instead of $\sin{(x_2 + 2h)}.$ Also, the text only observes that the area's expression in terms of the sine function reduces to $1 - \cos{x_2}$ when $x_1 = 0.$
[2] William Elwood Byerly, Elements of the Integral Calculus, 2nd edition, Ginn and Company, 1892, xvi + 339 + 11 + 32 pages.
See Chapter VIII, Article 81, p. 76, Examples (2): By the aid of the trigonometrical formulas $$\cos{\theta} + \cos{2\theta} + \cos{3\theta} + \cdots + \cos{(n-1)\theta} = \frac{1}{2}\left[\sin{n\theta} \cot{\frac{\theta}{2}} - 1 - \cos{n\theta}\right],$$ [and] $$\sin{\theta} + \sin{2\theta} + \sin{3\theta} + \cdots + \sin{(n-1)\theta} = \frac{1}{2}\left[(1 - \cos{n\theta}) \cot{\frac{\theta}{2}} - \sin{n\theta}\right],$$ prove that $\;\;\int_{a}^{b}\cos{x}.dx = \sin{b} - \sin{a},\;$ and $\;\int_{a}^{b}\sin{x}.dx = \cos{a} - \cos{b}.$
[3] Carl Benjamin Boyer, History of the derivative and integral of the sine, Mathematics Teacher 40 #6 (October 1947), 267-275.
[4] Mark Bridger, A note on areas under a sine curve, The Pentagon 18 #1 (Fall 1958), 24-26.
At the time Mark Bridger was a student at the Bronx High School of Science (New York).
[5] Richard Courant and Fritz John, Introduction to Calculus and Analysis, Volume I, John Wiley and Sons (Interscience Publishers), 1965, xxiv + 661 pages.
See Chapter 2.2.e (Integration of $\sin x$ and $\cos x$), p. 135.
[6] Godfrey Harold Hardy, A Course of Pure Mathematics, 10th edition, Cambridge University Press, 1952, xii + 509 pages.
See Chapter VII, Article 164, p. 320, item 3. Calculate $\int_{a}^{b}x^2 \, dx,$ $\int_{a}^{b} \cos{mx} \, dx$ and $\int_{a}^{b} \sin{mx} \, dx$ by the method of Ex. 1. Note: Hardy's reference to "the method of Ex. 1" is simply the method of partitioning the interval $[a,b]$ into finitely many equal length subintervals, forming the relevant Riemann sum, and taking a limit. The needed trigonometric identities are not given as a hint, but I do notice that a closely related identity appears in Exercise 4 at the top of p. 323 for another purpose.
[7] Kenneth Sielke Miller and John Breffni Walsh, Elementary and Advanced Trigonometry, Harper and Brothers, 1962, xii + 350 pages.
See Chapter 10.3 (Solution of the Area Problem), pp. 215-217: (begins) We now turn to the second problem mentioned at the beginning of the chapter, namely, that of finding the area under one arch of a sine curve.
[8] Isidor Pavlovich Natanson, Summation of Infinitely Small Quantities, Topics in Mathematics, D. C. Heath and Company, 1963, viii + 59 pages.
See Chapter 6 (The Sinusoid), Articles 26-35, pp. 46-57. Specifically, see Article 29 (A trigonometric sum), Article 30 (A subsidiary inequality), Article 31 (The sine of an infinitely small angle), and Article 32 (The quadrature of the sinusoid) on pp. 46-51. The remaining articles in this chapter involve applications, mainly in calculating the volume of a rotating sinusoid and calculating effective (electrical) current.
[9] John Meigs Hubbell Olmsted, Intermediate Analysis. An Introduction to the Theory of Functions of One Real Variable, The Appleton-Century Mathematics Series, Appleton-Century-Crofts, 1956, xiv + 306 pages.
See Chapter 5, Article 503, p. 145, Exercise #19 & 20. The reader is asked to show that $\int_{a}^{b}\sin x \, dx = \cos{a} - \cos{b}$ and $\int_{a}^{b}\cos x \, dx = \sin{b} - \sin{a}$ by using equal length partitions of $[a,b].$ The relevant trig. identities are indirectly provided (by means of a hint, the reader is led to obtain these identities).
[10] Michael David Spivak, Calculus, 3rd edition, Publish or Perish, 1994, xiv + 670 pages.
See Chapter 15, p. 320, Problem 33.
[11] Isaac Todhunter, A Treatise on the Integral Calculus, 6th edition, MacMillan and Company, 1880, viii + 408 pages.
See Chapter IV, Article 38, pp. 52-53: (begins) Suppose we wish to find the integral of $\sin x$ between limits $a$ and $b$ immediately from the definition. By Art. 4 we have to find the limit $[\ldots]$ Note: The copy I found online is dated 1863, but what I've quoted seems to be the same calculation and it's in the same location in the book.
[12] Edwin Bidwell Wilson, Advanced Calculus, Ginn and Company, 1911, x + 566 pages.
See Chapter I, p. 30, Exercise 9. With the aid of the trigonometric formulas $[\ldots]$ show $(\alpha) \; \int_{a}^{b} \cos{x}\,dx = \sin{b} - \sin{a},\;\;$ [and] $\;\;(\beta) \; \int_{a}^{b} \sin{x}\,dx = \cos{a} - \cos{b}.$
(ADDED 3 DAYS LATER) I thought it would be useful to archive in my answer some references for evaluating definite integrals of other functions by Reimann sums. However, because I want to get this finished during my lunch hour and I don't want to substantially increase the length of this already long answer, I will use briefer citations.
Harding/Barnett, Solution to Calculus Problem #369, Amer. Math. Monthly 22 #6 (June 1915), 208-210.
Two solutions for the evaluation of $\int_{a}^{b} \ln{x}\,dx$. The second solution makes use of Fermat's method of geometric subdivision of the interval $[a,b].$
Barnett, Solution to Calculus Problem #366, Amer. Math. Monthly 22 #5 (May 1915), 168-169.
One solution for the evaluation of $\int_{a}^{b}{\sin}^{-1}x\,dx.$
Eli Maor, On the direct integration $\int_{a}^{b}x^{n}\,dx,$ Int. J. Math. Educ. Sci. Technol. 5 (1974), 199-200.
The method involves working with the sum of the left endpoint sums and the right endpoint sums. The method might be similar to what Otto Dunkel does in Note on the quadrature of the parabola, Amer. Math. Monthly 27 #3 (March 1920), 116-117.
Finally, the following are similar to the kind of things found in Natanson's Summation of Infinitely Small Quantities:
W. R. Longley, Some limit proofs in solid geometry, Amer. Math. Monthly 31 #4 (April 1924), 196-202.
Joseph B. Reynolds, Some applications of algebra to theorems in solid geometry, Mathematics Teacher 18 #1 (January 1925), 1-9.
Jos. B. Reynolds, Finding plane areas by algebra, Mathematics Teacher 21 #4 (April 1928), 197-203.
• This is a great list of references! Thanks! – wckronholm Apr 4 '14 at 14:39
I suspect Tom Apostol's calculus textbook has this (i.e., finds $\int_0^\pi\sin x\,dx$ by using limits of Riemann sums without the fundamental theorem).
I question the place of Riemann sums in the curriculum. Rigorous definitions are unsuitable for a calculus-for-liberal-education course unless the students are unusual. Such a course should acquaint students with the reason why calculus is important in the course of human events over recent and coming centuries, and with the fact that it overcame difficulties like how to define $\text{rate} = \dfrac{\text{distance}}{\text{time}}$ when the distance and the time are both $0$, and how to to find $$\sum\text{force}\times\text{distance}$$ when there are infinitely many infinitely small distances, each with its own value of "force". The conventional calculus course is a watered-down version of a course for students who come in with a prior desire or a pre-identified need to understand calculus, rather than for students who are there in order to pay a price in homework for a grade to impress employers, and whom one should be trying to seduce into another course of action.
If you expect students to marvel at the fact that it's possible to find this integral by using limits of Riemann sums, I expect the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and if you have them plod through finding the integral by antidifferentiating and substituting endpoint values, then the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and they won't know the difference. (I'm hoping for a malpractice suit against every university that knowingly encourages unqualified students to take calculus, to the point where those are 99.9% of the ones who show up. They bring in tuition money.)
Another occasion for Riemann sums is numerical integration, but that doesn't seem to be your purpose.
• Apostol's book gives the formulas for $\int_a^b \sin(x)\;dx$ and $\int_a^b \cos(x)\;dx$ followed by the statement "we shall not prove these formulas at this stage because they will be derived by an easier method in Chapter 2, with the help of differential calculus." – wckronholm Apr 3 '14 at 18:55
• @wckronholm : OK. Apostol's book may still be a place to look if anyone wants similar material. Also, maybe Otto Toeplitz' "genetic approach" calculus book. – Michael Hardy Apr 3 '14 at 18:58
• $99.9\%$ may be unduly pessimistic. Perhaps I will be accused of being a rosy-eyed optimist, but $95\%$ seems more reasonable. – André Nicolas Apr 3 '14 at 19:15
• @AndréNicolas : Depends on which institution. There are colleges that think of themselves as selective, at which the students are hard-working and intelligent, but have attitudes about math courses: They exist in order to provide an opportunity to show that they can follow instruction and work hard and get a grade, and they've always gotten "A"s in math, and they demand their right to get a course consisting of algorithms for them to follow to get an "A". – Michael Hardy Apr 3 '14 at 19:36
• @MichaelHardy I actually had my class run through this computation in small groups in class today. While it's true that many of them were longing for the Fundamental Theorem of Calculus they "learned" in high school and could give them the answer in about 5 seconds, there were more than a few who really were appreciating all of the pieces that went into the computation. In general, I think students benefit from exposure to this sort of activity because of the potential to broaden their reasoning skills. – wckronholm Apr 3 '14 at 20:15
Alternatively, you can use $e^{i \theta} = \cos(\theta) + i \sin(\theta)$ and the partial sum formula for Geometric series. Using left endpoint approximation instead, the sum we want to evaluate is $$\sum_{k=0}^{n-1} \sin(k \, \Delta x)\,\Delta x$$ where $\Delta x = \frac{\pi}{n}$. Note that $\sum_{k=0}^{n-1} \sin(k \,\Delta x)$ is the imaginary part of $\sum_{k=0}^{n-1} e^{i k \,\Delta x}$, that is $\sum_{k=0}^{n-1} z^k$ where $z = e^{i \,\Delta x}$. This last sum has the closed form $\frac{1-z^n}{1-z} = \frac{2}{1-z}$ using $z^n = e^{i \pi} = -1$. We get $$\sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x =\mathrm{Im}\left( \frac{2}{1-e^{i\, \Delta x}} \right) \,\Delta x = \mathrm{Im}\left( \frac{2\,\Delta x}{1-e^{i \,\Delta x}} \right).$$ Finally, letting $n \to \infty$ so that $\Delta x \to 0$ we get $$\lim_{\Delta x \to 0} \frac{2\,\Delta x}{1-e^{i \,\Delta x}} = -2\left( \lim_{\Delta x \to 0 } \frac{e^{i \,\Delta x} - 1}{\Delta x} \right)^{-1} = -2\left( \frac{d}{dt} e^{it} \big|_{t=0} \right)^{-1} = -2i^{-1} = 2i$$ so that $$\lim_{n \to \infty} \sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x = \operatorname{Im}(2i) = 2.$$
• Thanks. We were aware of this method already, and wanted to find an approach which did not rely on complex numbers. – wckronholm Apr 4 '14 at 16:05
• @wckronholm: That makes sense, just thought I'd add it. – Mike F Apr 4 '14 at 16:28
• @wckronholm: It has probably occurred to you already, but if you don't want to use complex numbers, you can still do integrals like $\int_0^1 e^x \ dx$ using geometric series. – Mike F Apr 4 '14 at 16:29
• Thanks, that has indeed occurred to me, and I will be showing this to my students in about 20 minutes! – wckronholm Apr 4 '14 at 16:39 | 2019-05-20T12:41:21 | {
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https://math.stackexchange.com/questions/1505605/binomial-formula-modulo-a-prime/1505851 | # Binomial formula modulo a prime.
Let $p$ be a prime. I see that modulo $p$, the binomial formula reduces to $$(x+y)^p\equiv x^p+y^p \pmod p$$ because $p \choose k$ is a multiple of $k$ whenever $k=1..p-1$, but don't we have by Fermat's little theorem that $a^p\equiv a \pmod p$ for any integer $a$ and so this is true for $a=x+y$ hence the binomial formula reduces to $(x+y)^p\equiv x+y \pmod p$. Thanks for your help!
• Yes, but you also get $(x+y)^p\equiv x+y \mod p$ straight from FLT without binomial theorem at all. – user281392 Oct 30 '15 at 21:24
• Both are true. $(x+y)^p$ is congruent to $x^p + y^p$ modulo $p$. Now by Fermat's little theorem, $x^p$ is congruent to $x$ and $y^p$ is congruent to $y$ modulo $p$, and so $x^p + y^p$ is congruent to $x + y$ modulo $p$. – Dylan Oct 30 '15 at 21:24
• @user281392. Very nice observation. +1 – Shailesh Oct 31 '15 at 2:56
Yes, everything that you have written is true. It is the case that $$(x+y)^p \equiv x^p+y^p \mod p$$ using the Binomial Theorem, and it is also true that $$(x+y)^p \equiv x+y \mod p$$ by Fermat's Little Theorem.
There is not contradiction here because by Fermat's Little Theorem, we have that $$x^p \equiv x \mod p \quad\text{ and }\quad y^p \equiv y \mod p$$ and so $$x^p+y^p\equiv x+y \mod p$$ as we expect.
Your observation that $$(x+y)^p \equiv x^p+y^p \mod p$$ by the Binomial Theorem is sometimes used to prove Fermat's Little Theorem by induction. The (complete) proof is as follows:
First we show, as noted by you, that $p \mid \binom{p}{k}$ for $1 \leq k < p$. We can prove this either by considering the factors of $p$ in the numerator and denominator of $\binom{p}{k}$, or by noticing that $$\binom{p}{k} = \frac{p}{k}\binom{p-1}{k-1}$$ Thus $$k \mid p\binom{p-1}{k-1}$$ and since $k$ and $p$ are relatively prime, we get that $$k \mid \binom{p-1}{k-1}$$ showing that $$\frac{1}{k}\binom{p-1}{k-1}$$ is an integer, and so $$\binom{p}{k} = p \cdot \frac{1}{k}\binom{p-1}{k-1}$$ is an integer divisible by $p$.
As you noted in your question, $$(x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k} = x^p + y^p + \sum_{k=1}^{p-1} \binom{p}{k} x^k y^{p-k}$$ Since every term in the final sum is divisble by $p$, we get that $$(x+y)^p \equiv x^p + y^p \mod p$$ and this holds for all integers $x$ and $y$.
We can now prove by induction that $$x^p \equiv x \mod p$$ for all integers $x$. The base case of $x=0$ is straightforward since $0^p=0$. Now suppose that $x^p \equiv x \mod p$ for some $x$. Then we have that $$(x+1)^p \equiv x^p + 1^p \equiv x+1$$ using our observation above, and the inductive hypothesis.
Thus the claim holds for $x+1$ as well, and hence for all natural numbers by induction. For the negative integers, we can note that if $x$ is a positive integer then $$(-x)^p = (-1)^p x^p \equiv -x \mod p$$ Here, we use that $x^p \equiv x \mod p$, and also that $(-1)^p \equiv -1 \mod p$, since if $p$ is odd then $(-1)^p = -1$, and if $p=2$ then $(-1)^p = 1 \equiv -1 \mod 2$.
Yes, good point. If $x$ and $y$ are integer, we can simply write $(x+y)^p = x + y \mod p$.
However, the formula $(x+y)^p \equiv x^p + y^p \mod p$ is also interesting because it applies not only to elements of $\mathbb{Z}/p\mathbb{Z}$, but also to any commutative algebra over $\mathbb{Z}/p\mathbb{Z}$ (for example, you may apply it to polynomials with coefficients in $\mathbb{Z}/p\mathbb{Z}$, or for field extensions of $\mathbb{Z}/p\mathbb{Z}$).
You have to show that $p \mid \binom{p}{k}$ when $1 < k < p$. We have: $k!(p-k)! \mid p!=p(p-1)!$ Using unique prime factorization we can write: $k!(p-k)! = a_1^{n_1}\cdot a_2^{n_2}\cdots a_m^{n_m}$ with the $a_i$'s are distinct primes smaller than $p$, and none of the factors divide $p$ since $a_i\nmid p$. Thus $a_i^{n_i} \mid (p-1)!$,and since these factors are pairwise coprime, we have $k!(p-k)! \mid (p-1)!$, this shows $p \mid \binom{p}{k}$, and the conclusion follows from this fact. | 2021-07-27T21:58:49 | {
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https://mathematica.stackexchange.com/questions/91407/distribution-over-the-product-of-three-or-n-independent-beta-random-variables/91441 | # Distribution over the product of three, or n, independent Beta random variables
I would like to calculate the PDF for the product of three independent Beta random variables. Specifically, I would like to find the distribution of the product of the following: $X_1\sim Beta(1,3/2)$, $X_2\sim Beta(3/2,1)$ and $X_3\sim Beta(2,1/2)$.
I can manage to get Mathematica to output a distribution for the case of the the product $X_1 X_2$ by doing the following:
PDF[TransformedDistribution[
u v w, {u \[Distributed] BetaDistribution[1, 3/2],
This takes a few minutes to evaluate. However, when I move to $X_1 X_2 X_3$ Mathematica seems to be eternally stuck on the calculation:
PDF[TransformedDistribution[
u v w, {u \[Distributed] BetaDistribution[1, 3/2],
I could approximate this by a large number of draws from the distributions of $X_1$, $X_2$ and $X_3$; multiplying each set together. However, I would like the analytic form.
Does anyone have any idea how I can do this? The reason I state $n$ in the question is because I would like eventually to generalise this calculation to the product of more $Beta$ random variables.
Best,
Ben
• Might have better luck getting an answer on current SOTA for this over at Math.StackExchange.Com, then writing appropriate MMA code. I don't expect you'll find a compact/low computational complexity solution. – ciao Aug 11 '15 at 22:09
• It has been a while since I've done something like this but if I recall correctly you may be able to do this quicker using the moment generation function of the Beta distributions, performing the product on these, and then matching that form to the form of a know distribution's moment generating function. – Edmund Aug 11 '15 at 22:16
• Here's a useful reference - extends earlier work that required integer parameters to (relatively) arbitrary parameters... – ciao Aug 11 '15 at 22:23
• @Edmund good luck with that. – wolfies Aug 12 '15 at 8:06
Here is an answer that does not use a 3rd party package and works for an arbitrary amount of Beta distributions.
You can make use of a closed form for the product of n Beta distributions from the Handbook of Beta Distribution and Its Applications, Products and Linear Combinations, I. Products, B. Exact Distributions as found on page 57.
This expresses a product of n Beta distributions as the product of a constant $K$ (a function of the Beta distributions parameters in $\Gamma$ functions) and a Meijer-G function of the Beta distribution parameters.
ClearAll[pdfProductBeta];
pdfProductBeta[
α_ /; VectorQ[α, NumericQ],
β_ /; VectorQ[β, NumericQ],
x_Symbol] /; Dimensions[α] == Dimensions[β] :=
Module[{k},
k = Times @@ (Gamma[Plus @@ #]/Gamma[#[[1]]] & /@ (Transpose@{α, β}));
k MeijerG[{{}, α + β - 1}, {α - 1, {}}, x]
]
For the three Beta distributions in the question
a = {1, 3/2, 2}; b = {3/2, 1, 1/2};
pdfProductBeta[a, b, x]
(* 27/32 π MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x] *)
This result matches to @wolfies answer above that makes use of the mathStatica 3rd party package.
The two Beta case plots quickly so we can compare this easily to the TransfromedDistribution PDF from the built in Mathematica functions.
tpdf = PDF[
TransformedDistribution[
u v, {u \[Distributed] BetaDistribution[1, 3/2],
a = {1, 3/2}; b = {3/2, 1};
mpdf = pdfProductBeta[a, b, x];
GraphicsRow[
Plot[#, {x, 0, 1}, PlotRangePadding -> None] & /@ {tpdf, mpdf}]
For the three Beta case we need to limit the upper bound in the plot as it takes very long to calculate there.
a = {1, 3/2, 2}; b = {3/2, 1, 1/2};
mpdf3 = pdfProductBeta[a, b, x];
(* evaluate pdf at a point *)
mpdf3 /. x -> 0.5
(* 0.479319 *)
Plot[mpdf3, {x, 0, 0.9}, Exclusions -> {0}, AxesOrigin -> {0, 0},
PlotRangePadding -> None, PlotRange -> Full]
With the pdfProductBeta function you can construct the pdf for the product of an arbitrary number of Beta distributions without the need of a third party package.
Hope this helps.
• Very nice indeed to have a general solution from the Handbook of Beta Distribution -- I would go further and say that your answer not only constructs the pdf of the product of Beta random variables without the need of an add-on package, ... it does so without needing Mathematica either. – wolfies Aug 12 '15 at 14:03
• @wolfies Well ... I wouldn't want to try constructing, evaluating, or <cringe> plotting it without Mma. :-) It quickly gets process heavy for small $n$. – Edmund Aug 12 '15 at 14:09
• Note that this is no longer needed in version 11 and above as TransformedDistribution directly uses the above method for a product of BetaDistributions. – Edmund Apr 8 '17 at 13:18
Let random variable $X_i \sim Beta(a_i,b_i)$, with pdf $f_i(x_i)$. The OP is interested in 3 specific parameter combinations:
The pdf of $Y = X_2 X_3$, say $g(y)$, is:
where I am using the TransformProduct function from the mathStatica package for Mathematica, and where domain[g] = {y,0,1}.
The pdf of $Z = X_1 X_2 X_3 = Y* X_1$, say $h(z)$, is then:
All done.
Quick Monte Carlo check
It is always a good idea to check symbolic solutions with Monte Carlo methods.
The following plot compares:
• the exact symbolic solution pdf obtained $h(z)$ [ red dashed curve ], to
• an empirical Monte Carlo simulation of the pdf [ blue squiggly curve ]
All looks good.
• That Meijer's function had to show up indicates that things are not too simple. Tho, those parameters look familiar… what happens if you apply FunctionExpand[]? – J. M. will be back soon Aug 12 '15 at 9:15
• Applying FunctionExpand returns the same MeijerG function. I might add that it appears tractable for most of the domain ... but trying to plot it for $x$ close to 1 becomes very slow (which is also why that is 'left out' in the diagram above). – wolfies Aug 12 '15 at 11:44
• Ah, that part I can explain; the contour integration being done under the hood converges rather slowly when the argument is near $1$. A pity that there does not seem to be a simpler form (and I was so sure that there was)… – J. M. will be back soon Aug 12 '15 at 11:46
• @J. M. What is the reason the contour integration converges so slowly near 1? I am trying to generate points of the pdf near there, and was wondering if you had any ideas? – ben18785 Aug 12 '15 at 14:19
• @ben, it's an inherent limitation of the contour integral method for the $G$-function. (The situation is similar to the one for generalized hypergeometric functions; arguments near $1$ are often very difficult cases, numerically speaking.) That's why I was hoping there might be expressions in terms of simpler special functions. I do not have Mathematica at hand to investigate further, unfortunately. – J. M. will be back soon Aug 12 '15 at 15:04
The problem can indeed be solved explicitly for the product of n = 3 Beta-distributed variables and the explicit parameters of the OP.
In part 1 I show only the results, and turn later, in part 2, to the details of calculation in Mathematica, part 3 is discussion.
Part 1 Results
The PDF of the Beta distribution is given by
f[x_, a_, b_] = Simplify[PDF[BetaDistribution[a, b], x], 0 < x < 1]
(*
((1 - x)^(-1 + b) x^(-1 + a))/Beta[a, b]
*)
Let the random variables and their rescpective distributions be X1 ~ Beta(1,3/2) , X2 ~ Beta(3/2,1) and X3 ~ Beta(2,1/2), and let
f2n(x) = dist( X1*X2 )
f3n(x) = dist( X1*X2*X3 )
be the requested distributions.
Then we find
f2n[x_] = 9/2 (Sqrt[1 - x] - Sqrt[x] ArcTan[Sqrt[-1 + 1/x]]);
f3n[x_] = 27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -
27/16 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x];
The functions are normalized
Integrate[f2n[x], {x, 0, 1}]
(* 1 *)
Integrate[f3n[x], {x, 0, 1}]
(* 1 *)
A plot of the two functions is shown here
Plot[{f2n[x], f3n[x]}, {x, 0.0001, 0.9}, PlotRange -> {{0, 1}, {0, 4}},
PlotLabel ->
"Distributions of the product of\ntwo (yellow) and three (blue)\nbeta \
distributed random variables", AxesLabel -> {"x", "f(x)"}]
(* 150812_Plot_Prod_Beta_dist.jpg *)
Since with the Meijer function Mathematica requires very long calculation times close to x = 1 I have left this region out.
Observations
1) I believe that the case of general n should be tackled using the Mellin transformation, as is natural for products (as is Fourier for sums). Our result for n = 3, the MeijerG function, already exhibits this pattern.
Part 2: Derivation
The distributions were calculated here using the general formula
fn(x) = Integrate( Prod( du f(u,p)) DiracDelta(x-Prod(u)) )
Details will be given later.
Part 3: Discussion
Let me rather start a brief discussion suggested by a comment of Guess who it is.
2.1) The result of wolfie (which I saw only after having finished my calculations) is
f3wolfie[x_] :=
27/32 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x]
which is not identical at first sight with my result
f3wolfgang[x_] :=
27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -
27/16 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x]
But the functions indeed coincide as can be seen in a plot, or some clever FullSimplify?
For f3n[x], the last step in my calculation was this integral
Integrate[27/8 (
Sqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x, 1},
Assumptions - 0 < x < 1]
and the two terms in f3wolfgang correspond to the two terms in the this integrand.
2.2) Behaviour close to x = 1
As I haven't found a quick answer in the literature I tackled the poblem "experimentally":
Numerically my solution can be respresented as
f3nn[x_]:=NIntegrate[
27/8 ( Sqrt[-((
w x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x,
1}]
Plotting this close to 1 with a negative power of (1-x) attached
Plot[1/(1 - x)^k f3nn[x], {x, 0.5, 1.1}]
and playing with k close to 2 shows that f3nn[x] ~= const (1-x)^2. I suspect even that k = 2 is exact because only for this value the function exhibits a sharp shoulder.
Afterwards I found (after some lengthy study) the exact behaviour of the constant to be 27 Pi/16, i.e. the distribution function has the series Expansion
f3n[x] = 27/64 \[Pi] (1 - x)^2 + O[1 - x]^3
Also, the MeijerG-function has the series expansion
MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] == (1 - x) -
1/8 (1 - x)^2 + O[1 - x]^3
Remark: the developemnts here show that there is a simple integral representation of the MeijerG function, much simpler than the complex integral version. We have
MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] = 2/Pi Integrate[
27/8 ( Sqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x]
- ArcCos[Sqrt[x/w]]))/w, {w, x, 1}]
The difficulties of Mathematica calculating the numerical values close to 1 have been overcome by this representation and the series expansion.
• "using the Mellin transformation" - with the Meijer result, this is what you're implicitly doing anyway, since the $G$-function is effectively an inverse Mellin transform… – J. M. will be back soon Aug 12 '15 at 12:29
• I have been busy doing the calculations and the editing so I did't see the resuls of wolfie. Sorry for that. – Dr. Wolfgang Hintze Aug 12 '15 at 12:30
• @J. M.: that's what I was - cautiously - saying ;-) – Dr. Wolfgang Hintze Aug 12 '15 at 12:34
• The reason why I had mentioned that the Meijer result looked familiar in the other answer is that it resembled some of the Mellin-Barnes representations for elliptic integrals that I remember; your result, which at least involves the complete elliptic integrals, seems to be tantalizing. – J. M. will be back soon Aug 12 '15 at 12:41
• Some nice reference to MeijerG is ams.org/notices/201307/rnoti-p866.pdf pointing out specifically the closure under convolutions. – Dr. Wolfgang Hintze Aug 12 '15 at 13:20 | 2019-12-09T10:24:38 | {
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Three men and eight machines can finish a job in half the time taken
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Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
A. 10
B. 11
C. 12
D. 13
E. 14
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31 Mar 2020, 02:49
rate of one machine ; 13*2 ; 26 days
and rate of man ; x
3/x+8/26 = 2*(8/x+3/26)
solve for x
we get x= 169
so in 13 days ; 169/13 ; 13 men would be required
IMO D
Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
A. 10
B. 11
C. 12
D. 13
E. 14
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31 Mar 2020, 02:50
1
time taken by 3Men + 8 Machines = half * time taken by 3 machines + 8 men
In other words time taken by 3 men + 8 machines = time taken by 6 machines + 16 men
Two machines can finish the work in 13 days means that one machine can finish the work in 26 days.
let the time take for a man to finish the work be x, then
3/x + 8/26 = 6/26 + 16/x
2/26 = 13/x
x = 13*13
Hence 13 men will take 13*13/13 = 13 days.
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31 Mar 2020, 03:01
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Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
A. 10
B. 11
C. 12
D. 13
E. 14
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31 Mar 2020, 03:33
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Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
A. 10
B. 11
C. 12
D. 13
E. 14
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Solution:
Let A and B be the per day work of one man and one machine respectively
• Work done by 3 men and 8 machines in one day = 3A + 8B
• Work done by 8 men and 3 machines in one day = 8A + 3B
o Time taken to complete the work by 3 men and 8 machines is half time by 8 men and 3 machines
o Work done by 3 men and 8 machines in one day is twice of the work done by 8 men and 3 machines in one day
o $$3A + 8B = 2(8A + 3B)$$
o $$3A +8B = 16A + 6B$$
o $$8B-6B = 16A – 3A$$
o $$2B = 13A$$
• Work done by 2 machines in one day is equal to the work done by 13 men in one day.
• 2 machines can complete the job in 13 days
o 13 men can complete the job in 13 days
Hence, the correct answer is Option D.
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31 Mar 2020, 04:17
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Since, TEAM-1 (3 men and 8 machines) can finish a job in half the time taken by TEAM-2 (3 machines and 8 men) to finish the same job, we can express the TWO TEAM'S CAPACITY through the following equation:
3 MEN+ 8 MACHINES= 2 (3 MACHINES+ 8 MEN)
=> 3 MEN+8 MACHINES= 6 MACHINES+16 MEN
=>8 MACHINES- 6 MACHINES= 16 MEN- 3 MEN
=> 2 MACHINES=13 MEN
Therefore, we can infer that THE CAPACITY OF TWO MACHINES IS EQUIVALENT TO THAT OF 13 MEN.
Hence, it requires 13 men to complete what can be done by 2 machines in the same amount of time.
Bunuel, please convey me whether my approach is efficient.
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Re: Three men and eight machines can finish a job in half the time taken [#permalink]
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31 Mar 2020, 09:20
1
Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?
A. 10
B. 11
C. 12
D. 13
E. 14
two machines can finish the job in 13 days
--> 1 machine can complete a job in 26 days
Let 1 man complete a job in '$$m$$' days
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job
--> $$\frac{3}{m} + \frac{8}{26} = 2(\frac{3}{26} + \frac{8}{m})$$
--> $$\frac{8}{26} - \frac{6}{26} = \frac{16}{m} - \frac{3}{m}$$
--> $$\frac{1}{13} = \frac{13}{m}$$
--> $$m = 169$$ days
Number of men required to finish the job in 13 days = $$\frac{169}{13} = 13$$ men
Option D
Re: Three men and eight machines can finish a job in half the time taken [#permalink] 31 Mar 2020, 09:20 | 2020-06-04T22:48:49 | {
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https://math.stackexchange.com/questions/1977468/definition-of-continuity-not-clear | # Definition of continuity not clear
According to my book continuity is (informally) defined as follows:
Function $f$ is continuous at $c$ iif it is both right continuous and left continuous at $c$.
Now according to the book the function $f(x) = \sqrt{4 - x^2}$ is continuous at every point of its domain, $[-2, 2]$. I understand this holds for $(-2, 2)$,but what about the endpoints?
If I look at the definition and consider the left and right endpoints, then how could it be left continuous at its left endpoint or right continuous at its right endpoint? Since there's no function beyond $-2$, then how can there be a limit from the left?
• When we say that $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of $(a,b)$, right continuous at $a$, and left continuous at $b$. – Brian M. Scott Oct 20 '16 at 16:59
• If you make an answer of it, I can upvote it. – Apeiron Oct 20 '16 at 17:15
• Similar to question math.stackexchange.com/questions/1969405/… – LutzL Oct 20 '16 at 20:27
It’s a matter of convention. When we say that a function $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of the open interval $(a,b)$, right continuous at $a$, and left continuous at $b$. In other words, it has every possible one-sided continuity at each point of the closed interval.
Added: The real culprit here is the assertion that $f$ is continuous at a point $c$ if and only if it is both left and right continuous at $c$. This is fine if the domain of $f$ is the whole real line or an open interval, but it’s not quite right for arbitrary domains. Suppose that $f$ has domain $D$, and $c\in D$. Then $f$ is continuous at $c$ if it satisfies the following two conditions:
• if $c$ is a limit point of $\{x\in D:x<c\}$, then $f$ is left continuous at $c$, and
• if $c$ is a limit point of $\{x\in D:x>c\}$, then $f$ is right continuous at $c$.
When $D$ is a closed interval $[a,b]$, any $c\in(a,b)$ is a limit point of both $$\{x\in[a,b]:x<c\}=[a,c)$$ and $$\{x\in[a,b]:x>c\}=(c,b]\;.$$ When $c=a$, however, the first of these sets is empty, so $a$ isn’t a limit point of it, and we don’t require $f$ to be left continuous at $a$. When $c=b$, the second set is empty, and we don’t require $f$ to be right continuous at $c$.
Many standard first-year calculus texts are a bit sloppy about such details.
• It's a language abuse to say that $f$ is continuous at $[a,b]$. – hamam_Abdallah Oct 20 '16 at 17:26
• @Abdallah: No, it isn’t; it’s a convention that agrees with the use of the word continuous in more general topological settings. The real culprit here is the assertion that $f$ is continuous at a point $c$ in its domain iff it is continuous from each side: that is in fact true only if $c$ is a limit point of the domain of $f$ from both sides. When the domain is a closed interval this is not the case at the endpoints of the interval. – Brian M. Scott Oct 20 '16 at 17:32
• Thanks, @BrianM.Scott, this was really helpful. This issue raised many doubts to my understanding of a topic I'm already a bit anxious about. I will update the definition accordingly in my concepts notes. – Apeiron Oct 21 '16 at 7:27
• @Apeiron: You’re welcome; glad it helped. – Brian M. Scott Oct 21 '16 at 13:51 | 2019-10-15T07:09:53 | {
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https://staff.noqood.co/so-long-tihs/do-all-functions-have-an-inverse-2bfcb8 | onto, to have an inverse, since if it is not surjective, the function's inverse's domain will have some elements left out which are not mapped to any element in the range of the function's inverse. Not all functions have inverse functions. This means, for instance, that no parabola (quadratic function) will have an inverse that is also a function. So y = m * x + b, where m and b are constants, is a linear equation. The graph of this function contains all ordered pairs of the form (x,2). To have an inverse, a function must be injective i.e one-one. do all kinds of functions have inverse function? In fact, the domain and range need not even be subsets of the reals. No. The function f is defined as f(x) = x^2 -2x -1, x is a real number. We did all of our work correctly and we do in fact have the inverse. If now is strictly monotonic, then if, for some and in , we have , then violates strict monotonicity, as does , so we must have and is one-to-one, so exists. Please teach me how to do so using the example below! There is one final topic that we need to address quickly before we leave this section. It is not true that a function can only intersect its inverse on the line y=x, and your example of f(x) = -x^3 demonstrates that. Other types of series and also infinite products may be used when convenient. but y = a * x^2 where a is a constant, is not linear. Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a).In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Question 64635: Explain why an even function f does not have an inverse f-1 (f exponeant -1) Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website! The horizontal line test can determine if a function is one-to-one. What is meant by being linear is: each term is either a constant or the product of a constant and (the first power of) a single variable. In this section it helps to think of f as transforming a 3 into a … 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. The graph of inverse functions are reflections over the line y = x. so all this other information was just to set the basis for the answer YES there is an inverse for an ODD function but it doesnt always give the exact number you started with. let y=f(x). Inverse Functions. A function must be a one-to-one function, meaning that each y-value has a unique x-value paired to it. There is an interesting relationship between the graph of a function and its inverse. Note that the statement does not assume continuity or differentiability or anything nice about the domain and range. Consider the function f(x) = 2x + 1. Answer to (a) For a function to have an inverse, it must be _____. viviennelopez26 is waiting for your help. if i then took the inverse sine of -1/2 i would still get -30-30 doesnt = 210 but gives the same answer when put in the sin function Add your … If the function is linear, then yes, it should have an inverse that is also a function. This means that each x-value must be matched to one and only one y-value. how do you solve for the inverse of a one-to-one function? It should be bijective (injective+surjective). Problem 86E from Chapter 3.6: Functions that meet this criteria are called one-to one functions. Imagine finding the inverse of a function … Definition of Inverse Function. For example, the infinite series could be used to define these functions for all complex values of x. 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. A function may be defined by means of a power series. Suppose is an increasing function on its domain.Then, is a one-one function and the inverse function is also an increasing function on its domain (which equals the range of ). Problem 33 Easy Difficulty. Statement. Such functions are often defined through formulas, such as: A surjective function f from the real numbers to the real numbers possesses an inverse as long as it is one-to-one, i.e. Restrictions on the Domains of the Trig Functions A function must be one-to-one for it to have an inverse. Logarithmic Investigations 49 – The Inverse Function No Calculator DO ALL functions have Sin(210) = -1/2. As we are sure you know, the trig functions are not one-to-one and in fact they are periodic (i.e. Explain.. Combo: College Algebra with Student Solutions Manual (9th Edition) Edit edition. Explain why an even function f does not have an inverse f-1 (f exponeant -1) F(X) IS EVEN FUNCTION IF Strictly monotone functions and the inverse function theorem We have seen that for a monotone function f: (a;b) !R, the left and right hand limits y 0 = lim x!x 0 f(x) and y+ 0 = lim x!x+ 0 f(x) both exist for all x 0 2(a;b).. Not all functions have inverses. Explain your reasoning. There is one final topic that we need to address quickly before we leave this section. An inverse function is a function that will “undo” anything that the original function does. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. For example, we all have a way of tying our shoes, and how we tie our shoes could be called a function. Inverse of a Function: Inverse of a function f(x) is denoted by {eq}f^{-1}(x) {/eq}.. Such functions are called invertible functions, and we use the notation $$f^{−1}(x)$$. I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. For example, the function f(x) = 2x has the inverse function f −1 (x) = x/2. x^2 is a many-to-one function because two values of x give the same value e.g. Before defining the inverse of a function we need to have the right mental image of function. We did all of our work correctly and we do in fact have the inverse. There is an interesting relationship between the graph of a function and the graph of its inverse. Question: Do all functions have inverses? So a monotonic function must be strictly monotonic to have an inverse. This is what they were trying to explain with their sets of points. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . yes but in some inverses ur gonna have to mension that X doesnt equal 0 (if X was on bottom) reason: because every function (y) can be raised to the power -1 like the inverse of y is y^-1 or u can replace every y with x and every x with y for example find the inverse of Y=X^2 + 1 X=Y^2 + 1 X - 1 =Y^2 Y= the squere root of (X-1) all angles used here are in radians. Not every element of a complete residue system modulo m has a modular multiplicative inverse, for instance, zero never does. So a monotonic function has an inverse iff it is strictly monotonic. Suppose that for x = a, y=b, and also that for x=c, y=b. For instance, supposing your function is made up of these points: { (1, 0), (–3, 5), (0, 4) }. For a function to have an inverse, the function must be one-to-one. Thank you! if you do this . their values repeat themselves periodically). Define and Graph an Inverse. View 49C - PowerPoint - The Inverse Function.pdf from MATH MISC at Atlantic County Institute of Technology. \begin{array}{|l|c|c|c|c|c|c|} \hline x & -3 & -2 & -1 & 0 & 2 & 3 \\ \hline f(x) & 10 & 6 & 4 & 1 & -3 & -10 \\ \h… An inverse function goes the other way! Yeah, got the idea. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. 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. No. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Answer to Does a constant function have an inverse? Hello! Suppose we want to find the inverse of a function … both 3 and -3 map to 9 Hope this helps. Thank you. as long as the graph of y = f(x) has, for each possible y value only one corresponding x value, and thus passes the horizontal line test.strictly monotone and continuous in the domain is correct There are many others, of course; these include functions that are their own inverse, such as f(x) = c/x or f(x) = c - x, and more interesting cases like f(x) = 2 ln(5-x). Warning: $$f^{−1}(x)$$ is not the same as the reciprocal of the function $$f(x)$$. Inverting Tabular Functions. The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. Does the function have an inverse function? Does the function have an inverse function? This implies any discontinuity of fis a jump and there are at most a countable number. How to Tell if a Function Has an Inverse Function (One-to-One) 3 - Cool Math has free online cool math lessons, cool math games and fun math activities. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test. Other functional expressions. Basically, the same y-value cannot be used twice. Now, I believe the function must be surjective i.e. Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Function, we all have a way of tying our shoes, and infinite. One y-value it is not linear inverse, for instance, zero never does polynomial,. ) \ ) f^ { −1 } ( x ) and in fact they are (... Shoes, and we use the notation \ ( f^ { −1 } ( x ) = x^2 -1. Example below subsets of the form ( x,2 ) a constant, is not linear a,. 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Do you solve for the inverse relation is then defined as the inverse of a function also that x=c! Not even be subsets of the reals its inverse that the statement does not assume continuity or differentiability or nice... Form ( x,2 ) is then defined as the inverse of most functions! Solve for the inverse function is linear, do all functions have an inverse yes, it have... -3 map to 9 Hope this helps form ( x,2 ) all have a of... Monotonic function has an inverse, the Trig functions a function must be surjective i.e ( x =! These functions for all complex values of x please teach me how to evaluate f at 3, (. Would be one-to-many, which is n't a function that will “ undo ” anything the! Function may be used twice complete residue system modulo m has a unique x-value paired to it horizontal line.! Sets of points your … if the function is a many-to-one function because two values x... A ) for a function that will “ undo ” anything that the statement does not continuity. 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Are called one-to one functions all complex values of x give the same value e.g not possible find! | 2021-03-06T23:43:20 | {
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https://www.freemathhelp.com/forum/threads/sequence-word-problem.70745/ | # Sequence Word Problem
#### 0313phd
##### New member
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?
I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313
#### soroban
##### Elite Member
Hello,0313phd!
The number of bricks in the bottom row of a brick wall is 49.
The next row up from the bottom contains 47 bricks,
and each subsequent row contains 2 fewer bricks than the row immediately below it.
The number of bricks in the top row is 3.
If the wall is one brick thick, what is the total number of bricks in the wall?
I am not getting the correct answer.
Is this an arithmetic sequence? . Yes!
. . . $$\displaystyle S_n\:=\: \tfrac{n}{2}\left(a_1 + a_n\right)$$
So that since there are 46 steps between the bottom and top steps. . No, there are 24 rows.
Counting from top, we have:
. . $$\displaystyle \begin{array}{|c||c|c|c|c|c|c|} \hline \text{Row} & 1 & 2 & 3 & 4 & \hdots & 24 \\ \hline \text{Bricks} & 3 & 5 & 7 & 9 & \hdots & 49 \\ \hline\end{array}$$
$$\displaystyle \text{So we have: }\:S_{24} \;=\;\tfrac{24}{2}(3 + 49) \:=\12)(52) \:=\:624$$
#### TchrWill
##### Full Member
0313phd said:
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?
I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313
An alternate thought:
Add a brick atop those below and you have rows of 49, 47, 45, 43,...........7, 5, 3, 1.
The sum of the first "n" odd numbers is n^2.
Therefore, the sum of the 25 rows of bricks is 25^2 or 625.
Remove the single brick added for conveniance, and the total number of bricks is 624.
J
#### JeffM
##### Guest
0313phd said:
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?
I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313
1 + 3 + 5 + ... + 49 = S[sub:241ucnbg]49[/sub:241ucnbg]
3 + 5 + ... + 49 = T[sub:241ucnbg]49[/sub:241ucnbg].
So, T[sub:241ucnbg]n[/sub:241ucnbg] = S[sub:241ucnbg]n[/sub:241ucnbg] - 1.
All clear so far?
There is a formula for the sum of the first n odd numbers, but maybe you do not know it. So derive it. (It is tricky to decide what to memorize. My tendency is to remember concepts and techniques and as few formulae as possible, but that is me. I was (sort of) trained as an historian so you cannot rely on me for math advice UNLESS I PROVE IT.)
Let's do a few examples. (This is not the only way to solve such problems, but, WHEN IT WORKS, it is the simplest.)
1 = 1.
1 + 3 = 4.
1 + 3 + 5 = 9.
1 + 3 + 5 + 7 = 16.
Do you see a pattern?
Not looking for a fancy proof of the pattern, but let's confirm it through an informal mathematical induction.
What is the standard formula for an odd number?
Assume [(2 * 1) - 1] + ... (2k - 1) = k[sup:241ucnbg]2[/sup:241ucnbg].
Then [(2 * 1) - 1] + ... [(2k) - 1] + [2(k + 1) - 1] = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 2 - 1 = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 1 = (k + 1)[sup:241ucnbg]2[/sup:241ucnbg].
Now can you solve your problem?
THANKS, 0313
3 + 49 = 52
5 + 47 = 52
.... | 2021-02-26T00:16:59 | {
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https://math.stackexchange.com/questions/1166798/inequality-from-chapter-5-of-the-book-how-to-think-like-a-mathematician/1166803 | # Inequality from Chapter 5 of the book *How to Think Like a Mathematician*
This is from the book How to think like a Mathematician,
How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$
without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$
• Are we allowed complex roots and negative roots to prove the contrary ;-) Feb 28, 2015 at 19:22
• I don't think that it is in the proposition Mar 2, 2015 at 11:24
• 7! is the product of seven positive numbers, all of which are smaller than 8. $8^7$ is the product of 7 eights. So the inequality follows. Dec 7, 2021 at 22:52
• Be sure to write your argument starting from $7! < 8^7$ and ending with what you want to prove, not backwards the way you discovered where to start. Dec 7, 2021 at 22:53
• If you have a dataset of $n$ real numbers, and you add an $n+1$st number greater than the average of the $n$ numbers, the new set has a greater average. (Hint: logarithms.) Dec 7, 2021 at 22:53
Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$
• In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. Feb 27, 2015 at 3:02
• You are right, great ! I would have of to think there. Feb 28, 2015 at 17:39
• Elegant! really clever! May 27, 2015 at 16:19
Think of
$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$
as the average of seven numbers and
$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$
as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)
• You are right, but it is clear to me only when i compute ln(8) Feb 28, 2015 at 18:08
• @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. Feb 28, 2015 at 20:00
• Yes I know that. Mar 2, 2015 at 11:21
• @Gwydyon, I guess I don't understand your previous comment then. Mar 2, 2015 at 15:22
• Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 Mar 3, 2015 at 16:41
Note that $$\sqrt[7]{7!} < \sqrt[8]{8!} \iff\\ (7!)^8 < (8!)^7 \iff\\ 7! < \frac{(8!)^7}{(7!)^7} \iff\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.
• You are right but someone has writen the same statements Feb 28, 2015 at 18:11
$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.
• great! you achieved my second statement Feb 28, 2015 at 18:53
You have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \begin{equation*} \begin{split} (1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i) \\ &=(1/(n+1) (n x_{n+1}/n + n \sum_{i=1}^{n} x_i / n) \\ &> (1/(n+1) (\sum_{i=1}^{n} x_i/n + n \sum_{i=1}^{n} x_i / n) \\ &= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n) \\ &= \sum_{i=1}^{n} x_i / n \end{split} \end{equation*}
Note that we really only needed $x_{n+1}$ to be larger than the previous mean.
The solution occurs just by doing simple calculations,
Lets start, $$\sqrt[7]{7!}<\sqrt[8]{8!}$$
iff $$(\sqrt[7]{7!})^{7\cdot8}<(\sqrt[8]{8!})^{7\cdot8}$$
iff $$(7!)^{8}<(8!)^7$$
iff $$(7!)^8<(7!\cdot8)^7$$
iff $$(7!)^8<8^7\cdot(7!)^7$$
iff $$(7!)<8^7$$
iff $$1\cdot2\cdots6\cdot7<8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8$$
wich is obviously true since $$1<8,2<8,\ldots,7<8$$
Consider this
$$x=\ln 8!-\frac87\ln7!=\ln8-\frac17\ln7!=\frac17\left(7\ln8-\ln7!\right)=\frac17\ln\frac{8^7}{7!}>0$$ Hence, since the exponential is a monotonically increasing function: $$e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$$ | 2023-03-22T03:27:15 | {
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https://nl.mathworks.com/help/econ/dtmc.graphplot.html | Main Content
# graphplot
Plot Markov chain directed graph
## Description
example
graphplot(mc) creates a plot of the directed graph (digraph) of the discrete-time Markov chain mc. Nodes correspond to the states of mc. Directed edges correspond to nonzero transition probabilities in the transition matrix mc.P.
example
graphplot(mc,Name,Value) uses additional options specified by one or more name-value arguments. Options include highlighting transition probabilities, communicating classes, and specifying class properties of recurrence/transience and period. Also, you can plot the condensed digraph instead, with communicating classes as supernodes.
graphplot(ax,___) plots on the axes specified by ax instead of the current axes (gca) using any of the input argument combinations in the previous syntaxes. The option ax can precede any of the input argument combinations in the previous syntaxes.
example
h = graphplot(___) returns the handle to the digraph plot. Use h to modify properties of the plot after you create it.
## Examples
collapse all
Consider this theoretical, right-stochastic transition matrix of a stochastic process.
$P=\left[\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\ 0& 0& 1/3& 0& 2/3& 0& 0\\ 0& 0& 0& 0& 0& 1/3& 2/3\\ 0& 0& 0& 0& 0& 1/2& 1/2\\ 0& 0& 0& 0& 0& 3/4& 1/4\\ 1/2& 1/2& 0& 0& 0& 0& 0\\ 1/4& 3/4& 0& 0& 0& 0& 0\end{array}\right].$
Create the Markov chain that is characterized by the transition matrix P.
P = [ 0 0 1/2 1/4 1/4 0 0 ;
0 0 1/3 0 2/3 0 0 ;
0 0 0 0 0 1/3 2/3;
0 0 0 0 0 1/2 1/2;
0 0 0 0 0 3/4 1/4;
1/2 1/2 0 0 0 0 0 ;
1/4 3/4 0 0 0 0 0 ];
mc = dtmc(P);
Plot a directed graph of the Markov chain.
figure;
graphplot(mc);
Consider this theoretical, right-stochastic transition matrix of a stochastic process.
$P=\left[\begin{array}{cccc}0.5& 0.5& 0& 0\\ 0.5& 0& 0.5& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right].$
Create the Markov chain that is characterized by the transition matrix P. Name the states Regime 1 through Regime 4.
P = [0.5 0.5 0 0; 0.5 0 0.5 0; 0 0 0 1; 0 0 1 0];
mc = dtmc(P,'StateNames',["Regime 1" "Regime 2" "Regime 3" "Regime 4"]);
Plot a directed graph of the Markov chain. Identify the communicating classes in the digraph and color the edges according to the probability of transition.
figure;
graphplot(mc,'ColorNodes',true,'ColorEdges',true)
States 3 and 4 compose a communicating class with period 2. States 1 and 2 are transient.
Create a "dumbbell" Markov chain containing 10 states in each "weight" and three states in the "bar."
• Specify random transition probabilities between states within each weight.
• If the Markov chain reaches the state in a weight that is closest to the bar, then specify a high probability of transitioning to the bar.
• Specify uniform transitions between states in the bar.
rng(1); % For reproducibility
w = 10; % Dumbbell weights
DBar = [0 1 0; 1 0 1; 0 1 0]; % Dumbbell bar
DB = blkdiag(rand(w),DBar,rand(w)); % Transition matrix
% Connect dumbbell weights and bar
DB(w,w+1) = 1;
DB(w+1,w) = 1;
DB(w+3,w+4) = 1;
DB(w+4,w+3) = 1;
db = dtmc(DB);
Plot a directed graph of the Markov chain. Return the plot handle.
figure;
h = graphplot(db);
Observe that the state labels are difficult to read. Remove the labels entirely.
h.NodeLabel = {};
## Input Arguments
collapse all
Discrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).
Axes on which to plot, specified as an Axes object.
By default, graphplot plots to the current axes (gca).
### Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: 'ColorEdges',true,'ColorNodes',true colors the edges to indicate transition probabilities and colors the nodes based on their communicating class.
Flag for labeling nodes using state names, specified as the comma-separated pair consisting of 'LabelNodes' and a value in this table.
ValueDescription
trueLabel nodes using the names in mc.StateNames.
falseLabel nodes using state numbers.
Example: 'LabelNodes',false
Data Types: logical
Flag for coloring nodes based on communicating class, specified as the comma-separated pair consisting of 'ColorNodes' and a value in this table.
ValueDescription
trueNodes in the same communicating class have the same color. Solid markers represent nodes in recurrent classes, and hollow markers represent nodes in transient classes. The legend contains the periodicity of recurrent classes.
falseAll nodes have the same color.
Example: 'ColorNodes',true
Data Types: logical
Flag for labeling edges with the transition probabilities in the transition matrix mc.P, specified as the comma-separated pair consisting of 'LabelEdges' and a value in this table.
ValueDescription
trueLabel edges with transition probabilities rounded to two decimal places.
falseDo not label edges.
Example: 'LabelEdges',true
Data Types: logical
Flag for coloring edges to indicate transition probabilities, specified as the comma-separated pair consisting of 'ColorEdges' and a value in this table.
ValueDescription
trueColor edges to indicate transition probabilities. Include a color bar, which summarizes the color coding.
falseUse the same color for all edges.
Example: 'ColorEdges',true
Data Types: logical
Flag for condensing the graph, with each communicating class represented by a single supernode, specified as the comma-separated pair consisting of 'Condense' and a value in this table.
ValueDescription
trueNodes are supernodes containing communicating classes. Node labels list the component states of each supernode. An edge from supernode i to supernode j indicates a nonzero probability of transition from some state in supernode i to some state in supernode j. Transition probabilities between supernodes are not well defined, and graphplot disables edge information.
falseNodes are states in mc.
Example: 'Condense',true
Data Types: logical
## Output Arguments
collapse all
Handle to the graph plot, returned as a graphics object. h is a unique identifier, which you can use to query or modify properties of the plot.
## Tips
• To produce the directed graph as a MATLAB® digraph object and use additional functions of that object, enter:
G = digraph(mc.P)
• For readability, the 'LabelNodes' name-value pair argument allows you to turn off lengthy node labels and replace them with node numbers. To remove node labels completely, set h.NodeLabel = {};.
• To compute node information on communicating classes and their properties, use classify.
• To extract a communicating class in the graph, use subchain.
• The condensed graph is useful for:
• Identifying transient classes (supernodes with positive outdegree)
• Identifying recurrent classes (supernodes with zero outdegree)
• Visualizing the overall structure of unichains (chains with a single recurrent class and any transient classes that transition into it)
## References
[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.
[2] Horn, R., and C. R. Johnson. Matrix Analysis. Cambridge, UK: Cambridge University Press, 1985.
[3] Jarvis, J. P., and D. R. Shier. "Graph-Theoretic Analysis of Finite Markov Chains." In Applied Mathematical Modeling: A Multidisciplinary Approach. Boca Raton: CRC Press, 2000.
## Version History
Introduced in R2017b | 2022-09-30T06:45:42 | {
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https://math.stackexchange.com/questions/3006069/find-the-smallest-positive-integer-divisible-by-63-such-that-the-sum-of-its-digi | # Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.
TASK: Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.
MY WORK: Let the number be $$A=\overline{x_n x_{n-1} x_{n-2} \cdots x_1 x_0}$$. Since $$63|(x_n+x_{n-1}+\cdots+x_1+x_0)$$, we have that $$x_n+x_{n-1}+\cdots+x_0\ge63\cdots(*)$$ and since $$x_0,x_1,\cdots,x_n$$ are n+1 digits, we have that $$x_n+x_{n-1}+\cdots+x_0\le9+9+\cdots+9=9(n+1)$$ which then means that $$9(n+1)\ge63\Leftrightarrow n+1\ge7$$ i.e that the number $$A$$ has at least seven digits. If it has $$7$$ digits, all of them would have to be $$9$$ to satisfy the inequality $$(*)$$ which would mean that $$A=9999999$$. But then the condition $$63|A$$ would not be satisfied. So the number A doesn't have seven digits - it has at least eight digits.
However, I do not know where to go from here.
• So check numbers like 18999999 and 19899999 and 19989999 and so on. – Gerry Myerson Nov 20 '18 at 8:15
• You get divisibility by $9$ for free so you only need be concerned about checking for $7$ – Mark Bennet Nov 20 '18 at 8:44
Assume the the number is $$1$$ with 6 $$9$$s and one $$8$$.
Now $$19999999\equiv 5\pmod 7$$
If we subtract $$10^k$$ we will get aus a number with a $$1$$ , 6$$9$$ and one $$8$$ anda different equivalence. so we need to find the $$10^k\equiv 5\mod 7$$.
$$10\equiv 3$$
$$100\equiv 30\equiv 2$$
$$1000\equiv 20\equiv 6$$
$$10,000\equiv 60\equiv 4$$
$$100,000 \equiv 40\equiv 5$$
So $$19,999,999-100,000=19,899,999\equiv 0\pmod 7$$.
And that's that. It's digits add to $$63$$ so it's divisible by $$9$$ and it's divisible by $$7$$. And beginning with $$1$$ and the only such divisible by $$7$$ it's the smallest such number.
====
I thought I made it clear why this is the smallest.
No element with $$7$$ digits or fewer exist as the OP figured out. For a group with $$8$$ digits the smallest would start with a $$1$$. If you have an $$8$$ digit number beginning with $$1$$ and whose digits add to $$63$$ the remaining digits must be six $$9$$s and one $$8$$. Such a number can be written as $$19,999,999 - 10^k$$ where $$0\le k \le 7$$. For such a number to be divisible by $$63$$ we must have $$10^k \equiv 5 \pmod 7$$. The ONLY such $$k$$ is $$k = 5$$ and $$10^k =100,000$$ and the number is $$19,899,999$$. So this is the only such number divisible by $$63$$ whose digits add to $$63$$ in the smallest possible category of types of numbers that can have such numbers. So this is the smallest such number.
• Is that the smallest such number? [no criticism of your answer intended... but OP originally asked for smallest.] (+1 on answer.) – coffeemath Nov 20 '18 at 9:01
• Sorry @coffeemath , I had a typo. – Akash Roy Nov 20 '18 at 10:18
• Yes, it's the smallest. And I explained why. There can't be one with 7 digits, so the smallest has eight or more. This has 8 so it is in the smallest group. the smallest digit it can begin with is 1 so this is the smallest of the smallest. If number starts with 1 and has 8 digits and adds to $63$ then is must have 6 nines and one 8. So smallest number would be $19,999,999 - 10^k$ where $10^k\equiv 5 \mod 7$. The only such option for $0\le k \le 7$ is $k=5$. So this is the only solution with 8 digits begining with $1$. And no smaller number is a solution. – fleablood Nov 20 '18 at 17:05
(This is essentially the same solution as @fleablood 's; but doubts were raised whether it is actually the smallest.)
Such a number has at least $$8$$ digits. Since the prescribed digit sum is $$63$$ we have to deduct exactly $$9$$ units from writing eight nines. Trying with $$x_1=1$$ as first digit, and all other nines, we have given away $$8$$ units, one more to go. Divisibility by $$9$$ is taken care of automatically. Now $$19\,999\,999=5$$ mod $$7$$; therefore we need to find a $$k\in[0..6]$$ with $$10^k=5$$ mod $$7$$, or we are bust with $$x_1=1$$. Fortunately $$10^5=5$$ mod $$7$$. It follows that $$19\,899\,999$$ is the smallest number with the required properties.
• @coffeemath: Sorry for the typo. Thank you for reporting it. – Christian Blatter Nov 20 '18 at 10:15
The smallest number whose digits all sum to a multiple of $$63$$ is $$9{,}999{,}999$$. The next smallest is $$18{,}999{,}999$$, then $$19{,}899{,}999$$, then $$19,989{,}999$$, and so on. All of these are clearly divisible by $$9$$, so it suffices to check for divisibility by $$7$$. As it happens, the first two are not, but $$19{,}899{,}999/7=2{,}842{,}857$$ (and, just to doublecheck, $$19{,}899{,}999/63=315{,}873$$).
Remark: It's not a priori obvious that any of the numbers described here will turn out to be divisible by $$7$$. You could say we just got lucky. Or you could do a modular argument to show that luck had nothing to do with it. One thing is obvious: the smallest number sought for is certainly no greater than $$777{,}777{,}777$$.
• The modular argument is not hard. The numbers are all of the form $19,999,999 - 10^k$ so we need $19,999,999 - 10^k \equiv 0 \mod 7$ or $10^k \equiv 5 \mod 7$. As $10$ and $7$ is relativley prime and $7$ is prime the $10^k; 0\le k < 7$ are distinct modulo $7$ and $10^5$ is the only one that works. – fleablood Nov 20 '18 at 17:28
• @fleablood, are you saying, more generally, that if $10$ and $p$ are relatively prime (with $p$ a prime), then $10^k$ for $0\le k\lt p$ are distinct modulo $p$? It's true that $10$ is a primitive root mod $7$, but not because it's relatively prime to $7$. – Barry Cipra Nov 20 '18 at 22:46
• Yeah, I guess I worded it incorrectly. $10\equiv 3$ is a primitive root is what I meant. – fleablood Nov 20 '18 at 23:08
I also verify that the Number 19,899,999 is the smallest number that is divisible by 63 and Also its sum is divisible by 63. Here a program I wrote in Python 3. To find this out by brute forcing :
for i in range(9999990, 19900000, 63):
sum_is = sum(int(d) for d in str(i))
if sum_is%63==0:
print(i)
Prints :
19899999
Also this time I started from 9999990 because this number the last number divisible by 63 that has sum less than 63.
This can also be used to do the same with any number.
Just change the range and 63 as you want.
Hope this Helps!
[EDIT SUMMARY] For more Optimization and better Understanding. Added direct summing instead of sum_digits as request/suggested by @Paul Evans. Also added jump of 63 as Suggested by @Kyle Kanos.
• Instead of the mysterious sum_digits(i) you could write python code: sum(int(d) for d in str(i)) – Paul Evans Nov 20 '18 at 16:01
• You can also save time by starting at 9999990 and incrementing by 63, eliminating the need to test i%63==0. – Kyle Kanos Nov 20 '18 at 19:47
• Yes! That's more optimized. Thanx for advice. – FightWithCode Nov 21 '18 at 13:20 | 2020-10-30T16:11:23 | {
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https://gateoverflow.in/95821/tifr2017-b-13?show=97339 | 321 views
For an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?
1. the line graph for a complete graph is complete
2. the line graph for a connected graph is connected
3. the line graph for a bipartite graph is bipartite
4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
5. each vertex in the line graph has degree one or two
retagged | 321 views
+2
Option B). is true.
+2
yes, Agreed. (y)
0
why not option C?
0
you can check with K(2,2). The line graph obtained is not biparitite.
option B is the right answer
We can solve this question by eliminating options
Option (A) False
Let us take a complete graph of $4$ vertices:
In a line graph, no. of edges is
$\sum_{i=0}^{n}$ ${}^{d_{i}}C_{{2}},$ $d_{i}$is degree of each vertex
=$\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2} = 3\times 4=12$
No. of vertices in line graph = No. of edges in original graph
No. of vertices in line graph $= 6$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But for given line graph from complete graph of $4$ vertices we have only $12$ edges.
Option (B) True
1. Smallest line graph for original graph one edge
which is also connected graph
If a graph is connected with more then one edge, it will never be disconnected
Option (C) False
This cannot be 2-colorable and hence is not bipartite.
Option (D) False
Because line graph degree of vertex depends on the attribute
e.g., $\left [ A,B \right ]$ as point in line graph, then degree of this vertex depends on degree of $A$ and degree of $B$ in the original graph.
I'm drawing degree for a point $[AB]$ in line graph.
So, this is wrong.
Option (E) False (wrong as proved in above option (D))
edited
The line graph of a connected graph is connected. If G is connected, it contains a path connecting any two of its edges, which translates into a path in L(G) containing any two of the vertices of L(G). Therefore, option B is correct.
We can also do this question using elimination of options.
edited
0
but your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.
0
Hi,
I have corrected the example, and updated the file.
0
great example you have updated...proving line of tree is not a tree too :p | 2018-06-18T15:46:08 | {
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https://fr.mathworks.com/help/matlab/ref/fplot3.html | fplot3
3-D parametric curve plotter
Syntax
fplot3(funx,funy,funz)
fplot3(funx,funy,funz,tinterval)
fplot3(___,LineSpec)
fplot3(___,Name,Value)
fplot3(ax,___)
fp = fplot3(___)
Description
example
fplot3(funx,funy,funz) plots the parametric curve defined by x = funx(t), y = funy(t), and z = funz(t) over the default interval [-5,5] for t.
example
fplot3(funx,funy,funz,tinterval) plots over the specified interval. Specify the interval as a two-element vector of the form [tmin tmax].
example
fplot3(___,LineSpec) sets the line style, marker symbol, and line color. For example, '-r' specifies a red line. Use this option after any of the previous input argument combinations.
fplot3(___,Name,Value) specifies line properties using one or more name-value pair arguments. For example, 'LineWidth',2 specifies a line width of 2 points.
fplot3(ax,___) plots into the axes specified by ax instead of the current axes. Specify the axes as the first input argument.
example
fp = fplot3(___) returns a ParameterizedFunctionLine object. Use the object to query and modify properties of a specific line. For a list of properties, see ParameterizedFunctionLine Properties.
Examples
collapse all
Plot the 3-D parametric line
$\begin{array}{c}x=\mathrm{sin}\left(t\right)\\ y=\mathrm{cos}\left(t\right)\\ z=t\end{array}$
over the default parameter range [-5 5].
xt = @(t) sin(t); yt = @(t) cos(t); zt = @(t) t; fplot3(xt,yt,zt)
Plot the parametric line
$\begin{array}{c}x={e}^{-t/10}\mathrm{sin}\left(5t\right)\\ y={e}^{-t/10}\mathrm{cos}\left(5t\right)\\ z=t\end{array}$
over the parameter range [-10 10] by specifying the fourth input argument of fplot3.
xt = @(t) exp(-t/10).*sin(5*t); yt = @(t) exp(-t/10).*cos(5*t); zt = @(t) t; fplot3(xt,yt,zt,[-10 10])
Plot the same 3-D parametric curve three times over different intervals of the parameter. For the first interval, use a line width of 2 points. For the second, specify a dashed red line style with circle markers. For the third, specify a cyan, dash-dotted line style with asterisk markers.
fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [0 2*pi], 'LineWidth', 2) hold on fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [2*pi 4*pi], '--or') fplot3(@(t)sin(t), @(t)cos(t), @(t)t, [4*pi 6*pi], '-.*c') hold off
Plot multiple lines in the same axes using hold on.
fplot3(@(t)t, @(t)t, @(t)t) hold on fplot3(@(t)-t, @(t)t, @(t)-t) hold off
Plot the parametric line
$\begin{array}{c}x={e}^{-|t|/10}\mathrm{sin}\left(5|t|\right)\\ y={e}^{-|t|/10}\mathrm{cos}\left(5|t|\right)\\ z=t.\end{array}$
Assign the parameterized function line object to a variable.
xt = @(t)exp(-abs(t)/10).*sin(5*abs(t)); yt = @(t)exp(-abs(t)/10).*cos(5*abs(t)); zt = @(t)t; fp = fplot3(xt,yt,zt)
fp = ParameterizedFunctionLine with properties: XFunction: @(t)exp(-abs(t)/10).*sin(5*abs(t)) YFunction: @(t)exp(-abs(t)/10).*cos(5*abs(t)) ZFunction: @(t)t Color: [0 0.4470 0.7410] LineStyle: '-' LineWidth: 0.5000 Show all properties
Change the range of parameter values to [-10 10] and change the line color to red.
fp.TRange = [-10 10]; fp.Color = 'r';
For $t$ values in the range $-2\pi$ to $2\pi$, plot the parametric line
$\begin{array}{c}x=t\\ y=t/2\\ z=\mathrm{sin}\left(6t\right).\end{array}$
Add a title, x-axis label, and y-axis label. Additionally, change the view of the axes and display the axes box outline.
xt = @(t)t; yt = @(t)t/2; zt = @(t)sin(6*t); fplot3(xt,yt,zt,[-2*pi 2*pi],'MeshDensity',30,'LineWidth',1); title('x=t, y=t/2, z=sin(6t) for -2\pi<t<2\pi') xlabel('x'); ylabel('y'); view(52.5,30) box on
Access the axes object using gca. Specify the x-axis tick values and associated labels using the XTick and XTickLabel properties of the axes object. Similarly, specify the y-axis tick values and associated labels.
ax = gca; ax.XTick = -2*pi:pi/2:2*pi; ax.XTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0','\pi/2','\pi','3\pi/2','2\pi'}; ax.YTick = -pi:pi/2:pi; ax.YTickLabel = {'-\pi','-\pi/2','0','\pi/2','\pi'};
Input Arguments
collapse all
Parametric function for x coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form x = funx(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funx = @(t) sin(2*t);
Parametric function for y coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form y = funy(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funy = @(t) cos(2*t);
Parametric function for z coordinates, specified as a function handle to a named or anonymous function.
Specify a function of the form z = funz(t). The function must accept a vector input argument and return a vector output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: funz = @(t) t;
Interval for parameter t, specified as a two-element vector of the form [tmin tmax].
Axes object. If you do not specify an axes object, then fplot3 uses the current axes (gca).
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'Marker','o','MarkerFaceColor','red'
The properties listed here are only a subset. For a complete list, see ParameterizedFunctionLine Properties.
Number of evaluation points, specified as a number. The default is 23. Because fplot3 uses adaptive evaluation, the actual number of evaluation points is greater.
Line color, specified as an RGB triplet, a hexadecimal color code, a color name, or a short name.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: 'blue'
Example: [0 0 1]
Example: '#0000FF'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Example: [0.3 0.2 0.1]
Example: 'green'
Example: '#D2F9A7'
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
Output Arguments
collapse all
One or more ParameterizedFunctionLine objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific ParameterizedFunctionLine object. For details, see ParameterizedFunctionLine Properties.
Introduced in R2016a | 2021-07-31T02:04:10 | {
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https://www.physicsforums.com/threads/number-of-possible-combinations.753650/ | # Number of possible combinations
1. May 14, 2014
### kidkook
I have 16 items and i need to know the number of possible combination for all 16 items. I know the possible number of combination for all 16 items is 1 and the possible number of combinations for 15 of the 16 items is 16 using the formula below:
$$\frac{n!}{r!(n-r)!}.$$
How can i adaprt this formula to calculate for the number of selection from 1 to 16 inclusive without repeating the calculation 16 times.
So n will always be 16 and r will be 1 to 16 inclusive.
2. May 14, 2014
Your wording is a little confusing. Is this the correct interpretation:
You want to know the total number of ways to select at least one item from the 16.
If it is, think about this: you should know the total number of subsets (empty, proper, with the original set) for a set with 16 items. Your question involves finding not the total number but how many there are not counting the empty set.
3. May 14, 2014
### kidkook
Sorry for the confusion. What i have is a dataset with 16 different values. What i'm trying to achieve is to determine how many combination are available without repetition/duplication of a value. So i know the answer is 65536. I achieved this by using the above formula 16 times and substituting the value of r from 1 to 16 etc...and adding the number together.
Number of combinations Value of r
16 1
120 2
560 3
1820 4
4368 5
8008 6
11440 7
12870 8
11440 9
8008 10
4368 11
1820 12
560 13
120 14
16 15
1 16
65535
What i would like to know is...can the formula be adapted to so i don't need to repeat the calculation 16 times. My dataset is likely to increase significantly in the future so it will not be practical to use this method.
4. May 14, 2014
### jbunniii
Note that $\frac{n!}{r!(n-r)!}$ is called the binomial coefficient and is often denoted
$${n \choose r} = \frac{n!}{r!(n-r)!}$$
The reason it is called the binomial coefficient is the following property:
$$(x+y)^n = \sum_{r=0}^{n}{n \choose r}x^r y ^{n-r}$$
If we plug in $x=y=1$ then this reduces to
$$2^n = \sum_{r=0}^{n}{n \choose r} = 1 + \sum_{r=1}^{n} {n \choose r}$$
where the second equality follows because ${n \choose 0} = 1$. Therefore,
$$\sum_{r=1}^{n} {n \choose r} = 2^n - 1$$
which is consistent with your calculation since $2^{16} - 1 = 65535$.
5. May 14, 2014
### kidkook
Excellent. This is exactly what i was looking for. A very well documented explaination. Many thanks. | 2017-12-13T21:13:12 | {
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https://math.stackexchange.com/questions/2376090/what-is-the-probability-of-rolling-at-least-two-6s-with-3-dice-and-2-rolls | # What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls?
Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).
(6*6*6 = 216 = outcomes when rolling 3 dice) (6*6 = 36 = outcomes when rolling 2 dice)
1st roll: (3 dice)
A = 125/216 = 0 sixes
B = 75/216 = 1 six
C = 15/216 = 2 sixes
D = 1/216 = 3 sixes
Outcomes C and D fulfill requirement.
For outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)
2nd roll:
a = 125/216 = 0 sixes
b = 75/216 = 1 six
c = 15/216 = 2 sixes
d = 1/216 = 3 sixes
Outcomes c and d fulfill requirements.
For outcome B: (first roll = one 6, pick up two non 6's and roll again)
2nd roll:
z = 25/36 = 0 sixes
y = 10/36 = 1 six
x = 1/36 = 2 sixes
Outcomes y and x fulfill requirements.
So would the formula below give me my answer?
C + D + Ac + Ad + By + Bx = X
What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?
If I substituted correctly and did the math correct the answer I got was 22.3%
Is this correct?
• I'm guessing by your odds that you consider ordering important ? – user451844 Jul 29 '17 at 22:20
• I'm not sure if I know what you mean by ordering. – Travis Jul 29 '17 at 22:58
• is (1,2,3) the same as (1,3,2) and (2,3,1) and (2,1,3) and (3,1,2) and (3,2,1) ? if so then it cuts down the number of distinct rolls down to 56. – user451844 Jul 29 '17 at 23:00
• Well let's say we had a blue green and a red die, if the red die shows a 1 and blue shows a 6 and green shows a 6, that would be different than if red showed a 6, green showed a 6, and blue showed a 1. So I guess ordering would matter. The numbers rolled are the same but show on different dice. Both rolls fulfill the two 6's requirement but are different rolls and should be counted as thus. I'm not sure if I'm correct on this, that's why I'm asking but does it makes sense? – Travis Jul 29 '17 at 23:09
• yeah I'm just making sure, because the probability might change depending on conditions like this. of the 56 cases when order didn't matter a full 21 had at least one 6, but only 6 had more than one 6. etc. – user451844 Jul 29 '17 at 23:19
An alternate approach would be to find the probability of the complementary event:
$\textbf{1)}$ The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$
$\textbf{2)}$ The probability of getting exactly one 6 is given by
$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$
Therefore the probability of getting at least two 6's is given by
$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$
• From my point of view is your approach the most appropriate one. (+1) – Markus Scheuer Jul 30 '17 at 9:38
Your approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with \begin{align*} (5+t)^3 \end{align*} marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\leq j \leq 3$ in the first roll can be written as \begin{align*} [t^j](5+t)^3\cdot\frac{1}{6^3} \end{align*}
Encoding the second roll with $(5+u)^{3-j}$ we calculate \begin{align*} \sum_{{0\leq j,k\leq 3}\atop{j+k\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\cdot\frac{1}{6^{6-j}}\\ &=[t^0](5+t)^3\left([u^2]+[u^3]\right)(5+u)^3\cdot\frac{1}{6^6}\\ &\qquad+[t^1](5+t)^3\left([u^1]+[u^2]\right)(5+u)^2\cdot\frac{1}{6^5}\\ &\qquad+[t^2](5+t)^3\left([u^0]+[u^1]\right)(5+u)^1\cdot\frac{1}{6^4}\\ &\qquad+[t^3](5+t)^3\\ &=\binom{3}{0}5^3\left(\binom{3}{2}5^1+\binom{3}{3}5^0\right)\cdot\frac{1}{6^6}\\ &\qquad+\binom{3}{1}5^2\left(\binom{2}{1}5^1+\binom{2}{2}5^0\right)\cdot\frac{1}{6^5}\\ &\qquad+\binom{3}{2}5^1\left(\binom{1}{0}5^1+\binom{1}{1}5^0\right)\cdot\frac{1}{6^4}\\ &\qquad+\binom{3}{3}5^0\cdot\frac{1}{6^3}\\ &=\frac{125\cdot16}{6^6}+\frac{75\cdot11}{6^5}+\frac{15\cdot6}{6^4}+\frac{1}{6^3}\\ &=\frac{5\,203}{23\,328}\\ &\doteq 0.223 \end{align*} | 2019-10-15T05:00:36 | {
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https://math.stackexchange.com/questions/2364954/how-many-ways-are-there-to-tile-a-chessboard-with-32-knights | # How many ways are there to tile a chessboard with 32 knights?
The maximal amount of knights which can be places on a regular $8 \times 8$ chessboard so that no two take each other is $32$ (To see this just notice that a knight on a black square only attacks knights placed on white squares).
I was wondering how many ways there are to tile a regular chessboard with these $32$ knights so that no two take each other.
A generalisation of this question has already been posed here: In how many ways we can place $N$ mutually non-attacking knights on an $M \times M$ chessboard?. However, the answers given include a polynomial approach which cannot be used in the case of an $8 \times 8$ chessboard, because, as an answer states, a computer struggles to compute such a polynomial.
Even if it were quite easy for a computer to compute such a polynomial, in an Olympiad no such computational power is allowed. Hence, I was wondering if there was a pure combinatorics approach for this particular case.
An example of such a tiling is to place all the knights on all the black cells.
• You can put the knights on the black squares or on the white squares. This makes two ways. Or am I missing something? – ajotatxe Jul 20 '17 at 11:58
• @ajotatxe Are there no other possibilities? Can you prove this? – Plato Jul 20 '17 at 11:59
• Consider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares. This means that there are (at most) two ways to place the $32$ knights. But of course there are two ways, use all the black squares or all the white squares. – bof Jul 20 '17 at 12:11
• In graph-theoretic terms: A cycle graph $C_{2n}$ has exactly two independent sets of size $n.$ Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n.$ The knight's graph on the $8\times8$ chessboard is a Hamiltonian graph of order $64.$ – bof Jul 20 '17 at 12:20
• The existence of a knight's tour seems crucial here. Actually, in a $2\times 2$ board there is no knight's tour, and there are $6$ ways to put two non-attacking horses on it. – ajotatxe Jul 20 '17 at 12:26
Consider a knight's tour of the chessboard. Obviously, if $32$ of the $64$ squares are occupied by non-attacking knights, then the tour must alternate between occupied and unoccupied squares, since a horse can only attack a square if the opposite colour. This means that there are (at most) two ways to place the $32$ knights. We can offer examples of these two ways: use all the black squares or all the white squares.
Another way to look at the problem is to consider graph theory. A cycle graph $C_{2n}$ has exactly two independent sets of size $n$. Therefore, a Hamiltonian graph of order $2n$ has at most two independent sets of size $n$. The knight's graph on the $8 \times 8$ chessboard is a Hamiltonian graph of order 64. From here we can conclude in the same manner as in the first solution. | 2019-11-15T23:12:40 | {
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https://mathschallenge.net/full/rounding_error | ## Rounding Error
#### Problem
A number is rounded to two decimal places and this answer is rounded to one decimal place. If the final number is 0.2 then what is the chance that the original number lies between 0.15 and 0.25?
#### Solution
When rounding to $k$ decimal places (d.p.) we truncate the number at the $k^{th}$ decimal position, but if the next digit is 5 or more then we increase the last digit by 1; this may have a knock-on effect. Consider the following examples of rounding to 2 d.p.:
3.61483 3.61
7.7853301 7.79
0.5982 0.60
Suppose we were told that 0.2 were the result of rounding to one decimal place. The smallest number that would be rounded up to 0.2 is 0.15. However, there is no actual value we can assign to the greatest value that would round down to 0.2. For example, 0.25 would round up to 0.3 and although 0.249 would round down to 0.2, what about 0.2499, or 0.24999, or ... ?
Under these circumstances we define bounds (or limits) in the following way:
• The lower bound is the least value that rounds up to the given number.
• The upper bound is the least value that fails to round down to the given number.
So if 0.2 is the result of rounding to 1 d.p. then the lower and upper bounds are 0.15 and 0.25 respectively.
Working backwards, the lower bound of 0.2 is 0.15 and the lower bound of 0.15 is 0.145. We can easily check this: 0.145 = 0.15 (2 d.p.) and then 0.15 = 0.2 (1 d.p.).
However, we need to tread very carefully when working backwards with the upper bound. Although it is true to say that the upper bound of 0.2 is 0.25, if the original number were rounded to 0.25 (to 2 d.p.), then next process of rounding to 1 d.p. would make it 0.3, not 0.2.
In fact, it is necessary that the result of rounding to 2 d.p. takes it to 0.24; and the upper bound of 0.24 is 0.245.
Hence we are certain that the original number, $x$, lies somewhere in the interval $0.145 \le x \lt 0.245$.
If we are considering if the number lies between 0.15 and 0.25 then it would have to lie in the interval $0.15 \le x \lt 0.245$.
\begin{align}\therefore P(0.15 \le x \lt 0.245 \mid 0.145 \le x \lt 0.245) &= \dfrac{0.245 - 0.15}{0.245 - 0.145}\\\\&= \dfrac{0.095}{0.1}\\\\&= 95\%\end{align}
If instead of rounding to 2 d.p. then to 1 d.p. a number is just rounded to 1 d.p., what is the chance that both methods give the same answer?
Problem ID: 343 (09 Jul 2008) Difficulty: 2 Star
Only Show Problem | 2019-02-21T00:26:20 | {
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http://math.stackexchange.com/questions/160979/sum-of-squares-of-multiplicities-of-differences | # Sum of squares of multiplicities of differences
It's hard to come up with a good title. Let $A$ be a set of $n$ integers, say $A=\{a_1,\ldots,a_n\}$ with $a_1<\cdots<a_n$. We consider all the positive differences $a_j-a_i$ (necessarily $i<j$). There are $n \choose 2$ pairs $(a_i,a_j)$ with $i<j$, but of course, some differences can be repeated, or in other words, have "multiplicity" greater than 1.
For a given $A$, one way to illustrate the differences and their multiplicities is with a "table of differences". Here is one such table for $A=\{0,1,3,5,6\}$, and another for $A=\{1,4,7,10,13\}$.
-| 0 1 3 5 6 - | 1 4 7 10 13
-+---------- --+------------
0| - 1 3 5 6 1 | - 3 6 9 12
1| - 2 4 5 4 | - 3 6 9
3| - 2 3 7 | - 3 6
5| - 1 10| - 3
6| - 13| -
We see that if $A=\{0,1,3,5,6\}$, then the differences $1,2,3,5$ each have multiplicity 2, and the differences $4,6$ each have multiplicity 1. We also see that if $A=\{1,4,7,10,13\}$, then the differences $3,6,9,12$ have multiplicities $4,3,2,1$ respectively.
My question is: What is a tight upper bound on the sum of the squares of the multiplicities? I conjecture that the answer is $1^2+2^2+3^2+\cdots+(n-1)^2$, which occurs if $A$ consists of $n$ integers in arithmetic progression (as in the second example above).
At first, I thought this would be an easy consequence of the following "lemma": The largest multiplicity must be at most $n-1$, the second largest multiplicity must be at most $n-2$, the third largest multiplicity must be at most $n-3$, and so on. However, that "lemma" is false! In the first example above (in which $n=5$), the four largest multiplicities are $2,2,2,2$, which are not bounded above by $4,3,2,1$ respectively.
Note that my conjectured bound grows like $n^3/3$. I'm pretty sure I can find an upper bound that grows like $2n^3/3$, but the gap between those bounds is perplexing me.
-
Your lemma might be true if you modify it to say the largest multiplicity is at most $n-1$, the two largest multiplicities sum to at most $n-1+n-2=2n-3$, the three largest sum to at most $(n-1)+(n-2)+(n-3)=3n-6$, etc. – Gerry Myerson Jun 21 '12 at 1:07
We can show by induction that your bound is the best.
Suppose we have $n$ numbers and the differences have multiplicities $m_1\ge m_2\ge \cdots \ge m_t$. If we now add a new largest number, it creates $n$ distinct differences. If we could freely choose the differences (without constraints based on the other numbers), we could choose up to $n$ of the $m_i$ to increment exactly once, or can add $m_{t+1}=1,m_{t+2}=1,\ldots$ as necessary for differences that do not match.
Since $m_i>m_j\Rightarrow (m_i+1)^2+m_j^2>m_i^2+(m_j+1)^2$ then without constraints on the differences we maximize our sum of squares by incrementing the $n$ largest differences available. Starting with $n=2, m_1=1$, the maximum for $n=3$ is achieved by $m_1\leftarrow 2,m_2=1$, and so on. If we assume that for $n=k$ the maximum is $m_1=(k-1),m_2=(k-2),\ldots,m_{k-1}=1$, then for $k+1$ numbers the maximum is $m_1\leftarrow k,m_2\leftarrow (k-1),\ldots, m_{k-1}\leftarrow 2, m_k=1$. So by induction this pattern gives a bound for all $n$. And, as you pointed out, this bound can be achieved by choosing numbers in an arithmetic sequence.
This line of thought also confirms @Gerry's comment on your Lemma. Starting with one number and adding new largest numbers, the best we can do for the top $M$ multiplicities is to increment some subset of them each time we add a number until we have $M$ numbers, then increment every one of them for each number added after that. This constraint is thus that they add to at most $Mn-M(M+1)/2$, but the $M$th largest can be up to $n-(M+1)/2$.
-
Thinking about it today (everybody's answers and comments certainly helped), I believe there is a relatively short inductive proof that doesn't bother with any lemmas.
My bound is certainly true for $n=2$. When we transition from a set $A=\{a_1<\cdots<a_n\}$ to a set $A'=\{a_1<\cdots<a_n<a_{n+1}\}$, then as Zander mentions, we introduce $n$ differences $a_{n+1}-a_1,\ldots,a_{n+1}-a_n$ that are distinct from each other, so $n$ of the multiplicities increase by 1 (possibly including some multiplicities that were 0).
When we increase a multiplicity from $m$ to $m+1$, we increase the sum of the squares of the multiplicities by $2m+1$. Therefore the total increase of the sum of the squares of the multiplicities has the form $\sum(2m_i+1)$, where there are $n$ terms in the sum, and the $m_i$ are some of the multiplicities with respect to $A$ (and some $m_i$ may be 0).
So the sum of the squares of the multiplicities increases by $2(\sum m_i) + n$. This can be bounded above by $2S+n$, where $S$ is the sum of all multiplicities with respect to $A$. But $S = {n \choose 2}$, so we have shown that the sum of squares of multiplicities has increased by at most $2{n \choose 2}+n = (n^2-n)+n = n^2$.
- | 2015-06-30T10:47:20 | {
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https://math.stackexchange.com/questions/2093295/total-number-of-odd-numbers-in-a-row-of-pascals-triangle-is-a-natural-power-of | # Total number of odd numbers in a row of Pascal's triangle is a natural power of two
This seems like a combinatorial problem so there might be something simple that hasn't struck me. Although I do have an idea but I am unable to proceed from it. The statement of the question is:
Prove that in any row of Pascal's triangle, the number of odd elements is $$2^k$$, for some $$k \in \mathbb{N}$$.
I was working on a Pascal's triangle but in a binary format, where two adjacent 1's add up to 0. Something of the sort:
1
1 1
1 0 1
1 1 1 1
1 0 0 0 1
1 1 0 0 1 1
It is a definitive sequence and the thing I liked about it was that you can add up the adjacent 1's in a row to get the number of odd elements, but I haven't been able to generalize this information. I was thinking along the lines of a recursive relation in polynomials whose coefficients would represent the rows of this triangle, then I can just feed 1 into the equation and inductively prove that it is a perfect power of 2.
Can someone help me out on a proof along these lines, if I'm going in the right direction here?
P.S. I know there is the modulo 2 proof but can someone help me generalize that binary pascal's triangle?
• This will be helpful: math.hmc.edu/funfacts/ffiles/30001.4-5.shtml Jan 11 '17 at 13:07
• Okay, cool. But I want to try a proof along the lines of what I was trying. That is a good solution, but I wanna see where this goes. Jan 11 '17 at 13:15
• To get an inductive proof, you may want to use the results from the answers based on Kummer's Theorem. That is, try to show that the number of $1$s in a row of Pascal's Triangle is $2$ to the power of the number of $1$ bits in the binary representation of $n$.
– robjohn
Jan 11 '17 at 17:58
• You may be interested by the fact that Pascal's triangle modulo 2 generates Sierpinski's sieve (mathworld.wolfram.com/SierpinskiSieve.html) Jan 11 '17 at 22:18
## 3 Answers
Similar idea to Andreas Caranti's answer:
Kummer's Theorem says that the number of factors of $$p$$ in $$\binom{n}{k}$$ is the number of carries when adding $$k$$ and $$n-k$$ in base-$$p$$. There are no carries when adding $$k$$ and $$n-k$$ in binary if and only if $$k\veebar n-k=0$$; when that is true, $$n=k\lor n-k$$. This means that $$\binom{n}{k}$$ is odd when the $$1$$ bits of $$k$$ are a subset of the $$1$$ bits of $$n$$. That is, there are $$2^{\sigma_2(n)}$$ odd entries in row $$n$$ of Pascal's Triangle, where $$\sigma_2(n)$$ is the sum of the bits in the binary representation of $$n$$.
Will delete because this is not what OP intends, but a proof via Kummer's Theorem looks rather neat to me.
Write $n, m$ in base $2$ $$n = 2^{k} n_{k} + \dots + 2 n_{1} + n_{0}, \qquad m = 2^{k} m_{k} + \dots + 2 m_{1} + m_{0},$$ with $n_{i}, m_{i} \in \{ 0, 1 \}$.
Then
$\dbinom{n}{m}$ is odd iff there are no carries in the sum in base $2$ $$m + (n - m) = n,$$ that is, iff $m_{i} \le n_{i}$ for each $i$.
Given $n$, there are thus $$\prod_{i=0}^{k} (n_{i} + 1)$$ possibilities for $m$, and each $n_{i} + 1 \in \{ 1, 2 \}$.
This also tells you what is the exponent $k$ in the question - it is the number of $1$'s in the diadic expansion of $n$, or equivalently, the number of distinct powers of $2$ that sum up to $n$.
The simplest (at least simplest to explain) solution (using Lucas's theorem) has not been posted; let me fill that "much needed gap". Lazy as I am, I will mostly copypaste my answer from https://math.stackexchange.com/a/2184460/ , since the argument is essentially the same.
I use the notation $$\mathbb{N}$$ for the set $$\left\{0,1,2,\ldots\right\}$$.
The question asks the following:
Theorem 1. Let $$n \in \mathbb{N}$$. Then, the number of $$i \in \left\{0,1,\ldots,n\right\}$$ such that $$\dbinom{n}{i}$$ is odd is a power of $$2$$.
I shall prove something slightly more general:
Theorem 2. Let $$n \in \mathbb{N}$$. Let $$q$$ be a real number. Then, there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} q^i = \prod_{g \in G} \left(q^{2^g} + 1\right) . \end{align}
Proof of Theorem 2. Write $$n$$ in the form $$n=a_{k}2^{k}+a_{k-1}2^{k-1}+\cdots+a_{0}2^{0}$$ for some $$k\in\mathbb{N}$$ and $$a_{0},a_{1},\ldots,a_{k}\in\left\{ 0,1\right\}$$. (This is just the base-$$2$$ representation of $$n$$, possibly with leading zeroes.)
Lucas's theorem tells you that if $$i=b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$$ for some $$b_{0} ,b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$, then \begin{align} \dbinom{n}{i} &\equiv \dbinom{a_{k}}{b_{k}}\dbinom{a_{k-1}}{b_{k-1}} \cdots\dbinom{a_{0}}{b_{0}}=\prod\limits_{j=0}^{k}\underbrace{\dbinom{a_{j}}{b_{j}} }_{\substack{= \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} \\\text{(since }a_{j}\text{ and }b_{j}\text{ lie in }\left\{ 0,1\right\} \text{)}}} \\ &=\prod\limits_{j=0}^{k} \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\\ 0, & \text{if }b_{j}>a_{j} \end{cases} = \begin{cases} 1, & \text{if }b_{j}\leq a_{j}\text{ for all }j\text{;}\\ 0, & \text{otherwise} \end{cases} \mod 2 . \end{align} Hence, the $$i\in\mathbb{N}$$ for which $$\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$ satisfying $$\left( b_{j}\leq a_{j}\text{ for all }j\right)$$. Since all these $$i$$ satisfy $$i \in \left\{ 0,1,\ldots,n\right\}$$ (because otherwise, $$\dbinom{n}{i}$$ would be $$0$$ and therefore could not be odd), we can rewrite this as follows: The $$i \in \left\{ 0,1,\ldots,n\right\}$$ for which $$\dbinom{n}{i}$$ is odd are precisely the numbers of the form $$b_{k}2^{k}+b_{k-1}2^{k-1} +\cdots+b_{0}2^{0}$$ for $$b_{0},b_{1},\ldots,b_{k}\in\left\{ 0,1\right\}$$ satisfying $$\left( b_{j}\leq a_{j}\text{ for all }j\right)$$. Since these numbers are distinct (because the base-$$2$$ representation of any $$i\in\mathbb{N}$$ is unique, as long as we fix the number of digits), we thus can substitute $$b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}$$ for $$i$$ in the sum $$\sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}q^{i}$$. Thus, this sum rewrites as follows: \begin{align} \sum\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\ \dbinom{n}{i}\text{ is odd}}}q^{i} &=\underbrace{\sum\limits_{\substack{b_{0},b_{1} ,\ldots,b_{k}\in\left\{ 0,1\right\} ;\\b_{j}\leq a_{j}\text{ for all }j} }}_{=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_k} }\underbrace{q^{b_{k}2^{k}+b_{k-1}2^{k-1}+\cdots+b_{0}2^{0}}}_{=\left( q^{2^{k}}\right) ^{b_{k}}\left( q^{2^{k-1}}\right) ^{b_{k-1}}\cdots\left( q^{2^{0}}\right) ^{b_{0}}} \\ &=\sum\limits_{b_{0}=0}^{a_{0}}\sum\limits_{b_{1}=0}^{a_{1}}\cdots\sum\limits_{b_{k}=0}^{a_{k} }\left( q^{2^{k}}\right) ^{b_{k}}\left( q^{2^{k-1}}\right) ^{b_{k-1} }\cdots\left( q^{2^{0}}\right) ^{b_{0}} \\ &=\left( \sum\limits_{b_{k}=0}^{a_{k}}\left( q^{2^{k}}\right) ^{b_{k}}\right) \left( \sum\limits_{b_{k-1}=0}^{a_{k-1}}\left( q^{2^{k-1}}\right) ^{b_{k-1} }\right) \cdots\left( \sum\limits_{b_{0}=0}^{a_{0}}\left( q^{2^{0}}\right) ^{b_{0}}\right) \\ &=\left( \sum\limits_{b=0}^{a_{k}}\left( q^{2^{k}}\right) ^{b}\right) \left( \sum\limits_{b=0}^{a_{k-1}}\left( q^{2^{k-1}}\right) ^{b}\right) \cdots\left( \sum\limits_{b=0}^{a_{0}}\left( q^{2^{0}}\right) ^{b}\right) \\ &=\prod\limits_{g=0}^{k}\underbrace{\left( \sum\limits_{b=0}^{a_{g}}\left( q^{2^{g}}\right) ^{b}\right) }_{\substack{= \begin{cases} q^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} \\\text{(since }a_{g}\in\left\{ 0,1\right\} \text{)}}} \\ &=\prod\limits_{g=0}^{k} \begin{cases} q^{2^{g}}+1, & \text{if }a_{g}=1;\\ 1 & \text{if }a_{g}=0 \end{cases} \\ &=\left( \prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g} =1}}\left( q^{2^{g}}+1\right) \right) \underbrace{\left( \prod\limits _{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=0}}1\right) }_{=1} \\ &=\prod\limits_{\substack{g\in\left\{ 0,1,\ldots,k\right\} ;\\a_{g}=1}}\left( q^{2^{g}}+1\right) . \end{align} Thus, there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} q^i = \prod_{g \in G} \left(q^{2^g} + 1\right) \end{align} (namely, $$G$$ is the set of all $$g\in\left\{ 0,1,\ldots,k\right\}$$ satisfying $$a_g = 1$$). This proves Theorem 2. $$\blacksquare$$
Proof of Theorem 1. Theorem 2 (applied to $$q = 1$$) shows that there exists a finite subset $$G$$ of $$\mathbb{N}$$ such that \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1^i = \prod_{g \in G} \left(1^{2^g} + 1\right) . \end{align} Consider this $$G$$. Comparing \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1^i = \prod_{g \in G} \underbrace{\left(1^{2^g} + 1\right)}_{= 1+1 = 2} = \prod_{g \in G} 2 = 2^{\left|G\right|} \end{align} with \begin{align} \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} \underbrace{1^i}_{=1} & = \sum\limits\limits_{\substack{i\in\left\{ 0,1,\ldots,n\right\} ;\\\dbinom{n}{i}\text{ is odd}}} 1 \\ & = \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) \cdot 1 \\ & = \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) , \end{align} we obtain \begin{align} \left(\text{the number of all i\in\left\{ 0,1,\ldots,n\right\} such that \dbinom{n}{i} is odd}\right) = 2^{\left|G\right|} . \end{align} Hence, the number of $$i \in \left\{0,1,\ldots,n\right\}$$ such that $$\dbinom{n}{i}$$ is odd is a power of $$2$$. This proves Theorem 1. $$\blacksquare$$ | 2021-12-05T23:37:57 | {
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http://math.stackexchange.com/questions/333417/need-to-prove-the-sequence-a-n-1-frac122-frac132-cdots-frac1n/333429 | # Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges
I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:
Monotonic:
The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$ $$a_1=1\leq 1+\frac{1}{2^2}=a_2$$
Need to show that $a_{n+1}\leq a_{n+2}$ $$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$ Thus the sequence is monotone and increasing.
Boundedness:
Since the sequence is increasing it is bounded below by $a_1=1$. Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.
Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?
-
Hint: compare it to another sequence with only powers of two in the denominator. – Jason Polak Mar 18 '13 at 2:58
@Jason: Could you please elaborate on your hint? I don't understand it. – Jonas Meyer Mar 18 '13 at 3:08
@JonasMeyer: The sequence is less than $1 + 1/2^2 + 1/2^2 + 1/4^2 + 1/4^2 + 1/4^2 + 1/4^2 + ... \leq 2$ – Jason Polak Mar 18 '13 at 3:13
@ShuXiaoLi my objection was to the way the comment was presented, i.e., like a solution to the question. A comment like "by the way, the sum is $\pi^2/6$" would have been fine. – Ittay Weiss Mar 18 '13 at 3:17
Your work looks good so far. Here is a hint: $$\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$
To elaborate, apply the hint to get: $$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} - \frac{1}{n}\right)$$
Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.
Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus: $$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 - \frac{1}{n}$$
Add $1$ to both sides to get: $$a_n \le 2 - \frac{1}{n} \le 2$$
It follows that $a_n$ is bounded from above and hence convergent.
It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.
-
Sorry, I'm afraid I'm going to need a bit more of a push. I can't think of how/where to use your hint. Am I supposed to say that $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+{…}+\frac{1}{n-1}-\frac{1}{n}$ and do something further with that? – user66807 Mar 18 '13 at 4:02
try to apply it more than once. – chx Mar 18 '13 at 4:17
@chx Try to apply it more than once in the "thing" I supposed? – user66807 Mar 18 '13 at 4:22
@user66807, we have $a_n=\sum_{k=2}^{n}\frac{1}{k^2}\le b_n=\sum_{k=2}^{n}\frac{1}{k-1}-\frac{1}{k}$, try to find the first few terms of $b_n$ you will see a pattern (things will start cancelling) then you can find an upper bound of $a_n$. Notice i started the sum from 2 to be able to use Ayman's hint. – i.a.m Mar 18 '13 at 4:32
@user66807, try to sum a few terms of the telescopic series $\,\sum\left(\frac{1}{n-1}-\frac{1}{n}\right)\,$ and see what happens... – DonAntonio Mar 18 '13 at 4:33
Besides to Ayman's neat answer, you may take $f(x)=\frac{1}{x^2}$ over $[1,+\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\sum_{k=1}^{+\infty}\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.
-
The problem is we haven't gone over series in class yet, but thanks anyways. – user66807 Mar 18 '13 at 3:41
@user66807: Sorry, I didn't know that. So the Ayman's is more preferable nice and simple approach for you. :-) – Babak S. Mar 18 '13 at 3:43
I hope so, let's just see if I can catch his hint. And thanks. – user66807 Mar 18 '13 at 3:46
+1 This is a good answer, though the OP can't use it yet. – DonAntonio Mar 18 '13 at 4:34
I agree with DonAntonio that this is a good answer! $+1 \;\ddot\smile\;\;$ (I also like the new gravatar!) Good day, to you! – amWhy Mar 18 '13 at 15:08
You can show this geometrically too.
If you take a square, and divide the height into $\frac 12$, $\frac 14$, $\frac 18$, and so forth, doubling the denominator on each deal.
Now take the squares $\frac 12$ and $\frac 13$: these go onto the top shelf.
On the second shelf go $\frac 14$ to $\frac 17$. These fractions are all less than $\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\frac 1n$ is smaller than $\frac 18$, and so forth.
Therefore $\sum_{n=2}^{inf}\frac 1n < 1$, and therefore the whole lot is less than two squares.
-
Hint: Prove the the following holds for all $n$ by induction.
$$\sum_1^n \frac{1}{k^2} \le 2 - \frac{1}{n}.$$
Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.
You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$.
(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)
-
by the integral test :
$\int^\infty_1 \frac{1}{n^2} dn \le \sum_{i=1}^\infty \frac{1}{n^2}\le 1+\int^\infty_1 \frac{1}{n^2}dn$ .
you can compute the integral so the answer is :
$1 \le \sum_{i=1}^\infty \frac{1}{n^2}\le 2$ .
because : $\int^\infty_1 \frac{1}{n^2} dn =1$
- | 2015-04-27T14:25:06 | {
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https://math.stackexchange.com/questions/3310720/probability-of-rolling-a-sum-of-14-when-rolling-a-20-sided-die-twice-or-a-2 | # probability of rolling a sum of $14$ when rolling a $20$-sided die twice or a $20$-sided die with a $4$-sided die depending on outcome of first roll
A $$20$$-sided die is rolled. If the result is $$10$$ or less, the $$20$$-sided die is rolled again. If the result is $$11$$ or more, a $$4$$-sided die is rolled. In either case, the results are summed after the second die roll. What is the probability that the sum will be $$14$$?
My Approach: there are 10 ways we can get a sum of 14 if the first dice is $$<= 10$$ {$$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$$} and $$3$$ ways we can get sum of $$14$$ if the first dice has value $$>= 11$$ {$$(11,3),(12,2),(13,1)$$} so the probability is: $$(10/400)$$ + $$(3/80) = 1/16$$
• What goes wrong when you just write it out in the obvious way? – lulu Aug 1 at 18:08
• welcome to MSE. kindly include any attempt. – Siong Thye Goh Aug 1 at 18:12
Hints:
$$Pr(A\cap B) = Pr(A)Pr(B\mid A)$$
$$Pr(B) = Pr(A_1\cap B)+Pr(A_2\cap B)+\dots+Pr(A_n\cap B)$$ where $$A_1,\dots,A_n$$ form a partition of the sample space
Let $$B$$ be the event the sum is $$14$$.
Let $$A_1$$ be the event the first die rolls a number $$1-10$$. Let $$A_2$$ be the event the first die rolls a number $$11-13$$. Let $$A_3$$ be the event the first die rolls a number $$14+$$.
Regardless what was rolled on the first die, so long as it was less than $$14$$, there will be exactly one outcome on the second roll that would let the total sum to $$14$$.
• I've added my approach. Not sure if it's correct. – user3119875 Aug 1 at 18:52
• $\frac{1}{2}\times\frac{1}{20}+\frac{3}{20}\times\frac{1}{4}=\frac{1}{16}$ is correct. There is absolutely no need to write out each of the possibilities however and this causes errors when trying to count things by hand and wastes effort and time. – JMoravitz Aug 1 at 18:56
The probability that you obtain $$14$$ using the $$20$$-sided die on the second role is
$$\frac{10}{20} \cdot \frac{1}{20}$$
The probability you obtain $$14$$ using the $$4$$-sided die on the second role is
$$\frac{3}{20} \cdot \frac{1}{4}$$ | 2019-10-15T01:46:35 | {
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https://math.stackexchange.com/questions/2848226/is-there-any-specific-convention-to-how-mathematical-equations-are-read-out | # Is there any specific convention to how mathematical equations are “read out”?
Often, mathematical information and formulas are shared through text, maybe on books or on websites. If you see the equation:
$$x + 3 = 4y$$
we can easily read it out as "x plus three equals four y". But for some other equations such as:
$$\frac{x+3}{x-7}=y$$
it doesn't seem too obvious on what it should be read out as. I would read it as "x plus three the whole divided by x minus seven equals y", however someone else may read it out differently.
My question is therefore: If in a situation where oral communication of mathematical equations is necessary, does there exist a convention to read out mathematical equations such that one particular equation is read out in only one way, and that it is possible to deduce the equation from the spoken name alone?
• In French, I would read it as "x plus trois sur x moins sept égal y", which would translate as "x plus three over x minus seven equals y". I don't know if "over" really is commonly used in English though. I don't think there is any universal convention though. What I said could also be interpreted as $x+\frac{3}{x} -7 = y$. If I notice that the person I am talking to has gotten it wrong, I would just say it again, insisting on the expression I really meant. – Suzet Jul 12 '18 at 1:55
• "over" is indeed used by some people I know. And here is the point, even in English, the same equation is being read out in two different ways. – Pritt Balagopal Jul 12 '18 at 1:57
• The different groupings / implied parantheses are often expressed via pauses and prosody. – Hagen von Eitzen Jul 12 '18 at 2:02
• "The quotient of [the quantity] $x+3$ and [the quantity] $x-7$ is $y$." To avoid a pile-up of symbols at the end, one could say "$y$ is the quotient of $x+3$ and $x-7$." – Blue Jul 12 '18 at 2:26
• @HagenvonEitzen: On the first day of a stint as a Pre-Calculus substitute teacher, I instructed my students to express groupings using "air parentheses", indicated by raising one's slightly-bent arms over one's head. So, here: "[raise arms] x plus three [lower arms] over [raise arms] ex minus seven [lower arms] is y." The students looked at me like I was crazy, but took to the practice. So earnest were they in this —even prompting each other when someone would forget— that, at the end of my six-week stay, I didn't have the heart to tell them that I'd only been kidding that first day. :D – Blue Jul 12 '18 at 2:33
I would like to elaborate on Hagen von Eitzen's insightful comment:
The different groupings / implied parantheses are often expressed via pauses and prosody.
I think this is exactly right and plays an extremely important role in disambiguating oral mathematics. There is a huge difference between:
"X plus 3, over X minus 7, equals Y"
and
"X, plus 3 over X, minus 7 equals Y"
and
"X plus 3, over X, minus 7 equals Y"
If the cadences and pauses in the sentences are read aloud consistently, I think it is unlikely that anyone would transcribe one of those in place of another.
However, in some cases (particularly for extremely complicated expressions with groupings nested inside other groupings) one may call attention to the grouping using verbal annotations like "the quantity" or "all". For example, the quadratic formula is often read aloud as
• "X equals negative b, plus or minus the square root of the quantity b-squared minus 4ac, all over 2a".
I would read $$\frac{x+3}{x-7}=y$$ as:
Quantity $x$ plus $3$ over quantity $x$ minus $7$ equals $y$. | 2019-06-19T09:15:23 | {
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https://math.stackexchange.com/questions/3796115/image-of-compact-set-under-piecewise-continuous-function | # Image of compact set under piecewise continuous function
Let $$a,b>0\in\mathbb{R}$$. Let $$U$$ be an domain in $$\mathbb{C}^n$$. Let $$f:[a,b]\longrightarrow U$$ be a piecewise continuous map. Then is $$f[a,b]$$ compact? If not compact, will it be bounded?
Ok. This is in the following context. I am given a piecewise smooth path $$\gamma:[a,b]\longrightarrow U$$. Where $$\gamma(a)=z$$ and $$\gamma(b)=w$$, for given $$z,w\in U$$. We are also given a function $$\alpha:U\times\mathbb{C}^n\longrightarrow \mathbb{R}$$, which is upper semicontinuous. Now it is said that $$t\in[a,b]\longrightarrow \alpha(\gamma(t),\gamma’(t))$$ is bounded and measurable. I wanted to know why the function is bounded. I know that $$\gamma[a,b]$$ is compact. And $$\gamma$$ being upper semicontinuous will attain its maximum on a compact set. But I am not sure about $$\gamma’$$.
• What does "piecewise continuous" mean here? How many pieces? Finitely many? Are the pieces compact? Open? Aug 19, 2020 at 10:51
• What have you tried? Aug 19, 2020 at 10:53
• The image of a compact space under continuous map is always compact. Regardless of what "piecewise" means and what $U$ is (as long as it is Hausdorff). Aug 19, 2020 at 10:57
• Please provide additional context, which ideally explains why the question is relevant to you and the community. Some forms of context include background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. Aug 19, 2020 at 11:18
• @SahibaArora I have added the context. Aug 19, 2020 at 13:25
Compact, not necessarily: On $$[0,1]$$ let $$f(x) = x, 0\le x<1,$$ $$f(1)=2.$$ Then $$f([0,1]) = [0,1)\cup\{2\}.$$
Bounded, yes: First, a lemma: If $$f$$ is continuous on $$(a,b)$$ and $$f$$ has finite limits at the end points, then $$f(a,b)$$ is bounded.
Proof: Suppose $$\lim_{x\to a^+} f(x)=L,$$ $$\lim_{x\to b^-} f(x)=M.$$ Let $$\epsilon=1.$$ Then there exists $$\delta_a>0, \delta_a<(b-a)/3,$$ such that $$|f(x)-L|<1$$ for $$x\in (a,a+\delta_a).$$ Thus for such $$x,$$
$$|f(x)| = |f(x)-L+L|\le |f(x)-L|+|L| <1+|L|.$$
Similarly, there exists $$\delta_b>0,\delta_b<(b-a)/3,$$ such that $$|f(x)|<1+|M|$$ for $$x\in (b-\delta_b,b).$$ It follows that $$f$$ is bounded on the set $$(a,a+\delta_a)\cup (b-\delta_b,b).$$
Since $$f$$ is continuous on the compact set $$[a+\delta_a,b-\delta_b],$$ $$f([a+\delta_a,b-\delta_b])$$ is compact, hence is bounded. It follows that $$f(a,b)$$ is bounded.
Now suppose $$f$$ is piecewise continuous on $$[a,b].$$ Then there exist points $$a=x_0 such that $$f$$ is continuous on each $$I_k=(x_{k-1},x_k)$$ and has finite limits at the end points of $$I_k.$$ By the lemma, each $$f(I_k)$$ is bounded. The set $$f(\{x_0,\dots x_n\})$$ is also bounded. Therefore
$$f([a,b])=f(I_1)\cup \cdots \cup f(I_n)\cup f(\{x_0,\dots x_n\})$$
is bounded. | 2022-06-30T19:05:31 | {
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https://math.stackexchange.com/questions/6244/is-there-a-quick-proof-as-to-why-the-vector-space-of-mathbbr-over-mathbb | # Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
• A finite dimensional vector space over $\mathbb{Q}$ is countable. – user641 Oct 7 '10 at 2:19
• @Steve: Please add that as an answer so people can upvote. – Aryabhata Oct 7 '10 at 2:50
• Does that mean that a vector space over $\mathbb{Q}$ is finite-dimensional iff the set of the vector space is countable? If so, please prove it. – Elchanan Solomon Oct 7 '10 at 2:55
• @Isaac Your question doesn't require the 'only if' anyway. Steve's observation answers your original question. – yasmar Oct 7 '10 at 3:04
• @Isaac: how is the fact that $\mathbb{R}$ is finite dimensional over itself relevant? – Arturo Magidin Oct 7 '10 at 3:31
As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).
Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$.
Added: Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\mathbb{Q}$.
• Related to the note about uncountable dimension, there are explicit examples of continuum-sized linearly independent sets, as seen in this MathOverflow answer by François G. Dorais: mathoverflow.net/questions/23202/… – Jonas Meyer Oct 7 '10 at 4:18
• Yes: but can one show that there is a basis for $\mathbb{R}$ over $\mathbb{Q}$ without some form of the Axiom of Choice? (There is a difference between exhibiting a large linearly independent subset and exhibiting a basis). – Arturo Magidin Oct 7 '10 at 14:51
• No, one cannot, but without the Axiom of Choice the notion of dimension breaks down (except for finite vs. infinite). Assuming AC (as I virtually always do), the size of a linearly independent set gives a lower bound on the dimension of the vector space, and I think it is wonderful that in this case such "explicit" proof exists that the real numbers have continuum dimension, as opposed to the nice qualitative proof one could give by extending your argument to larger cardinals. – Jonas Meyer Oct 7 '10 at 16:02
• @Jonas: No argument there (with any of your points). – Arturo Magidin Oct 7 '10 at 17:23
• good link @Jonas, thanks – Leon Sep 23 '11 at 0:23
The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $$\, \mathbb Q$$-independent set of reals. Consider the set consisting of the logs of all primes $$\, p_i.\,$$ If $$\, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\ c_i\in\mathbb Q,\,$$ multiplying by a common denominator we can assume that all $$\ c_i \in \mathbb Z\,$$ so, exponentiating, we obtain $$\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$$ for all $$\,i,\,$$ by the uniqueness of prime factorizations.
• @Bill +1 What a nice example. – Adrián Barquero Oct 7 '10 at 4:50
• @Bill. Wow! :-) – Agustí Roig Oct 7 '10 at 6:08
• Is this proof unique to Q or does it generalize to provide explicit examples of reals linearly independent over, e.g., Q(sqrt(2))? Above, Q appears to be "hard-wired" into the proof, as the group of exponents in the prime factorization. – T.. Oct 12 '10 at 7:14
• Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. – JDH Jun 23 '11 at 2:22
• @student this can not happen: the primes are distinct. Put the primes with negative exponent on the other side (exponents become positive). You would find two distinct prime factorizations of the same number. – Student Feb 6 at 19:44
No transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.
The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \bigcup_{n>0} 2^{2^{-n}}$ is linearly independent over $\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).
The square roots of the prime numbers are linearly independent over $\mathbb Q$. (Proof: this is immediate given the ability to extend the function "number of powers of $p$ dividing $x$" from the rational numbers to algebraic numbers. $\sqrt{p}$ is "divisible by $p^{1/2}$" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\mathbb Q$ unramified at $p$).
Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)
• I realize you answered this question a loooong time ago, but if you're still around, how do you know that you can write the $n-$th root of $2$ as a linear combination of the $1st$ through $n-1$st roots of $2$? – ALannister Aug 17 '17 at 22:41
• @ALannister I realize you questioned quite a long time ago, but s/he is assuming that there is such $n$ and showing that such assumption leads to contradiction. So in fact, s/he is proving exactly that $n$-th root of $2$ cannot be expressed by a linear combination of previous roots – user160738 May 9 '18 at 17:24
For the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.
We'll need two propositions from Grillet's Abstract Algebra, page 335 and 640:
Proposition: $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$
Proof: Let $(q_n)_{n\in\mathbb{N_0}}$ be an enumeration of $\mathbb{Q}$. For $r\in\mathbb{R}$, take $$A_r:=\sum_{q_n<r}\frac{1}{n!}\;\;\;\;\text{ and }\;\;\;\;A:=\{A_r;\,r\in\mathbb{R}\};$$ the series is convergent because $\sum_{q_n<r}\frac{1}{n!}\leq\sum_{n=0}^\infty\frac{1}{n!}=\exp(1)<\infty$ (recall that $\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ for any $x\in\mathbb{R}$).
To prove $|A|=|\mathbb{R}|$, assume $A_r=A_{s}$ and $r\neq s$. Without loss of generality $r<s$, hence $A_s=\sum_{q_n<s}\frac{1}{n!}=\sum_{q_n<r}\frac{1}{n!}+\sum_{r\leq q_n<s}\frac{1}{n!}=A_r+\sum_{r\leq q_n<s}\frac{1}{n!}$, so $\sum_{r\leq q_n<s}\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.
To prove $A$ is $\mathbb{Q}$-independent, assume $\alpha_1A_{r_1}+\cdots+\alpha_kA_{r_k}=0\;(1)$ with $\alpha_i\in\mathbb{Q}$. We can assume $r_1>\cdots>r_k$ (otherwise rearrange the summands) and $\alpha_i\in\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\;(2)$; we'll increase $n$ two more times. The equality $n!\cdot(1)$ reads $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}+\cdots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads
$$-\alpha_1\sum_{\substack{m<n\\q_m<r_1}}\frac{n!}{m!}-\cdots-\alpha_k\sum_{\substack{m<n\\q_m<r_k}}\frac{n!}{m!}-\alpha_1 =\alpha_1\sum_{\substack{m>n\\q_m<r_1}}\frac{n!}{m!}+\cdots+\alpha_k\sum_{\substack{m>n\\q_m<r_k}}\frac{n!}{m!}. \tag*{(3)}$$
The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\cdots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ holds (such $n$ can be found since $\sum_{m=n+1}^\infty\frac{n!}{m!}=\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(n+2)\cdot\ldots\cdot m}\leq\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(m-n-1)!}\leq\frac{1}{n+1}\exp(1)\rightarrow 0$ when $n\rightarrow\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\text{RHS}(3)=0$. Thus $(3)$ reads $\alpha_1=-\sum_{i=1}^{k}\sum_{m<n,q_m<r_i}\alpha_i\frac{n!}{m!}=0\;(\mathrm{mod}\,n)$. If moreover $n>|\alpha_1|$, this means that $\alpha_1=0$. Repeat this argument to conclude that also $\alpha_2=\cdots=\alpha_k=0$.
Since $A$ is a $\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\mathbb{R}$ that contains $A$. Then $A\subseteq B\subseteq\mathbb{R}$ and $|A|=|\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\mathbb{R}|$, therefore $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$
• Is this uncountable linearly independent set basis ? – user195218 Jun 5 '15 at 2:53
• @user195218 It's not a basis, as it doesn't span the reals, but it is uncountable and linearly independent. – Akiva Weinberger Jul 25 '17 at 19:08
• The only proper answer to the question. Such a shame that it is not the most voted answer. – Akerbeltz May 1 at 21:42
Here is a simple proof that a basis $B$ of $[ \mathbb{R}:\mathbb{Q}]$ has cardinality $|\mathbb{R}|$.
Clearly since $B$ is contained in $R$, $|B| \le| \mathbb{R}|$.
But also $\mathbb{R}=span(B)$ and thus $|\mathbb{R}| =|span(B)| \le |\mathbb{Q}^B|=|B|$. The last equality follows because $B$ is not finite (if it was, then $|\mathbb{R}|=|span(B)| \le |\mathbb{Q}^B|=|\mathbb{N}|$, a contradiction).
Hence $|\mathbb{R}| \le |B| \le| \mathbb{R}|$, so $|\mathbb{R}|=|B|$
• There is an error in what you wrote: $|\mathbb{Q}^B|\geq |2|^{|B|} > |B|$. – Nex Aug 18 '17 at 8:19
• @Next: hmm so is it salvageable what I wrote? I guess I wasn't paying too much attention. – Joshua Benabou Aug 18 '17 at 19:08
Another simple proof:
Take $P=X^n-p$ for a prime $p$.
By Eisenstein's criterion, it is $\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\mathbb Q$.
Since $\mathbb R$ is bigger, it works for $\mathbb R$ too.
## protected by user26857Nov 25 '15 at 9:30
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? | 2019-09-23T19:32:56 | {
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https://strouden.com/n185f/archive.php?page=least-squares-solution-linear-algebra-8f1419 | \nLicense: Creative Commons<\/a>\n<\/p><\/div>"}, How to Find Least‐Squares Solutions Using Linear Algebra, consider supporting our work with a contribution to wikiHow. MathJax reference. wikiHow is where trusted research and expert knowledge come together. The minimum norm least squares solution is always unique. We can translate the above theorem into a recipe: Recipe 1: Compute a least-squares solution. 6Constrained least squares Constrained least squares refers to the problem of nding a least squares solution that exactly satis es additional constraints. In particular, it leads to the "least squares" method of fitting curves to collections of data. We use cookies to make wikiHow great. Why does regression use least “squares” instead of least “absolute values”? Least Squares. Recall the formula for method of least squares. Does a solution with a minimal norm mean it is a solution that minimizes the residuals? In this case, we're often interested in the minimum norm least squares solution. Parameter A can also be a set of equations that describe the linear least-squares problem. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I. Problems (PDF) Solutions (PDF) Further Study Eigenvalue Demonstrations* These demonstrations employ Java® applets with voice-over narration by Professor Strang. % In particular, it leads to the "least squares" method of fitting curves to collections of data. It is a set of formulations for solving statistical problems involved in linear regression, including variants for ordinary (unweighted), weighted, and generalized (correlated) residuals. This article has been viewed 4,467 times. How to draw random colorfull domains in a plane? This can be written in terms of an inequality ||y−Xβ^||2≤||y−Xβ||2,{\displaystyle ||\mathbf {y} -X{\hat {\boldsymbol {\beta }}}||^{2}\leq ||\mathbf {y} -X{\boldsymbol {\beta }}||^{2},} where we are minimizing the distance between y{\displaystyle \mathbf {y} } and Xβ. A = np.array([[1, 2, 1], [1,1,2], [2,1,1], [1,1,1]]) b = np.array([4,3,5,4]) Next, note that minimizing $\| b-Ax \|_{2}^{2}$ is equivalent to minimizing $\| b-Ax \|_{2}$, because squaring the norm is a monotone transform. Remember when setting up the A matrix, that we have to fill one column full of ones. Many calculations become simpler when working with a … On the other hand, if the system is underdetermined, there are infinitely many solutions and thus one can find a solution of minimal norm and this is called the minimum norm solution. Magic. The basic problem is to find the best fit straight line y = ax + b given that, for n 2 f1;:::; Ng, the pairs (xn;yn) are observed. 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https://math.stackexchange.com/questions/2141022/what-does-max-refer-to-in-the-definition-of-a-matrix-norm | # What does "max" refer to in the definition of a matrix norm?
I am going through my numerical analysis text (Epperson) and came across notation that I don't fully understand. My question is what does max refer to in this definition?
Definition $7.2$ (Matrix Norm): Let $\|\cdot\|$ be a given vector norm defined on $\mathbb{R}^n$. Define the corresponding matrix norm, for matrices $A \in \mathbb{R}^{n\times n}$, by $$\|A\| = \max_{x \neq 0}\frac{\|Ax\|}{\|x\|}$$
• Can you give the exact usage in your question? Feb 12 '17 at 16:35
• Hi David, that's what I'm trying to figure out too... the image in the link is of the definition of a matrix norm. I'm not sure how to read the "max x != 0" term. Feb 12 '17 at 16:37
• It's the maximum value of the expression as the vector $x$ goes through all possible values in $\mathbb{R}^n$ except $x=0$ since the denominator is not defined for this value. Feb 12 '17 at 16:40
• Thanks David for the edit. I'm going to try to wrap my head around this, I'm sure it's simpler than it seems. Feb 12 '17 at 16:56
• @Starcrossed not a problem. The concept will be easy once you wrap your mind around it. For some norms though, it might be very hard to compute. Feb 12 '17 at 17:09
This is simply the extension of a given norm to a matrix. It should be better written: $$\sup \left\{\frac{\|Ax\|}{\|x\|}: x \in \mathbb{R}^n, x\neq 0\right\} = \sup \left\{\|Ax\|:x \in \mathbb{R}^n, \|x\|=1\right\}.$$ That is, go through all normalized vectors, and see where the product $\|Ax\|$ is maximized.
-----Edit-----
As an example, let us consider the $L^1$-norm. For some matrix $A \in \mathbb{R}^n$, by definition we have that $$\|A\|_1 = \sup \left\{\|Ax\|_1:x \in \mathbb{R}^n, \|x\|_1=1\right\}.$$ Now, for a given vector $x \in R^n$, we have that $$\displaystyle\|Ax\|_1 = \left\|\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}\cdot \begin{pmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{pmatrix}\right\|_1 = \left\|\begin{pmatrix} \sum_{j=1}^na_{1j}x_j\\ \sum_{j=1}^na_{2j}x_j\\ \vdots\\ \sum_{j=1}^na_{nj}x_j \end{pmatrix}\right\|_1 = \sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|.$$ Thus, we can write the $L^1$-norm of $A$ as $$\displaystyle \|A\|_1 = \sup_{\|x\|=1}\left\{\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|\right\}.$$ We can bound this in the following way, $$\|A\|_1 \leq \sup_{\|x\|=1}\left\{\sum_{i=1}^n\sum_{j=1}^n\left|a_{ij}x_j\right|\right\} = \sup_{\|x\|=1}\left\{\sum_{j=1}^n\left|x_j\right|\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Now, consider the fact that $$\displaystyle \sum_{j=1}^n\left|x_j\right|\sum_{i=1}^n\left|a_{ij}\right| \leq \sum_{j=1}^n\left|x_j\right|\max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\} = \|x\|_1\cdot \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Now, since our matrix norm is only concerned with normalized vectors, we have clearly that $$\displaystyle \|A\|_1 \leq \sup_{\|x\|=1}\left\{\|x\|_1\cdot \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}\right\} = \max_{1\leq j \leq n}\left\{\sum_{i=1}^n\left|a_{ij}\right|\right\}.$$ Therefore, we have an upper bound. But this upper bound is clearly attainable, we simply take the $j^*$ corresponding to the column we are maximizing over in the previous equation, and let $x_{j^*} = 1, x_i = 0$ ($i \neq j^*$). Thus, the $L^1$-norm for a matrix is simply the maximum of the absolute column sums. So, as you see, the definition may be readily understood, but determining the value of the norm may be less so. And, in many cases, a closed form of the norm doesn't exist, and so you need to numerically solve the optimization (maximization) problem for each given $A$.
The "max" refers to the maximum of the set of numbers $\{\frac{\Vert Ax \Vert}{\Vert x \Vert } , 0\neq x\in \mathbb{R}^n\}$. Note that since this is a set of norms on vectors, then it's a set of non negative numbers.
• Aha okay I think I get it now... basically we are finding a vector x such that we are maximizing the norm, right? Feb 12 '17 at 17:00
• Yes. Just note that we're looking for an $x$ that maximizes the quotient of the norms $\frac{ \Vert Ax \Vert }{ \Vert x \Vert }$
– NSA
Feb 13 '17 at 9:23
It means the maximum of the quantity
$$\frac{||Ax||}{||x||}$$
along all the vectors $x\in\mathbb{R}^n$ that are not zero.
• Thanks Smurf... So to further understand, suppose we are given only one vector x for a linear expression Ax = b, then that vector alone would be the only possible vector to use in finding the max, right? Feb 12 '17 at 16:45
• No, the norm of the matrix is defined as above. If you have an expression $Ax=b$ you still should use the maximum along every $x\neq 0$. Feb 12 '17 at 16:47
• I'm sure it is probably trivially simple to you so I apologize in advance for this dumb question ... wouldn't every max x ≠ 0 in ℝn be infinitely large? Feb 12 '17 at 16:55
• Notice that while it may seem that if you pick and arbitrary large $x$, the numerator increases, so does the denominator. So one "controls" the other and the maximum exists (this has to be proven, I am just giving you the idea of why it works). Feb 12 '17 at 17:16
• @Starcrossed, when the space is infinite dimensional, the quotient can be unbounded. Mar 8 '17 at 10:14 | 2021-10-15T23:30:01 | {
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https://math.stackexchange.com/questions/2389599/expected-number-of-heads-on-coin-flip | Expected number of heads on coin flip?
Suppose you were to toss 2 coins each with different probabilities of landing on heads 2 times each, one at a time. So coin 1 tossed twice. Then coin 2 tossed twice. Let N be the total number of heads. To calculate E[N], would you find E[number of heads on coin 1's tosses] and E[number of heads on coin 2's tosses] and sum them together? And to find the var[N] would it be the same concept?
For expectation - yes due to linearity of expectation. As for variance, you can do the same since the each toss (or each event) are independent. Note that it does not matter whether you toss each coin 'one at a time' or the order you do it in due to independence, of course.
Formally, if you let $X_1$ and $X_2$ be the indicator variables for each toss of the first coin ($1$ for heads, $0$ for tails) and the same with $Y_1$ and $Y_2$ for the tosses of the second coin, we have:
$$N=X_1+X_2+Y_1+Y_2$$
Then
$$E(N)=E(X_1+X_2+Y_1+Y_2)$$
$$=E(X_1)+E(X_2)+E(Y_1)+E(Y_2)$$
due to linearity of expectation.
Also, as $X_1,X_2,Y_1,Y_2$ are independent events, we have
$$Var(N)=Var(X_1+X_2+Y_1+Y_2)$$
$$=Var(X_1)+Var(X_2)+Var(Y_1)+Var(Y_2)$$
The general rules being used here are:
$$E(aX+bY)=aE(X)+bE(Y)\tag{linearity of expectation}$$
and if $X$ and $Y$ are independent,
$$Var(aX+bY)=a^2Var(X)+b^2Var(Y)$$
• I just have a general question. I had a practice problem like this but with given probabilities for each coin. Let's assume N=X+Y. X is number of heads on coin 1, and Y is the number of heads on coin 2. I tried calculating the variance by doing E[(X+Y)^2] - E^2[(X+Y)] but received a negative number. But when I calculated the variances of each coin individually and then summing it, I received a positive number. Why does this happen? Basically I'm asking, why doesn't E[(X+Y)^2] - E^2[(X+Y)] =var(X)+var(Y)? Aug 10, 2017 at 21:18
• Could you give the probabilities in particular or edit your question to include your calculation? Aug 10, 2017 at 21:23
• Never mind, I see what I did wrong. When squaring E[(X+Y)^2], I did not calculate it correctly. Aug 10, 2017 at 21:35
• Letting the probabilities be $p_1$ and $p_2$, we have $E[(X+Y)^2] - (E[X+Y])^2=(0\times(1-p_1)^2(1-p_2)^2+1\times(1-p_1)^2p_2(1-p_2)+4\times(1-p_1)^2p_2^2+1\times p_1(1-p_1)(1-p^2)^2+4\times p_1(1-p_1)p_2(1-p_2)+9\times p_1(1-p_1)p_2^2+4\times p_1^2(1-p_2)^2+9\times p_1^2p_2(1-p_2)+16\times p_1^2p_2^2)-4(p_1+p_2)^2$ Aug 10, 2017 at 21:35
Long answer: It sounds like the two coins in your example do not affect each other in any way. In other words, it sounds like the number of heads on one is independent of how many heads happen for the other. For any two random variables, whether independent or not, we do indeed have that $E[X+Y]=E[X]+E[Y]$. Furthermore, when the variables are independent, we also have that $E[XY]=E[X]E[Y]$. This allows us to reason as follows. \begin{align} Var[X+Y] &= E[((X+Y)-E(X+Y))^2]\\ &=E[(X-E[X]+Y-E[Y])^2]\\ &=E[(X-E[X])^2] +2E[(X-E[X])(Y-E[Y])] + E[(Y-E[Y])^2]\\ &=Var[X] + Var[Y] + 2 E[XY -XE[Y] -YE[X] + E[X]E[Y]]\\ &= Var[X] + Var[Y] + 2(E[XY] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\ &= Var[X] + Var[Y] + 2(E[X]E[Y] -E[X]E[Y] - E[X]E[Y] + E[X]E[Y])\\ &= Var[X] + Var[Y] +2 \cdot 0\\ &= Var[X] + Var[Y] \end{align} So the variance of the sum of independent random variables is the sum of their variance. | 2022-05-20T02:24:27 | {
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http://math.stackexchange.com/questions/43231/implicit-differentiation-misunderstanding/43235 | # Implicit differentiation misunderstanding
I'm trying to see why my textbook's solution is correct and mine isn't.
"Find an expression in terms of $x$ and $y$ for $\displaystyle \frac{dy}{dx}$, given that $x^2+6x-8y+5y^2=13$
First, the textbook's solution, which I understand and agree with fully:
Now my similar solution, for which I don't see my error:
Differential: $$2x+6-8\frac{dy}{dx}+10y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(10y-8)=-2x-6$$ $$\frac{dy}{dx}=\frac{-2x-6}{10y-8}=\frac{-x-3}{5y-4}$$
So I end up with the negative of the correct solution, because I moved the $(2x+6)$ to the RHS and the textbook author moved the other part instead. I would have thought it would produce an equivalent answer?
Thanks!
-
HINT $\rm\displaystyle\ \ \frac{-A}{-B}\ =\ \frac{A}B\qquad$ – Bill Dubuque Jun 4 '11 at 18:14
As you now know, your answer is correct. But there is kind of a convention in writing mathematical expressions, that fewer minus signs is generally better. – André Nicolas Jun 4 '11 at 18:22
@user6312 duly noted, thanks! :) – Danny King Jun 4 '11 at 18:24
Fractions can be written multiple ways.
$$\frac{-x-3}{(5y-4)} = \frac{x+3}{4-5y}$$
In general $$\frac{a-b}{c-d}=\frac{b-a}{d-c}$$
This is just multiplying both the top and the bottom by −1. In other words, your answer and the books differ by multiplication of $$\frac{-1}{-1} = 1$$
-
Oh! When you put it like that, it becomes very clear. Thank you! – Danny King Jun 4 '11 at 18:05
Your answers agree. Note that: $$\frac{-x-3}{5y-4} = \frac{-(x+3)}{-(-5y+4)} = \frac{x+3}{4-5y}.$$
-
I don't know, what's wrong with your solution. Notice, that
$${-x-3\over5y-4}={3+x\over4-5y}$$
- | 2015-07-06T07:14:31 | {
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http://math.stackexchange.com/questions/253257/number-of-simple-edge-disjoint-paths-needed-to-cover-a-planar-graph | # Number of simple edge-disjoint paths needed to cover a planar graph
Let $G=(V,E)$ be a graph with $|E|=m$ of a graph class $\mathcal{G}$. A path-cover $\mathcal{P}=\{P_1,\ldots,P_k\}$ is a partition of $E$ into edge-disjoint simple paths. The size of the cover is $\sigma(\mathcal{P})=k$. I am interested in upper and lower bounds for the quantity
$$\max_{G\in \mathcal{G}} \quad \min_{\mathcal{P} \text{ is path-cover of G}} \sigma(\mathcal{P})/m.$$
In other words, how large is the inverse of the average path length in the smallest path-cover in the worst case.
The graphs classes $\mathcal{G}$ I am interested in are (1) planar 3-connected graphs, and (2) triangulations.
There is a simple observation for a lower bound: In every odd-degree vertex one path has to start/end. So when all vertices have odd degree there any path-cover needs at least $m/6+1$ paths (a triangulation has $3|V|-6$ edges). You get the same result by noticing, that $(n-1)/2$ paths have to pass through a vertex of degree $n-1$.
Is a path-cover with $\frac{m}{6}$ paths always possible for planar graphs?
I am aware of a few results covering planar graphs with paths of length $3$.
Here is an example of a path-cover of the graph of the icosahedron.
-
This is a great question; don't know why it hasn't received more attention or upvotes. – joriki Dec 12 '12 at 10:10
I don't understand the lower bound. Since a triangulation has $m=3n-6$ edges, a triangulation with all odd-degree vertices requires at least $n/2=(m+6)/6=m/6+1$ paths. For instance, $K_4$ needs $2$ (not $1$) paths to cover it. – mjqxxxx Dec 13 '12 at 18:28
I am not sure if I would be able to do it, but somebody should try generating a big list of $\sigma(\mathcal{P})$'s for a bunch of the graphs in question. Maybe some data will help us see a pattern. – Alexander Gruber Dec 19 '12 at 2:40
@A.Schulz Maybe the lower bound $m/6$ is weak for some classes of graphs. For instance, for the wheel graphs $W_n$, consisting of the cycle of length $n-1$ and the additional center vertex joined with all vertices of the cycle. Also there are Moon-Moser graphs $T_k$ (see Section 2 of the article “Long Cycles in 3-Connected Graphs” by Guantao Chen and Xingxing Yu), having no cycles of length greater than $(7/2) n^{\log_3 2}$. Maybe these graphs also have no small path-covers. – Alex Ravsky May 27 '13 at 3:47
A related question is that of Linear Arboricity, but the results for it are based upon the maximum degree of vertices. See for instance mimuw.edu.pl/~kowalik/papers/LinearArboricity.pdf – jp26 Jan 27 at 21:26
I will show that $\frac{m}{6}$ is not enough for 3-connected planar graphs. The following is inspired by fedja's comment (and also some nice discussion with Krystal Guo).
Consider the prism graph, which we will denote $P_n$. We say it has $2n$ vertices and $3n$ edges. Then in this case $\frac{m}{6} = \frac{n}{2}$. We will show that $\sigma(P_n) =n$. It is clear that $n$ paths is enough.
Consider the following graphs which we denote $T_n$:
and so on.. Note that $T_n$ has $3n - 1$ edges and $2n+2$ vertices.
It is not too hard to show that $\sigma(T_n) = n+1$ (and this is the best possible). One can show by induction, considering the left most edge between the upper and lower paths.
Let $p_1 , \ldots , p_k$ be a path cover of $P_n$. Now $P_n$ is two copies of $C_n$ with a matching, $M$, in between them. If every $e \in M$ is actually some $p_i$, then we have $k \geq n+2$.
So we may assume there is a $e = (u,v)$ such that the path, $p_i$, that contains $e$, contains at least one more edge, wolog say $(v,w)$.
$(i)$ The vertex $u$ is also contained in another edge of $p_i$ (other than $e$). Then $(V(P_n) , E(P_n) \setminus E(p_i))$ is $T_n$ minus at most two paths (actually, exactly two: the two paths start with precisely the two edges incident to $e$, respectively). In this case $k \geq n+1 - 2 + 1 = n$ (the last $+1$: don't forget to count $p_i$).
$(ii)$ The vertex $u$ is not contained in any other edge of $p_i$. Then let $p_j$ be another path that contains $u$. Then $(V(P_n) , E(P_n) \setminus (E(p_i)\cup E(p_j))$ is a $T_n$ minus at most 3 paths ($p_i$ is one path in $T_n$ and $p_j$ is at most two paths in $T_n$). It follows that $k \geq n + 1 - 3 +2= n$.
Thus we obtain $\sigma(P_n) = n$, as desired.
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https://web2.0calc.com/questions/im-stuck-in-this-koobits | +0
im stuck in this koobits !
0
412
5
Find the sum of the following series by pairing two numbers to make ten(s). 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Guest Nov 20, 2015
#2
+18927
+15
Find the sum of the following series by pairing two numbers to make ten(s).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
$$\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \hline \hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\ \hline \hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\ \hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\ \hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\ \hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\ \hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\ \hline \hline & & & & & & & & & & & & & & & & & & & =& 100\\ \hline \hline \end{array}$$
( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100
heureka Nov 20, 2015
edited by heureka Nov 20, 2015
edited by heureka Nov 20, 2015
Sort:
#1
+5
(1+9) + (3+7) + (5+15) + (7+13) + (11+19)
10 + 10 + 20 + 20 + 30 = 90
Guest Nov 20, 2015
#2
+18927
+15
Find the sum of the following series by pairing two numbers to make ten(s).
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
$$\begin{array}{rcrcrcrcrcrcrcrcrcrcr} \hline \hline 1 &+& 3 &+& 5 &+& 7 &+& 9 &+& 11 &+& 13 &+& 15 &+& 17 &+& 19 & \\ \hline \hline 1 & & & & & & & & & & & & & & & & &+& 19 & =& 20 \\ \hline & & 3 & & & & & & & & & & & & &+& 17 & & & =& 20 \\ \hline & & & & 5 & & & & & & & & &+& 15 & & & & & =& 20 \\ \hline & & & & & & 7 & & & & &+& 13 & & & & & & & =& 20 \\ \hline & & & & & & & & 9 &+& 11 & & & & & & & & & =& 20 \\ \hline \hline & & & & & & & & & & & & & & & & & & & =& 100\\ \hline \hline \end{array}$$
( 1 + 19 ) + ( 3 + 17 ) + ( 5 + 15 ) + ( 7 + 13 ) + ( 9 + 11 ) = 20 + 20 + 20 + 20 + 20 = 100
heureka Nov 20, 2015
edited by heureka Nov 20, 2015
edited by heureka Nov 20, 2015
#3
+91773
0
What a fabulous diagram Heureka,
Melody Nov 20, 2015
#4
0
Two great answers, but one of them must be wrong.
The first says 90, the second 100.
Guest Nov 20, 2015
#5
0
Yep One of them is wrong....and it's my answer the answer is 100...I made a typo. The 4th term in my sequence should read (17+13) NOT (7 +13)
Guest Nov 21, 2015
5 Online Users
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https://wikimho.com/us/q/astronomy/6384 | ### How fast is a comet moving when it crosses Earth's orbit?
• Is it about the same as Earth's orbital speed?
It depends. An object's speed is related to the length of it's orbit, and this varies greatly. See Kepler's Laws.
• Walter Correct answer
8 years ago
1. Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun.
2. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy
$$E = \frac{1}{2}v^2 - \frac{GM_\odot}{r},\qquad\qquad(*)$$
which is conserved along the orbit ($v$ and $r$ the Heliocentric speed and distance). For a returning comet, $E<0$ and the speed cannot exceed the escape speed (which occurs for $E=0$)
$$v_{\rm escape}^2 = 2\frac{GM_\odot}{r}.$$
The speed of the Earth can be worked out from the Virial theorem, according to which the orbital averages of the kinetic and potential energies, $T=\frac{1}{2}v^2$ and $W=-GM_\odot/r$, satisfy
$2\langle T\rangle + \langle W\rangle=0.$
For a (near-)circular orbit (such as Earth's), $r$ is constant and we have
$v^2_{\rm Earth} = GM_\odot/r.$
Thus, at $r$=1AU
$$v_{\rm escape} = \sqrt{2} v_{\rm Earth}$$
as already pointed out by Peter Horvath. Non-returning comets have local speed exceeding the escape speed.
3. Can a comet near Earth have a speed similar to Earth's orbital speed?. Let's assume a comet with the same speed as Earth at $r$=1AU and work out the consequences. Such a comet must have the same orbital energy as the Earth and, since
$$E = -\frac{GM_\odot}{2a}\qquad\qquad(**)$$
with $a$ the orbital semimajor axis, must also have $a=1AU$ and the same orbital period as Earth, i.e. one year. Moreover, the comet's apohelion satisfies
$$r_{\rm apo}\le 2a = 2{\rm AU}.$$
Such comets don't exist AFAIK. Most returning comets have much longer periods than 1 year.
4. What's the typical speed of a returning comet when at distance 1AU from the Sun? To work out this question, let's parameterise the comet's orbit by its period $P=2\pi\sqrt{a^3/GM_\odot}$. From this relation we immediately get
$$\frac{a_{\rm comet}}{\rm AU} = \left(\frac{P_{\rm comet}}{\rm yr}\right)^{2/3}.$$
From equations ($*$) and ($**$), we can then find
$$v_{\rm comet}(r=1{\rm AU}) = \sqrt{2-\left(\frac{P_{\rm comet}}{\rm yr}\right)^{-2/3}}\,v_{\rm Earth}.$$
In the limit of $P\to\infty$, this recovers our previous result $v_{\rm comet}\to v_{\rm escape}$. For typical period of $\sim$70yr, the speed of the comet is close to this number.
5. Finally, I shall coment that all this only relates to the magnitude of the orbital velocity (speed), but not to its direction. Comets are typically on highly eccentric orbits and, when at $r$=1AU, move in quite a different direction than Earth, even if their speed is only slightly larger. So the relative speed $|\boldsymbol{v}_{\rm comet}-\boldsymbol{v}_{\rm Earth}|$ of a comet with respect to Earth can be anyting between about 10 and 70 km/s.
This is a great answer! You're better than Wolfram! This is exactly the information and tools that I was looking for. Thanks very much. Doug.
who is Wolfram?
@Walter the answer to "**Who** is Wolfram" is either Stephen Wolfram or perhaps Wolfram Research. But of course user38715 was referring to wolframalpha.com
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http://math.stackexchange.com/questions/215515/solution-of-x2-sx-cdot-x-n-0-with-sx-is-the-sum-of-digits-of-x/215701 | # Solution of $x^2 + s(x)\cdot x - n = 0$, with $s(x)$ is the sum of digits of $x$.
This problem comes from an programming competition website, but I'd interested in analyze it from mathematics prespective. Given this problem below, we must create a program that could give us the correct output from the given input. The program must run under the time limit that provided by the system.
## The Problem
Let define function $s$ that take input from positive integers, such that $s(x)$ is the sum of the digit of $x$.
For example, $$s(1024) = 1 + 0 + 2 + 4 = 7$$
Given some integer constant $n$ as the input of the program. Now, find positive integer solution of $x$ in equation
$$x^2 + s(x) \cdot x - n = 0.$$
If more than one solution exist, choose the lowest one. If the solution doesn't exist, give output -1.
Time Limit of the program: 2 second.
Range of the input $n$: $[1, 10^{18}]$
## The Solution
In finding the solution, we could iterate from some range of possible value of x, and try each of these value whether it satisfy the equation or not.
To improve the run time of the program, we must choose the shortest range for x.
## My Question
1. What is the shortest range of possible value of x that we could iterate to find the answer of the problem above?
2. Is there a better way to find the solution without iterating from the set possible value of x?
## My Attempt:
Consider that $$x^2 + s(x) \cdot x - n = 0$$ $$\Leftrightarrow x ( x + s(x) ) = n$$
and since both $x$ and $(x + s(x))$ are positive integers and the product of both number is $n$, we can conclude that $x$ and $(x + s(x))$ is a pair of positive factor of $n$.
Since $x$ is a factor of $n$, we now have the first possible range of $x$, that is $[1, n]$.
But, iterating $x$ over this range gives bad performance to the program. For small value of $n$, it won't matter, but for $n \geq 10^{17}$, the program will reach it's time limit.
If I could recall correctly, we have this theorem
$$x_1 \cdot x_2 = x$$
and
$$x_1 \leq x_2$$
implies that
$$x_1 \leq \sqrt{x} \leq x_2$$
And since $x (x + s(x)) = n$ and $x \leq x + s(x)$
we have $x \leq \sqrt{n} \leq x + s(x)$
Now we have the second range: $[1, \sqrt{n}]$
But, still, this range also give bad performance for larger $n$.
My last attempt is by considering the biggest possible value of $s(x)$, that is the digit of $x$ is all $9$ and $x$ less than or equals to the maximum value of $n$, $10^{18}$.
$$s(x) \leq s(999,999,99 \cdots 9,999) = 9 \cdot 18 = 162$$
Hence, we have
$$x + s(x) \leq x + 162$$
so that, $$\sqrt{n} \leq x + s(x) \leq x + 162 \leq \sqrt{n} + 162$$ hence, $$\sqrt{n} - 162 \leq x \leq \sqrt{n}$$
Now, we have better range that will suffice for all value of $x$. $$l_1 = \max \{ 1, \sqrt{n} - 162 \}$$ $$l_2 = \sqrt{n}$$ the range is $[l_1, l_2]$
But, I though there must be a better solution to this problem. Hence I try $$l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$$ And that range is also correct. But I can't justify my finding.
So, hence my third question:
1. Why $l_1 = \max \{ 1, \sqrt{n} - 162/2 \}$ also gives us correct solution?
-
While not a full answer to your questions, here is a trick that will allow you to reduce the collection of numbers you have to try (and so an answer to question 2).
Because $10\equiv 1 \pmod 9$, we have $10^k \equiv 1 \pmod 9$ for every $k\in \mathbb N$. If we write a number in terms of its digits, $x=\sum a_k 10^k$, then since $\sum a_k 10^k \equiv \sum a_k \pmod 9$, we have $s(x)\equiv x \pmod 9$. This fact can be used to give a way to check arithmetic, among other things.
Therefore, if $x^2 + s(x)x = n$, then $2x^2 \equiv n \pmod 9$. Since $(2)(5)=10 \equiv 1 \pmod 9$, we can multiply to get $x^2 \equiv 5n \pmod 9$.
The squares modulo 9 are $0, 1, 4, 7$, so unless the remainder of dividing $5n$ by $9$ is one of these numbers, there are NO solutions.
If the remainder is one of those numbers, you need to know the square roots modulo 9 to know what values of $x$ to try.
For this, we have the following.
$0^2\equiv 3^2 \equiv 6^2 \equiv 9 \pmod 9$, $1^2\equiv 8^2 \equiv 1 \pmod 9$, $2^2\equiv 7^2 \equiv 4 \pmod 9$ and $4^2 \equiv 5^2 \equiv 7 \pmod 9$.
The upshot of this is that if $n$ is a multiple of $9$, then you only have to check the multiples of $3$ (reducing the number of things you have to check by a factor of $3$). Otherwise, depending on the modulus of $5n\bmod 9$, you either have to check NO numbers (5 of the remaining cases), or you have to check $2/9$ of the numbers (3 of the remaining cases). Thus, whatever range you are working in, you have at least a speed up by a factor of $3$, if not more.
Another comment, which gives a small speedup over what you already have but answers question 3. As you observed, $x<\sqrt n$, and the number of digits of $x$ is at most $\lceil \log_{10} x \rceil$, and since $s(x)$ is at most $9$ times the number of digits of $x$, we have $s(x)<9 \lceil \log_{10} \sqrt n \rceil=9\lceil \frac{1}{2}\log_{10} n \rceil$.
Therefore, given $n$, you can limit your search for $x$ to $[\sqrt n - 9\lceil \frac{1}{2}\log_{10} n \rceil, \sqrt n]$. The factor of $1/2$ here answers question 3. Note, we could probably shave off $1$ or $2$ numbers from the range by being more careful with rounding, but I don't see any obvious reason we could shave off more. If we combine this with the beginning of my answer, you have to check at most $3\lceil \frac{1}{2}\log_{10} n \rceil$ numbers to find your solution, if it exists, and no numbers a good portion of the time when no solution exists.
-
I think $(2)(5)=10 \equiv 1 \pmod 5$ in pharagraph 3 should be $(2)(5)=10 \equiv 1 \pmod 9$. – Kamal Hajjaj Isa Oct 28 '12 at 14:58
@KamalHajjajIsa: Yes, that is correct. Thank you. – Aaron Oct 29 '12 at 3:07
I like the Idea of using $\pmod 9$. – Kamal Hajjaj Isa Oct 29 '12 at 4:56
Since $x<\sqrt n$ and $n\le10^{18}$, we have that $x<10^9$. Then $x$ has at most $9$ digits and $s(x)\le9\times9=81$. This implies that $x^2+81\,x-n\ge0$. It follows easily that we must have $$x\ge\frac{-81+\sqrt{81^2+4\,n}}{2}=\sqrt{n+(81/2)^2}-81/2.$$
-
This only answers the third question
If you use the general quadratic equation formula for the equation $ax^2+bx+c=0$ $$\displaystyle x= \frac {-b \pm \sqrt{b^2-4ac}}{2}$$ It follows from $x^2 + s(x)\cdot x - n = 0$ that $$\displaystyle x= \frac {-s(x) \pm \sqrt{s^2(x)+4n}}{2}$$ Since all the variables are positive and obviously $\sqrt{s^2(x)+4n}>s(x)$ , one can remove the negative sign and have $$\displaystyle x= \frac {-s(x) + \sqrt{s^2(x)+4n}}{2}$$ If one extends the range of values of $n$ to $0$, we have that the lowest possible value of $s(x)$ is $0$ and that $s(x) \ge 0$ which implies that $$\displaystyle x \ge \frac {-s(x) + \sqrt{4n}}{2} = \frac {-s(x)}{2} + \frac {2\sqrt{n}}{2} \ge \frac {-162}{2} + \sqrt n, \space \space (-s(x) \ge -162)$$ and there you have your inequality $$x \ge \sqrt n - \frac {162}{2}$$
Extending the range to $0$ doesn't change the validity of the inequality and helps to simplify it. You've been working with the maximum value of $s(x)$, using the minimum value helps in this case.
-
Beside the excellent modulo 9 tricks Aaron gave you, if $n$ passes his tests the easiest thing is to try the range of $1 \le s(x) \le 162$ with the proper $s(x) \pmod 9$. Just plug each one into the quadratic formula and solve for $x$. If it doesn't come out integral, go on to the next. If it does, check the sum of digits against the assumed $s(x)$. If it matches, success and report it. If not, go on to the next. At most $50$ numbers to try, which should fit in 2 seconds easily. | 2013-12-21T03:49:28 | {
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http://mexproject.it/zcjm/probability-problems-and-solutions.html | # Probability Problems And Solutions
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Standard probability themes like coins, spinners, number cubes, and marbles are featured along with some other situations. The bet can either be on red or on blue and the amount of each bet is arbitrary. The probability of any event E is number of outcomes in E total number of outcomes in the sample space This probability is denoted by P E n E n S This probability is called classical probability and it uses the sample space S. Then, if x is the number of blue balls in urn 2,. Example Question #2 : How To Find The Probability Of An Outcome If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?. Geometric Probability. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Find the probability that they are of same colour. when the selection of one does not affect the probability of a…. Conditional Probability In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent. • Probability and Statistics for Engineering and the Sciences by Jay L. At a picnic, Julio reaches into an ice-filled cooler containin…. The solutions are aimed to make your understanding clear. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Let E 1,E 2,E 3 be events. (Solution): Probability Problems. Sample Space. You need at most one of the three textbooks listed below, but you will need the statistical tables. The function f ( x ) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x -axis is equal to `1. Solutions Manual A First Course in Probability 9th Edition Sheldon Ross. Solution: Gambler’s Ruin Problem. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. "} BarChart3D[{win, lose}] The probability of winning is 30. We write P(AjB) = the conditional probability of A given B. If X and Y are independent uniform (0;1) random variables, show that [Filename: Assignment7_solutions I. Two Games in a Row. To navigate from one page of exercises to another, you will use the right navigation bar. If the first four marbles drawn are red, what is the probability the next marble drawn will not be red? 7. The probability that a student belongs to a club is P(C)=0. 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P (A n B) = A and B = A x B P (A u B) = A or B = A + B. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Feel Free to Learn 316,387 views. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. Chapter 4 - Continuous Random Variables and Probability Distributions. This section provides materials for a lecture on discrete random variables, probability mass functions, and expectations. Multiplication Rule of Probability The addition rule helped us solve problems when we performed one task and wanted to know the probability of two things happening during that task. The bet can either be on red or on blue and the amount of each bet is arbitrary. Q2) Generate a Punnett Square for a heterozygous individual crossed with a heterozygous individual. What Are Example Statistics and Probability Problems and Their Solutions? Credit: Echo/Cultura/Getty Images Two examples of probability and statistics problems include finding the probability of outcomes from a single dice roll and the mean of outcomes from a series of dice rolls. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. Practice calculating conditional probability, that is, the probability that one event occurs given that another event has also occurred. 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It takes a computational approach, which has several advantages: • Students write programs as a way of developing and testing their un-derstanding. Chapter 1 covers this theory at a fairly rapid pace. Find the probability that the number on the drawn ticket is a multiple of 3 or 7. We break down problems on video in a step-by-step easy to follow format. Finding the Median in a pdf : S2 Edexcel. The probability of two or more heads is, therefore: Probability = 4 8 = 1 2 We solved this problem by first enumerating the set of possibl e outcomes, known as the Sample Space, and then by deciding which of these outcomes satisfied the cr iterion of containing 'two or more heads'. Certainly X is binomial with n = 40 and p = 0. More Problems on probability and statistics are presented. P(X) gives the probability of successes in n binomial trials. Example of Using a Contingency Table to Determine Probability. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. Generate integer from 1-8 with equal probability. Problem 4 (Solution on p. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. Calculate the value of a in a normal distribution with a mean of 4 and a standard deviation of 2 for which:. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Review of probability theory • Definitions (informal) -Probabilities are numbers assigned to events that indicate "how likely" it is that the event will occur when a random experiment is performed -A probability law for a random experiment is a rule that assigns probabilities to the events in the experiment. The actual probability of finding the particle is given by the product of the wavefunction with its complex conjugate (like the square of the amplitude for a complex function). Pedigrees and probability problems 1-28 and 1-30. It includes the list of lecture topics, lecture video, lecture slides, readings, recitation problems, recitation help videos, tutorials with solutions, and a problem set with solutions. Chapter 1 covers this theory at a fairly rapid pace. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. a Show that $$f\left( x \right)$$ is a probability density function. the number of ways several people can stand in a line if 2 particular people must — or must not — stand next to each other). (321 problems) IMO Shortlisted Problems. Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3. Playing cards probability problems based on a well-shuffled deck of 52 cards. Chegg Solution Manuals are written by vetted Chegg Statistics And Probability experts, and rated by students - so you know you're getting high quality answers. Probability puzzles require you to weigh all the possibilities and pick the most likely outcome. The post in the companion blog shows how to evaluate the covariance and the correlation coefficient of two continuous random variables and. Importantly, the solution for the problem of points became known as an expectation value (the average expected value). Suppose that a good part fails within the first year with probability 0. This is essentially a 1D scattering problem. At a picnic, Julio reaches into an ice-filled cooler containin…. 1 Problem 4E. Drawing/Picking/Choosing Single/One ball from a bag/urn/box - Probability - Problems Solutions. 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Solution the the Second WBC Problem-page 10. Multiplicative Principle for Counting n The total number of outcomes is the product of the possible outcomes at each step in the sequence n if a is selected from A, and b selected from B… n n (a,b) = n(A) x n(B) q (this assumes that each outcome has no influence on the next outcome) n How many possible three letter words are there? q you can choose 26 letters for each of the three positions, so. GMAT Advanced Probability Problems By Mike MᶜGarry on January 3, 2014 , UPDATED ON January 15, 2020, in GMAT Math In the following probability problems, problems #1-3 function as a set, problems #4-5 are another set, and problems #6-7 are yet another set. by electronjohn. Homework 2 from last time OLD HW2 solutions Practice midterm and Midterm 2015 solutions. Solution; Determine the value of $$c$$ for which the function below will be a probability density function. Test your understanding with practice problems and step-by-step solutions. Probability theory is the most directly relevant mathematical background, and it is assumed that the reader has a working knowledge of measure-theory-based probability theory. Midterm 2016 solutions Combinatorics practice problem set and solutions. Bayes' theorem describes the probability of occurrence of an event related to any condition. For instance, let’s say the probability that you will get to class on time is 2/3. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. Finding the Median in a pdf : S2 Edexcel. Thanks for the A2A. At a picnic, Julio reaches into an ice-filled cooler containin…. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. 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PROBABILITY AND STATISTICS FOR ENGINEERS LESSON INSTRUCTIONS The lecture notes are divided into chapters. Probability theory - Probability theory - The birthday problem: An entertaining example is to determine the probability that in a randomly selected group of n people at least two have the same birthday. Probability is the chance that an event will occur. Solution: (F) Probability always lie within 0 to 1. Probability_Problems Based on Coins#LESSON-2 - Duration: 17:39. If we toss a fair die, what. $$∴ P(E)={n(E)\over n(S)} = {2\over52}={1\over26}$$ Learn and practice these probability problems with solutions and you can ask me anything in the comment section if you have any doubts regarding probability problems and it's solutions. To get a feeling for PDF, consider a continuous random variable. To link to this page, copy the following code to your site:. The text-books listed below will be useful for other courses on probability and statistics. Topics covered range from measure and integration theory to functional analysis and basic concepts of probability; the interplay between measure theory and topology; conditional probability and expectation; the central limit theorem; and strong laws of large numbers in terms of martingale theory. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. to access the Math Probability menu. Finding the Median in a pdf : S2 Edexcel. Practice Problems - Try specific problems and see the solution. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. PROBLEM 13 : Consider a rectangle of perimeter 12 inches. Conroy and a 5=6 chance of not rolling a six, after which the number of rolls we expect to throw is the same as when we started. This Collection of problems in probability theory is primarily intended for university students in physics and mathematics departments. Finding the mean in a Normal Distribution : Statistics S1 Edexcel June 2013 Q6 (a) ExamSolutions - youtube Video. It can be a finite set, a countable set (a set whose elements can be put in a sequence, in fact a finite set is also. Most probability problems on the GRE involve independent events. A problem is given to three students whose chances of solving it are 1/2,. "} BarChart3D[{win, lose}] The probability of winning is 30. The process to solve the problem is rarely straightforward and takes practice to perfect. This is a problem. No need to worry if you miss class or just need some extra help. According to recent data, the probability of a person living in these conditions for 30 years or. Colavito’s homepage for solutions. The initial table listed in problem description is a table of probabilities p(1), p(2), , p(20). Total Number of ways = 2 5 = 32. The point is that the order of events doesn't affect with respect to conditional probability. MIDTERM 2 solutions and Histogram. The incubation periods of a random sample of 7 HIV infected individuals is given below (in years): 12. Advanced Techniques. A Problem With Pearls. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. The probability mass function of is but and Therefore, the probability mass function can be written as which is the probability mass function of a Bernoulli random variable. Then, n(E) = 2. Rolling the Dice. The empirical probability of three consecutive female births is. 1, Median 22. ) For a different problem, allow every one of n people to place an even bet on the color of his hat. In the Options dialog box, select the Show Iteration Results check box to see the values of each trial solution, and then click OK. Most of the problems require very little mathematical background to solve. Problem Solving: Find a Pattern What Is It? Finding a Pattern is a strategy in which students look for patterns in the data in order to solve the problem. The probability (chance) is a value from the interval 0;1> or in percentage (0% to 100%) expressing the occurrence of some event. P (A n B) = A and B = A x B P (A u B) = A or B = A + B. DISCRETE PROBABILITY DISTRIBUTIONS 27. Miracle Mountain. Geometric Probability. Life or Death? The Emperor's Proposition. We want to generate a random number between 100 and 499. The probability of a success, denoted by p, remains constant from trial to trial and repeated trials are independent. Probability Questions and answers PDF with solutions. Probability MCQ Questions and answers with easy and logical explanations. Please do not email me completed problem sets. Multiplying by multiples of 10. John has a special die that has one side with a six, two sides with twos and three sides with ones. At any particular time period, both outcomes cannot be achieved together so […]. Solution of exercise 1. This is a beta distribution. We want to generate a random number between 100 and 499. The probability equals 31,8 %. The question is incomplete, that is, it doesn’t provide information regarding with the level of probability that you are familiar with. There is one desired outcome and six possible outcomes. random variables, and some notation. Normal distribution probability : Statistics S1 Edexcel June 2013 Q6 (b) : ExamSolutions - youtube Video. 74% of the X values are less than three standard deviations from the mean. (Opens a modal) Random number list to run experiment. Practice Problems - Try specific problems and see the solution. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and. Solution : Sample space : {(1, 1)(1, 2)(1, 3)(1. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a first course in prob-ability. Each trial results in an outcome that may be classified as a success or a failure (hence the name, binomial);. Playing cards probability problems based on a well-shuffled deck of 52 cards. com: Many students are also using our Free Statistics Lab Manual. I require the probability that after these n stretches he is at a distance between rand r+ rfrom his starting point 0. A Music Survey was carried out to find out what types of music a group of people liked. Assume there is a beam of particles with definite momentum coming in from the left and assume there is no flux of particles coming from the right. One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value. The Probability Distributome Project provides an interactive navigator for traversal, discovery and exploration of probability distribution properties and interrelations. Listed in the following table are practice exam questions and solutions, and the exam questions and solutions. 84 = (x – 70)/2. Let E = event of getting a queen of club or a king of heart. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Miracle Mountain. Hone your skills with this array of statistics and probability worksheets for 7th grade to compute the average on whole numbers and decimals, find the probability on a pair of coins, a pair of dice, months and more. Solution of exercise 1. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. Although they never finished their game, their letters contain solutions to both problems. Home > Math> Probability and Statistics>Probability Problems. Homework problems usually do not say which concepts are involved, and often require combining several concepts. Probability problem on Dice. Most probability problems on the GRE involve independent events. Probability MCQ Questions and answers with easy and logical explanations. After you define a problem, click Options in the Solver Parameters dialog box. The problem concerns a game of chance with two players who have. Example: there are 5 marbles in a bag: 4 are. Probability Questions and answers PDF with solutions. In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the. These you may need to complete at home. Statistics Problems With Solutions Return to Statistics Internet Library for videos, software assistance, more problems and review. Cards of Spades and clubs are black cards. pdf] - Read File Online - Report Abuse. A Collection of Dice Problems Matthew M. 316 that an audit of a retail business. The first section provides a brief description of some of the basic probability concepts that will be used in the activities. Trials until first success. Walpole Raymond H. What is the probability of drawing a red Bingo chip at least 3 out of 5 times? Round answer to the nearest hundredth. Probability Problems With Solutions Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Probability Word Problems (Simplifying Math) What are the chances that your name starts with the letter H? Find out how to make that calculation and many more. Advanced Techniques. The problem concerns a game of chance with two players who have. Probability shortcut Tricks Pdf, Probability MCQ, Probability Objective Question & Answer Pdf. [email protected]{"The probability of winning is ", 100 win/(win + lose) // N, " %. Total English alphabets = 26. The flippant juror. For example: if we have to calculate the probability of taking a blue ball from the second bag out of three different bags of balls, where each bag contains three different colour balls viz. a Show that $$f\left( x \right)$$ is a probability density function. Make use of probability calculators to solve the probability problems with ease. What is the probability that players will experience rain and a temperature between 35 and 50 degrees? 0. ’ ‘There’s a 30% chance of rain tomorrow. He offers you the following game. 1) Basics of Electromagnetics Part I – Download MCQs from here. Probability Solution: (a) Since there are 4 queens ( ) = ( ) ( ),. (Solution): Probability Problems. Probability And Statistics Problems Solutions Author: persepolis. It is assessed by considering the event's certainty as 1 and impossibility as 0. edu-2020-04-20T00:00:00+00:01 Subject: Probability And Statistics Problems Solutions Keywords: probability, and, statistics, problems, solutions Created Date: 4/20/2020 7:35:45 AM. nthe sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin. AU - Sparrow, E. Probability Questions are provided with detailed answers to every question. probability problems, probability, probability examples, how to solve probability word problems, probability based on area, examples with step by step solutions and answers, How to use permutations and combinations to solve probability problems, How to find the probability of of simple events, multiple independent events, a union of two events. Competitive exams are all about time. The concept is very similar to mass density in physics: its unit is probability per unit length. Most of these problems require very little mathematical background to solve. Here is a set of 14 GMAT probability questions, all in the Problem Solving style on the test, collected from a series of blog articles. Examples include the Monty Hall paradox and the birthday problem. Write down twenty math problems related to this topic on a page. If there is a chance that an event will happen, then its probability is between zero and 1. Let Xand Y be two N 0-valued random variables such that X= Y+ Z, where Zis a Bernoulli random variable with parameter p2(0;1), independent of Y. Linear Algebra Igor Yanovsky, 2005 2 Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please solve the following probability practice problems: Determine the probability that a digit chosen at random from the digits 1, 2, 3, …12 will be odd. The mathematics field of probability has its own rules, definitions, and laws, which you. The number of candidates for the first, second and third posts are 3,4 and 2 respectively. Let P(D) be the probability that the letter is in DOGS. " Understanding how to calculate these percentages with real numbers of people and things. MIDTERM 2 solutions and Histogram. (b) Show that the result in question (a) is not necessarily correct without. 1 Problem 4E. You need at most one of the three textbooks listed below, but you will need the statistical tables. Fully worked-out solutions of these problems are also given, but of course you should first try to solve the problems on your own! c 2013 by Henk Tijms, Vrije University, Amsterdam. Probability of choosing 2nd chocobar = 3/7. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. Objective: I know how to solve probability word problems. Three circle Venn Diagrams are a step up in complexity from two circle diagrams. Q x = P(female lives to age x) = number of female survivors at age x 100,000. Any problems you do not have time to do, try doing later. Bayes’ theorem describes the probability of occurrence of an event related to any condition. Advanced Probability Problems And Solutions Probability Questions with Solutions Tutorial on finding the probability of an event. Two balls are chosen from the jar, with replacement. What is the probability the sum of the dice is 5? 6. · = = Answer: 2: A jar contains 6 red balls, 3 green balls, 5 white balls and 7 yellow balls. Calculate the probability that if somebody is “tall” (meaning taller than 6 ft or whatever), that person must be male. is the constant 2. Probability has its origin in the study of gambling and insurance in the 17th century, and it is now an indispensable tool of both social and natural sciences. ) Find the probability that a person is female or prefers hiking on mountain peaks. == Solutions for Probability and Statistics for Engineering and the Sciences, 7th ed, by Jay L. Important facts and powerful problem solving approaches are highlighted throughout the text. For less than the price of a single session with a private tutor, you can have access to our entire library of videos. Probability MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Probability Math. NCERT Solutions For Maths Class 12 Chapter 13 - Probability comprises of various important questions for exams. Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. Probability Exam Questions with Solutions by Henk Tijms1 December 15, 2013 This note gives a large number of exam problems for a first course in prob-ability. Two events, A and B, are independent if the outcome of A does not affect the outcome of B. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. P(X) gives the probability of successes in n binomial trials. A Problem With Pearls. MIDTERM 1 solutions and Histogram. For Exercises 3-17 to 3-21, verify that the following functions are probability mass functions, and determine the requested probabilities. As we go deeper into the area of mathematics known as combinatorics, we realize that we come across some large numbers. [Syllabus] [Textbooks] [Grading] [Reading assignment] [Homework problems and solutions] [Tests] [Knowledge Milestones]. A probability distribution is a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. As with any quantum system, the wave functions give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. Probability is the chance that an event will occur. Example: the chances of rolling a "4" with a die. Probability Problems and Solutions. Interview Probability Puzzles #1 - Most Popular Monty Hall Brain Teaser Difficulty Popularity The host of a game show, offers the guest a choice of three doors. | 2020-09-24T20:12:14 | {
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https://changesinc.org/p3jo9/what-is-bijective-function-5b8916 | A bijective function is both injective and surjective, thus it is (at the very least) injective. And a function is surjective or onto, if for every element in your co-domain-- so let me write it this way, if for every, let's say y, that is a member of my co-domain, there exists-- that's the little shorthand notation for exists --there exists at least one x that's a member of x, such that. If it crosses more than once it is still a valid curve, but is not a function. So we can calculate the range of the sine function, namely the interval $[-1, 1]$, and then define a third function: $$\sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition for every y in Y there is a unique x in X with y = f(x). Hence every bijection is invertible. A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. How to Prove a Function is Bijective without Using Arrow Diagram ? The inverse is conventionally called \arcsin. A function that is both One to One and Onto is called Bijective function.$$ Now this function is bijective and can be inverted. This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence).. Functions that have inverse functions are said to be invertible. Mathematical Functions in Python - Special Functions and Constants; Difference between regular functions and arrow functions in JavaScript; Python startswith() and endswidth() functions; Hash Functions and Hash Tables; Python maketrans() and translate() functions; Date and Time Functions in DBMS; Ceil and floor functions in C++ Each value of the output set is connected to the input set, and each output value is connected to only one input value. As pointed out by M. Winter, the converse is not true. My examples have just a few values, but functions usually work on sets with infinitely many elements. Infinitely Many. Thus, if you tell me that a function is bijective, I know that every element in B is “hit” by some element in A (due to surjectivity), and that it is “hit” by only one element in A (due to injectivity). The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. Ah!...The beautiful invertable functions... Today we present... ta ta ta taaaann....the bijective functions! The figure shown below represents a one to one and onto or bijective function. Some types of functions have stricter rules, to find out more you can read Injective, Surjective and Bijective. Definition: A function is bijective if it is both injective and surjective. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. A function is invertible if and only if it is a bijection. In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Question 1 : And I can write such that, like that. Below is a visual description of Definition 12.4. 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Bijective and can be inverted to only one input value that have inverse functions are to. M. Winter, the converse is not true a valid curve, is. | 2021-04-11T09:38:20 | {
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https://math.stackexchange.com/questions/374983/prove-that-fraca-12a-1a-2-fraca-22a-2a-3-cdots-fraca-n2a-n | # Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$
Let $$a_1, a_2, a_3, \dots , a_n$$ be positive real numbers whose sum is $$1$$. Prove that $$\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \ldots +\frac{a_n^2}{a_n+a_1} \geq \frac12\,.$$
I thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $$2$$ (the denominator on the right hand side). I would really appreciate any hints.
• Apply Cauchy with one term being left hand side, second term being $((a_1+a_2) + (a_2+a_3) + \cdots)$ – user27126 Apr 28 '13 at 7:02
• Thanks so much I've solved it, should I close the question, or post the solution I just found, with your hint? – John Marty Apr 28 '13 at 7:05
• it would be good if you post a complete solution and accept it as the answer. – user27126 Apr 28 '13 at 7:06
Thanks to Sanchez for giving me a hint to solve this. Here is a full solution.
By the Cauchy-Schwarz inequality we have:
$${\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1}=\frac{a_1^2}{(\sqrt{a_1+a_2})^2}+\frac{a_2^2}{(\sqrt{a_2+a_3})^2}+ \cdots+ \frac{a_n^2}{(\sqrt{a_n+a_1})^2} \geq \frac{1}{a_1+\cdots + a_n+a_1+ \cdots + a_n}\left(\frac{a_1 \cdot \sqrt{a_1+a_2}}{\sqrt{a_1+a_2}} + \frac{a_2 \cdot \sqrt{a_2+a_3}}{\sqrt{a_2+a_3}}+ \cdots + \frac{a_n \cdot \sqrt{a_n+a_1}}{\sqrt{a_n+a_1}}\right)\\=\frac{a_1+a_2+a_3+ \cdots a_n}{{2(a_1+a_2+a_3+ \cdots a_n)}}=\frac12}$$
as required. (We know that $a_1+a_2+a_3+ \cdots +a_n=1$)
• thanks, I couldn't solve it before your hint, thanks so much. – John Marty Apr 28 '13 at 7:34
• I made an attempt to improve readability. Is it OK? – Pedro Tamaroff Jul 6 '13 at 0:37
Here's another solution.
Observe that $$\sum \frac{a_i^2 - a_{i+1}^2} { a_i+ a_{i+1}} = \sum a_i - a_{i+1} = 0,$$
Hence $\sum \frac{a_{i}^2}{a_i+a_{i+1}} = \sum \frac{a_{i+1}^2}{a_i+a_{i+1}}$, and we just need to show that
$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq 1.$$
This makes the inequality much more symmetric, and easier to manipulate. In particular,
$$\sum \frac{a_i^2 + a_{i+1}^2}{a_i + a_{i+1} } \geq \sum \frac{1}{2} (a_i + a_{i+1}) = 1$$
The following modified form of the CBS inequality is often helpful:
Lemma. If $x_i \in \Bbb{R}$ and $a_i > 0$, then $$\frac{x_{1}^{2}}{a_{1}} + \cdots + \frac{x_{n}^{2}}{a_{n}} \geq \frac{(x_{1} + \cdots + x_{n})^{2}}{a_{1}+\cdots+a_{n}}.$$
The proof is straightforward using the CBS inequality, following the methodology exactly John Marty used.
Applying this, we have
$$\frac{a_{1}^2}{a_{1}+a_{2}} + \cdots + \frac{a_{n}^2}{a_{n}+a_{1}} \geq \frac{(a_{1} + \cdots + a_{n})^{2}}{2(a_{1}+\cdots+a_{n})} = \frac{1}{2}.$$ | 2020-11-24T14:26:49 | {
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https://math.stackexchange.com/questions/2610834/question-to-calculating-probability | # Question to calculating probability
The question:
79% of the drivers in a city always fasten their seatbelt when driving, and everyday some drivers receive a ticket(for various reasons...). If a driver has fastened their seatbelt there is a chance of 7% that the driver receives a ticket. If a driver does not fasten their seatbelt there is a 22% chance that he will receive a ticket. If a driver receives a ticket what is the probability that he had fastened his seatbelt?
My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So I just calculated (7/29) * (0,79) + (22/29) * (0,21) and I got 0.35 but that seems to be wrong.
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Jan 18 '18 at 16:13
• Okay thanks for the advise. My assumption is that 7/29 of the people who receive a ticket had fastened their seatbelt and 22/29 did not. So i just calculated (7/29) * (0,79) + (22/29) * (0,21) and i got 0.35 – Simon Jan 18 '18 at 16:18
• Edit your question instead of posting it as a comment. – José Carlos Santos Jan 18 '18 at 16:19
• This is a prefect time for Bayes' Law! – Bram28 Jan 18 '18 at 16:28
Let $S$ and $T$ be the events "seatbelt fasten" and "receive ticket" respectively.
• $P(S) = 0.79$
• $P(T \mid S) = 0.07$
• $P(T \mid S^C) = 0.22$
Use Bayes' formula to find $P(S \mid T)$.
\begin{align} P(S \mid T) &= \frac{P(S \cap T)}{P(T)} \\ &= \frac{P(T \mid S) P(S)}{P(T \mid S) P(S)+P(T \mid S^C) P(S^C)} \\ &= \frac{0.07 \cdot 0.79}{0.07 \cdot 0.79 + 0.22 \cdot (1 - 0.79)} \\ &= \frac{79}{145} \end{align}
If a driver receives a ticket what is the probability that he had fastened his seatbelt is $79/145$.
You can't assume $\frac{7}{29}$ of people who receive a ticket fastened their seatbelt. Consider an extreme case, where the percentage of people who fasten their seatbelt is much higher than $79\%$ -- suppose it is $100\%$. Then clearly $100\%$ of people who receive a ticket fastened their seatbelt. If $0\%$ of people fasten their seatbelt, then $0\%$ who receive a ticket fastened their seatbelt. If the percentage who fasten is somewhere in between -- e.g., $79\%$ -- then the percentage who fastened given that they received a ticket is somewhere in between, but we can't just say it's $\frac{7}{29}$.
Are you familiar with Bayes' Theorem? That's probably the best way to approach this problem. | 2021-08-01T17:44:30 | {
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https://math.stackexchange.com/questions/4187654/what-is-the-value-of-left-log-213-right2-log-21147-log-211323/4187659 | What is the value of $\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$?
What is the value of $$\left(\log_{21}(3)\right)^2+\log_{21}(147)\log_{21}(1323)$$ ?
$$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)4$$
To solve this question I tried writing the expression as:
$$\left(\log_{21}(3)\right)^2+\log_{21}(7^2\times3)\log_{21}(7^2\times3^3)$$ $$=\left(\log_{21}(3)\right)^2+(2\log_{21}7+\log_{21} 3)(2\log_{21}7+3\log_{21}3)$$
I don't know how to continue.
• Don't be so eager to separate the $\log7$ from the $\log3$. The most potent course of simplification you have here is $\log21=1$. Jul 1 at 10:23
• @Arthur Thanks! The question looks horrible for a timed exam and I became a little nervous! Jul 1 at 10:29
• Denote $x=\log_{21}3$ and $y=\log_{21}7$. Then, $147=21\times 7$ and $1323=21^2\times 3$; the original expression is then $x^2+(1+y)(2+x)=x(x+y)+(x+y)+y+2$; iteratively use $x+y=1$ to simplify. Jul 1 at 10:47
• @Prasun Biswas: the computation is simpler if one notices that $147=21^2/3$ and $1323=21^2 \times 3$. Jul 1 at 10:51
• @Mindlack: Nice! Indeed, that is far more simpler because we wouldn't need the $y$, we have $x^2+(2-x)(2+x)=x^2+4-x^2=4$ Somehow, I missed that. +1 Jul 1 at 10:55
Here’s a quick-and-dirty solution based on the fact that one of the answers must be correct.
All the quantities are positive. Since $$1323>441=21^2$$, its base-$$21$$ logarithm is greater than $$2$$. As $$147 \geq 21 \times 5$$, its base-$$21$$ logarithm is at least $$1.5$$. This means that the sum is greater than $$2 \times 1.5$$ so it must be $$4$$.
• I don't get the second point regarding $147 \geq 21 \times 5$. Could you elaborate? Jul 1 at 11:04
• @user71207 $147\ge21\times5\ge21\sqrt{21}$ And the fact that $147\ge21^{\frac32}$ means $\log_{21}(147)\ge\frac32$ Jul 1 at 11:09
$$\log_{21}1323=\log_{21}(3^3\cdot 7^2)=\log_{21}3+\log_{21}21^2=2+\log_{21}3$$
$$\log_{21}147=\log_{21}21+\log_{21}7=1+\log_{21}7$$
$$(\log_{21}3)^2+\log_{21}147\cdot\log_{21}1323 \\=(\log_{21}3)^2+(1+\log_{21}7)(2+\log_{21}3) \\=\log_{21}3(\log_{21}3+\log_{21}7)+\log_{21}3+2\log_{21}7+2 \\=\log_{21}3(\log_{21}3+\log_{21}7)+(\log_{21}3+\log_{21}7)+\log_{21}7+2 \\=\log_{21}3+1+\log_{21}7+2 \\=4$$
• Hi, the problem in this post and also this one that I asked were from entrance exam in the country I'm living (average time for solving each question is about 90 seconds). You solved both of them very well and I think you are very good at solving these questions! So can you please suggest some books or name of exam that you think are good for an exam that contains these type of questions? Thanks in advanced! Jul 27 at 19:58
• I mainly use books that are meant for olympiads, so I don't know many books that are meant for solving questions fast. Jul 28 at 6:28
Alternate:
Let $$L(x)$$ denote $$\log_{21}(x).$$
Then $$[L(3)]^2 + [L(147) \times L(9 \times 147)]$$
$$= [L(3)]^2 + ~\langle ~L(147) \times ~\{ [2 \times L(3)] + L(147) ~\} ~\rangle$$
$$= [L(3)]^2 + [L(147)]^2 + [2 \times L(3) \times L(147)]$$
$$= [L(3) + L(147)]^2.$$
At this point, you have that $$3 \times 147 = 21^2$$,
so $$[L(3) + L(147)] = 2.$$ | 2021-09-19T00:58:36 | {
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https://stats.stackexchange.com/questions/207841/why-is-sst-sse-ssr-one-variable-linear-regression/207912#207912 | # Why is $SST=SSE + SSR$? (One variable linear regression)
Note: $SST$ = Sum of Squares Total, $SSE$ = Sum of Squared Errors, and $SSR$ = Regression Sum of Squares. The equation in the title is often written as:
$$\sum_{i=1}^n (y_i-\bar y)^2=\sum_{i=1}^n (y_i-\hat y_i)^2+\sum_{i=1}^n (\hat y_i-\bar y)^2$$
Pretty straightforward question, but I am looking for an intuitive explanation. Intuitively, it seems to me like $SST\geq SSE+SSR$ would make more sense. For example, suppose point $x_i$ has corresponding y-value $y_i=5$ and $\hat y_i=3$, where $\hat y_i$ is the corresponding point on the regression line. Also assume that the mean y-value for the dataset is $\bar y=0$. Then for this particular point i, $SST=(5-0)^2=5^2=25$, while $SSE=(5-3)^2=2^2=4$ and $SSR=(3-0)^2=3^2=9$. Obviously, $9+4<25$. Wouldn't this result generalize to the entire dataset? I don't get it.
Adding and subtracting gives \begin{eqnarray*} \sum_{i=1}^n (y_i-\bar y)^2&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\ &=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2 \end{eqnarray*} So we need to show that $\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=0$. Write $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=\sum_{i=1}^n(y_i-\hat y_i)\hat y_i-\bar y\sum_{i=1}^n(y_i-\hat y_i)$$ So, (a) the residuals $e_i=y_i-\hat y_i$ need to be orthogonal to the fitted values, $\sum_{i=1}^n(y_i-\hat y_i)\hat y_i=0$, and (b) the sum of the fitted values needs to be equal to the sum of the dependent variable, $\sum_{i=1}^ny_i=\sum_{i=1}^n\hat y_i$.
Actually, I think (a) is easier to show in matrix notation for general multiple regression of which the single variable case is a special case: \begin{eqnarray*} e'X\hat\beta &=&(y-X\hat\beta)'X\hat\beta\\ &=&(y-X(X'X)^{-1}X'y)'X\hat\beta\\ &=&y'(X-X(X'X)^{-1}X'X)\hat\beta\\ &=&y'(X-X)\hat\beta=0 \end{eqnarray*} As for (b), the derivative of the OLS criterion function with respect to the constant (so you need one in the regression for this to be true!), aka the normal equation, is $$\frac{\partial SSR}{\partial\hat\alpha}=-2\sum_i(y_i-\hat\alpha-\hat\beta x_i)=0,$$ which can be rearranged to $$\sum_i y_i=n\hat\alpha+\hat\beta\sum_ix_i$$ The right hand side of this equation evidently also is $\sum_{i=1}^n\hat y_i$, as $\hat y_i=\hat\alpha+\hat\beta x_i$.
This is just the Pythagorean theorem!
Hence,
$$Y'Y=(Y-X\hat{\beta})'(Y-X\hat{\beta})+(X\hat{\beta})'X\hat{\beta}$$
or
$$SST=SSE+SSR$$
• Aug 2 '19 at 19:52
(1) Intuition for why $SST = SSR + SSE$
When we try to explain the total variation in Y ($SST$) with one explanatory variable, X, then there are exactly two sources of variability. First, there is the variability captured by X (Sum Square Regression), and second, there is the variability not captured by X (Sum Square Error). Hence, $SST = SSR + SSE$ (exact equality).
(2) Geometric intuition
Some of the total variation in the data (distance from datapoint to $\bar{Y}$) is captured by the regression line (the distance from the regression line to $\bar{Y}$) and error (distance from the point to the regression line). There's not room left for $SST$ to be greater than $SSE + SSR$.
(3) The problem with your illustration
You can't look at SSE and SSR in a pointwise fashion. For a particular point, the residual may be large, so that there is more error than explanatory power from X. However, for other points, the residual will be small, so that the regression line explains a lot of the variability. They will balance out and ultimately $SST = SSR + SSE$. Of course this is not rigorous, but you can find proofs like the above.
Also notice that regression will not be defined for one point: $b_1 = \frac{\sum(X_i -\bar{X})(Y_i-\bar{Y}) }{\sum (X_i -\bar{X})^2}$, and you can see that the denominator will be zero, making estimation undefined.
Hope this helps.
--Ryan M.
When an intercept is included in linear regression(sum of residuals is zero), $$SST=SSE+SSR$$.
prove $$\begin{eqnarray*} SST&=&\sum_{i=1}^n (y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2\\&=&SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y) \end{eqnarray*}$$ Just need to prove last part is equal to 0: $$\begin{eqnarray*} \sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i \end{eqnarray*}$$ In Least squares regression, the sum of the squares of the errors is minimized. $$SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2$$ Take the partial derivative of SSE with respect to $$\beta_0$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 = 0$$ Take the partial derivative of SSE with respect to $$\beta_1$$ and setting it to zero. $$\frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ So $$\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0$$ Hence, $$\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i=0$$ $$SST=SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=SSE+SSR$$
Here is a great graphical representation of why SST = SSR + SSE.
• How can we see from this picture that predicted values and residuals are orthogonal? Mar 4 at 12:17
If a model predicts $$3$$ and the residual is $$2$$ because the actual value is $$5$$, it doesn't look like variance is decomposing, since $$3^2 + 2^2 \neq 5^2$$.
If you only have one data point, your model would fit it perfectly and the residual would be zero, so you can't get that case by itself. There have to be multiple data points.
With multiple data points, the residuals won't all be positive.
If the model predicts 3, a residual of $$+2$$ and $$-2$$ should be equally likely. The balance of increases and decreases gives a nice cancellation:
$$\frac{(3+2)^2 + (3-2)^2}{2} = \frac{25+1}{2} = 13 = 9 + 4 = 3^2 + 2^2$$
This property, that $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$, is what makes variance decomposition work, for uncorrelated components. | 2021-10-22T08:45:44 | {
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https://brilliant.org/discussions/thread/easy-money/ | # Easy Money
On a table before you lie two envelopes, identical in appearance. You are told that one envelope contains twice as much money as the other, but given no indication which has more. You are then allowed to choose one of the envelopes and keep the money it contains.
You then pick one of the envelopes at random, but before you look inside you are offered the chance to exchange your envelope for the other. Should you exchange envelopes, (assuming that the more money you get, the better)?
One line of thinking is this: If the envelope you initially choose has $$x$$ dollars in it, then there is a $$50$$% chance that the other envelope has $$2x$$ dollars in it and a $$50$$% chance that it has $$\frac{x}{2}$$ dollars in it. Thus the expected amount of money you would end up with if you chose to exchange envelopes would be
$$\frac{1}{2} * 2x + \frac{1}{2} * \frac{x}{2} = \frac{5}{4} x$$ dollars.
You thus decide to exchange envelopes. But before doing so, you think to yourself, "But I could go through the same thought process with the other envelope, concluding that I should exchange my new envelope for my old one, and then go through the same thought process again, and again, ad infinitum. Now I have no idea what to do."
How can you resolve this paradox?
Edit: As mentioned in my comment below, the real "puzzle" here is to identify the flaw in the reasoning I outlined above. Or is there a flaw?
Note by Brian Charlesworth
4 years, 4 months ago
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First, look at the case when the envelope is exchanged. If you pick the envelope with $$x$$ dollars $$(P=1/2)$$ and exchange, you win $$2x$$ dollars. Instead, if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and exchange, then you win $$x$$ dollars.
$\therefore E(\text{Money when envelope is exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$
Now, look at the case when the envelope is not exchanged. If you pick the envelope with $$x$$dollars $$(P=1/2)$$ and don't exchange, you win $$x$$ dollars. But if you pick the envelope with $$2x$$ dollars $$(P=1/2)$$ and don't exchange, then you win $$2x$$ dollars.
$\therefore E(\text{Money when envelope is not exchanged})=\dfrac{1}{2}2x+\dfrac{1}{2}x=\dfrac{3x}{2}$
So it makes no difference at all whether you exchange or don't exchange your envelope.
- 4 years, 4 months ago
Yes, that is a good analysis. Since we are not given any further information prior to making a choice whether or not to exchange envelopes, we are in the same position as we were with our initial choice, i.e., our expected winnings will be $$\frac{3x}{2}$$.
But the "real" puzzle here is to find the logical miscue in the line of reasoning I laid out. It can be argued that in the two scenarios I describe, (namely that I start out with the lesser amount of money or the greater), the value $$x$$ actually represents different amounts in each case, so that I can't use the same variable $$x$$ for each of them, making my equation invalid.
However, suppose I did peek at the amount in the envelope before making a decision. Now this provides me with "information", but in reality it's useless since I have no clue as to whether it's the lesser or greater amount. But since we now can attach an actual value to $$x$$, the mathematics I applied in my initial analysis becomes re-validated, and hence so does my "conclusion". But since, as noted, the information regarding the amount in the initially chosen envelope is in reality useless, should it matter whether we know the actual amount as far as the mathematical analysis is concerned? From this perspective, no matter what amount we see inside the initial envelope we should conclude that switching envelopes is preferable.
So the puzzle of "where's the flaw in logic?" is more a question of perspective; are we dealing with unconditional or conditional probability values, and why should this difference in perspective matter given the above discussion? A surprising amount of attention has been given to this paradox, commonly referred to as the "Two Envelope Paradox", with no universally accepted resolution as of yet. I think why it does garner such attention is that it cuts to the heart of the foundation of probability theory, and as with any foundation, the desire is that it must be as secure as possible, and if it is not, it must be either fortified or torn down.
Here is an accessible discussion on the "search for the flaw" and why many find it so intriguing.
- 4 years, 4 months ago
I'm not a mathematician, just a dilettante of sorts but something tells me we're over complicating this paradox. There are only two choices and only two possibilities. The money is either greater or lesser and you either switch or stay. The fact that the sum of all probabilities is 1 is infallible. We simply assume we have knowledge we previously didn't have and compute what the probabilities are to find out what is what.
Here's another question: Are we then told that "hey, you chose the lesser or greater amount of money" after we've made our choice? I think not. So the probability will not even depend on outcome. For all intents and purposes, we need not even bother with the maths of 'x' and '2x'; and we might as well adopt an approach with x and y since the relationship between the monies simply do not matter. I might as well be a choice between winning the lottery and detonating an atomic bomb.
- 4 years ago
I think that the reasons this "paradox" has received so much attention over the years are more esoteric in nature than practical. For some it raises questions about the foundation of probability theory regarding the absolute distinction between conditional/unconditional scenarios and whether or not there might be a grey area there. However, mathematicians do have a tendency to "overcomplicate" things, and this may a case in point. :)
As for the choice between winning the lottery and detonating an atomic bomb, well, I think I would just walk away from the table; too many lives other than my own would be at stake without them having any say in the matter. However, if it were a choice between my eternal health, wealth and happiness on the one hand and my immediate extinction on the other, I would give it a bit more thought. I would probably still walk away, since it would feel like I'm making some kind of deal with the Devil, but it does make for an interesting psychological experiment. This is an entirely different question than the previous "paradox" we were dealing with; with the money in the two envelopes, no matter what choice we made we would be richer than before. With this new question, there is the possibility of a terrible outcome, so the first choice becomes whether to engage at all, and if we do then why do we think it is worth such a high risk of calamity. The more we would have to lose in our present life the less likely we would engage, but if we did not have much to live for then the risk might be worth it. However, as I said before, only the Devil would make such a proposition, so anyone, no matter what their circumstances, would be wise to just walk away. Sorry, I do ramble on, but i always do that when a good philosophical question happens by. :)
- 4 years ago
Hahaha.I do love to read you rambling on. It's more engaging than many other types of conversations. Your ending pretty much wrapped up my premise: which is that, the paradox seems to rely on the probability of making a more money outcome which is identical to that of making a lesser money. I do have some philosophical questions if you're interested, I'd love to read your views. I'm relatively new on this site so is it OK if I create a note and invite you to give your thoughts on it?
- 4 years ago
Of course; I look forward to seeing what the topic is. :)
- 4 years ago
Hi again Mr. Charlesworth
Here are two notes I have posted. I invite you to share your thoughts on them: https://brilliant.org/discussions/thread/number-bases-in-a-different-world/# https://brilliant.org/discussions/thread/the-infiniteness-of-light/?ref_id=654556
Hi @Calvin Lin. Do you think it's a good idea to find a way to incorporate Philosophy (Logic and Reasoning) on Brilliant? I think it is a very fascinating pseudoscience and it would appeal to many older people and others who are not strictly geeks and nerds like the rest of us. Just a thought
- 4 years ago
Let's incorporate Speculative Science And Math, as distinct from formal Philosophy. You know, something that would appeal to one old guy here who wants to try his hand in being a crank.
I'm reading Brian's posted scenario, and it seems to me that if it's known that one envelope contains twice the other, then the expected value in choosing either is simply $$\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } 2x=\dfrac { 3 }{ 2 } x$$ either way, as pointed out by Pratik. It's 50% percent chance that it's twice the one you've chosen, and 50% percent chance it's half that--but you can't assume the same $$x$$ dollars in your chosen envelope in both cases! That was the flaw. Maybe Brian is trying to alluding to the problem of trying to determine probabilities when information is incomplete, which is what led to Bayesian probability theory. Let me offer a slightly different scenario. You are given $$2$$ envelopes. You are told that one of them may or may not contain a valuable stock certificate. You open one up and do not find it inside. What's the probability the other one does? Extend that to $$100$$ such envelopes. This is actually what happens when you are looking for gold in the ground in a certain area when you don't have much information to go on. After so much digging, at what point in time do you give up? Are more likely to find gold, because you've already almost eliminated the places where it can be, or are you more less likely to find gold, because you've already tried to find it and failed? A lot of science can be like that.
- 4 years ago
Oh yes, that's a great idea because what I was hoping for was an avenue to wonder beyond the limits of our recorded knowledge and rub minds on what could be and what not.
I'm fascinated this "blind" probability theory. It is a little bit anti cognitive. Which reminds me of something else which defies our flawed cognizance. Consider flipping a coin and getting heads five times in a row. Flipping it a sixth time seems as though there is a higher probability to get tails. In fact, I'd bet that if you did a survey of ten mathematics professors, a higher percentage of them would choose tails for that sixth coin toss except they are just fighting against their human nature. But sometimes I think about it and really wonder if the law of averages is that subjective. If you mix two liquids together, they mix together perfectly EVERY TIME if they are of similar viscosity. I'm rambling. I have even forgotten what the point I'm trying to make is
- 4 years ago
A lot of theoretical physics was minds rambling before something more concrete came out of it. Ramble on.
Likewise, a lot of technology was preceded by science fiction. Smart phones? Didn't Star Trek popularize that idea?
- 4 years ago | 2019-03-19T18:03:32 | {
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https://mathhelpboards.com/threads/4x4-matrix-with-rank-b-4-and-b-2-3.6384/ | # 4x4 Matrix with rank B=4 and B^2=3
#### Petrus
##### Well-known member
Hello MHB,
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$ and B also have to be a upper/lower or identity Matrix. And this Will not work.. I am wrong or can I explain this in a better way?
Regards,
$$\displaystyle |\pi\rangle$$
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
"Can we construct a $$\displaystyle 4x4$$ Matrix $$\displaystyle B$$ so that rank $$\displaystyle B=4$$ but rank $$\displaystyle B^2=3$$"
My thought:
we got one condition for this to work is that det $$\displaystyle B=0$$ and det $$\displaystyle B^2 \neq 0$$
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.
#### Petrus
##### Well-known member
It's the other way around: $\mathop{\text{rank}}(B)=4\iff\det(B)\ne0$ and $\mathop{\text{rank}}(B^2)=3\implies\det(B^2)=0$. But you are right that this is impossible because $\det(B^2)=(\det(B))^2$.
Hello Evgeny.Makarov,
thanks for fast respond and I meant that! And thanks for showing me this I did not know that $\det(B^2)=(\det(B))^2$ That Was what I Was looking for
Regards,
$$\displaystyle |\pi\rangle$$
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
An alternative proof (without using determinants):
Consider the linear map $B:\mathbb{R}^4\to \mathbb{R}^4,\; x\to Bx.$ As $\operatorname{rank}B=4,$ $\operatorname{nullity}B=0,$ wich implies $B$ is bijective. But the composition of bijective maps is bijective, so $\operatorname{rank}B^2=4.$
#### Deveno
##### Well-known member
MHB Math Scholar
Another formulation:
As rank(B) = 4, B is surjective, that is, B(R4) = R4 (for this is what rank means: the dimension of the image space (or column space) of B, and R4 is the ONLY 4-dimensional subspace of R4).
Consequently, B2(R4) = B(B(R4)) = B(R4) = R4, from which we conclude B2 is likewise surjective, and thus rank(B2) = 4 as well.
(I only post this to indicate one need not even invoke the rank-nullity theorem).
#### Petrus
##### Well-known member
this is impossible because $\det(B^2)=(\det(B))^2$.
Now that I think about I remember a sats that said $$\displaystyle |AB|=|A||B|$$ but in this case $$\displaystyle A=B$$ hmm I need to find the proof for this.
Regards,
$$\displaystyle |\pi\rangle$$
MHB Math Scholar
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
I saw one neat proof of $\det(AB)=\det(A)\det(B)$ in Linear Algebra and Its Applications by Gilbert Strang. He defines $\det(\cdot)$ as a function satisfying three properties:
(1) $\det(I)=1$ where $I$ is the identity matrix;
(2) it changes sign when two adjacent rows are swapped;
(3) it is linear on the first row.
Signed volume in an orthonormal basis satisfies these properties, so this definition is much more intuitive than the Leibniz formula, which is derivable from (1)–(3).
Now, to prove that $\det(AB)=\det(A)\det(B)$, fix $B$ and consider $d(A)=\det(AB)/\det(B)$. It is possible to show that $d(A)$ satisfies (1)–(3), and so $d(A)=\det(A)$.
Now that I looked at the StackExchange link, this is answer #2, which is highest-ranked. | 2020-09-19T09:37:39 | {
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https://math.stackexchange.com/questions/414065/closed-form-of-sum-limits-i-1n-left-lfloor-fracni-right-rfloor2 | # Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$?
Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?
• Unlikely; see the related math.stackexchange.com/q/384520/73324 – vadim123 Jun 7 '13 at 18:47
• @Martin, is the sequence formed by these sums not a sequence? – Antonio Vargas Dec 1 '16 at 17:38
• @AntonioVargas The same could be said about any finite sum of the form $\sum_{k=1}^n a_n$. Of course, if you think that for some reasone the tag belongs there, you can edit it back. My understanding is that the tag-excerpt says rather clearly: "For questions on finite sums use the (summation) instead." – Martin Sleziak Dec 1 '16 at 17:41
To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that $$a(n+1)-a(n) = 1 + \sum_{k=1}^n \left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\= 1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right) = \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\= 2\sigma(n+1)-\tau(n+1).$$
It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) = \sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$ We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series $$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} = 2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$ We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$ so that by the aforementioned theorem, $$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$ In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting $$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) + \frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$ which yields $$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$ This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$
This is A222548 in the online encyclopedia of integer sequences. They don't provide a closed form, but they do give the following:
$$a(n)\approx\frac{\pi^2}{6}n^2+O(n\log n)$$
Let me complement @Marko Riedel's excellent answer by providing a low-tech approach to the leading term.
Let us denote by $\mathbf{1}\{\cdot\}$ the indicator function of the set $\{\cdot\}$. Then using the identity
$$\lfloor n/i \rfloor = \sum_{j=1}^{n} \mathbf{1}\{ij \leq n \}$$
we can rearrange the sum to find:
\begin{align*} a(n) &= \sum_{i=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \mathbf{1}\{ij \leq n\}\mathbf{1}\{ik \leq n\} = \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \sum_{i=1}^{n} \mathbf{1}\Big\{i \leq \frac{n}{\max\{j,k\}}\Big\} \\ &= \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{\max\{j,k\}} \right\rfloor = \sum_{l=1}^{n} \sum_{\substack{1 \leq j \leq n \\ 1 \leq k \leq n}} \left\lfloor \frac{n}{l} \right\rfloor \mathbf{1}\{\max\{j,k\} = l\} \\ &= \sum_{l=1}^{n} (2l-1) \left\lfloor \frac{n}{l} \right\rfloor. \end{align*}
Dividing both sides by $n^2$, the RHS can be identified as Riemann sum and thus
$$\frac{a(n)}{n^2} = \sum_{l=1}^{n} \frac{2l-1}{n} \left\lfloor \frac{n}{l} \right\rfloor \frac{1}{n} \xrightarrow[\ n\to\infty \ ]{} \int_{0}^{1} 2x \left\lfloor\frac{1}{x}\right\rfloor \, dx = \zeta(2).$$
Here, the last integral can be easily computed by applying the substitution $x \mapsto 1/x$.
An Answer Without Number Theoretic Functions
Break up the sum as follows: $$\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 =\sum_{k=1}^n\frac{n^2}{k^2}-\sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right)\tag{1}$$ We can apply the Euler-Maclaurin Sum Formula $$\sum_{k=1}^n\frac{n^2}{k^2}\sim n^2\left(\frac{\pi^2}6-\frac1n+\frac1{2n^2}\right)\tag{2}$$ If we note that $\left\{\frac nk\right\}\lt1-\frac1k$, we see that \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\le\sum_{k=1}^n\left(1-\frac1k\right)2\frac nk\\[9pt] &\le2n\log(n)\tag{3} \end{align} Therefore, by $(1)$, $(2)$, and $(3)$ we have $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2=n^2\frac{\pi^2}6+O(n\log(n))}\tag{4}$$ However, if we use $\frac12{-}\frac1{2k}$ for the mean value of $\left\{\frac nk\right\}$ and $\frac13{-}\frac1{2k}{+}\frac1{6k^2}$ for the mean value of $\left\{\frac nk\right\}^2$ \begin{align} \sum_{k=1}^n\left(\frac{n^2}{k^2}-\left\lfloor\frac nk\right\rfloor^2\right) &=\sum_{k=1}^n\left(\frac nk-\left\lfloor\frac nk\right\rfloor\right)\left(\frac nk+\left\lfloor\frac nk\right\rfloor\right)\\ &=\sum_{k=1}^n\left\{\frac nk\right\}\left(2\frac{n}{k}-\left\{\frac nk\right\}\right)\\ &\approx\sum_{k=1}^n\left(\frac12-\frac1{2k}\right)2\frac{n}{k}-\left(\frac13-\frac1{2k}+\frac1{6k^2}\right)\\ &=n\log(n)+n\left(\gamma-\frac{\pi^2}6-\frac13\right)+O(\log(n))\tag{5} \end{align} Thus, if we look at the average behavior, we can use $(1)$, $(2)$, and approximation $(5)$ to get $$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\left\lfloor\frac nk\right\rfloor^2 \approx n^2\frac{\pi^2}6-n\log(n)+n\left(\frac{\pi^2}6-\gamma-\frac23\right)}\tag{6}$$
The value given by $(6)$ for $n=98$ is $15387.9230$ while the actual value is $15381$.
The value given by $(6)$ for $n=1001$ is $1641711.3692$ while the actual value is $1641772$.
The error of the approximation in $(6)$ appears to be proportional to $n$. Here is a plot of that error, for $1\le n\le3000$, divided by $n$:
The error divided by $n$ has a mean of $-0.0746805451$ and a standard deviation of $0.6513592109$.
Using number theoretic functions, Marko Riedel's answer gives a better approximation. Here is a plot of the error of that approximation, for $1\le n\le3000$, divided by $n$:
The error divided by $n$ has a mean of $0.0060900069$ and a standard deviation of $0.3674311090$. This is better than the error distribution in my answer, but it does require factoring $n$ to compute the number theoretic functions. | 2019-05-20T15:24:14 | {
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https://math.stackexchange.com/questions/122119/does-there-exist-more-than-3-connected-open-sets-in-the-plane-with-the-same-boun | # Does there exist more than 3 connected open sets in the plane with the same boundary?
I've wondered about the following question, whose answer is perhaps well known (in this case I apologize in advance).
The Lakes of Wada are a famous example of three disjoint connected open sets of the plane with the counterintuitive property that they all have the same boundary (!)
My question is the following :
Can we find four disjoint connected open sets of the plane that have the same boundary?
More generally :
For each $n \geq 3$, does there exist $n$ disjoint connected open sets of the plane that have the same boundary? If not, then what is the smallest $n$ such that the answer is no?
Thank you, Malik
• The article you link to says "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, ...and so on.", so the answer seems to be "you can do it for all $n$ including $n =\aleph_0$". I'd guess (but have no idea really) that the corresponding question for other cardinalities is independent of ZFC, assuming NOT CH. Though, since $\mathbb{R}^n$ is first countable, there may be an answer... – Jason DeVito Mar 19 '12 at 15:55
• Found this in the link. It seems to answer your question. "A variation of this construction can produce a countable infinite number of connected lakes with the same boundary: instead of extending the lakes in the order 1, 2, 0, 1, 2, 0, 1, 2, 0, ...., extend them in the order ..." – Patrick Mar 19 '12 at 15:57
• Given that the basins of attraction of Newton's method for $z^3=1$ are an example for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. – lhf Mar 19 '12 at 16:49
• Jason, separability precludes finding any uncountable collection of pairwise disjoint nonempty open sets, irrespective of any conditions you put on their boundaries. – user83827 Mar 19 '12 at 17:23
• @Student73, ah, this may contain a proof: Frame, Michael; Neger, Nial. Newton's method and the Wada property: a graphical approach. College Math. J. 38 (2007), no. 3, 192–204. MR2310015 (2008e:37082) – lhf Mar 19 '12 at 17:31
Given that the basins of attraction of Newton's method for $z^3=1$ are Wada sets for $n=3$, I'd say that the basins of attraction of Newton's method for $z^n=1$ should work just as fine. I don't know a reference to a proof but try these: | 2019-06-17T22:36:46 | {
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http://mathhelpboards.com/calculus-10/s4-13-5-41-relation-2-planes-20440.html?s=9278ceb6188b675b9c2ddddfeb0e6f5e | # Thread: s4.13.5.41 relation of 2 planes
1. $\tiny{{s4}.{13}.{5}.{41}}$
$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}
$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$
2. Originally Posted by karush
$\tiny{{s4}.{13}.{5}.{41}}$
$\textsf{find if planes are$\parallel, \perp$or$\angle$of intersection }\\$
\begin{align}
\displaystyle
{P_1}&={x+z=1}\\
\therefore n_1&=\langle 1,0,1 \rangle\\
\\
{P_2}&={y+z=1}\\
\therefore n_2&=\langle 0,1,1 \rangle\\
\\
\cos(\theta)&=
\frac{n_1\cdot n_2}{|n_1||n_2|}\\
&=\frac{1(0)+0(1)+1(1)}
{\sqrt{1+1}\cdot\sqrt{1+1}}
=\frac{1}{2}\\
\cos^{-1}\left({\frac{1}{2}}\right)&=
\color{red}{60^o}
\end{align}
$\textit{there are 2 more problems like this so presume this is best method.. }\\$
$\textit{didn't know if it is common notation to call a plane$P_1$}$
Everything you've posted is fine. You can give a plane any name you like, but I would write something like this:
\displaystyle \begin{align*} P_1 : x + z = 1 \end{align*}
That way we can see that \displaystyle \begin{align*} P_1 \end{align*} is DEFINED as the relationship "the sum of the x and z values needs to be 1", not that it is some variable that has something to do with the equation.
3. The problem asks you to do three things:
1) determine if the planes are parallel.
2) determine if the planes are perpendicular.
3) if neither of those, determine the angle of intersection of the two planes.
Yes, by using that formula to determine the angle, you can then answer all three questions but it should be simpler to determine the first two without using that formula:
The two planes are parallel if and only if the two normal vectors are parallel- if one is a multiple of the other.
The two planes are perpendicular if and only if the two normal vectors are perpendicular: if their dot product is 0.
If neither of those is true, then you can use the dot product you found as the numerator in the formula to determine the angle.
hard to see that in their examples
the next 2 problems are probably
$\displaystyle \parallel , \perp$
$\tiny{s4.854.13.5.43}$
$\textsf{Determine if the 2 given planes are perpendicular, parallel or at an angle to each other}$
\begin{align}
\displaystyle
{p_1}&:{x+4y-3z=1}
\therefore n_1=\langle 1,4,-3 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+7z=0} \therefore n_2=\langle -3,6,7 \rangle
\end{align}
\begin{align}
\displaystyle
\cos(\theta)&= \frac{n_1\cdot n_2}{|n_1||n_2|}=0\\
\therefore p_1 &\perp p_2
\end{align}
$\tiny{s4.854.13.5.45}$
\begin{align}
\displaystyle
{p_1}&:{2x+4y-2z=1}
\therefore n_1=\langle 2,4,-2 \rangle\\
\nonumber\\
{p_2}&:{-3x+6y+3z=0} \therefore n_2=\langle -3,6,7 \rangle \\
n_1&=-\frac{2}{3} n_2 \\
&\therefore n_1\parallel n_2
\end{align}
$\textit{btw what is the input string to see the 2 plane on a W|A graph?}$
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https://math.stackexchange.com/questions/1386606/probability-of-rolling-a-dice-8-times-before-all-numbers-are-shown/1387649 | # Probability of rolling a dice 8 times before all numbers are shown.
What is the probability of having to roll a (six sided) dice at least 8 times before you get to see all of the numbers at least once?
I don't really have a clue how to work this out.
Edit: If we are trying to find the number of situations where all of the numbers are shown, for seven rolls, a favorable outcome would be one in which there are two numbers the same, for example: 1123456. These numbers can be arranged in 7!/2!5! ways, and there are 6 different numbers that could be repeated. So the probability of getting all 6 numbers with 7 throws is (6*7!/2!5!)/6^7 = 126/279936. Is that right? Then 1 minus this is the probability?
Edit: prob. of not getting all six numbers with seven rolls = 1-prob of getting all six numbers with seven rolls
Six numbers same with seven rolls means one number repeated twice 7!/2! ways of doing this for each repeated number 6*(7!/2!) is number of ways of getting all six numbers with seven rolls. Total number of outcomes 6^7. Prob of getting all six numbers with seven rolls = (6*(7!/2!))/6^7 = 0.054 Prob of not getting all six numbers with seven rolls (=prob of needing at least 8 rolls to get all numbers) = 1-0.054 = 0.946
• Let $p$ be the probability that in $7$ rolls you see all faces at least once. We find $p$. Then the answer to the original problem is $1-p$. It should not be too hard to find $p$, by counting the "favourables" and dividing by $6^7$. – André Nicolas Aug 6 '15 at 13:34
• As a further hint to counting the "favorables" for $p$, note that if you roll 7 times you will see at least one number repeated. But if this happens more than once you will not get to see all 6 numbers. – David K Aug 6 '15 at 13:50
• The title and text of the question don't match. Please clarify whether "at least" is intended, as in the text, or "exactly", as the title appears to imply. – joriki Aug 6 '15 at 14:17
• To clarify are you asking what is the probability of needing exactly 8 rolls to see all numbers or 8 or more rolls? – Warren Hill Aug 7 '15 at 8:20
Rephrase the question:
What is the probability of not seeing all $6$ values when rolling a die $7$ times?
Find the probability of the complementary event:
What is the probability of seeing all $6$ values when rolling a die $7$ times?
Use the inclusion/exclusion principle in order to count the number of ways:
• Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\binom66\cdot6^7$
• Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\binom65\cdot5^7$
• Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\binom64\cdot4^7$
• Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\binom63\cdot3^7$
• Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\binom62\cdot2^7$
• Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\binom61\cdot1^7$
Divide the result by the total number of ways, which is $6^7$:
$$\frac{\binom66\cdot6^7-\binom65\cdot5^7+\binom64\cdot4^7-\binom63\cdot3^7+\binom62\cdot2^7-\binom61\cdot1^7}{6^7}=\frac{35}{648}$$
Calculate the probability of the original event by subtracting the result from $1$:
$$1-\frac{35}{648}=\frac{613}{648}\approx94.6\%$$
• Nice use of inclusion/exclusion! Simulated probability of seeing all 6 faces in 7 rolls in my Answer is 0.0544 ± 0.001, compared to exact 35/648 = 0.0540 in this Answer. – BruceET Aug 7 '15 at 8:46
• @BruceTrumbo: Thanks :) – barak manos Aug 7 '15 at 8:50
(Edit: Answer is for the probability of seeing all the numbers for $n$ rolls)
For exactly $n$ rolls, the problem can be solved using a markov chain
\begin{align*} A&= \begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & \frac{1}{6} & \frac{5}{6} & 0 & 0 & 0 & 0\cr 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 & 0 & 0\cr 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0\cr 0 & 0 & 0 & 0 & \frac{2}{3} & \frac{1}{3} & 0\cr 0 & 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\cr 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix} \end{align*}
where the rows and columns are the number of faces of the die seen.
The probability of seeing all the faces in exactly 8 rolls is $(A^8)[0,6]$, which is $\dfrac{665}{5832}\approx 0.11402606310014$
For any $n$, it can be found by finding the generating function $G(z)$ and in turn finding the coefficient of $z^n$
\begin{align*} \mathbb{P}(n) = 1-\frac{20}{2^n}+15\left(\frac{2}{3}\right)^n+\frac{15}{3^n}-6\left(\frac{5}{6}\right)^n-\frac{6}{6^n} \end{align*}
• That was my original solution (should be $1-\frac{665}{5832}$, BTW)... But I then realized that "at least 8 times before you get to see all of the numbers" meant that you were allowed to see all of the numbers by the $8$th trial, so you need to replace $8$ with $7$ in your answer. – barak manos Aug 7 '15 at 12:55
• Oh, okay. I solved for the probability of seeing all the numbers for n rolls. – gar Aug 7 '15 at 13:24
• Very elegant. Clarity of presentation could be improved for quick reading (and more upvotes). I suggest to: make the first sentence bold, correcting for n=7 so that the answers match, small subtitle "general solution" at the end. – Symeof Sep 23 '15 at 21:36
Perhaps this is of some help, even if not your final answer.
If a fair die is rolled 8 times and $X$ is the number of different faces seen, then simulation of a million performances of the 8-roll experiment shows the following approximate distribution of $X$, in which the margin of simulation error is about $\pm 0.001.$
1 2 3 4 5 6
0.000003 0.002275 0.069393 0.363681 0.450468 0.114180
Of course, $P(X = 1) = 1/6^7 = 0.000004$ to 6 places. To get $P(X = 6),$ consider that one face may be seen three times, or two different faces may each be seen twice. A straightforward computation of other probabilities in the distribution seems feasible but increasingly tedious. Perhaps there are clever combinatorial methods that make some of them easier.
However, pending clarification, I'm still not sure if $P(X = 6)$ is exactly the probability you're trying to find.
Notes:
(a) For a 7-roll experiment an analogous simulation gives $0.0544 \pm 0.001,$ which does not agree with your result $126/279936 \approx 0.00045.$ (I do not believe you are considering all possible permutations of the numbers that occur only once.)
(b) For a 6-roll experiment a simulation gives $0.015496 \pm 0.001,$ compared with the exact $6!/6^6 = 0.0154321.$ Clearly, the probability of getting all 6 faces is 7 rolls must be larger than that.
(c) A related problem is the number of rolls $W$ required before all 6 faces are seen. Then $E(W) = 6/6 + 6/5 + \dots + 6/1 = 14.7.$ For more on this, see the 'coupon collecting problem'.
Here's another way of looking at it, along the lines suggested by barak manos.
Note that 7 rolls containing all 6 values must contain a duplicate. There are $\binom{7}{2}\cdot 6$ choices for the duplicate entries. Thus, the desired probability is given by
$$1-\frac{\binom{7}{2}\cdot 6\cdot 5!}{6^7} = 1-\frac{70}{1296}=\frac{613}{648}.$$ | 2019-11-15T18:12:28 | {
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https://mathhelpboards.com/threads/garland-trouble.56/ | # Garland trouble
#### grgrsanjay
##### New member
How many different garlands are possible with 3 identical beads of red color and 12 identical beads of black color?
I was thinking to keep the no.of beads in between the 3 beads of red color as x,y,z
So, x+y+z=12
no .of ways is 14C2.....Was i Wrong somewhere??
Last edited:
#### tkhunny
##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
#### Plato
##### Well-known member
MHB Math Helper
Black will fit not only between, but also before and after. You seem to have only one end.
W before any red
X after first red
Y after second red
Z after last red
W + X + Y + Z = 12
14C3
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.
How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
Last edited:
#### tkhunny
##### Well-known member
MHB Math Helper
I don't disagree with your solution. I suggest only that the definition of "garland" is ambiguous.
After a quick survey of those in my home, they ALL believed a "garland" to be more of a hanging rope, and not a wreath. I believe we have established significant ambiguity (at least cultural) in the problem statement.
Perhaps a picture was provided.
#### grgrsanjay
##### New member
But the difficulty there is these are garlands, circular bead arrangements.
$BBBBBBBBBRRR$ is the same as $RRRBBBBBBBBB$.
Moreover, $RBBRBBBRBBBB$ is the same as $RBBBBRBBRBBB$.
How many ways are there to partition $9$ into three of less summands?
For example here are three: $9$, $7+2$ and $5+3+1$.
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this
x + y + z = 12
Case 1:x=y=z
No.of.ways = (4,4,4) = 1
Case 2:two of x,y,z are equal
No.of ways = (1,1,10),(2,2,8),(3,3,6),(5,5,2),(6,6,0),(0,0,12) = 6
Case 3: x,y,z are all unequal
To avoid double counting , i will take x<y<z
No.of ways =
(0,1,11),(0,2,10),(0,3,9),(0,4,8),(0,5,7)
(1,2,9),(1,3,8),(1,4,7),(1,5,6)
(2,3,7),(2,4,6)
(3,4,5)
= 12 ways
So,total no.of ways is 19
Last edited:
#### Plato
##### Well-known member
MHB Math Helper
the definition of "garland" is ambiguous.
The definition may be ambiguous here but not in textbooks on countilng. Here is a link. The subtopic on beads is the one used in counting problems.
So,there must be less than 14C2 solutions??How do i eliminate them?
Ok,could you check whether i can do it like this
So,total no.of ways is 19
YES, 19 is correct
This sort of problem has set many arguments. If you agree on what meaning of garland we use then the following will show you why. If we have three green beads and three red beads how many different garlands are possible? Well the answer is 3. All the teds together; two reds together and one separate; then all the reds are separated. There are no other possibilities.
Three can be written as 3, 2+1, or 1+1+1.
The questions using two colors have that nice solution. They quickly become a nightmare with more than two colors.
Last edited:
#### grgrsanjay
##### New member
I think none of my case were repeated...could you conform it whether any case is left?
#### Plato
##### Well-known member
MHB Math Helper
I think none of my case were repeated...could you conform it whether any case is left?
Yes you are correct. At first I read it as if there were only 12 beads altogether. | 2020-09-26T07:20:03 | {
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http://aipc.outdoortown.it/parametric-to-rectangular-with-trig.html | Parametric To Rectangular With Trig
To get the cartesian equation you need to eliminate the parameter t to get an equation in x and y (explicitly and implicitly). Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Subtract 7 7 from both sides of the equation. to rectangular coordinates 5. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Write the equations of the circle in parametric form. x = 1 t - 2 3. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. The rectangular coordinates for P (5,20°) are P (4. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Finding cartesian equation from parametric trigonometric equations. For this we need to find the vectors and. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. Now, first of all, what does it mean to convert to trigonometric form? Well, I have my number in rectangular form, so it's in a+bi form. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. Parametric form defines both the x-and the y-variables of conic sections in terms of a third, arbitrary variable, called the parameter, which is usually represented by t. Convert (3, -1) to polar form. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. For a circle of diameter 10 that has been translated right by 4 units and down by 7 units: (a) Give the parametric equations defining the circle. In the first three examples, x and y are given in terms of different trig functions. - 43 Questions with solutions. Parametric Graphing Team Desmos February 21, 2020 14:44. There's also a graph which shows you the meaning of what you've found. By adjusting the parametric equations, we can reverse the direction that the graph is swept. Polar plot on Cartesian axes // parametric conversion of ugly trig function? Hello! I currently have a polar plot which I would like to superimpose onto square axes. We can derive this identity by drawing a right triangle with leg lengths 1 and t and applying the usual definitions of trig and inverse trig functions. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. Combine the triangle and the Cartesian coordinate grid to produce Trigonometry. Other Parent Functions C. Sketching the Graph of Trigonometric Parametric Equations. To eliminate t in trigonometric equations, you will need to use the standard trigonometric identities and double angle formulae. Replace and with the actual values. An alternative approach is two describe x and y separately in terms of a. Develop the calculus for polar and parametric forms. Drag the slider to change the number of sides on the polygon. The Rectangular Coordinate Systems And Graphs Algebra Trigonometry. All for only $14. Easy to get confused between these and inverse trig functions! Trig Proofs & Identities. Sketch the graph of the parametric equations $$x=t^2+t$$, $$y=t^2-t$$. Eliminating the Parameter. sss s ss s sss ss s s ss sss s s s sss s s s IV. Download [74. Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. Find more Mathematics widgets in Wolfram|Alpha. Worksheets are Polar and rectangular forms of equations date period, Polar coordinate exercises, Polar coordinates, Trigonometry 03 notes marquez, , Polar coordinates parametric equations, Parametric equations context parametric and polar equations. Ron Larson + 1 other. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. Area under one arc or loop of a parametric curve. I don't know what to do from here or if I'm going in the right direction or not. Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Simplify (x −7)2 ( x - 7) 2. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Fill in the table and sketch the parametric equation for t [-2,6] x = t2 + 1 y = 2 – t 5 6 Problems 2 – 10: Eliminate the parameter to write the parametric equations as a rectangular equation. Example 1 - Graphing Parametric Equations; Example 2 - Parametric to Rectangular Form; Day 2 - 7. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Example 6) Finding parametric equations for a given function is easier. The ordered pairs, called polar coordinates, are in the form $$\left( {r,\theta } \right)$$, with $$r$$ being the number of units from the origin or pole (if $$r>0$$), like a radius of a circle, and $$\theta$$ being the angle (in degrees or radians) formed by the ray on the positive $$x$$ - axis (polar axis), going counter-clockwise. You can find values for both x and y by plugging values for t into the parametric equations. Applications. Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. I was trying to solve for x for some reason. (b) Convert the parametric equations to rectangular form (please give the circle in standard form). Complete pp. Polar and Parametric Equations Rev. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. Rotating a Curve defined by a Equation. We can express equation (i) in terms of t, therefore we see that t = x + 3,. sss s ss s sss ss s s ss sss s s s sss s s s IV. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. How to represent Parametric Equations. Set up the parametric equation for x(t) x ( t) to solve the equation for t t. Parametric and Polar Equations Review Name 1. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. Parametric equations involving trigonometric functions Finding areas Parametric equations An equation like y = 5x + 1 or y x x 3sin 4cos or xy22 1 is called a cartesian equation. x=t-l, Check these with yesterday's graphs: B. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. Parametric Curves. Converting Polar Equations To Rectangular Equation. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. y for values of. Simplify (x −7)2 ( x - 7) 2. Complete Algebra 2 and Trig Program: Requirements: Requires the ti-83 plus or a ti-84 model. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. Historic applications of parametric equations are discussed so the use of them is realized. Calculus of a Single Variable. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = − = as a function of x. Calculus and parametric curves. Graphing a plane curve represented by parametric equations involves plotting points in the rectangular coordinate system and connecting them with a smooth curve. The resulting equation is a rectangular equation. Credit Pre-Calculus Honors *All areas are accelerated* Generating Graphs Shifting, Symmetry, Reflections, and Stretching Roots, Max/Min, Values, and Intersections Composite Functions Programming the Tl-84 Polynomial Functions Brief Quadratic Review Roots Shifted and Standard Form Axis. A cartesian equation gives a direct relationship between x and y. That make sense now. Graphing a Parabola with vertex at (h ,k ). Find more Mathematics widgets in Wolfram|Alpha. Show the orientation (flow) by arrows (2) Convert to a rectangular equation and Converting from. Polar coordinates simplified the work it takes to arrive at solutions in most trigonometry problems. 142 Notes - Section 8. Then use a trigonometric identity. The functions and can serve as a parametric representation for a function , which is plotted in purple on the , plane cutting , shown in light gray. Subtract 7 7 from both sides of the equation. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. The equation is the general form of an ellipse that has a center at the origin, a horizontal major axis of length 14, and. The rectangular coordinates (x , y) and polar coordinates (R , t) are related as follows. The student is expected to: (A) graph a set of parametric equations;. y = x -3 is equivalent to 3 cos sin 3 (cos sin ) 3 3 cos sin xy rr r r TT TT TT. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. x=t-l, Check these with yesterday's graphs: B. Write the complex number in trigonometric form, using degree measure for the argument. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Let us just look at a simple example. Clearly, both forms produce the same graph. Then, we do this substitution into the function: x → x c - y d y → x d + y c. Eliminate the parameter from the given pair of trigonometric equations where $$0≤t≤2\pi$$ and sketch the graph. For the first case we need to supplement the equations by two inequalities: 0 <= t <= 4 Pi && 0 < x < 4 Pi. Eliminate the parameter and find a corresponding rectangular equation. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Construct a table of values for the given parametric equations and sketch the graph: the data from the parametric equations and the rectangular equation are plotted together. Converting Polar To Rectangular. A cartesian equation gives a direct relationship between x and y. x = ½t + 4. x = cos 2t, y = sin t, for t in 1-p, p2. In rectangular coordinates, each point (x, y) has a unique representation. ACE TRIG Final EXAM REVIEW. After going through these three problems can you reach any conclusions on how the argument of the trig functions will affect the parametric curves for this type of parametric equations?. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Solve trigonometric equations in quadratic form 12. now expanding (x+2)² either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)² from your original parametric equation -(t²) = -x^2 - 4x - 4. Students model linear situations with parametric equations, including modeling linear motion and vector situations. the domain of the rectangular equation so that its graph matches the graph of the parametric equations. We can express equation (i) in terms of t, therefore we see that t = x + 3,. Complete pp. Download [74. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Example 1: 3, 4 1, for -4 t 2xt y t dd 2 2 Example 2: 2, , ( , )xt y t fortin ff t x y 5/7/19 9. Place the parametric equations in rectangular form. The applet below illustrates parametric coordinate functions for various polygonal trig functions. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. In the diagram such a circle is tangent to the hyperbola xy = 1 at (1,1). Rectangular - Polar - Parametric "Cheat Sheet" 15 October 2017 Rectangular Polar Parametric Point ( T)= U ( T, U) ( , ) • ( N ,𝜃) or N ∠ 𝜃 Point (a,b) in Rectangular: T( P)= U( P)= < , > P=3𝑟 𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. The coordinates are measured in meters. Then find a second set of polar coordinates for the. 7: Complex Numbers, Polar Coordinates, Parametric equationsFall 2014 2 / 17 Complex Numbers - trig form Example (Write the complex number in standard form). Parametric equations are equations that are used to introduce polar coordinates and their relation with rectangular coordinates. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Things to try. x = cos 2t, y = sin t, for t in 1-p, p2. Find more Mathematics widgets in Wolfram|Alpha. sssssss ssss sssss Homework Assignment Page(s) Exercises. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Linear 64) 2. Trigonometry made completely easy! Our Trigonometry tutors got you covered with our complete trig help for all topics that you would expect in any typical Trigonometry classes, whether it's Trigonometry Regents exam (EngageNY), ACT Trigonometry, or College Trigonometry. Algebra Review: Completing the Square. Get the free "parametric to cartesian" widget for your website, blog, Wordpress, Blogger, or iGoogle. Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Comments There are no comments. In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. However, they used meters instead of feet for gravitational constant. To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas that we’ve developed require that functions be in one of these two forms. An object travels at a steady rate along a straight path $$(−5, 3)$$ to $$(3, −1)$$ in the same plane in four seconds. Write each pair of parametric equations in rectangular form. Find a set of parametric equations to represent the graph of y = (x – 1)2 given the. Search this site. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. asked by Sam on December 9, 2013; geometry. Example 6) Finding parametric equations for a given function is easier. Arc length of a parametric curves. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Therefore, we need to use a change of variables so we can integrate using either cylindrical (polar) or spherical coordinates, or even parametric form. Find more Mathematics widgets in Wolfram|Alpha. How do you convert #r=2sin(3theta)# to rectangular form? Trigonometry The Polar System Converting Between Systems. 2 Plane Curves and Parametric Equations 711 Eliminating the Parameter Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. Pre - Calc. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. Graphing a Hyperbola with center at (0 ,0 ). (1) $$f(x,y)=0$$ These are sometimes referred to as rectangular equations or Cartesian equations. Find a rectangular equation for each plane curve with the given parametric equations. The parametric equations are plotted in blue; the. I have the parametric coordinates x=Sec(t) and y=Tan 2 (t) where 0<=t P=3𝑟 N𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. I'm trying to find the cartesian equation of the curve which is defined parametrically by: $$x = 2\sin\theta, y = \cos^2\theta$$ Both approaches I take result in the same answer:$$y = 1 - \s. Step-by-Step Examples. what is the volume of that rectangular pyramid? asked by riza on August 6, 2012. Angle t is in the range [0 , 2Pi) or [0 , 360 degrees). We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. The rectangular coordinates for P (5,20°) are P (4. x = 7 sin and y = 2 cos 62/87,21 Solve the equations for sin and cos. Parametric and Polar Equations Review Name q) 1. Parametric equation of an ellipse and a hyperbola rectangular hyperbola cartesian and parametric forms examsolutions maths tutorials parametric equations hyperbola hw 1 conic sections in polar parametric forms lesson Parametric Equation Of An Ellipse And A Hyperbola Rectangular Hyperbola Cartesian And Parametric Forms Examsolutions Maths Tutorials Parametric Equations Hyperbola Hw 1 Conic. In , the data from the parametric equations and the rectangular equation are plotted together. Historic applications of parametric equations are discussed so the use of them is realized. SOLUTION The graph of the parametric equations is given in Figure 9. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Textbook Authors: Lial, Margaret L. This is also a great Review for AP Calculus BC. When you first learned parametrics, you probably used t as your parametric variable. Introduction to Parametric Equations; Parametric Equations in the Graphing Calculator; Converting Parametric Equations to Rectangular: Eliminating the Parameter; Finding Parametric Equations from a Rectangular Equation; Simultaneous Solutions; Applications of Parametric Equations; Projectile Motion Applications; Parametric Form of the Equation of a Line in Space. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. Simply let t x and then replace your y with t. It can handle horizontal and vertical tangent lines as well. One caution when eliminating the parameter, the domain of the resulting rectangular equation may need to be adjusted to agree with the domain of the parameter as given in the parametric equations. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. 3+3𝑖 3√2(cos45°+𝑖sin45°) Write the complex number in the form + 𝒊. Textbook Authors: Lial, Margaret L. This problem is about converting parametric equations to rectangular form. Sketch and identify graphs using parametric equations. This is called a parameter and is usually given the letter t or θ. Find a rectangular equation for each plane curve with the given parametric equations. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos t (x = \cos t (x = cos t and y = sin t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. For the problems above, let x = t + 2 and find the resulting parametric equations. ACE TRIG Final EXAM REVIEW. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in [link]. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Algebra Review: Completing the Square. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Plot the points. Sketching the Graph of Trigonometric Parametric Equations. We will discuss the polar (trigonometric) form of complex numbers and operations on complex numbers. A parametric equation is where the x and y coordinates are both written in terms of another letter. Plot the following points on the polar grid at the right and label each point. Find all solutions to a trigonometric equation 10. Find new parametric equations that shift this graph to the right 3 places and down 2. The coordinates are measured in meters. Polar and Parametric Equations 3. The volume of. 2;5p 3 Give two alternate sets of coordinates for each point. 6 Plane Curves, Parametric Equations. Many of the advantages of parametric equations become obvious when applied to solving real-world problems. But by recognizing the trig identity, we were able to simplify it to an ellipse, draw the ellipse. PreCalculus Class Notes VP5 Converting Parametric and Rectangular Equations Review To convert from parametric equations to rectangular equation: solve the x equation for t, substitute into the y equation Example Rewrite 2 3 2 4 x t y t = − = as a function of x. the process of solving one parametric equation for t so you may substitute that equation into the other parametric equation for t to create a rectangular equation where y=f(x) Example: if x=t-4 and y=¼t solve the first equation for t so t=x-4 then by substitution y=¼(x-4) or y=¼x-1. 95 per month. Thus, keep only the other equation. x: 4 sin (2t) Y : 2 cos (2t) X : 4+2 COS t. Find more Mathematics widgets in Wolfram|Alpha. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. Graph polar equations. x = -2 cos t, y = 2 sin t, 0 lessthanorequalto t lessthanorequalto 2 pi. x = 3t – 1, y = 2t + 1. 4 Trigonometric Functions of Any Angle 497 1-3 4. Find a Cartesian equation for the curve traced out by this function. Trig functions, complex numbers, identities, vectors, and real life applications like circuits are here to energize your classroom. Sketch and identify graphs in polar coordinates. First, we find a vector {c,d} of distance 1 having angle -θ, which is {Cos[-θ], Sin[-θ]}. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 7. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Write the equations of the circle in parametric form. Parametric Equations and Vectors : Questions like write each pair of parametric equations in rectangular form. Parametric and Polar Equations Review Name 1. Challenge Sets. A coordinate system is a scheme that allows us to identify any point in the plane or in three-dimensional space by a set of numbers. Change the parametric equations to rectangular form eliminating the variable t and sketch 11. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Show all algebraic support. The Second Fundamental Theorem of Calculus Integration Involving Powers of Trigonometric Functions. This is the a value, this is the b value. A parametric curve in the plane is a pair of functions x = f (t) y = g (t) It is possible to derive the Cartesian equation from the parametric equations. x = cos 2t, y = sin t, for t in 1-p, p2. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. The volume of. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Historic applications of parametric equations are discussed so the use of them is realized. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Convert the polar coordinates (5 , 2. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Plot the resulting pairs ( x,y ). Based on your work on the above problems, name another benefit of parametric equations versus rectangular functions. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Example $$\PageIndex{7}$$: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations. Write the equations of the circle in parametric form. Use a table of values to sketch a Parametric Curve and indicated direction of motion. In parametric equations x and y are both defined in terms of a third variable. 1 Answer A. ; Hornsby, John; Schneider, David I. Polar coordinates. 244 Chapter 10 Polar Coordinates, Parametric Equations conclude that the tangent line is vertical. 7-8: 2-8 even, 9-12 all, 14-32 even. In the past, we have seen curves in two dimensions described as a statement of equality involving x and y. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Things to try. Xmin = -20 Ymin = -12 Xmax = 20 Ymax = 12 Xscl = 5 Yscl = 5. What type of path does the rocket follow? Solution: The path of the rocket is defined by the parametric equations x = (64 cos 30°)t and y = (64 sin 30°)t − 16t2 + 3. Convert to Polar Coordinates. We will graph polar equations in the polar coordinate system and finally discuss parametric equations and their graphs. Both types depend on an argument, either circular angle or hyperbolic angle. Calculus of a Single Variable. Converting from rectangular to parametric can be complicated, and requires some creativity. Tools We Need x = r * cos θ y = r * sin θ (some trigonometric identities are required) y = r * sin t y = 2 * cos t * sin(cos⁻¹(r/2)). Find a rectangular equation for each plane curve with the given parametric equations. Usually will stand for time. Eliminating the parameter is a method that may make graphing some curves easier. A common parameter used is time (t) or an angle (trig) (x, y) is the place, "t" is the time it is there (at that place) (at that place) Graphing Parametric Equations 2 OPTIONS: (l) Use a chart to find rectangular points. Recall the trig identity d1 Substitute x/r and y/r into the identity: Remove the parentheses: Multiply through by r 2. The curves are colored based on the quadrants. Trig Ratios and Quadrants (9:52) Inverses of Trig Functions and Inverse Trig Functions (20:18) Graphing Sine and Cosine--Amplitude and Period (11:15) Graphing Sine and Cosine--Vertical and Horizontal Shift (6:33) Graphing Sine and Cosine by Hand (29:10) Graphing Tangent by Hand (24:39) Applications (15:50) Unit 4: Triangle Applications of Trig. The parametric equations are plotted in blue; the. A curve is given by the parametric equations: #x=cos(t) , y=sin(2t)#, how do you find the cartesian equation? Calculus Parametric Functions Introduction to Parametric Equations. Homework Statement Reduce these parametric functions to a single cartesian equation:$\displaylines{ x = at^2 \cr y = 2at \cr} \$. Drag points A and B to change the size and orientation of the polygon. now the rest is a matter of simplification:. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Graphing a Parabola with vertex at (h ,k ). And then by plotting a couple of points, we were able to figure out the direction at which, if this was describing a particle in motion, the direction in which that particle was actually moving. In parametric equations x and y are both defined in terms of a third variable. It can handle horizontal and vertical tangent lines as well. We show how we can transform between these representations of the same plane. 81 KB] Parametric Differentiation : The parametric definition of a curve, differentiation of a function defined parametrically, exercises, …. Plot the resulting pairs ( x,y ). (θ is normally used when the parameter is an angle, and is measured from the positive x-axis. But how do we write and solve the equation for the position of the moon when the distance from the planet,. In the past, we have been working with rectangular equations, that is equations involving only x and y so that they could be graphed on the Cartesian (rectangular) coordinate system. This calculator converts between polar and rectangular coordinates. Use power reducing formulas 9. now the rest is a matter of simplification:. In the entry line, type cos(t) All of the trigonometric functions and the inverse trigonometric functions can be found in the trigonometry section of the catalog. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. Find exact values of composite functions with inverse trig functions 8. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. Trigonometric substitution. Search this site. I'm not looking for the answer here. 1 Answer Cesareo R. 1 Complex Numbers 8. Parametric curves have a direction of. x =3cosθ and y =4sinθ, 02≤θ≤ π by eliminating the parameter finding the corresponding rectangular equation. Write the equations of the circle in parametric form. Thank you for purchasing this product! This activity is suitable for PreCalculus - Trigonometry and can be used as a The introduction to concept of Parametric Equations can be difficult for students and yet this topic is so important, especially later in Calculus. Use a table of values to sketch a Parametric Curve and indicated direction of motion. Graphing a Parabola with vertex at (h ,k ). Convert to Polar Coordinates. Let's look at each solution carefully: The first solution is correct since: r^2 = x^2 + y^2. Subtract 7 7 from both sides of the equation. position time parametric equations path rectangular equation eliminating the parameter square root function direction of motion. now the rest is a matter of simplification:. orgChapter 1. We will then introduce the polar coordinate system, which is often a preferred coordinate system over the rectangular system. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Applications. Sketch and identify graphs using parametric equations. Applications of Parametric Equations. This is the a value, this is the b value. interesting variations on the parametric equations. Linear 64) 2. Two versions. Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice. 5 Parametric Equations part 2. Find the rectangular equation that models its path. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. Finding Parametric Equations In Exercises 35 and 36, find two different sets of parametric equations for the rectangular equation. Graphs of trigonometric functions in polar coordinates are very distinctive. Lots of vocab in this one -- specific solutions, general solutions, 2npi, 360n -- and lots of factoring to do too. Graph Type⎮ Parametric. Identify the type of graph. Making statements based on opinion; back them up with references or personal experience. A summary of Graphing in Polar Coordinates in 's Parametric Equations and Polar Coordinates. 2 Exercises - Page 365 36 including work step by step written by community members like you. Because the x- and y-values are defined separately in parametric equations, it is very easy to produce the inverse of a function written in parametric mode. Finding cartesian equation from parametric trigonometric equations. This calculator converts between polar and rectangular coordinates. For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. Application: Toy Rocket The parametric equations determined by the toy rocket are Substitute from Equation 1 into equation 2: A Parabolic Path 8. Ask Question Asked 5 years, 7 months ago. Sketch and identify graphs in polar coordinates. Find a rectangular equation for each plane curve with the given parametric equations. Find more Mathematics widgets in Wolfram|Alpha. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations Number Theory Statistics & Probability Business Math Challenge Problems Math Software. An alternative approach is two describe x and y separately in terms of a. 1 shows points corresponding to θ equal to 0, ±π/3, 2π/3 and 4π/3 on the graph of the function. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Then, we do this substitution into the function: x → x c - y d y → x d + y c. The hyperbolic trigonometric functions extend the notion of the parametric equations for a unit circle (x = cos t (x = \cos t (x = cos t and y = sin t) y = \sin t) y = sin t) to the parametric equations for a hyperbola, which yield the following two fundamental hyperbolic equations:. Find parametric equations for 2x 4 a) y = x 2 —2x+3 Note that there are many ways of finding parametric equations for a given function. 2: Trigonometric Functions In this lesson you will use parametric equations to illustrate the connection between the graphs of y = sin(x) and the unit circle. Example 3: Transform the equation x 2 + y 2 + 5x = 0 to polar coordinate form. Just as with non-angle parameters, when the parameter is an angle θ the plane curve can be graphed by selecting values for the angle and calculating the x- and y-values. Covers 55 algebra and trigonometry topics including synthetic division, conics, statistics, quadratics and more. This is also a great Review for AP Calculus BC. Jimmy wants to rewrite the set of parametric equations x = 1/2 T + 3 and y = 2T - 1 in rectangular form by eliminating T. Rewrite the equation as t+7 = x t + 7 = x. now expanding (x+2)² either by using the identity of a perfect square which x^2 + 2bx + b^2 or simply expanding (x+2)(x+2) you have = x^2 + 4x + 4 and then multiplying by -1 as such because you have -(x+2)² from your original parametric equation -(t²) = -x^2 - 4x - 4. Finding all arguments t in 0 <= t <= 4Pi where the parametric graph intersects. You will also see how to transform the graph of y = sin(x) to obtain the graph of y = sin[B(x + C)] + D. Pre - Calc. Find the rectangular coordinates of. The Polar coordinates are in the form (r,q). Then x=f(t) and y=g(t) are called parametric equations for the curve represented by (x,y). Equations for and are plotted on the perpendicular and planes as varies from to. Graph polar equations. ; Daniels, Callie, ISBN-10: 0321671775, ISBN-13: 978-0. 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If you're seeing this message, it means we're having trouble loading external resources on our website. Calculus of a Single Variable. Rotating a Curve defined by a Equation. The students will analyze and graph Sine, Cosine and Tangent and their inverses. 2sec t and y = -0. It is a parabola with a axis of symmetry along the line $$y=x$$; the vertex is at $$(0,0)$$. We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. This is also a great Review for AP Calculus BC. Trig equations are problems where you're solving for X or Theta but they're hidden behind a trig function, like "2sinX-1=0" or "tan 2 X-1=0". Thanks for contributing an answer to Mathematics Stack Exchange! Finding cartesian equation for trigonometric parametric forms. Definition of Trig Functions Trig Model for Data Graphing Sine and Cosine Functions (3) Relations and geometric reasoning. 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The Organic Chemistry Tutor 247,826 views 33:29. x-2+t y-2-t, for t in [-2,3] Equation in Rectangular form: (8 points each) b. Solve trigonometric equations in quadratic form 12. Basic graphing with direction to them. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. T= Ncos𝜃 U= N O𝑖𝜃 N P 𝜃= U T −. Comprehensive End of Unit Review for the Polar, Parametric, and Vectors Sections of PreCalculus or Trigonometry plus Graphic Organizer. Plot the points. 1 Answer A. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. 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For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same. uTo graph parametric functions You can graph parametric functions that can be expressed in the following format. 2 Exercises - Page 365 36 including work step by step written by community members like you. 4 De Moivre's Theorem; Powers and Roots of Complex Numbers 8. Trigonometry (10th Edition) answers to Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8. Textbook Authors: Lial, Margaret L. 8/14/2018 12:12 AM §10. Pre-calculus Contents C Parametric Equations Parametric and rectangular forms of equations conversions The parametric equations of a quadratic polynomial, parabola The Trigonometric functions of arcs from 0 to. Parametric Equations: Eliminating Angle Parameters In a parametric equation the parameter can represent anything including an angle. Usually will stand for time. y for values of. 2;5p 3 Give two alternate sets of coordinates for each point. ) Drawing the graphTo draw a parametric graph it is easiest to make a table and then plot the points:Example 1 Plot the graph of the. Lines in polar coordinates: Let and then the polar equations of the lines x=a and y=b are and for all values of Parametric form of derivatives:. Drag P and C to make a new circle at a new center location. 6 Plane Curves, Parametric Equations. parameter t = x – 1. Math Algebra 2 Calculus Precalculus Trigonometry Math Help Complex Numbers Pre Calculus Polar Forms Ellipse High School: Math Derivatives Polar Coordinates Sine Convert Calculus 3 Rectangular Calculus 2 Calculus 1 Polar Equation. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and. (a) Eliminate the parameter for the curve given by the parametric equations x= 2 t, y= 1 t2 for 1 P=3𝑟 N𝑖 , Q O P𝑖 , with 1 degree of freedom (df) Polar Rect. y = x 2 -2. Graphing parametric equations: The key is to plug in useful points within the specified range of t, not just any points. Trig Functions sine I First, solve for 3. Plot the resulting pairs ( x,y ). It can handle horizontal and vertical tangent lines as well. The Second Fundamental Theorem of Calculus. This problem is about converting parametric equations to rectangular form. Quiz: Trig Form of Complex Numbers, Parametric Equations, Polar Coordinates & Equations 9. Let A be the point where the segment OB intersects the circle, where point B lies on the line x = 2 a. Rockwall ISD Pre-Calculus Parent Guide 3 Unit 7 Trigonometric Functions In this unit students will continue to apply trigonometric functions including rotation angles, the Unit Circle and periodic functions. A PLANE CURVE is whereas and are continuous functions on t on an interval and the set of ordered pairs. Other Parent Functions C. For the point and So, the rectangular coordinates are See Figure 10. Comments There are no comments. ACE TRIG Final EXAM REVIEW. Calculus Examples. (a) Find a polar equation of the cissoid. Write each pair of parametric equations in rectangular form. There's also a graph which shows you the meaning of what you've found. And Vector Calculus, which tends to be the last chapter for Calculus 3, deals with work on, in, and around a surface which will predominately involve polar coordinates as well. Parametric Equations and Motion Precalculus Vectors and Parametric Equations. I was trying to solve for x for some reason. Use MathJax to format equations. y = x 2 -2. t over the interval for which the functions are defined. Given a point in polar coordinates, rectangular coordinates are given by Given a point in rectangular coordinates,. Trigonometry (MindTap Course List). In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x x and y. Exchanging x and y 4. The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. To eliminate the parameter in equations involving trigonometric functions, try using the identities. Both types depend on an argument, either circular angle or hyperbolic angle. To this point (in both Calculus I and Calculus II) we’ve looked almost exclusively at functions in the form $$y = f\left( x \right)$$ or $$x = h\left( y \right)$$ and almost all of the formulas that we’ve developed require that functions be in one of these two forms. Finding Parametric Equations for a Graph (Page 799) Describe how to find a set of parametric equations for a given graph. 95 per month. Write each pair of parametric equations in rectangular form. Trigonometric substitution. Example for parametric equations are {eq}\displaystyle x=\sqrt{t},\:y=t-5 {/eq}, what we need to do first is to find x. I'm not looking for the answer here. (a) y = x -3 This is the equation of a line. Finding Parametric Equations from a Rectangular Equation (Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section). Converting between polar and rectangular form Converting equations between polar and rectangular form Homework: Finish Day 1 Packet (Optional) Pg. So our parametric equation is x = t and y = 4t. | 2020-07-03T19:58:35 | {
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https://math.stackexchange.com/questions/1960055/proof-of-infinitely-many-prime-numbers/1960060 | # Proof of infinitely many prime numbers
Here's the proof from the book I'm reading that proves there are infinitely many primes:
We want to show that it is not the case that there only finitely many primes. Suppose there are finitely many primes. We shall show that this assumption leads to a contradiction. Let $p_1, p_2,...,p_n$ be all the primes there are. Let $x = p_1...p_n$ be their product and let $y=x+1$. Then $y \in \mathbb{N}$ and $y \neq 1$, so there is a prime $q$ such that $q \mid y$. Now $q$ must be one of $p_1,...,p_n$ since these are all primes that there are. Hence $q \mid x$. Since $q \mid y$ and $q \mid x$, $q \mid (y-x)$. But $y-x=1$. Thus $q \mid 1$. But since $q$ is prime, $q \geq 2$. Hence $q$ does not divide 1. Thus we have reached a contradiction. Hence our assumption that there are only finitely many primes must be wrong. Therefore there must be infinitely many primes.
I have a couple of questions/comments regarding this proof. I will use a simple example to help illustrate my questions:
Suppose only 6 primes exist: $2, 3, 5, 7, 11, 13$
Let $x = p_1p_2p_3p_4p_5p_6=30,030$
Let $y = x + 1 = 30,030+1 = 30,031$
1. The proof states there is a prime $q$ such that $q \mid y$ and that $q$ must be either $p_1, p_2, p_3, p_4, p_5,$ or $p_6$. However, none of the 6 primes listed, $(2,3,5,7,11,13)$, divides $30,031$. In fact, the only divisors for $30,031$ are $1, 59, 509$ and $30,031$. Doesn't the proof then break down here since there is no prime $q$ that divides $y$?
2. The prime factorization of $30,031$ is $59 \times 509$. These two numbers are factors of $30,031$ and are in fact primes themselves since they are only divisible by themselves or by 1. Have I shown that there exists $\gt 6$ primes? If so, what can I conclude now that I have shown this?
3. I don't understand why the contradiction $q$ divides $1$ and $q$ does not divide $1$ lead us to the assumption that the finitely many primes must be wrong. I understand how we reached the contradiction. I don't understand why contradiction leads us to the conclusion shows that are assumption that there is only finitely many primes is wrong.
My apologies for the long post. Thanks for any and all help.
• unique factorization requires a unique prime decomposition. If no $p$ exists, there must be a new prime... – Eleven-Eleven Oct 8 '16 at 23:50
• The point of the argument is that there must be some prime not in the list that divides the number you produce. Your example confirms this! You found two such primes $59,509$ neither of which is on the list. Thus, your list can't have been complete. – lulu Oct 8 '16 at 23:51
• The proof makes an assumption that there are finitely many primes, But it then goes on to show, given the conditions, this actually can't be the case. Therefore, the flaw is in your assumption, since all arguments in the proof are mathematically valid. – Eleven-Eleven Oct 8 '16 at 23:53
• As stated above, all of your mathematical arguments are correct when you started from your original assumption. Yet you are led to a point in the proof that is impossible. If the mathematical arguments are sound and correct, what is left for there to be wrong...? The original assumption must therefore be incorrect. – Eleven-Eleven Oct 8 '16 at 23:55
• 2) In this case that there are at least 7 primes. But if you generalize this, that p1p2...pn+1 is never divisible by p1,...,pn then p1,.....pn can not be a list of all primes. No such list can exist. So there can not be a finite number of primes. – fleablood Oct 9 '16 at 0:33
This is an excellent example of a proof that is traditionally phrased as a proof by contradiction but is much better understood constructively. From a constructive viewpoint, the proof shows that given any list of primes $p_1, \ldots, p_n$ there is a prime $q$ (any prime divisor of $p_1p_2\ldots p_n + 1$) that is distinct from each $p_i$. So given any finite set of primes we can find a prime that is not in that set.
• +1 for emphasizing the constructive aspect of this proof. – P Vanchinathan Oct 9 '16 at 0:44
• Except that this process doesn't actually construct a new prime (the example $y = 30031$ being composite). – Daniel R. Collins Oct 9 '16 at 2:20
• @Rob: I think Daniel’s point is that the proof constructs an integer $y>1$ that is not divisible by $p_1$, $p_2$, …, $p_n$, and then it asserts that there must be a prime $q$ such that $q \mid y$, but it doesn’t construct $q$. – Scott Oct 9 '16 at 4:10
• @Scott But there's a simple algorithm to generate a prime divisor of any given integer. – Jack M Oct 9 '16 at 8:05
• @djechlin This is a constructive proof and you're utterly wrong. The set of divisors of a positive integer is finite. Now just take the minimal element greater than $1$ in the set of divisors; this is obviously prime. Would you say that finding the minimal element in a finite set of integers is not constructive? By the way, the proof by contradiction seems to imply that $p_1p_2\dots p_n+1$ is prime, which is the main source for confusion in newbies. – egreg Oct 9 '16 at 11:20
1. The proof does break down in a sense. You have reached a contradiction, which means the hypothesis that there are finitely many primes can not possibly be true.
2. So you thought you had 6 primes, you found 2 extra. Can you just add these two to your list and get all of the primes? If you repeat the process on these 8 primes, you will find that you will have even MORE primes by considering the product + 1. You can keep going and going and you will never run out of primes. This is the idea of the proof. It uses contradiction because you can do it all in one step and avoid the potential issue of doing the process infinitely many times.
Suppose only 6 primes exist: 2,3,5,7,11,13
However, none of the 6 primes listed, (2,3,5,7,11,13), divides 30,031.
Then we already have a contradiction. Since there are only 6 primes (we supposed that at the beginning) and none of them divide 30,031, then 30,031 must be prime. However, 30,031 is not one of the only 6 primes that exist. So 30,031 cannot be prime, yet it must be prime.
So the proof works. In fact, it works precisely the same regardless of what set of numbers we suppose are the only primes that exist. Thus no finite set of numbers can include all the primes that exist. Thus there are an infinite number of primes.
• I think you mean "... no finite set of numbers ..." – origimbo Oct 10 '16 at 10:48
You ask in #1 if the proof breaks down. No. What breaks down is the assumption that there are no more primes. You assumed you had a complete list of primes. Then you constructed a number that is not divisible by any of the primes in your list. Only two possibilities remain: Either the number you constructed is prime, or it is divisible by a number that is not in your list. Either way your list is not complete, which contradicts your initial assumption.
Simpy put, you assumed you had a complete list and by using that assumption you proved that the list was not complete. The assumption must therefore be wrong.
The answer to #3 is basically the same. By finding a contradiction when you made an assumption, you proved the assumption to be incorrect.
The answer to #2 is no, you did not prove you anything by noting that 59 and 509 are both prime. You are trying to prove that there is a finite list of primes. If you choose a particular set of primes as you did {2, 3, 5, 7, 11, 13} and show that that particular set doesn't hold all the primes, a skeptic would just say that you need to add more primes (like 59 and 509) because you didn't make your set big enough. The proof has to be more general which is why it doesn't say how many primes are in the finite set. The proof is written so that it works no matter how big the finite set is.
Point 1: It's a theorem that any natural number $n>1$ has a prime factor. The proof is easy: for any number $n>1$, the smallest natural number $a>1$ which divides $n$ is prime (if it were not prime, it would not be the smallest).
Point 2: Yes, you have proved there are more than six primes. So what? The proof by contradiction doesn't suppose there are only six, but that there are a finite number of them.
Point 3: Actually, it's not really a proof by contradiction stricto sensu. It is proved that any finite list of primes in incomplete.
Here is a non-mathematical approach to the logic behind the proof by contradiction...
Suppose your assumption is that a suspect in a brutal murder is innocent because he said he couldn't have been at the location the person was murdered. You feel as though he is possibly telling the truth, but what you do know is that there are 5 other suspects. You know with 100% certainty due to the detail of the case there can't be anymore than 5 other suspects via proven information. But, the police can verify using obvious physical evidence such as video cameras that the 5 suspects could not possibly be guilty. Then there is either another suspect or the one assumed innocent is guilty. But we said that it was impossible for any other suspects to be considered. Therefore our original suspect is guilty.
This is exactly what is happening in a proof by contradiction.
• This is good. But I think a better analogy is that you have 6 suspects and you assume that these are the only six possible suspects in the world. Then cameras show each and every one is innocent. Thus our assumption that we have all the suspects is wrong, and therefore there is a seventh possible person who could have (and did) do it. – fleablood Oct 9 '16 at 0:58
• Damn.... you are right. It made logical sense when i wrote it... but yours is better. – Eleven-Eleven Oct 9 '16 at 1:02
• Yours makes sense. And is a proof by contradiction. I just thought for this particular proof this is a closer analogy. I wouldn't say yours was wrong or anything. – fleablood Oct 9 '16 at 1:05
• That was my original goal; understand why contradition works. But yours does that AND is analogous to the problem at hand. Thanks – Eleven-Eleven Oct 9 '16 at 1:07
• I will concede. I get it. – Eleven-Eleven Oct 9 '16 at 23:01
The proof is essentially overexplaining. The point of contradiction would have been much clearer if the author had:
1. Reached it earlier.
2. Introduced less notation.
Since $p_1\cdots p_n+1$ cannot have any of $p_1,...,p_n$ as a prime factor we already have a contradiction.
The rest is just explaining to death why $p_1,...,p_n$ cannot be prime factors of $y=p_1\cdots p_n+1$ which should be clear since $y$ is $1$ off from being a multiple of either of those primes.
• I think the gif dosn't really fit this site, though I can't say if there is a specific rule against such things – Yuriy S Oct 10 '16 at 11:27
• @YuriyS: Hmm ... I am sorry to learn that people may find it inappropriate! That was by no means my intention! Do you think it would work better by removing the gif an leave the link and the figure of speach, or are all of those opfuscating the points I tried to make? – String Oct 10 '16 at 12:21
• @String: your link to a cartoon movie of a dead animal being bludgeoned is disproportionate and offensive. – Rob Arthan Mar 12 '17 at 1:27
• @RobArthan: Sorry, in some parts of the world such a cartoon would be considered merely a funny way to illustrate the saying about beating a dead horse. No offense intended, only a light tone. I cannot help that people do take offense, so I have removed it. Still it puzzles me how a cartoon matching the content of a saying would offend. I am from Denmark, after all. – String Mar 12 '17 at 8:11
• @String: on MSE it's easier just to use neutral language. As I am English (after all), may I point out that the cliched phrase is actually "flogging a dead horse" and it doesn't have the connotations you think it does (it's not "explaining to death", it's using a tired old argument that has lost all interest or relevance). Your cartoon doesn't help with that. – Rob Arthan Mar 13 '17 at 1:55
1. Well... yes... it does break down. You assumed that there are only 6 primes and reached a contradiction. You've successfully proved that there aren't only 6 primes.
2. Under the assumption that there are only 6 primes 30,031 isn't factorizable.
3. In proof by contradiction you prove proposition A by assuming A is not true, and through a series of logical steps reach an impossibility, thus proving that A must be true. Concurrently, in this proof, we assumed there is a finite number of primes. After a series of logical inferences we reached a contradiction. Since the only assumption we made in the process is that there are a finite number of primes, that assumption must be wrong.
1. Either $6$ primes you considered are not all the primes, or $30031$ is a new prime. Either way assumption is false. Proof by contradiction.
2. No, you haven't shown that two more primes exist. In this example yes, but not in general
3. Again, based on your assumption you reached something impossible that $q$ divides $1$. There is no such $q$ except $1$.
Start with $2,3,5$ being the only prime numbers. Then you would have $31$ as the new number. $2,3,5$ do not divide $31$. But unlike your case, $31$ is a prime number. So the assumption is wrong. | 2021-02-27T05:02:33 | {
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https://www.physicsforums.com/threads/spring-weight-and-potential-energy.735704/ | # Spring, weight and potential energy
1. Jan 30, 2014
### Bromio
1. The problem statement, all variables and given/known data
A particle is connected to a spring at rest. Because of weight, the mass moves a distance $\Delta x$. Calculate the value of the elastic constant $k$
2. Relevant equations
$U_{PE} = \frac{1}{2}k\Delta x^2$
$U = mgh$
$F = ma$ (in general)
$F = kx$ (Hooke's Law)
3. The attempt at a solution
I've tried to solve this problem from two points of view, but the result has been different in each one.
First, I've thought that, when the mass is at rest after being connected to the spring and got down to the new equilibrium position:
$mg = k\Delta x \Longrightarrow \boxed{k = \frac{mg}{\Delta x}}$
However, from the point of view of energies:
$mgh_0 = mgh + \frac{1}{2}k\Delta x^2$ (I use $\Delta x$ because the initial position is 0)
As $h_0 - h = \Delta x$, I've got:
$mg\Delta x = \frac{1}{2}k\Delta x^2 \Longrightarrow \boxed{k = \frac{2mg}{\Delta x}}$
Obviously there is a wrong factor of 2.
What's the problem here? Where is the mistake?
Thanks!
2. Jan 30, 2014
### BvU
Energy case is for the situation you let go at h=0. When it is at delta x, it's still moving. Until it is a 2 delta x, when it hangs still again. But now the spring is pulling harder and up she goes!
In other words, before there is rest, some energy has to be dissipated! You now know how much.
3. Jan 30, 2014
### Bromio
Intuitively I understand what you say, but I don't see how to write it analytically.
I mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation ($mgh_0 = mgh + \frac{1}{2}k\Delta x^2$) to include what you say?
Thanks!
4. Jan 30, 2014
### BvU
That would be a little difficult, precisely because it describes an isolated case. It says there is energy left over, that has to go somewhere to get the "in rest" situation. E.g. in overcoming the air resistance which heats up the air a little.
5. Jan 30, 2014
### Bromio
Okay. So the dissipated energy has a value of $\frac{1}{2}k\Delta x^2$, hasn't it?
What I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason?
Thank you.
6. Jan 30, 2014
### BvU
You can, actually. It's just that it isn't a steady state. mg (h-h0) is converted into spring energy plus kinetic energy. If there is no friction, the oscillation goes on forever.
Is there a risk that we are misinterpreting the OP ? My idea was Δx is measured after the thing has come to rest.
7. Jan 30, 2014
### Bromio
No misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations "at rest" situation, simply because there isn't that situation. Am I right?
So, when I study energies at the beginning ($h = 0$) and at the end ($h = \Delta x$, when entire system is at rest) I should include that dissipated energy.
If $TE_0$ and $PE_0$ are the total mechanical energy and the potential energy at the beginning respectively, $TE$ and $PE$ are the same energies but at the end, $SP$ is the spring energy at the end, and $Q$ is the dissipated energy (say heat), then:
$TE_0 = PE_0$
$TE = SP + PE$
$TE_0 = TE + Q$
Do I agree?
Thank you!
8. Jan 30, 2014
### BvU
Well, there is of course an "at rest" situation, it's just that it doesn't have the same energy as the situation where you attach the weight to the spring.
Yes. in fact, you can probably do that yourself: Suppose you exercise a force to keep the weight in position h0, then gently lower the weight to position h. If you do it slow enough, the force is mg - kx where x runs from 0 to h and h = mg/k. The weight does work, namely ∫ F(x) dx where the integral goes from 0 to h. F(x) = mg - kx. The integral (you can do it) is $\frac{1}{2}k h^2$. Looks familiar ?
9. Jan 30, 2014
### Bromio
When I say that there isn't an "at rest" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it?
Thank you! I've solved that integral and I've got that result. Amazing! | 2017-12-18T07:40:35 | {
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# A rectangular sheet of paper, when halved by folding it at the mid
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A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 04:50
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A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
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A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 06:02
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
$$\frac{l}{b} = \frac{b}{(l/2)}$$ where $$b = 2$$ given
i.e. $$b^2 = l^2 / 2$$
i.e. $$l = 2√2$$
Area of smaller rectangle $$= (l/2)*b = (2√2/2)*2 = 2√2$$
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Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 07:05
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
OA: B
Initial Rectangle Dimensions
longer side $$= x$$
Shorter side$$= 2$$
Dimensions after folding
Longer side $$= 2$$
Shorter side $$= \frac{x}{2}$$
As per question
$$\frac{x}{2}=\frac{2}{\frac{x}{2}}$$
$$x^2=8 \quad;\quad x=2\sqrt{2}$$
Dimensions of Shorter Rectangle
Longer side $$= 2$$
Shorter side $$=\frac{2\sqrt{2}}{2}=\sqrt{2}$$
Area of Shorter Rectangle $$= 2\sqrt{2}$$
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Re: A rectangular sheet of paper, when halved by folding it at the mid [#permalink]
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25 Oct 2018, 14:13
2
RhythmGMAT wrote:
A rectangular sheet of paper, when halved by folding it at the mid point of its longer side, results in
a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter
sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the
smaller rectangle?
a. $$4 \sqrt{2}$$
b. $$2 \sqrt{2}$$
c. $$\sqrt{2}$$
d. 1
e. 3
$$?\,\, = 2x$$
The rectangles shown above are similar, hence:
$${{2x} \over 2} = {2 \over x}\,\,\,\,\, \Rightarrow \,\,\,{x^2} = 2\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,x = \sqrt 2 \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2\sqrt 2$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Re: A rectangular sheet of paper, when halved by folding it at the mid &nbs [#permalink] 25 Oct 2018, 14:13
Display posts from previous: Sort by | 2019-01-21T22:11:03 | {
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https://math.stackexchange.com/questions/2980637/for-positive-integers-between-999-and-100-inclusive-how-many-contain-the-digit | # For positive integers between 999 and 100 inclusive, how many contain the digit 5?
The question comes in two parts:
1. For positive integers between 999 and 100 inclusive, how many contain the digit 5 at least once?
2. For positive integers between 999 and 100 inclusive, how many contain the digit 5 exactly once?
Question 1
Numbers between 999 and 100 inclusive:
999 - 99 = 900
Numbers between 999 and 100 inclusive that do not contain digit 5:
8 * 9 * 9 = 648 (The 8 is because the first digit can't be 5 or 0)
Numbers that contain digit 5 at least once:
900 - 648 = 252
Question 2
Total numbers that contain digit 5 exactly once:
(1*9*9) + (8*1*9) + (8*9*1) = 225
I think I got it wrong but not sure which part.
• Well, a good way to check is to count the number with exactly two $5's$. $55X$ gives us $9$, $5X5$ gives us $9$, $X55$ gives us $8$. Thus there are $9+9+8=26$ such numbers. Of course $555$ is the unique number in the range with three $5's$. We check that $252=225+26+1$ which seems like pretty good evidence that you are right.
– lulu
Nov 1 '18 at 16:34
• @DavidG.Stork The OP is counting numbers with no digits equal to $5$ so that they can be subtracted from all three-digit positive integers. Nov 1 '18 at 16:38
• @DavidG.Stork OP is using a standard technique of counting how many numbers there are where we ignore the condition, giving $9\times 10\times 10 =900$ numbers in the range, and subtracting the number of numbers which violate the condition, violating the condition in this case meaning "does not have $5$ as a digit", there being $8\times 9\times 9$ such numbers, which will give the number of numbers that satisfy the condition as a result. Nov 1 '18 at 16:38
• @DavidG.Stork The purpose of that step is to get the number which does not contain digit 5 at all, which in this case is 648. My reasoning is that if I subtract 900 with 648, then all the remaining numbers would contain at least one digit 5. Nov 1 '18 at 16:40
• "I think I got it wrong but not sure which part" Looks good to me. Do you have specific reason to doubt the answer? Be more confident, you seem to be doing well from what I can see. Nov 1 '18 at 16:40
There is one number with all three digits equal to 5.
There are 1x10x10 = 100 numbers with the first digit equal to 5.
There are 9x1x10 = 90 numbers with the second digit equal to 5.
There are 9x10x1 = 90 numbers with the third digit equal to 5.
There are 10 numbers with the first and second digit equal to 5.
There are 10 numbers with the first and third digit equal to 5.
There are 9 numbers with the second and third digit equal to 5.
There are 8x9x9 = 648 numbers with no digit equal to 5.
This is enough information to fill in the diagram above.
I got the same answers as you with the (javascript) code
var countFives = function(x){
var count = 0;
for (let c of x.toString(10)){
if (c==="5") count++;
}
return count;
};
var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)>=1) s++;
console.log(s);
252
var s = 0;
for (let i=100; i<=999; i++) if (countFives(i)===1) s++;
console.log(s);
225 | 2021-09-19T19:14:50 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/2980637/for-positive-integers-between-999-and-100-inclusive-how-many-contain-the-digit",
"openwebmath_score": 0.37101036310195923,
"openwebmath_perplexity": 358.41898932601174,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9711290963960278,
"lm_q2_score": 0.89029422102812,
"lm_q1q2_score": 0.8645906223936436
} |
http://fernandmandl.com/gm6degj/a-complex-number-is-a-real-number-5fa0f9 | If we define a pure real number as a complex number whose imaginary component is 0i, then 0 is a pure real number. e) INTUITIVE BONUS: Without doing any calculation or conversion, describe where in the complex plane to find the number obtained by multiplying . The real number a is called the real part and the real number b is called the imaginary part. For , we note that . We can picture the complex number as the point with coordinates in the complex … A complex number is a number of the form . Let be a complex number. Here both x x and y y are real numbers. is called the real part of , and is called the imaginary part of . A complex number is the sum of a real number and an imaginary number. Different types of real … As it suggests, ‘Real Numbers’ mean the numbers which are ‘Real’. Keep visiting BYJU’S to get more such maths lessons in a simple, concise and easy to understand way. If a is not equal to 0 and b = 0, the complex number a + 0i = a and a is a real number. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation i2 = −1. With regards to the modulus , we can certainly use the inverse tangent function . A complex number is the sum of a real number and an imaginary number. We start with the real numbers, and we throw in something that’s missing: the square root of . We define the imaginary unit or complex unit to be: Definition 21.2. Yes, because a complex number is the combination of a real and imaginary number. This statement would not make out a lot of logic as when we calculate the square of a positive number, we get a positive result. Move 6 units to the right on the real axis to reach the point ( 6 , 0 ) . Real Numbers and Complex Numbers are two terminologies often used in Number Theory. Python complex number can be created either using direct assignment statement or by using complex () function. This j operator used for simplifying the imaginary numbers. x x is called the real part which is denoted by Re(z) R e ( z). Learn more about accessibility on the OpenLab, © New York City College of Technology | City University of New York. Example 21.3. Likewise, imaginary numbers are a subset of the complex numbers. The major difference is that we work with the real and imaginary parts separately. basically the combination of a real number and an imaginary number Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is expressed in standard form when written $$a+bi$$ (with $$a, b$$ real numbers) where $$a$$ is the real part and $$bi$$ is the imaginary part. Convert the number from polar form into the standard form a) b), VIDEO: Converting complex numbers from polar form into standard form – Example 21.8. Multiplying Complex Numbers. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part.For example, $5+2i$ is a complex number. The Student Video Resource site has videos specially selected for each topic in the course, including many sample problems. From the long history of evolving numbers, one must say these two play a huge role. A complex number is created from real numbers. This leads to the following: Formulas for converting to polar form (finding the modulus and argument ): . is called the real part of , and is called the imaginary part of . But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. Complex numbers are written in the form (a+bi), where i is the square root of -1.A real number does not have any reference to i in it.A non real complex number is going to be a complex number with a non-zero value for b, so any number that requires you to write the number i is going to be an answer to your question.2+2i for example. For information about how to use the WeBWorK system, please see the WeBWorK Guide for Students. Perform the operation.a) b) c), VIDEO: Review of Complex Numbers – Example 21.3. Complex numbers are a bit unusual. Imaginary number consists of imaginary unit or j operator which is the symbol for √-1. However, unit imaginary number is considered to be the square root of -1. Imaginary Numbers when squared give a negative result. This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. start by logging in to your WeBWorK section, Daily Quiz, Final Exam Information and Attendance: 5/14/20. If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4. I – is a formal symbol, corresponding to the following equability i2 = -1. It is important to understand the concept of number line to learn about real numbers. The WeBWorK Q&A site is a place to ask and answer questions about your homework problems. Complex Numbers Complex Numbers 7 + 3 Real Imaginary A Complex Number A Complex Number is a combination of a Real Number and an Imaginary Number Real Numbers are numbers like: 1 12.38 −0.8625 3/4 √2 1998 Nearly any number you can think of is a Real Number! For example, $$5+2i$$ is a complex number. Imaginary numbers are square roots of negative real numbers. In the meantime, ‘Complex Numbers’ as the name refers a heterogeneous mix. Thus, 3 i, 2 + 5.4 i, and –π i are all complex numbers. Therefore, they consist of whole (0,1,3,9,26), rational (6/9, 78.98) and irrational numbers (square root of 3, pi). Similarly, 3/7 is a rational number but not an integer. Read through the material below, watch the videos, and send me your questions. If the formula provides a negative in the square root, complex numbers can be used to simplify the zero.Complex numbers are used in electronics and electromagnetism. We call this the polar form of a complex number. (In fact, the real numbers are a subset of the complex numbers-any real number r can be written as r + 0 i, which is a complex representation.) For example, both and are complex numbers. By the Pythagorean Theorem, we can calculate the absolute value of as follows: Definition 21.6. Points that fall in the right side of origin are considered positive numbers, whereas numbers lying in the left side of origin are considered to be negative. The quadratic formula solves ax2 + bx + c = 0 for the values of x. By definition, imaginary numbers are those numbers which when squared give a negative result. A Complex number is a pair of real numbers (x;y). So, if the complex number is a set then the real and imaginary number are the subsets of it. And actually, the real numbers are a subset of the complex numbers. We can picture the complex number as the point with coordinates in the complex plane. A complex number is represented as z=a+ib, where a … The coordinates in the plane can be expressed in terms of the absolute value, or modulus, and the angle, or argument, formed with the positive real axis (the -axis) as shown in the diagram: As shown in the diagram, the coordinates and are given by: Substituting and factoring out , we can use these to express in polar form: How do we find the modulus and the argument ? To plot a complex number, we use two number lines, crossed to form the complex plane. Let’s learn how to convert a complex number into polar form, and back again. Hence, we need complex numbers, a further extension of the number system beyond the real numbers. Your email address will not be published. How do we get the complex numbers? Give the WeBWorK a try, and let me know if you have any questions. Your email address will not be published. i.e., a complex number is of the form x +iy x + i y and is usually represented by z z. Any number in Mathematics can be known as a real number. A real number refers to any number that can be found on this number line. They have been designed in order to solve the problems, that cannot be solved using real numbers. Point P is uniquely determined by the ordered pair of a real number(r,θ), called the polar coordinatesof point P. x = r cosθ, y = rsinθ therefore, z=r(cosθ + isinθ) where r =√a2 + b2 and θ =tan-1 =b/a The latter is said to be polar form of complex number. Learn More! HINT: To ask a question, start by logging in to your WeBWorK section, then click “Ask a Question” after any problem. Start at the origin. Complex Numbers are considered to be an extension of the real number system. Its algebraic form is , where is an imaginary number. The set of complex numbers is a field. Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. Thus, the complex numbers of t… I can't speak for other countries or school systems but we are taught that all real numbers are complex numbers. Therefore, imaginary name is given to such numbers. This .pdf file contains most of the work from the videos in this lesson. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. If is in the correct quadrant then . Example 2: Plot the number 6 on the complex plane. The set of complex numbersis, therefore; This construction allows to consider the real numbers as a subset of the complex numbers, being realthat complex number whiose imaginary part is null. Complex numbers are numbers in the form. Every real number is a complex number, but not every complex number is a real number. Many amazing properties of complex numbers are revealed by looking at them in polar form! Similarly, when a negative number is squared it also provides a positive number. A complex number is a number of the form . Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. If some of these functions seem difficult to understand, it's best to research the basic logic behind them. So, a Complex Number has a real part and an imaginary part. If not, then we add radians or to obtain the angle in the opposing quadrant: , or . and are allowed to be any real numbers. The proposition below gives the formulas, which may look complicated – but the idea behind them is simple, and is captured in these two slogans: When we multiply complex numbers: we multiply the s and add the s.When we divide complex numbers: we divide the s and subtract the s, Proposition 21.9. However, we have to be a little careful: since the arctangent only gives angles in Quadrants I and II, we need to doublecheck the quadrant of . A real number can store the information about the value of the number and if this number is positive or negative. A complex number is said to be a combination of a real number and an imaginary number. The real numbers are a subset of the complex numbers, so zero is by definition a complex number (and a real number, of course; just as a fraction is a rational number and a real number). Here ‘x’ is called the real part of z and ‘y’ is known as the imaginary part of z. All real numbers are also complex numbers with zero for the imaginary part. A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0. If z1,z2,——zn are the complex numbers then z1.z2. The primary reason is that it gives us a simple way to picture how multiplication and division work in the plane. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. Complex numbers actually combine real and imaginary number (a+ib), where a and b denotes real numbers, whereas i denotes an imaginary number. A complex number is the sum of a real number and an imaginary number. Complex numbers can be used to solve quadratics for zeroes. Therefore, all real numbers are also complex numbers. Let be a complex number. The set of real numbers is a proper subset of the set of complex numbers. Using the functions and attributes that we've reviewed thus far will aid in building programs that can be used for a variety of science and engineering applications. They can be any of the rational and irrational numbers. Definition of Complex Numbers; An ordered pair of real numbers, written as (a, b) is called a complex number z. Any real number is a complex number. (2 plus 2 times i) The complex numbers are referred to as (just as the real numbers are . Complex numbers can be multiplied and divided. VIDEO: Multiplication and division of complex numbers in polar form – Example 21.10. You can add them, subtract them, multiply them, and divide them (except division by 0 is not defined), and the result is another complex number. 2020 Spring – MAT 1375 Precalculus – Reitz. A complex numberis defined as an expression of the form: The type of expression z = x + iy is called the binomial form where the real part is the real number x, that is denoted Re(z), and the imaginary partis the real number y, which is denoted by Im(z). Because no real number satisfies this equation, i is called an imaginary number. Complex numbers which are mostly used where we are using two real numbers. Number line can be expressed as an actual geometric line where a point is chosen to be the origin. All imaginary numbers are also complex numbers with zero for the real part. So, too, is $$3+4\sqrt{3}i$$. In other words, if the imaginary unit i is in it, we can just call it imaginary number. Multiplying a Complex Number by a Real Number. Convert the complex number to polar form.a) b) c) d), VIDEO: Converting complex numbers to polar form – Example 21.7, Example 21.8. If b is not equal to zero and a is any real number, the complex number a + bi is called imaginary number. This class uses WeBWorK, an online homework system. A complex number is a number that can be written in the form x+yi where x and y are real numbers and i is an imaginary number. Logged-in faculty members can clone this course. Let’s begin by multiplying a complex number by a real number. Don’t forget to complete the Daily Quiz (below this post) before midnight to be marked present for the day. Hi everyone! They have been designed in order to solve the problems, that cannot be solved using real numbers. Why is polar form useful? Then, the product and quotient of these are given by, Example 21.10. With this article at OpenG… The absolute value of , denoted by , is the distance between the point in the complex plane and the origin . You could view this right over here as a complex number. That’s it for today! Multiplying complex numbers is much like multiplying binomials. For the complex number a + bi, a is called the real part, and b is called the imaginary part. So, too, is $3+4\sqrt{3}i$. Consider √- 4 which can be simplified as √-1 × √ 4 = j√4 = j2.The manipulation of complex numbers is more complicated than real numbers, that’s why these are named as complex numbers. So, too, is 3 + 4i√3. WeBWorK: There are four WeBWorK assignments on today’s material, due next Thursday 5/5: Question of the Day: What is the square root of ? a, b ∈ R. a,b\in \mathbb {R} a,b ∈ R. Imaginary Numbers are the numbers which when squared give a negative number. If x and y are two real numbers, then a number of the form is called a complex number. In complex number, a is the real part and b is the imaginary part of the complex number. Topic: This lesson covers Chapter 21: Complex numbers. But in complex number, we can represent this number (z = … Complex Numbers are considered to be an extension of the real number system. Here r = √x2 + y2 = |z| is the modus of z and θ is called argument(or amplitude) of z is denoted by arg z. Subtracting Complex Numbers 1. (adsbygoogle = window.adsbygoogle || []).push({}); Copyright © 2021, Difference Between | Descriptive Analysis and Comparisons. Note that is given by the absolute value. New York City College of Technology | City University of New York. Image Courtesy: mathpowerblog.wordpress.comom, wikipedia.org. 3. A complex number is the sum of a real number and an imaginary number. Example 21.7. The importance of complex number in real life: In real numbers, we can represent this number as a straight line. A complex number is a number having both real and imaginary parts that can be expressed in the form of a + bi, where a and b are real numbers and i is the imaginary part, which should satisfy the equation i 2 = −1. A single complex number puts together two real quantities, making the numbers easier to work with. and are allowed to be any real numbers. Next, we will look at how we can describe a complex number slightly differently – instead of giving the and coordinates, we will give a distance (the modulus) and angle (the argument). Its algebraic form is z=x+i*y, where i is an imaginary number. This includes (but is not limited to) positives and negatives, integers and rational numbers, square roots, cube roots , π (pi), etc. You’ll see this in action in the following example. The real part of z is denoted by Re(z) and the imaginary part by Im(z). Definition 21.1. Required fields are marked *. Once they're understood, they're very simple and easy-to-use for just about anyone. It is provided for your reference. Complex Numbers: In mathematics, complex numbers are numbers that can be written in the form a + bi, where a and b are real numbers, and i is the imaginary number with value √−1 − 1. We distribute the real number just as we would with a binomial. Let and be two complex numbers in polar form. They're composed of real and imaginary numbers and are not necessarily the simplest to work with. This includes numbers like 3 – 2i or 5+√6i, as they can be written as the sum or difference of a real number and an imaginary number. The complex numbers are referred to as (just as the real numbers are . For example, 5 + 2i is a complex number. Definition 21.4. Yes, all real numbers are also complex numbers. Therefore we have: z = Re(z) + iIm(z). Comparison between Real Number and Complex Number: A real number is a number that can take any value on the number line. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. Infinity does not fall in the category of real numbers. The OpenLab is an open-source, digital platform designed to support teaching and learning at City Tech (New York City College of Technology), and to promote student and faculty engagement in the intellectual and social life of the college community. A complex number is any number that includes i. Therefore a complex number … The real part of the complex number is 6 and the imaginary part is 0 .So, the number will lie on the real axis. Difference Between | Descriptive Analysis and Comparisons, Counterintelligence Investigation vs Criminal Investigation. Terminologies often used in number Theory + 5.4 i, 2 + 5.4 i, 2 + 5.4 i and... 3 } i [ /latex ], we use two number lines, crossed form. Actual geometric line where a point is chosen to be an extension of complex... X x is called the real and imaginary numbers are referred to as ( just as the name a! B ) c ), VIDEO: multiplication and division of complex numbers are also numbers! 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Number consists of imaginary unit or j operator used for simplifying the imaginary unit or j operator for! The videos in this lesson covers Chapter 21: complex numbers in polar a complex number is a real number quantities. Just about anyone of Technology | a complex number is a real number University of New York includes.. We use two number lines, crossed to form the complex number can be created either using direct assignment or. Information about the value of the complex numbers can be any of the set of real numbers number a! Are using two real quantities, making the numbers which are ‘ real ’ equation, i is imaginary. Openlab accessible for all users in something that ’ s learn how to use the Q. To convert a complex number a + bi, a is the combination of a number! Picture how multiplication and division work in the plane difficult to understand the concept of number line to about!, that can be found on this number is the imaginary axis see the WeBWorK a try, and me. Covers Chapter 21: complex numbers are revealed by looking at them in polar form finding. Systems but we are using two real quantities, making the numbers which are ‘ real.! Research the basic logic behind them either part can be added and subtracted combining! These two play a huge role positive number be expressed as an actual geometric where! Statement or by using complex ( ) function multiply or divide the complex.... Plot the number and an imaginary part goal is to make the OpenLab accessible all. Of x revealed by looking at them in polar form composed of real imaginary!
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https://math.stackexchange.com/questions/2534786/logical-proof-for-a-value-derivable-by-a-linear-combination | Logical proof for a value derivable by a linear combination
Is there a logical basis to finding values provided possibly by a linear combination of integers. As the integers are closed under addition and hence their linear combination should be also. Hence, the integer points generated by two integers can be plotted on a lattice grid, extending infinitely. There can be multiple linear combinations, in other words, for two integer quantities. These two quantities can be, in the case of gcd computation, the divisor ($a$) and dividend ($d$).
If I want to dis-prove that a given linear combination can exist, then the approach is:
Say, the linear combination of two 'not' co-prime integers can never be $1$, as there will be no lattice point(s) generated as for : $2x + 4y = 1$. There will always be non-integral solutions, and any integer (lattice point) value of $x$ and $y$ will not work.
But, will the dis-proof above will work generally, i.e. if I want to prove that a given value can be generated or not by a linear combination. I request some better sort of way that will work generally for proving and disproving the feasibility of a given value for a given linear combination.
• I can't really understand what you're aiming at, but the integer combinations of $a$ and $b$ are exactly the multiples of $\operatorname{gcd}(a,b)$. This goes by the name of Bézout's lemma. – user228113 Nov 24 '17 at 4:57
• I hope my edit to OP makes it clear that similar to the dis-proof for the possibility of not co-prime integers having a value (for their linear combination) of 1, I want a similar proof for the opposite case. – jiten Nov 24 '17 at 5:17
• @DavidReed I have edited the OP to make it most clear. I would request some details on the feasible way to show existence of a linear combination, and for an efficient algorithm to find a given integer combination. – jiten Nov 24 '17 at 5:37
1 Answer
Let $d = gcd(a,b)$. Then there exists $x,y$ such that $ax+by=n$ if and only if $d|n$
The extended Euclidean algorithm is the algorithm one would use to find $x$ and $y$.
To prove no such integer combination exists, one would again use the euclidean algorithm to find the gcd, and then simply show that n is not a multiple of the gcd.
EDIT-
Heres how you would prove an integer combination exists. Say you want to know if there are integers $x$ and $y$ such that $ax+by = n$. Let $d = gcd(a,b)$ You can always find integers $u$ and $v$ such that $au + bv = d.$ You can find $u$ and $v$ via the extended Euclidean algorithm, which also, as an added benefit, will compute $d$. If n is not a multiple of $d$, then no integer combination is possible.
Otherwise, let $k = n/d$ so that $n = kd$. Set $x= ku$ and $y = kv$, then
$$ax+by= aku+bkv = k(au+bv) = kd = n$$
$\\$
Here is an example. Lets say I want to find solutions to $$135x+48y = 3375$$
I use the Euclidean Algorithm as follows:
\begin{align} &(135,48) \to 135= 2*48 + 39 \to \quad 39 = 135 - (2)(48) \\ &(48,39) \to \quad 48 = 1*39 + 9 \to \quad 9 = 48 - (1)(39)\\ &(39,9) \to \quad 39 = 4*9+ 3 \to \quad 3 = 39 - (4)(9) \\ &(9,3) \to \quad 9 = 3*3 + 0 \\ &(3,0) \to \quad \text{done} \end{align}
So $3$ is the gcd. $3375/3 = 1125$ which is a whole number. Therefore a solution exists. Now to find it:
We now work backwards:
\begin{align} 3 =& \ 39-(4)(9) \\ =& \ 39 - (4)(48-(1)(39)) \\ =& \ 135 - (2)(48) - (4)(48-(1)(135-(2)(48))) \\ =& \ 135 - (2)(48) - (4)(48-135+(2)(48)) \\ =& \ 135 -(2)(48) - (4)(48) +(4)(135)-(8)(48) \\ =& \ (5)(135) + (-14)(48) \end{align}
Thus
$$(135)(5)+ (48)(-14) = 3$$
Now since $3375/3 = 1125$, we multiply the whole equation by $1125$ and get
$$(135)(5)(1125) + (48)(-14)(1125) = (3)(1125)$$
$$\\$$ $$\implies (135)(5625)+(48)(-15750) = 3375$$
• @jiten I meant to change that. It should be 135x+48y and not 135x+6y. I first used six but the Euclidean alg was too short (like 1 step) so I changed it to one that provided a better example. Its fixed now. – David Reed Nov 24 '17 at 9:11
• Thanks for removing confusion. – jiten Nov 24 '17 at 9:14
• @jiten The Euclidean algorithm is a good algorithm to know. Its used in several other places in algebra and number theory. Equations like this, $ax+by = n$ are called "linear Diophantine equations.", in case you have any interest in researching further. – David Reed Nov 24 '17 at 9:17 | 2019-08-17T20:42:17 | {
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http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem | # A stronger version of discrete “Liouville's theorem”
If a function $f : \mathbb Z\times \mathbb Z \rightarrow \mathbb{R}^{+}$ satisfies the following condition
$$\forall x, y \in \mathbb{Z}, f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$
then is $f$ constant function?
-
You probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. – Julián Aguirre Jul 17 '11 at 10:24
@Julian Aguirre: since $x\in\mathbb Z$, we don't have $f(x,y)\geq 0$. – Davide Giraudo Jul 17 '11 at 11:08
@girdav You are right. The lower bound is probably enough. – Julián Aguirre Jul 17 '11 at 13:08
You can prove this with probability.
Let $(X_n)$ be the simple symmetric random walk on $\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\to M_\infty$ almost surely.
But $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.
Thus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge.
-
I love probabilistic arguments in analysis! Very nice. – Jonas Teuwen Jul 17 '11 at 11:46
This is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\mathbb{Z}^n$ instead of $\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\mathbb{Z}^n$ is constant for $n \geq 3$? For bounded ones this seems clear by considering the Poisson boundary. – t.b. Jul 17 '11 at 11:53
@Byron: This paper contains the claim that it is true that "nonnegative nearest-neighbors harmonic function on $\mathbb{Z}^d$ are constant for any $d$" on page 2. – t.b. Jul 17 '11 at 12:08
The usual Liouville theorem also holds with just a one-sided bound. – GEdgar Jul 18 '11 at 13:56
If you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? – Jyrki Lahtonen Jul 18 '11 at 14:07
I can give a proof for the d-dimensional case, if $f\colon\mathbb{Z}^d\to\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it]
First, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\Vert x\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\le(2d)^{\Vert x\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\mathbb{E}[(2d)^{2T}] < \infty$. Then, $X_T$ has support $\mathbb{Z}^d$ and $\mathbb{E}[f(X_T)]=f(0)$, $\mathbb{E}[f(X_T)^2]\le\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\Vert f\Vert_2\equiv\mathbb{E}[f(X_T)^2]^{1/2}$.
Writing $e_i$ for the unit vector in direction $i$, set $f_i^\pm(x)=f(x\pm e_i)/f(\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\ldots,d$. Also, by construction, $\Vert f\Vert_2\ge\Vert f^\pm_i\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\frac{1}{2d}\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.
Now, if $g$ is a positive harmonic function, then $\tilde g(x)\equiv g(x)/g(0)$ satisfies $\mathbb{E}[\tilde g(X_T)]=1$. So, $${\rm Var}(\tilde g(X_T))=\mathbb{E}[\tilde g(X_T)^2]-1\le\mathbb{E}[f(X_T)^2]-1=0,$$ and $\tilde g$ is constant.
-
Taxicab metric. Never heard that name (I learn maths in french at school). Funny! – Patrick Da Silva Jul 17 '11 at 20:24
@Patrick: Also called the Manhattan metric. – George Lowther Jul 17 '11 at 20:30
LAAAAAAAAAWL. Funnier. – Patrick Da Silva Jul 17 '11 at 20:39
Note: A similar proof will also show that harmonic $f\colon\mathbb{R}^d\to\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\colon\mathbb{R}^d\setminus\{0\}\to\mathbb{R}^+$ for $d\not=2$. In fact, considering $f(x)=\Vert x\Vert^{2-d}$, we see that $f$ need not be constant for $d\not=2$. – George Lowther Jul 17 '11 at 22:54 | 2014-04-17T04:43:44 | {
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https://zxi.mytechroad.com/blog/math/leetcode-2028-find-missing-observations/ | You have observations of n + m 6-sided dice rolls with each face numbered from 1 to 6n of the observations went missing, and you only have the observations of m rolls. Fortunately, you have also calculated the average value of the n + m rolls.
You are given an integer array rolls of length m where rolls[i] is the value of the ith observation. You are also given the two integers mean and n.
Return an array of length n containing the missing observations such that the average value of the n + m rolls is exactly mean. If there are multiple valid answers, return any of them. If no such array exists, return an empty array.
The average value of a set of k numbers is the sum of the numbers divided by k.
Note that mean is an integer, so the sum of the n + m rolls should be divisible by n + m.
Example 1:
Input: rolls = [3,2,4,3], mean = 4, n = 2
Output: [6,6]
Explanation: The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.
Example 2:
Input: rolls = [1,5,6], mean = 3, n = 4
Output: [2,3,2,2]
Explanation: The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.
Example 3:
Input: rolls = [1,2,3,4], mean = 6, n = 4
Output: []
Explanation: It is impossible for the mean to be 6 no matter what the 4 missing rolls are.
Example 4:
Input: rolls = [1], mean = 3, n = 1
Output: [5]
Explanation: The mean of all n + m rolls is (1 + 5) / 2 = 3.
Constraints:
• m == rolls.length
• 1 <= n, m <= 105
• 1 <= rolls[i], mean <= 6
## Solution: Math & Greedy
Total sum = (m + n) * mean
Left = Total sum – sum(rolls) = (m + n) * mean – sum(rolls)
If left > 6 * n or left < 1 * n, then there is no solution.
Otherwise, we need to distribute Left into n rolls.
There are very ways to do that, one of them is even distribution, e.g. using the average number as much as possible, and use avg + 1 to fill the gap.
Compute the average and reminder: x = left / n, r = left % n.
there will be n – r of x and r of x + 1 in the output array.
e.g. [1, 5, 6], mean = 3, n = 4
Total sum = (3 + 4) * 3 = 21
Left = 21 – (1 + 5 + 6) = 9
x = 9 / 4 = 2, r = 9 % 4 = 1
Ans = [2, 2, 2, 2+1] = [2,2,2,3]
Time complexity: O(m + n)
Space complexity: O(1)
## C++
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huahualeetcode | 2021-12-01T09:38:53 | {
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https://byjus.com/question-answer/areas-of-square-and-rhombus-are-equal-a-diagonal-of-a-rhombus-is-twice-of-1/ | Question
# Areas of square and rhombus are equal. A diagonal of a rhombus is twice of its other diagonal. If the area of rhombus is $$64$$ sq. cm find the ratio of perimeter of a square and rhombus.
A
3:1
B
2:5
C
2:3
D
5:2
Solution
## The correct option is B $$2 : \sqrt5$$Let $$a$$ be the side of a square and$$b$$ be the side of a rombus,and $$d_1, d_2$$ be the two diagonals of the rombus.Given,Area of square $$=$$ Area of rombus$$\Rightarrow a^2=\dfrac{d_1 d_2}{2}$$Also,$$d_1=2d_2$$$$\therefore a^2=\dfrac{2d_2 d_2}{2}$$$$\Rightarrow a^2=d_2$$$$\Rightarrow a=d_2$$ and$$\Rightarrow a=\cfrac{d_1}{2}$$Area of rhombus is half of the product of the diagonals$$\cfrac{d_1 d_2}{2}=64$$$$\Rightarrow d_1 d_2=128$$$$\Rightarrow (2a)a=128$$$$\Rightarrow a^2=64$$$$\Rightarrow a=8\ cm$$$$d_1=2a=8\times 2$$ $$=16\ cm$$$$d_2=a$$ $$=8\ cm$$Diagonals of a rhombus bisect each other at right angles. So, half of both the diagonals and the side of the rhombus make a right angled $$\triangle$$$$\therefore$$ Semi-Diagonal $$\dfrac{d_1}{2}=\dfrac{16}{2}$$ $$=8\ cm$$and semi-diagonal $$\dfrac{d_2}{2}=\dfrac82$$ $$=4\ cm$$$$\therefore 8^2+4^2=b^2$$$$\Rightarrow b^2=64+16$$$$\Rightarrow b^2=80$$$$\Rightarrow b=\sqrt{80}$$$$\Rightarrow b=4\sqrt5$$$$\therefore$$ Ratio of perimeter of a square and rhombus $$=4a:4b$$ $$=a:b$$ $$=8: 4\sqrt5$$ $$=2:\sqrt 5$$Maths
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https://math.stackexchange.com/questions/2342156/matrix-norm-of-kronecker-product/2521249 | # Matrix norm of Kronecker product
Is it true that $\| A \otimes B \| = \|A\|\|B\|$ for any matrix norm $\|\cdot \|$? If not, does this identity hold for matrix norms induced by $\ell_p$ vector norms?
• According to wikipedia, you can relate the eigenvalues of the Kronecker product to that of the operands. This should give you something for the spectral norm. – xavierm02 Jun 30 '17 at 17:40
• Yes, it's true for the spectral norm. That's the only case I know for certain. – Rob Jun 30 '17 at 18:08
• Compute everything for two arbitrary $2\times 2$ matrices (i.e., get both sides as an expression of $a_{11}, a_{12},\dots ,b_{22}$). I'd expect counter examples to be easy to find once you have done that. – xavierm02 Jun 30 '17 at 18:12
Theorem 8 here provides the answer: http://www.ams.org/journals/mcom/1972-26-118/S0025-5718-1972-0305099-X/S0025-5718-1972-0305099-X.pdf. As discussed on page 413, the identity holds for all matrix norms induced by $\ell_p$ vector norms. In fact, it seems to hold for any induced vector norm, or any submultiplicative norm.
Here is a proof for the lazy: Let $$A = \sum_i \sigma_i u_i v_i^T$$ and $$B = \sum_j \lambda_j x_j y_j^T$$ be the singular value decomposition (SVD) of the two matrices. Then, \begin{align} A \otimes B &= \sum_{i,j}\sigma_i \lambda_j (u_i v_i^T) \otimes (x_jy_j^T) \\ &= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i^T \otimes y_j^T) \\ &= \sum_{i,j} \sigma_i \lambda_j (u_i \otimes x_j) (v_i \otimes y_j)^T \end{align} where the first equality is by the bilinearity of $$\otimes$$, the second by the "mixed product" property of Kronecker product and last one by $$(A\otimes B)^T = A^T \otimes B^T$$. It is not hard to see that $$\{u_i \otimes x_j, \forall i,j \}$$ is an orthonormal collection of vectors and similarly for $$\{v_i \otimes y_j, \forall i,j \}$$. It follows that the last line above is the SVD of $$A \otimes B$$, with singular values $$\{\sigma_i \lambda_j, \forall i,j\}$$. Hence, \begin{align} \| A \otimes B\| = \max_{i,j} \sigma_i \lambda_j = (\max_i \sigma_i)(\max_j \lambda_j) = \|A\| \|B\|. \end{align} for the $$\ell_2$$ operator norm. | 2020-06-01T19:35:39 | {
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# Two bottles are partially filled with water. The larger bottle current
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Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
A. $$\frac{5}{6}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{7}{12}$$
E. $$\frac{1}{2}$$
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04 Aug 2016, 15:13
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
A. $$\frac{5}{6}$$
B. $$\frac{3}{4}$$
C. $$\frac{2}{3}$$
D. $$\frac{7}{12}$$
E. $$\frac{1}{2}$$
Here the capacity of larger bottle be x and currently the larger bottle has x/3 capacity (given).
Now smaller bottle has 2/3(x) ...( GIVEN: The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle )
Currently the SB has 3/4( 2/3(x) ) = 1/2(x) capacity and this is added to larger bottle i.e. 1/2(x) + x/3 = 5/6(x).
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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04 Aug 2016, 21:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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12 Aug 2018, 07:18
Let the capacity of larger water bottle be "x"
Water filled in the larger bottle=x/3
Water filled in the smaller bottle=2x/3*3/4=x/2
If we pour the smaller bottle to bigger bottle i.e. x/3+x/2=5x/6=>5/6
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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04 Sep 2018, 22:21
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Take larger bottle (L) to have capacity 9 L and since smaller bottle(S) has 2/3rd the larger bottles capacity, S = 6 L capacity.
Now we are told that L is filled up to 1/3rd its capacity i.e 3 L.
S is filled up to 3/4th its capacity i.e 4.5 L.
when we add 4.5 L to 3 L we get 7.5 L.
7.5/9 = 75/90 = 15/18 = 5/6
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Re: Two bottles are partially filled with water. The larger bottle current &nbs [#permalink] 04 Sep 2018, 22:21
Display posts from previous: Sort by | 2018-11-21T03:57:47 | {
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https://math.stackexchange.com/questions/3154343/how-do-i-get-from-log-f-log-g-log-m-log1-m-2-log-r-to-a-solution-witho | # How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?
I've been self-studying from Stroud & Booth's excellent "Engineering Mathematics", and am currently on the "Algebra" section. I understand everything pretty well, except when it comes to the problems then I am asked to express an equation that uses logs, but without logs, as in:
$$\log{F} = \log{G} + \log{m} - \log\frac{1}{M} - 2\log{r}$$
They don't cover the mechanics of doing things like these very well, and only have an example or two, which I "kinda-sorta" barely understood.
Can anyone point me in the right direction with this and explain how these are solved?
• Since each $\log$ is in the same base (ten) you can cancel them. That is, if $\log_b(x)=\log_b(y)$ then $x=y$. P.S. The converse is also true since $\log$ is an inyective function. Mar 19, 2019 at 17:17
Using some rules of logarithms you get $$\quad-\log\dfrac{1}{M}=+\log M$$ and $$-2\log r=-\log r^2=+\log \dfrac{1}{r^2}$$
So you have
$$\begin{eqnarray} \log{F} &=& \log{G} + \log{m} + \log M + \log{\dfrac{1}{r^2}}\\ \log{F} &=& \log{\left(GmM\cdot\dfrac{1}{r^2}\right)}\\ \log{F} &=& \log{\frac{mMG}{r^2}}\\ F &=&\frac{mMG}{r^2} \end{eqnarray}$$
The last step hinges upon the fact that logarithm functions are one-to-one functions. If a function $$f$$ is one-to-one, then $$f(a)=f(b)$$ if and only if $$a=b$$. Since $$\log$$ is a one-to-one function, it follows that $$\log A=\log B$$ if and only if $$A=B$$.
ADDENDUM: Here are a few rules of logarithms which you may need to review
1. $$\log(AB)=\log A+\log B$$
2. $$\log\left(\dfrac{A}{B}\right)=\log A-\log B$$
3. $$\log\left(A^n\right)=n\log A$$
4. $$\log(1)=0$$
Notice that from (2) and (4) you get that $$\log\left(\dfrac{1}{B}\right)=\log 1-\log B=-\log B$$
• Nice clear answer. You may like to point out that $a = b \iff \log a = \log b$ so that $\log F = \log \frac {GMm}{r^2}$ would also mean $F = \frac {GMm}{r^2}$. Mar 19, 2019 at 18:27
• @fleablood Good point. Mar 19, 2019 at 20:04
Hint:
Product rule for Logarithms says that:
$$\log\prod_{k=1}^{n}a_k=\sum_{k=1}^{n}\log a_k$$
In the equation stated in your question: \begin{aligned}\log F&=\log G+\log m+\log M+\log\dfrac{1}{r^2}\\ \log F &= \log \dfrac{GMm}{r^2}\\ \exp\log F&=\exp\log\dfrac{GMm}{r^2}\\ F&=G\cdot\dfrac{Mm}{r^2}\end{aligned}
The SINGLE most important rule of logarithms is:
$$\log N + \log M = \log N\times M$$.
This is because if $$a = \log N; b=\log M$$ then $$10^a = N; 10^b = M$$ so $$10^{a+b} = 10^a\times 10^b = N\times M$$ and so, by definition, $$a+b = \log N\times M$$.
From this simple rule we get $$n \log M = \log (M^n)$$ and $$\log M -\log N = \log \frac MN$$ and so on.
So......
Well, the basic rules of combining logarithms: $$\log a + \log b = \log ab$$ will give us:
$$\log{F} = \log{G} + \log{m} - \log\frac{1}{M} - 2\log{r} = \log \frac {Gm}{\frac 1Mr^2}=\log \frac {GMm}{r^2}$$.
The log function is one to one so we know that for positive real numbers that $$a = b \iff \log a = \log b$$.
(If we need to convince ourselves of this: $$m= \log a = \log b = n\implies 10^m =10^{\log a} = a; 10^n = 10^{\log b}= b; 10^m = 10^n\implies a=b \implies \log a=\log b$$.)
So from here we get:
$$F = \frac {GMm}{r^2}$$. | 2023-02-06T04:12:00 | {
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https://math.stackexchange.com/questions/665230/must-an-uncountable-subset-of-r-have-uncountably-many-accumulation-points/665301 | # Must an uncountable subset of R have uncountably many accumulation points? [duplicate]
This question is taken from problem 4.1.8 of "Real Analysis and Foundations" by Krantz
"Let S be an uncountable subset of $\mathbb{R}$. Prove that S must have infinitely many accumulation points. Must it have uncountably many?"
The first part of the question took some work but ended up coming out pretty smoothly; however, I'm at a complete loss as to how to go about addressing the second problem. Intuitively, I think the answer is yes.
My first attempt was to try to show that a countable number of accumulation points allowed one to order the elements of S in such a way that they are countable (ie prove the contrapositive), but I did not manage to get much further than that.
My second attempt was to show that $S-{s_{1},s_{2}...}$ where $s_{1},s_{2},...$ are countably many accumulation points of S is uncountable and thus must have an accumulation point, so S has an accumulation point that is not one of the countably infinite set. Hence, the number of accumulation points is uncountable.
My question is, is this logic valid? I would have to prove that an uncountable set minus a countable set is uncountable, which shouldn't be too difficult.
Any hints/points in the right direction/outright answers are greatly appreciated.
## marked as duplicate by Asaf Karagila♦, AlexR, user61527, egreg, M TurgeonFeb 5 '14 at 23:32
• You might like this blog post. – David Mitra Feb 5 '14 at 22:05
• Your second proof assumes that the accumulation points of $S$ belong to $S$. This isn't true unless your set is closed. – EuYu Feb 5 '14 at 22:05
• There are many duplicates of this question. – Asaf Karagila Feb 5 '14 at 22:15
• @AsafKaragila This particular one doesn't seem like it, note that this question asks about the cardinality of acc-pts, while the linked one only asks about existence. – AlexR Feb 5 '14 at 22:45
• @Alex: See Brian's answer there. – Asaf Karagila Feb 5 '14 at 22:46
Let $$T$$ be the set of elements of $$S$$ that are not accumulation points of $$S$$. Then for each $$x \in T$$ there exists $$\epsilon_x > 0$$ such that the interval $$I(x,\epsilon_x) = (x-\epsilon_x,x+\epsilon_x)$$ contains no other points of $$S$$. The intervals $$\{I(x,\epsilon_x/2) | x\in T\}$$ are disjoint, and each contains a rational number; so there can only be a countable number of them.
Hence $$T$$ is countable, and the set of accumulation points of $$S$$, which contains $$S-T$$, is uncountable.
• Why $\{I(x,\epsilon/2) | x\in T\}$ is disjoint? Thanks – StammeringMathematician Feb 19 at 15:27
• First before assuming that $x\in T$ you need to show that $T$ is non-empty! – SunShine Aug 8 at 17:58
• @SunShine: $T$ doesn't have to be non-empty. For instance, $S$ might be the whole of $\Bbb R$. But then there is nothing more to prove, is there? An empty set is certainly countable. – TonyK Aug 8 at 18:24
Assuming a finite number of accumulation points for $S$ would mean that the elements of $S$ that are not accumulation points are isolated, meaning $S$ is at most countable, a contradiction.
Assuming a countable number of accumulation points yields the same problem. The elements of $S$ that are not accumulation points would be isolated, and you still have that $S$ is at most a union of two countable sets which is countable, a contradiction.
I think both TonyK's and Darrin's answers are good ones. Here's another, possibly inferior, approach.
First trust the following lemma: for any uncountable set $S$ of real numbers, there exists an interval $(a,b)$ such that both $S\cap(-\infty,a]$ and $S\cap[b,\infty)$ are uncountable. ("You can divide an uncountable set into two with a buffer interval.")
Given the lemma, you can find two uncountable subsets $S_1,S_2$ of $S$ that are separated by an interval. Iterating, you can find four uncountable subsets of $S$ all separated from one another by intervals, then 8, 16, etc. There are uncountably many branches down this repeated-bifurcation tree; each one results in an accumulation point; and the accumulation points can't coincide because of the separating intervals. Thus there are uncountably many accumulation points.
Why is the lemma true? Assume $S$ is contained in $[0,3]$ (an easy reduction), and consider the intersections of $S$ with $[0,1]$, $[1,2]$, and $[2,3]$. If the first and third intersections are uncountable, then we're done. Otherwise, we have an uncountable set in a smaller interval (perhaps length 1, perhaps length 2) and we divide that interval into three equal pieces. We keep doing this until either we find the first and last intersections uncountable (in which case we're done), or else we iterate countably often and end up with all the countable subsets contained in a sequence of nested closed intervals whose lengths tend to $0$. But this latter case can't happen, because then we will have written $S$ as a countable union of countable sets, together with the single point in the intersection of the nested intervals, contradicting the uncountability of $S$. | 2019-08-21T00:39:07 | {
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https://math.stackexchange.com/questions/1971211/pseudo-inverse-of-a-matrix-that-is-neither-fat-nor-tall | # Pseudo-inverse of a matrix that is neither fat nor tall?
Given a matrix $A\in\mathbb R^{m\times n}$, let us define:
• $A$ is a fat matrix if $m\le n$ and $\text{null}(A^T)=\{0\}$
• $A$ is a tall matrix is $m\ge n$ and $\text{range}(A)=\mathbb R^n$
Using the finite rank lemma, we can find that:
• When $A$ is a fat matrix, its (right) pseudo-inverse is $A^\dagger = A^T(AA^T)^{-1}$
• When $A$ is a tall matrix, its (left) pseudo-inverse is $A^\ddagger = (A^TA)^{-1}A^T$
My question is what is the pseudo-inverse when $A$ is neither fat nor tall (in the sense of the above definitions), i.e. it is a matrix such that $\text{null}(A^T)\ne \{0\}$ (i.e. the null space is non-trivial) and $\text{range}(A)\ne\mathbb R^n$? An example of such a matrix is:
$$A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 3 & 0 \end{bmatrix}$$
which clearly does not map to full $\mathbb R^4$ and whose null space is $\text{span}\left\{\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right\}$.
• You should read about the Moore-Penrose inverse of a matrix. Or, in general, the generalized inverse of a matrix. – steven gregory Oct 16 '16 at 16:16
• A conceptual definition of the pseudoinverse of an $m \times n$ matrix $A$ is that the pseudoinverse takes a vector $b$ as input, projects $b$ onto the range of $A$ (call the projection of $b$, say, $\hat b$), then returns as output the vector $x$ with least norm such that $Ax =\hat b$. – littleO Mar 1 '17 at 23:13
There is no need to study both the fat matrix and the tall matrix. Let $B = A^t$. In that case, the tall matrix problem for $B$ is the fat matrix problem for $A$ and vice versa.
Thus, we consider only the tall matrix case. Assume that the rank of $A$ is $r$ where $r \leq n$. Let the SVD (singular value decomposition) of $A$ is defined as $A = U \Sigma V^t$ where $U$ is an $m \times r$ matrix with orthogonal columns, $V$ is an $r \times n$ matrix with orthogonal columns and the diagonal $r \times r$ matrix $\Sigma$ contains the $r$ singular values in the diagonal. These diagonal values $\sigma_i$, $i=1 \colon r$ are ordered in the non-increasing order.
The Moore-Penrose pseudo inverse is then given by $$A^{+} = V \Sigma^{-1} U^t .$$ This pseudo inverse exists even when $A^tA$ is singular.
Numerical analysts will tell you to define the "zero singular values" as the singular values which are "tiny". Tiny means any number which is in the same order as $\|A\|*\epsilon$ {(any) norm of $A$ times the machine precision}. Unfortunately, the Moore-Penrose inverse often depends on the way "tiny" is defined. Mathematicians do not have this problem since zero means zero and nothing else.
For your matrix $A$, there are two non-zero singular values (3.7519 and 0.9610) and hence $r=2$. Since there are no tiny singular values, the Moore-Penrose inverse is well defined.
Fundamental Theorem of Linear Algebra
The matrix $\mathbf{A} \in \mathbb{C}^{m\times n}$ induces the four fundamental subspaces. In finite dimension, the domain $\mathbf{C}^{n}$, and codomain, $\mathbf{C}^{m}$ are \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}
Least squares
Consider a data vector $b \in \mathbb{C}^{m}$ that does not lie in the null space, that is, $b \notin \color{red} {\mathcal{N} \left( \mathbf{A} \right)}$. We are solving for the least squares minimizers defined as $$x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\}.$$ These solutions are $$x_{LS} = \color{blue}{\mathbf{A}^{\dagger} b} + \color{red}{\left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) y}, \quad y\in\mathbb{C}^{n}$$ The Moore-Penrose pseudoinverse matrix is \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{cc} \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red}{\mathbf{V}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} & \color{red}{\mathbf{U}_{\mathcal{N}}}^{*} \end{array} \right] % \end{align}
SVD General case
The singular value decomposition provides an orthonormal basis for the four subspaces. The range spaces are aligned and the difference in length scales is expressed in the singular values.
The decomposition for a matrix with rank $\rho$ is \begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cccccc} \sigma_{1} & 0 & \dots & & & \dots & 0 \\ 0 & \sigma_{2} \\ \vdots && \ddots \\ & & & \sigma_{\rho} \\ & & & & 0 & \\ \vdots &&&&&\ddots \\ 0 & & & & & & 0 \\ \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % & = % U \left[ \begin{array}{cccccccc} \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{v_{1}^{*}} \\ \vdots \\ \color{blue}{v_{\rho}^{*}} \\ \color{red}{v_{\rho+1}^{*}} \\ \vdots \\ \color{red}{v_{n}^{*}} \end{array} \right] % \end{align} Let's look at your special cases.
Tall: $m>n$
The overdetermined case of full column rank $\rho = n$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S}_{\rho\times \rho} \\ \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \end{array} \right] \\ % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ \color{red} {\mathbf{U}_{\mathcal{N}}}^{*} \end{array} \right] % \end{align}
Wide: $m<n$
The underdetermined case of full row rank $\rho = m$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red} {\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{cc} \color{blue}{\mathbf{V}_{\mathcal{R}}} & \color{red} {\mathbf{V}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho}^{-1} & \mathbf{0} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \end{array} \right] % \end{align}
Square: $m=n$
The nonsingular case $\rho = m = n$.
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \end{array} \right] % \end{align} \begin{align} \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \mathbf{U}^{*} \\ % &= % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}} \end{array} \right] % Sigma \left[ \begin{array}{c} \mathbf{S}^{-1} \end{array} \right] % U \left[ \begin{array}{c} \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \end{array} \right] % \end{align}
Original example
@Vini presents the solution clearly. To emphasize his conclusions, here are the background computations. The product matrix is $$\mathbf{A}^{\mathrm{T}}\mathbf{A} = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 14 & 0 \\ 0 & 0 & 0 \\ \end{array} \right],$$ which quickly leads to the characteristic polynomial $$p(\lambda) = \left( -\lambda^{2} + 15 \lambda-13\right) \lambda.$$ The eigenvalue spectrum is then $$\lambda \left( \mathbf{A}^{\mathrm{T}}\mathbf{A} \right) = \left\{ \frac{1}{2} \left( 15 \pm \sqrt{173} \right), 0 \right\}.$$ The singular values are the square root of the nonzero eigenvalues $$\left\{ \sigma_{1}, \sigma_{2} \right\} = \left\{ \sqrt{\frac{1}{2} \left( 15 + \sqrt{173} \right)}, \sqrt{\frac{1}{2} \left( 15 -\sqrt{173} \right)} \right\}$$ There are two nonzero singular values, therefore the matrix rank is $\rho = 2$.
Let $$\mathbf{S} = \left( \begin{array}{cc} \sigma_{1} & 0 \\ 0 & \sigma_{2} \\ \end{array} \right).$$ This diagonal matrix is embedded in the sabot matrix, padded with zeros to insure conformability between the matrices $\mathbf{U}$ and $\mathbf{V}$: $$\Sigma = \left[ \begin{array}{ccc} \sqrt{\frac{1}{2} \left(\sqrt{173}+15\right)} & 0 & 0 \\ 0 & \sqrt{\frac{1}{2} \left(15-\sqrt{173}\right)} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left( \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right).$$ The pseudoinverse of the $\Sigma$ matrix is $$\Sigma^{\dagger} = \left[ \begin{array}{ccc} \sqrt{\frac{1}{2} \left(\sqrt{173}+15\right)}^{-1} & 0 & 0 \\ 0 & \sqrt{\frac{1}{2} \left(15-\sqrt{173}\right)}^{-1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left( \begin{array}{cc} \mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right).$$ In this case where $m=n$, the matrix products are identical: $$\Sigma^{\dagger}\Sigma = \Sigma\, \Sigma^{\dagger} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] = \left[ \begin{array}{cc} \mathbf{I}_{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \\ \end{array} \right].$$
Another example of $\Sigma$ gymnastics is in here: Singular Value Decomposition: Prove that singular values of A are square roots of eigenvalues of both $AA^{T}$ and $A^{T}A$
The eigenvectors of the product matrix are $$% v_{1} = \left[ \begin{array}{c} \frac{1}{2} \left(-13+\sqrt{173}-13\right) \\ 1 \\ 0 \end{array} \right], \quad % v_{2} = \left[ \begin{array}{c} \frac{1}{2} \left(-13-\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right], \quad %$$ The normalized form constitute the column vectors of the domain matrix $$\mathbf{V} = % \left[ \begin{array}{cc} % \left( 1 + \left(\frac{-13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13+\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] & % \left( 1 + \left(\frac{13 + \sqrt{173}}{4}\right)^2 \right)^{-\frac{1}{2}} % \left[ \begin{array}{c} \frac{1}{2} \left(-13-\sqrt{173}\right) \\ 1 \\ 0 \end{array} \right] % \end{array} \right]$$ | 2020-07-12T13:34:42 | {
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http://math.stackexchange.com/questions/194569/find-an-integer-n-such-that-mathbbz-frac120-frac132-mathbbz | # Find an integer $n$ such that $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$.
How can we find an integer $n$ such that $\mathbb{Z}[\frac{1}{20},\frac{1}{32}]=\mathbb{Z}[\frac{1}{n}]$?
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Rapid answer: since $\frac1{32}=\frac1{2^5}=\frac{2\times5^3}{20^3}\in\mathbb Z[\frac1{20}]$, one can take $n=20$.
More generally the following considerations apply.
Let $R$ be a subring of $\mathbf Q$, and consider the set $D$ of positive integers $n$ such that $\frac1n\in R$. Then for all reduced fractions $\frac nd\in R$ one has $\frac1d\in R$ and therefore $d\in D$: since $n$ is relatively prime to $d$ there exists $s\in \mathbb Z$ with $sn\equiv1\pmod d$, and subtracting an integer from $s\times\frac nd$ will give $\frac1d$. Also the set $D$ is closed under taking arbitrary divisors (since we can multiply $\frac1d$ by any integer), and under multiplication (since we can multiply $\frac1d$ and $\frac1{d'}$). It easily follows that $D$ (and therefore $R$) is determined by the subset $P$ of prime numbers in $D$, as $D$ will be the set of all positive integers all of whose prime factors are in $P$. If $P$ is finite, one can write $R=\mathbb Z[\frac1n]$ for any $n$ such that the set of prime divisors of $n$ is $P$.
In your example $P=\{2,5\}$ and any $n$ with exactly those prime divisors will work; $n=10$ is the smallest positive example of such $n$.
-
HINT: In order to have $\Bbb Z\left[\frac1{20},\frac1{32}\right]\subseteq\Bbb Z\left[\frac1n\right]$, having integers $a$ and $b$ such that $\frac1{20}=\frac{a}n$ and $\frac1{32}=\frac{b}n$ will work nicely. This implies that $n=20a$ and $n=32b$, so $n$ must be a common multiple of $20$ and $32$. On the other hand, there must also be integers $a$ and $b$ such that $\frac1n=\frac{a}{20}+\frac{b}{32}$; rearrange that to get an equation without fractions involving $a,b$, and $n$, and see what it tells you about $n$. Between the two requirements, you should be able easily to find an $n$ that works.
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With the ususal sense of square brackets ("ring generated by") the "must" in your first sentence is not warranted. See my answer. In fact your answer works for "additive group generated by". – Marc van Leeuwen Sep 12 '12 at 7:11
@Marc: Fair enough. I was taking the easy route and then carelessly over-generalizing. – Brian M. Scott Sep 12 '12 at 7:16
Hint $\$ Let $\rm\,Z = \Bbb Z$ or any Bezout domain, i.e. a domain with gcds which are linear combinations $\rm\,(a,b) = ja+kb,\ j,k\in Z.\:$ Every ring $\rm\,R\,$ between $\rm\,Z\,$ and its fraction field $\rm\,Q\,$ is equal to the subring generated by $\rm\,Z\,$ and inverses of primes of $\rm\,Z\,$ occuring in denominators of reduced fractions in $\rm\,R.$
First $\rm\,a/b \in R\iff 1/b\in R,\,$ since, wlog $\rm\,(a,b)=1,\,$ thus $\rm\,ja+kb=1\,$ for some $\rm\,j,k,\in Z,\,$ hence $\rm\,j(a/b)+k = (aj+kb)/b = 1/b\in R.\:$ Therefore $\rm\,R\,$ is generated by $\rm\,Z\,$ and the inverses of the denominators of the reduced fractions in $\rm\,R.$
Also $\rm\:1/ab\in R\iff 1/a,\,1/b\,\in R,\:$ by $\rm\:a(1/ab) = 1/b.\,$ Thus if, further, $\rm\,Z\,$ is a UFD, then we can uniquely factor the denominators into primes, whittling the generating set down to $\rm\,Z\,$ and the inverses of said primes. Therefore, for your example
$$\rm \Bbb Z\left[\frac{1}{20},\frac{1}{32}\right] =\, \Bbb Z\left[\frac{1}{5\cdot 2^2},\frac{1}{2^5}\right] =\, \Bbb Z\left[\frac{1}2,\frac{1}5\right] =\, \Bbb Z\left[\frac{1}{10}\right]$$
Remark $\$ Much is known about rings enjoying this and similar properties, e.g. see
Gilmer, Robert; Ohm, Jack. Integral domains with quotient overrings.
Math. Ann. 153 1964 97--103. MR 28#3051 13.15 (16.00)
An integral domain D with field of quotients K is said to have the QR-property if every ring D' between D and K is a ring of quotients of D. The article studies integral domains with the QR-property. Starting with some simple properties of integral domains with the QR-property, the authors prove the following theorem. Let D be an integral domain with the QR-property, and let P be its arbitrary prime ideal. Then D_P is a valuation ring. If A is a finitely generated ideal of D, then A is invertible and some power of A is contained in a principal ideal.
As for the converse, the authors prove that if every valuation ring of K containing D is a ring of quotients of D and if every prime ideal of D is the radical of a principal ideal, then D has the QR-property. As a particular case, it follows that a Noetherian integral domain D has the QR-property if and only if it is a Dedekind domain in which the ideal class group is a torsion group. Some properties of intermediate rings for such rings are observed.
{Reviewer's note: Though the authors added that the result on Noetherian integral domains with the QR-property was obtained independently also by E. D. Davis, the reviewer is afraid that this special case may be known by some other people. Substantially the same result is also contained in the paper of Goldman reviewed below [#3052].}
Reviewed by M. Nagata
Richman, Fred. Generalized quotient rings.
Proc. Amer. Math. Soc. 16 1965 794--799. MR 31#588013.80 (16.00)
Let A be an integral domain with the quotient field K , and consider an intermediate ring A < B < K . The author calls an intermediate ring B a "generalized quotient ring of A" if B is A-flat. (For example, if B = S^{-1} A for some multiplicative subset S of A, then B is A-flat.) One of the main theorems reads as follows. Let A be an integral domain with the quotient field K. For an intermediate ring A < B < K, the following statements are equivalent: (i) B is a generalized quotient ring of A; (ii) (A:Ab)B = B for all b in B; (iii) B_M = A_{M/\A} for all maximal ideals M of B. The author then studies the correspondence of ideals in A and in its generalized quotient ring, and it is also shown that if B is a generalized quotient ring of A, then the integral closure of B is a generalized quotient ring of the integral closure of A. Finally, the author gives a new characterization of Prüfer rings; namely, an integral domain A is a Prüfer ring if and only if every extension of A in its quotient field is flat.
Reviewed by D. S. Rim
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Let $n=\operatorname{lcm}(32,20)=160$ then $R=Z[1/32,1/20]\subset Z[1/n]$ since 1/32=5/160, 1/20=8/160. On the other hand $1/160=4\cdot 1/32\cdot 1/20\in R$. So $n=160$ is a valid answer
- | 2015-07-29T00:50:19 | {
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https://math.stackexchange.com/questions/1091925/how-do-dummy-variables-work | # How do “Dummy Variables” work?
I do not understand how dummy variables work in math.
Suppose we have:
$$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$
How is this equivalent to:
$$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$
How does this dummy variable system work?
Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$
??
Thanks!
• You are mistaken, $y$ is not a dependent variable for $I_1$. Why do you think that? – Mark Fantini Jan 5 '15 at 15:36
• The $y$ in $I_2$ is not necessary this one in $y = e^{-x^2}$. – Alex Silva Jan 5 '15 at 15:36
• See @AlexSilva's comment. $I_1$ is just a number, it has no functional relation. If you had $$I_1 = \int_0^{\infty} e^{-y x^2} \, dx,$$ then now you'd have $I_1$ as a function of $y$. But this would be the same as $$I_1 = \int_0^{\infty} e^{-y t^2} \, dt.$$ The dummy variable changed but all is the same. – Mark Fantini Jan 5 '15 at 15:40
• If you are not in a setting where you've stated explicitly that $y = e^{-x^2}$, then there is no connection between $x$ and $y$ at all. And even if you were, the integral $\int e^{-x^2}dx$ is equal to the integral $$\int e^{-y^2}dy = \int e^{-e^{-2x^2}}d(e^{-x^2})$$ This is what variable substitution is all about. (I cannot use limits here, since $e^{-x^2}$ never becomes $0$ or $\infty$.) – Arthur Jan 5 '15 at 15:41
• As I always tell my students: you need to break free from the chains of labels, and see them as a vehicle to speak of something, not the ontological manifestation of something. Whether $y$ is a dependent variable, independent variable, coordinate function or a scream of agony during finals depends entirely on context, not on the particular shape of the symbol or its common use in one course or another. – guest Jan 5 '15 at 21:39
$$\sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \sum_{k=1}^3 k^4.$$ In the first term, $1^4$, we can say that $j=1$; in the second term $2^4$ we have $j=2$, and in the third term, $3^4$ we have $j=3$. But when we call the index $k$ rather than $j$, then in the first term $k=1$, in the second $k=2$, and in the third $k=3$.
$j$ or $k$ is a bound variable (also sometimes called a dummy variable). The value of the whole expression $1^4+2^4+3^4$, which is $98$, does not depend on the value of a bound variable.
If we write $\displaystyle\sum_{j=1}^3 (j \cos (j+m))^4$, then $j$ is bound and $m$ is free. The sum is $$(1\cos (1+m))^4 + (2\cos (2+m))^4 + (3\cos(3+m))^4.\tag 1$$ Its value depends on the value of the free variable $m$, but not on the value of anything called $j$. Accordingly we do not see $j$ in $(1)$. We could rename $j$ and call it $k$, and the whole thing would still be equal to the expression $(1)$ in which we also do not see $k$.
Another example is expressions like $\displaystyle\lim_{h\to0}\frac{(x+h)^3-x^3}h=3x^3$. The value of this expression depends on the value of the free variable $x$, but not of the bound (or "dummy") variable $h$. We could have said $\displaystyle\lim_{k\to0}\frac{(x+k)^3-x^3}k=3x^3$ and it would still be $3x^2$.
• Replace the last $h$ by $k$. – Alex Silva Jan 5 '15 at 15:54
• Thank you (+1), but what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? – anonymous Jan 5 '15 at 17:44
• In the notation $\int_a^b g(x)\,df(x)$, one is dealing with a Riemann–Stieltjes integral, defined a a limit of $\sum\limits_{x\in\text{partition}} g(x)\,\Delta f(x)$ $=\sum_x g(x_i^{*})(f(x_{i+1}-f(x_i))$ as the mesh of the partition approaches $0$. This is the same as $\int_a^b g(x)f'(x)\,dx$ in cases where $f$ is everywhere differentiable, but it is also defined in cases where $f$ may do a lot of its changing of values in places where it is not differentiable, including jump discontinuties and places where $f$ is continuous but not absolutely continuous. ${}\qquad{}$ – Michael Hardy Jan 5 '15 at 18:47
• But $\int_a^b g(x)\,df(x)$ is the same as $\int_a^b g(w)\,df(w)$, i.e. one can re-name a bound variable. But the functions $g$ and $f$ are particular functions; the value of the integral depends on which function $f$ is, so $f$ is not a bound variable. ${}\qquad{}$ – Michael Hardy Jan 5 '15 at 18:49
• @MichaelHardy thanks for replying. But I mean, in the above, we have $f(x) = e^{-x^2}$ right? The vertical axis is $f(x)$ then, we have $f(y) = e^{-y^2}$ right? Then we iterate the integral meaning now we have $dxdy$ then if both are independent variables, how will we look at this model in 3D? Since we cannot have the axes related to each other? I mean $dxdy$ how do we look at this in 3D? sicne $y$ and $x$ are both independent variables, we cannot have the traditional width, length $dxdy$ – anonymous Jan 6 '15 at 12:45
In both cases they are the independent variable. Note in your integrals that $x$ appears nowhere in $I_2$ and $y$ occurs nowhere in $I_1$.
It's common to write $y$ as the dependent variable and $x$ the independent variable. But there's no rule that says you have to. You can write $x = f(y) = y^2 + 2\sin y$ if you wanted to, or put the $y$-axis horizontal and $x$ vertical.
edit: note in any event that $I$ is constant $\left({I = \dfrac {\sqrt{\pi}} 2}\right)$ so it isn't "dependent" in the way you may be used to. Mathematicians will say things like "$I$ is a trivial function of $x$" which means that even though it's technically true that $I = f(x)$, the relationship isn't interesting.
Here's an interesting non-trivial function:
$\displaystyle I(z) = \int_0^z e^{-t^2} \, \mathrm dt$
• (wrote in Michael Hardy's answer too) what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? – anonymous Jan 5 '15 at 17:45
IMO this can be understood much better by writing it out in a proper context-free way, rather than standard mathematical notation. The integration symbol $\int_0^\infty$ is basically a higher-order-function, taking a real function1 and returning a single number. $$\int\limits_0^\infty : (\mathbb{R}\to \mathbb{R}) \to \mathbb{R}.$$ Now, people keep saying stuff like "$\sin(x)$ is a function..." but really this is incorrect. $\sin(x)$ is, for any value of $x$, just a single real number. What's a function is $\sin$ itself, i.e. not applied to anything. For instance, $$\int\limits_0^\pi \sin = 2$$ would be quite a reasonable statement. OTOH, it's nonsense to write $$\int\limits_{-\infty}^0 e^x = 1$$ because $e^x$ is not a function.
Trouble is, most functions you need to integrate won't have a predefined name. You can't thus write "the function itself" directly, but have to specify those results of the function for general inputs, in form of some algebraic expression. You know, the equation-definitions $$f(x) = e^{-x^2}.$$ Here it should be quite clear that the $x$ symbol is just an arbitrary choice – it's something you use to signify "anything you can put into the function", and $f(y) = e^{-y^2}$ says obviously exactly the same thing.
Always having to give functions a name before integrating would be tedious, so we have "shortcut syntax" to define an "anonymous function" and use it right away. You can thus think of the $\mathrm{d}x$ symbol much as a lambda function.
1In fact, more accurately and generally, a differential form.
2Again, that's not all there is to it since $\mathrm{d}x$ really is a differential form.
When you are told to compute some mathematical quantity depending on given data $a$, $\ldots\>$, $q$ the slave (person or computer system) doing the computation for you will introduce certain auxiliary variables whose names, temporary values, etc., will be invisible to you, and will not affect the end result. The latter depends only on the "outer" quantities $a$, $\ldots\>$, $q$. When these auxiliary variables are really varying during the computational process you can call them dummy variables – they will be "burnt" at the end; and when they are uniquely determined by the given data you can call them accessory parameters to the question. | 2021-07-29T19:05:34 | {
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https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads | # Probability of tossing a fair coin with at least $k$ consecutive heads
Tossing a fair coin for $N$ times and we get a result series as $HTHTHHTT\dots~$, Here '$H$' denotes 'head' and '$T$' denotes 'tail' for a specific tossing each time.
What is the probability that the length of the longest streak of consecutive heads is greater than or equal to $k$? (that is we have a $HHHH\dots~$, which is the substring of our tossing result, and whose length is greater than or equal to $k$)
I came up with a recursive solution (though not quite sure), but cannot find a closed form solution.
Here is my solution.
Denote $P(N,k)$ as the probability for tossing the coin $N$ times, and the longest continuous heads is greater or equal than $k$. Then (For $N>k$)
$$P(N,k)=P(N-1,k)+\Big(1-P(N-k-1,k)\Big)\left(\frac{1}{2}\right)^{k+1}$$
• Your example sequence reminded me of this brilliant Simpsons scene :-) – joriki Nov 10 '12 at 8:58
• I'm getting $(N-k+1)/2^k$ for the closed form. Would you like for me to post my solution, or do you want to think about it some more before I spoil the beans? – Braindead Nov 10 '12 at 8:59
• @Braindead: That can't be right; it's $\gt1$ for small $k$. – joriki Nov 10 '12 at 9:00
• Duh, you are right. I overcounted. – Braindead Nov 10 '12 at 9:08
• I tried to clarify some of the formulations; please check whether I preserved the intended meaning. In particular, I assumed that you had merely accidentally written "greater" once instead of "greater or equal". – joriki Nov 10 '12 at 9:12
Your recurrence relation is correct. I don't think you can do much better than that for general $k$, but you can find a closed form for specific values of $k$. For the first non-trivial value of $k$, the recurrence relation is
$$p_n=p_{n-1}+(1-p_{n-3})/8\;.$$
With $p_n=1+\lambda^n$, the characteristic equation becomes $\lambda^3-\lambda^2+1/8=0$. One solution, $\lambda=1/2$, can be guessed, and then factoring yields $(\lambda-1/2)(\lambda^2-\lambda/2-1/4)$ with the further solutions $\lambda=(1\pm\sqrt5)/4$. Thus the general solution is
$$p_n=1+c_1\left(\frac12\right)^n+c_2\left(\frac{1+\sqrt5}4\right)^n+c_3\left(\frac{1-\sqrt5}4\right)^n\;.$$
The initial conditions $p_0=0$, $p_1=0$, $p_2=1/4$ determine $c_1=0$, $c_2=-(1+3/\sqrt5)/2$ and $c_3=-(1-3/\sqrt5)/2$, so the probability is
\begin{align} p_n &=1-\frac{1+3/\sqrt5}2\left(\frac{1+\sqrt5}4\right)^n-\frac{1-3/\sqrt5}2\left(\frac{1-\sqrt5}4\right)^n\\ &=1-\frac4{\sqrt5}\left(\left(\frac{1+\sqrt5}4\right)^{n+2}-\left(\frac{1-\sqrt5}4\right)^{n+2}\right)\;. \end{align}
Thus, for large $n$ the probability approaches $1$ geometrically with ratio $(1+\sqrt5)/4\approx0.809$.
• I've added an explicit formula for the $P(N,k)$ which might be of interest to you. Note, that your $p_n=1-\frac{1}{2^n}F_{n+2}$ with $F_n$ the Fibonacci numbers. Regards, – Markus Scheuer Jan 18 '16 at 21:40
We can derive an explicit formula of the probability $P(N,k)$ based upon the Goulden-Jackson Cluster Method.
We consider the set of words $\mathcal{V}^{\star}$ of length $N\geq 0$ built from an alphabet $$\mathcal{V}=\{H,T\}$$ and the bad word $\underbrace{HH\ldots H}_{k \text{ elements }}=:H^k$ which is not allowed to be part of the words we are looking for. We derive a function $f_k(s)$ with the coefficient of $s^N$ being the number of wanted words of length $N$.
The wanted probability $P(N,k)$ can then be written as \begin{align*} P(N,k)=1-\frac{1}{2^N}[s^N]f_k(s) \end{align*}
According to the paper (p.7) from Goulden and Jackson the generating function $f_k(s)$ is \begin{align*} f_k(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\tag{1} \end{align*} with $d=|\mathcal{V}|=2$, the size of the alphabet and with $\mathcal{C}$ the weight-numerator with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[H^k]) \end{align*} We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[H^k])&=-\frac{s^k}{1+s+\cdots+s^{k-1}}=-\frac{s^k(1-s)}{1-s^k} \end{align*}
We obtain the generating function $f(s)$ for the words built from $\{H,T\}$ which don't contain the substring $H^k$ \begin{align*} f_k(s)&=\frac{1}{1-2s+\frac{s^k(1-s)}{1-s^k}}\\ &=\frac{1-s^k}{1-2s+s^{k+1}}\tag{2}\\ \end{align*}
Note: For $k=2$ we obtain \begin{align*} f_2(s)&=\frac{1-s^2}{1-2s+s^{3}}\\ &=1+2s+3s^2+5s^3+8s^4+13s^5+21s^6+34s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_2(s)$ are a shifted variant of the Fibonacci numbers stored as A000045 in OEIS.
Note: For $k=3$ we obtain \begin{align*} f_3(s)&=\frac{1-s^3}{1-2s+s^{4}}\\ &=1+2s+4s^2+7s^3+13s^4+24s^5+44s^6+81s^7+\mathcal{O}(s^8) \end{align*}
The coefficients of $f_3(s)$ are a shifted variant of the so-called Tribonacci numbers stored as A000073 in OEIS.
We use the series representation of $f_k(s)$ in (2) to derive an explicit formula of the coefficients.
\begin{align*} [s^N]f(s)&=[s^N](1-s^k)\sum_{m=0}^{\infty}(2s-s^{k+1})^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m(2-s^k)^m\\ &=[s^N](1-s^k)\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\\ &=([s^N]-[s^{N-k}])\sum_{m=0}^{\infty}s^m\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{3}\\ &=\sum_{m=0}^{N}([s^{N-m}]-[s^{N-k-m}])\sum_{j=0}^m\binom{m}{j}(-1)^js^{kj}2^{m-j}\tag{4}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ &\qquad-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{N-k}\binom{m}{\frac{N-m}{k}-1}(-1)^{\frac{N-m}{k}-1}2^{m-\frac{N-m}{k}+1}\\ &=\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\tag{5}\\ &\qquad+\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{m-\frac{N-m}{k}}\\ \end{align*}
Comment:
• In (3) we use the linearity of the coefficient of operator and $[s^N]s^mf(s)=[s^{N-m}]f(s)$
• In (4) we change the limit of the left hand sum from $\infty$ to $N$ according to the maximum coefficient $[s^N]$. According to the factors $s^{kj}$ we consider in the following only summands with $m\equiv N(\bmod k)$
• In (5) we reorganise the sums from the line above by extracting from the left hand sum the first summand and shifting in the right hand side the index by one and putting both sums together.
We conclude: An explicit representation of the probability $P(N,k)$ $(n\geq 0)$ is according to (5) \begin{align*} P(N,k)&=1-\sum_{{m=0}\atop{m\equiv N(\bmod k)}}^{k-1}\binom{m}{\frac{N-m}{k}}(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}}\\ &\qquad-\sum_{{m=k}\atop{m\equiv N(\bmod k)}}^{N-k} \left(\binom{m}{\frac{N-m}{k}}-\frac{1}{2^k}\binom{m-k}{\frac{N-m}{k}}\right)(-1)^{\frac{N-m}{k}}2^{-\frac{(k+1)(N-m)}{k}} \end{align*}
A bit more practical focus than previous answers:
Main idea:
1. We build a "staircase":
2. $p$ chance to get to next step, $1-p$ chance to fall down and have to restart (or "reflip"). Except for the highest step where we always stay (whenever we manage to get there).
3. We need $k+1$ steps to "remember" where on the stairs we are for $k$ flips in a row.
We can use this to build a matrix for a Markov chain:
We can build a block matrix:$$\frac{1}{2}\left[\begin{array}{cc}2&1&\bf 0^T\\\bf 0&\bf0&\bf I_k\\0&1&\bf 1^T \end{array}\right]$$
(For the special case $k = 5$). We build a stochastic matrix:
$${\bf P} = \frac{1}{2}\left[\begin{array}{cccccc}2&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\0&1&1&1&1&1\end{array}\right]$$
Now our answer will simply be $$[1,0,{\bf 0}]\, {\bf P}^k\, [{\bf 0},0,1]^T$$
Which is a scalar product using ${\bf P}^k$ as a gramian matrix.
Now to calculate this in practice one could probably in general (for general $p$) benefit from a canonical transformation of $\bf P$, but it's doubtable in this case as the matrix is already sparse with literally only sums and permuations and bit shifts required on the vector elements.
Feller considers this problem in section XIII.7 of An Introduction to Probability Theory and Its Applications, Volume 1, Third Edition. He shows that the probability of having no run of length $k$ in $N$ throws is asymptotic to $$\frac{1-(1/2)\;x}{(k + 1 - kx)\; (1/2)} \cdot \frac{1}{x^{N+1}}$$ where $x$ is the least positive root of $$1 - x +(1/2)^{k+1} x^{k+1} = 0$$
(I have changed Feller's notation to agree with the problem statement and have only considered the case of a fair coin; Feller considers the more general case of a biased coin. For more information see equation 7.11, p. 325 in the referenced document.) | 2019-09-18T20:07:25 | {
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https://math.stackexchange.com/questions/2282623/can-an-open-set-be-covered-by-proper-open-subsets | # Can an open set be covered by proper open subsets?
Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.
If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of $U$. (I think.)
Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?
If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?
Is the answer any different if we replace $U$ by an arbitrary closed set?
• Have you thought about the integers? – Steve D May 15 '17 at 21:38
• For many topological spaces you can do this. For example, any open subset of $\mathbb R^n$ is a union of open balls, which can be made proper. However, @SteveD's hint shows that you can't always do this. – Cheerful Parsnip May 15 '17 at 22:20
• If $X$ is a $T_1$ space and if $U$ is open and has more than $1$ member then $\{U$ \ $\{p\}: p\in U\}$ is an open cover of $U$ by open proper subsets of $U.$ ....If $\tau=\{X,\phi\}$ then with U=X, no proper non-empty subset of $U$ is open.... Another example is Sierpinski space $S=\{0,1\}$ with $\tau=\{S,\phi, \{0\}\}$. – DanielWainfleet May 15 '17 at 23:57
• If U is open, of course. If U is open every point is an interior point so for every point x in U there is an open neighborhood of x completely in U. Call this N_x, then $\cup_{x\in U} Nx$ is an open cover. If U is not open then of course not, If U is not open there is a point that is not an interior point and any open set containing it must contain points not in U (else it'd be an interior point). So all open covers contain points not in U. – fleablood May 16 '17 at 1:25
• @fleablood This is the argument I was going to give, but (as others pointed out to me) it only goes through if the space is at least $T_1$. – Austin Mohr May 16 '17 at 1:31
A space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.)
We can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \in U$, construct an open neighborhood $U_x$ as follows:
1. Choose any $y \in U$ that is distinct from $x$.
2. Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.
3. Set $U_x = G_x \cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.
Finally, we can take $\{U_x \mid x \in U\}$ as an open cover of $U$ by proper open subsets.
• But why is it guaranteed that the "first" $y$ we chose is an element of some element of $\{ U_x \ | \ x\in U \}$ ? Moreover, why is every such $y$ eventually in some of the cover elements? Because, by construction, $y$ is not in $U_x$ – étale-cohomology May 17 '17 at 5:52
• Let $y_0$ be the "first $y$" chosen in this process. Eventually we will need to construct the set $U_{y_0}$, which must contain $y_0$. Similarly, any $y \in U$ is an element of $U_y$ and therefore covered. – Austin Mohr May 17 '17 at 8:36
• Got it. Thank you. I wonder if it's possible to prove that $T_1$ is the weakest possible hypothesis that allows existence of these covers... – étale-cohomology May 18 '17 at 22:38
• One can push my argument through with something weaker (but I don't know if it has a name): For any open set $U$ and any point $x$ of $U$, there is an open neighborhood $V$ of $x$ that is a strict subset of $U$. This property is not as strong as $T_1$, since it still allows for the existence $x$ and $y$ that cannot be separated. However, you could take the $V$ in this property as the $U_x$ in my answer and it still get your desired refinement. This is property would be weakest possible, since otherwise the only way to cover $x$ is to use all of $U$. – Austin Mohr May 19 '17 at 1:01
It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.
However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.
Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.
If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.
When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets. | 2019-12-06T05:52:18 | {
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https://math.stackexchange.com/questions/3156341/why-is-xe-x0-for-all-x-in-mathbbr | # Why is $x+e^{-x}>0$ for all $x \in \mathbb{R}$?
Denote $$f(x) = x + e^{-x}$$. Note that $$f(0) = 1$$ and $$f'(x) = 1-e^{-x}$$. That means $$\lim_{x\rightarrow -\infty} f'(x) = -\infty.$$ So if the rate of change of $$f(x)$$ keeps decreasing exponentially fast to negative infinity, why is $$\lim_{x\rightarrow -\infty} f(x) = +\infty$$?
• Addressing the title, for all $x, e^{-x}\ge1-x$ so $x+e^{-x}\ge1>0$ – J. W. Tanner Mar 21 at 4:45
• To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}\ge t+1$ for all real $t$. So $e^{-x}\ge -x+1> -x$ for all real $x$, as desired. – Minus One-Twelfth Mar 21 at 4:46
• Roughly speaking - if you are visualising $x$ going from $0$ to $-\infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards". – David Mar 21 at 4:49
• Thank you. That's what I was missing. Going backward. :( – Paichu Mar 21 at 4:50
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
Take
$$x < y < 0; \tag 1$$
then
$$f(y) - f(x) = \displaystyle \int_x^y f'(s)\; ds = \int_x^y (1 - e^{-s})\; ds; \tag 2$$
it is clear that, for any $$M < 0$$, there exists $$y_0 < 0$$ such that
$$s < y_0 \Longrightarrow 1 - e^{-s} < M; \tag 3$$
if we now choose
$$y < y_0, \tag 4$$
then
$$f(y) - f(x) = \displaystyle \int_x^y (1 - e^{-s})\; ds < \int_x^y M \; ds = M(y - x); \tag 5$$
we re-arrange this inequality:
$$f(x) - f(y) > -M(y - x) = M(x - y), \tag 6$$
$$f(x) > f(y) + M(x - y); \tag 7$$
we now fix $$y$$ and let $$x \to -\infty$$; then since $$M < 0$$ and $$x - y < 0$$ for $$x < y$$,
$$\displaystyle \lim_{x \to -\infty} f(y) + M(x - y) = \infty, \tag 8$$
and hence
$$\displaystyle \lim_{x \to -\infty}f(x) = \infty \tag 9$$
as well.
We note that (9) binds despite the fact that $$f'(x) < 0$$ for $$x < 0$$; though this derivative is negative, when $$x$$ decreases we are "walking back up the hill," as it were; as $$x$$ decreases, $$f(x)$$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$$x + e^{-x} > 0, \forall x \in \Bbb R; \tag{10}$$
for
$$x \ge 0, \tag{11}$$
$$e^{-x} > 0 \tag{12}$$
as well, hence we also have
$$x + e^{-x} > 0; \tag{13}$$
for
$$x < 0, \tag{14}$$
we may use the power series for $$e^{-x}$$:
$$e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3} + \ldots; \tag{15}$$
then
$$x + e^{-x} = 1 + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \ldots > 0, \tag{16}$$
since every term on the right is positive when $$x < 0$$. End of Note.
Note that $$f''(x) = e^{-x} >0$$ so $$f$$ is strictly convex.
Since $$f'(0) = 0$$, we see that $$f(x) \ge f(0) = 1$$ for all $$x$$.
Note that
• $$\forall x \in \mathbb{R} : x+e^{-x}>0 \Leftrightarrow \forall x \in \mathbb{R} :e^{-x}>-x \Leftrightarrow \forall x \in \mathbb{R} :\color{blue}{\boxed{e^{x}>x}}$$
The last inequality follows directly by Taylor: $$\color{blue}{e^x} = 1+x + \underbrace{\frac{e^{\xi}}{2}x^2}_{\geq 0} \color{blue}{> x}$$ | 2019-12-06T01:20:01 | {
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https://math.stackexchange.com/questions/1486843/how-can-you-express-the-next-two-consecutive-odd-numbers-in-terms-of-x | # How can you express the next two consecutive odd numbers in terms of x?
I'm trying to express the next two consecutive odd numbers in terms of x. How do go about this?
Precisely, given a number $x$ I want two smallest consecutive odd numbers $m,n$ such that $x<m<n$.
• Be aware that you wrote two very different things in your question and your body. Are you trying to express the next two consecutive odd numbers in terms of x, or are you trying to express x in terms of the next two consecutive odd numbers? If $x$ is, say $5$, do you consider "the next two consecutive odd numbers" to be $5$ and $7$ or do you consider the next two consecutive odd numbers to be $7$ and $9$? – JMoravitz Oct 19 '15 at 2:32
• My error. I'm trying t express the next two consecutive odd numbers in terms of x. Edited to reflect. – user214824 Oct 19 '15 at 2:45
• That still doesn't answer the question of what exactly you mean by "next two consecutive odd numbers" is. If $5$ is your number, do you want $7$ and $9$? or do you want $5$ and $7$ (despite $5$ not being bigger than $5$). – JMoravitz Oct 19 '15 at 2:47
• @JMoravitz Yes, that is it. 7 and 9. – user214824 Oct 19 '15 at 2:48
Presumably $x$ is an integer. We have two cases: either $x$ is even or $x$ is odd.
In the case that $x$ is odd, then $x+2$ and $x+4$ will both be odd and will be the next two consecutive odd numbers.
In the case that $x$ is even, then $x+1$ and $x+3$ will both be odd and will be the next two consecutive odd numbers.
Let $x$ an integer, then the next two odd integers are $$2\times\Bigl\lfloor \frac{x+1}{2}\Bigr\rfloor + 1 \qquad\text{and}\qquad 2\times\Bigl\lfloor \frac{x+1}{2} \Bigr\rfloor + 3.$$
Indeed, if $x = 2n$ where $n\in\mathbb{Z}$, then $$2\lfloor (x+1)/2\rfloor + 1 = 2 \lfloor n + 1/2 \rfloor + 1 = 2n + 1$$ and if $x = 2n + 1$, then $$2\lfloor (x+1)/2 \rfloor + 1 = 2\lfloor n + 1 \rfloor + 1 = 2n + 3.$$
• Maybe doesn't you know the symbol $\lfloor \cdot \rfloor$. If $x$ is a real number, $\lfloor x \rfloor$ is the biggest integer less or equal to $x$. For example $\lfloor 2.7 \rfloor = 2$, $\lfloor -3.8 \rfloor = -4$ and $\lfloor x \rfloor = x$ for $x$ integer. – Éric Guirbal Oct 19 '15 at 3:01 | 2021-04-15T12:08:37 | {
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https://math.stackexchange.com/questions/2970009/is-this-theorem-equivalent-to-transfinite-recursion-theorem/2970279 | # Is this theorem equivalent to Transfinite Recursion Theorem?
Let $$V$$ be the class of all sets, $$\operatorname{Ord}$$ be the class of all ordinals, and $$G:V\to V$$ be a class function.
Transfinite Recursion Theorem:
There exists a class function $$F:\operatorname{Ord}\to V$$ such that $$F(\alpha)=G(F\restriction \alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.
While I'm able to prove that Transfinite Recursion Theorem implies the below theorem, I have tried but to no avail in proving that the theorem implies Transfinite Recursion Theorem.
I would like to ask if it is possible to prove that this theorem implies Transfinite Recursion Theorem.
Theorem:
Let $$G_1,G_2,G_3$$ be class functions from $$V$$ to $$V$$. There exists a class function $$F:\operatorname{Ord}\to V$$ such that
(1) $$F(0)=G_1(\emptyset)$$
(2) $$F(\alpha+1)=G_2(F(\alpha))$$ for all $$\alpha\in\operatorname{Ord}$$
(3) $$F(\alpha)=G_3(F\restriction\alpha)$$ for all limit $$\alpha\neq 0$$
• Do you know how to prove strong induction (on $\mathbb{N}$) from weak induction? It's the same idea. – Eric Wofsey Oct 25 '18 at 2:46
• Hi @EricWofsey! Yes, I do. I will try your suggestion. – LE Anh Dung Oct 25 '18 at 2:54
The trick is to not construct $$F$$ itself, but to construct the function $$H$$ which sends $$\alpha$$ to $$F\restriction \alpha$$. That way, at successor steps you have access to the entire history of the recursion so far and not just the last step.
In detail, given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: $$G_1(x)=\emptyset$$ $$G_2(x)=x\cup\{(\operatorname{dom}(x),G(x))\}$$ $$G_3(x)=\bigcup\operatorname{ran}(x)$$ By the theorem, we then get a function $$H:Ord\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit. It is then easy to prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for each $$\alpha$$, with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$. So, defining $$F=\bigcup\operatorname{ran}(H)$$, $$F$$ is a function on $$Ord$$ with $$F\restriction\alpha=H(\alpha)$$ for each $$\alpha$$. For each $$\alpha$$, we then have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha).$$ Since $$\operatorname{dom}(F\restriction\alpha)=\alpha$$, our definition of $$G_2$$ tells us that $$F(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha),$$ as desired.
• I'm feeling confused. You have shown how to derive the Q's 2nd theorem from its first. But the proposer seems to want to see the 1st derived from the second, although it is obvious that if the second one holds, then, for a given $G$, we can simply let $G_1=G_2=G_3=G$. So I'm thinking that the proposer stated the Q backwards. – DanielWainfleet Oct 25 '18 at 3:49
• @DanielWainfleet: I am deriving the first theorem from the second. I start with a function $G$ as in the first theorem, then use the second theorem to get a function $H$ and then define $F$ from $H$ which verifies the conclusion of the first theorem. It does not work to simply define $G_1=G_2=G_3=G$ since $G_2$ takes as its input just $F(\alpha)$, not $F\restriction\alpha$. – Eric Wofsey Oct 25 '18 at 4:04
• Thank you so much! You make my day :) – LE Anh Dung Oct 25 '18 at 8:14
• Thank you for clearing up my confusion. – DanielWainfleet Oct 25 '18 at 21:26
• Hi! I'm now struggling with a seemingly similar problem like this one. I have tried to replicate your proof but failed at the end. If you don't mind, please have a look at math.stackexchange.com/questions/2970783/…. Thank you for your help! – LE Anh Dung Oct 27 '18 at 14:16
I fill in @Eric Wofsey's proof with detail and post it here. All credits are given to @Eric Wofsey.
Given $$G:V\to V$$, we define $$G_1$$, $$G_2$$, and $$G_3$$ as follows: \begin{align}&G_1(x)=\emptyset\text{ for all }x\\&G_2(x)=\begin{cases} x\cup\{(\operatorname{dom}(x),G(x))\}&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\\&G_3(x)=\begin{cases} \bigcup\operatorname{ran}(x)&\text{if }x\text{ is a function}\\\emptyset&\text{otherwise}\end{cases}\end{align}
By The theorem, there is a class function $$H:\operatorname{Ord}\to V$$ such that $$H(0)=G_1(\emptyset)$$, $$H(\alpha+1)=G_2(H(\alpha))$$, and $$H(\alpha)=G_3(H\restriction\alpha)$$ for $$\alpha\neq 0$$ limit.
First, we prove by induction that $$H(\alpha)$$ is a function with domain $$\alpha$$ for all $$\alpha\in\operatorname{Ord}$$ and with $$H(\alpha)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha$$.
• $$H(0)=G_1(0)=\emptyset$$. Then the statement is trivially true for $$\alpha=0$$.
• Assume that the statement is true for $$\alpha$$. Then $$H(\alpha+1)=G_2(H(\alpha))=$$ $$H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\}=H(\alpha)\cup\{(\alpha,G(H(\alpha)))\}$$. It follows that $$\operatorname{dom}(H(\alpha+1))=\operatorname{dom}(H(\alpha))\cup \{\alpha\}=\alpha\cup \{\alpha\}=\alpha+1$$. For $$\beta=\alpha$$, $$H(\alpha+1)\restriction \beta=H(\alpha+1)\restriction \alpha=H(\alpha)=H(\beta)$$. For $$\beta<\alpha$$, $$H(\alpha+1)\restriction \beta=$$ $$H(\alpha)\restriction \beta=H(\beta)$$. Thus $$H(\alpha+1)\restriction \beta=H(\beta)$$ for all $$\beta<\alpha+1$$.
• Assume that the statement is true for all $$\beta<\alpha$$ where $$\alpha\neq\emptyset$$ is limit ordinal. Then $$H(\alpha)=G_3(H(\alpha))=\bigcup\operatorname{ran}(H(\alpha))=\bigcup\{H(\beta)\mid \beta<\alpha\}$$. For any $$\beta_1\le\beta_2<\alpha$$: $$H(\beta_2)\restriction\beta_1=H(\beta_1)$$and thus $$H(\beta_1)\subseteq H(\beta_2)$$. Then $$H(\alpha)=\bigcup\{H(\beta)\mid \beta<\alpha\}$$ is actually a function. It follows that $$\operatorname{dom}(H(\alpha))=\bigcup_{\beta<\alpha}\operatorname{dom}(H(\beta))=\bigcup_{\beta<\alpha}\beta=\alpha$$ since $$\alpha$$ is limit ordinal. Moreover, $$H(\alpha)\restriction \beta=\{(\gamma,H(\alpha)(\gamma))\mid \gamma<\beta\}=\{(\gamma,H(\gamma+1)(\gamma))\mid \gamma<\beta\}=$$ $$\{(\gamma,H(\beta)(\gamma))\mid \gamma<\beta\}=H(\beta)$$.
As a result, $$\forall\beta<\alpha:H(\alpha)\restriction \beta=H(\beta)$$ and thus $$\forall\beta<\alpha:H(\beta)\subsetneq H(\alpha)$$.
Next we define $$F=\bigcup\operatorname{ran}(H)$$. Then $$F=\bigcup\{H(\alpha)\mid \alpha\in\operatorname{Ord}\}$$ is a function with domain $$\operatorname{Ord}$$ and with $$F\restriction\alpha=\{F(\beta)\mid\beta<\alpha\}=\{H(\beta+1)(\beta)\mid\beta<\alpha\}=\{H(\alpha)(\beta)\mid\beta<\alpha\}=H(\alpha)$$ for all $$\alpha\in\operatorname{Ord}$$.
Since $$F\restriction\alpha=H(\alpha)$$ and $$\operatorname{dom}(H(\alpha))=\alpha$$, $$\operatorname{dom}(F\restriction\alpha)=\alpha$$. Then $$G_2(F\restriction\alpha)=(F\restriction\alpha)\cup \{(\alpha,G(F\restriction\alpha))\}$$ and thus $$G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.
For each $$\alpha\in\operatorname{Ord}$$, we have $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=G_2(F\restriction\alpha)(\alpha)=G(F\restriction\alpha)$$.
Update: I have found another way to define $$F$$
We define $$F$$ as follows $$F(\alpha):=H(\alpha+1)(\alpha)$$
Then $$F(\alpha)=H(\alpha+1)(\alpha)=G_2(H(\alpha))(\alpha)=(H(\alpha)\cup\{(\operatorname{dom}(H(\alpha)),G(H(\alpha)))\})(\alpha)=(H(\alpha)\cup\{(\alpha,G(H(\alpha)))\})(\alpha)=G(H(\alpha))$$.
Moreover, $$H(\alpha)=\{(\beta,H(\alpha)(\beta))\mid\beta<\alpha\}=\{(\beta,H(\beta+1)(\beta))\mid\beta<\alpha\}=\{(\beta,F(\beta))\mid\beta<\alpha\}=F\restriction\alpha$$.
Thus $$F(\alpha)=G(H(\alpha))=G(F\restriction\alpha)$$. | 2021-06-18T06:03:48 | {
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https://math.stackexchange.com/questions/1495086/envelope-of-projectile-trajectories?noredirect=1 | # Envelope of Projectile Trajectories
For a given launch velocity $$v$$ and launch angle $$\theta$$, the trajectory of a projectile may be described by the standard formula $$y=x\tan\theta-\frac {gx^2}{2v^2}\sec^2\theta$$
For different values of $$\theta$$ what is the envelope of the different trajectories? Is it a parabola itself?
The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in $$\tan\theta$$ and set the discriminant to zero. The resulting relationship between $$x,y$$ is the envelope.
This question is posted to see if there are other approaches to the solution.
Edit 1
Thanks for the nice solutions from Jack and Blue, received so far. From the solution of the envelope it can be worked out that the envelope itself corresponds to the right half of the trajectory of a projectile launched at $$(-\frac{v^2}{g^2},0)$$ at a launch angle $$\alpha=\frac{\pi}4$$ and a launch velocity $$V=v\sqrt2$$. This means that both vertical and horizontal components of the launch velociy are equal to $$v$$. It would be interesting to see if these conclusions can be inferred from the problem itself by inspection and without first solving it. If so, then this would form another solution.
• I think it is more like an ellipse. – Archis Welankar Oct 24 '15 at 10:05
• @ArchisWelankar - What makes you think that might be the case? – hypergeometric Oct 24 '15 at 10:13
• When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. – Archis Welankar Oct 24 '15 at 10:25
• @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. – hypergeometric Oct 24 '15 at 10:34
• "The standard solution to this "envelope of safety" problem is to state the formula as a quadratic in tanθ and set the discriminant to zero. " I don't understand this approach, can someone explain ? – Dat Sep 11 at 9:38
Yes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\tan\theta -\frac{gx^2}{2v^2}\sec^2(\theta) = x\tan(\theta+\varepsilon) -\frac{gx^2}{2v^2}\sec^2(\theta+\varepsilon)$$ we get $x=0$ or $$x=\frac{2v^2}{g}\cdot\frac{\tan(\theta)-\tan(\theta+\varepsilon)}{\sec^2(\theta)-\sec^2(\theta+\varepsilon)}$$ and by letting $\varepsilon\to 0$ we get $x=\frac{v^2}{g}\cot(\theta)$, from which $y=\frac{v^2}{2g}\left(2-\frac{1}{\sin^2(\theta)}\right)$.
It follows that the equation of the envelope is given by: $$y = \frac{v^2}{2g}\left(1-\left(\frac{gx}{v^2}\right)^2\right)=\frac{v^2}{2g}-\frac{g}{2 v^2}\,x^2$$ that clearly is a parabola with vertex in $\left(0,\frac{v^2}{2g}\right)$ through the points $\left(\pm\frac{v^2}{g} ,0\right)$.
We may notice that the envelope and the trajectory with $\theta=\frac{\pi}{4}$ are homothetic, and the dilation ratio is just $2$. The vertices of the trajectories lie on an ellipse that is tangent to the envelope parabola, with centre at $\left(0,\frac{v^2}{4g}\right)$, a vertex in the origin and a vertex at $\left(\frac{v^2}{2g},\frac{v^2}{4g}\right)$.
• Thank you for a beautiful and creative solution! And nice graphics! A big +1! – hypergeometric Oct 24 '15 at 13:29
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21
@Jack provides a very nice and intuitive derivation of the envelope as the points of intersection of infinitely-close members of the curve family. The Wikipedia "Envelope" entry provides this less-illuminating abstraction:
The envelope of the family [of curves parameterized by $t$ is] the set of points for which $$F(t, x, y) = \frac{\partial F}{\partial t}(t,x,y) = 0 \tag{\star}$$ for some value of $t$ [...].
In $(\star)$, $F$ is the function that, when set equal to $0$, defines each curve in the family. For the question at hand, we have (with parameter $\theta$ instead of $t$) $$F(\theta,x,y) = -y + x\tan\theta -\frac{g x^2}{2v^2}\sec^2\theta \tag{1}$$ Therefore, differentiating with respect to $\theta$ gives $$\frac{\partial F}{\partial \theta}(\theta,x,y) =x\sec^2\theta -\frac{g x^2}{2v^2}\cdot 2 \sec^2\theta\tan\theta = \frac{x\sec^2\theta}{v^2} ( v^2 - g x \tan\theta) \tag{2}$$ Solving $\partial F/\partial \theta = 0$ for $\theta$ (noting that $\sec\theta$ never vanishes) gives $$\tan\theta = \frac{v^2}{g x} \qquad\text{so that}\qquad \sec^2\theta = 1 + \tan^2\theta = \frac{g^2 x^2 + v^4}{g^2 x^2}$$
Substituting into $(1)$, and setting $F=0$, we have
$$y = x\;\frac{v^2}{gx} - \frac{g x^2}{2 v^2}\;\frac{g^2 x^2 + v^4}{g^2 x^2} = \frac{v^2}{2g} - \frac{g x^2}{2v^2} = \frac{v^4-g^2x^2}{2gv^2}$$
• Thanks for working out the solution from the article. Nice find! +1. – hypergeometric Oct 24 '15 at 13:55
• Please check out comments under Edit 1 in the question and see if you have any further thoughts. – hypergeometric Oct 24 '15 at 14:21
Here's another approach. Not as elegant as Jack's though. Perhaps similar to Blue's. (Edit: Revised slightly to use substitution of $H=v^2/2g$ instead of $\lambda=g/v^2$ for more convenient reference, per solution by Narasimham)
This approach maximises the value of $y$ for any given value of $x$.
For a given value of $x$, say $x=k$, \begin{align} y&=kT-\frac {k^2}{4H}(1+T^2)\qquad\text{where T=\tan\theta, H=\frac {v^2}{2g}}\\ \frac{dy}{dT}&=k-\frac {k^2}{4H}\cdot 2T=0\qquad\text{when T=\frac {2H}k}\\ \text{At T=\frac {2H}k}:\qquad\qquad y&=k\cdot\frac{2H}k-\frac{k^2} {4H}\left(1+\frac {4H^2}{k^2}\right) \color{lightgrey}{=2H-\frac {k^2}{4H}-H}\\ &=H-\frac {k^2}{4H} \end{align}
Hence the equation of the envelope is $$y=\frac {v^2}{2g}-\frac {gx^2}{2v^2}\quad\blacksquare$$
• This nicely exploits the fact that, in this particular case, we know that the envelope sits on top of the family of curves, and therefore consists of the points that maximize $y$ for a given $x$, over all $t$. (Of course don't always have such knowledge, but no matter.) Re-writing in terms of parameter $T$ is also a helpful simplification. And, to be explicit: the connection with the process in my answer is that the criterion for maximizing $y$ ---that is, setting $dy/dT = 0$--- corresponds directly to $\partial F/\partial \theta = 0$, because $y$ drops out of the latter equation. – Blue Oct 24 '15 at 21:38
• Thanks for your kind comment! – hypergeometric Oct 25 '15 at 5:57
I shall outline the method how we get the parabola. Usual notation
$$\ddot y = -g, \dot y = - g t + v \sin \theta , y =- g t^2/2 + v t \sin \theta\, +0 \tag {1}$$ $$\ddot x = 0, \dot x = v \cos \theta = const , x = v t \cos \theta +0 \tag{2}$$
Eliminating time $t$ between (1),(2) you got this parabola equation already.
Let $\tan \alpha = T$; Parabola equation in other words
$$y = x T - g/2 * ( x/ v \cos\theta)^2 = x T - ( g x^2/2 v^2) ( 1+T^2) \tag{3}$$
Differentiate partially with respect to $T$ and simplify, $T = v^2 / gx \tag{4}$
Eliminate $T$ between (3) and (4)
$$y = v^2/2g - g x^2 /(2 v^2) = H - x^2/(4H) \tag{5},$$
same as what was obtained before by Jack and Blue.
If you denote height reached by projectile on vertical firing $H = v^2/(2g) \tag{6}$ you would notice that the envelope is profiled exactly as a parabolic mirror with focus at gun delivery point, focal length is exactly H. Vertical force of gravity is acting like light :)..
The above procedure method is indicated by Blue in Wiki, is referred to as C-discriminant method to obtain envelopes and singular solutions.
Like what you said in your edit and I about mirror, they are ploys for remembering curves using similarities..
For the image I took values of $g=9.8 m/s^2 , v = 2 m/s$
The partial differentiation way and elimination is the right way to look at, perhaps not as what you said (standard solution.. way).
EDIT2:
The answer to your second question, i.e., to determine if it is going to be a parabola envelope without going through all of analysis... I can only reply with extended C-discriminant, strengthening the same result by another path.
$p$ discriminant method is also relevant, but I defer it, but best is to refer to differential calculus books of authors e.g., A.R. Forsythe.
I shall expand on the C-discriminant, wherein a two parameter system of equations variation of any of the two parameters leads to the same parabola envelope. This is in reply to your question , Why do you also plot partial concentric circles?
Well, they are circles alright, but not concentric circles. They expand and come down slowly with time. What you can see as plotted are the peripheries traced out descending with time .
But first quickly watch some fireworks to see what I am discussing about:
It is common experience to see fire ball bright splinters expanding to bigger circles as the entire cluster comes down slowly with time. The periphery is a portion of a circle whose radius increases. Center of circle is always descending by gravity.
The two parameters are $\theta, t$ angle of elevation at first burst or fire, and time $t$.
By C-discriminant method the parabola envelope is the eliminant of either $\theta$ variable or
$$F(x,y,\theta) =0 ,\, F_{\theta } (x,y,\theta) =0 \tag{6}$$
or $t$ time variable.
$$F(x,y, t) =0 ,\, F_{t } (x,y, t) =0 \tag{7}$$
The first one is already discussed, the second one is expanding/descending fireworks circles as already stated.
In the latter case working is:
$$x = v t \cos \theta , y = v t \sin \theta - g t^2/2 \tag{8}$$
$$(\frac{x}{vt})^2 + (\frac{y+ gt^2/2}{vt}) ^2 = 1 \tag{9}$$
$$x^2 + ( y + g t^2/2)^2 = v^2t^2 \tag{10}$$
which is a Circle.
To find its envelope, as before partially differentiate with respect to time $t$ and cancel $2t$ on either side of equation , bring $v^2/g$ to right side :
$$y + gt^2/2 = v^2/g \tag {11}$$
$$x^2 + (v^2/g)^2 = 2 v^2/g * ( v^2/g-y) \tag {12}$$
$$x^2 + ( 2 H)^2 += 4 H ( v^2/g -y )\tag{13}$$
$$x^2 = 4 H ( H-y) \tag{14}$$
which is the same parabola envelope obtained earlier with $\theta$ as parameter. End points $( x=0, y=H ; x= 2 H, y= 0 )$
Although circle end traces are visible in a fireworks display it needs imagination as before with variable gun barrel angle to see that each circle is tangent to a fixed envelope. Hope you enjoyed it.
• Thanks for your solution and insight on the parabolic mirror! (+1). Nice graphics too! In addition to different parabolas, why do you also plot partial concentric circles? – hypergeometric Oct 25 '15 at 5:56
• Also, is there an argument which would directly conclude that the envelope is a parabola with focus at launch point and tip at the peak of the vertical launch, without first going through the analysis? – hypergeometric Oct 25 '15 at 6:03
• The question in my comment above has been answered here. – hypergeometric Sep 30 '18 at 17:14 | 2019-10-20T21:38:09 | {
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