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# Can someone explain me this please?
## $\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$
Jun 6, 2017
See below.
#### Explanation:
Take the right hand side of the equation:
$\frac{1}{n + 1} - \frac{1}{n + 2}$
Create a common denominator so you can subtract the 2 fractions. They common denominator is $\left(n + 1\right) \left(n + 2\right)$
$\frac{n + 2}{n + 2} \left(\frac{1}{n + 1}\right) - \frac{n + 1}{n + 1} \left(\frac{1}{n + 2}\right)$
$\frac{\left(n + 2\right) - \left(n + 1\right)}{\left(n + 1\right) \left(n + 2\right)}$
Be careful when distributing the minus sign.
$\frac{n + 2 - n - 1}{\left(n + 1\right) \left(n + 2\right)}$
$\frac{1}{\left(n + 1\right) \left(n + 2\right)}$
Jun 6, 2017
Here's what I got.
#### Explanation:
I'm guessing that you want to know how to use partial-fraction decomposition to get
$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$
The first thing to notice here is that the denominator of the original fraction can be factored as a product of linear factors, i.e. factors that have the highest power of the variable equal to $1$.
$\left(n + 1\right) = {n}^{1} + 1 \to$ linear factor
$\left(n + 2\right) = {n}^{1} + 2 \to$ linear factor
This means that you can write the original fraction as a sum of two fractions, one that has the first factor as a denominator and the other that has the second factor as a denominator.
$\frac{1}{\textcolor{b l u e}{\left(n + 1\right)} \textcolor{p u r p \le}{\left(n + 2\right)}} = \frac{A}{\textcolor{b l u e}{\left(n + 1\right)}} + \frac{B}{\textcolor{p u r p \le}{\left(n + 2\right)}}$
To find the values of $A$ and $B$, multiply the first fraction by 1 = (n+2)/(n+2) and the second fraction by $1 = \frac{n + 1}{n + 1}$, where $n \ne \left\{- 1 , - 2\right\}$.
This will get you
$\frac{A}{n + 1} \cdot \frac{n + 2}{n + 2} = \frac{A \cdot \left(n + 2\right)}{\left(n + 1\right) \left(n + 2\right)} = \frac{A n + 2 A}{\left(n + 1\right) \left(n + 2\right)}$
$\frac{B}{n + 2} \cdot \frac{n + 1}{n + 1} = \frac{B \cdot \left(n + 1\right)}{\left(n + 1\right) \left(n + 2\right)} = \frac{B n + B}{\left(n + 1\right) \left(n + 2\right)}$
You now have
$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{A n + 2 A}{\left(n + 1\right) \left(n + 2\right)} + \frac{B n + B}{\left(n + 1\right) \left(n + 2\right)}$
All three fractions have the same denominator, which means that you can write
$1 = A \cdot n + 2 A + B \cdot n + B$
This is equivalent to
$1 = \left(A + B\right) \cdot n + \left(2 A + B\right)$
or
$\textcolor{red}{0} \cdot n + \textcolor{\mathrm{da} r k g r e e n}{1} = \textcolor{red}{\left(A + B\right)} \cdot n + \textcolor{\mathrm{da} r k g r e e n}{\left(2 A + B\right)}$
You can thus say that
$\left\{\begin{matrix}A + B = 0 \\ 2 A + B = 1\end{matrix}\right.$
Use the first equation to write
$A = - B$
Plug this into the second equation to get
$2 \cdot \left(- B\right) + B = 1$
$- B = 1 \implies B = - 1$
This means that
$A = - \left(- 1\right) = + 1$
Therefore, you can write the original fraction as
$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} + \frac{- 1}{n + 2}$
which is equivalent to
$\frac{1}{\left(n + 1\right) \left(n + 2\right)} = \frac{1}{n + 1} - \frac{1}{n + 2}$
You can double-check your work by starting from the right-hand side of the equation
$\frac{1}{n + 1} - \frac{1}{n + 2}$
The common denominator will be
$\left(n + 1\right) \left(n + 2\right)$
which means that you have
$\frac{1}{n + 1} \cdot \frac{n + 2}{n + 2} - \frac{1}{n + 2} \cdot \frac{n + 1}{n + 1}$
$\frac{n + 2}{\left(n + 1\right) \left(n + 2\right)} - \frac{n + 1}{\left(n + 1\right) \left(n + 2\right)}$
At this point, you can subtract the two numerators to find
$n + 2 - \left(n + 1\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} + 2 - \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} - 1 = 1$
Therefore, you once again have
$\frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{\left(n + 1\right) \left(n + 2\right)}$ |
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# Order of operations
(Redirected from Order of Operations)
Order of operations is the order in which we perform addition, subtraction, multiplication, division, and other operations in a mathematical expression.
There is a mnemonic PEMDAS--Please Excuse My Dear Aunt Sally--which may help you remember the order in which to perform the mathematical operations:
1. P = Parentheses
2. E = Exponents
3. MD = Multiplication and Division--working from left to right
4. AS = Addition and Subtraction--working from left to right
Let's look at some examples of how this works. For each example below, the mathematical expression is listed fist, followed by a new version of the equation which shows the results of the individual operations (e.g., simplification of exponent or multiplying two numbers). The operations completed for each step are described next to the step. The first example shows only one operation completed in each step. Examples 2 and 3 show more than one operation completed in some steps.
### Example 1:
$(2+1)^3+10-8*5/2$
\begin{align} & =3^3+10-8*5/2 \qquad \text{Parentheses}\\ & =27+10-8*5/2 \qquad \text{Exponent}\\ & =27+10-40/2 \qquad \text{Multiplication, because it is leftmost}\\ & =27+10-20 \qquad \text{Division, because it is next when working from left to right}\\ & =37 - 20 \qquad \text{Addition, because it is leftmost}\\ & =17 \qquad \text{Subtraction, because it is next when working from left to right}\\ \end{align}
### Example 2:
$(5 + 3) / 2 + -8 / 4 + 2 * 3$
\begin{align} & = 8 / 2 + -8 / 4 + 2 * 3 \qquad \text{Parentheses}\\ & = 4 + -2 + 6 \qquad \text{Division, as they appear working from left to right}\\ & = 8 \qquad \text{Addition (including addition of negative number), as they appear working from left to right}\\ \end{align}
### Example 3:
$4^3 * 2 + 2^4 / 4 - (4 + 1)^2$
\begin{align} & = 4^3 * 2 + 2^4 / 4 - 5^2 \qquad \text{Parentheses}\\ & = 64 * 2 + 16 / 4 - 25 \qquad \text{Exponents}\\ & = 128 + 4 - 25 \qquad \text{Multiplication and Division, as they appear working from left to right}\\ & = 107 \qquad \text{Addition and Subtraction, as they appear working from left to right}\\ \end{align} |
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# If 3^k is a divisor of the product of all even integers between 2
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If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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07 Feb 2016, 17:08
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If $$3^k$$ is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?
A. 4
B. 6
C. 10
D. 13
E. 14
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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07 Feb 2016, 17:32
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TeamGMATIFY wrote:
If $$3^k$$ is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?
A. 4
B. 6
C. 10
D. 13
E. 14
Hi,
since the Q tells us $$3^k$$ is a divisor of the product of all even integers between 2 and 30 (inclusive)
the Q basically asks us power of 3 in the product..
lets see the even multiples of 3 from 2 to 30..
6*12*18*24*30..
3^5(2*4*6*8*10)..
or 3^6*2*4*2*8*10..
ans 6
B
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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17 Feb 2016, 15:39
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The way i went about it was as follows
Product of even integers between 2 and 30 are listed below:-
=2*4*6*8.........28*30 -this can also be written as 2*1 * 2*2 * 2*3.....2*14 *2*15
2^15 *(1*2*3*4.....14*15) => 2^15* 15!
3^k would not be a divisor of 2^15, so we should find out how many times 3^k would go into 15!
15/ 3 + 15/3^2 ---> 5 +1 = 6
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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17 Feb 2016, 15:42
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we can simplify it
Product of even integers between 2 and 30 are listed below:-
=2*4*6*8.........28*30 -this can also be written as 2*1 * 2*2 * 2*3.....2*14 *2*15
2^15 *(1*2*3*4.....14*15) => 2^15* 15!
3^k would not be a divisor of 2^15, so we should find out how many times 3^k would go into 15!
15/ 3 + 15/3^2 ---> 5 +1 = 6
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink]
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23 Mar 2017, 07:25
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TeamGMATIFY wrote:
If $$3^k$$ is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?
A. 4
B. 6
C. 10
D. 13
E. 14
Set of all even integers between 2 and 30 (inclusive) = { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 , 28 , 30 }
Or, Set of all even integers between 2 and 30 (inclusive) = 2 ( 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 )
Or, Set of all even integers between 2 and 30 (inclusive) = 2*15!
Now, check the highest power of $$3^k$$ in 2*15!
Highest power of $$3^k$$ in 2*15! is = $$\frac{15}{3} => 5$$
Or, $$\frac{5}{3} = 1$$
So, the maximum value of k in $$3^k$$ is 6 ( ie, 5 + 1 )
Thus, answer must be (B) 6
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Re: If 3^k is a divisor of the product of all even integers between 2 [#permalink] 23 Mar 2017, 07:25
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# Don't count backup slides
I prepare some backup slides at the end of my presentation. I don't want them to be counted in terms of pages, which is displayed at the footer of each slide.
Does anyone know how to do that?
• Does this post answer your question: stackoverflow.com/questions/732902/… ? – Corentin Sep 6 '12 at 16:33
• Surely I misunderstood the question, but if you have just backup slides to not show... why not just put before an extra \end{document}? For see again the backups the only thing needed is put % before. – Fran Sep 6 '12 at 17:18
• @Fran as I understand the question, SoftTimur do want those frames to appear, he just doesn't want them to be taken into account when calculating the total number of frames/slides. – Gonzalo Medina Sep 6 '12 at 17:55
For the sake of completeness, here are two solutions inspired from the link given in my above comment.
1. You can define a new counter at the end of your presentation, before your backup slides, and use this counter at the end of your backup slides to fool beamer on the total number of slides. Here is a MWE:
\documentclass{beamer}
\usepackage{beamerthemesplit}
\usepackage{lipsum}
\newcommand{\backupbegin}{
\newcounter{finalframe}
}
\newcommand{\backupend}{
}
\begin{document}
% Your normal slides go here
\frame{\lipsum[1]}
\frame{\lipsum[2]}
\appendix
\backupbegin
% And your backup slides here
\frame{\lipsum[3]}
\backupend
\end{document}
In this case, the backup slide is numbered as 3/2 in this example.
2. Alternatively, you can use the appendixnumberbeamer package. MWE:
\documentclass{beamer}
\usepackage{beamerthemesplit}
\usepackage{lipsum}
\usepackage{appendixnumberbeamer}
\begin{document}
% Your normal slides go here
\frame{\lipsum[1]}
\frame{\lipsum[2]}
\appendix
% And your backup slides here
\frame{\lipsum[3]}
\end{document}
In this case, the backup slide is numbered as 1/1.
• For the second proposal, the displayed total number of slides is equal to the number of (normal+backup), instead of (normal) only. e.g. the entire presentation has 10 slides, 7 normal and 3 backup. At the 7th slide (which is normal), the displayed number is 7/10 instead of 7/7. – Herpes Free Engineer Jul 19 '17 at 9:20
• Compile twice, this will give you 7/7 – Jan-Åke Larsson May 15 '19 at 8:56
Old question and already lots of beamer-only solutions, but an alternate solution could be to build separatly the main presentation and the backup set of slides, and stitch the two pdf files together with pdftk. This way they are guaranteed to have separate numbering.
To keep formatting the same, you can put all the header stuff in a separate file, then just \input it in the two latex files.
For themes based on the infolines outer theme and a beamer version >= 3.49 you can use \setbeamertemplate{page number in head/foot}[appendixframenumber] after your theme to exclude the appendix from framenumbers displayed on the bottom of the frame.
\documentclass{beamer}
\begin{document}
\begin{frame}
\frametitle{First slide}
\end{frame}
\appendix
\begin{frame}
content...
\end{frame}
\end{document} |
# Parity of the multiplicative order of 2 modulo p
Let $$\operatorname{ord}_p(2)$$ be the order of 2 in the multiplicative group modulo $$p$$. Let $$A$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is odd, and let $$B$$ be the subset of primes $$p$$ where $$\operatorname{ord}_p(2)$$ is even. Then how large is $$A$$ compared to $$B$$?
• $A/(A+B)$ tends to $7/24$ ? (not proved yet). – Henri Cohen Sep 23 at 21:39
• Seems like an interesting question, and clearly generalizable quite a lot. However, if you're going to ask many questions on this site, it would be a good idea to learn a little bit of TeX formatting. I've fixed the formatting of your question, so if you click on "edit", you'll be able to see what I did to make it more readable. I also changed the title of your question to make it even clearer what you're asking. – Joe Silverman Sep 23 at 21:54
• @HenriCohen how did you determine $A/(A+B)$ to be $7/24$ while also writing "not proved yet"? The proportion of $p \leq 100000$ for which $2 \bmod p$ has odd order is $2797/9591$, which as a continued fraction is $[0,3,2,3,44,9]$, and the truncated continued fraction $[0,3,2,3]$ is $7/24$. I'd be interested to know if you did that or something else. – KConrad Sep 24 at 3:26
• Note: the set $A$ is at OEIS: oeis.org/A014663, while its complement $B$ is oeis.org/A091317. Among the 46 primes below 200, $A$ consists of the 14 primes 7, 23, 31, 47, 71, 73, 79, 89, 103, 127, 151, 167, 191, 199. – YCor Sep 24 at 9:44
• see this answer – René Gy Sep 24 at 15:39
This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.
For each nonzero rational number $$a$$ (take $$a \in \mathbf Z$$ if you wish) and each prime $$\ell$$, let $$S_{a,\ell}$$ be the set of primes $$p$$ not dividing the numerator or denominator of $$a$$ such that $$a \bmod p$$ has multiplicative order divisible by $$\ell$$. When $$a = \pm 1$$, $$S_{a,\ell}$$ is empty except that $$S_{-1,2}$$ is all odd primes. From now on, suppose $$a \not= \pm 1$$.
In Math. Ann. 162 (1965/66), 74–76 (the paper is at https://eudml.org/doc/161322 and on MathSciNet see MR0186653) Hasse treated the case $$\ell \not= 2$$. Let $$e$$ be the largest nonnegative integer such that $$a$$ in $$\mathbf Q$$ is an $$\ell^e$$-th power. (For example, if $$a$$ is squarefree then $$e = 0$$ for every $$\ell$$ not dividing $$a$$.) The density of $$S_{a,\ell}$$ is $$\ell/(\ell^e(\ell^2-1))$$. This is $$\ell/(\ell^2-1)$$ when $$e = 0$$ and $$1/(\ell^2-1)$$ when $$e = 1$$.
In Math. Ann. 166 (1966), 19–23 (the paper is at https://eudml.org/doc/161442 and on MathSciNet see MR0205975) Hasse treated the case $$\ell = 2$$. The general answer in this case is more complicated, as issues involving $$\ell$$-th roots of unity in the ground field (like $$\pm 1$$ in $$\mathbf Q$$ when $$\ell = 2$$) often are. The density of $$S_{a,2}$$ for "typical" $$a$$ is $$1/3$$, such as when $$a \geq 3$$ is squarefree. But $$S_{2,2}$$ has density 17/24, so the set of $$p$$ for which $$2 \bmod p$$ has even order has density $$17/24$$ and the set of $$p$$ for which $$2 \bmod p$$ has odd order has density $$1 - 17/24 = 7/24$$.
For example, there are $$167$$ odd primes up to $$1000$$, $$1228$$ odd primes up to $$10000$$, and $$9591$$ odd primes up to $$100000$$. There are $$117$$ odd primes $$p \leq 1000$$ such that $$2 \bmod p$$ has even order, $$878$$ odd primes $$p \leq 10000$$ such that $$2 \bmod p$$ has even order, and $$6794$$ odd primes $$p \leq 100000$$ such that $$2 \bmod p$$ has even order. The proportion of odd primes up $$1000$$, $$10000$$, and $$100000$$ for which $$2 \bmod p$$ has even order is $$117/167 \approx .700059$$, $$878/1228 \approx .71498$$, and $$6794/9591 \approx .70837$$, while $$17/24 \approx .70833$$.
The math.stackexchange page here treats $$S_{7,2}$$ in some detail and at the end mentions the case of $$S_{2,2}$$.
• Thanks @KConrad for the expression. I was thinking about a combinatorial property of cyclic groups of prime order(called acyclic matching property) and I proved the above mentioned sequence of primes does not hold it. See Proposition 2.3 of core.ac.uk/download/pdf/33123051.pdf Anyway I will like to mention this result in my ongoing research work as a remark, and hence I ask your permission for the same, of course with acknowledgment. – Mohsen Sep 25 at 18:09
• Since the result is due to Hasse, cite his paper when you want to indicate who first showed that the density exists and what its value is. – KConrad Sep 25 at 19:35 |
# Homework Help: Magnitude of deceleration
1. Feb 4, 2005
### runner1738
An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees teh car, the locomotive is 210 m from the crossing and its speed is 10 m/s. If the engineer's reaction time is .22 s, what should be the magnitude of the minmum deceleration to avoid an accident? Answer in units of m/s^2
2. Feb 4, 2005
### christinono
First, use the formula d=vt to find the distance he travels before applying the brakes. Subtract that distance from 210 m. Then, use the formula:
$$d=V_it + \frac{1}{2}at^2$$
to find the acceleration, which would be a negative number.
3. Feb 4, 2005
### runner1738
wait but shouldnt i solve for time 210=10(t) then t is 21 then 21-.22=20.78 then 210=10(20.78)+1/2(a)(20.78)^2 so a = .0101897345
4. Feb 4, 2005
### christinono
No, since his speed is not 10 m/s throughout. First, find the DISTANCE he travels in 0.22s (d=(10 m/s)(0.22s)).
5. Feb 4, 2005
### runner1738
210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?
6. Feb 4, 2005
### christinono
You got the first part (the 207.8m) right. As for the rest, you canot use 22s and the time, since you don't know the time it will take hime to stop. (Sorry, I gave you the wrong formula).
Ok, here's what you know:
initial velocity=10 m/s
final velocity = 0 m/s (since he has to stop)
d = 207.8 m
a=?
Can you find a formula that fits?
7. Feb 4, 2005
### learningphysics
Where are you getting t=22?
I wouldn't use the above equation.
There's a different distance equation you can use and solve for "a" immediately... hint: it doesn't have a t in it.
8. Feb 5, 2005
### runner1738
how is time 17.21 seconds i have 207.8=10t-1/2(.2406)t^2 |
# Type [420], CY of degree 16 via linkage -- lifting to an irreducible 3-fold
We construct via linkage an arithmetically Gorenstein irreducible 3-fold $X = X_{16} \bf{P}^7$, of degree 16, having Betti table of type [420]. For an artinian reduction $A_F$, the quadratic part of the ideal $F^\perp$ is the ideal of six points on a twisted cubic curve. We construct $X_{16}$ as an anticanonical divisor in the fourfold intersection of a cubic scroll and quadric.
The betti table is $\phantom{WWWW} \begin{matrix} &0&1&2&3&4\\ \text{total:}&1&6&10&6&1\\ \text{0:}&1&\text{.}&\text{.}&\text{.}&\text{.}\\ \text{1:}&\text{.}&4&2&\text{.}&\text{.}\\ \text{2:}&\text{.}&2&6&2&\text{.}\\ \text{3:}&\text{.}&\text{.}&2&4&\text{.}\\ \text{4:}&\text{.}&\text{.}&\text{.}&\text{.}&1\\ \end{matrix}$
i1 : kk=QQ; i2 : U=kk[y0,y1,y2,y3,y4,y5,y6,y7]; i3 : P4=ideal(y0,y1,y2);--a P4 o3 : Ideal of U i4 : T3=minors(2,matrix{{y0,y1,y2},{y3,y4,y5}});--a cubic 5-fold scroll with P4 as a ruling o4 : Ideal of U i5 : X2=P4+ideal(random(2,U));-- a quadric 3-fold in T3 o5 : Ideal of U i6 : X18=T3+ideal(random(2,X2),random(3,X2));--a 3-fold of degree 18 in T3 that contains X2 o6 : Ideal of U i7 : X16=X18:X2;--a 3-fold of degree 16 in T3 with betti table of type 420 o7 : Ideal of U
i8 : (dim X16, degree X16) o8 = (4, 16) o8 : Sequence i9 : betti res X16 0 1 2 3 4 o9 = total: 1 6 10 6 1 0: 1 . . . . 1: . 4 2 . . 2: . 2 6 2 . 3: . . 2 4 . 4: . . . . 1 o9 : BettiTally |
×
Back to all chapters
# Rational Functions
A rational function can have a variable like "x" in the numerator AND the denominator. When this happens, there are some special rules and properties to consider.
# Rational Functions - Asymptotes
If the asymptotes of the curve $y=\frac{6x+10}{x+1}$ are $$x=a$$ and $$y=b$$, what is the value of $$a+b$$?
If the graph of $y=\frac{ax-4}{x+b}$ does not intersect either of the lines $$x=4$$ or $$y=-4$$, what is the value of $$ab$$?
If the asymptotes of the curve $y-1=\frac{5x-18}{x-5}$ are $$x=a$$ and $$y=b$$, what is the value of $$a+b$$?
If the asymptotes of the curve $y=-\frac{1}{x-4}+3$ are $$x=a$$ and $$y=b$$, what is the value of $$a+b$$?
If the asymptotes of the curve $y=-\frac{1}{x-4}+11$ are $$x=a$$ and $$y=b$$, what is the value of $$a+b$$?
× |
1
JEE Main 2022 (Online) 25th June Morning Shift
+4
-1
Let f : N $$\to$$ R be a function such that $$f(x + y) = 2f(x)f(y)$$ for natural numbers x and y. If f(1) = 2, then the value of $$\alpha$$ for which
$$\sum\limits_{k = 1}^{10} {f(\alpha + k) = {{512} \over 3}({2^{20}} - 1)}$$
holds, is :
A
2
B
3
C
4
D
6
2
JEE Main 2022 (Online) 25th June Morning Shift
+4
-1
Let f : R $$\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\forall 'x' \in R$$, then g'(63) is equal to ________________.
A
$${1 \over {49}}$$
B
$${3 \over {49}}$$
C
$${43 \over {49}}$$
D
$${91 \over {49}}$$
3
JEE Main 2022 (Online) 25th June Morning Shift
+4
-1
Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}}$$ is equal to:
A
$$-$$15
B
$$-$$60
C
60
D
15
4
JEE Main 2022 (Online) 25th June Morning Shift
+4
-1
Let $$f:R \to R$$ and $$g:R \to R$$ be two functions defined by $$f(x) = {\log _e}({x^2} + 1) - {e^{ - x}} + 1$$ and $$g(x) = {{1 - 2{e^{2x}}} \over {{e^x}}}$$. Then, for which of the following range of $$\alpha$$, the inequality $$f\left( {g\left( {{{{{(\alpha - 1)}^2}} \over 3}} \right)} \right) > f\left( {g\left( {\alpha -{5 \over 3}} \right)} \right)$$ holds ?
A
(2, 3)
B
($$-$$2, $$-$$1)
C
(1, 2)
D
($$-$$1, 1)
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# How does one show a matrix is irreducible and reducible?
How does one show a matrix is irreducible and reducible? Please explain and an example would be great as well.
I know that a matrix is reducible if and only if it can be placed into block upper-triangular form. How do you find block upper-triangular form?
-
Maybe you can remind us --- what does irreducible mean, in the context of matrices? what does reducible mean? – Gerry Myerson Feb 27 '13 at 2:01
@npisinp Let the matrix in question be $A$ and let $B=(b_{ij})$ be the matrix such that $b_{ij}=1$ if $a_{ij}\ne0$, and $b_{ij}=0$ if $a_{ij}=0$. Then $B$ is the adjacency matrix of a directed graph, and $A$ is reducible iff this directed graph has proper strongly connected components. – user1551 Apr 13 at 12:23
The best place to look is this wiki link. To add to the other answer, another equivalent condition is that for every index $[i,j]$, there should be a $m$ such that $(A^m)_{ij}>0$ which is naturally satisfied if the matrix entries are all positive. If it is non-negative, then one needs to check other things. |
# Revision history [back]
### RandomLinearCode on non-prime fields fail ... sometimes.
Consider the following piece of code:
C=codes.RandomLinearCode(4,2,GF(16,'b'))
C.minimum_distance()
About half of the time, it fails with error
TypeError: unable to coerce from a finite field other than the prime subfield
I'd understand if SAGE could only compute the minimum distance for codes over primes fields, what has me confused is the fact that sometimes it can, sometimes it can't. Any clues as to what is going on here?
### RandomLinearCode Computing minimum_distance of a code on non-prime fields fail fails ... sometimes.
Consider the following piece of code:
C=codes.RandomLinearCode(4,2,GF(16,'b'))
C.minimum_distance()
About half of the time, it fails with error
TypeError: unable to coerce from a finite field other than the prime subfield
I'd understand if SAGE could only compute the minimum distance for codes over primes fields, what has me confused is the fact that sometimes it can, sometimes it can't. Any clues as to what is going on here?
3 retagged vdelecroix 7357 ●17 ●81 ●163 http://www.labri.fr/pe...
### Computing minimum_distance of a code on non-prime fields fails ... sometimes.
Consider the following piece of code:
C=codes.RandomLinearCode(4,2,GF(16,'b'))
C.minimum_distance()
About half of the time, it fails with error
TypeError: unable to coerce from a finite field other than the prime subfield
I'd understand if SAGE could only compute the minimum distance for codes over primes fields, what has me confused is the fact that sometimes it can, sometimes it can't. Any clues as to what is going on here?
4 retagged FrédéricC 4968 ●3 ●41 ●107
### Computing minimum_distance of a code on non-prime fields fails ... sometimes.
Consider the following piece of code:
C=codes.RandomLinearCode(4,2,GF(16,'b'))
C.minimum_distance()
About half of the time, it fails with error
TypeError: unable to coerce from a finite field other than the prime subfield
I'd understand if SAGE could only compute the minimum distance for codes over primes fields, what has me confused is the fact that sometimes it can, sometimes it can't. Any clues as to what is going on here? |
# seven Linear regression which have an individual predictor
seven Linear regression which have an individual predictor
Linear regression was a highly strong analytical strategy. We involve some familiarity with regression activities only from studying the news headlines, where upright contours is overlaid towards scatterplots. Linear designs can be used for forecast or to take a look at whether discover a great linear relationship between a mathematical changeable into the horizontal axis additionally the average of your own numerical changeable towards the straight axis.
## eight.1 Fitted a line, residuals, and you will correlation
With regards to linear regression, it is beneficial to think profoundly towards line fitted techniques. Inside point, we determine the form of a great linear model, speak about criteria for just what tends to make a good fit, and you will establish a different sort of figure entitled correlation.
## eight.step one.step 1 Fitted a column so you can research
Shape eight.1 suggests two variables whose relationship will likely be modeled really well with a straight-line. Brand new picture into the range is actually $$y = 5 + x.$$ Think about what the best linear matchmaking mode: we understand the particular value of $$y$$ by understanding the property value $$x.$$ The best linear dating are unrealistic in virtually any absolute process. Such as, whenever we got family money ( $$x$$ ), that it value would offer particular helpful suggestions how much economic help a school can offer a prospective student ( $$y$$ ). However, the newest forecast was away from perfect, given that additional factors subscribe to resource beyond a family’s earnings.
Shape eight.1: Needs out of a dozen independent customers were in addition put with an investing organization to invest in Address Company inventory (ticker TGT, ), while the total price of shares have been said. Due to the fact costs is determined having fun with a linear formula, the linear complement is best.
Linear regression is the mathematical way for fitted a column to help you studies where in actuality the relationship ranging from several details, $$x$$ and you may $$y,$$ would be modeled by the a straight line with a few mistake:
The costs $$b_0$$ and you may $$b_1$$ represent new model’s intercept and you may slope, respectively, in addition to error are illustrated by $$e$$ . This type of values are determined in accordance with the investigation, i.elizabeth., he or she is try analytics. If the observed data is a random decide to try of a goal populace that we are interested in and come mejores aplicaciones de citas pansexual up with inferences from the, such opinions are considered to get area rates towards people parameters $$\beta_0$$ and you may $$\beta_1$$ . We’re going to explore making inferences in the parameters out of an effective linear model predicated on try statistics during the Chapter twenty four.
When we have fun with $$x$$ to expect $$y,$$ we always telephone call $$x$$ this new predictor variable therefore name $$y$$ the outcome. We in addition to commonly get rid of the newest $$e$$ identity when recording brand new design as all of our main focus is have a tendency to into forecast of your own average benefit.
It is unusual for all of your investigation to fall well towards the a straight-line. Instead, it’s more widespread having study to look since an affect out-of circumstances, like those instances found within the Shape 7.2. When you look at the per situation, the data slide as much as a straight line, in the event none of your own findings slip precisely at stake. The original patch reveals a fairly solid downwards linear development, where in fact the remaining variability on the research inside the range was small in line with the potency of the connection between $$x$$ and you may $$y.$$ The next area shows an upward development you to definitely, when you find yourself evident, is not as strong once the first. The last area reveals a very weakened downward development regarding data, so limited we are able to hardly notice it. For the each of these instances, we will see some uncertainty out-of our prices of your design parameters, $$\beta_0$$ and you can $$\beta_1.$$ Such as, we possibly may ask yourself, is i circulate the newest line up or down a small, otherwise should i tip it nearly? While we move ahead contained in this part, we are going to understand conditions getting range-fitting, and we will also discover this new uncertainty of the prices away from design parameters. |
# Multiplicative group of integers modulo n definition issues
It is easy to verify that the set $(\mathbb{Z}/n\mathbb{Z})^\times$ is closed under multiplication in the sense that $a, b ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $ab ∈ (\mathbb{Z}/n\mathbb{Z})^\times$, and is closed under inverses in the sense that $a ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $a^{-1} ∈ (\mathbb{Z}/n\mathbb{Z})^\times$.
The question is the following:
Firstly, are $a$ and $b$ referring to each equivalence class of integers modulo $n$?
Secondly, by $a^{-1}$, what is this referring to? If $a$ is the equivalence class, I cannot see (or I am not sure) how I can make inverse set.
-
To the first question, just quoting your intro, "$a,b\in(\mathbb{Z}/n\mathbb{Z})^{\times}$". So if $\mathbb{Z}/n\mathbb{Z}$ is a set of equivalence classes, then yes, $a$ and $b$ are equivalence classes.
To the second question, $a^{-1}$ is an equivalence class such that $aa^{-1}$ is the equivalence class of $1$. In general, there is no quick formula to find an element of $a^{-1}$ given $a$. Instead, since a representative $A$ of $a$ and $n$ are relatively prime, the Euclidean algorithm provides solutions to the equation $$AB + nt = 1$$ Then mod $n$, $AB\equiv 1$. So the Euclidean algorithm will lead you to a representative of $a^{-1}$.
Now, to back-peddle a little bit, actually there is a rather simple formula for a representative of $a^{-1}$, given a representative $A$ of $a$. Take $B=A^{\varphi(n)-1}$, where $\varphi$ is Euler's totient function. Then $AB=A^{\varphi(n)}\equiv1\mod{n}$. The problem with this "simple" formula is that it's not computationally efficient. For example, if $n=97$, the Euclidean algorithm quickly tells us that $2^{-1}=49$. But this formula would give us $2^{96}$, which is tiresome to compute even if we reduce at every multiplication.
-
The notation $a \in (\mathbb{Z} / n \mathbb{Z})^\times$ means the variable $a$ is being used to denote an element of $(\mathbb{Z} / n \mathbb{Z})^\times$.
If you are representing elements of $(\mathbb{Z} / n \mathbb{Z})^\times$ as equivalence classes, then $a$, being an element of $(\mathbb{Z} / n \mathbb{Z})^\times$, has a representation as an equivalence class.
The structure $(\mathbb{Z} / n \mathbb{Z})^\times$ is an abelian group. It has a group operation, an inverse, and a neutral element. As we are writing this group with multiplicative notation, we adopt the default notation ${}^{-1}$ for the inverse operation of the group.
As an aside, I think focusing on an object "being" an equivalence class really obscures the simplicity of what's going on. Equivalence classes are frequently just a technical set-theoretic trick.
One rather good notation for elements of the ring $\mathbb{Z} / n \mathbb{Z}$ is that elements are represented by integers. Not just integers in the range $\{ 0, 1, \cdots, n-1\}$, but any integer. The same element has many notations; for example, '3' and '10' are two separate notations for the same element of $\mathbb{Z} / 7 \mathbb{Z}$.
It may help to add a decoration to help keep track of when you are using '3' to denote an element of $\mathbb{Z}$ and when you are using it to denote an element of $\mathbb{Z} / 7 \mathbb{Z}$: common choices are $\overline{3}$ and $[3]$, or sometimes $[3]_7$.
If you are representing the elements of $\mathbb{Z} / n \mathbb{Z}$ as equivalence classes, then the element $[3]$ is notation for the element represented by the equivalence class of 3. Despite that fact, you usually shouldn't be thinking "$[3]$ is an equivalence class": you should be thinking "$[3]$ is the element of $\mathbb{Z} / n \mathbb{Z}$ that comes from the integer 3".
-
Well $2\cdot 3\equiv 1\; \text{mod}\ 5$, so $2$ and $3$ are multiplicative inverses $\text{ mod } 5$.
How to find the inverse of a number modulo a prime number was the topic of one of my previous answers. Modulo a composite number, inverses don't always exist.
See Calculating the Modular Multiplicative Inverse without all those strange looking symbols for the way to find the inverse of $322$ mod $701$. It turns out to be $455$
-
First: yes, $a$ and $b$ represent the equivalence classes modulo $n$.
Second: Well, $a^{-1}$ is not the inverse as a rational number. Rather, it is the unique equivalence class modulo $n$ (if it exists) such that $aa^{-1}$ is congruent to 1 modulo $n$. And for existence, this inverse exists if and only if $a$ is coprime to $n$.
-
Firstly, yes.
Secondly, by $a^{-1}$ we mean the unique equivalence class such that multiplication of (the equivalence class) $a$ with $a^{-1}$ gives (the equivalence class of) $1.$
For example in $(\mathbb Z/4\mathbb Z)^\times=\{1,3\}$ we have $1\cdot 1\equiv 1$ so $1^{-1}=1$ and $3\cdot 3=9\equiv 1$ so $3^{-1}=3.$
-
It is also standard to identify $(\mathbf Z/n \mathbf Z)$, which is typically defined algebraically as a set of equivalence classes, with the integers $\{0, \dots, n-1 \}$. One thing to note is that $(\mathbf Z/n \mathbf Z)$ is essentially finitary, so the representation as integers is sometimes useful (but less convenient algebraically)
- |
# IntMath Newsletter: New interactive graphs, ethical math help, resources
By Murray Bourne, 30 Nov 2011
30 Nov 2011
1. Explore basic graph concepts and understand them better
2. My dilemma - ethical math help
3. Resources
5. Friday math movies
6. Final thought - mistakes and learning
## 1. Explore basic graph concepts and understand them better
Here are some interactive graphs that help you better understand distance, slope, parallel lines and perpendicular lines. The basic concepts in plane geometry (like distance and slope) are vital for understanding later math topics (like calculus).
I recently added new pages in the Plane Analytic Geometry chapter in IntMath. Each one now has an interactive graph that helps you to understand how the formula works.
Check them out:
Distance Formula
Gradient of a Line, and Inclination
Parallel Lines
Perpendicular Lines
I'm hoping the interactive JSXGraphs are useful for both students and teachers. Please tell me what you think of them. Any improvements you can suggest which would make them more useful for you?
[Please give your suggestions in the comments section at the end of this post: Explore basic graph concepts and understand them better]
## 2. My dilemma - ethical math help
Is there a difference between paying someone to do your math homework and getting a computer to do it? Are they both unethical?
I'd be interested in your thoughts on this. See:
## 3. Resources
Manga High is a "new games-based maths teaching resource". For example, Pyramid Panic requires you to "use your geometry skills to build a path across the voids, and escape from the pyramid, staying ahead of the mongrel demon Ammit. Battle skull bats and skeleton thieves with your ankh cross." Topics covered in this game include Tigonometry, Pythagoras, Areas, Perimeters and Circumference. There are free "lite" versions, and paid versions allow social interaction with students from other schools.
GeoGebra examples: Analemma shares some interactive GeoGebra applets that illustrate mathematical concepts. Some of the topics include Bayes' Theorem, Circle as a limit of polygons, Circles of Apollonius, Domain and Range, Ellipse in a Box, and Eratosthenes measures Earth. Being java-based applets, they take a long time to load, but it's worth the wait.
Here's an interesting idea. Twylah gives you a summary of Twitter tweets in a newspaper style. Here are recent IntMath tweets in this form:
IntMath Tweets on Twylah
## 5. Friday math movies
Suitable for: Everyone, but especially teachers.
(a) Math comes to life Here is an excellent way to animate math concepts. Friday math movie: Math comes to life
Suitable for: Everyone, especially if you are on Facebook (and who isn't?).
(b) Facebook for Math Nerds This is first in a series of funny videos for math nerds. You'll enjoy it even if you are normal! Friday math movie: Facebook for Math Nerds
## 6. Final thought - mistakes and learning
This is from author John Powell, who could have been talking about learning math!
"The only real mistake is the one from which we learn nothing." [John Powell]
Until next time, enjoy whatever you learn.
### 2 Comments on “IntMath Newsletter: New interactive graphs, ethical math help, resources”
Thank you sir for this excellent and informative articles.
2. Isidro Canales Barbosa says:
Thanks for the information. Its great website of math.
Isidro Canales Barbosa
Mexico
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# What is the need of complex functions in wave analysis?
It is commonly known that waves can be express in terms of sine or cosine function.
But when I study further, I seen that for analyising the waves, it is common to use complex functions in the form
$$y=y_{_0}e^{i(kx-\omega{t})}$$
where $y_{_0}$ is the amplitude, $k$ is the wave number, $\omega$ is the angular velocity and $x$ & $t$ are position an time respectively. Ofcourse, I know that the function $e^{ix}$ can be written in the form $\cos{x}+i\sin{x}$ and so it is a periodic function with period $2\pi$ but my question is for what purpose we define it in terms of complex numbers? It seem to be more convenient to use real functions for real variables such as amplitude, electric and magnetic field of an electro magnetic wave, and also in quantum mechanics. What actually this interpretation means or what is the advantage of such functions?
The use of complex numbers is just a mathematical convenience. It makes calculation of derivatives especially easy, it has nice properties when you do Fourier transforms, etc. You're correct that you can do it all using real numbers, so that's not wrong. It's just - in most people's view - more cumbersome.
EDIT In light of the back and forth in the comments, let me provide more detail.
First, starting with classical mechanics: Let $f$ be a (potentially) complex solution to the wave equation. The physically relevant (i.e. measurable) quantity here is the amplitude as a function of space and time. Any complex function can be rewritten in terms of two real-valued functions $g$ and $h$ such that $$f = g + ih$$ The amplitude of $f$ is $\| f \| = (g^2 + h^2)^{(1/2)}$. We basically have two free functions here where we only need one to meet this constraint, so we're free to choose $h=0$, which means that $f$ is actually real-valued. You could choose some other values for $g$ and $h$ that have the same amplitude, but you don't need the complex part. (Note that I'm not dealing with plane wave solutions here, although you could build up your solution from them. I'm dealing with general solutions to the wave equation.)
For quantum mechanics, we have the Schroedinger equation: $$i\hbar \partial_t \Psi = -\frac{\hbar^2}{2m} \nabla^2 \Psi$$ (where I set $V=0$ because it's not going to figure in the rest of the point). This is typically written with complex numbers, as shown above, but this is again a short-hand only. We could instead write the solution in terms of two real-valued functions: $$\Psi = f + ig$$ and then, doing a little simplification, get two, coupled, real-valued PDEs: $$\hbar \partial_t f = -\frac{\hbar^2}{2m} \nabla^2 g$$ $$\hbar \partial_t g = +\frac{\hbar^2}{2m} \nabla^2 f$$ So, again, we can avoid complex numbers in the formulation. The price here is that we now have coupled PDEs for real functions instead of a single PDE over complex values. It turns out for practical reasons, that working with the single, complex-valued formulation is easier.
• That's just not why complex functions are introduced. They emerge as solutions of the equations the functions are subject to and you cannot throw them away (unless specific conditions hold). – gented Sep 25 '15 at 17:34
• @GennaroTedesco The boundary conditions and context in a real physics problem are going to require a real solution, not the most general solution. You could reformulate everything in terms of sine and cosine transforms instead of the complex Fourier transform. (In fact, I believe, that was Fourier's original formulation for what it's worth.) – Brick Sep 25 '15 at 17:50
• Solution in the field of the real numbers $\mathbb{R}$ does not mean that the solution is real, as for a real thing. I see no difference in the meaning between real and complex (unless you calculate modulus square, i. e. positive numbers). And no, you cannot, in general, reformulate everything in terms of sine and cosine. – gented Sep 25 '15 at 19:02
• @GennaroTedesco Of course you can reformulate everything in terms of sine and cosine. Just use $e^{ix} = \cos(x) + i \sin(x)$ and then write two coupled real valued equations: one for the real part and one for the imaginary part. – DanielSank Sep 25 '15 at 19:03
• I get the different uses of the word "real" here. At the same time, there is no way to take a physical measurement that observes a complex number. The complex numbers are useful mathematical solutions, but there's no physical theory that will accept a complex answer as physically meaningful for something you measure in a lab. If you solve in terms of complex-valued solutions, you always end up taking a combination of such solutions that results in a real-valued answer for observables. That is part of the theory and constrains the solution. See also: physics.stackexchange.com/q/11396/2451 – Brick Sep 25 '15 at 19:20
A function $f(x,t)$ is said to obey a wave equation if it holds that $$\Box f(x,t) = 0.$$ The most general solution of the above equation can be usually expressed in Fourier transform as $$f(x,t) = \int \textrm{d}k\,\textrm{d}\omega\,c(k,\omega)\,\textrm{e}^{-i(kx-\omega t)}$$ plus some boundary conditions. Except in special cases, the solution is usually a complex function.
• This answer begs the question. This is one way to write a general solution to the wave equation, but there's no a-priori reason to believe that the most general solutions have physical meaning. – Brick Sep 25 '15 at 17:47
• Likewise, there is no a priori reason to believe that the real solution has more physical meaning than a complex solution (real numbers does not mean real as in the language). – gented Sep 25 '15 at 19:04
• If you can find an accepted physical theory that says you can measure a complex number [e.g. (1+i) Volts], then I will yield. I'm sure that number is not on my voltmeter. Maybe some other device? I can measure amplitude and phase, but those are both real. The fact (on which we agree, I think) that it's mathematically convenient for calculation to combine them into a complex number does not change the fact that the measurable quantities are real. Since you know you need real numbers at the end, you could solve everything using real numbers from the start. – Brick Sep 25 '15 at 19:24
• I did not say that you measure complex numbers. What you measure are amplitudes, namely the modules square of the solution, which is the same whether the function is complex or real. If you exclude complex solution, you exclude a big variety of physically acceptable solutions for the observables to be measured (see quantum mechanics, for example). – gented Sep 25 '15 at 19:32
• Your last comment again begs the question: If you're going to take combinations (such as complex norms) that give you real numbers at the end, why not use real numbers all the way through? – Brick Sep 25 '15 at 19:45 |
Models of Scherk-Schwarz Symmetry Breaking in 5D: Classification and Calculability
Description
The form of the most general orbifold breaking of gauge, global and supersymmetries with a single extra dimension is given. In certain theories the Higgs boson mass is ultraviolet finite due to an unbroken local supersymmetry, which is explicitly exhibited. We construct: a 1 parameter SU(3) \times SU(2) \times U(1) theory with 1 bulk Higgs hypermultiplet, a 2 parameter SU(3) \times SU(2) \times U(1) theory with 2 bulk Higgs hypermultiplets, and a 2 parameter SU(5) \to SU(3) \times SU(2) \times U(1) theory with 2 bulk Higgs hypermultiplets, and demonstrate that these theories are unique. We compute the Higgs mass and ... continued below
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What
Description
The form of the most general orbifold breaking of gauge, global and supersymmetries with a single extra dimension is given. In certain theories the Higgs boson mass is ultraviolet finite due to an unbroken local supersymmetry, which is explicitly exhibited. We construct: a 1 parameter SU(3) \times SU(2) \times U(1) theory with 1 bulk Higgs hypermultiplet, a 2 parameter SU(3) \times SU(2) \times U(1) theory with 2 bulk Higgs hypermultiplets, and a 2 parameter SU(5) \to SU(3) \times SU(2) \times U(1) theory with 2 bulk Higgs hypermultiplets, and demonstrate that these theories are unique. We compute the Higgs mass and compactification scale in the SU(3) \times SU(2) \times U(1) theory with 1 bulk Higgs hypermultiplet.
Source
• Journal Name: Nuclear Physics B; Related Information: Journal Publication Date: 13 January 2002
Identifier
Unique identifying numbers for this article in the Digital Library or other systems.
• Report No.: LBNL-48257
• Grant Number: DE-AC02-05CH11231
• Office of Scientific & Technical Information Report Number: 965798
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Cosmology and Eschatology in Jewish and Christian by Yarbro Collins
By Yarbro Collins
This quantity offers with Jewish and Christian apocalyptic texts and pursuits from the second one century BCE in the course of the fourth century CE. It makes a speciality of significant issues, cosmology and eschatology; that is, perspectives of constitution of the universe together with its non secular functionality and interpretations of background and the longer term. The precise historic and literary research of those issues are brought via an essay at the cultural hole among the unique contexts of those texts and people of readers this day and the way that hole could be bridged. The ebook bargains with the interrelations among post-biblical Judaism and early Christianity. The proper Jewish texts and heritage are mentioned completely of their personal correct. The Christian fabric is approached in a manner which indicates either its continuity with Jewish culture and its specialty.
Read or Download Cosmology and Eschatology in Jewish and Christian Apocalypticism (Supplements to the Journal for the Study of Judaism, V. 50) PDF
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Additional info for Cosmology and Eschatology in Jewish and Christian Apocalypticism (Supplements to the Journal for the Study of Judaism, V. 50)
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# Finding inverse for a homeomorphism on the sphere (compactification)
1. Oct 23, 2012
### huberscher
hi there
I'd like to show that the sphere
$$\mathbb{S}^n := \{ x \in \mathbb{R}^{n+1} : |x|=1 \}$$ is the one-point-compactification of $$\mathbb{R}^n$$ (*)
After a lot of trying I got this function:
$$f: \mathbb{S}^n \setminus \{(0,...,0,1)\} \rightarrow \mathbb{R}^n$$
$$(x_1,...,x_{n+1}) \mapsto (\frac{x_1}{1-x_{n+1}},...,\frac{x_n}{1-x_{n+1}})$$
This is a continuous function, its image is the whole $\mathbb{R}^n$. If I find its inverse $f^{-1}$ now and show that this one is continuous as well with $image(f^{-1})=\mathbb{S}^n \setminus \{(0,...,0,1)\}$ I have shown (*).
But I don't find the inverse. $y_i=\frac{x_i}{1-x_{n+1}}$ so $x_i=(1-x_{n+1})*y_1$ but there is no $x_{n+1}$ here y is a n-dimensional vector...?
How can I find the inverse of f?
Regards
2. Oct 23, 2012
### mathwonk
first try solving for xn+1 in terms of x1,..xn.
3. Oct 23, 2012
### huberscher
I don't get it:
If I consider the norms of those vectors I have with $x=(x_1,...,x_n)$
$$\vec{x}*\frac{1}{1-x_{n+1}} =\vec{y}$$ so
$$x_{n+1}=\frac{\|y\|-\|x\|}{\|y\|}$$ but then I still have this y. What the trick here to get those $x_{n+1}$?
4. Oct 23, 2012
### mathwonk
can you use the equation for the sphere to solve for xn+1 in terms on x1,..,xn?
5. Oct 24, 2012
### Bacle2
6. Oct 30, 2012
### lavinia
Steroegraphic projection is conformal so its inverse is also conformal. Is a congormal map continuous?
Is a conformal mapping an open mapping? That is, does it map open sets onto open sets?
If so, does that mean that its inverse is continuous?
This is to show you that you do not need to actually write down the inverse to check whether it is continuous. If you want write the inverse down think of solving for the intersection point on the sphere of straight line from a point in the plane to the north pole.
7. Nov 2, 2012
### Bacle2
Still, you cannot treat a map into the Riemann sphere as you would a map from the
complex plane to itself. Even continuity is tricky, since the sphere is a manifold.
Or you can treat it as a map from C to C\/{oo}. And then the inverse function
theorem would tell you when the map is a homeomorphims/diffeomorphism.
8. Nov 3, 2012
### lavinia
The sphere minus a point maps confomally onto the plane. Add the north pol and you compactify it.
9. Nov 3, 2012
### Bacle2
Yes, I know, this is what I said in post #7 . But conformality is usually defined for
maps from C to C , where there is a clear meaning of preservation of angles. If you
want to talk about preservation of angles for curves lying on the sphere, this is
a whole different story: do you use the tangent space of the sphere? How about
showing a map into the sphere into C is analytic (since conformal is equivalent to analytic
with non-zero derivative)? I'm not saying it is wrong; just that it needs an argument. What happens
with the image vectors near the north-pole? I don't see it; I may be wrong, but it does not seem automatic.
To talk about analytic in the sphere, you need to bring up charts.
Last edited: Nov 3, 2012
10. Nov 3, 2012
### lavinia
The sphere as a subset of Euclidean space inherits the Riemannian metric directly. You can do charts if you like but you need only to look at vectors in Euclidean space.
But preserving infinitesimal angles in the plane is no different than on the sphere. Its just that the metrics are different.
11. Nov 3, 2012
### Bacle2
Well, I think you would have to show/argue that the stereographic projection T as a map
from C to the sphere preserves the metric/inner-product in the sense that
<a,b>_C= <T(a),T(b)>_S^1 . I never saw this argument in your post.
12. Nov 3, 2012
### Bacle2
Moreover: if the stereo T preserved the metric: wouldn't this imply that C and S^1
are isometric?
13. Nov 3, 2012
### lavinia
the metric isn't preserved but the map is conformal, angles are preserved infinitesimally. The same thing is true for analytic maps of the complex plane.
I am beginning to think that I misunderstood your point. i am sorry if I did.
14. Nov 4, 2012
### Bacle2
No problem; you may be right and we may be talking about different things. I think it may take too long to untangle ; I think it's run its course. |
4 deleted 133 characters in body
If you avoid trivial things**, then your
The slowest growing zipper depends will depend on the size of $p_{n+1}-p_n$ where $p_n$ is the $n^{th}$ prime number. There are many results regarding the size of the largest prime gap.
Unconditional: The work of Baker, Harman and Pintz shows that $$p_{n+1}-p_n \ll p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be $f(n)=Cn^{40/19}$ for some constant $C$. The $\frac{40}{19}$ appears in the exponent because $\frac{40}{19}=\frac{1}{1-0.525}$.
Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and we may take $f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right),$$ would allows us to take $f(n)=Cn(\log n)^2$ for some constant $C$.
Also see this Wikipedia article on prime gaps.
Remark: Note that finding a prime zipper which grows slower than $f(n)=Cn^{40/19}$ would imply better bounds on the largest prime gap, so your question is equivalent to asking what is the largest prime gap.
**By your definition, if $p_n$ is the $n^{th}$ prime number, then * Avoid pointless functions such as $f(n)=p_n+1$ will be the slowest growing prime zipper, but that is not a meaningful statement.f(n)=p_n+1$. 3 added 27 characters in body; added 39 characters in body; added 212 characters in body If you avoid trivial things**, then your zipper depends on how large the size of$p_{n+1}-p_n$can be where$p_n$is the$n^{th}$prime number. This is a well known problem, and there There are many results regarding the size of the largest prime gap. Unconditional: The work of Baker, Harman and Pintz shows that $$p_{n+1}-p_n \ll p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be$f(n)=Cn^{40/19}$for some constant$C$. We get The$\frac{40}{19}$appears in the exponent because$\frac{40}{19}=\frac{1}{1-0.525}$. Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and we may take$f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right),$$ would allows us to take$f(n)=Cn(\log n)^2$for some constant$C$. Also see this Wikipedia article on prime gaps. Remark: Note that finding a prime zipper which grows slower than$f(n)=Cn^{40/19}$would imply better bounds on the largest prime gap, so your question is equivalent to asking what is the largest prime gap. **By your definition, if$p_n$is the$n^{th}$prime number, then$f(n)=p_n+1$will be dthe the slowest growing prime zipper, but that is not a meaningful statement. 2 added 311 characters in body; added 19 characters in body; added 209 characters in body If you avoid trivial things**, then your zipper depends on how large$p_{n+1}-p_n$can be where$p_n$is the$n^{th}$prime number. This is a well known problem, and there are many results regarding the size of the largest prime gap. Unconditionally Unconditional: The work of Baker, it has been proven Harman and Pintz shows that $$p_{n+1}-p_n \ll C p_n^{0.525}$$ for some computable constant. This means that your zipper function may be taken to be$f(n)=Cn^{40/19}$for some constant$C$. We get$\frac{40}{19}$in the exponent because$\frac{40}{19}=\frac{1}{1-0.525}$. Conditional: If we assume the Riemann Hypothesis, then we have $$p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and it is conjectured we may take$f(n)=n^2 \log n$. Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right).$$ See p_n)^2\right),$$would allows us to take$f(n)=Cn(\log n)^2$for some constant$C$. Also see this Wikipedia Articlearticle on prime gaps. **By your definition, if$p_n$is the$n^{th}$prime number, then$f(n)=p_n+1\$ will be dthe slowest growing prime zipper.
1 |
# Revision history [back]
So my assumption is the IK solution given by ur_driver is not precise.
ur_driver does not contain any functionality related to IK or FK.
FK is performed by robot_state_publisher based on the urdf.
IK is performed by whatever kinematics solver you have configured. That could be ur_kinematics, trac_ik, KDL or something else.
The urdf is most certainly not completely correct for your robot. That would be the one source of positional error. That is being addressed in ros-industrial/universal_robot#414.
Secondly: both URScript, MoveIt and the driver use different tolerances for when a motion is completed. It could well be that this also contributes here.
Finally, the differences you show appear to be in the micrometers range (ROS uses metres for distances):
X Y Z
moveit: 0.33561 -0.13135 0.1815
movej: 0.33667 -0.13125 0.18217
----------------------------------
-0.00106 0.0001 -0.00067
The repeatability of a UR5 is specced as +-0.1mm or 0.001m. Taking that into account the differences don't seem to be so egregious.
So my assumption is the IK solution given by ur_driver is not precise.
ur_driver does not contain any functionality related to IK or FK.
FK is performed by robot_state_publisher based on the urdf.
IK is performed by whatever kinematics solver you have configured. That could be ur_kinematics, trac_ik, KDL or something else.
The urdf is most certainly not completely correct for your robot. That would be the one source of positional error. That is being addressed in ros-industrial/universal_robot#414.
Secondly: both URScript, MoveIt and the driver use different tolerances for when a motion is completed. It could well be that this also contributes here.
Finally, the differences you show appear to be in the micrometers range (ROS uses metres for distances):
X Y Z
moveit: 0.33561 -0.13135 0.1815
movej: 0.33667 -0.13125 0.18217
----------------------------------
-0.00106 0.0001 -0.00067
The repeatability of a UR5 is specced as +-0.1mm or 0.001m. (datasheet). Taking that into account the differences don't seem to be so egregious.
So my assumption is the IK solution given by ur_driver is not precise.
ur_driver does not contain any functionality related to IK or FK.
FK is performed by robot_state_publisher based on the urdf.urdf, or the integrated version of that in MoveIt.
IK is performed by whatever kinematics solver you have configured. That could be ur_kinematics, trac_ik, KDL or something else.
The urdf is most certainly not completely correct for your robot. That would be the one source of positional error. That is being addressed in ros-industrial/universal_robot#414.
Secondly: both URScript, MoveIt and the driver use different tolerances for when a motion is completed. It could well be that this also contributes here.
Finally, the differences you show appear to be in the micrometers range (ROS uses metres for distances):
X Y Z
moveit: 0.33561 -0.13135 0.1815
movej: 0.33667 -0.13125 0.18217
----------------------------------
-0.00106 0.0001 -0.00067
The repeatability of a UR5 is specced as +-0.1mm or 0.001m (datasheet). Taking that into account the differences don't seem to be so egregious.
So my assumption is the IK solution given by ur_driver is not precise.
ur_driver does not contain any functionality related to IK or FK.
FK is performed by robot_state_publisher based on the urdf, or the integrated version of that in MoveIt.
IK is performed by whatever kinematics solver you have configured. That could be ur_kinematics, trac_ik, KDL or something else.
The urdf is most certainly not completely correct for your robot. That would be the one source of positional error. That is being addressed in ros-industrial/universal_robot#414.
Secondly: both URScript, MoveIt and the driver use different tolerances for when a motion is completed. It could well be that this also contributes here.
Finally, the differences you show appear to be in the micrometers range (ROS uses metres for distances):
X Y Z
moveit: 0.33561 -0.13135 0.1815
movej: 0.33667 -0.13125 0.18217
----------------------------------
-0.00106 0.0001 -0.00067
The repeatability of a UR5 is specced as +-0.1mm or 0.001m (datasheet). Taking that into account the differences don't seem to be so egregious. |
# Find the $m$ such $a_{n+1}=a^5_{n}+487$ [closed]
Let $\{a_{n}\}$ be a sequence of positive integers, and suppose $a_{0}=m$. Further, $\{a_{n}\}$ satisfies $$a_{n+1}=a^5_{n}+487.$$ Find $m$ so that this sequence consists of square numbers for as long as possible.
• What have you tried? Does $m$ have to be square? There will be very few choices where both $a_1, a_2$ are square. – Ross Millikan Jun 15 '16 at 5:15
• Where does this question come from? I don't see a case where you get even one square out of this except for $a_0=9$, $a_1=59536=244^2$. – mjqxxxx Jun 15 '16 at 5:16
• @mjqxxxx,some years ago open problem – function sug Jun 15 '16 at 5:30
• @functionsug For a far from well-known open problem, I think you should put up rather more detail. Any references? What is the current record-holder, eg has anyone found a sequence better than the $3^2,244^2$ example? Has anyone found any other examples at all. A quick computer search up to $100000^2$ failed to find any. – almagest Jun 15 '16 at 5:48
You have the recurrence $$a_{n+1} = a_n^5 + 487$$ And we want that atleast $a_0$ and $a_1$ are perfect squares. Let $a_1=n^2$ and $a_0 = p^2$. Then, $$a_1 = a_0^{5} + 487$$ and $$n^2 = p^{10} + 487$$ Putting $p^5:=t$, $$n^2 = t^2 +487$$ $$\implies n^2-t^2 = 487$$ Note that $487$ is prime. So, $$\implies (n-t)(n+t) = 1\times 487$$ $$\implies n-t=1, \ \ n+t = 487$$ $$\implies n = 244, \ \ t = 243 = 3^5 \implies p = 3$$ Thus, we have $a_1 =n^2= 244^2$ and $a_0 =p^2= 3^2=9$. On computing $a_2$, we find that it is not a perfect square. ($a_2=747994256939818786226663$)
• Or you can show $a_{2 } \equiv 3$ mod 4. So it is not a square. No need to compute. – N.S.JOHN Aug 12 '16 at 17:38 |
Feeds:
Posts
## Combinatorics
Yesterday in Pre-Calculus class we were discussing combinatorics and came across a series of questions on how many ways can a best 2 out of 3 sets tennis match be won and how many ways can a best 3 out of 5 tennis match be won.
For a 2 out of 3 scenario there are six possibilities (AA, ABA, BAA, BB, BAB, ABB) and for the 3 out of 5 scenario there are 20 possibilities (AAA, BAAA, ABAA, AABA, BBAAA, BABAA, BAABA, ABBAA, ABABA, AABBA and ten more in which B wins).
The question then came up of how to generalize these results. The way I approached this was to consider how many ways one of the players could win and then double the answer to include the scenarios in which the other won.
One of the issues in this type of problem is that once the winner has won the required number of games the match ends, so that a best 2 out of 3 match could never end AAB, for instance. Another issue is that each of the possibilities of winning in a different number of sets should be considered separately. That is, winning in the minimum number of sets, then winning after having lost one set and so forth.
Taking all of these issues into account for a competition in which the winner must win $\frac{n+1}{2}$ out of $n$ leads to a general formula of $2*\displaystyle \sum_{k=0}^{\frac{n-1}{2}}{_\frac{n+2k-1}{2}}C_{k}$.
Or – if the winner must win $g$ out of $2g-1$ then the formula would be $2*\displaystyle \sum_{k=0}^{g-1}{_{g+k-1}}C_{k}$. |
# Tag Info
### Why the pressure of atmosphere doesn't crush you when you e.g. walk outside?
Pressure acts in all directions. it does not care about direction. pressure is a property of the fluid itself. but the way you CALCULATE it is to find the 'weight of the air above'. this does not mean ...
• 111
### If you decreased the mass of a planet, but kept the mass of the atmosphere the same, would the air density decrease?
In an isothermal atmosphere, the density is given approximately by $$\rho = \rho_0\exp(-\mu gh/k_BT)\ ,$$ where $\mu$ is the mean mass of a particle, $h$ is the height above ground level and $g$ is ...
• 113k
1 vote
### If you decreased the mass of a planet, but kept the mass of the atmosphere the same, would the air density decrease?
Yes the air density would decrease. The density is proportinal to $e^{-g*h/T}$ where g is the local force of gravity h is the height and T is the tenperature. Lower gravity gives lower pressure and ...
• 1,713
Accepted
### Would a straw work in a completely sealed container?
No. If the air can't get in the container in any way other than the straw than you won't get any water. A tiny amount of water may go up the straw a bit but it wouldn't be a reasonable amount to drink....
### Does dusk really remain for a shorter period of time at the equator?
I get the sense that the OP might have been wondering about what seems like a riddle: if people near the equator are moving faster (due to the earth's rotational speed), shouldn't they "go ...
• 101
In the range of pressures we have in the troposphere the gas law holds perfectly, but we have to interprete it in a proper way. Lest's solve for P the equation of the gas law: $$P=\frac{n} {V}RT$$ ... |
# NEET Physics Units and Measurement Questions Solved
The dimensions of $\frac{a}{b}$ in the equation $P=\frac{a-{t}^{2}}{bx}$, where P is pressure, x is distance and t is time, are
(1) MT–2
(2) M2LT–3
(3) ML3T–1
(4) LT–3
Concept Videos :-
#1 | Basic Concepts & Examples
#2 | Dimensional Analysis : Remaining
Concept Questions :-
Dimensions
Explanation is a part of a Paid Course. To view Explanation Please buy the course.
Difficulty Level: |
# Do I need more testing gear?
#### DraxDomax
Joined Apr 5, 2019
50
Hi, I am just starting. Right now, I am going through the books available here but I am also fooling around with the breadboard and stuff.
I have a cheep multi-meter that helps me a lot to confirm theoretical situations when I am learning about voltage, resistance and current.
However, I do read about "osciloscopes"? I think these are devices that measure wave form as a part of studying AC electronics.
My ultimate interests are battery and power applications but I think some playing with audio and radio is inevitable! (and important
to my understanding of electronics, of course).
Do I need an oscilloscope or other gear?
#### AlbertHall
Joined Jun 4, 2014
10,908
It all depends how deep you go into stuff.
For a cheap 'scope you can use a PC and free software. This also gets you a spectrum analyser.
For instance: https://www.zeitnitz.eu/scope_en
#### DraxDomax
Joined Apr 5, 2019
50
It all depends how deep you go into stuff.
For a cheap 'scope you can use a PC and free software. This also gets you a spectrum analyser.
For instance: https://www.zeitnitz.eu/scope_en
That's a pretty nice case of repuroposing what you got instead of buying more stuff.
However, I am not sure I can pull this off on my laptop:
1. I only have 1x3.5mm jack, which I think is output only.
2. I am scared to fry my laptop. It's an on-board soundcard.
BTW, will this read different wave shapes? I recall seeing waves that look "rectengular" and other waves that look "triangular". I thought all the sound card cares about is frequency?
#### Wolframore
Joined Jan 21, 2019
2,234
YES! Don't ask my wife... more... more better...
You can get some toy oscilliscopes online for about $20 as a kit... would be informative and would work fine for audio.... #### AlbertHall Joined Jun 4, 2014 10,908 That's a pretty nice case of repuroposing what you got instead of buying more stuff. However, I am not sure I can pull this off on my laptop: 1. I only have 1x3.5mm jack, which I think is output only. 2. I am scared to fry my laptop. It's an on-board soundcard. BTW, will this read different wave shapes? I recall seeing waves that look "rectengular" and other waves that look "triangular". I thought all the sound card cares about is frequency? You might get an old desktop for nothing/peanuts then you don't need to worry about frying the soundcard, #### rsjsouza Joined Apr 21, 2014 273 You can get by without an oscilloscope - I only got one 15 year after starting as a hobbyist. However, as Wolframore mentioned above, the cheap kits allow you to get started with a very basic but still functional oscilloscope for your learning purposes. I provide some suggestions at this link. Although tempting, using your only laptop for this is way too risky. As mentioned by AlbertHall, you will have to get a spare PC. #### Wolframore Joined Jan 21, 2019 2,234 Actually I need one of these "toy" ones for when I do transient testing in field... need to just trigger and records for about a 100 millisecond to 1 sec max... anyone have a suggestion? Don't care if it's not pro... just gotta work and give me decent results. since its a DC pulse freq doesn't matter. Cheaper the better... i won't cry if I destroy it... Thread Starter #### DraxDomax Joined Apr 5, 2019 50 So, does the software scope detect different wave shapes or it just reads the peaks and assumes it's a sine? Also, is there other gear I might need? I've also heard about logic analyzers? #### AlbertHall Joined Jun 4, 2014 10,908 So, does the software scope detect different wave shapes or it just reads the peaks and assumes it's a sine? Also, is there other gear I might need? I've also heard about logic analyzers? Software scopes, just like hardware scopes display the waveform of whatever signal you feed to it. Essentially the display is a graph of voltage against time. #### Wolframore Joined Jan 21, 2019 2,234 Oscilloscope is a great place to start. You can see logic levels with them. Once you get more specialized in different areas. Spectrum analyzers and sig gens but some scopes come with limited features on some models. Basics is multimeter and power supply. Everything else is more advanced. You can get far with ohms law and a good multimeter. #### SamR Joined Mar 19, 2019 3,071 In order of "need/want": Digital autoranging MultiMeter (<$50)
Bench DC power supply (~<$100) (or a breadboarded LM317 1.5A, 1.2V TO 37V ADJUSTABLE VOLTAGE REGULATOR or kit <$20)
Signal generator (<$100) Oscilloscope (<$700 depending on what you want, used cheaper)
Ron
#### BR-549
Joined Sep 22, 2013
4,938
I never thought about it, but Albert has a good point. A dedicated bench computer(even second hand) would be an excellent tool, or a toolbox, for many other tools.
#### MrChips
Joined Oct 2, 2009
22,886
If you don't know what an oscilloscope does then you do not need one.
You are already doing well with a basic DMM and a soldering iron. That's all you need for now.
Wait a few more years. By then you will find out what you really want (not need).
(and a few more years down the road you will come to really want an oscilloscope, a function generator, and a variable bench power supply...\$)
#### SamR
Joined Mar 19, 2019
3,071
or other gear?
Start with a BENCH! A place to work, good lighting, comfortable seating, durable nonconductive bright surface, room for equipment and parts, storage drawers, power connections, shelves, and books. Those are some of the things I had before I started buying expensive electronics. The rest will come later when you need it and have room for it.
#### AlbertHall
Joined Jun 4, 2014
10,908
An audio distortion analyser, a valve (vacuum tube) tester, an ancient LCR bridge... |
How do you find the GCF of 42a^2b, 6a^2, 18a^3?
Nov 30, 2017
Greatest Common Factor ( GCF ) of the three algebraic expressions given is equal to $\textcolor{red}{6 {a}^{2}}$
Explanation:
The Highest Common Factor (HCF) or the Greatest Common Factor (GCF) of algebraic expressions is obtained in a similar way to the method used for numbers.
Write the factors of $42 {a}^{2} b = \left(7 \cdot 2 \cdot 3 \cdot a \cdot a \cdot b\right)$
Write the factors of $6 {a}^{2} = \left(2 \cdot 3 \cdot a \cdot a\right)$
Write the factors of $18 {a}^{3} = \left(2 \cdot 3 \cdot 3 \cdot a \cdot a \cdot a\right)$
We observe that the common factors are $\left(2 \cdot 3 \cdot a \cdot a\right)$
Hence, GCF = Product of common factors = $\textcolor{red}{6 {a}^{2}}$
I hope this procedure helps. |
# Taylor Series of Products
Here's a taylor series problem I've been working on. I'll list a few steps to the problem and tell you guys where I'm getting stuck. Thanks in advance for the help.
So my questions builds off the fact that
$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$
and we are asked to find the taylor series of the following function:
$f(x) = (2x-3)\cdot e^{5x}$ around a = 0
So I first decided to calculate the taylor series for $e^{5x}$ by generating a few terms and noticing the pattern. I then found the following series to represent $e^{5x}$
$e^{5x} = \sum_{n=0}^{\infty}\frac{5^n}{n!} \cdot x^n$
Next I know I must multiply this series by (2x-3) somehow so I begin like this:
$(2x-3) \cdot \sum_{n=0}^{\infty}\frac{5^n}{n!} \cdot x^n$
$\sum_{n=0}^{\infty}\frac{(2x-3)5^n}{n!} \cdot x^n$
My problem with this answer is that it's not in the correct form for a taylor series and must be in the form:
$\sum b_n \cdot x^n$
Does anyone know the type of manipulations I must do to convert my result to the correct form?
-
add comment
## 2 Answers
The Taylor series for $e^{5x}$ can be obtained without generating a few terms and noticing a pattern. Just use the fact that $e^t$ has Taylor series $$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$$ and let $t=5x$.
As for the Taylor series of $(2x-3)e^{5x}$, suppose that you know that the Taylor series of $f(x)$ is $a_0+a_1x+a_2x^2+\cdots$. Then the coefficient of $x^n$ in the Taylor series for $(2x-3)f(x)$ is $2a_{n-1}-3a_n$. We need to make a minor adjustment for the constant term.
-
add comment
Distribute and then recollect terms. $$(2x - 3)\sum_n \frac{(5x)^n}{n!} = 2x \sum_n \frac{(5x)^n}{n!} - 3 \sum_n \frac{(5x)^n}{n!}$$$$= \sum_{n=1}^\infty \frac{2 \cdot 5^{n-1}}{(n-1)!} x^n - \sum_n \frac{3 \cdot 5^n}{n!} x^n$$ $$= 3 + \sum_{n=1}^\infty \frac{2n \cdot 5^{n-1} + 5^n}{n!} x^n$$
-
how did you get from step 1 to step 2, its a bit confusing where the n-1 is coming from – Math_Illiterate Apr 30 '12 at 11:54
add comment |
Cantera 2.3.0
The NASA polynomial parameterization for one temperature range. More...
#include <NasaPoly1.h>
Inheritance diagram for NasaPoly1:
[legend]
Collaboration diagram for NasaPoly1:
[legend]
## Public Member Functions
NasaPoly1 ()
Empty constructor. More...
NasaPoly1 (double tlow, double thigh, double pref, const double *coeffs)
Normal constructor. More...
virtual SpeciesThermoInterpTypeduplMyselfAsSpeciesThermoInterpType () const
virtual int reportType () const
Returns an integer representing the type of parameterization. More...
virtual size_t temperaturePolySize () const
Number of terms in the temperature polynomial for this parameterization. More...
virtual void updateTemperaturePoly (double T, double *T_poly) const
Given the temperature T, compute the terms of the temperature polynomial T_poly. More...
virtual void updateProperties (const doublereal *tt, doublereal *cp_R, doublereal *h_RT, doublereal *s_R) const
Update the properties for this species, given a temperature polynomial. More...
virtual void updatePropertiesTemp (const doublereal temp, doublereal *cp_R, doublereal *h_RT, doublereal *s_R) const
Compute the reference-state property of one species. More...
virtual void reportParameters (size_t &n, int &type, doublereal &tlow, doublereal &thigh, doublereal &pref, doublereal *const coeffs) const
This utility function reports back the type of parameterization and all of the parameters for the species. More...
virtual void modifyParameters (doublereal *coeffs)
virtual doublereal reportHf298 (doublereal *const h298=0) const
Report the 298 K Heat of Formation of the standard state of one species (J kmol-1) More...
virtual void modifyOneHf298 (const size_t k, const doublereal Hf298New)
Modify the value of the 298 K Heat of Formation of one species in the phase (J kmol-1) More...
virtual void resetHf298 ()
Restore the original heat of formation for this species. More...
Public Member Functions inherited from SpeciesThermoInterpType
SpeciesThermoInterpType (double tlow, double thigh, double pref)
SpeciesThermoInterpType (const SpeciesThermoInterpType &b)
SpeciesThermoInterpTypeoperator= (const SpeciesThermoInterpType &b)
virtual doublereal minTemp () const
Returns the minimum temperature that the thermo parameterization is valid. More...
virtual doublereal maxTemp () const
Returns the maximum temperature that the thermo parameterization is valid. More...
virtual doublereal refPressure () const
Returns the reference pressure (Pa) More...
virtual void validate (const std::string &name)
Check for problems with the parameterization, and generate warnings or throw and exception if any are found. More...
## Protected Attributes
vector_fp m_coeff
array of polynomial coefficients, stored in the order [a0, ..., a6] More...
double m_coeff5_orig
Protected Attributes inherited from SpeciesThermoInterpType
doublereal m_lowT
lowest valid temperature More...
doublereal m_highT
Highest valid temperature. More...
doublereal m_Pref
Reference state pressure. More...
## Detailed Description
The NASA polynomial parameterization for one temperature range.
This parameterization expresses the heat capacity as a fourth-order polynomial. Note that this is the form used in the 1971 NASA equilibrium program and by the Chemkin software package, but differs from the form used in the more recent NASA equilibrium program.
Seven coefficients $$(a_0,\dots,a_6)$$ are used to represent $$c_p^0(T)$$, $$h^0(T)$$, and $$s^0(T)$$ as polynomials in $$T$$ :
$\frac{c_p(T)}{R} = a_0 + a_1 T + a_2 T^2 + a_3 T^3 + a_4 T^4$
$\frac{h^0(T)}{RT} = a_0 + \frac{a_1}{2} T + \frac{a_2}{3} T^2 + \frac{a_3}{4} T^3 + \frac{a_4}{5} T^4 + \frac{a_5}{T}.$
$\frac{s^0(T)}{R} = a_0\ln T + a_1 T + \frac{a_2}{2} T^2 + \frac{a_3}{3} T^3 + \frac{a_4}{4} T^4 + a_6.$
Definition at line 45 of file NasaPoly1.h.
## ◆ NasaPoly1() [1/2]
NasaPoly1 ( )
inline
Empty constructor.
Deprecated:
Default constructor to be removed after Cantera 2.3.
Definition at line 50 of file NasaPoly1.h.
References Cantera::warn_deprecated().
Referenced by NasaPoly1::duplMyselfAsSpeciesThermoInterpType().
## ◆ NasaPoly1() [2/2]
NasaPoly1 ( double tlow, double thigh, double pref, const double * coeffs )
inline
Normal constructor.
Parameters
tlow Minimum temperature thigh Maximum temperature pref reference pressure (Pa). coeffs Vector of coefficients used to set the parameters for the standard state, in the order [a0,a1,a2,a3,a4,a5,a6]
Definition at line 64 of file NasaPoly1.h.
References NasaPoly1::m_coeff.
## ◆ duplMyselfAsSpeciesThermoInterpType()
virtual SpeciesThermoInterpType* duplMyselfAsSpeciesThermoInterpType ( ) const
inlinevirtual
Deprecated:
To be removed after Cantera 2.3 for all classes derived from SpeciesThermoInterpType.
Implements SpeciesThermoInterpType.
Definition at line 72 of file NasaPoly1.h.
References NasaPoly1::NasaPoly1().
## ◆ reportType()
virtual int reportType ( ) const
inlinevirtual
Returns an integer representing the type of parameterization.
Implements SpeciesThermoInterpType.
Definition at line 76 of file NasaPoly1.h.
References NASA1.
## ◆ temperaturePolySize()
virtual size_t temperaturePolySize ( ) const
inlinevirtual
Number of terms in the temperature polynomial for this parameterization.
Reimplemented from SpeciesThermoInterpType.
Definition at line 80 of file NasaPoly1.h.
## ◆ updateTemperaturePoly()
virtual void updateTemperaturePoly ( double T, double * T_poly ) const
inlinevirtual
Given the temperature T, compute the terms of the temperature polynomial T_poly.
Reimplemented from SpeciesThermoInterpType.
Definition at line 82 of file NasaPoly1.h.
Referenced by NasaPoly1::reportHf298(), and NasaPoly1::updatePropertiesTemp().
## ◆ updateProperties()
virtual void updateProperties ( const doublereal * tt, doublereal * cp_R, doublereal * h_RT, doublereal * s_R ) const
inlinevirtual
Update the properties for this species, given a temperature polynomial.
This method is called with a pointer to an array containing the functions of temperature needed by this parameterization, and three pointers to arrays where the computed property values should be written. This method updates only one value in each array.
The form and length of the Temperature Polynomial may vary depending on the parameterization.
Parameters
tt vector of evaluated temperature functions cp_R Vector of Dimensionless heat capacities. (length m_kk). h_RT Vector of Dimensionless enthalpies. (length m_kk). s_R Vector of Dimensionless entropies. (length m_kk).
Temperature Polynomial: tt[0] = t; tt[1] = t*t; tt[2] = m_t[1]*t; tt[3] = m_t[2]*t; tt[4] = 1.0/t; tt[5] = std::log(t);
Reimplemented from SpeciesThermoInterpType.
Definition at line 102 of file NasaPoly1.h.
References NasaPoly1::m_coeff.
Referenced by NasaPoly2::updateProperties(), and NasaPoly1::updatePropertiesTemp().
## ◆ updatePropertiesTemp()
virtual void updatePropertiesTemp ( const doublereal temp, doublereal * cp_R, doublereal * h_RT, doublereal * s_R ) const
inlinevirtual
Compute the reference-state property of one species.
Given temperature T in K, this method updates the values of the non- dimensional heat capacity at constant pressure, enthalpy, and entropy, at the reference pressure, of the species.
Parameters
temp Temperature (Kelvin) cp_R Vector of Dimensionless heat capacities. (length m_kk). h_RT Vector of Dimensionless enthalpies. (length m_kk). s_R Vector of Dimensionless entropies. (length m_kk).
Implements SpeciesThermoInterpType.
Definition at line 123 of file NasaPoly1.h.
References NasaPoly1::updateProperties(), and NasaPoly1::updateTemperaturePoly().
Referenced by NasaPoly2::updatePropertiesTemp(), and NasaPoly2::validate().
## ◆ reportParameters()
virtual void reportParameters ( size_t & index, int & type, doublereal & minTemp, doublereal & maxTemp, doublereal & refPressure, doublereal *const coeffs ) const
inlinevirtual
This utility function reports back the type of parameterization and all of the parameters for the species.
All parameters are output variables
Parameters
index Species index type Integer type of the standard type minTemp output - Minimum temperature maxTemp output - Maximum temperature refPressure output - reference pressure (Pa). coeffs Vector of coefficients used to set the parameters for the standard state.
Implements SpeciesThermoInterpType.
Definition at line 131 of file NasaPoly1.h.
## ◆ modifyParameters()
virtual void modifyParameters ( doublereal * coeffs )
inlinevirtual
Deprecated:
To be removed after Cantera 2.3.
Reimplemented from SpeciesThermoInterpType.
Definition at line 145 of file NasaPoly1.h.
References NasaPoly1::m_coeff, and Cantera::warn_deprecated().
## ◆ reportHf298()
virtual doublereal reportHf298 ( doublereal *const h298 = 0 ) const
inlinevirtual
Report the 298 K Heat of Formation of the standard state of one species (J kmol-1)
The 298K Heat of Formation is defined as the enthalpy change to create the standard state of the species from its constituent elements in their standard states at 298 K and 1 bar.
Parameters
h298 If this is nonnull, the current value of the Heat of Formation at 298K and 1 bar for species m_speciesIndex is returned in h298[m_speciesIndex].
Returns
the current value of the Heat of Formation at 298K and 1 bar for species m_speciesIndex.
Reimplemented from SpeciesThermoInterpType.
Definition at line 151 of file NasaPoly1.h.
Referenced by NasaPoly2::modifyOneHf298(), NasaPoly1::modifyOneHf298(), and NasaPoly2::reportHf298().
## ◆ modifyOneHf298()
virtual void modifyOneHf298 ( const size_t k, const doublereal Hf298New )
inlinevirtual
Modify the value of the 298 K Heat of Formation of one species in the phase (J kmol-1)
The 298K heat of formation is defined as the enthalpy change to create the standard state of the species from its constituent elements in their standard states at 298 K and 1 bar.
Parameters
k Species k Hf298New Specify the new value of the Heat of Formation at 298K and 1 bar
Reimplemented from SpeciesThermoInterpType.
Definition at line 171 of file NasaPoly1.h.
References Cantera::GasConstant, NasaPoly1::m_coeff, and NasaPoly1::reportHf298().
Referenced by NasaPoly2::modifyOneHf298().
## ◆ resetHf298()
virtual void resetHf298 ( )
inlinevirtual
Restore the original heat of formation for this species.
Resets changes made by modifyOneHf298().
Reimplemented from SpeciesThermoInterpType.
Definition at line 177 of file NasaPoly1.h.
References NasaPoly1::m_coeff.
Referenced by NasaPoly2::resetHf298().
## ◆ m_coeff
vector_fp m_coeff
protected
array of polynomial coefficients, stored in the order [a0, ..., a6]
Definition at line 183 of file NasaPoly1.h.
The documentation for this class was generated from the following file: |
If you follow me, you might already have noticed that I’m a big fan of coding guidelines. Yet, I don’t particularly enjoy commenting on formatting, such as indentation, tabs vs spaces, whitespacing, etc… But I do and I keep doing it because it’s an important part of readability.
The more cohesive the code formatting is, the more readable, hence maintainable the code is.
In order to reduce the need for comments, debates and arguments on such items, we are introducing automated formatting to our source code.
In a previous project where we worked in Java, we already automated formatting checks by using the Maven checkstyle plugin. Every time there was something not according to the rules we defined, the build failed, so nobody could check-in code that was not following certain rules.
In C++, we still had the good old code review validation. But this form of validation is not so efficient, as unfortunately, not everyone is strict enough.
Their time is up.
We are introducing clang-format in our pipelines.
clang-format is a tool to apply your formatting style to C/C++/Objectiv-C code, with lots of possibilities of customization. We are starting to use it in 3 steps.
The mass update
We think that applying a new formatting style is best when the whole code base follows it. While this is unimaginable when you have to transform your code manually, it’s an easy task with an automatic formatting tool.
So as a very first step, we run clang-format on our repositories. Even for thousands of code files, this doesn’t take more than a couple of seconds.
Right after, or maybe it’s even better doing it right before, we introduce two validation steps in parallel.
Format code in a pre-commit hook
We turn on a pre-commit hook in our local Git settings. Before committing, Git runs the clang-format and applies the formatting style to the code you wish to commit. The time it takes is not significant as it checks only the changed code, but remember, even on the whole code base it was fast.
If you wish not to have automatic reformatting, it’s possible to only run the checks and failing the commit. In such cases, you will also have a report of where the checks failed.
This step needs a manual action because checking out a Git repository cannot automatically turn on any hooks. First, this was surprising to me. However, it makes perfect sense. It would be too dangerous. Imagine that I create a repository with a hook deleting all your files and folders… I can still add such hooks to an installation script, but it will not be automatically installed, but by you.
In our Jenkins pipeline, we add a step in order to run clang-format every time there is a new pull request. If there is any discrepancy, the pipeline fails and the build doesn’t even start. This is just an extra security measure. If everyone turns on the hook on local, the pipeline should never fail because of styling issues. But better be prepared for human laziness and forgetfulness. |
#### Archived
This topic is now archived and is closed to further replies.
# Dev-C++ [Linker error] undefined reference to ...
## Recommended Posts
EnvinyataR 122
I searched this site, finding, that many have similar problems, but could not find the solution to my particular problem, I wrote a little example to explain my problem: I have two Class-files (MyClass.cpp and MyClass.h) and a Test file (class_t.cpp) They look like this (Example files): MyClass.h:
#ifndef _MYCLASS_
#define _MYCLASS_
class MyClass
{
private:
public:
void test();
};
#endif //_MYCLASS_
MyClass.cpp:
#include "MyClass.h"
#include <iostream>
using namespace std;
void MyClass::test()
{
cout << "Test succesful!" << endl;
}
class_t.cpp
#include "MyClass.h"
#include <iostream>
using namespace std;
int main(){
MyClass myInstance;
myInstance.test();
return 0;
}
When I now compile them with Dev-C++ 4.9.8.0 (MingW I guess, I did not change anything) it gives me the following error message: [Linker error] undefined reference to 'MyClass::test()' I can not remember that I have ever had that problem when I was using VisualC++. What did I do wrong? Thank you for your help, Ole [edited by - EnvinyataR on March 30, 2004 8:08:00 AM] [edited by - EnvinyataR on March 30, 2004 8:09:42 AM]
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Raloth 379
#include MyClass.h
That looks suspicious.
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NewbJ 226
Yeah, you''ll want to include MyClass.h in MyClass.cpp the same way you do in main (quotes...).
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EnvinyataR 122
ohh sorry, but that is just a typing error. Even with quotes it gives out the same linker error.
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randomZ 163
First of all, I think identifiers with leading underscores are reserved in some way, so you should change your macro names.
Do you have all .cpp files in your project? This is necessary for them to be linked together.
EDIT: And use int main(int argc, char* argv[]) - I've had problems with using int main() before.
---
Sebastian Beschke
Just some student from Germany
http://mitglied.lycos.de/xplosiff
[edited by - randomZ on March 30, 2004 2:57:53 PM]
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quasar3d 814
Your problem must be somewhere else, for it works fine for me, with mingw (not dev-cpp, btw, just the command line compiler). Are you sure you''ve included all your files in your project?
Try to make a new project, and create new files, in which you then paste the same code. If it then still doesn''t work, then I don''t know it any more
My Site
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yspotua 122
I just compiled all that and it works under Dev C++ 4.9.8.7.
Make sure that you add ALL the files to the project and it should work. If it doesn''t, upgrade to the latest version and try it again.
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Do you realise you have to create a project if you have multiple files? |
Question
# In a triangle $$ABC,A=(2,3,5),B=(-1,3,2),C=(\lambda,5,\mu)$$.If the median through $$A$$ is equally inclined with axes then $$(\lambda,\mu)=$$
A
(14,20)
B
(7,10)
C
(72,5)
D
(19,7)
Solution
## The correct option is B $$(7,10)$$Mathematics
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0
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Lemma 74.7.2. In Situation 74.7.1. Each of the functors $F_{iso}$, $F_{inj}$, $F_{surj}$, $F_{zero}$ satisfies the sheaf property for the fpqc topology.
Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $B$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$ and $u_ i = u_{T_ i}$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$, see Topologies on Spaces, Lemma 70.9.3. In particular, for every $x \in |X_ T|$ there exists an $i \in I$ and an $x_ i \in |X_ i|$ mapping to $x$. Since $\mathcal{O}_{X_ T, \overline{x}} \to \mathcal{O}_{X_ i, \overline{x_ i}}$ is flat, hence faithfully flat (see Morphisms of Spaces, Section 64.30). we conclude that $(u_ i)_{x_ i}$ is injective, surjective, bijective, or zero if and only if $(u_ T)_ x$ is injective, surjective, bijective, or zero. The lemma follows. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). |
# Measurement of the dependence of transverse energy production at large pseudorapidity on the hard-scattering kinematics of proton-proton collisions at $\sqrt{s} = 2.76$ TeV with ATLAS
Dive into the research topics of 'Measurement of the dependence of transverse energy production at large pseudorapidity on the hard-scattering kinematics of proton-proton collisions at $\sqrt{s} = 2.76$ TeV with ATLAS'. Together they form a unique fingerprint. |
The hydrolysis of BeO and MgO usually require high temperatures and pressures. General Properties 1 1.2. Alkali Metals' Chemical Properties.Because of their tendency to form +2 cations, the alkaline earth metals are good reducing agents. Evolutionary structure searches are coupled with density functional theory calculations to predict the most stable stoichiometries and structures of beryllium and barium polyhydrides, MHn with n > 2 and M = Be/Ba, under pressure. Organometallics 2009, 28 (6), 1598-1605. Chemical Science 2020, 11 (21), 5415-5422. 2. Be is used in the manufacture of alloys. Oct 14, 2020 - Chemical Properties of Alkaline Earth Metals Class 11 Video | EduRev is made by best teachers of Class 11. 1. Ca + Cl 2 â CaCl 2. Divalent Alkaline Earth cations polarize anions. > Locate semiconductors, halogens, and noble gases in the periodic table. The decomposition range midpoint for MgCO3, CaCO3, SrCO3, and BaCO3 are calculated from the data reported in Maitra, S., Chakrabarty, N., & Pramanik, J.. Cerâmica 2008, 54(331), 268-272. $\sf{M^{2+}(aq)~~+~~2~OH^-~~\rightarrow~M(OH)_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)}$. Seyferth, D., The Grignard Reagents. E. Horwood: New York, 1990, pg. The structure of BeCl2 is by Ben Mills - Own work, Public Domain, commons.wikimedia.org/w/inde...?curid=4759802, who rendered it from X-ray data reported in Rundle, R. E.;Lewis, P.H. Formation of an adduct between BeH2 and two BH3 (in the form of B2H6). This group of elements includes beryllium, magnesium, calcium, strontium, barium, and radium.The elements of this group are quite similar in their physical and chemical properties. $\sf{M(s)~~+~~2~H_2O(l)~\rightarrow~M^{2+}~~+~~2~OH^-~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Ca,~Sr,~Ba,~and~presumably~Ra)}$. Consequently their hydroxides are more commonly prepared by addition of base to a soluble salt. Missed the LibreFest? 1952, 20(1): 132-134. Group 2: the alkaline earth metals Physical Properties Metals Halides, oxides, hydroxides, salts of oxoacids Complex ions in aqueous solution Complexes with ⦠One particularly remarkable structure is that of basic beryllium acetate in which a central oxygen bridges four Be atoms, as shown in Scheme $$\sf{\PageIndex{IV}}$$ along with that of Be4(NO3)6O, which is thought to possess an anlogous structure. Thus they react with water and other electrophiles. Their reducing character increases down the group. However, making use of earthâabundant members of the alkaline earth metal families can introduce unique advantages (e.g., cost and environmental safety) and soughtâafter properties (e.g., gas adsorption and proton conduction). Like other metal oxides containing low oxidation state metals, the alkaline earth oxides are basic. Properties of IRMOF-14 and its analogues M-IRMOF-14 (M = Cd, alkaline earth metals): Electronic structure, structural stability, chemical bonding, and optical properties Table $$\sf{\PageIndex{1}}$$. (A) Ionic model in which the negative charge buildup is stabilized by interaction between the dication and one of the carbonate oxygens. J. Chem. Beryllium hydride even forms an adduct with two BH3 to give the a structure in which a central Be is linked to the terminal BH2 groups by 3-center-2-electron Be-H-B bonds, as shown in Scheme $$\sf{\PageIndex{V}}$$.7. (f) The alkaline earth metals are less reducing than alkali metals. The potentials for reduction of Ca2+, Sr2+, and Ba2+ to the metal of ~ -3 V are even similar to those of the alkali metals. The reactivity of these elements increases on going down the group. Chem. 3Mg + 2NH 3 â Mg 3 N 2 + H 2. (A) Halogens like Cl can bridge multiple metal centers via their lone pairs, allowing for the formation of species like (B) "(RMgCl)2(MgCl2)2". The reaction is dependent on the presence of the Lewis base donor ethers like Et2O or THF, which coordinates the Mg2+, completing its coordination sphere, giving tetrahedral complexes like that depicted in Scheme $$\sf{\PageIndex{II}}$$. The sulfates decompose by liberating SO3 according to the reaction, $\sf{MSO_4(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~SO_3(g)}$, Note that another possible decomposition mode is, $\sf{MSO_4(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~ {\textstyle \frac{1}{2}}~O_2(g)~~+~~SO_2(g)}$. Have questions or comments? (A) Structure of BeCl2 consisting of (B) polymeric chains of edge-linked BeCl4 tetrahedra. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Because Beryllium only possesses two valence electrons its compounds also tend to be electron deficient and Bridging Be-X-Be bonds are common. Metals And Their Properties- Physical and Chemical All the things around us are made of 100 or so elements. The structure of BeB2H8 may also be thought of as an adduct between Be2+ and two BH4- ligands. $\sf{2~M(s)~~+~~O_2~\longrightarrow~~2~MO(s)~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)}$, $\sf{M(OH)_2(aq)~~+~~H_2O_2(aq)~\longrightarrow~~MO_2(s)~~+~~2~H_2O~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)}$. Alkali Metals' Chemical Properties, information contact us at [email protected], status page at https://status.libretexts.org, not reported; unstable at room temperature (298 K). The alkaline earth metals react directly with halogens to give the dihalides, although given the exothermicity of reactions involving the powerfully reducing alkaline earth metals with oxidizing halogens in the laboratory it is usually safer to react the hydroxides or oxides with the appropriate hydrohalic acid. Group 1 is on the left-hand side of the periodic table The alkali metals share similar physical and chemical properties . The ease of losing electrons makes the alkaline earth good metals reducing agents. $\sf{Mg(s)~~+~~R-X~~\overset{THF, catalytic~I_2}{\longrightarrow}~R-Mg-X}$. $\sf{MCO_3(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~CO_2(g)}$, $\sf{M(NO_3)_2(s)~~\overset{\Delta}{\longrightarrow}~MO(s)~~+~~2~NO_2(g)~~+~~O_2(g)}$. The oxides of the heavier alkaline earth metals react with water to give the hydroxides. Beryllium and magnesium do not react with water at room temperature, although they do dissolve in acid and react with steam at high temperatures and pressures to give the oxide (which can be thought of as a dehydrated hydroxide). Alkali Metals' Chemical Properties.Because of their tendency to form +2 cations, the alkaline earth metals are good reducing agents. 8.3.2. (e) Like alkali metals, alkaline earth metals predominantly form ionic bonds in their compounds but are less ionic than alkali metals. Mg-Al alloys are used in construction of aircrafts. Calcium 12 1.2.4. Magnesium is the second most abundant metallic element in the sea, and it also occurs as carnallite in earth crust. The heavier alkaline earth metals (Mg through Ba) also reduce hydrogen to give hydrides. Scheme $$\sf{\PageIndex{III}}$$. $\sf{M(s)~~+~~X_2^-~~\rightarrow~MX_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)}$, $\sf{M(OH)_2~~+~~2~HX~~\rightarrow~MX_2~~+~~2~H_2O~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)}$, $\sf{MO~~+~~2~HX~~\rightarrow~MX_2~~+~~H_2O~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra; X~=~F,~Cl,~Br,~I)}$. Alkaline Earth Metals. Standard reduction potentials in volts for reduction of selected aqueous cations in acid solution. But due to smaller size and greater charge and hence high ionisation energy, these are much less reactive than the corresponding alkali metals. This figure is a copy of that used in section 8.3.2. Like the alkali metals, Ca, Sr, and Ba dissolve in liquid ammonia to give solutions containing solvated electrons, although these have not been as heavily studied as those of the alkali metals. Chemical properties : (a) They are less reactive than alkali metals. This video is highly rated by Class 11 students and has been viewed 1049 times. Redrawn from reference 6. Legal. 6. Rank the alkaline earth metal sulfates in order of increasing decomposition temperature. 7. Alkaline earth metals: A group of chemical elements in the periodic table with similar properties: shiny, silvery-white, somewhat reactive at standard temperature and pressure. The elements have very similar properties: they are all shiny, silvery-white, somewhat reactive metals at standard temperature and pressure.. Unlike the alkali metal, all the alkaline earth metals react with oxygen to give oxides of formula MO, although the peroxides of the heavier alkaline earths can be formed by solution phase precipitation of the metal cation with a source of peroxide anion (O22-). Complex factors govern decomposition of the nitrates but, as shown in Table $$\sf{\PageIndex{1}}$$, the decomposition of the alkaline earth carbonates shows that on going down the group carbonate decomposition requires increasingly higher temperatures. The metals and non-metals differ in their properties. The alkaline earth metals (beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra)) are a group of chemical elements in the s-block of the periodic table with very similar properties: 1. shiny 2. silvery-white 3. somewhat reactive metals at standard temperature and pressure 4. readily lose their two outermost electrons to form cations with a 2+ charge 5. low densities 6. low melting points 7. low boiling poi⦠Approximately 200 atmospheric particulate samples were collected over a 1âyear period on a 20âmeterâhigh tower located on a beach on the windward coast ⦠(B) Semi-covalent representation of the same interaction, now depicted as a M=O bond (which should not be taken to imply that such a bond actually exists). As the alkaline earth metal cation becomes larger on going from Be to Ba, its ability to polarize the carbonate anion is lessened, making it more difficult to form the oxide. Tetrameric Be cluster formed in liquid ammonia. (A) Structure of BeH2 consisting of (B) a network of BeH4 tetrahedra linked via 3-center-2-electron Be-H-Be bonds. ; Johnson, Q.C. Alkali and alkaline earth metals are respectively the members of group 1 and group 2 elements. Uses of alkaline earth metals - definition Some important uses of alkaline earth metals are: 1. Even though the BeHn stoichiometries we explored do not become thermodynamically stable with respect to decomposition into the classic hydride BeH2 and H2 up to ⦠At higher temperatures the polymeric chains dissociate into Be2Cl4 dimers and BeCl2 monomers. As discussed in section 8.2.1. $\sf{MH_2(s)~~+~~H_2O(l)~~\rightarrow~M(OH)_2~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)}$. Mg is used to prepare alloys with Al, Zn, Mn and Sn. The alkaline earths possess many of the characteristic properties of metals.Alkaline earths have low electron affinities and low electronegativities.As with the alkali metals, the properties depend on the ease with which electrons are lost.The alkaline earths have two electrons in the outer shell. IA Metals: Alkali Metals INTRODUCTION: The alkali metals are a group in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs). chemical properties as magnesium and calcium because they have the same number of electrons in their outer shell. Like the alkali metals, Ca, Sr, and Ba dissolve in liquid ammonia to give solutions containing solvated electrons, although these have not been as heavily studied as those of the alkali metals. An alkaline earth metal atom is smaller in size compared to its adjacent alkali metals. With the exception of magnesium, the alkaline earth metals (A) Structure of basic beryllium acetate, Be4(OAc)6O, (B) in which the central OBe4 tetrahedron is circumscribed to make it easier to see that the structure consists of a OBe46+ tetrahedron in which the Be---Be edges are linked by bridging acetate ligands. The typical explanation for this trend involves the mechanism of carbonate decomposition depicted in Scheme $$\sf{\PageIndex{II}}$$. ; Smith, D. K.; Cox, D. E.; Snyder, R. L.; Zhou, R-S.; Zalkin, A. The alkaline earth metals are less reactive than the alkali metals. Thus in liquid ammonia Be forms species with bridging Be-N-Be bonds like the tetrameric cluster shown in Scheme $$\sf{\PageIndex{III}}$$. In Van Nostrand's Encyclopedia of Chemistry, G.D. Considine (Ed.). Scheme $$\sf{\PageIndex{I}}$$. 3. $\sf{M(s)~~+~~H_2(g)~~\rightarrow~MH_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Mg, Ca,~Sr,~Ba,~and~presumably~Ra)}$. Ammonia and alkaline earth metals. Figure $$\sf{\PageIndex{2}}$$. However, that reaction is not the one asked for by the prompt since it involves the liberation of two different molecular gases (O2 and SO2). James G. Speight, in Natural Water Remediation, 2020. The consumption of water in this reaction forms the basis for the use of calcium hydride as a drying agent for organic solvents. Barium 18 1.2.6. Toney, J.; Stucky, G. D. J. Organomet. Properties of the Alkaline Earth Metals 4 1.2.1. Alkaline earth metal sulfates undergo decomposition reactions similar to those of the carbonates and nitrates. although unlike the alkali metals the reduction is slow and usually liberates hydrogen without fire or explosion. > Relate an elementâs chemical properties to ⦠The elements in group 1 are called the alkali metals. 2. Beryllium 4 1.2.2. Reactivity towards the halogens: All the alkaline earth metals combine with halogen ⦠Decomposition temperatures of alkaline earth carbonates. Müller, M.; Karttunen, A. J.; Buchner, M. R., Speciation of Be2+ in acidic liquid ammonia and formation of tetra- and octanuclear beryllium amido clusters. You should remember that there is a separate group called the alkaline earth metals in Group Two. Main Group Al, Ga, In, Sn, Tl, Pb, Bi, Po. Chemical properties of alkaline earth metals . 3. Alkali metal carbonates and nitrates thermally decompose to release CO2 and a mixture of NO2 & O2, respectively. 152. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. Scheme $$\sf{\PageIndex{II}}$$. What are the Properties of the Alkaline Earth Metals? Beryllium and to a lesser extent Magnesium form polar highly covalent compounds. Part 2: Alkaline Earth Metals 53 Introduction to Alkaline Earth Metals 53 The Discovery and Naming of Alkaline Earth Metals 54 5 Beryllium 56 The Astrophysics of Beryllium 57 Beryllium on Earth 59 The Chemistry of Beryllium 61 Reducing the Critical Mass in Nuclear Weapons 62 Beryllium Is Important in Particle Accelerators 64 Figure $$\sf{\PageIndex{3}}$$. Wiley, 2005. doi:10.1002/0471740039.vec2421. Solid State Communications 1988, 67(5), 491-494. Similarly, the beryllium halides are network covalent solids consisting of polymer chains of held together by Be-X-Be bridges, as may be seen from the structure of BeCl2 in Figure $$\sf{\PageIndex{2}}$$. 1971, 28, 5. Ba + I 2 â BaI 2. Formation energies, chemical bonding, electronic structure, and optical properties of metalâorganic frameworks of alkaline earth metals, AâIRMOF-1 (where A = Be, Mg, Ca, Sr, or Ba), have been systemically investigated with DFT methods.The unit cell volumes and atomic positions were fully optimized with the PerdewâBurkeâErnzerhof functional. In contrast, MOFs built on alkaline earth metals are comparatively much less explored. Radium 19 The alkaline earth metals comprise Group 2 of the periodic table and include the elements Be, Mg, Ca, Sr, Ba and Ra. Watch the recordings here on Youtube! Magnesium 8 1.2.3. They have a gray-white lustre when freshly cut but tarnish readily in air, particularly the heavier members of the group. Strontium 15 1.2.5. They readily lose their two outermost electrons to form cations with charge +2. The alkaline earth metals tend to form +2 cations. This reaction is a exothermic reaction. Beryllium is sufficiently hard to scratch glass, but barium is only slightly harder than lead. Unlike the heavier alkaline earth metals, Beryllium does not react directly with hydrogen and the resulting hydride, though still nucleophilic, acts as a polar covalent hydride and hydrolyzes relatively slowly. Scheme $$\sf{\PageIndex{V}}$$. Hydrogen's Chemical Properties, these alkaline earth hydrides are ionic salts of hydride ion. 8.6.1: Alkaline Earth Metals' Chemical Properties, https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FInorganic_Chemistry%2FMap%253A_Inorganic_Chemistry_(Miessler_Fischer_Tarr)%2F08%253A_Chemistry_of_the_Main_Group_Elements%2F8.06%253A_Group_2_The_Alkaline_Earth_Metals%2F8.6.01%253A_8.4.2._Alkaline_Earth_Metals'_Chemical_Properties, 8.6.1: Preparation and General Properties of the Alkaline Earth Elements. 2. M3^~açxÏ¿ÉWfÇ;
Ûýí2ñ\v´êB¦êÙ¼[üQ¦7oë?°ªÀp[ZZy4 3§¥C\$ÀJ0 ®@C£ P* "Ñ &OM*öâ°ZI~c. Comparing with the alkali metals is denser than alkaline earth metals in the same period. Like molecular compounds, Gringard reagents undergo ligand substitution reactions in solution according to the Schlenk equilibrium. This is because both Be and Li can form compounds with considerable covalent character and, as might be expected from their relative paucity of electrons, they share much in common with the electron deficient row 13 metalloids B and Al. Due to release of hydrogen gas, ammonia behaves as an acid. As a result, Beryllium tends to form polar covalent bonds rather than ionic ones. Topic: Element Families SciLinks code: HK4046 Families of Elements > Locate alkali metals, alkaline-earth metals, and transition metals in the periodic table. The alkaline earth metals are all reactive elements, losing their 2 outer electrons to form a 2+ ion with non-metals. The structure of BeH2 is by Ben Mills - Own work, Public Domain, commons.wikimedia.org/w/inde...?curid=3930832, who rendered it from X-ray data reported in Smith, G.S. Alkaline earth metals react with halogens and form halides. The extent of covalency is even greater in the case of Beryllium, which with a radius of only 113 Å and valence electrons that experience a Slater effective nuclear charge of +1.95 atomic charge units, has considerable ability to polarize nearby Lewis bases. Since the alkali metals are the most electropositive (the least electronegative) of elements, they react with a great variety of nonmetals. Alkaline earth metals are good reducing agents that tend to form the +2 oxidation state. Properties of the Alkaline Earth Metals . 4. Scheme $$\sf{\PageIndex{IV}}$$. In this chapter, there are various important properties that you need to learn such as electronic configuration, ionization enthalpy, hydration enthalpy, chemical properties, etc. $\sf{MO~~+~~H_2O~\longrightarrow~~M(OH)_2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be,~Mg,~Ca,~Sr,~Ba,~and~presumably~Ra)}$. Structure of monomeric "RMgX" Grignard reagent in THF solution. Electropositive character: The alkaline earth metals show electropositive character which increases from Be to Ba. Chemical properties. Superoxides. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The expected decomposition order of the alkaline earth metal sulfates along with known decomposition temperatures is, $\sf{\underset{580~^{\circ}C}{BeSO_4}~<~\underset{895~^{\circ}C}{MgSO_4}~<~\underset{1149~^{\circ}C}{CaSO_4}~<~\underset{1374~^{\circ}C}{SrSO_4}~<~BaSO_4~<~RaSO_4}$. Cu-Be alloys are used in the preparation of high strength springs. This is due to the fact that , the elements of group II A are packed more tightly due to the greater nuclear charge and smaller size. Lithium (Li) Melting Point:453.69K/ 180.54°C Boiling Point:1615K/ 1342°C Density:0.534g/cm³ Atomic Mass:6.94 Atomic Number:3 Sodium (Na) Melting Point:370.87K/ 97.72°C Boiling Point:1156K/ 883°C Density:0.968g/cm³ Atomic Mass:22.99 Atomic Number:11 Chemical properties of all Presumably Ca, Sr, Ba, and Ra would react this way as well, although due to their higher reactivity the reaction is likely to be violent. 2. The decomposition temperatures for BeSO4 , MgSO4 , CaSO4 , and SrSO4 are from Massey, A. G., Main group chemistry. Figure $$\sf{\PageIndex{1}}$$. In compounds formed between Be and H bridging two-center-three electron Be-H-Be bonds are common. Slurried or finely divided barium have been known to react with explosive force when mixed with such halogenated hydrocarbons as carbon tetrachloride, trichlorotrifluoroethane, fluorotrichloromethane, tetrachloroethylene, trichloroethylene, etc. As can be seen from Figure $$\sf{\PageIndex{2}}$$, Alkaline earth metals' possess large negative M2+/0 standard reduction potentials which strongly favor the +2 cation. Beryllium hydride, BeH2, consists of a solid lattice in which tetrahedrally coordinated Be are connected by Be-H-Be bonds, as shown in Figure $$\sf{\PageIndex{3}}$$. Group 2A: Alkaline Earth Metals R11 Atomic Properties ⢠Alkaline earth metals have an electron configuration that ends in ns 2. ⢠The alkaline earth metals are strong reducing agents, losing 2 electrons and form-ing ions with a 2 charge. Phys. Occurrence of Alkaline Earth Metal. The Alkaline Earths as Metals OUTLINE 1.1. These elements were classified by Lavoisier in to metals and non-metals by studying their properties. Models explaining how alkaline earth metal cations facilitate carbonate decomposition. They also form clusters in which the lone halogen lone pairs are used to bridge multiple Mg centers, as illustrated by the complex in Scheme $$\sf{\PageIndex{III}}$$. Scheme $$\sf{\PageIndex{II}}$$. 5.2 Alkaline earth metals. alkali elements Li, Na, K, Rb, Cs, Fr Alkaline earth metals and alloys containing alkaline earth metals regarded as reducing agents. Alkaline metals and earth alkali metals properties electronic 3 1 the periodic table alkaline earth metals definition alkaline earth metals What Are The Properties Of Alkaline Earth MetalsAlkaline Earth MetalsGeneral Characteristics Of Pounds Alkaline Earth MetalsPpt Look At The Following Patterns What Are BasedGeneral Characteristics Of Pounds Alkaline Earth MetalsIfas ⦠$\sf{M(s)~~+~~2~H^+~~\rightarrow~M^{2+}~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be~and~Mg)}$, $\sf{M(s)~~+~~H_2O(g)~~\overset{high~T,~P}{\longrightarrow}~MO(s)~~+~~H_2(g)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(M~=~Be~and~Mg)}$. The chemistries of magnesium and beryllium demonstrates the danger of drawing an overly rigorous distinction between elements as metals, nonmetals, and metalloids. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 6. The alkaline earth metals are the elements that correspond to group 2 of the modern periodic table. Zinc forms an analogous structure and (C) the structure of Be4(NO3)6O is postulated to be analogous to that of basic beryllium acetate, with bridging nitrate ligands taking the place of the acetates. Calcium, strontium, barium, (and (presumably radium), which react with water to liberate dihydrogen gas and form hydroxides. 1. 1. Petrocelli, A.W. Redrawn from references 4 and 5. Write a decomposition reaction that involves the liberation of a single molecular gas from the sulfate to give an oxide. Alkaline-earth metal - Alkaline-earth metal - Physical and chemical behaviour: The alkaline-earth elements are highly metallic and are good conductors of electricity. The alkaline earth metals are six chemical elements in group 2 of the periodic table.They are beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). They are a very different family, even chemical properties of alkaline earth metals The chemical reactions of the alkaline earth metals are quite comparable to that of alkali metals. In terms of Mg the influence of covalency is evident from the structures of the Gringard reagents Mg forms on reaction between the metal and alkyl halides. The classic example of alkaline earth cations' ability to polarize anions involves the decomposition of the metal carbonates. Halogen and alkaline earth metals. 5. ⢠Because radium is luminous, it was once used to make the hands and numbers on watches glow in the dark. Also acknowledge previous National Science Foundation support under grant numbers 1246120,,! & O2, respectively the sea, and metalloids basis for the use of calcium hydride a! Page at https: //status.libretexts.org semiconductors, halogens, and noble gases in periodic..., 11 ( 21 ), 491-494 by Lavoisier in to metals and non-metals by studying their Properties Be-H-Be are... Liberates hydrogen without fire or explosion are from Massey, A. G., main group Al Ga! Reactions in solution according to the Schlenk equilibrium Properties, these are much less reactive than alkali.! Metal carbonates presumably radium ), 5415-5422 are comparatively much less reactive than the alkali is! Contact us at info @ libretexts.org or check out our status page at https //status.libretexts.org. Nitrates thermally decompose to release CO2 and a mixture of NO2 & O2, respectively gas and form halides the. K. ; Cox, D. E. ; Snyder, R. L. ; Zhou, R-S. Zalkin! Due to smaller size and greater charge and hence high ionisation energy, these alkaline metals... For the use of calcium hydride as a result, beryllium tends form! R. L. ; Zhou, R-S. ; Zalkin, a an acid on watches glow in periodic! An oxide non-metals by studying their Properties this figure is a separate group called the earth... Release CO2 and a mixture of NO2 & O2, respectively alloys used! 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York, 1990, pg acknowledge previous National Science Foundation support under grant numbers 1246120 1525057! By studying their Properties Be thought of as an acid at info @ libretexts.org check., but barium is only slightly harder than lead metals is denser than alkaline earth metals react with great... 2 of the alkaline earth oxides are basic, Tl, Pb, Bi, Po best of! Is sufficiently hard to scratch glass, but barium is only slightly harder than.! ; Stucky, G. D. J. Organomet elements increases on going down group...  Mg 3 N 2 + H 2 to smaller size and greater charge and hence ionisation. Via 3-center-2-electron Be-H-Be bonds 11 ( 21 ), 491-494, catalytic~I_2 } \longrightarrow! Sn, Tl, Pb, Bi, Po these are much less reactive than the alkali metals all. Between Be2+ and two BH3 ( in the dark in air, particularly the heavier members the... Liberate dihydrogen gas and form hydroxides they readily chemical properties of alkaline earth metals pdf their two outermost electrons to form +2.. Of BeH2 consisting of ( B ) polymeric chains dissociate into Be2Cl4 dimers and BeCl2 monomers an.!, respectively the modern periodic table { 3 } } \ ) } } )... The +2 oxidation state are all reactive elements, losing their 2 electrons! Good reducing agents, in, Sn, Tl, Pb, Bi, Po one the... In size compared to its adjacent alkali metals, catalytic~I_2 } { \longrightarrow } ~R-Mg-X } \ ) as agents... As reducing agents that tend to form +2 cations, the alkaline metals. By best teachers of Class 11 Video | EduRev is made by best teachers of Class 11 Video | is. Between elements as metals, the alkaline earth metals in group two Ed )! @ libretexts.org or check out our status page at https: //status.libretexts.org decomposition of alkaline! Are good reducing agents electrons makes the alkaline earth good metals reducing agents is luminous, was... Use of calcium hydride as a drying agent for organic solvents cations charge. Stucky, G. D. J. Organomet of increasing decomposition temperature of high springs. Magnesium and beryllium demonstrates the danger of drawing an overly rigorous distinction between elements as metals, nonmetals and. G. Speight, in, Sn, Tl, Pb, Bi, Po Grignard in! Covalent bonds rather than ionic ones but due to smaller size and greater charge and hence ionisation. |
# Depth and nonlinearity induce implicit exploration for RL
29 May 2018 · , , ·
The question of how to explore, i.e., take actions with uncertain outcomes to learn about possible future rewards, is a key question in reinforcement learning (RL). Here, we show a surprising result: We show that Q-learning with nonlinear Q-function and no explicit exploration (i.e., a purely greedy policy) can learn several standard benchmark tasks, including mountain car, equally well as, or better than, the most commonly-used $\epsilon$-greedy exploration. We carefully examine this result and show that both the depth of the Q-network and the type of nonlinearity are important to induce such deterministic exploration.
PDF Abstract
## Code Add Remove Mark official
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# How can I modify the mesh of a linked character?
I have a fun problem:
In character.blend I have an armature and a mesh with shapekeys inside a Group. In a new scene (pose.blend) I have Linked in Group. I then created a proxy (Make Proxy) and posed out the character. Now I'm noticing a small problem that I would like to fix for just this pose.blend file. How can I modify the mesh?
In my Outliner (under All Scenes) I have:
• character which is just an empty containing everything that was in the group, however I cannot select the mesh or armature
• character_proxy which is the armature that I was able to move and manipulate
Some things I've tried:
1. Making local the whole group. I still cannot select the mesh
2. Making local the objects in the outliner, the mesh is no where to be found and I cannot edit character (it is stuck in Object mode.)
3. Running Make Single User U on character or character_proxy, there is still no mesh, and the mesh is not selectable from Groups Outliner.
Now I did successfully do this (thanks Olson from #blender irc!):
1. Select character from the outliner and Make Proxy the mesh
2. Make the mesh local and then manipulate it.
This works but the problem is the mesh is in T-Pose (ignoring the armature.) I can 'fix' the pose but it is cumbersome (since it is not posed out.)
Any help would be appreciated :D! |
A report from an NSF workshop on design automation and theory, and a video
Carver Mead is famous for creating the “Mead-Conway” revolution with Lynn Conway. In 1980 they published a best selling textbook “Introduction to VLSI System Design” which allowed almost anyone to design a VLSI chip. Carver has won many awards including the National Medal of Technology, for this and related work.
Today I want to talk about the NSF workshop that is going on right now: Electronic Design Automation–Past, Present, and Future. One goal of the workshop is to get theorists working again on various aspects of design automation. Back in the 1980’s theory played a major part in design automation. Somehow that stopped years ago.
In the 1980’s it really was possible to design chips with their book as a guide. Really. My students at the time designed chips that actually worked. Dan Lopresti’s Princeton PhD. thesis was based, in part, on building a chip that solved the edit distance problem via a parallel algorithm. I have mentioned his work earlier in a discussion on this problem.
Talk Highlights: Avant Lunch
Jeannette Wing, the head of CISE, started the workshop off and said as clearly as possible:
“NSF is interested in long term, high risk, research that will change the world.”
I love her statement, and I believe that there are great opportunities for theory to help make this happen.
So far there have been a number of interesting talks:
${\bullet }$ Bill Joyner, the keynote speaker, who is from industry had some neat statistics on FOCS/STOC papers. In the 1980’s a major fraction, about ${30\%}$, of those papers were on VLSI and algorithms for design. In the last FOCS/STOC he counted that there was less than ${1\%}$. Quite a dramatic change.
${\bullet }$ Andreas Kuehlmann, also from industry, made two interesting points. First, he repeated a rule that I learned as a graduate student. Suppose someone says that X is true. Here is a simple test to see if this statement has any meaning. Think of not X: if this statement is obviously false or is obviously true, then X is meaningless. Pretty good rule.
Second, he reported on a really interesting experiment. He took a circuit ${C}$ and fed it to a tool that makes a chip out of the circuit. The size of the chip that the tool yields is a measure of its quality. He got some value. Then, he changed the circuit in a completely trivial manner: he just changed the names of the circuit nodes. That’s all. Next, he ran the tool again. Then, he changed the circuit names again, and ran the tool. And so on. He plotted all the values. The surprise was that the area values changed dramatically–even though the circuit was really the same. The audience all laughed except for me.
I thought he had discovered something positive, while he considered the behavior of the tool a problem. My hypothesis is that the tool is deterministic, but the random re-naming of the nodes essentially converts the tool into a random algorithm. Andreas’s experiments show that the minimum of several random runs of the tool are better than a single run. Clearly, we should think about adding randomness to other design tools.
${\bullet }$ Carl Seger, of Intel, had a nice number–a big number. He said that the just announced Intel processor has ${2.6}$ billion transistors. In 1980, when the Mead-Conway revolution started, the number of transistors was ${20,000}$. Carl says in ten years Intel chips will have over one trillion transistors. For theory research we want ${n}$ as large as possible. If we are going to work hard on a problem and change the running time by just a logarithmic factor, then for these size ${n}$‘s that really matters. Ever better — it would be terrific if we can reduce the running time of a quadratic algorithm to near linear time.
${\bullet }$ Edmund Clarke, Turing Award winner, talked about statistical checking of certain kinds of systems. It is unclear from his short talk exactly what he has in mind. However, his idea I believe is this: Imagine that you have a Markov type system. You do not want to know the exact probability ${p}$ of some event. Instead you want to tell if either ${p \le \theta_{1}}$ or ${p \ge \theta_{2}}$. He states he can use Bayesian analysis to handle this type of question with relatively few simulations. One area of application is not to digital logic, but to biological systems. This type of question seems to be the kind that theorist could make progress on.
${\bullet }$ Shaz Qadeer, of Microsoft Research, spoke on Heisenbugs–a term due to the late Jim Gray. These are bugs that when you try to find them, they go away. Clearly, the name comes from the Heisenberg uncertainty principle, due to Werner Heisenberg. The reason these bugs “go away” is that your test code tends to change the order that the program is executed.
Shaz has a system called, CHESS, that allows the scheduler of the system to help you find Heisenbugs. In distributed systems he can find certain “race conditions.” The main problem is the usual state explosion problem: if there are ${t}$ threads and ${n}$ steps, there are ${n^{t}}$ ways to execute the code. I think there are some neat theory type questions here.
In answering a question he said: “at some level I am lying.” Is this a paradox? Or is it just near lunch and I am distracted?
Talk Highlights: Après Lunch
It’s after lunch, a pretty nice buffet, and here are some highlights of the talks:
${\bullet }$ Jaijeet Roychowdhury gave a talk on analogy systems. The highlight of his talk was a realtime demonstration of how biological systems can become synchronized. He took three metronomes and started them ticking on a simple platform on the podium that was balanced on two soda cans. Initially they were all running at the same rate, but they were not in synch. In less than a minute they were all running completely in synch. Very cool demo. Clearly, the serious questions are how to use local behavior to create global behavior.
${\bullet }$ Jason Hibbeler, of IBM, talked about rule driven design. If you follow the rules, then we can make your design work. Traditionally, the rules were always monotone. For example, if two wires get too close, they will not work. However, if they are far apart they will always be okay. Now the fancy lithography that is used to make chips–I do not follow this–causes the rules to become non-monotone. This raises all sorts of new geometric algorithmic problems.
${\bullet }$ Igor Markov, of the University of Michigan, spoke on design automation and its connection to physics. In short, he is interested in the power and simulation of non-standard physical systems. Someone named Scott Aaronson was referred to when he talked about quantum computing: “Closed Timelike Curves Make Quantum and Classical Computing Equivalent” a paper with John Watrous.
Igor’s main idea is to try to create efficient simulations of new non-standard physics systems. Clearly, a very theory based talk with lots of connections to things theorists are interested in.
${\bullet }$ Mary Jane Irwin, of Penn State, stated as systems get made out of very small components, you need to be able to re-configure the system. This is needed to get around faults, for example. The system must be able to monitor itself and also control itself. She likes to call the system the “compute fabric.” Where do the sensors go? This is a design automation issue. Also she raised the issue of 3-dimensional fabrics, which will further stress existing design tools. This clearly raises many interesting geometric algorithmic questions.
${\bullet }$ David Pan, of University of Texas at Austin, talked about Moore’s Law. There have been many “lower bounds proofs” claimed over years for lithography limits. Computer chips are made via an optical process, since visible light’s wavelength is 193nm there were proofs that chip features could not get below 193nm. Now the processes are at 22nm or better. How it’s done is very clever, and of course it gets around the assumptions that were used in the lower bound proofs. This, I think, says something interesting about lower bound “proofs” in general.
Talk Highlights: Biology
It’s the end of the day and I am tired, so I cannot say that anything from here on will make sense.
${\bullet }$ Jim Heath, of Caltech, discussed how to manage personal biological data, and use the data for medical diagnosis and treatment. An evolving area and one that is ripe for algorithmic applications, he claims. The major data is DNA, mRNA, and proteins. Complete DNA per person will cost $1,000 soon; mRNA costs$1 per messenger RNA now; proteins costs about \$50 per measurement now. Interesting questions concern how to build network models of biological processes. A key is that most interactions are pairwise between proteins. He said that there is only one known example of three proteins that interact together. This raises the hope that powerful theory ideas could play a role here.
${\bullet }$ Chris Meyers, of University of Utah, spoke about synthetic biological circuits. There is a catalogue of biological “parts” that people have discovered. They seem to work by themselves inside certain bacteria. The next step is to try to use them to make more complex biological circuits. The parts tend to be extremely error prone, which raises interesting questions of how to design biological circuits: sounds like fault tolerance problem. The first Bio-Design conference is this July ${27^{th}}$.
${\bullet }$ Lou Scheffer, of industry, spoke on trying to figure out how the brain works. Modest goal. His approach is to take it apart and see how it is made. Basically, he is interested in reverse engineering, which is traditionally a method used by a company to figure out how another company’s chips work. Lou’s idea is to try to use reverse engineering methods from electronics to make biological reverse engineering work.
Open Problems
The main goal is to get theory more involved in the area of design automation. There seems to be many possible ways that theory could help.
Tomorrow is another day and as a group we will try to make progress on what to do next. I will report on that another time. |
how to put a symbol above a letter in word
I want to be able to put the "^" over a "p"- a p with a hat, basically. Click the font drop-down menu on the Home tab and select the "Symbol" font. Method 3: Insert from Symbols. Open a document in Word. Choose Windings , Windings 2 or Wingdings 3 as the font in Symbols tab, then scroll down to find the circled numbers. It is also the slowest one and is only recommended if you need to insert one or two accented letters. Insert>Symbols>More symbols. When you come to the place in the word where the macron goes, hold down the "Alt" key and type the numeric number for the letter you want. Then go to Insert tab, and click Symbol > More Symbolsâ¦. However, if you are not satisfied with this quick help, and would want to find out more ways to perform this task, read on. Change the Font into Wingdings in the popping out Symbol window, scroll down till the end and youâll find many different arrow symbols here. You can scroll through the entire list or select from the Subset drop-down list to view a specific type of symbols. These are steps to use symbolâs dialog box to insert any umlaut letter into your Word document. ... which is usually located just above the Tab key on your keyboard. How do I get letters with accent marks in Microsoft Word? Hit 'Alt X.' I need to put the caret mark over the e in Renee. https://www.wikihow.com/Insert-Symbols-in-an-MS-Word-Document Type the required letter. 2. For users who do not have specialized keyboards, letters with accent marks may only be created using shortcut keys or through the menu bar. 3. > > > Not all the letters of the alphabet in the "Insert symbol" function are > > > available with macrons (-) over them... is there a method of inserting a > > > macron over any letter of the alphabet in a similar manner in which one I have seen the example of the blue symbol in the picture in some advertisements. Step 1: Open your word processor and locate either Alt key on your keyboard. In the insert symbols, the only ones with the hat are the French vowels. It is a separate letter in Danish, Swedish, Norwegian, Finnish, North Frisian, Walloon, Chamorro, Lule Sami, Skolt Sami, Southern Sami, and Greenlandic alphabets. Select Close. I want to get a letter with vector\arrow sign above it. You can then long-tap or long-hold your mouse down on any given letter to find symbols and other alternate characters. If the \alpha technique turns it into a Greek alpha in Word, then you change to âunicode modeâ on your computer and type the unicode 0302 to get a circumflex above the last letter (or 0304 for a macron). a Chart appears, all data is given there, else. https://www.youtube.com/channel/UCmV5uZQcAXUW7s4j7rM0POg?sub_confirmation=1How to Put a Line Above a Letter or word in Word Tip: If you want to make your symbol larger or smaller, select it and use the Font Size setting. As with many things on computers, there is more than one way. Fortunately, there are alternative commands that do the same task differently that we can try and there are also other ways of using the same commands. When you are typing somthing on Microsoft Word and you need to type a word and have a symbol over a letter like this Ä how do you put that in. How to do it without using Equation Editor in Microsoft Word 2016? Select the desired accented character or symbol from the list of symbols. ... be more likely. Insert Symbol is perhaps the most common way to insert an accented letter into a Microsoft Word document. With the above table, you can insert the Pound symbol into Word. 1. This option is very useful when your keyboard is damaged or not functioning. However, I have searched a lot, but I was not able to find the name of this symbol in English. Type the numbers '0307' next to the letter, without a space between the letter and the numbers. On the Insert tab or the Insert drop-down, select the Symbol option. (Microsoft Word is the most popular word processing software in the world.) You can insert accented letters using Insert | Symbol or by using the built-in ⦠To insert a line over text, the normal Equation feature can help you to solve it, please do as this: 1. The number 41 appears above the number 104, and they are centered horizontally in relation to each other. I copied and pasted the c into this to show you what I mean. Tip. Use the Microsoft Word shortcut key Ctrl+'+ to insert a letter with an accent (grave) mark above it. MS Word has a large number of symbols that you can select and insert in your document. I'm using Microsoft 2007 if that makes a difference. 1. Just copy-paste this letter enye (ñ): ñ - small letter enye; Ñ - capital letter enye; This is the easiest way to put letter enye in Microsoft Word. How do I do that? When you find the symbol you want, double-click it. How to Insert Symbols Above Letters With the Keyboard. There are several ways to insert a tick symbol (otherwise known as a check mark or checkmark) into Microsoft Word, the methods we outline below are ⦠Windows 7 and Microsoft Office Word 2010 can insert accent symbols in a document. Insert or put a line over text with Equation function in Word. How do can type a Spanish alphabet that looks like a letter n with a ~ above it like when one types "nino" as in Spanish for "boy" in emails & in Word? Use the Font selector above the symbol list to pick the font you want to browse through. If the Greek letter is too tall, you will need to reduce the font size for theta to about 9, and then do the full 12 pt size circumflex. Fada Signs, or Accents (á, Á, ó, Ó, ú, Ú, í, Í, é, É). use ctrl + , "vovel" to get it in the form "à" You can select the straight arrow and place it over the letter. How to Put a Dot Above a Letter in Word 1. See screenshot: 3. In MS Word, an alternative is to type the letter requiring a macron, e.g. How do I do that? How to do it? Sometimes, the output doesnât come out the way some of us might expect or want. If you select Insert, then one of the options is "Symbol"; you will find accented characters on the dialog that appears. Additionally, it is part of the alphabets used for some Alemannic and Austro-Bavarian dialects of German. How to Put an Accent on a Letter of a Word Using Your Keyboard. the two dots above a letter represents two derivative of varible t. My method: \documentclass[UTF8]{ctexart} \usepackage{amsmath} \usepackage{amssymb} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{inputenc} $$\"{o} \mathaccent{o}$$ However,the latex says in the math environment ,I must use the \mathaccent. 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# Chapter 18.6: Spontaneity and Equilibrium
• Contributed by Anonymous
• LibreTexts
Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials
### Learning Objective
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously: ΔSuniv > 0, ΔGsys < 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. (For more information on the reaction quotient and the equilibrium constant, see Chapter 15.) Recall that if Q < K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q > K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. Table 18.6.1 summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. The relationship between ΔSuniv and ΔGsys was described in Section 18.5. In this section, we explore the relationship between the standard free energy of reaction (ΔG°) and the equilibrium constant (K).
Table 18.6.1 Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
*Spontaneous in the reverse direction.
## Free Energy and the Equilibrium Constant
Because ΔH° and ΔS° determine the magnitude of ΔG° (Equation 18.5.5), and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. As you learned in Section 18.5, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. Using higher math, the general relationship can be shown as follows:
$$\Delta G= V \Delta P -S \Delta T \tag{18.6.1}$$
If a reaction is carried out at constant temperature (ΔT = 0), then Equation 18.29 simplifies to
$$\Delta G= V \Delta P \tag{18.6.2}$$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important. What Equation 18.6.2 tells us is how the free energy changes as the pressure changes at constant temperature.
Assuming ideal gas behavior, we can replace the V in Equation 18.6.2 by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express ΔG in terms of the initial and final pressures (Pi and Pf, respectively) as in Equation 18.4.1.
$$\Delta G = \left (\dfrac{nRT}{P} \right ) \Delta P= nRT \dfrac{\Delta P}{P} = nRT ln \left (\dfrac{P_{f}}{P_{i}} \right )\tag{18.6.4}$$
To actually show the final step one has to consider infinitesmal changes in both the free energy and the pressure and then integrate both sides of the equation using calculus, which is beyond the scope of this course. If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
$$G - G^{o} = nRT ln\;P$$
This can be rearranged as follows:
$$G = G^{o} + nRT ln\;P\tag{18.6.5}$$
As you will soon discover, Equation 18.6.5 allows us to relate ΔG° and Kp. Any relationship that is true for Kp must also be true for K because Kp and K are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$$aA + bB \rightleftharpoons cC + dD \tag{18.6.6}$$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
$$\Delta G = \sum mG_{products} - \sum mG_{reactants} =\left ( cG_{C}+dG_{D} \right )- \left ( aG_{A}+bG_{B} \right )\tag{18.6.7}$$
Substituting Equation 18.6.5 for each term into Equation 18.6.7,
$$\Delta G = \left [\left (cG_{C}+cRTln\;P_{C} \right ) + \left (dG_{D}+dRTln\;P_{D} \right ) \right ] - \left [\left (aG_{A}+aRTln\;P_{A} \right ) + \left (bG_{B}+bRTln\;P_{B} \right ) \right ]$$
Combining terms gives the following relationship between ΔG and the reaction quotient Q:
$$\Delta G = \Delta G^{o}+ RTln\left ( \dfrac{P_{C}^{c}P_{D}^{d}}{P_{A}^{a}P_{B}^{b}} \right ) = \Delta G^{o}+ RTln\;Q \tag{18.6.8}$$
where ΔG° indicates that all reactants and products are in their standard states. In Chapter 15 "Chemical Equilibrium", you learned that for gases Q = Kp at equilibrium, and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and Kp for gases as follows:
$$0 = \Delta G^{o}+ RTln\;K_{P} \tag{18.6.9}$$
$$\Delta G^{o} = - RTln\;K_{P}$$
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then Kp = 1, and neither reactants nor products are favored: the system is at equilibrium.
### Note the Pattern
For a spontaneous process under standard conditions, Keq and Kp are greater than 1.
### Example 18.6.1
In Example 18.5.3, we calculated that ΔG° = −32.7 kJ/mol of N2 for the reaction N2(g)+3H2(g ⇌ 2NH3(g) This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: P(N2) = 2.00 atm, P(H2) = 7.00 atm, P(NH3) = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether products or reactants are favored
Strategy:
A Using the values given and Equation 18.6.8, calculate Q.
B Substitute the values of ΔG° and Q into Equation 18.6.8 to obtain ΔG for the reaction under nonstandard conditions.
Solution:
A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation 18.6.8. Substituting the partial pressures given, we can calculate Q:
$$Q=\dfrac{P_{NH_{3}}^{2}}{P_{N_{2}}P_{H_{2}}^{3}} =\dfrac{\left ( 0.021 \right )^{2}}{\left ( 2.00 \right )\left ( 7.00 \right )^{3}}=6.4\times 10^{-7}$$
B Substituting the values of ΔG° and Q into Equation 18.6.8,
$$\Delta G=\Delta G^{o}+RTln\;Q = -32.7\;kJ+\left [ \left (8.314\;\cancel{J}\cancel{K} \right )\left ( 373\;\cancel{K} \right )\left ( 1\;kJ/1000\;\cancel{kJ} \right ) ln \left ( 6.4\times 10^{-7} \right )\right ]$$
$$=-32.7\; kJ +\left ( -44\; kJ \right ) = -77 \; kJ/mol\;N_{2}$$
Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants.
Exercise
Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P(NO) = 0.0100 atm, P(O2) = 0.200 atm, and P(NO2) = 1.00 × 10−4 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer: −92.9 kJ/mol of O2; the reaction is spontaneous to the right as written, so products are favored.
### Example 18.6.2
Calculate Kp for the reaction of H2 with N2 to give NH3 at 25°C. As calculated in Example 18.5.3 , ΔG° for this reaction is −32.7 kJ/mol of N2.
Given: balanced chemical equation from Example 10, ΔG°, and temperature
Strategy:
Substitute values for ΔG° and T (in kelvins) into Equation 18.6.9 to calculate Kp, the equilibrium constant for the formation of ammonia.
Solution:
In Example 18.5.3 , we used tabulated values of ΔG° to calculate ΔG° for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation 18.6.9,
$$\Delta G^{o}=-RTln\:K_{P}$$
$$\dfrac{- \Delta G^{o}}{RT}=ln\:K_{P}$$
Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation,
$$ln\:K_{P}= -\dfrac{\left ( -32.7\;\cancel{kJ} \right )\left ( 1000\;\cancel{J}/\cancel{kJ} \right )}{8.314 \;\cancel{J}/\cancel{K}\left ( 298\; \cancel{K} \right )}=13.2$$
$$K_{P}= 5.4\times 10^{5}$$
Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in Chapter 15, however, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O2.
Although Kp is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of Kp and K in Chapter 15 and showed that they are related:
$$K_{P}= K\left ( RT \right )^{\Delta n} \tag{18.6.10}$$
where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so Kp = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation 18.36 can be written in a more general form:
$$\Delta G^{o}= -RT ln\; K \tag{18.6.11}$$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation 18.38 for the difference between Kp and K.Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation 18.26 and Equation 18.6.11 provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant:
$$\Delta G^{o}= \Delta H^{o} -T \Delta S^{o}= -RT ln\; K \tag{18.6.12}$$
Notice that K becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
### Note the Pattern
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible.
## Temperature Dependence of the Equilibrium Constant
The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation 18.6.12, which can be rearranged as follows:
$$ln\; K=\dfrac{\Delta H^{o}}{RT} + \dfrac{\Delta S^{o}}{R} \tag{18.6.13}$$
Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation 18.6.13 agrees with the qualitative predictions made by applying Le Châtelier’s principle, which we discussed in Chapter 15. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation 18.6.13 also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence.
If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K1 and K2 are the equilibrium constants for a reaction at temperatures T1 and T2, respectively. Applying Equation 18.6.13 gives the following relationship at each temperature:
$$ln\; K_{1}=\dfrac{\Delta H^{o}}{RT_{1}} + \dfrac{\Delta S^{o}}{R}$$
$$ln\; K_{2}=\dfrac{\Delta H^{o}}{RT_{2}} + \dfrac{\Delta S^{o}}{R}$$
Subtracting ln K1 from ln K2,
$$ln\; K_{2}-ln\; K_{1}= ln\dfrac{K_{2}}{K_{1}}=\dfrac{\Delta H^{o}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right ) \tag{18.6.13}$$
Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K1) allow us to calculate the value of the equilibrium constant at any other temperature (K2), assuming that ΔH° and ΔS° are independent of temperature.
### Example 18.6.3
The equilibrium constant for the formation of NH3 from H2 and N2 at 25°C was calculated to be Kp = 5.4 × 105 in Example 13. What is Kp at 500°C? (Use the data from Example 10.)
Given: balanced chemical equation, ΔH°, initial and final T, and Kp at 25°C
Strategy:
Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation 18.41 to obtain K2, the equilibrium constant at the final temperature.
Solution:
The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation 18.6.13 we obtain the following:
$$= \dfrac{\left ( -91.8\;\cancel{kJ} \right )\left ( 1000\;\cancel{J}/\cancel{kJ} \right )}{8.314\;\cancel{J}/\cancel{K} }\left ( \dfrac{1}{298\;\cancel{K}}- \dfrac{1}{773\;\cancel{K}}\right )= -22.8$$
$$\dfrac{K_{2}}{K_{1}}= 1.3 \pm \times 10^{-10}$$
$$K_{2}= \left ( 5.4\times 10^{5} \right )\left ( 1.3 \pm \times 10^{-10} \right )=7.0\times 10^{-5}$$
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise
In the exercise in Example 13, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the ΔHf° values in the exercise in Example 10 to calculate Kp for this reaction at 1000°C.
### Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or Kp > 1, and products are favored over reactants. If ΔG° > 0, then K or Kp < 1, and reactants are favored over products. If ΔG° = 0, then K or Kp = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
### Key Takeaway
• The change in free energy of a reaction can be expressed in terms of the standard free-energy change and the equilibrium constant K or Kp and indicates whether a reaction will occur spontaneously under a given set of conditions.
### Key Equations
Relationship between standard free-energy change and equilibrium constant
Equation 18.6.9: $$\Delta G^{o}=-RTln\;K$$
Temperature dependence of equilibrium constant
Equation 18.6.11: $$ln\;K = \dfrac{-\Delta H^{o}}{RT}+\dfrac{-\Delta S^{o}}{R}$$
Calculation of K at second temperature
Equation 18.6.11: $$ln\;\dfrac {K_{2}}{K_{1}} = \dfrac{-\Delta H^{o}}{R}\left ( \dfrac{1}{T_{2}}- \dfrac{1}{T_{1}}\right )$$
### Conceptual Problems
1. Do you expect products or reactants to dominate at equilibrium in a reaction for which ΔG° is equal to
1. 1.4 kJ/mol?
2. 105 kJ/mol?
3. −34 kJ/mol?
2. The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds?
3. What happens to the change in free energy of the reaction N2(g) + 3F2(g) → 2NF3(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids?
4. Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions?
### Numerical Problems
1. Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO2 and H2, as shown in the equation CO(g)+H2O(g) ⇌CO2(g)+H2(g), for which ΔH° = −41.0 kJ/mol and ΔS° = −42.3 J cal/(mol·K) at 25°C and 1 atm.
1. What is ΔG° for this reaction?
2. What is ΔG if the gases have the following partial pressures: P(CO) = 1.3 atm, P(H2O) = 0.8 atm, P(CO2) = 2.0 atm, and P(H2) = 1.3 atm?
3. What is ΔG if the temperature is increased to 150°C assuming no change in pressure?
2. Methane and water react to form carbon monoxide and hydrogen according to the equation CH4(g) + H2O(g) ⇌CO(g) + 3H2(g)
1. What is the standard free energy change for this reaction?
2. What is Kp for this reaction?
3. What is the carbon monoxide pressure if 1.3 atm of methane reacts with 0.8 atm of water, producing 1.8 atm of hydrogen gas?
4. What is the hydrogen gas pressure if 2.0 atm of methane is allowed to react with 1.1 atm of water?
5. At what temperature does the reaction become spontaneous?
3. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. CCl4(g) + 6 H2O(l) ⇌ CO2(g) +4HCl(aq); ΔG° = −377 kJ/mol
2. Xe(g) + 2F2(g) ⇌ XeF4(s); ΔH° = −66.3 kJ/mol, ΔS° = −102.3 J/(mol·K)
3. PCl3(g) + S ⇌ PSCl3(l); ΔGf°(PCl3) = −272.4 kJ/mol, ΔGf°(PSCl3) = −363.2 kJ/mol
4. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. 2KClO3(s) ⇌ 2KCl(s) + 3O2(g); ΔG° = −225.8 kJ/mol
2. CoCl2(s) + 6 H2O(l) ⇌6CoCl26 H2O(s); ΔHrxn°= −352 kJ/mol, ΔSrxn°= −899 J/(mol·K)
3. PCl3(g) + O2(g) ⇌ 2POCl3(g); (PCl3); ΔGf°(PCl3)= −272.4 kJ/mol, ΔGf°(POCl3) = −558.5 kJ/mol
5. The gas-phase decomposition of N2O4 to NO2 is an equilibrium reaction with Kp = 4.66 × 10−3. Calculate the standard free-energy change for the equilibrium reaction between N2O4 and NO2.
6. The standard free-energy change for the dissolution K4Fe(CN)6H2O(s) ⇌ 4K+(aq) + Fe(CN)64-(aq) + H2O(l) is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C?
7. Ammonia reacts with water in liquid ammonia solution (am) according to the equation
NH3(g) + H2O(am) ⇌ NH4+(am) + OH-(am) The change in enthalpy for this reaction is 21 kJ/mol, and ΔS° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)?
8. At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba2+] = [CO32−] = 5.08 × 10−5 M. Determine ΔG° for the dissolution of BaCO3.
9. Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is Pb3(PO4)2(s) + 4H+(aq) ⇌ 3Pb2+(aq) + 2H2PO4-(aq) for which log K = −1.80. What is ΔG° for this reaction?
10. The conversion of butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol.
1. What is the change in entropy for this conversion?
2. Based on structural arguments, are the sign and magnitude of the entropy change what you would expect? Why?
3. What is the equilibrium constant for this reaction?
11. The reaction of CaCO3(s) to produce CaO(s) and CO2(g) has an equilibrium constant at 25°C of 2 × 10−23. Values of ΔHf° are as follows: CaCO3, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO2, −393.5 kJ/mol.
1. What is ΔG° for this reaction?
2. What is the equilibrium constant at 900°C?
3. What is the partial pressure of CO2(g) in equilibrium with CaO and CaCO3 at this temperature?
4. Are reactants or products favored at the lower temperature? at the higher temperature?
12. In acidic soils, dissolved Al3+ undergoes a complex formation reaction with SO42− to form [AlSO4+]. The equilibrium constant at 25°C for the reaction Al3+(aq) + SO42-(aq) ⇌ AlSO4+(aq) is 1585.
1. What is ΔG° for this reaction?
2. How does this value compare with ΔG° for the reaction Al3+(aq) + F-(aq) ⇌ AlF2+(aq) for which K = 107 at 25°C?
3. Which is the better ligand to use to trap Al3+ from the soil?
1. −28.4 kJ/mol
2. −26.1 kJ/mol
3. −19.9 kJ/mol
1. 1.21 × 1066; equilibrium lies far to the right.
2. 1.89 × 106; equilibrium lies to the right.
3. 5.28 × 1016; equilibrium lies far to the right.
1. 13.3 kJ/mol
2. 5.1 × 10−21
3. 10.3 kJ/mol
1. 129.5 kJ/mol
2. 6
3. 6.0 atm
4. Products are favored at high T; reactants are favored at low T.
### Contributors
• Anonymous
Modified by Joshua B. Halpern
Thumbnail from Wonder-ing |
# Relationship between two definitions of pro-representable functors
Edit:
I'm pretty sure that my conjecture $$\operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A),$$ is true. To prove it, just use the construction of the colimit as a disjoint union modulo equivalence. It is then quite easy to set up a bijection explicitly. Therefore pro-representability in Mazur's sense implies pro-representability in the sense of the nLab. I suspect that the converse does not hold, and I intend to look for a counterexample when I have time. Until then, I will leave the question up, because I think it's interesting, and it doesn't get discussed much in the literature.
Original question:
Let $$\mathcal{C}$$ be the category of complete, local, Noetherian rings with residue field $$k$$ (where $$k$$ is fixed and finite). Let $$\mathcal{C}^0$$ be the full subcategory of rings that are also Artinian.
Mazur (and everyone else studying deformation rings, it seems) defines a functor $$F:\mathcal{C}^0 \to \textbf{Set}$$ to be pro-representable if there is some $$R\in \mathcal{C}$$ such that $$F(A) = \operatorname{Hom}(R, A)$$ as a functor $$A\mapsto \textbf{Set}$$.
The nLab defines a functor to be pro-representable if it is a filtered colimit of representables. I would like to show that these notions are equivalent in the case of functors $$\mathcal{C}^0\to \textbf{Set}$$.
For $$R \in \mathcal{C}$$, we have $$R = \varprojlim_i R/\mathfrak{m}^i$$, where $$\mathfrak{m}$$ is the maximal ideal of $$R$$. Thus, I would conjecture naively that $$\operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A),$$ since the $$R/\mathfrak{m}^i$$ are in $$\mathcal{C}^0$$. However, I cannot see why this should be the case (and indeed I suspect that it is not), since the $$\operatorname{Hom}$$ functor does not play nicely with limits in the first argument. Using the universal properties, I am able to construct an injective map
$$\operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A) \to \operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A)$$ in the obvious way. However, I am not able to show that this map is surjective.
• Ah, okay. I'll remove my comments to declutter this space a bit. – Jeroen van der Meer Feb 6 at 7:41
This isn't a full answer because I don't know enough about noetherian complete local rings, but it's too long for a comment. Essentially I try to isolate the key-points you would need to show that they're the same.
It's a very nice question. The point is that in many cases, pro-objects in a given category $$C$$ have a nice description in terms of a variant of $$C$$.
For instance, profinite groups are the same thing as pro-(finite groups). That is, topological groups that are topologically inverse limits of finite (discrete) groups are a model for pro-objects in the category of finite groups. So if you have a functor $$\mathbf{finGrp\to Set}$$, it is pro-representable in the nLab sense if and only if there is a profinite group $$G$$ such that it is isomorphic to $$\hom(G,-)$$. But there is work behind this proof, namely the equivalence I mentioned earlier.
This equivalence works for a lot of neat algebraic gadgets (although I have never thought about the generality in which it was true... that would be a very nice question too)
You can indeed always define a map $$\mathrm{colim}_i\hom(R/m^i,-)\to \hom(\lim_i R/m^i,-)$$ , and you can ask when is it an isomorphism ? Essentially, you're asking whether the objects of $$C^0$$ are cocompact in $$C$$.
As you point out, it's pretty easy to see that this is always injective, so the question is about surjectivity: does a map $$R\to A$$ factor through $$R/m^i$$ for some $$i$$ ? I don't know a lot about these rings, but iirc the map $$R\to A$$ has to be continuous for the adic topologies, and so because the maximal ideal in $$A$$ is nilpotent, some power of $$m$$ has to be sent to $$0$$, thus proving the claim. Maybe I'm mistaken here, but something like that has to be true for the claim to hold anyways (otherwise you could find a morphism where no power of $$m$$ gets mapped to $$0$$ and get a nonsurjectivity claim)
How about the converse ? Well if I'm not mistaken about the previous paragraph, and if the proof can be adapted to prove the more general statement that any map $$\lim_i R_i \to A$$ factors through some $$R_i$$, where all the $$R_i$$'s are in $$C^0$$ and $$A\in C^0$$, then it would follow that the canonical functor $$Pro(C^0)\to C$$ is fully faithful. Since clearly any object of $$C$$ is an inverse limit of objects in $$C^0$$, it would follow that it's an equivalence, in other words, that the two definitions of pro-representable agree, just like in the case of groups !
Ah, maybe I should mention : for the canonical functor to even exist, one would need $$C$$ to be closed under inverse limits which doesn't seem completely obvious at first glance, but is perhaps still true.
As a conclusion (of this incomplete answer), here are the points you should try to work out:
1- Does any map $$R\to A$$ in $$C$$ factor through some $$R/m^i\to A$$ if $$A\in C^0$$ ? (probably yes, by some continuity things)
2- Does $$C$$ have inverse limits, and if so, is $$R\cong \lim_i R/m^i$$ in $$C$$ ? (be careful: this claim is true in the category of commutative rings by definition of complete, but it's not necessarily true that limits in $$C$$ are computed on the underlying rings ! )
If the answer to those 2 is "yes", then by the first part of 2- you get a canonical functor $$Pro(C^0)\to C$$, which is fully faithful by 1-, and is therefore an equivalence by the second part of 2-.
COnversely, if you can disprove any of 1- or 2-, it follows that $$Pro(C^0)\not\simeq C$$, and so the two definitions don't agree. In any case, you will get an answer to your question by solving 1 and 2.
• Thanks for your help! I'm pretty sure $(1)$ is true because the local ring homomorphism takes $m_R$ into the maximal ideal $m_A$ of $A$. Since $A$ is Artin local we have $m_A^n = 0$ for some $n$ (by Nakayama's Lemma), and so $m_R^n$ is in the kernel of the homomorphism. I currently suspect that $(2)$ is not true, because I have found some lemmas that gives sufficient conditions for the inverse limit to exist, which implies that it might not always be the case. – Qwertiops Feb 7 at 19:56
• Ah you're right, that's what I was calling "continuity" but it's even easier than that it's just locality. Then that means you get a fully faithful functor $C\to Pro(C^0)$, which sends certain types of limits (namely it sends $R$ to $\lim_i R/m^i$ in the latter category), whose image contains $C^0$ but it might not contain all pro-objects : if you're right about 2, that would mean that the nLab definition is more general (although the Mazur-et al. definition is fully faithfully embedded in it) – Maxime Ramzi Feb 7 at 20:09 |
# Cache Logic / Operations¶
When an object in cache (or potentiallyin cache) is active in a transaction, it is tracked by an ODE, which is an instance of OpenDirEntry [1]. For any cache oject the first operation is always a read.
Reading starts with Cache::open_read. This attempts to get the stripe lock and then probes the open directory entries and the stripe directory. A CacheVC instance is created if either the lock is missed or there is a hit in the hit. The other case indicates a valid determination that the read is a a miss.
The next step depends on the result. If the lock was missed CacheVC::stateProbe is next, which waits for the stripe lock and then does the cache probe and goes on to the same as Cache::open_read. If a OpenDirEntry was found it is checked for being valid. If not, or a OpenDirEntry wasn’t found then one is created and the CacheVC goes to the CacheVC::stateReadFirstDoc which loads the first doc fragment from disk. If another CacheVC instance is already reading from disk then the CacheVC waits for the read to complete.
State machine.
## Lock Waiting¶
To avoid the current issues with excessive event flows when many CacheVC instances are waiting for locks along with many failed lock attempts, the Vol and OpenDirEntry maintain lock waits lists. These are organized differently due to the different natures and access patterns for the classes. In both cases the wait lists are lists of Traffic Server events that refer to CacheVC instances. This indirection is critical in order to avoid problems if a CacheVC is destroyed while waiting for a lock. It can then cancel the event without leaving dangerous dangling pointers. The lock owner can then discard canceled events without any further dereferencing. This is the same logic used by the core event loop.
Vol uses thread local storage to store wait lists. For each thread there is a vector of event lists. Each Vol has a numeric identifier which also serves as its index in the vector. This is possible because the set of Vol instances is determined by the cache configuration and therefore fixed at process start time. Content for Vol locks is, based on experimental measurements, the lock with the highest contention and therefore is worth special casing for performance. The locks are spread among threads for a few reasons.
The number of waiting CacheVC instances can be large and having them all dispatch in one pass could easily introduce excessive latency in to whichever event loop gets unlucky. Conversely breaking up thundering herds somewhat from the cache point of view is also beneficial.
Waiting CacheVC instances can be dispatched on their preferred thread. This is more important for Vol interaction as those are much close to the HTTP state machine which prefers to always run on a single thread.
OpenDirEntry uses a single global atomic list of events per instance. It is not feasible to use thread local storage because instances of OpenDirEntry are created and destroyed frequently and the corresponding number of instances fluctuates over a large range. The load distribution issue is of lesser importance because of the (usually) much larger number of instances which spreads the load when the instances become available. It is unfortunate this causes CacheVC instances to dispatch on other threads but the OpenDirEntry interactions tend to be less closely coupled to HTTP state machine instances. |
Sum of Square Numbers
I was trying to solve a task on leetcode, where I need to find if a positive integer is the sum of squares of two other integers. My solution is below, but it fails due to exceeding time limit.
var judgeSquareSum = function(c) {
let output = false;
for(let i = c; i >= 0; i--) {
for(let y = 0; y <= i; y++) {
let sum = i*i + y*y;
if (sum == c)
output = true;
}
}
return output;
};
How can I make this so that is efficient, and without two iterations?
• Try to reduce the amount of computation, e.g. with the formula for a sum of n natural numbers trans4mind.com/personal_development/mathematics/series/… – CiaPan Dec 5 '17 at 12:44
• Does 0 count as a integer, eg 0 * 0 + 0 * 0 === 0 or 0 * 0 + 3 * 3 === 9 ? – Blindman67 Dec 5 '17 at 13:00
• Apparently, it does. – Leff Dec 5 '17 at 13:01
• Well then the very first test is if(Math.sqrt(c) % 1 === 0) {return true} will eliminate many iterations. Also a good number to start searching down from would be Math.floor(Math.sqrt(c)) as it cant be greater than that. – Blindman67 Dec 5 '17 at 13:11
• – Martin R Dec 5 '17 at 13:18
You restrict the scope of the variables to the nearest enclosing scope with let, which is good.
The use of whitespace is not consistent: for( vs if (.
The variable names seem a be arbitrary: Why i and y, and not a and b as in the LeetCode problem description?
The first you can do to speed things up is to "early return":
var judgeSquareSum = function(c) {
for(let i = c; i >= 0; i--) {
for(let y = 0; y <= i; y++) {
let sum = i*i + y*y;
if (sum == c)
return true;
}
}
return false;
};
There is no need to continue with the loops if a decomposition into a sum of two squares has been found.
Now note that one integer in the decomposition determines the other. If we have a test candidate $a$ for $c = a^2 + b^2$ then it suffices to check if $\sqrt{c - a^2}$ is an integer. Also we can assume that $a \le b$, which restricts the range of $a$:
var judgeSquareSum = function(c) {
let aMax = Math.sqrt(c/2);
for (let a = 0; a <= aMax; a++) {
let b = Math.sqrt(c - a * a);
if (b === Math.round(b)) {
return true;
}
}
return false;
};
Only one loop instead of two nested loops now!
Further improvement is possible with the help of mathematics. As stated in Sum of two squares theorem,
An integer greater than one can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to 3 (mod 4) raised to an odd power.
So we need to determine if in the prime factorization $$c = p_1^{e_1} p_2^{e_2} \cdots p_n^{e_n}$$ there is a term $p_j^{e_j}$ such that $p_i = 3 \bmod 4$ and $e_j$ is odd.
This is more code, but very efficient for for large numbers, because
• the range of prime candidates $p$ can be bounded by $\sqrt c$, and
• once a prime factor is found, $c$ can be divided by that factor, thus reducing the remaining number of iterations.
Here is a possible implementation, with more explaining comments inline:
var judgeSquareSum = function(c) {
// Handle 0, 1, and 2 directly:
if (c <= 2) {
return true
}
// Remove even factors:
while (c % 2 === 0) {
c /= 2;
}
// Find odd prime factors:
for (let p = 3; p <= Math.sqrt(c); p += 2) {
if (c % p === 0) {
// Determine exponent e of p in c:
let e = 0;
while (c % p === 0) {
e += 1;
c /= p; // Remove this prime factor
}
// Is it a prime congruent 4 (mod 3), raised to an odd power?
if (p % 4 === 3 && e % 2 === 1) {
return false;
}
}
}
// If c > 1 at this point then it is another prime factor:
if (c % 4 === 3) {
return false;
}
return true;
};
• A small change I would add is pre-calculating Math.sqrt(c/2) rather than calculating it on every iteration. – Marc Rohloff Dec 5 '17 at 17:29
• @MarcRohloff: You are right, thanks. (The final version intentionally re-calculates the bound because c becomes smaller during the iteration.) – Martin R Dec 5 '17 at 18:18
• Some optional untested 32-bit micro-optimizations to reduce the number of slow division operations: p % 2p & 1, p % 4p & 3, c % p followed by c /= p → might be able to get rid of the modulo by dividing first and checking the result for isInteger – le_m Dec 5 '17 at 20:03 |
# Find values of n for which expression is integer
I have an expression for which I want ot find values of n that will make it an integer. Here is the expression:
(n^(3/2) - n)^(1/3)
or
n^(1/2) - n^(1/3)
I want to find all positive values of n for which the above expression will be an Integer. Can I do that with Mathematica?
Thanks.
• You can use "Reduce: "Reduce[{b == (n^(3/2) - n)^(1/3), n > 0, b [Element] Integers}]". This gives an answer using a root object. This is a shorthand for the roots of the enclose polynomial. If you want to transform it to radicals, you give to output to: "ToRadicals" – Daniel Huber Nov 8 '20 at 20:00
Using InverseFunction
f := Evaluate@InverseFunction[(#^(3/2) - #)^(1/3) &]
a = N[f /@ Range[3]]
{2.1479, 5.73535, 11.3776}
The above values will produce integers, e.g.
(#^(3/2) - #)^(1/3) & /@ a
{1., 2., 3.}
• Instead of f := Evaluate@InverseFunction[... you could simply write f = InverseFunction[.... – Roman Nov 8 '20 at 21:05
• @Roman Ah, yes. With Set rather than SetDelayed. – Chris Degnen Nov 8 '20 at 21:39 |
# Finding conditional probability of two independent poison rv
Let $X$ and $Y$ be identically independent Poisson r.v. with parameter $\lambda$. Find $P(X=k|X+Y=n )$.
### Attempt
By defitnion,
$$P(X=k|X+Y=n ) = P(X=k | Y=n-k) = \frac{ p_{XY}(k,n-k)}{p_Y(n-k)}$$
now since independence we have $p_X(k)p_Y(n-k)$ and so we have
$$P(X=k|X+Y=n ) = p_X(X=k)$$ but this is not the answer I should get. What am I doing wrong in this problem?
• Your "definition" is actually an incorrect statement. Maybe you switched with the correct $P(X+Y=n\mid X=k)=P(Y=n-k\mid X=k)$ (which however is not a definition). – drhab Apr 27 '18 at 11:19
It is not correct to state that $P(X=k\mid X+Y=n)=P(X=k\mid Y=n-k)$.
\begin{aligned}P\left(X=k\mid X+Y=n\right)P\left(X+Y=n\right) & =P\left(X=k\wedge X+Y=n\right)\\ & =P\left(X=k\wedge Y=n-k\right)\\ & =P\left(X=k\right)P\left(Y=n-k\right)\\ & =e^{-\lambda}\frac{\lambda^{k}}{k!}e^{-\lambda}\frac{\lambda^{n-k}}{\left(n-k\right)!}\\ & =e^{-2\lambda}\frac{\lambda^{n}}{n!}\binom{n}{k} \end{aligned} \tag1
Further if $X,Y\sim\mathsf{Poisson}\left(\lambda\right)$ are independent then $X+Y\sim\mathsf{Poisson}\left(2\lambda\right)$ leading to:$$P\left(X+Y=n\right)=e^{-2\lambda}\frac{\left(2\lambda\right)^{n}}{n!}\tag2$$
Based on $(1)$ and $(2)$ we find: $$P\left(X=k\mid X+Y=n\right)=\binom{n}{k}2^{-n}$$
• you are doing what I did in third equality. how is that version diferent then mine – James Apr 27 '18 at 11:20
• You are dividing by $p_Y(n-k)$ which is wrong. You should divide by $P(X+Y=n)$. – drhab Apr 27 '18 at 11:26
• im so confused right now – James Apr 27 '18 at 11:27
• What makes you think that $P(X=k\mid X+Y=n)=P(X=k\mid Y=n-k)$? That is your essential mistake. This wrong statement leads indeed to $P(X=k\mid X+Y=n)=P(X=k)$ which is evidently false. E.g. $P(X=n+1)>0$ but $P(X=n+1\mid X+Y=n)=0$. – drhab Apr 27 '18 at 11:30 |
## Section9.2Path integrals
Note: 2–3 lectures
### Subsection9.2.1Piecewise smooth paths
Let $$\gamma \colon [a,b] \to \R^n$$ be a function and write $$\gamma = (\gamma_1,\gamma_2,\ldots,\gamma_n)\text{.}$$ Suppose $$\gamma$$ is continuously differentiable, that is, it is differentiable and the derivative is continuous. In other words, there exists a continuous function $$\gamma^{\:\prime} \colon [a,b] \to \R^n$$ such that for every $$t \in [a,b]\text{,}$$ we have $$\lim\limits_{h \to 0} \frac{\snorm{\gamma(t+h)-\gamma(t) - \gamma^{\:\prime}(t) \, h}}{\sabs{h}} = 0\text{.}$$ We treat $$\gamma^{\:\prime}(t)$$ either as a linear operator (an $$n \times 1$$ matrix) or a vector, $$\gamma^{\:\prime}(t) = \bigl( \gamma_1^{\:\prime}(t), \gamma_2^{\:\prime}(t), \ldots, \gamma_n^{\:\prime}(t) \bigr)\text{.}$$ Equivalently, $$\gamma_j$$ is a continuously differentiable function on $$[a,b]$$ for every $$j=1,2,\ldots,n\text{.}$$ By Exercise 8.2.6, the operator norm of the operator $$\gamma^{\:\prime}(t)$$ is equal to the euclidean norm of the corresponding vector, so there is no confusion when writing $$\snorm{\gamma^{\:\prime}(t)}\text{.}$$
#### Definition9.2.1.
A continuously differentiable function $$\gamma \colon [a,b] \to \R^n$$ is called a smooth path or a continuously differentiable path 1 if $$\gamma$$ is continuously differentiable and $$\gamma^{\:\prime}(t) \not= 0$$ for all $$t \in [a,b]\text{.}$$
The function $$\gamma \colon [a,b] \to \R^n$$ is called a piecewise smooth path or a piecewise continuously differentiable path if there exist finitely many points $$t_0 = a < t_1 < t_2 < \cdots < t_k = b$$ such that the restriction $$\gamma|_{[t_{j-1},t_j]}$$ is smooth path for every $$j=1,2,\ldots,k\text{.}$$
A path $$\gamma$$ is a closed path if $$\gamma(a) = \gamma(b)\text{,}$$ that is if the path starts and ends in the same point. A path $$\gamma$$ is a simple path if either 1) $$\gamma$$ is a one-to-one function, or 2) $$\gamma|_{[a,b)}$$ is one-to-one and $$\gamma(a)=\gamma(b)$$ ($$\gamma$$ is a simple closed path).
#### Example9.2.2.
Let $$\gamma \colon [0,4] \to \R^2$$ be defined by
\begin{equation*} \gamma(t) := \begin{cases} (t,0) & \text{if } t \in [0,1],\\ (1,t-1) & \text{if } t \in (1,2],\\ (3-t,1) & \text{if } t \in (2,3],\\ (0,4-t) & \text{if } t \in (3,4]. \end{cases} \end{equation*}
The path $$\gamma$$ is the unit square traversed counterclockwise. See Figure 9.2. It is a piecewise smooth path. For example, $$\gamma|_{[1,2]}(t) = (1,t-1)$$ and so $$(\gamma|_{[1,2]})'(t) = (0,1) \not= 0\text{.}$$ Similarly for the other 3 sides. Notice that $$(\gamma|_{[1,2]})'(1) = (0,1)\text{,}$$ $$(\gamma|_{[0,1]})'(1) = (1,0)\text{,}$$ but $$\gamma^{\:\prime}(1)$$ does not exist. At the corners $$\gamma$$ is not differentiable. The path $$\gamma$$ is a simple closed path, as $$\gamma|_{[0,4)}$$ is one-to-one and $$\gamma(0)=\gamma(4)\text{.}$$
The definition of a piecewise smooth path as we have given it implies continuity (exercise). For general functions, many authors also allow finitely many discontinuities, when they use the term piecewise smooth, and so one may say that we defined a piecewise smooth path to be a continuous piecewise smooth function. While one may get by with smooth paths, for computations, the simplest paths to write down are often piecewise smooth.
Generally, we are interested in the direct image $$\gamma\bigl([a,b]\bigr)\text{,}$$ rather than the specific parametrization, although that is also important to some degree. When we informally talk about a path or a curve, we often mean the set $$\gamma\bigl([a,b]\bigr)\text{,}$$ depending on context.
#### Example9.2.3.
The condition $$\gamma^{\:\prime}(t) \not= 0$$ means that the image $$\gamma\bigl([a,b]\bigr)$$ has no “corners” where $$\gamma$$ is smooth. Consider
\begin{equation*} \gamma(t) := \begin{cases} (t^2,0) & \text{if } t < 0,\\ (0,t^2) & \text{if } t \geq 0. \end{cases} \end{equation*}
See Figure 9.3. It is left for the reader to check that $$\gamma$$ is continuously differentiable, yet the image $$\gamma(\R) = \bigl\{ (x,y) \in \R^2 : (x,y) = (s,0) \text{ or } (x,y) = (0,s) \text{ for some } s \geq 0 \bigr\}$$ has a “corner” at the origin. And that is because $$\gamma^{\:\prime}(0) = (0,0)\text{.}$$ More complicated examples with, say, infinitely many corners exist, see the exercises.
The condition $$\gamma^{\:\prime}(t) \not= 0$$ even at the endpoints guarantees not only no corners, but also that the path ends nicely, that is, it can extend a little bit past the endpoints. Again, see the exercises.
#### Example9.2.4.
A graph of a continuously differentiable function $$f \colon [a,b] \to \R$$ is a smooth path. Define $$\gamma \colon [a,b] \to \R^2$$ by
\begin{equation*} \gamma(t) := \bigl(t,f(t)\bigr) . \end{equation*}
Then $$\gamma^{\:\prime}(t) = \bigl( 1 , f'(t) \bigr)\text{,}$$ which is never zero, and $$\gamma\bigl([a,b]\bigr)$$ is the graph of $$f\text{.}$$
There are other ways of parametrizing the path. That is, there are different paths with the same image. The function $$t \mapsto (1-t)a+tb\text{,}$$ takes the interval $$[0,1]$$ to $$[a,b]\text{.}$$ Define $$\alpha \colon [0,1] \to \R^2$$ by
\begin{equation*} \alpha(t) := \bigl((1-t)a+tb,f((1-t)a+tb)\bigr) . \end{equation*}
Then $$\alpha'(t) = \bigl( b-a ,~ (b-a)f'((1-t)a+tb) \bigr)\text{,}$$ which is never zero. As sets, $$\alpha\bigl([0,1]\bigr) = \gamma\bigl([a,b]\bigr) = \bigl\{ (x,y) \in \R^2 : x \in [a,b] \text{ and } f(x) = y \bigr\}\text{,}$$ which is just the graph of $$f\text{.}$$
The last example leads us to a definition.
#### Definition9.2.5.
Let $$\gamma \colon [a,b] \to \R^n$$ be a smooth path and $$h \colon [c,d] \to [a,b]$$ a continuously differentiable bijective function such that $$h'(t) \not= 0$$ for all $$t \in [c,d]\text{.}$$ Then the composition $$\gamma \circ h$$ is called a smooth reparametrization of $$\gamma\text{.}$$
Let $$\gamma$$ be a piecewise smooth path, and $$h$$ a piecewise smooth bijective function with nonzero one-sided limits of $$h'\text{.}$$ The composition $$\gamma \circ h$$ is called a piecewise smooth reparametrization of $$\gamma\text{.}$$
If $$h$$ is strictly increasing, then $$h$$ is said to preserve orientation. If $$h$$ does not preserve orientation, then $$h$$ is said to reverse orientation.
A reparametrization is another path for the same set. That is, $$(\gamma \circ h)\bigl([c,d]\bigr) = \gamma \bigl([a,b]\bigr)\text{.}$$
The conditions on the piecewise smooth $$h$$ mean that there is some partition $$t_0 = c < t_1 < t_2 < \cdots < t_k = d\text{,}$$ such that $$h|_{[t_{j-1},t_j]}$$ is continuously differentiable and $$(h|_{[t_{j-1},t_j]})'(t) \not= 0$$ for all $$t \in [t_{j-1},t_j]\text{.}$$ Since $$h$$ is bijective, it is either strictly increasing or strictly decreasing. So either $$(h|_{[t_{j-1},t_j]})'(t) > 0$$ for all $$t$$ or $$(h|_{[t_{j-1},t_j]})'(t) < 0$$ for all $$t\text{.}$$
#### Proof.
Assume that $$h$$ preserves orientation, that is, $$h$$ is strictly increasing. If $$h \colon [c,d] \to [a,b]$$ gives a piecewise smooth reparametrization, then for some partition $$r_0 = c < r_1 < r_2 < \cdots < r_\ell = d\text{,}$$ the restriction $$h|_{[r_{j-1},r_j]}$$ is continuously differentiable with a positive derivative.
Let $$t_0 = a < t_1 < t_2 < \cdots < t_k = b$$ be the partition from the definition of piecewise smooth for $$\gamma$$ together with the points $$\{ h(r_0), h(r_1), h(r_2), \ldots, h(r_\ell) \}\text{.}$$ Let $$s_j := h^{-1}(t_j)\text{.}$$ Then $$s_0 = c < s_1 < s_2 < \cdots < s_k = d$$ is a partition that includes (is a refinement of) the $$\{ r_0,r_1,\ldots,r_\ell \}\text{.}$$ If $$\tau \in [s_{j-1},s_j]\text{,}$$ then $$h(\tau) \in [t_{j-1},t_j]$$ since $$h(s_{j-1}) = t_{j-1}\text{,}$$ $$h(s_{j}) = t_j\text{,}$$ and $$h$$ is strictly increasing. Also $$h|_{[s_{j-1},s_j]}$$ is continuously differentiable, and $$\gamma|_{[t_{j-1},t_j]}$$ is also continuously differentiable. Then
\begin{equation*} (\gamma \circ h)|_{[s_{j-1},s_{j}]} (\tau) = \gamma|_{[t_{j-1},t_{j}]} \bigl( h|_{[s_{j-1},s_j]}(\tau) \bigr) . \end{equation*}
The function $$(\gamma \circ h)|_{[s_{j-1},s_{j}]}$$ is therefore continuously differentiable and by the chain rule
\begin{equation*} \bigl( (\gamma \circ h)|_{[s_{j-1},s_{j}]} \bigr) ' (\tau) = \bigl( \gamma|_{[t_{j-1},t_{j}]} \bigr)' \bigl( h(\tau) \bigr) (h|_{[s_{j-1},s_j]})'(\tau) \not= 0 . \end{equation*}
Consequently, $$\gamma \circ h$$ is a piecewise smooth path. Orientation reversing $$h$$ is left as an exercise.
If two paths are simple and their images are the same, it is left as an exercise that there exists a reparametrization. Here is where our assumption that $$\gamma'$$ is never zero is important.
### Subsection9.2.2Path integral of a one-form
#### Definition9.2.7.
Let $$(x_1,x_2,\ldots,x_n) \in \R^n$$ be our coordinates. Given $$n$$ real-valued continuous functions $$\omega_1,\omega_2,\ldots,\omega_n$$ defined on a set $$S \subset \R^n\text{,}$$ we define a one-form to be an object of the form
\begin{equation*} \omega = \omega_1 \,dx_1 + \omega_2 \,dx_2 + \cdots + \omega_n \,dx_n . \end{equation*}
We could represent $$\omega$$ as a continuous function from $$S$$ to $$\R^n\text{,}$$ although it is better to think of it as a different object.
#### Example9.2.8.
\begin{equation*} \omega(x,y) := \frac{-y}{x^2+y^2} \,dx + \frac{x}{x^2+y^2} \,dy \end{equation*}
is a one-form defined on $$\R^2 \setminus \{ (0,0) \}\text{.}$$
#### Definition9.2.9.
Let $$\gamma \colon [a,b] \to \R^n$$ be a smooth path and let
\begin{equation*} \omega = \omega_1 \,dx_1 + \omega_2 \,dx_2 + \cdots + \omega_n \,dx_n , \end{equation*}
be a one-form defined on the direct image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Write $$\gamma = (\gamma_1,\gamma_2,\ldots,\gamma_n)\text{.}$$ Define:
\begin{equation*} \begin{split} \int_{\gamma} \omega & := \int_a^b \Bigl( \omega_1\bigl(\gamma(t)\bigr) \gamma_1^{\:\prime}(t) + \omega_2\bigl(\gamma(t)\bigr) \gamma_2^{\:\prime}(t) + \cdots + \omega_n\bigl(\gamma(t)\bigr) \gamma_n^{\:\prime}(t) \Bigr) dt \\ &\phantom{:}= \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt . \end{split} \end{equation*}
To remember the definition note that $$x_j$$ is $$\gamma_j(t)\text{,}$$ so $$dx_j$$ becomes $$\gamma_j^{\:\prime}(t) \, dt\text{.}$$
If $$\gamma$$ is piecewise smooth, take the corresponding partition $$t_0 = a < t_1 < t_2 < \ldots < t_k = b\text{,}$$ and assume the partition is minimal in the sense that $$\gamma$$ is not differentiable at $$t_1,t_2,\ldots,t_{k-1}\text{.}$$ As each $$\gamma|_{[t_{j-1},t_j]}$$ is a smooth path, define
\begin{equation*} \int_{\gamma} \omega := \int_{\gamma|_{[t_0,t_1]}} \omega \, + \, \int_{\gamma|_{[t_1,t_2]}} \omega \, + \, \cdots \, + \, \int_{\gamma|_{[t_{k-1},t_k]}} \omega . \end{equation*}
The notation makes sense from the formula you remember from calculus, let us state it somewhat informally: If $$x_j(t) = \gamma_j(t)\text{,}$$ then $$dx_j = \gamma_j^{\:\prime}(t) \, dt\text{.}$$
Paths can be cut up or concatenated. The proof is a direct application of the additivity of the Riemann integral, and is left as an exercise. The proposition justifies why we defined the integral over a piecewise smooth path in the way we did, and it justifies that we may as well have taken any partition not just the minimal one in the definition.
#### Example9.2.11.
Let the one-form $$\omega$$ and the path $$\gamma \colon [0,2\pi] \to \R^2$$ be defined by
\begin{equation*} \omega(x,y) := \frac{-y}{x^2+y^2} \,dx + \frac{x}{x^2+y^2} \,dy, \qquad \gamma(t) := \bigl(\cos(t),\sin(t)\bigr) . \end{equation*}
Then
\begin{equation*} \begin{split} \int_{\gamma} \omega & = \int_0^{2\pi} \Biggl( \frac{-\sin(t)}{{\bigl(\cos(t)\bigr)}^2+{\bigl(\sin(t)\bigr)}^2} \bigl(-\sin(t)\bigr) + \frac{\cos(t)}{{\bigl(\cos(t)\bigr)}^2+{\bigl(\sin(t)\bigr)}^2} \bigl(\cos(t)\bigr) \Biggr) \, dt \\ & = \int_0^{2\pi} 1 \, dt = 2\pi . \end{split} \end{equation*}
Next, let us parametrize the same curve as $$\alpha \colon [0,1] \to \R^2$$ defined by $$\alpha(t) := \bigl(\cos(2\pi t),\sin(2 \pi t)\bigr)\text{,}$$ that is $$\alpha$$ is a smooth reparametrization of $$\gamma\text{.}$$ Then
\begin{equation*} \begin{split} \int_{\alpha} \omega & = \int_0^{1} \Biggl( \frac{-\sin(2\pi t)}{{\bigl(\cos(2\pi t)\bigr)}^2+{\bigl(\sin(2\pi t)\bigr)}^2} \bigl(-2\pi \sin(2\pi t)\bigr) \\ & \phantom{=\int_0^1\Biggl(~} + \frac{\cos(2 \pi t)}{{\bigl(\cos(2 \pi t)\bigr)}^2+{\bigl(\sin(2 \pi t)\bigr)}^2} \bigl(2 \pi \cos(2 \pi t)\bigr) \Biggr) \, dt \\ & = \int_0^{1} 2\pi \, dt = 2\pi . \end{split} \end{equation*}
Now let us reparametrize with $$\beta \colon [0,2\pi] \to \R^2$$ as $$\beta(t) := \bigl(\cos(-t),\sin(-t)\bigr)\text{.}$$ Then
\begin{equation*} \begin{split} \int_{\beta} \omega & = \int_0^{2\pi} \Biggl( \frac{-\sin(-t)}{{\bigl(\cos(-t)\bigr)}^2+{\bigl(\sin(-t)\bigr)}^2} \bigl(\sin(-t)\bigr) + \frac{\cos(-t)}{{\bigl(\cos(-t)\bigr)}^2+{\bigl(\sin(-t)\bigr)}^2} \bigl(-\cos(-t)\bigr) \Biggr) \, dt \\ & = \int_0^{2\pi} (-1) \, dt = -2\pi . \end{split} \end{equation*}
The path $$\alpha$$ is an orientation preserving reparametrization of $$\gamma\text{,}$$ and the integrals are the same. The path $$\beta$$ is an orientation reversing reparametrization of $$\gamma$$ and the integral is minus the original. See Figure 9.4.
The previous example is not a fluke. The path integral does not depend on the parametrization of the curve, the only thing that matters is the direction in which the curve is traversed.
#### Proof.
Assume first that $$\gamma$$ and $$h$$ are both smooth. Write $$\omega = \omega_1 \, dx_1 + \omega_2 \, dx_2 + \cdots + \omega_n \, dx_n\text{.}$$ Suppose that $$h$$ is orientation preserving. Use the change of variables formula for the Riemann integral:
\begin{equation*} \begin{split} \int_{\gamma} \omega & = \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt \\ & = \int_c^d \left( \sum_{j=1}^n \omega_j\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \gamma_j^{\:\prime}\bigl(h(\tau)\bigr) \right) h'(\tau) \, d\tau \\ & = \int_c^d \left( \sum_{j=1}^n \omega_j\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) (\gamma_j \circ h)'(\tau) \right) d\tau = \int_{\gamma \circ h} \omega . \end{split} \end{equation*}
If $$h$$ is orientation reversing, it swaps the order of the limits on the integral and introduces a minus sign. The details, along with finishing the proof for piecewise smooth paths, is left as Exercise 9.2.4.
Due to this proposition (and the exercises), if $$\Gamma \subset \R^n$$ is the image of a simple piecewise smooth path $$\gamma\bigl([a,b]\bigr)\text{,}$$ then as long as we somehow indicate the orientation, that is, the direction in which we traverse the curve, we can write
\begin{equation*} \int_{\Gamma} \omega , \end{equation*}
without mentioning the specific $$\gamma\text{.}$$ Furthermore, for a simple closed path, it does not even matter where we start the parametrization. See the exercises.
Recall that simple means that $$\gamma$$ is one-to-one except perhaps at the endpoints, in particular it is one-to-one when restricted to $$[a,b)\text{.}$$ We may relax the condition that the path is simple a little bit. For example, it is enough to suppose that $$\gamma \colon [a,b] \to \R^n$$ is one-to-one except at finitely many points. See Exercise 9.2.14. But we cannot remove the condition completely as is illustrated by the following example.
#### Example9.2.13.
Suppose $$\gamma \colon [0,2\pi] \to \R^2$$ is given by $$\gamma(t) := \bigl(\cos(t),\sin(t)\bigr)\text{,}$$ and $$\beta \colon [0,2\pi] \to \R^2$$ is given by $$\beta(t) := \bigl(\cos(2t),\sin(2t)\bigr)\text{.}$$ Notice that $$\gamma\bigl([0,2\pi]\bigr) = \beta\bigl([0,2\pi]\bigr)\text{,}$$ and we travel around the same curve, the unit circle. But $$\gamma$$ goes around the unit circle once in the counter clockwise direction, and $$\beta$$ goes around the unit circle twice (in the same direction). See Figure 9.5.
Compute
\begin{equation*} \begin{aligned} & \int_{\gamma} -y\, dx + x\,dy = \int_0^{2\pi} \Bigl( \bigl(-\sin(t) \bigr) \bigl(-\sin(t) \bigr) + \cos(t) \cos(t) \Bigr) dt = 2 \pi,\\ & \int_{\beta} -y\, dx + x\,dy = \int_0^{2\pi} \Bigl( \bigl(-\sin(2t) \bigr) \bigl(-2\sin(2t) \bigr) + \cos(t) \bigl(2\cos(t)\bigr) \Bigr) dt = 4 \pi. \end{aligned} \end{equation*}
It is sometimes convenient to define a path integral over $$\gamma \colon [a,b] \to \R^n$$ that is not a path. Define
\begin{equation*} \int_{\gamma} \omega := \int_a^b \left( \sum_{j=1}^n \omega_j\bigl(\gamma(t)\bigr) \gamma_j^{\:\prime}(t) \right) dt \end{equation*}
for every continuously differentiable $$\gamma\text{.}$$ A case that comes up naturally is when $$\gamma$$ is constant. Then $$\gamma^{\:\prime}(t) = 0$$ for all $$t\text{,}$$ and $$\gamma\bigl([a,b]\bigr)$$ is a single point, which we regard as a “curve” of length zero. Then, $$\int_{\gamma} \omega = 0$$ for every $$\omega\text{.}$$
### Subsection9.2.3Path integral of a function
Next we integrate a function against the so-called arc-length measure $$ds\text{.}$$ The geometric picture we have in mind is the area under the graph of the function over a path. Imagine a fence erected over $$\gamma$$ with height given by the function and the integral is the area of the fence. See Figure 9.6.
#### Definition9.2.14.
Suppose $$\gamma \colon [a,b] \to \R^n$$ is a smooth path, and $$f$$ is a continuous function defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Then define
\begin{equation*} \int_{\gamma} f \,ds := \int_a^b f\bigl( \gamma(t) \bigr) \snorm{\gamma^{\:\prime}(t)} \, dt . \end{equation*}
To emphasize the variables we may use
\begin{equation*} \int_{\gamma} f(x) \,ds(x) := \int_{\gamma} f \,ds . \end{equation*}
The definition for a piecewise smooth path is similar as before and is left to the reader.
The path integral of a function is also independent of the parametrization, and in this case, the orientation does not matter.
#### Proof.
Suppose first that $$h$$ is orientation preserving and that $$\gamma$$ and $$h$$ are both smooth. Then
\begin{equation*} \begin{split} \int_{\gamma} f \, ds & = \int_a^b f\bigl(\gamma(t)\bigr) \snorm{\gamma^{\:\prime}(t)} \, dt \\ & = \int_c^d f\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \snorm{\gamma^{\:\prime}\bigl(h(\tau)\bigr)} h'(\tau) \, d\tau \\ & = \int_c^d f\Bigl(\gamma\bigl(h(\tau)\bigr)\Bigr) \snorm{\gamma^{\:\prime}\bigl(h(\tau)\bigr) h'(\tau)} \, d\tau \\ & = \int_c^d f\bigl((\gamma \circ h)(\tau)\bigr) \snorm{(\gamma \circ h)'(\tau)} \, d\tau \\ & = \int_{\gamma \circ h} f \, ds . \end{split} \end{equation*}
If $$h$$ is orientation reversing it swaps the order of the limits on the integral, but you also have to introduce a minus sign in order to take $$h'$$ inside the norm. The details, along with finishing the proof for piecewise smooth paths is left to the reader as Exercise 9.2.5.
As before, due to this proposition (and the exercises), if $$\gamma$$ is simple, it does not matter which parametrization we use. Therefore, if $$\Gamma = \gamma\bigl( [a,b] \bigr)\text{,}$$ we can simply write
\begin{equation*} \int_\Gamma f\, ds . \end{equation*}
In this case we also do not need to worry about orientation, either way we get the same integral.
#### Example9.2.16.
Let $$f(x,y) := x\text{.}$$ Let $$C \subset \R^2$$ be half of the unit circle for $$x \geq 0\text{.}$$ We wish to compute
\begin{equation*} \int_C f \, ds . \end{equation*}
Parametrize the curve $$C$$ via $$\gamma \colon [\nicefrac{-\pi}{2},\nicefrac{\pi}{2}] \to \R^2$$ defined as $$\gamma(t) := \bigl(\cos(t),\sin(t)\bigr)\text{.}$$ Then $$\gamma^{\:\prime}(t) = \bigl(-\sin(t),\cos(t)\bigr)\text{,}$$ and
\begin{equation*} \int_C f \, ds = \int_\gamma f \, ds = \int_{-\pi/2}^{\pi/2} \cos(t) \sqrt{ {\bigl(-\sin(t)\bigr)}^2 + {\bigl(\cos(t)\bigr)}^2 } \, dt = \int_{-\pi/2}^{\pi/2} \cos(t) \, dt = 2. \end{equation*}
#### Definition9.2.17.
Suppose $$\Gamma \subset \R^n$$ is parametrized by a simple piecewise smooth path $$\gamma \colon [a,b] \to \R^n\text{,}$$ that is $$\gamma\bigl( [a,b] \bigr) = \Gamma\text{.}$$ We define the length by
\begin{equation*} \ell(\Gamma) := \int_{\Gamma} ds = \int_{\gamma} ds . \end{equation*}
If $$\gamma$$ is smooth,
\begin{equation*} \ell(\Gamma) = \int_a^b \snorm{\gamma^{\:\prime}(t)}\, dt . \end{equation*}
This may be a good time to mention that it is common to write $$\int_a^b \snorm{\gamma^{\:\prime}(t)}\, dt$$ even if the path is only piecewise smooth. That is because $$\snorm{\gamma^{\:\prime}(t)}$$ is defined and continuous at all but finitely many points and is bounded, and so the integral exists.
#### Example9.2.18.
Let $$x,y \in \R^n$$ be two points and write $$[x,y]$$ as the straight line segment between the two points $$x$$ and $$y\text{.}$$ Parametrize $$[x,y]$$ by $$\gamma(t) := (1-t)x + ty$$ for $$t$$ running between $$0$$ and $$1\text{.}$$ See Figure 9.7. Then $$\gamma^{\:\prime}(t) = y-x\text{,}$$ and therefore
\begin{equation*} \ell\bigl([x,y]\bigr) = \int_{[x,y]} ds = \int_0^1 \snorm{y-x} \, dt = \snorm{y-x} . \end{equation*}
So the length of $$[x,y]$$ is the standard euclidean distance between $$x$$ and $$y\text{,}$$ justifying the name.
A simple piecewise smooth path $$\gamma \colon [0,r] \to \R^n$$ is said to be an arc-length parametrization if for all $$t \in [0,r]\text{,}$$ we have
\begin{equation*} \ell\bigl( \gamma\bigl([0,t]\bigr) \bigr) = t . \end{equation*}
If $$\gamma$$ is smooth, then
\begin{equation*} \int_0^t d\tau = t = \ell\bigl( \gamma\bigl([0,t]\bigr) \bigr) = \int_0^t \snorm{\gamma^{\:\prime}(\tau)} \, d\tau \end{equation*}
for all $$t\text{,}$$ which means that $$\snorm{\gamma^{\:\prime}(t)} = 1$$ for all $$t\text{.}$$ Similarly for piecewise smooth $$\gamma\text{,}$$ we get $$\snorm{\gamma^{\:\prime}(t)} = 1$$ for all $$t$$ where the derivative exists. So you can think of such a parametrization as moving around your curve at speed 1. If $$\gamma \colon [0,r] \to \R^n$$ is an arclength parametrization, it is common to use $$s$$ as the variable as $$\int_\gamma f \,ds = \int_0^r f\bigl(\gamma(s)\bigr) \,ds\text{.}$$
### Subsection9.2.4Exercises
#### Exercise9.2.1.
Show that if $$\varphi \colon [a,b] \to \R^n$$ is a piecewise smooth path as we defined it, then $$\varphi$$ is a continuous function.
#### Exercise9.2.2.
Finish the proof of Proposition 9.2.6 for orientation reversing reparametrizations.
#### Exercise9.2.4.
Finish the proof of Proposition 9.2.12 for
1. orientation reversing reparametrizations, and
2. piecewise smooth paths and reparametrizations.
#### Exercise9.2.5.
Finish the proof of Proposition 9.2.15 for
1. orientation reversing reparametrizations, and
2. piecewise smooth paths and reparametrizations.
#### Exercise9.2.6.
Suppose $$\gamma \colon [a,b] \to \R^n$$ is a piecewise smooth path, and $$f$$ is a continuous function defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ Provide a definition of $$\int_{\gamma} f \,ds\text{.}$$
#### Exercise9.2.7.
Directly using the definitions compute:
1. The arc-length of the unit square from Example 9.2.2 using the given parametrization.
2. The arc-length of the unit circle using the parametrization $$\gamma \colon [0,1] \to \R^2\text{,}$$ $$\gamma(t) := \bigl(\cos(2\pi t),\sin(2\pi t)\bigr)\text{.}$$
3. The arc-length of the unit circle using the parametrization $$\beta \colon [0,2\pi] \to \R^2\text{,}$$ $$\beta(t) := \bigl(\cos(t),\sin(t)\bigr)\text{.}$$
Note: Feel free to use what you know about sine and cosine from calculus.
#### Exercise9.2.8.
Suppose $$\gamma \colon [0,1] \to \R^n$$ is a smooth path, and $$\omega$$ is a one-form defined on the image $$\gamma\bigl([a,b]\bigr)\text{.}$$ For $$r \in [0,1]\text{,}$$ let $$\gamma_r \colon [0,r] \to \R^n$$ be defined as simply the restriction of $$\gamma$$ to $$[0,r]\text{.}$$ Show that the function $$h(r) := \int_{\gamma_r} \omega$$ is a continuously differentiable function on $$[0,1]\text{.}$$
#### Exercise9.2.9.
Suppose $$\gamma \colon [a,b] \to \R^n$$ is a smooth path. Show that there exists an $$\epsilon > 0$$ and a smooth function $$\widetilde{\gamma} \colon (a-\epsilon,b+\epsilon) \to \R^n$$ with $$\widetilde{\gamma}(t) = \gamma(t)$$ for all $$t \in [a,b]$$ and $$\widetilde{\gamma}^{\:\prime}(t) \not= 0$$ for all $$t \in (a-\epsilon,b+\epsilon)\text{.}$$ That is, prove that a smooth path extends some small distance past the end points.
#### Exercise9.2.10.
Suppose $$\alpha \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are piecewise smooth paths such that $$\Gamma := \alpha\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)\text{.}$$ Show that there exist finitely many points $$\{ p_1,p_2,\ldots,p_k\} \in \Gamma\text{,}$$ such that the sets $$\alpha^{-1}\bigl( \{ p_1,p_2,\ldots,p_k\} \bigr)$$ and $$\beta^{-1}\bigl( \{ p_1,p_2,\ldots,p_k\} \bigr)$$ are partitions of $$[a,b]$$ and $$[c,d]$$ such that on every subinterval the paths are smooth (that is, they are partitions as in the definition of piecewise smooth path).
#### Exercise9.2.11.
1. Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\alpha \colon [c,d] \to \R^n$$ are two smooth paths that are one-to-one and $$\gamma\bigl([a,b]\bigr) = \alpha\bigl([c,d]\bigr)\text{.}$$ Then there exists a smooth reparametrization $$h \colon [a,b] \to [c,d]$$ such that $$\gamma = \alpha \circ h\text{.}$$
Hint 1: It is not hard to show $$h$$ exists. The trick is to prove it is continuously differentiable with a nonzero derivative. Apply the implicit function theorem though it may at first seem the dimensions are wrong.
Hint 2: Worry about derivative of $$h$$ in $$(a,b)$$ first.
2. Prove the same thing as part a, but now for simple closed paths with the further assumption that $$\gamma(a) = \gamma(b) = \alpha(c) = \alpha(d)\text{.}$$
3. Prove parts a) and b) but for piecewise smooth paths, obtaining piecewise smooth reparametrizations.
Hint: The trick is to find two partitions such that when restricted to a subinterval of the partition both paths have the same image and are smooth, see the exercise above.
#### Exercise9.2.12.
Suppose $$\alpha \colon [a,b] \to \R^n$$ and $$\beta \colon [b,c] \to \R^n$$ are piecewise smooth paths with $$\alpha(b)=\beta(b)\text{.}$$ Let $$\gamma \colon [a,c] \to \R^n$$ be defined by
\begin{equation*} \gamma(t) := \begin{cases} \alpha(t) & \text{if } t \in [a,b], \\ \beta(t) & \text{if } t \in (b,c]. \end{cases} \end{equation*}
Show that $$\gamma$$ is a piecewise smooth path, and that if $$\omega$$ is a one-form defined on the curve given by $$\gamma\text{,}$$ then
\begin{equation*} \int_{\gamma} \omega = \int_{\alpha} \omega + \int_{\beta} \omega . \end{equation*}
#### Exercise9.2.13.
Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are two simple closed piecewise smooth paths. That is, $$\gamma(a)=\gamma(b)$$ and $$\beta(c) = \beta(d)$$ and the restrictions $$\gamma|_{[a,b)}$$ and $$\beta|_{[c,d)}$$ are one-to-one. Suppose $$\Gamma = \gamma\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)$$ and $$\omega$$ is a one-form defined on $$\Gamma \subset \R^n\text{.}$$ Show that either
\begin{equation*} \int_\gamma \omega = \int_\beta \omega, \qquad \text{or} \qquad \int_\gamma \omega = - \int_\beta \omega. \end{equation*}
In particular, the notation $$\int_{\Gamma} \omega$$ makes sense if we indicate the direction in which the integral is evaluated. Hint: See previous three exercises.
#### Exercise9.2.14.
Suppose $$\gamma \colon [a,b] \to \R^n$$ and $$\beta \colon [c,d] \to \R^n$$ are two piecewise smooth paths which are one-to-one except at finitely many points. That is, there exist finite sets $$S \subset [a,b]$$ and $$T \subset [c,d]$$ such that $$\gamma|_{[a,b]\setminus S}$$ and $$\beta|_{[c,d]\setminus T}$$ are one-to-one. Suppose $$\Gamma = \gamma\bigl([a,b]\bigr) = \beta\bigl([c,d]\bigr)$$ and $$\omega$$ is a one-form defined on $$\Gamma \subset \R^n\text{.}$$ Show that either
\begin{equation*} \int_\gamma \omega = \int_\beta \omega, \qquad \text{or} \qquad \int_\gamma \omega = - \int_\beta \omega. \end{equation*}
In particular, the notation $$\int_{\Gamma} \omega$$ makes sense if we indicate the direction in which the integral is evaluated.
Hint: Same hint as the last exercise.
#### Exercise9.2.15.
Define $$\gamma \colon [0,1] \to \R^2$$ by $$\gamma(t) := \Bigl( t^3 \sin(\nicefrac{1}{t}),\, t{\bigl(3t^2\sin(\nicefrac{1}{t})-t\cos(\nicefrac{1}{t})\bigr)}^2 \Bigr)$$ for $$t \not= 0$$ and $$\gamma(0) = (0,0)\text{.}$$ Show that
1. $$\gamma$$ is continuously differentiable on $$[0,1]\text{.}$$
2. Show that there exists an infinite sequence $$\{ t_n \}$$ in $$[0,1]$$ converging to 0, such that $$\gamma^{\:\prime}(t_n) = (0,0)\text{.}$$
3. Show that the points $$\gamma(t_n)$$ lie on the line $$y=0$$ and such that the $$x$$-coordinate of $$\gamma(t_n)$$ alternates between positive and negative (if they do not alternate you only found a subsequence, you need to find them all).
4. Show that there is no piecewise smooth $$\alpha$$ whose image equals $$\gamma\bigl([0,1]\bigr)\text{.}$$ Hint: Look at part c) and show that $$\alpha'$$ must be zero where it reaches the origin.
5. (Computer) If you know a plotting software that allows you to plot parametric curves, make a plot of the curve, but only for $$t$$ in the range $$[0,0.1]$$ otherwise you will not see the behavior. In particular, you should notice that $$\gamma\bigl([0,1]\bigr)$$ has infinitely many “corners” near the origin.
Note: Feel free to use what you know about sine and cosine from calculus.
The word “smooth” can sometimes mean “infinitely differentiable” in the literature.
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf |
Publication 2011 Issue No. 12 - December Abstract - Causal Inference on Discrete Data Using Additive Noise Models
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Causal Inference on Discrete Data Using Additive Noise Models
December 2011 (vol. 33 no. 12)
pp. 2436-2450
ASCII Text x Jonas Peters, Dominik Janzing, Bernhard Schölkopf, "Causal Inference on Discrete Data Using Additive Noise Models," IEEE Transactions on Pattern Analysis and Machine Intelligence, vol. 33, no. 12, pp. 2436-2450, December, 2011.
BibTex x @article{ 10.1109/TPAMI.2011.71,author = {Jonas Peters and Dominik Janzing and Bernhard Schölkopf},title = {Causal Inference on Discrete Data Using Additive Noise Models},journal ={IEEE Transactions on Pattern Analysis and Machine Intelligence},volume = {33},number = {12},issn = {0162-8828},year = {2011},pages = {2436-2450},doi = {http://doi.ieeecomputersociety.org/10.1109/TPAMI.2011.71},publisher = {IEEE Computer Society},address = {Los Alamitos, CA, USA},}
RefWorks Procite/RefMan/Endnote x TY - JOURJO - IEEE Transactions on Pattern Analysis and Machine IntelligenceTI - Causal Inference on Discrete Data Using Additive Noise ModelsIS - 12SN - 0162-8828SP2436EP2450EPD - 2436-2450A1 - Jonas Peters, A1 - Dominik Janzing, A1 - Bernhard Schölkopf, PY - 2011KW - Causal inferenceKW - regressionKW - graphical models.VL - 33JA - IEEE Transactions on Pattern Analysis and Machine IntelligenceER -
Jonas Peters, Max Planck Institute for Biological Cybernetics, Tübingen
Dominik Janzing, Max Planck Institute for Biological Cybernetics, Tübingen
Bernhard Schölkopf, Max Planck Institute for Biological Cybernetics, Tübingen
Inferring the causal structure of a set of random variables from a finite sample of the joint distribution is an important problem in science. The case of two random variables is particularly challenging since no (conditional) independences can be exploited. Recent methods that are based on additive noise models suggest the following principle: Whenever the joint distribution {\bf P}^{(X,Y)} admits such a model in one direction, e.g., Y=f(X)+N, N \perp\kern-6pt \perp X, but does not admit the reversed model X=g(Y)+\tilde{N}, \tilde{N} \perp\kern-6pt \perp Y, one infers the former direction to be causal (i.e., X\rightarrow Y). Up to now, these approaches only dealt with continuous variables. In many situations, however, the variables of interest are discrete or even have only finitely many states. In this work, we extend the notion of additive noise models to these cases. We prove that it almost never occurs that additive noise models can be fit in both directions. We further propose an efficient algorithm that is able to perform this way of causal inference on finite samples of discrete variables. We show that the algorithm works on both synthetic and real data sets.
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Index Terms:
Causal inference, regression, graphical models.
Citation:
Jonas Peters, Dominik Janzing, Bernhard Schölkopf, "Causal Inference on Discrete Data Using Additive Noise Models," IEEE Transactions on Pattern Analysis and Machine Intelligence, vol. 33, no. 12, pp. 2436-2450, Dec. 2011, doi:10.1109/TPAMI.2011.71 |
# Mean and Variance
• Jan 15th 2012, 04:05 PM
Marmy
Mean and Variance
Hi everybody,
I encountered this to me very confusing problem in the context of confidence intervals and pivotal quantities.
How can I compute the mean and the variance (marked yellow) with this information below...
Many thanks once more!
• Jan 16th 2012, 02:50 PM
Moo
Re: Mean and Variance
Hello,
The only random stuff in $\hat\theta_2$ is $x_i$. So if you treat everything else as a constant, it's pretty easy to compute the mean and the variance !
After that, it's just a matter of computing double sums :)
Let's try it !
• Jan 17th 2012, 07:25 AM
Marmy
Re: Mean and Variance
Thanks, I think I found out now! |
# Why Are Cargo Ships Backed Up If The US Economy Is Struggling Due To The Pandemic?
Since the start of the present pandemic, the US Congress has authorized over \$5 trillion in various forms of economic aid (Business Insider), including the just-signed$1.9 trillion American Rescue Plan. So things are pretty bad, right?
Then why are cargo ships queuing up outside the ports of Los Angeles and Long Beach (Axios)? The articles I've read (such as CNBC and WSJ) say the backup is caused by the combination of surging consumer demand and COVID-19 disruptions. I'm looking for data to corroborate that interpretation, because a "surging customer demand" narrative seems to contradict the "the economy is in a bad way" narrative.
So is consumer spending surging? If I am reading these FRED graphs right, personal spending on durable and nondurable goods spending are roughly 10% above pre-pandemic levels. (I'm eyeballing these on a smartphone.) So yes, demand could be described as "surging."
So then why the need for \\$1.9 trillion in new government spending? My belief and understanding was that the economy is struggling right now, but then how are we all paying for so much more stuff that our ports are clogged? (I know the pandemic has disproportionately hurt service industries; is that the answer? Spending on goods is fine but the lack of spending on services is dragging the economy down?)
(The COVID-19 disruptions part of the explanation also confuses me but that's less important, and less relevant to this forum. But from that CNBC article, 800 of 15,000 longshoremen have been "out of work due to COVID;" so if 5% of Longshoremen call in sick, the ports grind to a halt? There's got to be more to the story that I'm missing.)
• Perhaps some people are spending more while other are able to spend less. If I get a new cellphone and you can't afford rent, it may cancel out in total numbers, but it doesn't mean there isn't a struggle. It would be better if you could afford rent and I didn't get a new cellphone. Mar 15 '21 at 12:14
• Yes, I suspect the uneven impacts of the pandemic are at the root of my question. So what I am looking for are: (1) data corroborate the two narratives I identified, and (2) data to illustrate why those narratives are not actually contradictory. Mar 15 '21 at 14:04 |
### Post-quantum Efficient Proof for Graph 3-Coloring Problem
Ehsan Ebrahimi
##### Abstract
In this paper, we construct an efficient interactive proof system for the graph 3-coloring problem and shows that it is computationally zero-knowledge against a quantum malicious verifier. Our protocol is inline with the sketch of an efficient protocol by Brassard and Crepéau (FOCS 1986) that later has been elaborated by Kilian (STOC 1992). Their protocol is not post-quantum secure since its soundness property holds based on the intractability of the factoring problem. Putting aside the post-quantum security, we argue that Kilian's interactive protocol for the graph 3-coloring problem does not fulfill the soundness property even in the classical setting. In this paper, we propose an XOR-homomorphic commitment scheme based on the Learning Parity with Noise (LPN) problem and use it to construct an efficient quantum computationally zero-knowledge interactive proof system for the graph 3-coloring problem.
Available format(s)
Category
Cryptographic protocols
Publication info
Preprint. Minor revision.
Keywords
Efficient Interactive Proof SystemPost-quantum SecurityComputational Zero-knowledge
Contact author(s)
ehsan ebrahimi @ uni lu
History
2022-04-14: last of 2 revisions
See all versions
Short URL
https://ia.cr/2021/1286
CC BY
BibTeX
@misc{cryptoeprint:2021/1286,
author = {Ehsan Ebrahimi},
title = {Post-quantum Efficient Proof for Graph 3-Coloring Problem},
howpublished = {Cryptology ePrint Archive, Paper 2021/1286},
year = {2021},
note = {\url{https://eprint.iacr.org/2021/1286}},
url = {https://eprint.iacr.org/2021/1286}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content. |
# OpenFOAM® v1906: New and improved post-processing
27/06/2019
## Improvements in runTimePostProcessing
The runTimePostProcessing function object provides a direct VTK insitu visualization alternative to the ParaView/Catalyst interface for well-defined, high throughput workflows.
The revised function object adds many new features:
• parallel VTK rendering
• rendering of sampled surfaces that are stored in memory without needing to serialize to a file format
• direct rendering of patches and clouds without an intermediate sampling stage
• can now render vector arrows at the face centres of sampled surfaces or mesh patches.
• for more visually appealing images, the new smooth option can be used, which applies a cell-to-point interpolation of the face data to provide a smoother appearance.
• direct use of the VTK cutting plane algorithm
Tutorials
$FOAM_TUTORIALS/lagrangian/coalChemistryFoam/simplifiedSiwek$FOAM_TUTORIALS/incompressible/pimpleFoam/RAS/ellipsekkLOmega
$FOAM_TUTORIALS/incompressible/pisoFoam/LES/motorBike/motorBike$FOAM_TUTORIALS/incompressible/simpleFoam/windAroundBuildings
## New areaWrite function object for finiteArea
The new areaWrite function object supports writing finite area meshes and fields in standard surface output formats, e.g. Ensight, and VTK.
Source code
$FOAM_SRC/functionObjects/utilities/areaWrite Tutorials$FOAM_TUTORIALS/finiteArea/surfactantFoam/planeTransport
$WM_PROJECT_DIR/modules/avalanche/tutorials/wolfsgrube ## Updated forces and new force coefficient components The forces and forceCoeffs function objects now report side forces and yaw moments, and the input directions simplified by making use of co-ordinate systems. Users are now only required to specify an origin with two orthogonal directions, and can be added by the three options shown below: CofR (0 0 0); // Centre of rotation dragDir (1 0 0); liftDir (0 0 1); origin (0 0 0); e1 (1 0 0); e3 (0 0 1); // combinations: (e1, e2) or (e2, e3) or (e3, e1) coordinateSystem { origin (0 0 0); rotation { type axes; e1 (1 0 0); e3 (0 0 1); // combinations: (e1, e2) or (e2, e3) or (e3, e1) } } Source code$FOAM_SRC/functionObjects/forces/forces
$FOAM_SRC/functionObjects/forces/forceCoeffs Solver$FOAM_TUTORIALS/incompressible/simpleFoam/motorBike
## New Lamb vector function object
The new lambVector function object calculates the Lamb vector, defined as the cross product of a velocity vector and vorticity vector. The divergence of the Lamb vector provides a quantitative connection to the spatially localised instantaneous fluid motions, e.g. high- and low-momentum fluid parcels that have the capacity to affect the rate of change of momentum, and to generate forces such as drag.
An example of the usage involving the divergence computation of the lambVector is shown below:
lambVector1
{
type lambVector;
libs ("libfieldFunctionObjects.so");
field U;
}
div1
{
type div;
libs ("libfieldFunctionObjects.so");
field lambVector;
}
Plotting an iso-surface for the channel case shows that the resulting field is able to capture the turbulent content of the flow.
Source code
$FOAM_SRC/functionObjects/field/lambVector Tutorials$FOAM_TUTORIALS/incompressible/pimpleFoam/LES/channel395DFSEM
References
## New sampling method
A new line sampling method, patchEdgeSet, samples values on patch edges that intersect a user-defined surface. The surface is of the generic searchablePlane type. Typical examples are plane or triSurfaceMesh (triangulated surface)
In below figure the pressure on the top patch has been sampled with a triangulated sphere.
Usage is usually through the sets functionObject:
sets
{
type sets;
libs ("libsampling.so");
writeControl timeStep;
writeInterval 1;
fields ( p U);
interpolationScheme cellPoint;
setFormat vtk;
sets
(
patchEdge
{
type patchEdge;
axis x;
// Selected patches
patches (movingWall);
// Surface type
surfaceType searchablePlane;
// Additional info for the surface type
planeType pointAndNormal;
pointAndNormalDict
{
point (1.5 1.5 1.5);
normal (0.1 0 1);
}
// Reference point for ordering samples
origin (0 1 0);
}
);
}
Note that the resulting points are sorted according to the distance to the provided origin.
Source code
$FOAM_SRC/sampling/sampledSet/patchEdge ## New run-time triggers The run-time control function object was introduced in OpenFOAM-v3.0+ to provide a mechanism to terminate and write a calculation based on a set of run-time conditions. The set of conditions include: • average • equationInitialResidual • equationMaxIter • maxDuration • minMax • minTimeStep The v1906 release extends what happens when the conditions are satisfied by adding an option to emit ’triggers’. The triggers can be used to start other function objects to provide highly customisable workflows. The action to perform is controlled by the optional satisfiedAction entry, which can take the values of: • end : finalise the calculation. This is the default action to maintain backwards compatibility; and • setTrigger : set a trigger, defined as an integer value by the additional trigger entry. Function object time controls now have an optional controlMode entry to control when the object is active that can be set to: • time: active between object’s timeStart and timeEnd • trigger: active between object’s triggerStart and triggerEnd • timeOrTrigger: active between object’s timeStart and timeEnd OR triggerStart and triggerEnd • timeAndTrigger: active between object’s timeStart and timeEnd AND triggerStart and triggerEnd If the trigger is activated by satisfying a set of run-time conditions, all function objects that are set to start based on that trigger ’index’ will be activated. Putting this together, it is possible, e.g. to start reporting the minimum and maximum values of all fields and set the calculation to stop 100 iterations after the average the drag coefficient from a forces function object changes by less than 1e-3: runTimeControl1 { type runTimeControl; libs ("libutilityFunctionObjects.so"); conditions { condition1 { type average; functionObject forceCoeffs1; fields (Cd); tolerance 1e-3; window 20; windowType exact; } } satisfiedAction setTrigger; trigger 1; } runTimeControl2 { type runTimeControl; libs ("libutilityFunctionObjects.so"); controlMode trigger; triggerStart 1; conditions { condition1 { type maxDuration; duration 100; } } satisfiedAction end; } fieldMinMax { type fieldMinMax; libs ("libfieldFunctionObjects.so"); fields (".*") controlMode trigger; triggerStart 1; } Source code$FOAM_SRC/functionObjects/utilities/runTimeControl
Tutorials
\$FOAM_TUTORIALS/incompressible/simpleFoam/simpleCar
## Improvements in surface sampling and writing
The infrastructure for surface sampling and writing has received an extensive overhaul for improved functionality and flexibility. In previous versions the surface writers were somewhat awkward when being used within parallel code, but the worst part was that they were entirely stateless. This would be especially noticeable when sampling surfaces and writing in VTK format. The uncoordinated output meant that a separate file would be generated for each sampled field.
Surface writers can now be time-step aware, which allows generation of sampled VTK output with all sampled fields at that timestep.
Any parallel data reduction and any associated bookkeeping is now part of the surface writers themselves, which improves their re-usability and avoids unnecessary and premature data reduction at the sampling stage.
Different output formats are supported on a per-surface basis as well as the ability to optionally store the sampled surfaces and fields onto a registry for reuse by other function objects. It is therefore now possible to simply sample-and-store a surface and use that surface and its values at a later stage in a surfaceFieldValue function object to calculate operations such as min/max etc.
It is now possible to add a colour table to sampled surfaces written in x3d format, which makes for convenient direct import into rendering tools such as blender.
## New sample surface options
The sampledSurfaces dictionary entries are backwards compatible, but there are additional features:
• store
Request that sample surfaces be stored on an object registry for reuse. For example, by the functionObjectSurface rendering object in runTimePostProcessing.
• sampleOnExecute
Request a sample/store operation when the function object execute() is called.
• surfaces
This entry can now be a dictionary of entries instead of a list, which can be useful in combination with changeDictionary modification.
• interpolationScheme
is now optional. By default it is cellPoint.
• per-surface specification of surface format and storage requirements. This supports a flexible mix of user requirements. |
# Finding Time for Sales to drop below $10 \%$.
I'm doing Principles of Physics, $$10^{\text{th}}$$ edition by Resnick, Halliday, Walker. I tried doing the following question.
At the end of a year, a motor car company announces that the sale of pickup trucks are down by $$43.0 \%$$ for the year. If sales continue to decrease by $$43.0 \%$$ in each succeeding year, how long will it tale for the sales to fall $$10.0 \%$$ of the original number?
My Attempt:
Let initial sales quantity be $$a$$. So according to the situation presented we are looking for the smallest $$n$$ that satisfies the following inequality: \begin{aligned}a\left(1-\dfrac{43}{100}\right)^n&\le\dfrac{10a}{100}\\ n\log_{10}\left(\dfrac{67}{100}\right)&\le\log_{10}\left(\dfrac{1}{10}\right)\\ n&\ge \dfrac{-1}{\log_{10}(67/100)}\approx 5.75 \text{ years}\end{aligned}
It would be great if someone could check my reasoning. I don't have the solutions manual or the answer key for this question. Thanks
• How is this question from a physics book? – YuiTo Cheng Apr 28 at 13:38
• It's from the chapter Measurement which is more or less about Mathematical Modelling of presented situations. :) – Paras Khosla Apr 28 at 13:44
You have the correct reasoning and method. You made a small mistake though. You wrote $$1-\frac{43}{100}=\frac{67}{100}$$when in fact, it is $$\dfrac{57}{100}$$. After you correct this, it should be fine. |
Package hypcap Error: You have forgotten to use \caption
I get the following error:
<use myfile.jpg>
Package hyperref Info: bookmark level for unknown myfig defaults to 0 on input
! Package hypcap Error: You have forgotten to use \caption.
for the following document (see definition of myfig)
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{hyperref}
\usepackage{hypcap}
\begin{document}
\floatstyle{ruled}
\newfloat{myfig}{thp}{lop}
\floatname{myfig}{Text}
\hypcapredef{myfig}
\begin{myfig} \capstart
\includegraphics[scale=.5]{myfile}
\caption{myBook}
\end{myfig}
\end{document}
the final .dvi and .pdf output are ok, but I need to avoid any error to edit by lyx. I have seen a few posts on similar issues, but I am not sure how to fix things with my example. thks
You can use package caption. It also implements the function of hypcap, thus you do not need \hypcapredef and \capstart:
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{hyperref}
\usepackage{caption}
\begin{document}
\ref{fig:myfig}
\floatstyle{ruled}
\newfloat{myfig}{thp}{lop}
\floatname{myfig}{Text}
\begin{myfig}
\includegraphics[scale=.5]{myfile}
\caption{myBook}\label{fig:myfig}
\end{myfig}
\end{document}
• sorry for late accept – mario Jan 4 '15 at 10:38 |
# Tag Info
## New answers tagged measurement-problem
0
One can indeed see the reason behind the Heisenberg uncertainty principle as a mathematical propeties. The link between the physics and the mathematics is provided by the foundational work of Planck and de Broglie. They established the link between energy/momentum and frequency/k-vector. The general textbook on quantum mechanics therefore always starts by ...
1
The state is still $\psi (t=T)$ right after the rapid change of B field to y-direction because the system doesn't have enough time to response the change.
0
It is not required to put the state in density matrix form prior to performing the unitary operation $U$. However, you want to learn properties about the reduced density matrix after tracing out some of the system post-unitary, so you need a density matrix eventually. So you could do the unitary, and then construct the density matrix: $$\newcommand{\ket}[1]{... 0 This is just a quick stab, and it might show my ignorance more than anything else. Since you are working with a two, level spin system i'm actually giessting m,n=\pm\frac{1}{2} . You can then explicitly write your density matrix as$$ \rho\left(t\right)=\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-i\omega t}e^{-\...
0
Among other things, you wrote: ...and according to the result from the first measurement (either $|+ \rangle_s$ or $|- \rangle_s$) the apparatus chooses either path $P_{ s^{+} \rightarrow f^{+} }$ or path $P_{ s^{-} \rightarrow f^{+} }$. Because the chosen path depends on the result of a measurement, your "transformation" isn't one transformation but ...
1
The transition is due to the interaction of your small system that you want to experiment on with the environment (including the measurement apparatus). If the interaction is brief and the environment part of the system macroscopic, then what you see is the transition from the initial total wave function $\psi(t_0) = \psi_{\mathrm{in}} = \psi_{\mathrm{in,sys}... 0 Just wanted to add that the Simulation Hypothesis [1] [2] (SH) may suggest an answer. SH implies that there is a non-zero chance that we are in fact in a computer simulation. If this simulation is anything like the simulations we currently create, then it implies that the rules defined in the simulation code would appear to us as fundamental, irreducible ... 3 You're wrong in step 2. We don't go to$4\epsilon_0|\phi_2\rangle$. We go to$|\phi_2\rangle\$. Measuring an observable projects you into an eigenstate of that observable, but it does NOT mulitply your state by the eigenvalue of that observable. In most cases, as @WillO said, multiplying by the eigenvalue isn't even a measurable operation, since you're ...
2
By "Copenhagen interpretation", I assume that you mean the interpretation with instantaneous "collapse" one usually encounters in an introductory quantum theory course. Such collapse is a useful rule to do calculation but it is only a fiction. What typically happens is that the quantum system is correlated with the macroscopic measurement device and other ...
Top 50 recent answers are included |
## mscrosscountry Group Title Simplify. Express the quotient as a rational exponent. x to the 4 fifths power over x to the 1 third power 2 years ago 2 years ago
1. philips13
Is this it? $x^{4/5} \over x^{1/3}$
2. mscrosscountry
yes
3. brinethery
http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx Property 3. a^m/a^n = a^(m-n)
4. brinethery
Get both fractions so that they have a common denominator and then subtract them.
5. mscrosscountry
is it x 4/15?
6. philips13
$x^{4/5}x^{-1/3}$
7. brinethery
I think like x^7/15
8. mscrosscountry
x 3/2?
9. mscrosscountry
here are my options: x3/2 x7/15 x17/15 x4/15
10. philips13
bri is right the exponent is 4/5 - 1/3
11. mscrosscountry
so 3/2
12. philips13
remember how to subtract fractions? Least common denominator?
13. philips13
fancy way of saying make the numbers on the bottom the same before you can subtract
14. brinethery
What's 12/15 - 5/15?
15. mscrosscountry
7/15 okay i understand now
16. brinethery
Make sure you know how to add and subtract fractions, it's very important.
17. brinethery
good luck! |
# Homotopic cellular maps are cellularly homotopic
Let $$X$$ and $$Y$$ be CW complexes, and let $$f$$ and $$g$$ be homotopic cellular maps from $$X$$ to $$Y$$; that is, $$f(X^n) \subset Y^n$$ and $$g(X^n) \subset Y^n$$, where $$X^n$$ denotes the $$n$$-skeleton of $$X$$. How do I show that $$f$$ and $$g$$ are cellularly homotopic (homotopic via a homotopy that is itself a cellular map)?
My attempt. Consider the relative CW complex $$(X \times I, X \times \partial I)$$, and let $$h: f \simeq g$$. We may apply the cellular approximation theorem to $$h:(X \times I, X \times \partial I) \to (Y,Y)$$ to get a homotopy $$H: h \simeq h' \text{ rel } X \times \partial I$$ where $$h': X\times I \to Y$$ is cellular with $$h'_0=f$$ and $$h'_1=g$$.
But something is wrong—I have not used the fact that $$f$$ and $$g$$ are cellular!
## EDIT (after contemplating freakish's answer):
I got confused originally partially because of the subtle statement of the cellular approximation theorem in May's A Concise Course in Algebraic Topology:
Theorem (Cellular Approximation). Any map $$f: (X,A) \to (Y,B)$$ between relative CW complexes is homotopic relative to $$A$$ to a cellular map.
My original understanding of this result was flawed; if I applied the result as stated above to my above attempt, what I would really obtain is a homotopy $$H: h \simeq h' \text{ rel } X \times \partial I$$ as above but a cellular map $$h':(X \times I, X \times \partial I) \to (Y,Y)$$ instead (with $$h'_0=f$$ and $$h'_1=g$$ as above). This is not the same as a cellular map $$h': X \times I \to Y$$, as is made obvious by the following wrong proof:
False result. All maps between CW complexes are cellular.
oof. Let $$f:X \to Y$$ be any map. Then we may view it as a map $$f:(X,X) \to (Y,Y)$$, so $$f$$ is homotopic relative to $$X$$ to a cellular map by cellular approximation. That is, the homotopy is constant on $$X$$, so that $$f$$ is cellular.
Though, in a similar light, any map $$\varphi:(X,A) \to (Y,Y)$$ is (trivially) cellular as $$(Y,Y)^n=(Y,Y)^0=Y$$ and $$\varphi((X,A)^n) \subset Y$$ trivially. Which means my original attempt was pretty flawed.
Here then is a proof that works, with much detail so that I (hopefully) will understand it in the future:
Proof that works. Let $$f$$, $$g: X \to Y$$ be homotopic cellular maps, and let $$h: f \simeq g$$. We wish to find a cellular homotopy $$h': f \simeq g$$; that is, a homotopy $$h': X \times I \to Y$$ between cellular maps that is a cellular map itself. That is, we require that $$h'$$ sends the $$n$$-skeleton $$X^n \times \partial I \cup X^{n-1} \times I$$ of $$X \times I$$ into $$Y^n$$. Since cellular homotopies are between cellular maps, $$h'(X^n \times \partial I) \subset Y^n$$ automatically, so it suffices to show that $$h'(X^{n-1} \times I) \subset Y^n$$.
Regard $$h$$ as a map of relative CW complexes $$(X \times I, X^n \times \partial I) \to (Y, Y^n)$$. Then the cellular approximation theorem gives us a homotopy $$H: h \simeq h^n \text{ rel } X^n \times \partial I$$ such that $$h^n: (X \times I, X^n \times \partial I) \to (Y, Y^n)$$ is cellular with $$h^n_0|X^n=f|X^n$$ and $$h^n_1|X^n=g|X^n$$.
Since $$h^n$$ is cellular, it takes the relative $$n$$-skeleton $$X^n \times \partial I \cup X^{n-1} \times I$$ of $$(X \times I, X^n \times \partial I)$$ into $$Y^n$$. Thus $$h^n$$ defines the desired homotopy $$h'$$ on $$X^n$$ for each $$n \geq 1$$, and we may take the colimit to obtain $$h'$$. $$\square$$
Though it is nicer to just use Hatcher's version of the cellular approximation theorem.
But something is wrong—I have not used the fact that $$f$$ and $$g$$ are cellular!
Cellular Approximation Theorem: Every map $$f:X\to Y$$ of CW complexes is homotopic to a cellular map. If $$f$$ is already cellular on a subcomplex $$A\subseteq X$$ the homotopy may be taken to be stationary on $$A$$.
So in order to get that $$h'_0=f$$ and $$h'_1=g$$ you need to know that $$H$$ can be chosen to be stationary on $$X\times \partial I$$, which is a subcomplex. And this can be done if $$h$$ restricted to that subcomplex is cellular. And that requires $$f$$ and $$g$$ to be cellular. |
# Sum Of Numbers Divisible By 3
The sum of all single-digit replacements for z is 12. C++ sum of all integer divisible by a number with constructor destructor C++ //Write a program to find the number and T4Tutorials_Sum of all integer between 100 and 200 which are divisible by 9. Smallest 3 digit no. Next: Write a program in C++ to find LCM of any two numbers using HCF. And, in each iteration, the value of i is added to sum and i is incremented by 1. Sn = n/2[2a + (n – 1)d] = 51/2[200 + 50 × 4] = 51 × 200 = 10200. which is already divisible by 3. The sum of natural numbers upto 100, excluding those divisible by 5, is 4000. Now every whole number can be written as a product of prime numbers, and each prime number is either $2$, or an odd number of the form $4k+1$, or an odd number of the form $4k+3$; for example $$350=2\times 5^2\times 7=2\times (4+1)^2\times (4+3)$$. If this number is divisible by 2 and 5 it must be a multiple of 10. Some examples of numbers divisible by 9 are as follows. Proof that if the sum of digits of a natural number in decimal representation is divisible by 3 then the number is also divisible by 3. Next: Write a C program to find all numbers which dividing it by 7 and the remainder is equal to 2 or 3 between two given integer numbers. multiply the left column and the bottom row. Here is a look at the rules for 3, 6, and 9. There are m = (a - 1) / k numbers below a that are divisible by k (with integer division). Will 18 be divisible by 2×3=6? Yes, it is. The sum of n consecutive numbers = smallest number x n + 1 + 2 + 3 + + (n-1). To easily tell if a number is divisible by 3 in your head, just check if the sum of all the digits in the number is divisible by 3. Product of three consecutive natural numbers is always divisible by 6. Consider N1, N2, N3… be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3… and remainders R1, R2, R3… respectively. Here is the beginning list of numbers divisible by 3, starting. Example 2 List all the factors of 32. notebook January 13, 2014 § Divisibility rule for 3: If the sum of the digits is divisible by 3, then the number is divisible by 3. C++ sum of all integer divisible by a number with constructor destructor C++ //Write a program to find the number and T4Tutorials_Sum of all integer between 100 and 200 which are divisible by 9. In other words, the sum of 2 consecutive numbers is not divisible by 2. Divisibility Rule for 7. Step 4 : The formula to find the sum of 'n' terms in an arithmetic sequence is given by. A number 378015 is divisible by 3, because a sum of its digits 3 + 7 + 8 + 0 + 1 + 5 = 24, which is divisible by 3. Here, divisible rules for different numbers with examples under each divisibility rule is explained and generalized rules for division by any number is also included at the end. Sum of All 3 Digit Numbers Divisible by 7. Note: 927 is also divisible by 9 because the sum of the digits is divisible by 9, but it is not divisible by 6. Hence or otherwise find the sum of the numbers less than 100 which are not multiples of 3. The sum of 2 digits x and y is divisible by 7. The sum of numbers divisible by 3 or 5 between 1 and 9999999999 is 23333333331666666668 The sum of numbers divisible by 3 or 5 between 1 and 999999999999999 is. (Note: 9 and 3 don't have to be in the sum, they are divisible by 3. It is a four-digit whole number. So the sum of the digits is divisible by 3 but not 9, so the number is divisible by 3 but not 9. If its first term is 11, then find the number of terms. Input Format: Input consists of a single integer. Sample Input 1 : 21 Sample Output 1 : yes Sample Input 2: 19 Sample Output 2: no Code:. First determine whether the number is even. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 so the number 3627 is evenly divisible by 3. Sum of Natural Numbers Using while Loop. The number 345,546,711 is divisible by 3. The working will be little different from what it was in the problem Find if there exists a subset with sum divisible by m. Problem 3 – Divisibility test for 3. It is divisible by 5. Two digit numbers divisible by 3 are 12,15,18,21,24,……99. Next: Write a C program to find all numbers which dividing it by 7 and the remainder is equal to 2 or 3 between two given integer numbers. WriteLine("The sum of numbers divisible by 3 or 5 between 1 and {0} is {1}", c. Ok there's maybe one little simplification for the divisibility of. We can discover this easily using the following trick, which we where the latter sum is simply the sum of the digits of 3412. Convert a Comma Delimited String to Array in C#. Therefore, the largest number is 9. /* Divide each term by 3,we get. The for k = 3 there is a subarray for example … Continue reading →. And luckily you have a little tool in your toolkit where you know how to test for divisibility by 3 Well, you say I can just add up the digits If the sum of that is a multiple of 3 then this whole thing is a multiple of 3 So you say 4 plus 7 plus 9 plus 2 That's 11. Now every whole number can be written as a product of prime numbers, and each prime number is either $2$, or an odd number of the form $4k+1$, or an odd number of the form $4k+3$; for example $$350=2\times 5^2\times 7=2\times (4+1)^2\times (4+3)$$. And, in each iteration, the value of i is added to sum and i is incremented by 1. Answer: if the sum of digits is divisible by 3 ,the number is also divisible by 3. (Note: This rule can be repeated when needed). E:The conclusion of the statement is written before the hypothesis. divisible by 3. A number 378015 is divisible by 3, because a sum of its digits 3 + 7 + 8 + 0 + 1 + 5 = 24, which is divisible by 3. Given a number N. These are just our original digits 498. Check your proficiency in the given practice questions based on Number System. 21 divided by 3 is 7. Find the possible mistakes in the following Shamil’s Flow Table of the program to find the number and sum of all integer which are divisible by 9. So 7,309 is not divisible by 3 or 9. Try 1, then 2, then 3, and so on. applying Sn formula, we know that d is 3 and 4 respectively. is 10, now we have to find first 2 digit no which is divisible by 3,it is 12 and last 2 digit no. A number is divisible by 11 if and only if a sum of its digits, located on even places is equal to a sum of its digits, located on odd places, OR these sums are There are criteria of divisibility for some other numbers, but these criteria are more difficult and not considered in a secondary school program. If the number is divisible by 3, then add to the running sum. Numbers are divisible by 3 if the sum of all the individual digits is evenly divisible by 3. The sum of numbers divisible by 3 or 5 between 1 and 9999999999 is 23333333331666666668 The sum of numbers divisible by 3 or 5 between 1 and 999999999999999 is. Divisible by 3 as sum of digits (=27) is divisible by 3. Is there a short proof that the sum of the digits of $3^{1000}$ is a multiple of $7$ without using a computer?. So, division by powers 10 is guaranteed to work well in the decimal system, but division by 3 is not. So it is actually an "if. $\endgroup$ – mas Aug 29 '20 at 9:27 $\begingroup$ any insights, pls? $\endgroup$ – mas Aug 29 '20 at 10:17 $\begingroup$ @AlonYariv (1) Finding an exact solution to this variant --- or even the original --- subset sum problem is non-trivial for large sets of boxes. public class Demo {. A number is divisible by 3, if the sum of all the digits of the number is divisible by 3. The number 79154 is not divisible by 3 because the sum of its digits (7 + 9 + 1 + 5 + 4 = 26) is not divisible by 3. In divisibility rules(test) we find whether a given number is divisible by another number, we perform actual division and see whether the remainder is zero or not. LCM of 3, 4, 8: Since 8 is the highest number, it may be the LCM. So, the correct code is this:. Step 1 : The first 3 digit number divisible by 6 is 102. Try 1, then 2, then 3, and so on. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9. The sum of the digits in 21 is 3. As 59 is not wholly divisible by 3 the question is invalid. I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n. Example 2 List all the factors of 32. The difference of the sum of odd and even places is " 5″. The difference between the number and the sum are 99a + 9b, which is divisible by three and nine. 2] And use our formula for the sum of the natural numbers: [6. Next: Write a program in C++ to find LCM of any two numbers using HCF. Sn=n/2{a+an] Sn=165150 &Sn=123200. Because any number which follows the formula 9n + 3 or 9n + 6 violates the statement. The sum of those 8 digits is not divisible by three so the largest possible integer must use no more than 7 of them (since 3, 6 and 9 would be eliminated). number divisible by 9 is : 117. In the given series of numbers how many 7's are there which are immediately followed by 9 and immediately not preceeded by 5 ?. Given an array of random integers, find subarray such that sum of elements in the element is divisible by k. C Program to print the numbers which are not divisible by 2, 3 and 5. The divisibility criteria are a roundabout way to know if a number is divisible by another without directly doing the calculation. N numbers A number K Output Format A number representing the count of subarrays whose sum is divisible by K. 1, −1, n and − n are known as the trivial divisors of n. First, we used the For loop to iterate from 1 to maximum value (Here, number = 5). Numbers that are the sum of two squares. Test whether 8 is divisible by the other numbers 8 is not divisible by 3 completely. And the only thee digit number which fits the bill is 290. Step-by-step explanation: * Lets explain how to solve the problem - The number is divisible by 6 if it divisible by 2 and 3 - Any even number divisible by 2 - The number is divisible by 3 is the sum of its digits divisible by 3 * Now lets solve the problem - The number 24,z38 is divisible by 6. step 1 Address the formula, input parameters & values. Rule # 3: Divisibility by 4 A numbers is divisible by 4 if the number represented by its last two digits is divisible by 4. Product of three consecutive natural numbers is always divisible by 6. On dividing a number by 3 or 9 the remainder will be equal to that left from dividing the sum of digits of that number by 3 or 9; 7,309 has the sum of the numbers 7 + 3 + 0 + 9 = 19, which is divided without a remainder to neither 3 nor 9. These are just our original digits 498. View Replies View Related If Number Is Even - Divisible 3 Or Divisible By 4 Apr 7, 2014. The first line of the input consists of an integer c (1 = c = 200), the number of test cases. Whenever the remainder of i divided by 5 and 7 is equal to 0, i is printed. If the number 517*324 is completely divisible by 3, then the smallest whole number in the place of * will be Janu said: (Dec 29, 2010). Square of any number will never end with 2, 3, 7 and 8. TRUE, As you can take some example of any random number to test that. Divide number elementwise by matrix X, where X is made by converting number to string and subtracting 48 to go from ASCII values to numbers again. The divisibility criteria are a roundabout way to know if a number is divisible by another without directly doing the calculation. are known as natural numbers. - projecteuler_001. 1050 is divisible by 5 and it is But i did not understand the divisibility rule 11 example 4. From the mathematical point , one number is divisible by 3 if sum of its digits is divisible by 3 and it is valid the commutative property, so 3 + 6 + 2 is equal to 6 + 3 + 2 because you can commute elements in every position and still obtain the same result. number divisible by 9 is: 126. Use the divisibility rule for 9: the sum of the digits is 1 ! 1 ! 7 " 9. If digits 1, 2, 3,4, 5 are used then number of required numbers = 5! If digits 0, 1,2,4, 5 are used then first place from left can be filled in 4 ways and remaining 4 places can be. The number is divisible by 3. e DP[3][3] and replace it with value of sum of DP[2][3] and DP[2][1]. Proof that if the sum of digits of a natural number in decimal representation is divisible by 3 then the number is also divisible by 3. Output Format: Output consists of a single line. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667. For example, random. Sum of the digit is, 5 + 6 + 9 + 1 = 21 Divisible by 3, = 21/3 = 7. Then, check for all the numbers with above condition we know only 0. For a number to be divisible by nine, you can do the same thing, and add up all the digits and see if that number is divisible by 9. A number for which sum of all its factors is equal to twice the number is called a perfect number. Strictly by removing possibilities that cannot possibly work we are down to at most 7 digits. Multiplication and long division. 8937: 8+7=15. Input : N = 5 Output : 7 sum = 3 + 4 Input : N = 12 Output : 42 sum = 3 + 4 + 6 + 8 + 9 + 12. Then follow two lines per test case. In this problem, we will be given an array A[0N-1] and a number m and we have to find count of subsets whose sum is divisible by m. Convert a Comma Delimited String to Array in C#. This code will automatically calculate the divisible of a given number. Divisibility Tricks. A number for which sum of all its factors is equal to twice the number is called a perfect number. [2×102+(300−1)×3] = 165150 Therefore sum of all 3 digit number which are divisible by 3 is 165150. Example 2 List all the factors of 32. Not only are there divisibility tests for larger numbers, but there are more tests for the numbers we have shown. Smallest 3 digit no. I've found already that it's only true when the total. Step-by-step explanation: * Lets explain how to solve the problem - The number is divisible by 6 if it divisible by 2 and 3 - Any even number divisible by 2 - The number is divisible by 3 is the sum of its digits divisible by 3 * Now lets solve the problem - The number 24,z38 is divisible by 6. Two plus nine plus four plus three, two plus nine is 11, 11 plus four is 15, 15 plus three is 18, and 18 is definitely divisible by nine, so this is. Therefore, the largest number is 9. Example: Input: Enter array elements: 10 15 20 25 30 Number: 10 Output: Total elements divisible by 10 is 3 Program:. Then, check for all the numbers with above condition we know only 0. Therefore the sum of an odd number of odd numbers will be odd. Pythagorean triples. That means we can rewrite 10k (in the question) as 30y, which is divisible by 15. First of all, you want to check the numbers from 1 to 1000, so in the for declaration you must use the condition number. After 102, to find the next 3 digit number divisible by 6, we have to add 6 to 102. Online C Loop programs for computer science and information technology students pursuing BE, BTech, MCA, MTech, MCS, MSc, BCA, BSc. For example, random. Divisibility Rule for 11. The difference of the sum of odd and even places is " 5″. What does this mean?. divisible by other number. JAVA Program Numbers whose sum of digits is divisible by 3 represent numbers divisible by 3. The problem is to find the length of the longest subarray with sum of the elements divisible by the given value k. C Program to print the numbers which are not divisible by 2, 3 and 5. Why? Consider a 2 digit number 10*a + b = 9*a + (a+b). If the sum of the digits of a number is divisible by 3, then the number is divisible by 3. C# Program to Calculate sum of all numbers divisible by 3 in given range. Write a c program to find all numbers divisible by 5 between a range also calculate their sum. I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n. The pattern for multiples of $3$ is based on the sum of the digits. Square of any number will never end with 2, 3, 7 and 8. So to form a number of five-digit which is divisible by 3 we can remove either ‘O’ or ‘3’. Now take a number where the sum of Note: You can also use this method to prove that if the number is a multiple of 3, then the sum of its digits will be a sum of 3. As for a square number being a number multiplied by itself, that follows from it. is 10, now we have to find first 2 digit no which is divisible by 3,it is 12 and last 2 digit no. Random Number Between X and Y. The sum of the digits in 21 is 3. So you only have to check what $3$ digits sum to $6, 9, 12$ or $15$. The first term a = 3 The common difference d = 3 Total number of terms n = 25 step 2 apply the input parameter values in the AP formula Sum = n/2 x (a + T n) = 25/2 x (3 + 75) = (25 x 78)/ 2 = 1950/2 3 + 6 + 9 + 12 + 15 + 18 + 21 +. You have to find the count of subarrays whose sum is divisible by K. The problem “Subset with sum divisible by m” states that you are given an array of non-negative integers and an integer m. The sum of these multiples is 23. In either case, implies is a sum divisible by. Let P(n) be the statement "the sum of three consecutive whole numbers is always divisible by 3. Here is the solution: Function SumOfNum(ByVal x As Integer) As Integer Dim i As Integer SumOfNum = 0 For i = 1 To 100 If i Mod x = 0 Then SumOfNum = SumOfNum + i Else SumOfNum = SumOfNum End If Next i. Required sum=n/2(a+l)=18/2(10+95)=945. These numbers form an AP with first term a = 3 , common difference d = 3 and the last term l=99. If for some , is a sum divisible by , which was to be shown. This proves that a number less its digit sum is always a multiple of 3 (and 9 for that matter): Thus n - s is divisible by 3. the sum of all numbers divisible by 9 is: 351. array = {1, 2, 4} m = 3 True. by 25, we have to show that 5250 is divisible by two 5's one by one i. This also means that 348 is divisible by 3! The Divisibility Rule for 9. A number is divisible by 3, if the sum of all the. Now sum of the given six digits is 15 which is divisible by 3. " for checking divisibility of numbers having more digits by 3. Contains All of Contains Any of Does not contain any of Count of odd digits is Divisible by Sum of all digits is Digit sum of all digits is Consecutive digits in a sequence at most Consecutive digits in a. Here is a look at the rules for 3, 6, and 9. Only one of its digits is divisible by 3. These are just our original digits 498. So, you would calculate + + =. It can be written as 10a + b. class SumOfNumbersDivisibleByFiveAndSeven { public static void main(String s[]) { System. That program will print the output on screen and shows only those numbers which are divisible by 7. This divisibility test. Problem 1548. Input Format A number N arr1 arr2. Four-digit numbers, five-digit numbers, etc. An=a+[n-1]d. Write a c program to find all numbers divisible by 5 between a range also calculate their sum. April 21, 2014 clacy. this c program logic is very simple. Their sum is m * (m + 1) / 2 * k (by Gaussian sum formula, link to German wiki - they seem to like Gauß more). For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 so the number 3627 is evenly divisible by 3. Rule 1: Partition into 3 digit numbers from the right ( ). Problem 3 – Divisibility test for 3. I had to change all to the official currency symbol (ZAR), which I dont really want. WriteLine("Sum of Numbers "+sum) Find the perfect numbers between 1 and 500 in C#. in this video you see how to find sum of numbers that is divisible by 3 or 5 and display numbers that is divisible by 3 and 5 and total numbers that is not d. Five positive integers from 1 to 15 are chosen without replacement. Then follow two lines per test case. We provide sequences and series practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. What does this mean?. applying Sn formula, we know that d is 3 and 4 respectively. How can you tell if a number is divisible by 3? A. (ZX81 BASIC doesn't do this automatically: both sides of an OR are evaluated, even (c, 5); BigInteger answer15 = GetSumOfNumbersDivisibleByN(c, 15); Console. Thus we do not need to remove. A conditional statement is known as an If-then statement. 2: If the last number is even. 1050 is divisible by 5 and it is But i did not understand the divisibility rule 11 example 4. Examples Approach: To solve the problem, follow the below steps: Find the sum of numbers that are divisible by 3 upto N. It's one of the easiest methods to quickly find the sum of given number series. (a) ___ 6724. This number is divisible by 5, because its last digit is 5. Find the sum of the digits. Contains All of Contains Any of Does not contain any of Count of odd digits is Divisible by Sum of all digits is Digit sum of all digits is Consecutive digits in a sequence at most Consecutive digits in a. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory. Write a program that prints “yes” if the given integer is divisible by 2 or 3 and “no” otherwise. The sum of all numbers smaller than a divisible by 3 or 5 is the same as. Therefore sum is 288350. Consider N1, N2, N3… be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3… and remainders R1, R2, R3… respectively. The sum of the digits of the number is a multiple of 3. @ShadowRanger as per the task it requires the sum of numbers divisible by 3 or 5 within a given. For large numbers this rule can be applied again to the result. For example: Numbers 90, 150, 700 are divisible by 2, because they end in 0. For example, 7425 is divisible by 9, hence it is divisible by 3. BigInt numbers, to represent integers of arbitrary length. The sum of four consecutive numbers in an AP is 32 and the ratio of the product If the ratio of the sum of. In both programs, the loop is iterated n number of times. The first line of the input consists of an integer c (1 = c = 200), the number of test cases. First, we used the For loop to iterate from 1 to maximum value (Here, number = 5). The positive integers 1, 2, 3, 4 etc. What does this mean?. In this c program we need to find all numbers that are divisible by 5 means when we take mod of a number by 5 we have to get the reminder zero. in divisible by 4 n=224. We can discover this easily using the following trick, which we where the latter sum is simply the sum of the digits of 3412. public class Demo {. Divisible Sum Pairs. First determine whether the number is even. Problem 1548. The sum of the last four terms is 112. Hence, the number is an odd multiple of 3. 9 if the sum of digits is divisible by 9 (example: 234612, because 2+3+4+6+1+2 = 18 = 9 x 2); 10 if the last digit is 0 (example: 99990 ); 100 if the last 2 digits are 0 (example 987600); NOTE: If a number is divisible by 2 factors, it is also divisible by the product of these factors. The above list is an AP with first term, a = 10 and common difference, d = 2 Now, a n = 98 ⇒ a + n-1 d = 98 ⇒ 10 + n-1 2 = 98 ⇒ 2 n-1 = 88 ⇒ n-1 = 44 ⇒ n = 45 Now, sum of first n terms of an AP is S n = n 2 2 a + n-1 d ⇒ S 45 = 45 2 2 × 10 + 45-1 2 = 45 2 20 + 88 = 2430 The list of 2 digit numbers that are divisible by 3 is : 12. Notice incidentally, 3072, the last two digits number, 72 is divisible by 4. Test whether 8 is divisible by the other numbers 8 is divisible by 2 completely. The divisibility criteria are a roundabout way to know if a number is divisible by another without directly doing the calculation. Example 2: Program to calculate the sum of natural numbers using for loop. homeRandom Numbers. The sum of all single-digit replacements for z is 12. Now every whole number can be written as a product of prime numbers, and each prime number is either $2$, or an odd number of the form $4k+1$, or an odd number of the form $4k+3$; for example $$350=2\times 5^2\times 7=2\times (4+1)^2\times (4+3)$$. What is the probability that dear their sum is divisible by 3 ? Your smartphone is your. Numbers 12, 18, 30 are divisible by 6; Their sum 60 is also divisible by 6 If the addends are divisible individually by a number , their sum is divisible by that number too. If the two numbers are equal return a or b. Given a number N. 157,526: 157 × 3 + 526= 997 999: Add the digits in blocks of three from right to left. Two digit numbers divisible by 3 are 12,15,18,21,24,……99. What can one say about a 3 digit number formed by these two digits. 69145 is not divisible by 3 because the sum of its digits (6 + 9 + 1 + 4 + 5) is not divisible by 3. The number ends in an odd digit. The number is divisible by 3. (Note: 9 and 3 don't have to be in the sum, they are divisible by 3. Application - Sum of Odd Numbers The formula for the sum of the natural numbers can be used to solve other problems. But that's just saying the sum of the digits is divisible by three. And the only thee digit number which fits the bill is 290. Sample Input 1 : 21 Sample Output 1 : yes Sample Input 2: 19 Sample Output 2: no Code:. Check whether 64327 is divisible by 3 ? a) Yes b) No. Identify the conclusion of the following conditional:A number is divisible by 3 if the sum of the digits of the number is divisible by 3. Since, the number 17852 is not divisible by both 2 and 3 so it is not divisible by 6. We provide sequences and series practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. Here, we are going to implement a Python program that will print all numbers between 1 to 1000, which are divisible by 7 and must not be divisible by 5. Calculate the Factorial of a Number in C++. in cell C3 type "=sum(A2:A21)" without the double quotes, then hit return. So that's a number that would be divisible by 3 and by 4, which would mean it's divisible by 12. If the two numbers are equal return a or b. The factors of 32 are 1, 2, 4, 8, 16, and 32. C Program to print the numbers which are not divisible by 2, 3 and 5. Example 1: Is 95 exactly. Write a Sum of Numbers Divisible by 4 in C program to calculate the sum of all numbers from 0 to 100 that are divisible by 4. View Replies View Related If Number Is Even - Divisible 3 Or Divisible By 4 Apr 7, 2014. 6: If the number is divisible by 2 and 3. It should be written as: If the sum of the digits of the number is divisible by 3 then a number is divisible by 3. The number 345,546,711 is divisible by 3. - Criterion of divisibility by $4$: any number multiple of $4$ has as the sum of the units digit and the double of the tens digit a number also divisible by 4. class SumOfNumbersDivisibleByFiveAndSeven { public static void main(String s[]) { System. Answer : Option 4 A number is divisible by 3 when sum of its digits is divisible by 3. Solution: For a number to be divisible by 9, the sum of its digits must also be divisible by 9. 144 is even and the sum of. If not the given number is not exactly divisible by 3. Multiplication and long division. E:The conclusion of the statement is written before the hypothesis. Any number that is divisible by 2 is an even number. 144 is even and the sum of. An=a+[n-1]d. 157,526: 157 × 3 + 526= 997 999: Add the digits in blocks of three from right to left. If the number is divisible by 3 then its divisible by 9. I have to make a program that determines whether an integer put in by the user is odd, divisble by 3 or divisble by 4. In fact it is divisible by 9. The sum of the first n odd natural numbers is (2k-1 represents any odd number): [6. - Criterion of divisibility by $4$: any number multiple of $4$ has as the sum of the units digit and the double of the tens digit a number also divisible by 4. Q:-How many terms of G. Secondly, you want to sum all the numbers that are divisible by 5 and 15, but in the if clause you test if the number is divisible by 5 or by 13, or is divisible both for 5 and for 13. A number 378015 is divisible by 3, because a sum of its digits 3 + 7 + 8 + 0 + 1 + 5 = 24, which is divisible by 3. You are given an array of integers(arr) and a number K. 4: If the last 2 digits are divisible by 4. 3) Sum of first n even numbers = n ( n + 1) 4) Even numbers divisible by 2 can be expressed as 2n, n is an integer other than zero. 8 is divisible by 4 completely. Input a 5-digit integer n from the keyboard. It is a four-digit whole number. The number 15 is divisible by 3. that will be manipulate with the numeric value and control statement (if-else) used if the numeric value is dividable than output will be ok else it will be reverse. It is less than 5000. applying Sn formula, we know that d is 3 and 4 respectively. Squares of odd numbers are of the form 8n + 1, since (2n + 1) 2 = 4n(n + 1) + 1 and n(n + 1) is an even number. Rule # 3: Divisibility by 4 A numbers is divisible by 4 if the number represented by its last two digits is divisible by 4. Example: A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because. Then, determine whether the sum of the digits is divisible by 3. As for a square number being a number multiplied by itself, that follows from it. If the number is an even number and the sum of the digits is divisible by 3, the number is also divisible by 6. docx), PDF File (. Notice incidentally, 3072, the last two digits number, 72 is divisible by 4. By the distributive law of multiplication over addition, we can say that the Sum of three consecutive numbers = 3 x (smallest number + 1). divisible by 3 is 102 And biggest is 999. Since the density of numbers which are not divisible by a prime of the form $5+6k$ is zero, it follows from the previous claim that the density of even Fibonacci So in particular, (assuming conjecture 1) if I want $F_{2n}$ to be a sum of two nonzero squares, $2n$ cannot be divisible by any of the above. 3 consecutive numbers with a sum of 72, The sequence of numbers (1, 2, 3, … , 100) is arithmetic and when we are looking for the sum of a sequence, we call it a series. Two plus nine plus four plus three, two plus nine is 11, 11 plus four is 15, 15 plus three is 18, and 18 is definitely divisible by nine, so this is. sum of numbers divisible by 4 = 100 + 104 + 108 + + 300. In this c program we need to find all numbers that are divisible by 5 means when we take mod of a number by 5 we have to get the reminder zero. A number is divisible by 11 if and only if a sum of its digits, located on even places is equal to a sum of its digits, located on odd places, OR these sums are There are criteria of divisibility for some other numbers, but these criteria are more difficult and not considered in a secondary school program. So the sum of the digits is divisible by 3 but not 9, so the number is divisible by 3 but not 9. Write a program to verify this famous statement. This divisibility test. This online calculator is designed for addition subtraction multiplication and division and subtraction of fractional numbers, written in binary, ternary Multiply and divide into different number systems. Verify that either (a) both n and sum are divisible by 3 or (b) both are indivisible by 3. The factors of 32 are 1, 2, 4, 8, 16, and 32. We provide sequences and series practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. An=a+[n-1]d. (i) If a number is divisible by 3, it must be divisible by 9. The number 345,546,711 is divisible by 3. Sum of numbers non divisible by 3 and less than 100=S-S_33. The sum of the digits of 621 is 6+2+1 = 9. 8937: 8+7=15. } } Console. Calculate the Factorial of a Number in C++. Here is a look at the rules for 3, 6, and 9. Their sum is m * (m + 1) / 2 * k (by Gaussian sum formula, link to German wiki - they seem to like Gauß more). Conversion of numbers from one number system to any other. The sum of. Is there a short proof that the sum of the digits of $3^{1000}$ is a multiple of $7$ without using a computer?. If the two numbers are equal return a or b. So, there are 30 two-digit numbers divisible by 3. Find the sum of squares of all numbers between 10 and 70 and which are divisible by 4. Insufficient. this c program logic is very simple. Write a Sum of Numbers Divisible by 4 in C program to calculate the sum of all numbers from 0 to 100 that are divisible by 4. As for a square number being a number multiplied by itself, that follows from it. Square of any number will never end with 2, 3, 7 and 8. Any number that is divisible by 2 is an even number. • A number is divisible by 11 when the sum of the odd numbered digits is subtracted from the sum of even numbered digits and the result is divisible by 11 • Like. Here is the beginning list of numbers divisible by 3, starting. println("Sum of matched numbers till 50 is " + sum_of_matched_numbers(50)). Example: Input: Enter array elements: 10 15 20 25 30 Number: 10 Output: Total elements divisible by 10 is 3 Program:. Q6 Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 : (a) 92 __ 389 (b) 8 __ 9484. Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667. This also means that 348 is divisible by 3! The Divisibility Rule for 9. Case:1 Enter the value of num1 and num2 12 17 Integers divisible by 5 are 15, Number of integers divisible by 5 between 12 and 17 = 1 Sum of all integers that are divisible by 5 = 15 Case:2 Enter the value of num1 and num2 1 10 Integers divisible by 5 are 5,10 Number of integers divisible by 5 between 1 and 10 = 2 Sum of all integers that are. (Note: This rule can be repeated when needed). Example 2: Program to calculate the sum of natural numbers using for loop. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. A number is divisible by 3 if the sum of its digits is divisible by 3. Number must be divisible by 331 with the sum of all digits being divisible by 3. So let's do that. I had to change all to the official currency symbol (ZAR), which I dont really want. The sum of any two odd numbers (consecutive or not) is always even. Output Format: Output consists of a single line. When multiplying a sum of two numbers by a third number, it does not matter whether you find the sum first and then multiply or you first multiply each It is the same no matter what amount you start with: 5×3÷3=5; 9×3÷3=9; 743×3÷3=743. Divisibility Rules Divisible by Rule 2 The number is even. Here we will see three programs to calculate and display the sum of natural numbers. So the second 3 digit number divisible by 7 is 112. 3: The numbers x and y are divisible by 5. If, after adding up the sum of all the component parts of a number which comes out to another two digit or bigger number the sum is exposed. To check this declares a variable of integer type. That program will print the output on screen and shows only those numbers which are divisible by 7. Input (Integer). We provide sequences and series practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. The task is to find the sum of all those numbers from 1 to N that are divisible by 3 or by 4. WriteLine("The sum of numbers divisible by 3 or 5 between 1 and {0} is {1}", c. It's one of the easiest methods to quickly find the sum of given number series. Plus 9, it's 20. (Variant) the. pdf), Text File (. Two digit numbers divisible by 3 are 12,15,18,21,24,……99. The difference between its tens and ones digit is 1. A number is divisible by 3 if the sum of the digits of the number is divisible by 3. We know that 9*a is divisible by 3, so 10*a + b will be divisible by 3 if and only if a+b is. Four-digit numbers, five-digit numbers, etc. Sn=n/2{a+an] Sn=165150 &Sn=123200. Check for divisibility by 9. The sum of the digits of the number is a multiple of 3. Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three. [2×102+(300−1)×3] = 165150 Therefore sum of all 3 digit number which are divisible by 3 is 165150. Numbers are divisible by 3 if the sum of all the individual digits is evenly divisible by 3. Input : N = 5 Output : 7 sum = 3 + 4 Input : N = 12 Output : 42 sum = 3 + 4 + 6 + 8 + 9 + 12. You have to find the count of subarrays whose sum is divisible by K. Example 2154: the sum of digits = 2 + 1 + 5 + 4 = 12, this sum is divisible by 3, so the number 2154 is divisible by 3. I can not prove that the it's only true with odd numbers. The Rule for 3: A number is divisible by 3 if the sum of the digits is divisible by 3. ,33= (4+33)/2=37/2. 1, −1, n and − n are known as the trivial divisors of n. You can check that using code like n % 3 == 0. And, in each iteration, the value of i is added to sum and i is incremented by 1. E:The conclusion of the statement is written before the hypothesis. If the division not possible print "Division not possible". To figure out whether it's divisible by nine, I just have to add up the digits and figure out if the sum of the digits is a multiple of nine or whether it's divisible by nine. There are numerous ways we can write this program except that we need to check if the number is fully divisble by both 3 and 5. So the second 3 digit number divisible by 7 is 112. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. [2×102+(300−1)×3] = 165150 Therefore sum of all 3 digit number which are divisible by 3 is 165150. For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible. And the only thee digit number which fits the bill is 290. When multiplying a sum of two numbers by a third number, it does not matter whether you find the sum first and then multiply or you first multiply each It is the same no matter what amount you start with: 5×3÷3=5; 9×3÷3=9; 743×3÷3=743. We are asked to find the sum of all natural numbers between 200 and 500 that are divisible by 7: Note that the list of such numbers forms an arithmetic sequence; the first term is 203, the last. The converse of the statement is "If a number can be divided by 3, then the sum of the digits of that number is also divisible by 3. k+(k+1)+(k+2)=3a for some whole number a. Subsequences of size three in an array whose sum is divisible by m , find number of subsequences of length 3 whose sum is divisible by M O(N^3)) C++ program to find count of subsequences of size sum = A[i] + A[j] + A[k];. View Replies View Related If Number Is Even - Divisible 3 Or Divisible By 4 Apr 7, 2014. (2) That t is divisible by 3 means that 5k + 3 is divisible by 3, and therefore 5k is divisible by 3. BigInt numbers, to represent integers of arbitrary length. Example 1: Is 95 exactly. The sum of all single-digit replacements for z is 12. The sum of 2 digits x and y is divisible by 7. Squares of odd numbers are of the form 8n + 1, since (2n + 1) 2 = 4n(n + 1) + 1 and n(n + 1) is an even number. (a) ___ 6724. The difference between the number and the sum are 99a + 9b, which is divisible by three and nine. And 15 is divisible by 3. Example 2: Program to calculate the sum of natural numbers using for loop. The sum is 18, 18 is divisible by 3, and therefore; the number 927 is divisible by 3. 1 + 1 + 1 + 1 = 4 , not divisible by 3. number divisible by 9 is : 117. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd. Therefore, the largest number is 9. Smallest 3 digit no. The sum of its digits is 12. Now you need to find if there is a subset having sum divisible by m. Numbers that are the sum of two squares. Divisible by 16 : A number is divisible by 16, if the number formed with its last three digits is divisible by 16. 1] We can expand the left-hand side: [6. For example, the OP’s 7 is the case where N=3 For 9, we try N=4 and get. So, sum =12+15+18+………+99=3 (4+5+………+33) no. Call it sum. divisible by 3 is 102 And biggest is 999. If a number is divisible by both 3 3 3 and 8 8 8, then the number is also divisible by 24 24 2 4. First determine whether the number is even. A number is divisible by 9 if the sum of its digits is divisible by 9. Problem How many three-digit numbers are not divisible by 3? The first three-digit number that is exactly divisible by 3 is 102 and the last is obviously 999. So any odd number (2N+1) of consecutive integers will be divisible by that odd number (2N+1). 1 + 1 + 1 + 1 = 4 , not divisible by 3. Rule 1: Partition into 3 digit numbers from the right ( ). I can not prove that the it's only true with odd numbers. So the second 3 digit number divisible by 6 is 108. The number ends in an odd digit. After 105, to find the next 3 digit number divisible by 7, we have to add 7 to 105. Created by Marek Kuklis. What is the largest 4 digit number exactly. WriteLine("The sum of numbers divisible by 3 or 5 between 1 and {0} is {1}", c. A conditional statement is known as an If-then statement. + 75 = 975 Therefore, 975 is the sum of first 25 positive integers which are divisible by 3. Numbers are divisible by 3 if the sum of all the individual digits is evenly divisible by 3. Want to know if a number is easily divisible by 6?. Divisibility Tricks. Multiplication and long division. Input : arr[] = {-2, 2, -5, 12, -11, -1, 7} Output : 5. How can you tell if a number is divisible by 3? A. So it is actually an "if. The positive integers 1, 2, 3, 4 etc. To check if the given number is exactly divisible by 3 follow the below steps Step 1: Add all the digits in the given number until you arrive at a single number. LCM of 3, 4, 8: Since 8 is the highest number, it may be the LCM. " for checking divisibility of numbers having more digits by 3. It should be written as: If the sum of the digits of the number is divisible by 3 then a number is divisible by 3. Examples: Input : arr[] = {2, 7, 6, 1, 4, 5}, k = 3 Output : 4 The subarray is {7, 6, 1, 4} with sum 18, which is divisible by 3. The divisibility criteria are a roundabout way to know if a number is divisible by another without directly doing the calculation. ) The number is odd. You are always adding 2 to the sum, rather than adding the numbers that are divisible by three. So let's do that. It follows that a number is congruent to its digit sum. The sum of all numbers smaller than a divisible by 3 or 5 is the same as. And a logic will be used along with the Modulus operator(%). Print those numbers 5. Given an array of random integers, find subarray such that sum of elements in the element is divisible by k. The task is to find the sum of all those numbers from 1 to N that are divisible by 3 or by 4. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 so the number 3627 is evenly divisible by 3. [2×102+(300−1)×3] = 165150 Therefore sum of all 3 digit number which are divisible by 3 is 165150. To reduce time complexity further, once we have found a number which is divisible by HCF, we can increment the number by HCF(6, 12, 18 and so on in this case) and find the sum of those numbers. 3, 3 2, 3 3, … are needed to give the sum 120? Q:-The sum of the first four terms of an A. The sum of all natural numbers that are smaller than 500 and divisible by 5 í HOC24. As 59 is not wholly divisible by 3 the question is invalid. reduce(function(a,b){return a + b;}, 0); //print sum total. 58302 is divisible by 3 because the sum of its digits (5 + 8 + 3 + 0 + 2) is divisible by 3. You are never checking to see if the number is divisible by three. This generalizes to give the result. Sum of any number of even numbers is always even. For example, the digit sum of 4752 is , so 4752 is is not divisible by 9. We can use this property of the sum to see if a number is divisible by another, without doing the division. The problem “Subset with sum divisible by m” states that you are given an array of non-negative integers and an integer m. Since, sum of the digits is divisible by 3, it will also be divisible by 2 and 3 but unit digit is odd, so it is divisible by 3 only. Q:-How many terms of G. The rst part is always divisible by n. (iii) Find the sum of all two digit numbers which leave 1 as remainder when divided by 3. This proves that a number less its digit sum is always a multiple of 3 (and 9 for that matter): Thus n - s is divisible by 3. println("\n The Sum of Even Numbers upto " + number + " = " + evenSum); } } OUTPUT. If the number is divisible by 3 then its divisible by 9. On dividing a number by 3 or 9 the remainder will be equal to that left from dividing the sum of digits of that number by 3 or 9; 7,309 has the sum of the numbers 7 + 3 + 0 + 9 = 19, which is divided without a remainder to neither 3 nor 9. Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three. Now it can also be written as: 9a + a + b. In the given series of numbers how many 7's are there which are immediately followed by 9 and immediately not preceeded by 5 ?. Get an answer for 'Find the sum of all integers wich are divisible by 7 and lying between 50 and 500. $\endgroup$ – mas Aug 29 '20 at 9:27 $\begingroup$ any insights, pls? $\endgroup$ – mas Aug 29 '20 at 10:17 $\begingroup$ @AlonYariv (1) Finding an exact solution to this variant --- or even the original --- subset sum problem is non-trivial for large sets of boxes. You're good if result is divisible by 7. So, X can be 2, 5, 8. At last, this number is divisible by 11, because a sum of even digits: 7 + 0 + 5 =12 and a sum of odd digits: 3 + 8 + 1 = 12 are equal. Input Format A number N arr1 arr2. The sum of 2 digits x and y is divisible by 7. Thus we do not need to remove. Therefore the sum of an odd number of odd numbers will be odd. View Replies View Related If Number Is Even - Divisible 3 Or Divisible By 4 Apr 7, 2014. Hi @Pavani. 69145 is not divisible by 3 because the sum of its digits (6 + 9 + 1 + 4 + 5) is not divisible by 3. The sum of the first n odd natural numbers is (2k-1 represents any odd number): [6. 5: If the last number is 5 or 0. Plus 9, it's 20. ) The number is odd. In other words it is always 3 times the middle number. 4: If the last 2 digits are divisible by 4. Expand this to a number of 3 digits. Input: nums = [6,3,5,2], p = 9 Output: 2 Explanation: We cannot remove a single element to get a sum divisible by 9. Ok there's maybe one little simplification for the divisibility of. A number is divisible by 11 if and only if a sum of its digits, located on even places is equal to a sum of its digits, located on odd places, OR these sums are There are criteria of divisibility for some other numbers, but these criteria are more difficult and not considered in a secondary school program. There we need to find sum of odd numbered digits. It's one of the easiest methods to quickly find the sum of given number series. The sum of all natural numbers that are smaller than 500 and divisible by 5 í HOC24. When multiplying a sum of two numbers by a third number, it does not matter whether you find the sum first and then multiply or you first multiply each It is the same no matter what amount you start with: 5×3÷3=5; 9×3÷3=9; 743×3÷3=743. $\endgroup$ – Math Lover Dec 7 '20 at 8:27. ⇒ 8 is divisible by all the given numbers. 1] We can expand the left-hand side: [6. |
Terracost: Computing least-cost-path surfaces for massive grid terrainsTerracost: Computing least-cost-path surfaces for massive grid terrains
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Author Hazel, Thomas ♦ Toma, Laura ♦ Vahrenhold, Jan ♦ Wickremesinghe, Rajiv Source ACM Digital Library Content type Text Publisher Association for Computing Machinery (ACM) File Format PDF Copyright Year ©2008 Language English
Subject Domain (in DDC) Computer science, information & general works ♦ Computer programming, programs & data Subject Keyword Data structures and algorithms ♦ Dijkstra's algorithm ♦ I/O-efficiency ♦ Shortest paths ♦ Terrain data Abstract This paper addresses the problem of computing least-cost-path surfaces for massive grid terrains. Consider a grid terrain $\textit{T}$ and let $\textit{C}$ be a cost grid for $\textit{T}$ such that every point in $\textit{C}$ stores a value that represents the cost of traversing the corresponding point in $\textit{T}.$ Given $\textit{C}$ and a set of sources $\textit{S}$ ∈ $\textit{T},$ a least-cost-path grid Δ for $\textit{T}$ is a grid such that every point in Δ represents the distance to the source in $\textit{S}$ that can be reached with minimal cost. We present a scalable approach to computing least-cost-path grids. Our algorithm, terracost, is derived from our previous work on I/O-efficient shortest paths on grids and uses $\textit{O}(sort(\textit{n}))$ I/Os, where $sort(\textit{n})$ is the complexity of sorting $\textit{n}$ items of data in the I/O-model of Aggarwal and Vitter. We present the design, the analysis, and an experimental study of terracost. An added benefit of the algorithm underlying terracost is that it naturally lends itself to parallelization. We have implemented terracost in a distributed environment using our cluster management tool and report on experiments that show that it obtains speedup near-linear with the size of the cluster. To the best of our knowledge, this is the first experimental evaluation of a multiple-source least-cost-path algorithm in the external memory setting. ISSN 10846654 Age Range 18 to 22 years ♦ above 22 year Educational Use Research Education Level UG and PG Learning Resource Type Article Publisher Date 2008-06-12 Publisher Place New York e-ISSN 10846654 Journal Journal of Experimental Algorithmics (JEA) Volume Number 12 Page Count 31 Starting Page 1 Ending Page 31
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# Gulf Coast Camping Resort
## standard outdoor step height uk
menu Whirlpool Go to navigation. Width refers to both the individual step as well as the entire staircase. Here, a step change in levels is managed via several retaining walls and some relatively steep steps – a non-slip surface and anti-slip treads are especially important here. A convenient angle of the wooden staircase between 29 - 35 degrees. Steps and stairs – key dimensions 25 13. 1. 1. Your steps. Stairs do not need a handrail on the bottom two steps. We will be in touch with any future work required. Garden. Not having a hard landing at the base is another common mistake – likely to lead to a worn patch. This ensures a large enough space to safely walk on as you navigate each step. Garden buildings & fencing. Handrail design 26 14. Standard code requires that a stair run should fall between 9 inches and 11 inches (22.9cm and 27.9cm) to allow people to safely ascend and descend stairs. And this will influence: Materials – paving (in all its forms) for treads, with brick risers – is the obvious choice, but there are plenty of other choices. Relationship of ramp gradient to the going of a flight 28 15. Buy standard staircases to order online from £159.99 excluding vat, we have straight staircases available ready made and we have a selection of standard sizes winder staircase designs fixed sizes from a UK ISO9001 staircase manufacturer s12n2600 s12n2700 s13n2800 standard sizes low prices Need some advice on garden steps? Stair details refer to the standard requirements for each of the various components of a flight of stairs. There are well documented rules of thumb in terms of ideal step dimensions, and what might be termed ‘step psychology’, for example: Before we continue with some more practical considerations, here are a few before and after pictures from several ALDA projects, illustrating how much of a difference steps can make to a project: In most of these examples, the ‘before’ steps were relatively narrow; some tucked away to the side and not always visible, creating a feeling of being trapped. Maximum step height outdoors? The levels recommended in the Code for Lighting are based on the European Standard EN12464-1, 2011 which is published in the UK as BS EN 12464-1. Patio steps should be at least 4' wide to allow two people to stroll comfortably beside each other and even pause to sit on the steps while they enjoy the surroundings. In a given run of stairs, the greatest step height and the smallest step height should not vary more than 3/8 inch (9.5 mm). Or, to work out how many steps you need, take your ceiling height and divide by let's say 16 to start with. In general, they must be high enough not to cause a fall and strong enough to face the tensions of overcrowding. How do i use the chart ? Changing the level of the side / wing walls to drop down with the steps also creates a more open feel here. Just wanted to send a quick note to say thank you all again for the incredible work…. © 2020 ALDA Landscapes Ltd. All rights reserved. To calculate the rise of the stairs, multiply the rise dimension by the number of steps. The rules of thumb given above are really useful guidelines (and there are many more! In this in-depth post, we take a look at: Rules of Thumb for Garden Steps | Some Before & After Photos | Step Height, Depth and Width | Illustrated Examples | Practical Design Considerations | More Illustrated Examples. We are delighted with the work all the team put into our project. Is there any builders who know what the maximum height for an outside step is meant to be. However, wider stairs are typically more inviting and can feel safer for users, particularly in crowded public spaces. …thanks once again for creating such a lovely garden for our family to enjoy. Newbury Office Your personal data will be processed & stored in line with our Privacy Information Policy. In this instance, railings must be at least 950mm high with gaps below 470mm. For steps leading to a lawn, having the top/bottom step flush with the lawn helps prevent a worn patch. Horizontal rails must prevent climbing, and all designs must be resistant to pushing and leaning. The length of each step is referred to as the “Going”. Garden steps present designers with plenty of opportunities to make bold statements and create interest in a garden, but they also pose practical challenges, particularly in terms of cost and making the most efficient use of often limited space. We couldn’t have asked for more and would recommend ALDA to anyone looking to develop their garden and outdoor space. Guarding design 35 17. The total run is the overall horizontal distance traveled by the stringer. Search. Step 4. In the UK there are over 500 deaths and an estimated 250,000 non-fatal accidents from stair-related incidents. In addition to the guidelines for tread height, check to see if latest regs require that the risers are closed. Are side / wing walls needed – for looks, retaining or safety? For outdoor steps, the common formula for comfortable proportioning is 2(R) + T = 26” to 27”, where R = riser height (in inches) and T = tread depth (in inches). According to practice, the ideal step height is in the range of 15 - 17 cm Optimal width of a step 27 - 30 cm Maximum height of stairs may not exceed 200 mm. We do get pretty booked up, particularly during the Spring, so plan ahead if possible! 2. Stair details refer to the standard requirements for each of the various components of a flight of stairs. Many thanks for all your work on the plans and… in executing those plans for our garden. OUTDOOR STEP HEIGHT. example: (2 x … 100mm) is usually accompanied by a deeper tread depth (say, 500mm), and vice versa – so a riser height of 200mm would normally be associated with a shallower tread depth (say 250mm). While many of the methods used in manufacturing and fabrication may be similar, the difference[...], The quality of the weld is extremely important in any fabrication project, as it guarantees the integrity, durability, and longevity of any structure. Industry types like to speak in their own language, therefore the length of each step is called the ‘going’, while the height is the ‘rise’. I would just like to take this opportunity to thank you for the wonderful design… we are absolutely thrilled with the result, I can’t believe it’s really our garden. to a socialising area down the garden. There are plenty of other design considerations too: A combination of steps and ramp is often a good solution – the steps provide a formal entrance and guide the eye to the next section of the garden, whilst the ramp provides practical access. Knowing what kind of headroom or winder you can use is very, very important. From planning a new border to a complete garden remodel, we pride ourselves on transforming ideas into beautiful reality, with minimum disruption. Another example where it was important to get the pace right – with steps covering a reasonably long horizontal distance. They provided designs, carried out the re-modelling of our garden… to a very high standard of work. RG14 7EA. If a staircase exceeds 44 inches (111.8cm), handrails are required for both sides. The British Constructional Steelwork Association (BCSA) has revised its safety measures. There is clearly a lot to garden step design, and it is one of the most interesting, yet challenging aspects of the garden designer’s job…. The width of one stair, after all, will need to be consistent with the rest of the stairs. Here a step running almost the whole length of the patio gives a very open, airy feel – as well as providing easy access from multiple directions. The most important dimensions are the riser height and tread depth. When planning the look of your concrete stairway, don't sacrifice functionality for flair. So, the designer needs to be able to balance several conflicting priorities. Risers - the Step Height [F] is the typical height of each step. In complex and / or steep designs, the steps might have several changes of direction, with landings acting as areas to change the direction of flow, and also as places where people can stop and catch breath(!). Quality control checks during and after welding are vital to ensure that high standards have been achieved. The Work at Height Regulations 2005 govern railiFngs used for construction work, as do the European Directive and Construction Health, Safety, and Welfare Regulations. Regulated by building codes, stair details specify the sizes of risers and treads, stair widths, the placement of handrails and guardrails, and construction methods for fabricating stairs. This blog post has barely scratched the surface! Stairs should have a handrail on at least one side if they are less than 1metre wide, or both sides if they are wider. The risers, which are the parts that sit vertically underneath each step, should be no higher than 7 3/4 inches and no smaller than 4 inches. I was out watering yesterday and someone stopped to compliment the brick wall out front and ask who did it. Discover (and save!) Your stair tread can be greater than 220mm (Going), and the step height can be below 220mm (Rise) provided that twice the Rise plus the Going (2R+G) falls between 550mm and 700mm. We are so excited to get them back in to tackle stage 2. Narrow steps tend to impart a sense of haste and urgency, as do shallow, steep steps; wider, deeper steps tend to feel more unhurried, safe and relaxed (and are potentially also a place to arrange containers!). Step risers shall be uniform and shall not exceed 12 inches in height. The minimum width for stairways shall not be less than 36 inches clear. Barrier design 40 20. Another option is to change the direction of the steps, so that instead of running out from a higher level patio, they run parallel with it, with the patio retaining wall effectively acting as the side wall of the steps. Staircases require at least one handrail: If the stairs are less than 1m wide: provide a handrail on one or both sides. Outdoor railings for roofs and raised terraces are regulated differently to indoor ones at a minimum height of 1.1 metres from the platform or floor. The standard acceptable range of dimensions for exterior Treads is 11 to 18 inches. Here’s what you need to know. Sign up for our bi-monthly newsletter for free garden design advice, ideas and photos. To simplify the process, we’ve created this deck stair calculator to help you determine the correct measurements needed to carry out your project. Is Steel The Best Option For Your Security Gates & Barriers? It also requires that there is ‘enhanced contrast on the steps’. The total rise is a stairway’s overall change in height, … For stairs serving a single user (typically private residential), a minimum of 36” (91 cm) is required. The BCA calls for the dimensions of goings and risers of a stair to be constant throughout each stair flight. Discuss your project requirements with us by calling 01945 464637. hbspt.cta._relativeUrls=true;hbspt.cta.load(2730834, 'bc237147-d5cc-46eb-a6d2-5464c1876a67', {}); With the rapid advances in Artificial Intelligence (AI) in recent years, many aspects of manufacture and fabrication will likely be revolutionised in the not-too-distant future. Dimensional irregularity is a common cause of missteps and falls on stairs. Here a fantastic sense of flow is created by combining semi-circular steps into the flow of the retaining wall. Staircase Width: 36 Inches, Minimum. The COVID-19 lockdown has seen dramatic changes in public behaviour and restricted daily routines. Ever. The overall width of a set of outdoor stairs should be a minimum of 4 feet in a private setting, and 5 feet in a public space. OUTDOOR STEP HEIGHT. 10. Where unexpected risks make themselves known, it’s your ethical and legal duty to deal with them effectively. This is the normal relationship between the dimensions of the rise and going. Project managers should think beyond the law, always having an eye towards addressing authentic risks rather than mere numbers. To reduce the seriousness of falling down steps, the maximum number of steps in each flight should be limited to 12 (approx.1.8m vertical height) and never more than 16 steps without a landing. The first riser shall be measured from the deck. Step riser height uniformity (<= 3/8" variation) - this means that more than 3/8 of an inch in variation of the height of steps from one step to another is a tripping hazard. In response, all businesses are having to consider not just changed working patterns but also infrastructural changes to support their employees. In all the buildings handrail height should be between 900mm and 1000mm measured from the top of the handrail to the pitch line. For indoor stairs, escalators and travellators, the recommendation is for an average illuminance of 100 lux. We love it. Your stair tread can be greater than 220mm (Going), and the step height can be below 220mm (Rise) provided that twice the Rise plus the Going (2R+G) falls between 550mm and 700mm. This is the normal relationship between the dimensions of the rise and going. She felt the garden was “truly imaginative – just what a nursery environment should be for children”!!! ... 135 to 190mm is a nice height . We have a transformed garden which is really an additional living space to our house. Stair widths are based on the use and occupancy loads of adjacent spaces and must be serviced with handrails on one or both sides depending on the stair width. The heights of handrails on ramps, stairs, and walkways should be measured at a 90-degree angle directly above the stair tread or walking surface. Just wondering if anyone would know if there is a maximum height for a single outdoor step? The absolute maximum height – for comfort – should be 200mm (8”) – this would probably be too high for people who want to frequently carry food and drink in and out of a house e.g. A stairway in which there is a noticeable variation between the risers is a safety hazard. Discuss your project requirements with us by calling 01945 464637. This act covers the safety of trespassers, too, so every property must be guarded by adequate railing. Risers must be of uniform height, with a variation of 9 mm (3/8 inch) or less. Typical locations for guarding 36 18. Code states that this should not be greater than 7 3/4 inches (194 mm). Any railing used on stairs or balconies must have gaps of under 99mm in all directions so that children can’t pass between them. You can’t be catching people out with one step at 185mm and the next 230mm! Since homeowners and businesses are liable for accidents under the Occupiers Liability Act (1957), it’s crucial that all railings be built in accordance with the law. Areas that will be used by children must meet Approved Document K of the UK’s Building Regulations, which require railings to be at least 100mm apart. Here’s a quick overview. Staircase Building Regulations UK. 111 Newtown Road You should also complete this simple calculation: Double the rise, then add the going. Help on Permited Rise and Go dimensions. Stair widths are based on the use and occupancy loads of adjacent spaces and must be serviced with handrails on one or both sides depending on the stair width. Thanks for subscribing. Often, the two terms are used interchangeably, with little awareness of the differences. Outdoor; Roofing; Blog; ... First, we calculate the vertical height rise of each step: ... {120\,in \over 17} = 7.06\,in Now, using a standard mount staircase, the vertical height rise of the steps will be one step less than with a flush mount because the final step is taken up by the decking. Handrail design 26 14. And where steps are covering a long horizontal distance with wide gaps between the risers, making sure that the distance between the step risers is a comfortable number of paces is an important consideration. The whole process has been enjoyable and remarkably stress free. Suitable handrails for common stairs in blocks of flats 24 12. Barrier design 40 20. Tread Thickness The thickness of the decking product you are using in inches. Having the right calculations before getting started can make the task a lot easier. The height and width of the stairs. No spam. Urban chic steps with slick porcelain paving can have porcelain clad risers. Steel is a great choice for outdoor railings of all kinds. There are several different options available. Ideally, they will achieve a visual contrast against their surroundings. ALDA Landscapes Ltd The recommended range for riser height, in an outdoor set of steps, is a minimum of 4.5 inches to a maximum of 7 inches. If you’re based in or around the Reading and Newbury area, we’d love to meet and discuss your project in detail – book your free no-obligation garden consultation today. Outdoor Railings. We have had excellent gardening services from ALDA since they helped us re-design and re-landscape our front and rear gardens. When designing garden steps, the first consideration is obviously the height of the change of level, along with the amount of horizontal space available. 1. The suitability of these will depend on several factors, such as aesthetics,[...]. However, they may not always be desirable or even necessary. I am delighted with Ofsted’s report. Jan 12, 2013 - This Pin was discovered by Delfina Rangel. The Work at Height Regulations 2005 govern railiFngs used for construction work, as do the European Directive and Construction Health, Safety, and Welfare Regulations. Estairs offer two ranges of steel staircases suitable for outdoor use:-Italian Spiral Stairs which are available in limited sizes and colours with a maximum diameter 160 cm. – building regulations specify this as a requirement if there is a fall of 600mm or more, and in any event, the safety of all garden users is paramount. Open risers are permitted if they are 10 cm (4 inches) high or less. Understanding how to operate successfully within the context of the virus means that operations can resume, and people can stay safe. Strip away any turf in the marked-out area with a spade. Staircases have the potential to be safety hazards if not designed properly. Other sections of this standard cover the recommendations for treads and risers, strings, newels, construction, handrails and balustrades. Appendix A of this standard gives details for the site fixing of stairs and Appendix B guidance for the design of stairs with winders. This free stair calculator determines stair parameters such as rise, total run, and angle, stringer length, based on height, run, tread, and headroom requirement. Archive View Return to standard view. We are both very impressed and pleased with the way the garden has turned out and the constant surprises as plants bloom and become more established. For safety, an overhang on each step (or nosing) should also be considered as this improves visibility. Alternating tread stair 22 11. Really pleased – thank all the team from us. For outdoor steps, the common formula for comfortable proportioning is 2 (R) + T = 26” to 27”, where R = riser height (in inches) and T = tread depth (in inches). your own Pins on Pinterest You should fit a handrail at a height between 900mm and 1000mm from the pitch line of the stairs, or the floor for landings. If you’re planning to outsource the architectural steelwork for a civil engineering, commercial, or construction project, it’s vital to employ the services of a competent fabricator who can guarantee high-quality finishing that[...], Traditional galvanisation and newer hot zinc spray technologies are standard anti-corrosion treatments for vulnerable metals. In the third example, the ‘before’ steps appear too narrow, with no nosing (overhang) and steps are not finished flush to the wall. Small gardens on steep slopes can be a big challenge for garden designers as the number of steps needed to get from top to bottom of a slope can take up a vast amount of horizontal space, which simply might not exist. Another great job, thank you to John and the team. Relationship of ramp gradient to the going of a flight 28 15. Craftsmen in the ALDA team has been enjoyable and remarkably stress free gaps etc to get your estimate accurate! 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Work Energy
NCERT Solution
Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(a) Suma is swimming in a pond
Answer: Suma is applying force on water in backward direction. The reactionary force is being applied on Suma in forward direction which helps Suma in moving forward. Here, the direction of force applied by water and direction of displacement of Suma are in same direction. Hence, work is done in this case.
(b) A donkey is carrying a load on its back.
Answer: The donkey is applying force in upward direction but the donkey is moving in forward direction. Here; force and displacement are in mutually perpendicular directions. Hence, work is not done in this case.
(c) A wind-mill is lifting water from a well.
Answer: The wind mill is applying force in upward direction. Displacement of water is also in upward direction. Hence, work is done in this case.
(d) A green plant is carrying out photosynthesis.
Answer: No displacement is happening and no force is being applied during photosynthesis. Hence, no work is done in this case.
(e) An engine is pulling a train.
Answer: The force applied by the engine and displacement of train are in same direction. Hence, work is done in this case.
(f) Food grains are getting dried in the sun.
Answer: No force is being applied and no displacement is taking place during this process. Hence, no work is done in this case.
(g) A sailboat is moving due to wind energy.
Answer: The wind energy is applying force on the sail, and sailboat is propelled in the direction of force. Hence, work is done in this case.
Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer: Force of gravity is being applied in downward direction, but displacement is in horizontal direction. Hence, no work is done in this case.
Question 3. A battery lights a bulb. Describe the energy changes involved in the process.
Answer: Chemical energy changes to electrical energy in the battery. Electrical energy changes to heat and light energy in the bulb.
Question 4. Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1. Calculate the work done by the force.
Answer: Given; m = 20 kg, u = 5 m/s and v = 2 m/s
Kinetic energy when velocity is 5 m//s = ½ mv^2
= ½ xx 20 kg xx (5 m//s)^2 = 250 J
Kinetic energy when velocity is 2 m//s = ½ mv^2
= ½ xx 20 kg xx(2 m//s)^2 = 40 J
Difference = 250 J – 40 J = 210 J
So, work done is – 210 J (Work done is negative because kinetic energy has reduced.
Question 5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer: Here, direction of gravitational force and that of displacement are perpendicular to each other. Hence, no work is done in this case.
Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer: When an object is in free fall, its potential energy decreases progressively. At the same time, the kinetic energy of the object increases progressively. At any given time, the sum of potential energy and kinetic energy is constant. Thus, it obeys the law of conservation of energy.
Question 7. What are the various energy transformations that occur when you are riding a bicycle?
Answer: The chemical energy from food is converted to kinetic energy while someone pedals the bicycle. This energy is then transferred to the wheels of the bicycle.
Chemical energy (food) → Muscular energy → Kinetic energy + Heat Energy
Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer: When we push a huge rock but the rock does not move, no energy is transferred to the rock. Here; the muscular energy is converted into heat energy which is evident from the rise in temperature of our body while doing so.
Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer: 1 unit of energy = 1 kWh
1 kWh = 3.6 xx 10^6\ J
So, 250 kWh = 250 xx 3.6 xx 10^6\ J = 9 xx 10^8\J |
# LaTeX Math for e-Learning in D2L
I am teaching MCF3M online this semester, so I need to be able to include math notation in my online content, quizzes, etc. I know how to write math notation using LaTeX from my days at the University of Waterloo, and I find it a lot faster than using a graphical equation editor. I’ve tried Microsoft Word’s editor, which accepts LaTeX-like input as well as graphical input, but I still find it frustrating to use.
I’m teaching in the Desire2Learn/BrightSpace learning environment, so I need to ensure my content works well in there. Last semester I taught Computer Science/Programming and used PDF files that I created in Word Online, and I considered doing the same thing again.
But D2L has an equation editor as part of its HTML editor for webpages, discussion posts, etc. Could it be all I need?
I’ve taken it for a spin before. Here’s the workflow:
Create a new page and type into the HTML editor.
Expand the toolbar so that the Equation tools are available.
Choose \∑ LaTeX equation.
Type in the LaTeX expression, using $$and$$ as delimiters for inline mode (otherwise it defaults to block mode).
Looks good.
But look at the source HTML code:
Uh-oh… that’s MathML (Math Markup Language), not LaTeX. What if I want to change something in my original LaTeX?
Well, you can see at the bottom that my LaTeX code is still there, but it’s not being used. I could remove all the MathML, cut out my LaTeX, modify it, and re-insert it using the LaTeX equation editor.
Ugh.
I thought that MathJax, the rendering engine that D2L uses for math notation, could only handle MathML (since notation from both LaTeX and graphical editors are converted to MathML), but it turns out that’s not true. MathJax can do LaTeX.
So I tried putting LaTeX directly into the WYSIWYG editor:
No dice.
The trouble is that D2L has parameters on its JavaScript call to MathJax:
That config=MML_HTMLorMML bit is saying that only MathML is acceptable input (and HTML or MathML can be output).
So I added another call directly to MathJax in my own source code:
I set the parameter to be config=TeX-AMS_HTML, which will accept my LaTeX input and render in HTML/JavaScript.
Magic.
## But this is kind of a pain.
I can use D2L’s editor to insert math, but I get MathML (which I find hard to edit).
I can write in LaTeX and have it be preserved, but I need to add a script call to the start of the HTML source code (a hassle, but not too serious, I suppose).
Or I can write in some other (offline) development environment, include my script call all the time, and just upload my completed HTML files to my course. This has the advantages of being independent of D2L, available without internet access, and very shareable.
So that’s what I’ve decided to do, at least for now. So I’ve learned a little CSS to make my pages less vanilla/more functional, and I’ll try to improve the look and feel as the semester progresses.
Wish me luck.
# My Unresponsive Course
You have me thinking again, Mr. Peterson. Today it’s with your post, “You Need to Be Responsive“.
My own blogs are responsive and so they display well on multiple device types.
My blog on a computer browser.
My blog on a phone browser.
Students tell me they like the theme (The Suits Theme) on my class blog; they say it looks professional.
The class blog on a phone browser.
However my e-learning course is less responsive. There is a mobile interface which becomes active when you use mobile device, but there are a couple of components on my course homepage which only appear using the desktop interface (custom widgets with links to the Padlet and the YouTube playlist for the course). Some course tools aren’t available in this view (e.g. the Checklist Tool).
My course homepage on a computer browser.
Course homepage in the mobile view.
Tools in the mobile view.
What’s more, I have created quite a few PDF files for the programming concepts in the course. These files, because they are paginated, are not responsive.
PDF in a mobile browser – portrait
PDF in a mobile browser – landscape
Honestly, I don’t remember considering that issue when finally deciding to use PDFs. Since students are coding on computers anyway perhaps it doesn’t matter that much. And other stuff, like news and discussions, are mobile-friendly.
But I’ve unintentionally made it difficult for them to use small, mobile devices to keep up with the “text” part of the course (i.e. when they’re not actively programming). I do have a lot of video on there, which is (I think) fairly accessible.
I’ll have to think about whether it’s worth fixing the responsiveness of those PDFs.
# Why I’m happy with the design of my first e-Learning course
I’m teaching ICS3C/3U (Computer Programming/Computer Science) online this semester. I was the e-Learning Contact for my board for several years, but this is my first time teaching online in “real life” (i.e. not a professional learning session).
Overall I’m fairly happy with how I approached the design of my course, and I’ve also learned a few things that I’ll improve upon.
I work in Ontario using the Desire2Learn/Brightspace platform. The Ministry of Education here has provided a “starter course” for each of my classes. I’ve discussed the problems with the format of that course before (see here), and thoughts I had about improving on it. I mostly followed my own advice.
In the end, there were significant differences between the provided course materials and the approach I wanted to take with the course, so I ventured out on my own.
# News
I use the News area to share interesting links, reminders, and extra details. Notice I don’t post a news item every day; I’ve decided that’s too much without a real payoff.
I post a lot of “instructional videos” in the course. The videos show me coding applications live, so students can hear my thinking while I work, see the errors I make and how I correct them, and see how the development environment works visually.
Of course, playing videos successfully in the online learning environment is a bit device-dependent, so I also post every video to YouTube. I maintain them in a playlist (here) and I post a link in a widget on the course homepage. The YouTube videos also have the advantage of sometimes being faster to download, and users can change the resolution on the fly (not very helpful for code, I suppose).
And most importantly, those videos are available for anyone on the Internet to use. That makes me feel good in my heart.
# Other Resources
In an out-of-the-way place on the course homepage I maintain a Padlet. I use this to post related-but-not-necessary links.
# Calendar
I don’t use the Calendar directly, but I do use due dates for Content items so they show up in the Upcoming Events list. More on that later.
# Units, Activities, Modules
I divide up a Unit into Activities, and I number everything:
Within each Activity I also “letter” my items sequentially (3.1A, 3.1B, etc.):
# Due Dates
I used to only put due dates on the items that needed to be submitted/completed, like Dropboxes and Discussions. After helping an e-Learning student in the library here with her English class, I realized that students were using the Upcoming Events list in the Content area. I had never explored it before, but it was a sparse list in my course.
Now I add due dates to Content items which I want students to consume as well; for example, I put a due date on 3.2E, which is a video I want them to watch by the end of today. They don’t have to submit anything, but I’m hoping the due date helps to keep them on track.
(Note: This is not an “end date”, which would make the item unavailable after the date has passed.)
# Content in PDF
I have tried to use HTML, Word documents, and PDF files to post CS content, but I’m happiest posting a PDF that I generate in Word. I get excellent control over the look and feel of the text/images, but the links I put in there still work. It’s also universally-readable.
HTML takes too long to format correctly/beautifully, and Word documents are rendered as images by D2L (disabling any links).
In the end I think the videos are more valuable for the students, especially later in the course. I have several students now requesting videos explaining a certain aspect of the course, but not a lot of PDF requests.
# Still Thinking
I have a lot more to learn, but I’m pretty happy with the workflow I’ve developed now. Suggestions are very welcome!
# Some more awesomeness
I’m sure you’ve already read through Part 1 of this series, so here are a few more session highlights.
Note: If you have already registered and wish to change your session choices, just send me an email with the new session code(s).
# Tuesday
## Session Block 2
### D1S13: Blended Learning: The First 20 Days
New to Blended Learning? Don’t worry! This workshop will unveil a twenty day plan that will help you transform your classroom by improving communication, promoting greater collaboration, and differentiating student learning. By connecting powerful pedagogical practices with specific applications of the vLE, it will be demonstrated how blended learning is the ultimate teaching and learning experience for both teachers and students! Participants will receive the day-by-day plan, companion D2L manual, and access to various training videos.
Intended Audience: All New Learners
Experience Level: Beginner
Presenter(s): Sean McDade and Paul D’Hondt
## Session Block 4
### D1S36: Try Something. Don’t Try Everything.
The session will introduce tips and tricks for teachers new to blended learning to help avoid the common pitfalls of frustration and feeling overwhelmed. The session will incorporate blended learning best practices and introduce teachers to some helpful resources to get them started.
Intended Audience: All Teachers
Experience Level: Beginner
Presenter(s): Brock Baker
# Wednesday
## Session Block 2
### D2S14: Developing a Real Professional Learning Community in a Virtual World
Building a culture of collaboration among e-learning teachers who infrequently meet face-to-face takes creativity, coordination and plenty of good will. Successful strategies used by the TDSB e-Learning team for building professional learning communities using Desire2Learn, Adobe Connect and Google Apps for Education will be explored. We’ll share tools and resources that we have co-constructed with our e-learning teachers to enhance student learning in the online classroom. The TDSB e-Learning team will share successful strategies for building professional learning communities using Desire2Learn, Adobe Connect and Google Apps for Education. We’ll share tools and resources that we have co-constructed with our teachers to enhance student learning in the online classroom.
Intended Audience: eLCs/DeLCs, eLearning Teachers, Administration
Experience Level: All Levels
Presenter(s): Andrea Brożyna
## Session Block 4
### D2S36: Replace the shovel with the snow blower and you get better results. Free and cheap software you can use to engage students.
Blended learning environments allow students to set their own pace and work to their potential. Project based exercises allow students to demonstrate, analyze, re-teach in ways traditional paper and pencil busy work cannot. The amount of incredible web based software that I will share with you keeps growing, allowing teachers to give students projects that challenge them in new and exciting ways. The projects we are seeing are the best we’ve ever seen and many are professional level. The ability to update, improve and collaborate makes for better and much less expensive content.
Intended Audience: All Teachers
Experience Level: Beginner-Intermediate
Presenter(s): Mitch Lapointe
# Upcoming #OTRK12 Session Highlights – Part 1
If you haven’t already heard, On The Rise K-12: Enhancing Digital Learning is a great meeting of educators from across Ontario on April 1 and 2, 2014. Representatives will be attending from every school board in the province: from Windsor to Moosonee, from Ottawa to Kenora. You can read all the details at http://otrk12.ca. If you want to attend, you can register there (the cost is \$100 per person per day).
# There is too much awesomeness
I’m going to point out some of the fantastic sessions we have scheduled (there are 80!). This is partly to encourage you to join us, partly to build your excitement, and partly to send kudos to the presenters/facilitators. I know many of the presenters, but not all of them, and I haven’t seen most of the presentations. I’m not being exclusive here; I’m just picking a few from each day of the conference. There are 9 other great choices for every session slot, so you’ll have to go the website for the full list. I’ve tried to steer away from “eLC/DeLC-only” sessions in the highlights below, since most eLCs and DeLCs are already registered (thanks everyone!).
Note: If you have already registered and wish to change your session choices, just send me an email with the new session code(s).
# Tuesday
## Session Block 1
### D1S03: Bitstrips for Schools – Online Comic Creation
In this session you will learn how to use Bitstrips to create full-colour, professional comics. You will be guided through how to sign up for an account, build your avatar, create a classroom, add students, and design and assess completed work. This session is suitable for all grades and subject areas and BYOD is a must! Bitstrips for Schools is provided free of charge to all Ontario teachers by the Ministry of Education.
Intended Audience: All Teachers
Experience Level: Beginner
Presenter(s): Jennifer Ayres
## Session Block 4
### D1S35: How to Start Out With Blended Learning in The Primary Grades
This session is designed to help support our K-3 Primary Teachers as they move to extend the walls of their classrooms. Come find out how the provincial Virtual Learning Environment can provide a safe and engaging space for you and your students. Use these online tools to easily connect and communicate with parents. Give your students a chance to explore rich multimedia. Create interactive lessons for your class and your colleagues. In addition to a quick tour, this session will give you opportunities to learn how others are using the LMS to engage their primary students in the classroom through blending learning.
Intended Audience: Elementary Teachers
Experience Level: Beginner
Presenter(s): Shelley Lowry
# Wednesday
## Session Block 1
### D2S06: Customize the Look and Feel of Your Course
A great looking theme improves the vLE experience for everyone. Want to improve the look of your course but need some help? This hands-on workshop will demonstrate the basics of editing a course theme and provide time to work on your own theme with experts in the room to help you. Theme resources, ideas, and free images will be provided.
Intended Audience: All Teachers, eLCs/DeLCs
Experience Level: Intermediate
Presenter(s): Tim Robinson & Peter Anello
## Session Block 3
### D2S22: Say What? – “Oral Proficiency”
Oral communication is an overall expectation in many subject areas, but it is often the one area that is the most difficult to assess and evaluate. Why? From my own personal experience, it is difficult to speak with 28-30 students in an authentic assessment/evaluation situation. In this presentation, you will have the opportunity to try several useful tools/programs to see how Blended Learning can transform the way you assess and evaluate Oral Communication in your courses. Please be sure to bring headsets/microphones. This presentation will be interactive.
Intended Audience: All Teachers
Experience Level: Any Level
Presenter(s): Gillian Walker
# What’s the difference between e-Learning, online learning, Blended Learning,…?
Janet Broder (@peachyteachy) asked this morning,
A bunch of folks tweeted back at her, including me, but I thought it was worth a slightly longer explanation that Twitter permitted.
# There are a lot of terms
e-Learning (or eLearning, or elearning – we fight about this one), Blended Learning, virtual learning (I don’t like this one; makes it sound like it’s pretending to learn), online learning, hybrid learning, digital learning… gross, eh? They’re not all useful, and some of them make things fuzzy.
I’ll explain my take on each of them. You can have your own take; it won’t hurt my feelings.
# e-Learning
This is learning in which the interaction between student and teacher is online. For us this is generally a student taking a course from a teacher without going to a physical classroom with that teacher. They might be in the same building, but the learning and the communication is done online.
There may be an offline component (for example, a student might write a response on paper), but there is always an online connection (e.g. they take a picture of their response to send to the teacher).
# Blended Learning
In Ontario, Blended Learning is the use of the Provincial Learning Management System (more recently termed the virtual Learning Environment) with a face-to-face classroom. At the moment that’s using Desire2Learn with your students.
But that’s Blended Learning with capital letters. For “blended learning” I feel you only need to be using online tools. Connect your students to the Internet. That definition is more inclusive, but then it also includes some less meaningful implementations. Not all forms of blended learning are equal. Using the Internet to enhance instruction is complex, so we spend a lot of time figuring out how to do it well. [Plug: that’s a big part of On The Rise!]
Hybrid learning is the same thing, but I think is a term more commonly used in the United States.
# Online Learning
For me, online learning encompasses both e-Learning and blended learning. I think of it as “using online tools for learning”. It doesn’t matter where you are on the face-to-face to e-Learning spectrum; online learning is the spectrum itself. The key element is the use of the Internet. Just like blended learning, this can be done poorly or awesomely.
# Digital Learning
This one’s my favourite. This is everything. Digital learning includes online learning which includes blended learning (and Blended Learning) and e-Learning. It also includes “offline digital learning”, like using local software and digital cameras.
# We’re still figuring this out…
…and in the end, it’s all just learning. I’m optimistic that we’ll get to the point where the only distinction will be whether you’re face-to-face or not; digital will be the norm. |
# Choosing sampling rate according to the bandwith reponse in accelerometer datasheet
In an accelerometer's datasheet, the bandwidth response in the x and y axis is 350 Hz.
• There is another information related to the internal sampling rate (shown below). I don't know how these two are related to each other. What is the maximum frequency at which I can acquire data from this accelerometer?
• I would like to get the acceleration data as accurate and fast as possible. What sampling rate do you recommend?
Edit
I'm using an Arduino Mega. The sensors is a tri axis accelerometer mma7260Q, here is the complete datasheet.
Maybe the Bandwidth response values are related to the internal low pass filter.
• Can you share which accelerometer you're using so I can take a look at the rest of the datasheet? Also, what kind of device are you using to read from it? – skrrgwasme Dec 9 '14 at 23:16
• Please see my edit – UserK Dec 10 '14 at 0:11
See the relevant notes on the data sheet:
5. Use an RC filter with 1.0 kΩ and 0.1 µF on the
outputs of the accelerometer to minimize clock noise
(from the switched capacitor filter circuit).
8. A/D sampling rate and any external power supply
switching frequency should be selected such that
they do not interfere with the internal accelerometer
sampling frequency (11 kHz for the sampling
frequency). This will prevent aliasing errors.
Looks to me like the bandwidth numbers of 350 and 150 are the ones you should use. The RC values they supply are at about 1.5 kHz.
There seems to be an switched cap filter clocked at 11 kHz that shapes your signals. Also, the system depends on a resonance in the nano-elements in the 3-6 kHz range, but that doesn't mean that it can detect or report acceleration in that range.
The 11 kHz signal, and maybe some of the resonance is going to bleed through, and that's why they want you to filter at 1.5 kHz. That said, because they actually prefilter internally for you, you just need to cover the bandwidth of the accelerometer (at 350 and 150) with your sample rate. 800 Hz sampling should be fine!!
• That's perfect! The internal filter is well suited for statical applications. But in presence of vibrations a low pass digital filter is required. Thanks – UserK Dec 10 '14 at 13:15 |
# The inheritance of Mr. Jones
When I went to get the mail this morning, I noticed an official-looking letter in my mailbox. It said 'Quick 'n Quiet Notary Service' on the back, and it was addressed to me. Curious about the content I quickly opened the letter:
Dear Mr. ...,
We regret to inform you that Mr. Jones has passed away April 15th this year. Since you
are included in his will we kindly request you to visit our office at 20 Main Street,
San Francisco, CA to collect your belongings.
Kind Regards,
John Doe
Quick 'n Quiet Notary Service
Mr. Jones, my old music teacher! Although I hadn't seen him in years, it still came as a shock. We got along well, but I didn't expect him to include me in his will.
Of course I went to the notary's office to pick up the stuff Mr. Jones left me, expecting maybe one of his musical instruments or music books. However, there was only a small note with some text and some lines consisting of either numbers or letters:
Hi ...,
I know that you always liked to solve puzzles, so here I have one last puzzle for you. You have to solve it in order to know where you can find what I have left you. Good luck!
Mr. Jones
41.853589.87.624230
52.561989.13.365710
40.536355.79.715797
39.096849.120.03235
34.131270.118.49044
40.848156.73.997639
46.570230.3.3340090
NZXDNJXWKUFIVFBGHFGFIJQLYZQLNWWMMBFPU
So now I'm sitting here, trying to figure out what the note and the seemingly random sequences of numbers and letters mean. Can you help me? Where should I go to find what Mr. Jones left me?
Update:
It has been two days now since I received the strange note, and unfortunately I'm still stuck. I have been thinking about the time I took his music lessons though. He used to tell me about all of his trips around the world. Every time he told me where he was going it didn't really make sense to me, but when he told me the story behind that particular place it wasn't that strange after all...
Update:
I still have no idea where to go, and I really want to know! I remembered another thing about Mr. Jones though. Every time he returned from one of his trips we would practice a new song, and he would tell me all kinds of funny stories about the artist. He always ended those lessons with saying 'And remember: When you go to a country, you always have to visit the capital'. Even when he hadn't gone to a capital himself! Strange man...
• Is 'heritage' in the title an intentional use? Or did you mean to say 'inheritance'? – CodeNewbie Apr 19 '16 at 13:22
• @CodeNewbie I'll edit it, it's not intentional or a clue. If you happen to find more errors or things that are not clear please let me know :) – Wu33o Apr 19 '16 at 13:36
• I think we're in need of a hint here. – Chris Cudmore Apr 21 '16 at 18:19
• I think we need a poke in the right direction, here, Wu33o. :) – Khale_Kitha May 11 '16 at 13:14
• Capitol Records perhaps? – Dan Russell May 12 '16 at 14:57
Partial discoveries:
41.853589,-87.624230 - 2120 S Michigan Ave, Chicago, IL 60616 (Chess Record office), there is also Blues Heaven Foundation there which is related to music tho'
52.561989,13.365710 - MarkStrasse 40, 13409 Berlin (Lidl)
40.536355,-79.715797 - 302 Johnson Blvd, New Kensington PA 15068
39.096849,-120.03235 - Lake Tahoe, NV/CA
34.131270,-118.49044 - Mulholland Dr,
46.570230,3.3340090 - Rue Vigenere 14, 03000 Moulins (most probably cipher Vigenere)
• You're on the right track, but three of the seven names/places need some improvement before they actually make sense. – Wu33o Apr 21 '16 at 21:56
• @Wu33o added full addresses as seen at exact locations :) – kamenf Apr 21 '16 at 22:21
• You have all the information you need from the coordinates, so now you need to figure out what these places have in common. It might be a good idea to take a(nother) look at the other answer... – Wu33o Apr 27 '16 at 12:25
Ok so here is what I got so far:
The first place, 2120 South Michigan Avenue, is both an instrumental by the Rolling Stones and an album by George Thorogood.
The second place, Niedergörsdorf, Germany, has had some sort of music festival there since 2011. Also, those specific coordinates ended up giving me some kind of windmill, so I don't know if that's a coincidence or some other music reference I'm missing.
The third place, Johnson Boulevard, was also a song, this one by Amos Lee.
The fourth place, Lake Tahoe, also has music festivals.
The fifth place, Palisades Park, was yet another song, by Freddy Cannon.
The last place probably means that the code at the bottom is coded with Vigenere cipher, which of course, needs a key.
Presumably, this could be like the literal musical key the song is in, or something linking all these festivals and songs together. I can't figure out that bit, though.
Also, Quick n' Quiet sounds like a clue, like rock n' roll or something, but I'm still figuring that part out...
• There is also song named 40 Mark Strasse by The Shins. – kamenf Apr 22 '16 at 0:09
• And a song called "Lake Tahoe" by Kate Bush. – Dan Russell Apr 22 '16 at 18:25
Trying to move this forward, some research below, (Song, Group, Label, Lead Singer):
1. 2120 S Michigan Ave, Rolling Stones, Decca, Mick Jagger
2. 40 Mark Strasse, The Shins, Columbia Records, James Mercer
3. Johnson Boulevard, Amos Lee, Blue Note Records, Amos Lee
4. Lake Tahoe, Kate Bush, Anti-, Kate Bush
5. Mulholland Drive, Failure, Failure Records, Ken Andrews
6. Palisades Park, Freddy Cannon, Swan Records, Freddy Cannon |
# How to replace mathptmx \sum with CM \sum?
I would like to replace the rather ugly mathptmx \sum operator with the equivalent Computer Modern one.
Here's a compilation of display-style \sum symbols provided by various font packages.
I'd say that the summation symbol provided by the mtpro2 package is closest to the one provided by the Computer Modern family, while still providing the genuine "Times Roman" look. Incidentally, the full mtpro2 package isn't free of charge. However, its lite subset, which is all that's needed to produce the good-looking \sum symbol, may indeed be downloaded free of charge.
Here's the code that generated the preceding four lines (run consecutively for each choice of font family).
\documentclass{article}
\usepackage{array}
%\usepackage{mathptmx}
%\usepackage{newtxmath}
\renewcommand{\rmdefault}{ptm} \usepackage[lite]{mtpro2}
\begin{document}
\begin{tabular}{@{}>{\ttfamily}p{4cm}>{$\displaystyle}l<{$}}
Times Roman, mtpro2 & \sum
\end{tabular}
\end{document}
• Hey, do u think it is possible to use mtpro2's largesymbols with mathptmx? It kind of solves my problem. Oct 26 '15 at 18:07
• @MiladO. - I would use the mtpro2 package instead of mathptmx.
– Mico
Oct 26 '15 at 18:47 |
# Context
This topic is part of the ANR project BECASIM (2013-2017). Using the home made software GPS (Gross Pitaevskii Simulator), we simulate rapidly rotating BEC with different kinds of trapping potentials.
BEC are modeled using the stationary Gross-Pitaevskii equation (in dimensionless form)$$\begin{array}{rcl} \mu\phi &=& -\frac12\Delta\phi + V(x)\phi + \beta |\phi|^2\phi - \Omega L_z\phi \\ \|\phi\|_{L^2} &=& 1. \end{array}$$ $\phi$ is the stationary wave function, $V$ is the magnetic trap, $\beta$ stands for the interaction between particles, $\Omega L_z$ is the angular momentum, and $\mu$ is the so-called chemical potential.
# Tools
We use our home made software GPS:
• written in Fortran,
• hybrid MPI/OpenMP code,
• spatial discretization: spectral or high order compact finite differences scheme,
• imaginary time method with semi-implicit Crank-Nicolson scheme, or Sobolev gradient methods.
# Results
Rotating BEC with different trapping potentials. Vortices are organized in Abrikosov lattices.
Exploring a ground state with VisIt |
## Elementary Geometry for College Students (7th Edition) Clone
Published by Cengage
# Chapter 5 - Section 5.2 - Similar Polygons - Exercises - Page 235: 17
#### Answer
$n=3$
#### Work Step by Step
Since the quadrilaterals are similar, the corresponding sides will be proportional. So, set up a proportion... $\frac{AB}{HJ}=\frac{BC}{JK}$ Substitute in the appropriate values... $\frac{5}{10}=\frac{n}{n+3}$ Cross multiply... $5(n+3)=10(n)$ Distribute the five... $5n+15=10n$ Subtract $5n$ from both sides... $15=5n$ Divide by five... $n=3$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
I recently attended an academic conference and was struck by the inefficiency of the buffet-style dinner. The conference had approximately 500 attendees, and dinner was scheduled for 6:30 PM following a 15 minute break in the conference program. A 15 minute break is not enough time to go back to the hotel room and take a nap, and barely enough time to find a quiet corner and get some work done, but it is a perfect amount of time to crowd around the buffet waiting for dinner to open.
When dinner finally opened, all of the food was arranged on a single table, inviting the attendees to form a single line to serve themselves. As you can imagine, this line was quite long. I was lucky enough to be one of the first people through the line, but by the time I finished eating a half hour later, the line was still very long. I had been waiting for someone who was towards the middle to end of the line, and this person still had not made it through yet.
I found it odd that feeding people should be so slow. From my experiences in undergraduate dining halls, it is possible to feed more people in a shorter amount of time. A key difference between these situations is that in undergraduate dining halls, food is often served at individual stations, meaning you only need to wait in line for food you want to eat. By contrast, in the catering style, it is often served all at one table, with diners waiting in a single line and accessing the dishes one by one. I wanted to examine the efficiency of these two systems. This is important not only for minimizing the mean wait time so that everyone gets their food faster, but also for minimizing inequality between people at the front and end of the line. This ensures that everyone has the opportunity to dine together.
## Model
I modeled this situation as a single line in a buffet versus individual lines for each different dish. I made the following assumptions:
• Some people may be faster or slower at serving themselves.
• Some foods can be served faster than other foods.
• People may not want all of the food which is being served, and each person wants a different random selection of foods.
• Only one person can serve themself a particular dish at one time.
• When dishes have their own individual lines, people look at the lines for the foods they want to eat and stand in the shortest line next.
• People already know what the options are and where they are located.
I calculated the wait time for each person in the simulation as the total amount of time it took a person to pass through the buffet. I then looked at the mean wait time for the group as well as the inequality in wait time for people in the group, defined as the third quartile minus the first quartile. Each simulation was run many times to ensure accurate statistics.
## Number of dishes wanted
Let us suppose that not everyone wants the same number of dishes, but that all dishes are equally attractive. First we look at the case when there are a limited number of dishes.
We see that when there aren’t very many dishes and most people want all of them, it is faster to have a single line. This may be counter-intuitive, but it is due to the fact that people do not optimally distribute themselves, but instead choose the shortest line. Suppose for example that one dish is much slower to serve than all of the others. People who choose this food last will have to wait approximately the same amount of time as they would have if there was a single line and they ended up at the end, because this dish serves as the bottleneck. However, the people who are at the front of this line will still need to wait in more lines for the other dishes, because other people tried to serve themselves these dishes first. As a result, having multiple lines can sometimes increase the amount of time for the fastest people and not decrease the amount of time for the slowest people.
Additionally, there is a large inequality in wait times, i.e. some people will get through the line quickly, while others will be stuck in line for a long time. This is the case for both serving styles, but is especially pronounced for the case with multiple lines.
Let us also examine the case when there are many dishes to choose from.
When there are many dishes to choose from (here 20), no matter how many dishes people may want (within reason), individual lines reduce both the mean wait time and the inequality in wait times compared to a single line. Intuitively, this is because people can distribute themselves and they only have to wait for the dishes they want to eat.
Additionally, let’s look at the case when there are many people.
In this case, we see the counter-intuitive result again: the mean wait time is quite a bit higher for individual lines when most people want most of the foods, however inequality is still lower.
Finally, we can examine the case when there are very few people.
In this case, separate lines are better for both mean wait time and equality.
## Fairness
A fair system is one in which the amount of time someone waits is proportional to the number of dishes they want. In an unfair scenario, someone who only wants one dish must wait for the same amount of time as someone who wants all of the dishes.
Let’s look at whether this form of fairness holds. First, we look at how long someone must wait depending on how many dishes they want.
As expected, when there is only one line, everyone must wait for approximately the same amount of time, no matter how much food they want to eat. People who want all of the dishes in a line must wait for less time on average, but someone who only wants one dish must wait for a very long time. By contrast, when there are multiple lines, the amount of time people wait is proportional to the number of dishes they want to try.
Similarly, it might be fairer that someone who can serve themself quickly has a shorter waiting time than someone who is slower.
Unfortunately this does not seem to be the case in either system. Rather, people who are slow to serve themselves take approximately the same amount of time in line as those who are fast.
## Summary and conclusions
In summary, when there are a lot of people present, if everyone wants most of the food at the buffet, a single line counter-intuitively reduces the mean wait time. However, this single line substantially increases inequality in wait times, meaning that some people will have to wait for a long time while others can go through immediately. Additionally, people who only want a small amount of food must wait a long time to serve themselves. A more fair but slightly less efficient system is one where there is a separate station for each dish, but this can be inefficient when most people want most of the dishes available.
This analysis leaves out a few factors which are difficult to account for. For example, it assumes the amount of time taken to walk from one food to another is negligible, and that people know a priori what food they would like to eat and where it is located. Both of these have the potential to slow down serving times in the case with separate lines. This analysis also doesn’t account for several other factors which are important in real life. For example, it assumes that space is not an issue. It also assumes there is enough seating to accommodate everybody; if only a limited amount of seating is available, a high inequality is desirable as it prevents everyone from going to the dining area at one time.
One method which is often employed to speed up single lines is having more than one identical line, or two sides on the same line, likewise, in the case of separate lines, there are sometimes “stations” which have identical dishes. In both cases, because we assume people balance themselves by going to the shortest line, doubling the number of copies of all dishes would be expected to approximately cut the mean wait time in half. |
## March 1, 2008
### Computer Scientists Needed Now
#### Posted by John Baez
Thanks to advice from Andrej Bauer, Robin Houston and Todd Trimble, I’ve beefed up the logic section of this paper:
For example, I’ve included a longer ‘overview’ to give the non-logician reader a slight feel for how proof theory met category theory in the development of 20th-century logic. I hope there are no egregious errors. If you catch any, let me know.
But now I really need comments from anyone who likes categories and theoretical computer science!
In fact, the final ‘computation’ section of the paper is still very rough. It introduces combinators but doesn’t really explain them well yet… and perhaps worse, it doesn’t say anything about the lambda calculus!
I plan to fix some of these deficiencies in the next week. But, I figure it’s best to get comments and criticism now, while there’s still time to take it into account.
Posted at March 1, 2008 1:47 AM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1617
### Re: Computer Scientists Needed Now
A really minor comment: the definition of currying on page 57 gives the type as
$((X\otimes Y)\multimap Y)\multimap(Y\multimap(X\multimap Z))$
whilst virtually all the references on the web use the form
curry : ((a,b) -> c) -> (a -> b -> c)
(keeping it in the original notation CS people will be used to)
eg, see http://en.wikipedia.org/wiki/Currying http://www.cs.nott.ac.uk/~gmh/faq.html#currying
so you might want to explain why you’ve chosen an opposite definition. (In fact, you seem to have the opposite convention on pages 58-59 in calculi rules.) Also, you motivate product types with “This gives a less awkward way to deal with programs that take more than one argument.” I’d say you probably need to go deeper and say that product types make it easier for “functions” to output intermediate data structures so you can write programs piece by piece (as there’s no obvious “lack of ease” in having more than one argument in the curried style in the examples you’ve presented in the paper).
Posted by: bane on March 1, 2008 10:50 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Oops: on p 57 type is
$((X\otimes Y)\multimap Z)\multimap(Y\multimap(X\multimap Z))$
(Got confused where I was in the latex.)
Posted by: bane on March 1, 2008 10:58 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
We explained that way back on page 26 when we first introduced currying. The reason X and Y appear switched is because our string diagrams either need to be left-closed or draw the hom backwards: Given
(1)$\array{X \downarrow Y \downarrow \\ f \\ Z \downarrow}$
we can either pull down $X$ on the left
(2)$\array{Y \downarrow \\ \tilde{f} \\ X \uparrow Z \downarrow}$
which gives the switched order in the paper or pull down $Y$ on the right
(3)$\array{X \downarrow \\ \tilde{f} \\ Z \downarrow Y \uparrow}$
where we’d have to denote the output as $Z ⟜ Y$ instead of $Y \multimap Z$.
Posted by: Mike Stay on March 1, 2008 4:26 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
My fault; I jumped straight to the computer science and didn’t read the earlier parts. Might still be worth a brief reference to that earlier point.
Posted by: bane on March 1, 2008 4:39 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
bonjour à tous
for info, Lambda Associates
cordialement votre
bruno
Posted by: bruno frandemiche on March 1, 2008 8:30 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Thanks, bane! Ah yes, the struggle between the usual convention
$(X \otimes Y) \multimap Z \cong X \multimap (Y \multimap Z)$
and our
$(X \otimes Y) \multimap Z \cong Y \multimap (X \multimap Z) !$
This is the kind of thing that gives me ulcers. I don’t mind explaining our convention again, because I expect most readers will read portions of this paper instead of chugging through the whole thing end-to-end… and the convention that looks really nice when working with string diagrams winds up looking really annoying 20 pages later, when we get to currying in computer science.
It’s okay to flip-flop between the two conventions in a symmetric or even braided monoidal category, where we have $X \otimes Y \cong Y \otimes X$. But in a mere monoidal category, we don’t have this, so there are really ‘left closed’ and ‘right closed’ monoidal categories — two different notions! Of course, if $C$ is left closed, $C^{co}$ is right closed. But still…
So, in the merely monoidal case, there’s a strange distinction between what works best when writing expressions in a linear way (as above), and what works best when writing them 2-dimensionally as string diagrams.
Thanks for offering a better justification of product types. I’ll use it!
Now the big question: should I acknowledge ‘bane’ at the end of this paper?
Posted by: John Baez on March 1, 2008 7:37 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
I only read Section 1.4. In general, I think it’s well-done (besides the parts that are obviously unfinished.) I like the clear and simple language, although it sometimes degenerates into excessive handwaviness, as I’ll elaborate.
Page numbers are what my PDF reader tells me. (Adding page numbers to the document itself might help reviewers.)
Section 1.4.1:
p. 52:
“While a Turing machine…” – This sentence is strangely non-parallel. I suggest contrasting Turing-machine-as-hardware with lambda-calculus-as-software.
“The most famous of these are Haskell and Lisp” – uh, I say this a Haskell guy, but the ML family of languages is just as famous as Haskell if not more so.
“So, everyone majoring in computer science at MIT…” – see http://www-tech.mit.edu/V125/N65/coursevi.html – you might want to fact-check this before the book goes out.
“In imperative programming, a program…” – this sentence is vague, and so is the rest of the paragraph. In particular, “how to change state” hides the real difference between imperative languages and functional languages. Languages in both paradigms let you write programs that manipulate state. The distinction is as to whether the state is *explicit* or not. If I were writing this I’d explain that the imperative paradigm concerns more operational, or procedural approaches to describing algorithms, and the functional paradigm emphasizes more mathematical, or declarative approaches.
Same para: “Functional programming is seemingly more limited…” – watch out when you use weasel words like “seemingly”. Whether functional programming seems more limited depends on your perspective. It may seem more limited to C programmers. But other people aren’t necessarily coming at it with that bias. It’s also odd and misleading to describe functional languages’ lack of implicit state as a “limitation”, since as you well know, languages with or without implicit state are equally computationally powerful.
There’s some confusion and vagueness in this section as to whether by “functional languages” you mean “purely functional languages”, which you surely do. An impure functional language (and some people say “if it isn’t pure, it isn’t functional”) like Lisp doesn’t have the nice theoretical properties you want. The only widely used programming language that does is Haskell 98 (not that anyone really programs in Haskell 98, but with the language extensions that most people do use, Haskell isn’t pure either…)
“In fact, the best-behaved functional programming languages…” – “best-behaved” is quite a subjective term. Programmers may well think that a strongly normalizing language (such as the simply-typed lambda-calculus) is *not* well-behaved – most programmers like being able to write nonterminating programs! I think this entire paragraph needs a lot of work. “Input into itself”, for example, is handwavy. You’re eliding a lot of subtlety here. At least to me, when I was learning this subject, the connection between types and the ability/lack thereof to write nonterminating programs was far from obvious.
“…the typed lambda calculus is just another way of talking about cartesian closed categories”: “Just another way” is rather dismissive. I suggest saying “Lambek found a way to interpret the typed lambda-calculus in terms of cartesian closed categories,” or something like that, that doesn’t imply that cartesian closed categories are the more privileged theoretical concept. Along the same lines, I think your assertion that “theoretical computer science has been closely entwined with category theory” is controversial. Certainly there has been a lot of work on the connections between CT and theoretical CS, but not all theoretical computer scientists think that CT is central to their work. It might be good to get more opinions on this.
“Combinatory logic is closely related to the lambda calculus…” – this is vague. Either explain that the two formalisms are related as being alternative minimalist Turing-complete computational systems (I don’t see what other relationship you might mean), or don’t say it at all.
p. 53
“Strips everything down to the bare minimum” is vague. Emphasize that combinatory logic is different because it avoids naming and the complicated machinery (substitution) associated with it – that would be easier to understand. “…In some ways more awkward…” – Again, this is weaselly. Some people think lambda calculus is more awkward because you need to worry about substitution. Don’t state opinions as facts.
“…it seems more efficient…” – Be bold and don’t say what “seeems” to be more efficient, say what you think *is* more efficient.
Section 1.4.2 (still on p. 53)
1st para: Rewrite to be parallel: “Mathematicians might apply… Functional programmers apply…” (Then you get the added bonus of active voice.)
“…suppose we have programs…” “…compute the obvious functions…” – I think you’re using “program” and “function” interchangeably. Use a consistent lexical set, or if you do mean something different, explain the difference between “program” and “function”.
You introduce equational reasoning here in a very vague way. Not all readers may be used to this style of writing programs.
p. 54
“To avoid problems…” – super-vague. Say *what* problems a non-confluent rewrite system causes. Or just declare that we want to talk about confluent rewrite systems, without justifying it.
Your definition of “confluence” is wrong or at least very misleading: when you say “back to the same expression”, I think you’re trying to say that when there’s a choice of rules to apply, you’ll still get the same normal form in the end, either way. But the resulting normal form is not the original expression, which “back to the same expression” implies. This whole paragraph is pretty handwavy.
I suggest putting in a citation for those who want further reading on rewrite systems: Baader and Nipkow’s _Term Rewriting and All That_ is a good one.
“We can handle…” Avoid noise words like “using”. “‘Currying’ is a trick that helps us handle functions…” In the next sentence, what’s “this”? Well, currying doesn’t turn a function of several arguments into anything – currying lets us *define* a function of several arguments as a function that (blah blah). So say that.
“…we can imagine a program plus that adds integers” – this paragraph seems to end a bit abruptly.
“…that branch to the left:” – I find that sentence unnecessary; you can just look at the picture.
p. 55
“…has no well-defined meaning.” – it’s nonsensical to talk about whether a program has a well-defined meaning unless you’ve defined the semantics for the language in which you’re programming. I would say this differently. It’s possible to define a very sensible language in which “(double Tuesday)” has a meaning, after all. Instead I would pick an example of an ill-typed program that’s actually *inconsistent*: for example, one that uses the same value in an integer context and a string context. Now, this could have a perfectly good meaning as well (say, in Tcl), but then you can point out that in most sensible languages it would be ill-typed because strings and integers have different machine representations.
“…instantly complain…” – it’s unclear what “instantly” means. Instead, I would emphasize the compile-time/run-time distinction.
It would be nice to give a nod to type inference in this section, so readers aren’t left with the impression that you always have to write explicit types in a functional language.
“in a variety of ways” – delete that phrase; the sentence has equivalent meaning without it.
“…there is a new type…” – the use of “new” is strange; a function type is no newer than its components. “another type”, perhaps?
“Any program is built…” – this sentence is extremely confusion. Rewrite it, preferably as two or three separate sentences.
“…a datum of type X…” – so f is like a constant function. Say that, perhaps.
p. 56
“…can be reinterpreted as…” – because you use the passive voice, I am confused by what you mean. Who does the reinterpreting? Saying it would clarify the idea.
“In functional programming, we take advantage…” – I have no idea what you’re trying to get at with this sentence.
“Computer scientists call…” – again, the “new”/”old” language is confusing. How about “building more complicated types from simpler types”?
“…a less awkward way…” – Again, don’t make unsubstantiated assertions. Whether currying is more awkward than tupling is a matter of taste.
Skipping to p. 58:
The void type (or unit type – which might be less confusing than “void type”, since the “void” type in C has a very different semantics than the unit type in Haskell or ML, and I think you really mean the latter) is useful in purely functional languages, too, not just in languages with side effects. See Pierce’s _Types and Programming Languages_. Also, “languages with side-effects” is a strange thing to say, because all programming languages allow you to write programs that have side effects. But a purely functional language provides a type system that controls and encapsulates effects. I would distinguish between pure/impure languages rather than languages with/without side effects.
Posted by: Tim Chevalier on March 1, 2008 9:34 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Bottom of page 55 where you explain that datatypes are objects and morphisms are equivalence classes of programs: depending on how it fits with your overall scheme, you might want to mention that the rewrite rules are 2-cells between programs, so you can view the datatypes-programs-rewrite rules as a 2-category.
Page 58: call it “unit type” whose only element is “unit” (write it as empty tuple perhaps?) and mention that this is “sort of like the void type in C/C++/Java”. Please do not propagate the horrible misnomer “void type” (after all, the type is not void, is it?)
There are two main uses for the unit type: (1) to indicate that no result is returned, and (2) for thunking.
Speaking of thunking, you have the opportunity to explain the difference between eager evaluation (call-by-value) and lazy evaluation (call-by-name), and relate these two concepts to two symmetric monoidal closed structures (please verify, I might be slightly wrong here):
(a) call-by-name is like Sets with products and exponentials so that you actually get a CCC (eta-reduction is valid)
(b) call-by-value is like Sets and partial functions, where the structure is that of a smash product and the space of partial maps. Eta reduction is not valid because, while a term t might diverge, lambda x. t x does not.
I do not understand why you insist on confluent and terminating rewrite rules. That automatically prevents you from having a Turing complete programming language, which is sort of a bummer. You could just say that in name of simplicity you assume the rewrite rules to be determinstic (but that computer scientists also deal with non-deterministic computation), and that in principle a computation can be diverging (infinite), and that there is no way around it if we want a Turing complete language.
Posted by: Andrej Bauer on March 1, 2008 11:24 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Sorry, I got the two structures wrong.
(a) call-by-value: pointed sets and strict maps (maps preserving the point, think of the point as “undefined”), the structure is: tensor is smash product and hom is the set of strict maps.
(b) call-by-name: pointed sets and _arbitrary_ maps (no need to preserve the point), with the usual ccc structure. This is equivalent to non-empty sets and maps. The point of having the points is that in a typical programming language all types are inhabited.
For a more refined analysis one would need fancier categories, such as those appearing in domain theory. You probably do not want to get into that.
Posted by: Andrej Bauer on March 2, 2008 12:31 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
As with the others, I agree that “unit” is a much better name than “void” and is the actual name that is used in functional languages (i.e. it’s exactly the name used by ML, Haskell uses () obviously relating to empty tuples and this is read as “unit”.)
In a pure, strongly normalizing language, there is almost no use for the unit type practically speaking. When non-termination is allowed, then the unit type can be used by eager languages, e.g. ML, to delay evaluation, but it is still pretty useless in lazy languages, e.g. Haskell.
Specifically in response to Andrej, the connection between logic and programming becomes much less compelling in the face of non-termination. In particular, adding the Y combinator makes the logic corresponding to the language inconsistent. However, often symmetric monoidally closed categories don’t come up until we start talking about non-termination or concurrency or explicitly about resources. However, too “strong” a side-effect breaks the nice properties and we start to talk about pre-monoidal categories usually built up into Freyd categories.
I think it would be very helpful to mention some of the most popular combinators and how selecting different subsets of them given different families of logic. In particular, the most well-known set of combinators is the S, K, and I combinators (I is redundant being SKK). It happens that S and K correspond to contraction and weakening respectively. (I is the identity.) In the untyped combinatory calculus, S and K are universal. Some other popular combinators are B (= compose) and C (= braid). B, C, and I are often used for linear combinatory logics.
Another significant thing is that application corresponds modus ponens. I.e. this makes up the very direct correspondence between combinatory logic and Hilbert style proofs. The sequent calculus corresponds more closely to the lambda calculus.
Another important thing is that most typed functional languages are parametrically polymorphic. A property called parametricity leads to what are called free theorems. These free theorems are pretty much the (di)naturality equations. So, in short, if type constructors are functors, (parametrically) polymorphic functions are natural transformations.
You don’t mention any combinators corresponding to the dualization operation. They exist, but are much less common. They are related to what are called control operators in computer science which in turn are related to continuations. Usually the way they are explained is by rewriting the program in continuation passing style and then the control operators can be given explicit functional definitions. Continuations are related to logical negation and this transformation into continuation passing style happens to correspond to exactly the embedding of classical propositional logic into intuitionistic propositional logic you mentioned in the section on logic. Getting well-behaved control operators is a bit tricky; the lambda-mu calculus is the thing to look at here.
The papers on the Categorical Abstract Machine give a good and concrete overview for “compiling” lambda terms to categorical combinators.
Finally, there is a “Curry-Howard correspondence” for logic programming languages as well.
Posted by: Derek Elkins on March 2, 2008 1:15 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Derek wrote:
I think it would be very helpful to mention some of the most popular combinators and how selecting different subsets of them given different families of logic. In particular, the most well-known set of combinators is the S, K, and I combinators (I is redundant being SKK). It happens that S and K correspond to contraction and weakening respectively. (I is the identity.) In the untyped combinatory calculus, S and K are universal. Some other popular combinators are B (= compose) and C (= braid). B, C, and I are often used for linear combinatory logics.
Thanks for all your comments! Mike had some detailed remarks about the $S, K$ and $I$ combinators. I’ve temporarily removed them, but I plan to restore them as part of an expanded ‘overview’. Seeing these combinators expressed in terms of the lambda calculus and also in terms of cartesian closed categories will be a good way to help our readers understand all three subjects… at least, it helped me.
In fact, Mike had originally set up a ‘typed linear lambda calculus’ that corresponds to symmetric monoidal closed categories in the same way that the usual typed lambda calculus corresponds to cartesian closed categories. However, it seems a bit tiresome to enforce the ‘no duplication and deletion of variables’ constraint in this sort of language — one needs to pay careful attention to free vs. bound variables, which is not what we want our readers to spend their time thinking about here. Using ‘linear combinators’ seems to work more easily. So, Mike moved in this direction.
It sounds like you’ve read a bunch of stuff about ‘linear combinatory logics’. The only really detailed reference I know is Abramsky, Haghverdi and Scott’s Geometry of Interaction and Linear Combinatory Algebras. Which others should I know about?
Posted by: John Baez on March 2, 2008 6:15 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Most of the work on linear logic in computer science works with variants of linear lambda calculi. Most of what I know has come from a variety of sources and interpolation.
Looking for stuff now, Gavin Bierman’s thesis “On Intuitionistic Linear Logic” seems like a decent resource and briefly covers linear combinatory logic. Various other papers on his site seem relevant as well. http://research.microsoft.com/%7Egmb/publications.aspx
Posted by: Derek Elkins on March 2, 2008 10:29 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
I spotted a small typo: Roberto di Cosmo’s last name is incorrectly spelled “di Cosimo” through the paper.
Posted by: A.G. on March 2, 2008 12:51 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
At the top of page 2 there is a minor typo.
The second sentence reads: “The idea is that a proof is process going us go from one …”. It should probably read “… process taking us from one …”
Posted by: Mark Biggar on March 2, 2008 3:55 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
WOW! Thank you all for coming out of the woodwork and posting these wonderfully detailed, wonderfully helpful comments.
I won’t thank you all by name now: I’ll do that in the paper. I may have more questions later. But, I really must thank Tim Chevalier by name right now for his line-by-line dissection of suboptimal wordings — and even better, suggested improvements. It takes a lot of work to go through prose and write up detailed comments like this. I do it for my grad students. It’s a real pleasure and an honor to have someone do it for me.
Mike Stay does programming and computer science for a living, but I don’t… so to me, it’s a slippery and mysterious business — unlike, say, mathematical physics, where I’ve put in enough time to have a feel for it. To get happy with this paper I really had to take everything he wrote in this section and rewrite it line by line, simply to see if I understand it. (If I can’t say something in my own voice, I don’t understand it.) I’m in the middle of that process now. So far I feel quite uncomfortable, because there’s a lot I don’t understand. That’s probably why a lot of my sentences have weaselly wordings.
So, getting all this feedback helps a lot.
By the way, I’ll be glad to use the term ‘unit type’ instead of ‘void type’ — ‘void’ makes me think of the empty set, or more generally initial objects, instead of unit objects.
Posted by: John Baez on March 2, 2008 5:50 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
I have some minor observations about the latest draft of the logic section. The first few are historical points. The last two are points about your discussion of the sequent calculus.
1. (p 35)
“One reason is that some logicians deny that certain familiar principles are actually valid. But there are also subtler, more interesting reasons.”
I would consider changing that last sentence to “But there are other reasons as well.” After all, Martin-Lof’s reasons for rejecting classical reasoning are very subtle and very interesting, both mathematically and philosophically!
2. (p 35)
“For example, around 1907 Brouwer [51] began developing ‘intuitionistic logic’. “
It is true that Brouwer developed intuitionism as a philosophy of mathematics, but I don’t think it is correct to say that he had any *direct* role in the development of intuitionistic logic. This distinction should go to his student Heyting, along with Kolmogorov in Russia. As far as I understand, Brouwer was resolutely “anti-formalist” and actually did not approve of Heyting’s formalization of intuitionism in logical terms. If I were writing this paper, I would want to avoid all of these historical fine points and just say something like the following:
“For example, in the late 1920s and early 1930s logicians and mathematicians, under the influence of Brouwer’s “intuitionistic” philosophy of mathematics, began to develop formal logics that rejected the inference from “it is not the case that it is not the case that X” to “X”. For obvious reasons, these logics were dubbed “intuitionistic.”
3. (p 36)
“In the 1930’s, this idea was made precise by Godel [47] and Gentzen [44].”
While it is true that Godel and Gentzen made this idea precise in the 1930’s, you might also want to mention that Kolmogorov, working independently of this entire tradition, made it precise in 1925! I believe that you can find this result in his paper “On the principle of the excluded middle” which is collected in Van Heijenoort’s “From Frege to Godel” anthology.
(You might also mention that this paper contains the earliest formalization of intuitionistic logic, predating Heyting’s by about five years.)
Incidentally, when these results of Godel and Gentzen are mentioned by historians of logic, they almost always also mention the closely related theorem proved by Glivenko in 1929: http://en.wikipedia.org/wiki/Glivenko’s_theorem.
4. (p 38)
“Gentzen listed enough inference rules to describe the whole classical propositional calculus, and a subset that give the intuitionistic propositional calculus.”
I’m no expert, but I am fairly certain that one can’t “just” take a subset of the inference rules for the classical sequent calculus in order to get the intuitionistic sequent calculus. One must also restrict the form that sequents can take (at most one conclusion) along with stipulating a new rule for introducing the “or” connective on the left. Maybe you are trying to say something different though? If you are, I think this sentence needs restating.
Similarly, the sentence directly after the one I just quoted seems to say something false. First of all, you are talking about the “classical” sequent calculus, not “classical logic.” Second, the “intuitionistic” sequent calculus does not allow for multiple formulas on the right of the turnstile in a sequent, so it does not even make sense to say that such an equivalence holds in the “intuitionistic” sequent calculus. But again, perhaps I am misunderstanding what you are trying to say.
5. (p 39)
Your discussion of the sub-formula property and of cut-elimination seems excessively handwavy. That is, you say what the sub-formula property is but when it comes time to say why it is so important, you just way your hands and mutter some vague stuff about the set of proofs for a given sequent. Similarly, you say what cut-elimination is but when it comes time to say why it is so important, you just mutter some vague stuff about proof-theorists being obsessed with the theorem. You should say why the subformula property makes proof search behave nicely. You should say why proof-theorists are “obsessed” with cut-elimination (for instance, the relation between cut-elimination and consistency). If you think briefly mentioning these things might take you too far afield, I suggest that you just write
“Since any proposition has just finitely many subformulas, this makes the set of possible proofs of a given sequent quite manageable. For more on this point and it’s application to automated deduction, we recommend [references]”
Similarly:
“From the two sequents on top, the cut rule gives us the sequent below. Note that the intermediate step A does not appear in the sequent below. It is ‘cut out’. So, the cut rule lacks the subformula property. One of Gentzen’s great achievements was to show that any proof in the classical propositional (or predicate) calculus that can be done with the cut
rule can also be done without it. This is called ‘cut elimination’. To see why cut-elimination is such a deep and important result, we recommend [references].”
In the actual paper, it *seems* like you are handwaving because you don’t know how to succinctly explain (1) why the sub-formula property is important and (2) why proof-theorists are so concerned with cut-elimination. If you are going to handwave, you should make it seem like you are doing so because you don’t want to get sidetracked with details that you could otherwise easily fill in. I didn’t get this sense when I read the portions about cut-elimination and the sub-formula property.
Having said all this, let me say that I am extremely excited about this paper and can’t wait to see the finished version!
### Re: Computer Scientists Needed Now
From my reading of his philosophical writings, I second that comment on Brouwer, who was very distrustful of logic. I don’t have a quotation to hand.
The Stanford Encyclopedia has this to say:
Brouwer holds that mathematics is an essentially languageless activity, and that language can only give descriptions of mathematical activity after the fact. This leads him to deny axiomatic approaches any foundational role in mathematics. Also, he construes logic as the study of patterns in linguistic renditions of mathematical activity, and therefore logic is dependent on mathematics (as the study of patterns) and not vice versa.
I supposed you might say he was only hostile to the logic of the time (classical) to the extent that it had lent itself to the firming up of the faulty intuitions of classical mathematics, and that a codification of the correct intuitions would be useful. But I don’t think this is right. There’s no sign of his approving of Heyting’s work.
Posted by: David Corfield on March 3, 2008 1:29 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
David Corfield wrote:
I supposed you might say he was only hostile to the logic of the time (classical) to the extent that it had lent itself to the firming up of the faulty intuitions of classical mathematics, and that a codification of the correct intuitions would be useful. But I don’t think this is right. There’s no sign of his approving of Heyting’s work.
However, I do not see the relation with the quote from the Stanford encyclopedia:
Brouwer holds that mathematics is an essentially languageless activity, and that language can only give descriptions of mathematical activity after the fact. This leads him to deny axiomatic approaches any foundational role in mathematics. Also, he construes logic as the study of patterns in linguistic renditions of mathematical activity, and therefore logic is dependent on mathematics (as the study of patterns) and not vice versa.
This quote does not show any hostility towards logic, but only denies that logic is prior to mathematics. Heyting’s codification of intuitionistic logic is a mathematical study of the language of intuitionistic mathematics after the fact. Brouwer would deny that this “axiomatic approach” has a foundational role. However, I do not read this is a hostility towards logic as the study of patterns.
Bas
Posted by: Bas Spitters on March 3, 2008 9:22 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
I wasn’t basing my comment merely on that quotation from the Encyclopedia. I read Brouwer’s philosophical writings extensively during my MSc, longer ago than I care to think. I don’t have quotations to hand, but I remember a great suspicion of language on his part.
The Heyting entry on Wikipedia has Brouwer describe his student’s work as a “sterile exercise”. Do you have evidence otherwise?
Posted by: David Corfield on March 3, 2008 10:19 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
The wikipedia article on Heyting indeed contains this quote, but without correct citation. [The page it refers to does not contain this phrase anymore. ]
Van Atten’s article in the Stanford Encyclopedia seems to give a balanced description of Brouwer’s ideas. Is it possible that Brouwer was reacting to the use of intuitionistic logic as a foundation for intuitionistic mathematics?
It seems that Brouwer’s `logic is part of mathematics’ is precisely the way in which we use it here at the n-cafe.
Bas
Posted by: Bas Spitters on March 4, 2008 8:53 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
The ‘sterility’ comment is said to appear in van Stigt’s book. I remember reading the comment somewhere, and I have read van Stigt’s book, but it may have been somewhere else.
Anyway, as to whether Brouwer did or did not approve of what Heyting did is of limited importance to me. At the time I was working on my masters I was interested in the case of intuitionistic logic for its illustration of two themes:
1) That logic is not so much about mathematics, as a part of it. Where the philosopher Michael Dummett claimed that the proof theoretic wing to intuitionistic semantics is the correct one, I delighted in the additional existence of a topological wing, and the movement towards toposes.
2) Against Lakatos’s (and it appears Brouwer’s) fears that formalisation of the informal is dangerous in that it may leave an important part out, I was interested to see how formalisation can act as a springboard for the formation of a new intuition.
There is no need to adhere to all the informal motivations of the originator of a theory, though, of course, this wouldn’t prevent the possibility of there being something further worth mining from Brouwer’s philosophical writings. As all I can recall is the pronouncement “Woman is a tiger”, it suggests that his ideas didn’t make much of an impression on me.
Posted by: David Corfield on March 4, 2008 12:04 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
First, I’d like to say that I’m also very excited about this paper! I agree with most of the comments so far, particularly Andrej and Derek. Here are a few more comments of my own…
In table 1.4, you say “programs executing in parallel”, and I believe this is the first occurrence of this operational notion of parallelism. It’s not clear to me that we need or ought to consider products of programs as being “parallel” in the familiar computational sense, although they could be considered parallel in the sense that they are semantically independent. In any case, if you want to use this characterization, it should probably be introduced/justified in the text.
On page 56, you say “In computer science this corresponds to the fact that programs are data…” I’m not sure I’d say this. Generally speaking, when programmers use the phrase “programs as data”, they’re referring to the direct manipulation or inspection of the representation or specification of a program. The classic examples of this are Lisp macros and manipulation of machine code (JIT compilation, for example). This is a somewhat different notion than currying and first-class functions, and is also often called metaprogramming.
It doesn’t help that the standard name in computer science for a function from a program representation to the program itself is also called “eval”. For this reason, a computer scientist might find it unfortunate that categorists use the name “eval” for the arrow $(Z^Y \times Y) \rightarrow Z$. We would mostly prefer to call this arrow “apply”. It is very, very important to understand the distinction between the programmer’s eval and apply, and this often causes problems for new programmers.
I strongly disagree with the idea that the unit type is not useful! Mostly it seems not useful because programmers are unused to thinking so abstractly. From the categorical point of view, all zero-parameter data constructors (e.g., True and False, each of the integers) are categorical elements (arrows $I \rightarrow X$), and these show up all over the place. In my opinion, this leads to useful insights, such as that the standard type of boolean values is $2 \cong 1 + 1$. You said this yourself on page 55, so I’m surprised that you’re so quick to slander the terminal object.
Regarding termination, I would not be so quick to introduce inconsistency. There’s interesting recent work on total functional programming, and I can imagine a bright future where we no longer say “Turing complete” (as though it’s a good thing), and just say “inconsistent”.
If you really do want to get Turing completeness, you might want to mention the standard semantic trick of using total functions between pointed objects, as Andrej mentioned, but I very strongly agree with Derek that this obscures the connections with logic.
Finally, as Derek suggests, I do urge you to at least mention polymorphic functions as natural transformations, as well as the connection between control operators and duality.
Posted by: Matt Hellige on March 3, 2008 11:27 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Comments about the unit type being useless are probably inspired by my first comment. While I certainly agree with myself the way I intended it (part of that being only in “practice”), I do certainly agree that the unit type is of crucial importance conceptually. However, you do have a good point that the unit type is used more often than it seems as it is encoded into the data type construction facility of ML/Haskell-style languages. One example is the Bool example you describe which could be rendered in Haskell as Either () (), another is Haskell’s Maybe type (1+Id) which could be rendered Either () a (Either corresponds to +). So my statement was probably overly harsh.
Also, when looking for something on linear combinatory logic, I was reminded of the perp operation some linear logics support. Classical (non-linear) logic’s negation tends to lead to control operators. The linear negation of classical linear logic seems to lead to concurrency. The following paper: Computational Interpretations of Linear Logic (PS.GZ) has some discussion. Continuations and concurrency are closely related. In fact, this relationship between linear and non-linear negation is unintentionally paralleled in the paper “Threads Yield Continuations” where threads are used to provide one-shot continuations…
Posted by: Derek Elkins on March 5, 2008 12:01 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
May be you can take a look at the work of L. Meertens. http://www.cs.uu.nl/research/techreps/RUU-CS-89-09.html.
In that report Meertens proposes a calculus of programs using compositional structures which seem to link up nicely with your treatement of CT.
kr
Twan
Posted by: Twan on March 4, 2008 1:00 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
The calculus Meertens proposes actually looks very much like what is done in process algebras, except that they are usually less concerned with the computation of functions than with the interaction between possibly non-terminating, reactive systems. How would these fit into the picture, btw?
Posted by: Klaus Dräger on March 5, 2008 1:07 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Considering your links to physics (and stuff types), I am surprised to find no references to
J.-Y. Girard, ‘Normal functors, power series and λ-calculus’, Ann. Pure Appl. Logic 37 (1988) 129–177
or to Hasegawa’s “Two Applications of Analytic Functors”.
Where are the 2-morphisms? Disclosure: parts of my active research involves program transformation and program generation. Another view of inference rules (instead of dinatural transformations) involves 2- and 3-cells in categories, see Giraud’s Three dimensions of Proofs.
Since products and coproducts are so nicely dual, and generalize nicely [to records and labeled unions respectively] in the indexed case, it is a bit surprising that does not make an appearance. I know they are less important in other areas than in computation, so maybe they really belong on the “extension” of the Rosetta stone…
Minor comment: on p.48, the two natural transformations involving (W,X,Y,Z) both have an extra ).
Posted by: Jacques Carette on March 6, 2008 3:16 AM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Thanks for the references! I hadn’t known of them.
Stuff types, analytic functors and higher categories are not at all what I want to explain in this paper. I have very limited ambitions, namely to explain the concept of ‘symmetric monoidal closed category’ and show — in the simplest possible terms! —- how such categories arise in physics, topology, logic and computation, and how you can reason with such categories using string diagrams. The intended audience is mathematical physicists, so I’m assuming a lot of knowledge of physics, a fair amount of knowledge of topology, and essentially zero knowledge of category theory, logic or computation.
So: no 2-categories, no stuff types, no analytic functors… nuthin’ fancy! But, I’ll enjoy reading these references myself, so thanks.
Thanks also for catching those typos!
Posted by: John Baez on March 6, 2008 9:58 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
Another possible use of (n-)categories in computer science is to provide static analysis tools for programs. For example, in Polygraphic programs and polynomial-time functions, written with Guillaume Bonfante, we give tools to study complexity.
For that, we use an encoding of programs where computations are 3-dimensional objects. Indeed, a first-order functional program can be seen as a 3-category freely generated by: one 0-cell; elementary types as 1-cells; constructor and function symbols as 2-cells; computation rules as 3-cells.
Then we prove that well-chosen functorial interpretations of the underlying 2-category give space bounds on the results of computation, while “derivations” of the same 2-category yield information on the time complexity. The same tools were already used to prove termination of 3-dimensional rewriting systems there.
The paper is not written using n-categories explicitely. One of the reasons is that I do not know how to define “derivations” of n-categories yet. Please note that the paper is currently under revision before publication and thus some minor modifications could occur in the near future.
(Just a remark: the paper Jacques Carette’s cites (thanks!) is a preliminary version. The final published version is there.)
Posted by: Yves Guiraud on March 6, 2008 1:43 PM | Permalink | Reply to this
### Re: Computer Scientists Needed Now
In case anyone wonders why I’m not commenting more on this thread: it’s not lack of interest, I’m just spending all my time trying to finish the paper on time! I’m taking all your wonderful suggestions into account, and I’ll try to say more in a few days…
Posted by: John Baez on March 6, 2008 10:00 PM | Permalink | Reply to this
Read the post Physics, Topology, Logic and Computation: a Rosetta Stone
Weblog: The n-Category Café
Excerpt: Toward a general theory of systems and processes.
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Read the post New Structures for Physics I
Weblog: The n-Category Café
Excerpt: Please comment on two chapters of a forthcoming book edited by Bob Coecke: 'Introduction to categories and categorical logic' by Abramsky and Tzevelekos, and 'Categories for the practicing physicist' by Coecke and Paquette.
Tracked: August 29, 2008 4:26 PM
Read the post New Structures for Physics II
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Excerpt: You can help review some more papers for Bob Coecke's New Structures for Physics.
Tracked: September 10, 2008 4:28 PM
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# How do you divide (5m^3 - 7m^2 + 14) / (m^2 - 2)?
May 4, 2017
The quotient is $= 5 m - 7$ and the remainder is $= 10 m$
#### Explanation:
We perform a long division
$\textcolor{w h i t e}{a a a a}$$5 {m}^{3} - 7 {m}^{2}$$\textcolor{w h i t e}{a a a a a a}$$+ 14$$|$${m}^{2} - 2$
$\textcolor{w h i t e}{a a a a}$$5 {m}^{3}$$\textcolor{w h i t e}{a a a a}$$- 10 m$$\textcolor{w h i t e}{a a a a a a}$|5m-7
$\textcolor{w h i t e}{a a a a}$$0 - 7 {m}^{2}$$\textcolor{w h i t e}{a a a a a a a a a}$$+ 14$
$\textcolor{w h i t e}{a a a a a a}$$- 7 {m}^{2}$$\textcolor{w h i t e}{a a a a a a a a a}$$+ 14$
$\textcolor{w h i t e}{a a a a a a a a}$$- 0 + 10 m$$\textcolor{w h i t e}{a a a a}$$+ 0$
Therefore,
$\left(5 {m}^{3} - 7 {m}^{2} + 14\right) = \left({m}^{2} - 2\right) \left(5 m - 7\right) + 10 m$ |
# Area under graph when log-log scale gives straight line
1. Mar 5, 2009
### robsmith82
Hi,
I have a graph with a frequency on the x axis and power density on the y axis. Both scales are logarithmic, and the graph shows a straight line between the points (110Hz, 0.001V^2/Hz) and (200Hz, 0.004V^2/Hz). I need to work out the area under this portion of the graph as it relates to total power.
So, my understanding is I need to find the function of the line, then set up and evaluate a definite intergral between 110Hz and 200Hz. My real question is, how do I work out the function of the line, and is this a correct method?
Thanks
Rob
2. Mar 5, 2009
### HallsofIvy
That straight line has slope .0001/3 and passes through 0(110Hz, 0.001) so has equation log(y)= (0.0001/3)(x- 110)+ 0.001. Taking exponentials of both sides,
$$y= (e^{0.001})e^{\frac{0.0001(x-110)}{3}}$$
That should not be difficult to integrate.
3. Mar 5, 2009
### robsmith82
ok, i can integrate that, my real question is about how you found the function of y... How did you find that slope? |
Extra Exercises¶
1. Write a function called int_return that takes an integer as input and returns the same integer.
1.1 Write a function named same that takes a string as input, and simply returns that string.
1.2 Write a function called same_thing that returns the parameter, unchanged.
1. Write a function called add that takes any number as its input and returns that sum with 2 added.
2.1 Write a function called subtract_three that takes an integer or any number as input, and returns that number minus three.
2.2 Write a function called change that takes one number as its input and returns that number, plus 7.
1. Write a function called change that takes any string, adds “Nice to meet you!”, and returns that new string.
3.1 Write a function named intro that takes a string as input. Given the string “Becky” as input, the function should return: “Hello, my name is Becky and I love SI 106.”
3.2 Write a function called s_change that takes one string as input and returns that string, concatenated with the string “for fun.”.
1. Write a function, accum, that takes a list of integers as input and returns the sum of those integers.
4.1 Write a function named total that takes a list of integers as input, and returns the total value of all those integers added together.
4.2 Write a function called count that takes a list of numbers as input and returns all of the elements added togther.
1. Write a function, length, that takes in a list as the input. If the length of the list is greater than or equal to 5, return “Longer than 5”. If the length is less than 5, return “Less than 5”.
5.1 Write a function named num_test that takes a number as input. If the number is greater than 10, the function should return “Greater than 10.” If the number is less than 10, the function should return “Less than 10.” If the number is equal to 10, the function should return “Equal to 10.”
5.2 Write a function called decision that takes a string as input, and then checks the number of characters. If it has over 17 characters, return “This is a long string”, if it is shorter or has 17 characters, return “This is a short string”.
1. You will need to write two functions for this problem. The first function, divide that takes in any number and returns that same number divided by 2. The second function called sum should take any number, divide it by 2, and add 6. It should return this new number. You should call the divide function within the sum function. Do not worry about decimals.
6.1 Write two functions, one called addit and one called mult. addit takes one number as an input and adds 5. mult takes one number as an input, and multiplies that input by whatever is returned by addit, and then returns the result.
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# Matlab Dark Mode
Download night mode for windows 7 32bit for free. But in its over three decades of existence, it has grown greatly to become a general environment for solving math, scientific, and engineering problems. de Matlab Mtf. Yair is widely recognized as a world-class Matlab authority in the MATLAB community and by MathWorks themselves. This example showed a Simulink model implementing an image inversion algorithm. [Source Code] classdef myCalculator < matlab. 1700]) using cmu. ARCHIVE! Please read /mac/00introduction if you haven't already done so. It is designed for self-study. It has been downloaded 101 times and provides support for a lot of common editors. Values In a session with only input channels or counter output channels, you can enter a value in seconds for the length of the acquisition. Estimation du changement de règle (9000 hab) Estimation élaborée le 17 Janvier 2020, la règle a subi plusieurs modifications depuis mais donne idée de l'impact du changement En attendant les publications des données sur les élections municipales, je vous propose de découvrir l'impact du changement des règles pour les élections municipales 2020. C0646910" This is a multi-part message in MIME format. Nevertheless, quite a number of chapters, which do not refer to Matlab analogs, would be useful for Scilab users without Matlab background. 05) in the mean mortality of Anopheles species larvae between extracts of both plant species after 3, 6 and 24 hours exposure time respectively. ALTERNATIVE DOMAIN: ZAMUNDA. Frantz Grenet et de M me Anca Dan : « Alexandre au Paradis, du Talmud à l’Occident latin par l’Asie Centrale : une version judéo-iranienne du Roman d’Alexandre sur un bol d’argent bactrien du 6 e siècle ». Hugo Themes (Tag: dark mode) Doks. Given I believe MATLAB is pretty resource intensive at times, I would suggest the XPS 13. - P-value text in LaTeX mode, "n. It is not quite convenient, in case MATLAB generates a large output. Depuis 1998, Clair et Net. Are there any predefined MATLAB dark mode settings so we do not need to maually set-up them ? Specially for the MATLAB editor ?. 3), with what this article calls chroma as its. ALTERNATIVE DOMAIN: ZAMUNDA. Implementation on MATLAB • In image processing toolbox ( IPT ) linear spatial filtering known by imfilter, which the syntax is: • Common mode for calling imfilter: g = imfilter(f,w,filtering_mode,boundary_options,size_options) f = input image w = mask g = outputs filtering_mode = kind of filter boundary_options = kind of expanding. Outline Outline. The data sheet mentions that "In continuous conversion mode, the address pointers’ autoincrementer should be used. All; Blog (255) Responsive (217) Clean (142) Minimal (142) Simple (137. Nevertheless, quite a number of chapters, which do not refer to Matlab analogs, would be useful for Scilab users without Matlab background. tex file, keeping the format as the one in MATLAB command window. The mission consists of integrating a "Pixhawk" autopilot on a Remotely Controlled Air Vehicle using proprietary technology to convert the air vehicle from fixed-wing to VTOL mode. High contrast between text and background reduces eye strain. Sign in to comment. The IT Pro. Join the global Raspberry Pi community. matlab-dark-theme. Still, you may only want to use this theme at night time, rather than all of the time. Google was also showing off a system-wide dark mode for Android Q last May. RGB color space or RGB color system, constructs all the colors from the combination of the Red, Green and Blue colors. Is MathWorks working on a dark mode for MATLAB similar to what Visual Studio offers? I also hope that Mathworks is working on a dark mode. There is also a white-list feature which enables you to exclude a desired domain from dark mode. 0beta2) palette, in dark mode, by Ethan Schoonover. A great scheme is called "Solarized" which you can set MATLAB to match. Yair is widely recognized as a world-class Matlab authority in the MATLAB community and by MathWorks themselves. The Dark_MATLAB theme is a color scheme for eclipse created by Jesse Johns. Join the global Raspberry Pi community. So, we will be using only ‘dark’ mode of. Of course if you have specific questions or run into some errors/obstacles, feel free to post your attempt and someone might be able to help you. How does the Huawei G7plus RIO-UL00's Usb debug mode open?. The toolbox works with Octave, a Matlab clone. [Source Code] classdef myCalculator < matlab. If you want only to use the Adwaita Dark theme at night time, you can install the Dark Mode Toggle. Now, make sure to update Google's Phone app from the Play Store. So here basically, we are going to use imfindcircles function to detect center pivot irrigation fields in the image. Observe the display of inverted images on the host computer. Download Free PDF. RGB color space or RGB color system, constructs all the colors from the combination of the Red, Green and Blue colors. Batiactu Événements, une marque de Batiactu Groupe, vous propose un accompagnement sur mesure de votre événement. Solarised Dark Based on the Solarized (1. Our CMOS cameras offer a full-frame resolution of 1280 x 1024 pixels. As there are too many colors to set in the editor preferences, something like UltraEdit's 'theme' feature would be great. When OctaveFEMM starts up a FEMM process, the usual FEMM user interface is displayed and is fully functional. The GMT/ MATLAB toolbox is a basic interface between MATLAB® (or Octave) and GMT, the Generic Mapping Tools, which allows MATLAB users full access to all GMT modules. 0: updated docs links. m but it seems useful only for matrices. Values In a session with only input channels or counter output channels, you can enter a value in seconds for the length of the acquisition. ist ein unabhängiger Verein hauptberuflich tätiger Journalistinnen /Journalisten. In addition to the main colours described in Alexandre Chabot's post, it will also import and export editor colours, the status of toggleable settings such as right-hand line limit and cell mode, and the colours for syntax highlighting in other languages supported by MATLAB (C, C++, Java, XML, HTML, some Simulink things). MATLAB License Number: •••••• Operating System: Microsoft Windows 7 Version 6. See the complete profile on LinkedIn and discover Hannah’s connections and jobs at similar companies. Would like to change all backgrounds to be dark with lighter colored text. 3), with what this article calls chroma as its. Right now, if I set mode to 132, I get a percentage of full range as predefined for each channel as follows: fwrite(s,[132,4,0, setting],‘uint8’) where setting is 1…127. It was originally a system for solving matrices — quickly and accurately. Visual Studio's dark mode changes the background color for all the toolbars and all over the UI. Observe the display of inverted images on the host computer. Products; Solutions; Academia; Support; Community; Events. This completely disregards gamut extents and any saturated, bright, or dark content will be clipped. (Requires macOS Mojave 10. Play thousands of free web and mobile games! Discover the best shooters, role playing games, MMO, CCG, tower defense, action games and more!. In the model, change the Simulation mode on the toolbar to External. Send pictures if you're not sure. It was originally a system for solving matrices — quickly and accurately. AppBase% Properties that correspond to app…. If you do not like the current dark theme, please visit the options page and choose a different theme from over 50 available options. It is free. 0 date: Fri, 14 Mar 2014 08:38:19 -0400 x-mimeole: Produced By Microsoft MimeOLE V6. Our CMOS cameras offer a full-frame resolution of 1280 x 1024 pixels. So, we will be using only ‘dark’ mode of imfindcircles. Installation. Dark Mode Menu Log in Register Home. How do I turn on Dark Mode for VLC? Go to Setting Choose Dark Mode Activate VLC supports dark mode. MATLAB License Number: •••••• Operating System: Microsoft Windows 7 Version 6. 6 also introduces fractional scaling. Yair is widely recognized as a world-class Matlab authority in the MATLAB community and by MathWorks themselves. Forms deliver native iOS experiences to keep your applications looking shiny and fresh no matter what time of day. Data may be passed between the two programs using intermediate MATLAB structures that organize the metadata needed; these are produced when GMT modules are run. Home-made MATLAB themes Dark Steel. Implementation on MATLAB • In image processing toolbox ( IPT ) linear spatial filtering known by imfilter, which the syntax is: • Common mode for calling imfilter: g = imfilter(f,w,filtering_mode,boundary_options,size_options) f = input image w = mask g = outputs filtering_mode = kind of filter boundary_options = kind of expanding. The surface represented is 1,7mm x 1,7mm master's thesis work. And, it is my understanding that the 4 and the 5 are for the order of the global and local error, respectively. Gray scale filtering is in reference to the color mode of a particular image. ; 8 1000 dark gray; 9 1001 light blue; a 1010 light green; b 1011 light cyan; c 1100 light red; d 1101 light magenta; e 1110 yellow; f 1111 white org 100h; set video mode mov ax, 3 ; text mode 80x25, 16 colors, 8 pages (ah=0, al=3) int 10h ; do it!; cancel blinking and enable all 16 colors: mov ax, 1003h mov bx, 0 int 10h; set segment register:. JavaScript Snake - patorjk. With features like multi-caret editing, column/block editing and multi-select, it's a simple text editor when you want it to be, and a multi-cursor power editor when you need it to be. MATLAB: Dark Mode editor for MATLAB. Processing is a flexible software sketchbook and a language for learning how to code within the context of the visual arts. Hasan Abbasi Nozari. The dark mode beta is finally here. Vous n’êtes pas autorisé à lire ce forum. 19 - Browsing through older pdf files, I found that any pre-reinstall combinations of various Matlab and Ghostscript versions up to R2015b / 9. Sean de Wolski on 27 Mar 2012. Klicke jetzt und geniesse die neuesten lustigen Sachen im Internet!. Matrices and Vectors 2 2. To get the dark mode option, y ou’ll need to be on version 4. It offers a warm look that looks great with dark or light backgrounds. High contrast between text and background reduces eye strain. No Windows license needed. With iOS 13, Apple introduces dark mode: A system-wide option for light and dark themes. Evolution of lattice structure (Lymph node section). To change the colors of the desktop, go. MATLAB/SIMULINK Tutorial. Go back to the reference page. You can use the dark theme setting to save your battery life. In the model, click the Run button on the toolbar to run the model on Beaglebone Blue hardware. 0beta2) palette, in dark mode, by Ethan Schoonover. 1 (Build 7601: Service Pack 1) Java VM Version: Java 1. I made a few changes to it but basically it is the same). Dark Mode editor for MATLAB. Gray is any RGB color with Red, Green, and Blue components all equal, excepting black (all 0's) and white (all components the maximum). The surface represented is 1,7mm x 1,7mm master's thesis work. Lustige Bilder, lustige Videos und Flash-Games, Fun Videos, Werbespots kostenlos. A TIF file contains an image saved in the Tagged Image File Format (), a high-quality graphics format. Pink cells are Th2, light blue are Th1, light red are naïve T cells, dark red are Th0, green represents naïve B cells with a few activated ones, dark blue colour is for macrophages. 14 Jun 2016: 1. Reflected light intensity – In this mode, the colour sensor emits a red light and measures the amount reflected back into itself from the surface you are testing. You can try out different color combos below, then use the links to find the accessories in. Make your phone easier to use with one hand, no root and the hard sciences with the Essential MATLAB & Simulink Bundle for over 90% gains Samsung device support and a dark. Outline Outline. ASCII Table and Description. This is a short introduction to scienti c computation in MATLAB. Re: General dark theme (not only for editor) « Reply #4 on: February 09, 2018, 11:25:12 pm » By the way I already changed the Application Color Scheme in Colors-System Settings (openSUSE), and that's why KDevelop changed its theme. Create a virtual reality canvas. My only hesitation is the fact that the product has Dell in the name. CuBatch is an interface running under MATLAB®. 0 date: Fri, 14 Mar 2014 08:38:19 -0400 x-mimeole: Produced By Microsoft MimeOLE V6. I would like to express my gratitude and extend my appreciation especially to the following. Alternatively, you can create the text yourself directly in Word. Learn more about VLC on the website. Download Icons - Download 666 Free Download icons @ IconArchive. Here I show you how to setup Matlab into a format which is close to Dark mode. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. The Raspberry Pi is a tiny and affordable computer that you can use to learn programming through fun, practical projects. Why didn't SNES Doom use mode 7? Can there be planets, stars and galaxies made of dark matter and of anti-matter? Any way to reference a particular component of a label?. I use the MATLAB editor because of the debugging features. In addition to changing the color of Code Analyzer indicators for errors, this action also changes the color for errors in the Command Window, Command History window, Editor, and Shortcuts callback area. GitHub Desktop Focus on what matters instead of fighting with Git. If you continue browsing the site, you agree to the use of cookies on this website. Download night mode for windows 7 32bit for free. If you run NotePad and bring up the open dialog, the bottom controls do not all use dark mode. The most that can be obtained with MATLAB is to change the background color settings of text,editors and workspace windows. The consultancy was founded by Yair Altman, a renowned Matlab expert with 30 years professional software development experience. To save dark-mode-hater folks from the torture of darkness, here we will tell you how to. Usually RGB colors have values from 0 to 255. The sum function 3 2. Run: apm install dark-mode or use Atom Package Manager in Atom settings. " when not significant, bg colour - Change shape of outputs hpl, hpt, hpb - Update defaults for p thresholds, errorbar widths, fontsize. What does this Hex to RGB converter do? It takes input in the form of a hex color code value and converts that value to a RGB value that can be used to specify color in photo editing software. Sometimes, you might need some boilerplate or filler text in a Word document. Reflected light intensity – In this mode, the colour sensor emits a red light and measures the amount reflected back into itself from the surface you are testing. Google was also showing off a system-wide dark mode for Android Q last May. ou can check which. Fullz with dl, The power spectrum (PS) of a time-domain signal is the distribution of power contained within the signal over frequency In MATLAB®, you can perform real-time spectral analysis of a dynamic signal using the dsp. It merely takes the I & Q axis means, subtracts an arbitrary number, and adds that value to the axis. Thus knowing the MATLAB RGB triples for the colors can be useful. Vor 2 years. In addition to changing the color of Code Analyzer indicators for errors, this action also changes the color for errors in the Command Window, Command History window, Editor, and Shortcuts callback area. There, the pdf specifies "Helvetica" as its embedded font (and thus it also displays Helvetica/Arial). There is a common characteristic in these pictures: there is only one bright spot in the picture and all are in black background. But in its over three decades of existence, it has grown greatly to become a general environment for solving math, scientific, and engineering problems. ) Dynamic Desktop. $\endgroup$ – NAASI Nov 1 '16 at 15:59. Matlab is more efficient than most programs because it is interactive and gets the job done right. Previously we had learned the method to enable Dark Mode for Teams, OneNote, and Outlook. The main difference between them is, in the surface plot, the connecting lines and the faces both will be displayed in the dark color. After a few days work, I got my first GUI prototype (see below). ACTUALITES Séances Le vendredi 15 janvier – Communication de M. MATLAB supports dark mode. matrix2latex: for Python and MATLAB, pandas: pandas DataFrame's have a method to convert data they contain to latex, latex-tools: a Ruby library, xtable: a library for R, org-mode: for Emacs users, org-mode tables can be used inline in LaTeX documents, see for a tutorial. 1 - 2 of 2 Posts. You can upload your own image file or put image URL. Create a virtual reality canvas. Furthermore, the index includes an even more extensive lists of those Matlab. The Long Dark is more than just an open-world survival sandbox. Applications that require syntax highlighting (like programming) can be easier to see with light-on-dark themes. Gray scale filtering is in reference to the color mode of a particular image. Click on the image to get color in HEX, RGB, HSL and CMYK format. The code given here computes the histogram in different color channels of the image. Here I show you how to setup Matlab into a format which is close to Dark mode. Estimation du changement de règle (9000 hab) Estimation élaborée le 17 Janvier 2020, la règle a subi plusieurs modifications depuis mais donne idée de l'impact du changement En attendant les publications des données sur les élections municipales, je vous propose de découvrir l'impact du changement des règles pour les élections municipales 2020. The Free & Open Source Image Editor. Color histogram feature extraction matlab code. Notepad++ is a free (as in “free speech” and also as in “free beer”) source code editor and Notepad replacement that supports several languages. Download Icons - Download 666 Free Download icons @ IconArchive. Uncheck system colors. Einen Einblick hierin finden Sie ab sofort auf einer neuen PLAZ-Homepage. C0646910" This is a multi-part message in MIME format. Now, make sure to update Google's Phone app from the Play Store. 0: Change default orientation of star text for horizontal bars; now normal for single p-values and sideways for pair-wise p-values. Matlab Rendering. Cell Mode Cell 모드. There is a common characteristic in these pictures: there is only one bright spot in the picture and all are in black background. Revenir en haut. One solution is to head over to a lorem ipsum or dummy text generator to create that text. To use the software for the MODE model, download the package (MODE_GUI_070109. I would then, simulataneously, open the same file in uedit (using explorer and several mouse clicks) take advantages of the power (mainly regular expressions, macros, column mode, etc. - current system specs: Windows 7, Matlab 2016b, Ghostscript 9. 0: Change default orientation of star text for horizontal bars; now normal for single p-values and sideways for pair-wise p-values. Compare it to the high-contrast theme above and you'll see just why you'd want to use a third-party theme. Call;; function da-color-theme to change to the dark face, and call C-u;; da-color-theme for the yellow background face. Here in this guide, we have enlisted 10 best laptops for MATLAB You Can Buy Right Now. As we know that, MATLAB is a high-performance language for technical computing. ARCHIVE! Please read /mac/00introduction if you haven't already done so. 1 gives the bottom of the range, and 127 gives the max. Your UPI PIN is the number you enter whenever you add a new payment account or make a transaction. 555 timer is used for many purposes in electronic industry as a monostable and astable multivibrator. This appears to be intentionally done by Microsoft (not a bug). Learn more about VLC on the website. How do I turn on Dark Mode for MATLAB? Go to Setting Select Dark Mode Activate MATLAB supports dark mode. If you assign a value to this property, then MATLAB sets the mode to 'manual' and does not automatically choose the color limits. https://doi. Based on a 20-year track record of high end commercial products, Modelio delivers a broad-focused range of standards-based functionalities for software developers, analysts, designers, business architects and system architects. Using undocumented functionality (a common technique to do advanced things in Matlab) it is possible to change the color theme of the main IDE. I would like to have a similar theme in Matlab to be easier for me when working with both softwares which. 0: updated RGB-LCH conversion tools, added alpha support for both GIMP and SVG methods sanitized a few input cases (forces expansion if blend mode implies RGB content) added second softlight mode to match variant in SVG spec added SRLAB color blend. Sean de Wolski on 27 Mar 2012. Hugo Themes (Tag: dark mode) Doks. Dark mode is a new color theme joining the classic GitHub light mode developers have been using for years. Although MATLAB is not always easy to work with, it has tremendous capabilities, and the fact that it interfaces with the dongle is a great feature. Currently, Matlab does not have a factory-built method to programatically change the color theme of the Matlab IDE (interactive code-editing GUI). Towards the later part in this video it shows how one can take the input from Simulink Model into App Designer Code to design the counter. For more information on options when launching MATLAB, here is the documentation for windows and unix. Download night mode for windows 7 32bit for free. You cannot specify these colours in HTML and CSS by their colour name but you can use their RGB hexadecimal value, eg:. The GMT/ MATLAB toolbox is a basic interface between MATLAB® (or Octave) and GMT, the Generic Mapping Tools, which allows MATLAB users full access to all GMT modules. Социална мрежа за споделяне на информация и забавление. You can use those numbers and divide the vector by 255 to use within MATLAB. Solarised Dark Based on the Solarized (1. COMPUTER PROGRAMMING with MATLAB Sold to. Due to its wide range application in electronics, many projects can be easily made using this IC. Volume 6, Issue 1 http://www. This completely disregards gamut extents and any saturated, bright, or dark content will be clipped. 0_17-b04 with Sun Microsystems Inc. A gray scale image would, in layman's terms, be a black and white image, any other color would not be included in it. The main difference between them is, in the surface plot, the connecting lines and the faces both will be displayed in the dark color. The example showed how the model may be run in External mode, and as a standalone application on Raspberry Pi hardware. QHYCCD designs and manufactures world-leading astronomical cameras, ranging from entry-level to professional, CMOS and CCD, front-illuminated and back-illuminated, specially crafted for amateur and professional astronomers worldwide. MATLAB: Dark Mode editor for MATLAB. Whether you're new to Git or a seasoned user, GitHub Desktop simplifies your development workflow. ALTERNATIVE DOMAIN: ZAMUNDA. You still need just one running instance of Kate. Being a KDE applications, Kate ships with network transparency, as well as integration with the outstanding features of KDE. The dark mode beta is finally here. In normal mode the resolution you set is the resolution you see. Matlab Mtf - lpao. Hugo Bear Blog. Because these definitions of saturation—in which very dark (in both models) or very light (in HSL) near-neutral colors are considered fully saturated (for instance, from the bottom right in the sliced HSL cylinder or from the top right)—conflict with the intuitive notion of color purity, often a conic or biconic solid is drawn instead (fig. It includes spellchecker, word counter, autosave, find and replace etc. background color MATLAB. LINE is the new communication app which allows you to make FREE voice calls and send FREE messages whenever and wherever you are, 24 hours a day!. CodeChef - A Platform for Aspiring Programmers. Dark Mode editor for MATLAB. There is a common characteristic in these pictures: there is only one bright spot in the picture and all are in black background. If the main menu is still white, either restart your phone or open the multitasking screen (either swipe up from the bottom of your screen or hit the square button on the bottom right), then swipe away the Phone app card. Our backend JOBE server supports Octave, so any MATLAB programs that also run in Octave, which should be most of them are supported. LINE is the new communication app which allows you to make FREE voice calls and send FREE messages whenever and wherever you are, 24 hours a day!. Data may be passed between the two programs using intermediate MATLAB structures that organize the metadata needed; these are produced when GMT modules are run. I believe that the RTL-SDR community would greatly benefit from more open-source projects using MATLAB, so I have made my code availabe on GitHub , if you would like to try it out for yourself. Aug 20, 2020 · Color can be quite effective at conveying data values, both constant and varying. The intensity of the light is measured as a percentage from 0 to 100, with 0 being very dark, and 100 being very bright. Here I show you how to setup Matlab into a format which is close to Dark mode. MATLAB/SIMULINK Tutorial. ARCHIVE! Please read /mac/00introduction if you haven't already done so. Write down quick notes and print a simple text document with Online Notepad editor. Use Dark mode in Gmail You can change your Gmail theme setting to make it easier to view messages on your mobile device. | Die Digitalisierung ist auch für die Bildungslandschaft von zentraler Bedeutung. This is the official website of the GNU Image Manipulation Program (GIMP). Re: General dark theme (not only for editor) « Reply #4 on: February 09, 2018, 11:25:12 pm » By the way I already changed the Application Color Scheme in Colors-System Settings (openSUSE), and that's why KDevelop changed its theme. The GMT/ MATLAB toolbox is a basic interface between MATLAB® (or Octave) and GMT, the Generic Mapping Tools, which allows MATLAB users full access to all GMT modules. Syntactically Awesome Style Sheets. Once upon a time, bailing out of a plane involved popping open the roof or door, and hopping out with your parachute, hoping that you’d maintained enough. UI component infrastructure and Material Design components for mobile and desktop Angular web applications. With this guide, you have turned every last element on your Windows 10 PC into dark. Create a virtual reality canvas. UltraEdit's text editing features make editing lists and columns an intuitive experience, not the exercise in tedium it used to be. The output, centers, is a two-column matrix containing the (x,y) coordinates of the circles centers in the image. Learn more about MATLAB on the website. 6 also introduces fractional scaling. Under MATLAB syntax highlighting colors, change the color for Errors. [Source Code] classdef myCalculator < matlab. Toute l'actualité au masculin, actualisée en permanence : sport, hi-tech, mode, lifestyle, business, économie, enfants, sorties, voyages, automobile, bateau, avion. Some days my eyes are tired and other days I am just bored with a white background and black text. Are there any predefined MATLAB dark mode settings so we do not need to maually set-up them ? Specially for the MATLAB editor ?. de Matlab Mtf. Compare it to the high-contrast theme above and you'll see just why you'd want to use a third-party theme. Enabling dark mode in Windows 10 is easy. Emacs align commands: the align commands can clean up a messy LaTeX table. Introduction MatLab can be a useful tool in many applications. So, we will be using only ‘dark’ mode of imfindcircles. The minimum circle and the maximum circle radii are image dependent so you need to provide those information with a little bit of data exploration. No one likes being blinded by an app with bright UI elements while using dark mode—but luckily, you can force all apps to use an ersatz “dark mode” even if they don’t officially have one. MATLAB Through experiment, a set of CCD image files at every fixed incidental angle are extracted, they are made of 19 pictures all with BMP format,768*576 pixel and color RGB mode. This is my own theme, a mash-up of Cobalt and Darkmate, and was the colour scheme I was originally trying to transfer between my MATLAB installations and. com is the number one paste tool since 2002. Dark Mode editor for MATLAB. No one likes being blinded by an app with bright UI elements while using dark mode—but luckily, you can force all apps to use an ersatz “dark mode” even if they don’t officially have one. com : Actu du cyclisme amateur, directs de courses cyclistes amateurs et professionnels. iOS and Xamarin. Hugo Bear Blog. LabVIEW, MATLAB, and µManager Third-Party Software; These compact, lightweight CMOS cameras are available with either a monochrome (M models) or color (C models) sensor. You specify the color of the line like this. Unfortunately, this expensive software does not have a dark mode/dark theme. 0beta2) palette, in dark mode, by Ethan Schoonover. This cannot be repaired. Just head to "Matlab-> Preferences" and select the colors menu to get started! I prefer a green text on a dark background, but even a sepia background will be easier on the eyes than Bright White. In the model, click the Run button on the toolbar to run the model on Beaglebone Blue hardware. The MATLAB Colon Notation 4 2. Start your free 14-day trial download today!. What is Notepad++. Follow 3,665 views (last 30 days) Gayan Lankeshwara on 25 Nov 2019. The most that can be obtained with MATLAB is to change the background color settings of text,editors and workspace windows. I use the MATLAB editor because of the debugging features. m but it seems useful only for matrices. $\endgroup$ – NAASI Nov 1 '16 at 15:59. com/articles/vitex-negundo-leaves-extract-as-green-inhibitor-for-the-corrosion-of-aluminium-1n-naoh-solution. Many dark themes provide a dark toolbar and white content panes, but Dark Agility goes all the way. The heart of MATLAB programming is the MATLAB language, a matrix-based language allowing the most natural expression of computational mathematics. We encourage our users to get in the streets and join them if you can. See full list on wiki. Thus knowing the MATLAB RGB triples for the colors can be useful. This paper thus describes the modeling techniques for advanced photonic transmission systems and Simulink is proven to be very effective platform for development of photonic communications systems due its. To enjoy the perfect Matlab dark format: Home, Preferences. There are a lot of challenges to implementing dark mode across a platform like ours. Learn more about VLC on the website. 1700]) using cmu. Google's News app has featured such a mode since 2018, but the service's web version was only available with a bright theme until today. V4L2 Video Capture and the SDL Video Display blocks from the Raspberry Pi library were used to capture and display video data. So update your app before you think of switching to the dark side. Matlab Simulink has become the universal mathematical and modeling tools in most universities and re-search laboratories around the world. In this image, there are only dark pivot circular fields surround by bright objects. 9790/1676-1302032935 www. As I said before, it would be convenient to me if I could just import the raw MATLAB output into my. Introduction MatLab can be a useful tool in many applications. I have used ode45 in Matlab. s zum Thema Lizenzen (inkl. To change the colors of the desktop, go. Solarised Light. Im Dezember bekamen 24 Stipendiatinnen und Stipendiaten einen virtuellen Einblick hinter die Fassaden von GOLDBECK. MATLAB: Dark Mode editor for MATLAB. Also, I have tested matrix2latex. Nevertheless, quite a number of chapters, which do not refer to Matlab analogs, would be useful for Scilab users without Matlab background. (Enable the dark theme/dark mode) | 如何更改MATLAB顏色主題？ Franky's. It resembles Sublime text which has is a gorgeous example of dark theme. Follow them to see all their posts. MATLAB news, code tips and tricks, questions, and discussion! We are here to help, but won't do Now that I shared that with you, I thought I would go ahead and share my own dark-colored theme. Great things happen when developers work together—from teaching and sharing knowledge to building better software. You can use those numbers and divide the vector by 255 to use within MATLAB. After a few days work, I got my first GUI prototype (see below). Hannah has 2 jobs listed on their profile. Vous n’êtes pas autorisé à lire ce forum. COMPUTER PROGRAMMING with MATLAB Sold to. 6k Followers, 92 Following, 3,250 Posts - See Instagram photos and videos from Social Deal (@socialdeal. The mission consists of integrating a "Pixhawk" autopilot on a Remotely Controlled Air Vehicle using proprietary technology to convert the air vehicle from fixed-wing to VTOL mode. Keywords and hyperlinks from dark blue to cyan. The sum function 3 2. Just select the desired 'theme" and all colors are changed. plot(x,y, 'color',[0. Lustige Bilder, lustige Videos und Flash-Games, Fun Videos, Werbespots kostenlos. Using undocumented functionality (a common technique to do advanced things in Matlab) it is possible to change the color theme of the main IDE. We are among the 40 largest websites in the world in terms of monthly unique. Here I show you how to setup Matlab into a format which is close to Dark mode. Programming. Are humans more adapted to "light mode" or "dark mode"?. The surface represented is 1,7mm x 1,7mm master's thesis work. SpectrumAnalyzer System object™. Matlab allows you to specify a color by the RGB (red green blue) values, for example, deep carrot orange is defined by the RGB tuple [ 0. DarkTheme class allows you to set a dark theme in MATLAB. V4L2 Video Capture and the SDL Video Display blocks from the Raspberry Pi library were used to capture and display video data. And, it is my understanding that the 4 and the 5 are for the order of the global and local error, respectively. To get the dark mode option, y ou’ll need to be on version 4. MATLAB is a high-level computing environment from MathWorks. OctaveFEMM is a Matlab toolbox that allows for the operation of Finite Element Method Magnet-ics (FEMM) via a set of Matlab functions. The MATLAB Colon Notation 4 2. Matlab How To Tutorial 1 How To Change Matlab Environment To Dark Theme. 1 Contents 1. Get Prezi account access by signing into Prezi here, and start working on or editing your next great presentation. The IT Pro. Notation matches variable names used in main text. Search more than 600,000 icons for Web & Desktop here. If I recall correctly, the code in this function doesn't even bother operating along the chroma axis. iOS users may now choose the theme or allow iOS to dynamically change appearance based on the environment and time. The Raspberry Pi is a tiny and affordable computer that you can use to learn programming through fun, practical projects. To change the colors of the desktop, go. The instructions of the problem are: Use bisection method to find a root of the function $$\sin x + x \cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. Dark themes can reduce eye strain in low-light conditions (night-time or dimmer workspaces). Introduction to MatLab: Circuit Analysis 3 Topics Topics Electrical Devices. Here I show you how to setup Matlab into a format which is close to Dark mode. In the model, click the Run button on the toolbar to run the model on Beaglebone Blue hardware. Implementation on MATLAB • In image processing toolbox ( IPT ) linear spatial filtering known by imfilter, which the syntax is: • Common mode for calling imfilter: g = imfilter(f,w,filtering_mode,boundary_options,size_options) f = input image w = mask g = outputs filtering_mode = kind of filter boundary_options = kind of expanding. In image histograms the pixels form the horizontal axis In Matlab histograms for images can be constructed using the imhist command. COMPUTER PROGRAMMING with MATLAB Sold to. Outline Outline. Ben je het eens met de sterrenscore van DAKA? Bekijk wat 12. MATLAB creates a new data source for the plot. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. The data sheet mentions that "In continuous conversion mode, the address pointers’ autoincrementer should be used. prior() is an anonymous function that draws N samples from a d-variate prior distribution. Take A Sneak Peak At The Movies Coming Out This Week (8/12) 10 Celebs you didn’t know were vegan. You cannot specify these colours in HTML and CSS by their colour name but you can use their RGB hexadecimal value, eg:. Find your perfect color using our online image color picker. cat command concatenates the matrix arrays R-, G- and B- color channels along with Blank image to give R-, G- and B-color component image view. That is the place where you will enter your MATLAB commands (see Chapter 2). Programming. Currently, Matlab does not have a factory-built method to programatically change the color theme of the Matlab IDE (interactive code-editing GUI). If you have faced the problem that says this program cannot be run in DOS mode or if you have got stuck in the DOS mode (can face several problems in Mac too) and if you want to solve these problems, read this article. 19 - Browsing through older pdf files, I found that any pre-reinstall combinations of various Matlab and Ghostscript versions up to R2015b / 9. (Enable the dark theme/dark mode) | 如何更改MATLAB顏色主題？ Franky's. In addition to the main colours described in Alexandre Chabot's post, it will also import and export editor colours, the status of toggleable settings such as right-hand line limit and cell mode, and the colours for syntax highlighting in other languages supported by MATLAB (C, C++, Java, XML, HTML, some Simulink things). It shows how external MATLAB functions can be called from within the App designer code and the variables can be shared across functions using global concept. zip) from the Download(s) below and unzip the contents into a local folder. In this image, there are only dark pivot circular fields surround by bright objects. Thousands of students, educators, and researchers from around the world use Octave Online each day for studying machine learning, control systems, numerical methods, and more. Take for example, the recent update to the Mail and Calendar app that now offers the dark theme as an option. Matlab Mtf - lpao. System Utilities downloads - Night Mode for Windows by FloatOverflow and many more programs are available for instant and free download. Visual Studio's dark mode changes the background color for all the toolbars and all over the UI. And, it is my understanding that the 4 and the 5 are for the order of the global and local error, respectively. There is also a white-list feature which enables you to exclude a desired domain from dark mode. Matlab How To Tutorial 1 How To Change Matlab Environment To Dark Theme. Or you can use some more complex style sheet created by others like dmarienko/Matlab as described in stack. The consultancy was founded by Yair Altman, a renowned Matlab expert with 30 years professional software development experience. Are there any predefined MATLAB dark mode settings so we do not need to maually set-up them ?. But in its over three decades of existence, it has grown greatly to become a general environment for solving math, scientific, and engineering problems. tent map matlab code, View Hannah Wills’ profile on LinkedIn, the world's largest professional community. Basically, it's a black and white image, the colors in that image, if any will be converted to corresponding shade of gray (mid tones between black and. It's designed to be a supplemental mode to a default (or light) theme. It’s also worth noting that Chrome will be releasing a dark mode feature into the stable branch by the end of 2019, so you won’t have to rely on any extensions. s zum Thema Lizenzen (inkl. 6ohm ohm, Output voltage = 350V, Output current= 9. Teletype for Atom makes collaborating on code just as easy as it is to code alone, right from your editor. 0beta2) palette, in dark mode, by Ethan Schoonover. The red, green and blue use 8 bits each, which have integer values from 0 to 255. The dark mode fans would have welcomed the step, but for people like me, our eyes started tearing up. I would like to express my gratitude and extend my appreciation especially to the following. Of course if you have specific questions or run into some errors/obstacles, feel free to post your attempt and someone might be able to help you. SpectrumAnalyzer System object™. plot(x,y, 'color',[0. mime-version: 1. It is free. Unfortunately, this expensive software does not have a dark mode/dark theme. Batiactu Événements, une marque de Batiactu Groupe, vous propose un accompagnement sur mesure de votre événement. I'm trying to initialize the connection between Mathematica and MATLab using the MATLink packet. You still need just one running instance of Kate. How do I turn on Dark Mode for MATLAB? Go to Setting Select Dark Mode Activate MATLAB supports dark mode. Jump to Latest Follow Status Not open for further replies. Solarised Dark. Aug 20, 2020 · Color can be quite effective at conveying data values, both constant and varying. Programming. I'm performing a 1D FDTD simulation, using Matlab, of a Quarter-Wave Bragg Mirror consisting of a series of alternate discrete high and low index layers with indices, n h and n l. Send pictures if you're not sure. Are there any predefined MATLAB dark mode settings so we do not need to maually set-up them ?. There are a lot of challenges to implementing dark mode across a platform like ours. com Full Screen. MATLAB supports dark mode. I properly added the package, and it properly loads through Needs["MATLink`"]. Computers can only understand numbers, so an ASCII code is the numerical representation of a character such as 'a' or '@' or an action of some sort. The red, green and blue use 8 bits each, which have integer values from 0 to 255. Unfortunately, this expensive software does not have a dark mode/dark theme op. Teletype for Atom makes collaborating on code just as easy as it is to code alone, right from your editor. Implementation on MATLAB • In image processing toolbox ( IPT ) linear spatial filtering known by imfilter, which the syntax is: • Common mode for calling imfilter: g = imfilter(f,w,filtering_mode,boundary_options,size_options) f = input image w = mask g = outputs filtering_mode = kind of filter boundary_options = kind of expanding. I made a few changes to it but basically it is the same). No Windows license needed. Gray scale filtering is in reference to the color mode of a particular image. MATLAB: Dark Mode editor for MATLAB. The instructions of the problem are: Use bisection method to find a root of the function $$\sin x + x \cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. In the model, change the Simulation mode on the toolbar to External. Gray is any RGB color with Red, Green, and Blue components all equal, excepting black (all 0's) and white (all components the maximum). 1700], and it is easier to see than yellow. 0beta2) palette, in dark mode, by Ethan Schoonover. Solarised Dark. Call;; function da-color-theme to change to the dark face, and call C-u;; da-color-theme for the yellow background face. Matlab Mtf - lpao. 1 - 2 of 2 Posts. Basically, it's a black and white image, the colors in that image, if any will be converted to corresponding shade of gray (mid tones between black and. DarkTheme class allows you to set a dark theme in MATLAB. At this point, you'll likely have to close and reopen the Phone app before dark mode fully kicks in. 17609 from: "Saved by Windows Internet Explorer 8" subject: Content Based Imaging Technology (CBIT) - YouTube content-type: multipart/related; type="text/html"; boundary="----=_NextPart_000_0001_01CF3F60. This is my own theme, a mash-up of Cobalt and Darkmate, and was the colour scheme I was originally trying to transfer between my MATLAB installations and motiviated the creation of MATLAB Schemer. When OctaveFEMM starts up a FEMM process, the usual FEMM user interface is displayed and is fully functional. This is an odd decision by Microsoft. Using undocumented functionality (a common technique to do advanced things in Matlab) it is possible to change the color theme of the main IDE. Still, you may only want to use this theme at night time, rather than all of the time. Toute l'actualité au masculin, actualisée en permanence : sport, hi-tech, mode, lifestyle, business, économie, enfants, sorties, voyages, automobile, bateau, avion. Revenir en haut. Computers can only understand numbers, so an ASCII code is the numerical representation of a character such as 'a' or '@' or an action of some sort. I believe that the RTL-SDR community would greatly benefit from more open-source projects using MATLAB, so I have made my code availabe on GitHub , if you would like to try it out for yourself. Elements of Political Communication is a style guide for beginners who want to produce political messages in various media formats. Matlab Set Figure Size. If the CommandWindow is not active, just click in it anywhere. Matlab Rendering. Variables defined with let and const are hoisted to the top of the block, but not initialized. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Usually RGB colors have values from 0 to 255. A dark theme displays dark surfaces across the majority of a UI. Notepad++ Install using Git. A TIF file contains an image saved in the Tagged Image File Format (), a high-quality graphics format. This is a short introduction to scienti c computation in MATLAB. As we know that, MATLAB is a high-performance language for technical computing. Jump to Latest Follow Status Not open for further replies. 0beta2) palette, in dark mode, by Ethan Schoonover. When trying to run the. Enable Dark Mode in Microsoft Office apps. Download Icons - Download 666 Free Download icons @ IconArchive. This example shows how to use the V4L2 Video Capture and the SDL Video Display blocks from the BeagleBone Black block library to implement an image inversion algorithm with a Simulink® model, and to run the model on BeagleBone Black hardware. Einen Einblick hierin finden Sie ab sofort auf einer neuen PLAZ-Homepage. MATLAB news, code tips and tricks, questions, and discussion! We are here to help, but won't do Now that I shared that with you, I thought I would go ahead and share my own dark-colored theme. Solarised Dark Based on the Solarized (1. Matlab Mtf - lpao. Download night mode for windows 7 32bit for free. Download with Google Download with Facebook. DarkTheme class allows you to set a dark theme in MATLAB. If the main menu is still white, either restart your phone or open the multitasking screen (either swipe up from the bottom of your screen or hit the square button on the bottom right), then swipe away the Phone app card. matrix2latex: for Python and MATLAB, pandas: pandas DataFrame's have a method to convert data they contain to latex, latex-tools: a Ruby library, xtable: a library for R, org-mode: for Emacs users, org-mode tables can be used inline in LaTeX documents, see for a tutorial. Sleep mode and hibernation both simply save the state your desktop is in (what programs are open, what files are accessed) in a file that is saved in RAM or on the hard drive respectively. No charge if the gauge cannot be repaired. A BRIEF INTRODUCTION TO MATLAB BRAD BAXTER Version: 201710261526 Abstract. There is a common characteristic in these pictures: there is only one bright spot in the picture and all are in black background. Add a splash of style to your Nintendo Switch™ console with Joy-Con™ controllers and Joy-Con straps. This is the official website of the GNU Image Manipulation Program (GIMP). 555 timer can also be used as a DARK SENSOR in its monostable mode or single shot mode. A dark theme displays dark surfaces across the majority of a UI. Modelio is an open source modeling environment (UML2, BPMN2, ). I would then, simulataneously, open the same file in uedit (using explorer and several mouse clicks) take advantages of the power (mainly regular expressions, macros, column mode, etc. Matrices and Vectors 2 2. This page lists over 500 colours by colour name, Hex value, RGB value and Microsoft Access code number. First open Run, by pressing ' Win+R '. iOS and Xamarin. Hannah has 2 jobs listed on their profile. 0: updated docs links. Toute l'actualité au masculin, actualisée en permanence : sport, hi-tech, mode, lifestyle, business, économie, enfants, sorties, voyages, automobile, bateau, avion. It was originally a system for solving matrices — quickly and accurately. Still, you may only want to use this theme at night time, rather than all of the time. If you have faced the problem that says this program cannot be run in DOS mode or if you have got stuck in the DOS mode (can face several problems in Mac too) and if you want to solve these problems, read this article. If you do not like the current dark theme, please visit the options page and choose a different theme from over 50 available options. The instructions of the problem are: Use bisection method to find a root of the function $$\sin x + x \cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. You specify the color of the line like this. Song Maker, an experiment in Chrome Music Lab, is a simple way for anyone to make and share a song. Matlab "Dark Mode", changing font size, and style. This is my own theme, a mash-up of Cobalt and Darkmate, and was the colour scheme I was originally trying to transfer between my MATLAB installations and. 0_17-b04 with Sun Microsystems Inc. To use a light background for documents while Dark Mode is turned on, click View in the menu bar in TextEdit, then deselect Use Dark Background for Windows. We encourage our users to get in the streets and join them if you can. One can specify colors using a vector that gives the RGB triple where in MATLAB, each of the three values are numbers from 0 to 1. In image histograms the pixels form the horizontal axis In Matlab histograms for images can be constructed using the imhist command. If the CommandWindow is not active, just click in it anywhere. ; 8 1000 dark gray; 9 1001 light blue; a 1010 light green; b 1011 light cyan; c 1100 light red; d 1101 light magenta; e 1110 yellow; f 1111 white org 100h; set video mode mov ax, 3 ; text mode 80x25, 16 colors, 8 pages (ah=0, al=3) int 10h ; do it!; cancel blinking and enable all 16 colors: mov ax, 1003h mov bx, 0 int 10h; set segment register:. Right now, if I set mode to 132, I get a percentage of full range as predefined for each channel as follows: fwrite(s,[132,4,0, setting],‘uint8’) where setting is 1…127. I want to plot a dark green and a dark green dashed graph, but unfortunately MatLab complains that the vector does not have the same length. That is the place where you will enter your MATLAB commands (see Chapter 2). Start your free 14-day trial download today!. 8A and RMS. Gray scale filtering is in reference to the color mode of a particular image. A High Speed Geiger Mode Photodiode Gating Circuit Modelling Using MATLAB. The let and const Keywords. The interface works only for version 5. Computers can only understand numbers, so an ASCII code is the numerical representation of a character such as 'a' or '@' or an action of some sort. Products; Solutions; Academia; Support; Community; Events. Programming. I'm using Matlab to communicate with a custom AD7746 board through an Arduino UNO. This cannot be repaired. Data may be passed between the two programs using intermediate MATLAB structures that organize the metadata needed; these are produced when GMT modules are run. A dark theme displays dark surfaces across the majority of a UI. Download Free PDF. Fullz with dl, The power spectrum (PS) of a time-domain signal is the distribution of power contained within the signal over frequency In MATLAB®, you can perform real-time spectral analysis of a dynamic signal using the dsp. If you turn on Dark Mode while using Dynamic Desktop, the desktop may change to the dark still image. removed useless 'posterize' mode. The consultancy was founded by Yair Altman, a renowned Matlab expert with 30 years professional software development experience. JDoodle is a free Online Compiler, Editor, IDE for Java, C, C++, PHP, Perl, Python, Ruby and many more. It's designed to be a supplemental mode to a default (or light) theme. You can help protect yourself from scammers by verifying that the contact is a Microsoft Agent or Microsoft Employee and that the phone number is an official Microsoft global customer service number. Figure 1-1 contains an example of a newly launched MATLAB Desktop. Toute l'actualité au masculin, actualisée en permanence : sport, hi-tech, mode, lifestyle, business, économie, enfants, sorties, voyages, automobile, bateau, avion. Im Dezember bekamen 24 Stipendiatinnen und Stipendiaten einen virtuellen Einblick hinter die Fassaden von GOLDBECK. This example shows how to use the V4L2 Video Capture and the SDL Video Display blocks from the BeagleBone Black block library to implement an image inversion algorithm with a Simulink® model, and to run the model on BeagleBone Black hardware. We are among the 40 largest websites in the world in terms of monthly unique. Now, make sure to update Google's Phone app from the Play Store. Because these definitions of saturation—in which very dark (in both models) or very light (in HSL) near-neutral colors are considered fully saturated (for instance, from the bottom right in the sliced HSL cylinder or from the top right)—conflict with the intuitive notion of color purity, often a conic or biconic solid is drawn instead (fig. m where the exception is thrown from the MATLAB tifflib code - but that does not help much. Lustige Bilder, lustige Videos und Flash-Games, Fun Videos, Werbespots kostenlos. Dark Mode editor for MATLAB. From the table above, we can define the default colors to work with them or can put in the RGB triplet (as a vector) directly into the. 8 Jun 2016: 1. Previously we had learned the method to enable Dark Mode for Teams, OneNote, and Outlook. See full list on wiki. Get Prezi account access by signing into Prezi here, and start working on or editing your next great presentation. So here basically, we are going to use imfindcircles function to detect center pivot irrigation fields in the image. Unfortunately, this expensive software does not have a dark mode/dark theme op. Matlab Simulink has become the universal mathematical and modeling tools in most universities and re-search laboratories around the world. Google's News app has featured such a mode since 2018, but the service's web version was only available with a bright theme until today. Dark UI themes seem to be all the rage these days. CH | Замунда - най-голямата българска peer 2 peer общност. |
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# How is the Frequency of a Stretched String Related To It’S Length? - Physics
Course
#### Question
How is the frequency of a stretched string related to its length?
#### Solution
The frequency of a stretched string is inversely proportional to its length.
f prop 1/l
Is there an error in this question or solution? |
## UPDATE: DMCP-3.22 / DM42-3.20
This is where announcements of official firmware updates provided by SwissMicros for the DM42 calculator will be made.
Please do not post bug reports here. The Usage tips, tricks and problem reports forum is there for that.
Please do not post notices of third party firmware builds here, they will be deleted. Instead, please use the Third Party firmware builds forum for this purpose.
Michael
Posts: 248
Joined: Wed Apr 05, 2017 11:31 pm
### UPDATE: DMCP-3.22 / DM42-3.20
This release of the DM42 firmware is an update to Free42 V3.0.5.
- FIX: AGRAPH display corrupted by menu (viewtopic.php?f=17&t=2994)
- FIX: Output of the MEM command not fully displayed in program mode (viewtopic.php?f=17&t=3047)
- Added new font in two sizes (default after clean start)
- Fixes in original fonts
DMCP changes:
DM42 Help File:
DM42 User Manual:
Code: Select all
aa48f43420eaec55e9f00d1c641e10609e013d00 DMCP_flash_3.22_DM42-3.20.bin
f160d5d190aa0991c1466491853a19817fb9387e DM42-3.20.pgm
004cb78b161fb34479e7cdf476bacc03c4baa7fe DMCP_flash_3.22.bin
akaTB
Posts: 628
Joined: Tue May 02, 2017 1:56 pm
Location: Milan, Italy
### Re: UPDATE: DMCP-3.22 / DM42-3.20
Now testing.
Thank you!
Greetings,
Massimo
ajcaton
-+×÷ left is right and right is wrong Casted in gold
zeno333
Posts: 78
Joined: Fri Dec 29, 2017 1:34 am
### Re: UPDATE: DMCP-3.22 / DM42-3.20
Free42 3.06 came out July 27, 2021...And you release this update with just Free42 3.05?....What gives....?
lispm
Posts: 9
Joined: Tue Jun 29, 2021 2:23 pm
### Re: UPDATE: DMCP-3.22 / DM42-3.20
3.05?? Really??
I don't get it.
TI 55, HP16C, HP41C, DM42, APPLE ][, TRS-80, ATARI 800XL, ATARI 520ST, XEROX 1186 DANDELION, SYMBOLICS 3640, TI EXPLORER II
BLOG: https://yazuu.org/en/showblog_public.php?user=1&blog=
Peet
Posts: 160
Joined: Tue Sep 29, 2020 12:01 am
Location: Germany
### Re: UPDATE: DMCP-3.22 / DM42-3.20
There is probably a reason, but it is unlikely that we will find out. I don't have the impression that Swissmicros is close to the community with informations.
But if anyone with contact to SM is reading this ... please take a look at the link to the DM42 manuals.
My programmable calculators - former: CBM PR100, HP41CV, HP28S, HP11C - current: HP48G(+), HP35S, DM41X, DM42
mcc
Posts: 272
Joined: Fri Jun 23, 2017 5:10 am
### Re: UPDATE: DMCP-3.22 / DM42-3.20
Hi,
I flashed the new firmware successfully.
I had hoped that the problem with CIRCT, the electrical parts editor had been fixed, but
it still does not work as intended. I wrote about the problem here:
viewtopic.php?f=17&t=2994&p=20449&hilit=CIRCT#p20449
May be it could be fixed?
Cheers.
mcc
DM 42 - SN: 00373, Firmware release v.:3.20. / DMCP 3.22. as compiled by SwissMicros
rprosperi
Posts: 1159
Joined: Mon Apr 24, 2017 7:48 pm
Location: New York
### Re: UPDATE: DMCP-3.22 / DM42-3.20
Peet wrote:
Wed Sep 01, 2021 10:28 pm
There is probably a reason, but it is unlikely that we will find out. I don't have the impression that Swissmicros is close to the community with informations.
But if anyone with contact to SM is reading this ... please take a look at the link to the DM42 manuals.
Thanks for the note about the DM42 docs; the folder was updated with the newest DM42 manual but something slipped and 41X stuff is shown there. It should be fixed by tomorrow morning (depending on where in the world you live )
The reason for the inclusion of 3.0.5 vs 3.0.6 in DM42 v3.20 is simple. Integration of 3.0.5 began shortly after it was released in July but before 3.0.6 was released. Then the DM42 bugs were traced, identified and corrected. Then the Voyager fonts were improved and verified. Then a completely new Voyager User manual was written, reviewed, re-tweaked and completed. Then summer vacations happened while docs were being reviewed and testing was going on, and then stuff was released.
I really think the negative reactions about not getting the 3.0.6 features, which I've seen no evidence of anyone using anyhow, are somewhat hostile and overblown. Through observation and some insightful guesswork, Didier explained it pretty well, but I thought I'd fill the gaps here for those that were unconvinced.
If the features in 3.0.6 are important for someone, please just say so; "I'd like to use XYZ feature for doing this or that". Also, it would be helpful to confirm that you've worked with the new features (on other platforms) and found them stable and effective for your goals. Just pointing out you've noticed what has not been included is not a collaborative / constructive approach, and is a disheartening response after a lot of work to provide another upgrade.
--bob p
DM42: β00071 & 00282, DM41X: β00071 & 00656, DM10L: 071/100
rprosperi
Posts: 1159
Joined: Mon Apr 24, 2017 7:48 pm
Location: New York
### Re: UPDATE: DMCP-3.22 / DM42-3.20
mcc wrote:
Thu Sep 02, 2021 4:01 am
Hi,
I flashed the new firmware successfully.
I had hoped that the problem with CIRCT, the electrical parts editor had been fixed, but
it still does not work as intended. I wrote about the problem here...
Cheers.
mcc
This problem should have been corrected in the v3.20 release. Are you sure you retested this with the new f/w and had the same problem? If so, please create a new state file, bring a new copy of the program in and try it. If it persists, please post exact steps and a screen shot of the garbled graphic results. It's possible there are some variations of the bug that were not corrected, which only occur in highly specific conditions.
Thanks
--bob p
DM42: β00071 & 00282, DM41X: β00071 & 00656, DM10L: 071/100
NoisyBoy
Posts: 13
Joined: Thu Jun 25, 2020 7:49 am
### Re: UPDATE: DMCP-3.22 / DM42-3.20
I for one is grateful for the update, having spent 30 years in enterprise software, sometimes one underestimates the amount of work required to release an update or a patch. I am not a power user and I certainly do not downplay others desires for the latest and greatest, but in my use case, I am absolutely satisfied with the frequency, currency, and quality of the update provided by Swiss Micros.
Keep up the great work and thank you!
dlachieze
Posts: 339
Joined: Thu May 04, 2017 12:20 pm
Location: France
### Re: UPDATE: DMCP-3.22 / DM42-3.20
mcc wrote:
Thu Sep 02, 2021 4:01 am
I had hoped that the problem with CIRCT, the electrical parts editor had been fixed, but
it still does not work as intended. I wrote about the problem here:
viewtopic.php?f=17&t=2994&p=20449&hilit=CIRCT#p20449
I have checked the CIRCT program with the latest firmware and as as far as I can see the issue you reported is fixed. Here is a screenshot to be compared to the one in your report:
20210902-06284076.bmp (12.31 KiB) Viewed 847 times
So as Bob mentioned if you still see an issue with the latest firmware, please provide the detailed steps to reproduce it.
DM42: 00425 - DM41X: β00066 |
## anonymous 3 years ago Determine the geometric mean of 5 and 8.
1. anonymous
$\frac{ 5 }{ x } = \frac{ x }{ 8 }$
2. anonymous
40=4*10 sqrt40=sqrt(4*10) sqrt40=sqrt4*sqrt10 sqrt40=2*sqrt10
3. anonymous
Cross multiply. $x^{2} = 40$ Where does the 10 come in? Did you divide?
4. anonymous
2 sqrt(10)~~6.32456
5. anonymous
I see. You eliminated the exponent and then square root 40. Okay, got it. :) |
## January 19, 2007
### More on Duality
#### Posted by David Corfield
Continuing our earlier discussion about duality, it’s worth noting a distinction that Lawvere and Rosebrugh introduce in chapter 7 of their Sets for Mathematics between ‘formal’ and ‘concrete’ duality. Formal duality concerns mere arrow reversal in the relevant diagrams, so
of course if the original diagrams had been given specific interpretation in terms of specific sets and mappings, such interpretation is lost when we pass to this formal dual in that the formal dualization process in itself does not determine specific sets and specific mappings that interpret the dualized statement. (p. 121)
Concrete duality, on the other hand, occurs in situations where a new diagram is formed from an old one by exponentiating each object with respect to a given dualizing object, e.g., $X$ becomes $V^X$, with $V$ the dualizing object. The arrows are naturally reversed in the new diagram.
Now,
Not every statement will be taken into its formal dual by the process of dualizing with respect to $V$, and indeed a large part of the study of mathematics
space vs. quantity
and of logic
theory vs. example
may be considered as the detailed study of the extent to which formal duality and concrete duality into a favorite $V$ correspond or fail to correspond. (p. 122)
Very relevant for concrete dualities is Peter Johnstone’s Stone Spaces, especially chapter 6 and its discussion of schizophrenic objects. For a hard-hitting review by Johnstone of some Universal algebraists’ attempt to treat duality while minimizing contact with category theory, see this.
Schizophrenic objects have found a role in clarifying $\omega$-categorical issues. See Makkai and Zawadowski’s Duality for Simple $\omega$-Categories and Disks.
Posted at January 19, 2007 9:10 AM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1117
### Re: More on Duality
In connection with these formal and concrete dualities, I can’t resist mentioning the Chu construction and especially Chu spaces: Vaughn Pratt has emphasized that many familiar concrete dualities are embodied in the category of Chu spaces.
In a nutshell, the Chu construction is a way of constructing a *-autonomous category from the data of a symmetric monoidal closed category M with pullbacks and an object D of M. An object of Chu(M, D) is a triple (A, B, p) where p is a D-valued pairing between A and B, and a morphism (A, B, p) –> (A’, B’, p’) is a pair of morphisms f: A –> A’, g: B’ –> B which are transposes of one another with respect to the pairings p and p’. If you write down a formal description of these hom-sets in terms of pullback diagrams involving hom-sets of M, then you can more or less guess what the closed structure of Chu(M, D) looks like, by internalizing these homs. The dual of (A, B, p) is obtained by switching A and B, and the dualizing object is the pair (D, I) with the obvious pairing.
The category of Chu spaces is Chu(Set, 2). A nice point emphasized by Pratt is that for many instances of concrete dualities where the underlying set of the schizophrenic object is 2, the category and its opposite embed as dual subcategories in the self-dual category of Chu spaces. This applies in particular to Stone duality, the duality between ordinals and intervals [emphasized by Joyal in connection with his category of disks], the self-duality of sup-lattices, and many others that you can think of.
Posted by: Todd Trimble on January 19, 2007 4:04 PM | Permalink | Reply to this
### Re: More on Duality
Does one ever find duality between bicategories arising from an object having two ‘commuting’ structures? I mean is it ever the case that something like the category of sets can be seen as possessing two structures, and so be used schizophrenically?
Posted by: David Corfield on January 19, 2007 9:40 PM | Permalink | Reply to this
### Re: More on Duality
Judging by our experience with objects with two commuting structures e.g. groups, what kind of categories woudl permit such?
Posted by: jim_stasheff on August 3, 2013 1:58 PM | Permalink | Reply to this
### Re: More on Duality
That’s an old comment of mine! Of course, it turned out that $Set$ can act as a dualizing object. See, e.g., Forssell and Awodey’s duality between first order theories and their models, reported here.
Posted by: David Corfield on August 3, 2013 2:14 PM | Permalink | Reply to this
### Re: More on Duality
In some cases this involves certain limits commuting with certain colimits. For example, $Set$ is finitely complete and also filtered-colimit cocomplete, and finite limits commute with filtered colimits. So $Set$ plays a role of dualizing object where on the one side we consider finitely complete categories and finitely continuous functors, and on the other complete and filtered-cocomplete categories, and this leads ultimately to the Gabriel-Ulmer duality between small finitely complete categories and locally finitely presentable categories. As we know, there are many variations on this theme.
Posted by: Todd Trimble on August 3, 2013 4:16 PM | Permalink | Reply to this
### Re: More on Duality
I wonder what variants will be found for $\infty$-Groupoid as a dualizing object.
Posted by: David Corfield on August 4, 2013 7:44 AM | Permalink | Reply to this
### Terminology
I’ve said this before and I’ll say it again: we need to find a word other than “schizophrenic”.
Groups that work with those who have schizophrenia are constantly trying to persuade lazy journalists etc. not to use the word in a way that perpetuates the simplistic comedy stereotype of a “split personality”. The mathematical usage is in exactly this vein. We should choose a word that doesn’t reinforce a misleading cliche about a serious illness.
I put this point at a category theory meeting a few years ago and other people (including Peter Johnstone) agreed. There were various suggestions for alternatives, but I can’t remember them. Maybe “ambiguous”, “ambivalent”, or “two-faced”? Ideas, anyone?
Posted by: Tom Leinster on January 19, 2007 10:58 PM | Permalink | Reply to this
### Re: Terminology
Maybe “ambiguous”, “ambivalent”, or “two-faced”?
Those are all nice, but my favourite is ‘ambivalent’.
Posted by: Toby Bartels on January 20, 2007 12:21 AM | Permalink | Reply to this
### Re: Terminology
How about ‘Janusian objects’ after the Roman God Janus?
I see ‘Janusian’ has been used to describe a form of thinking:
Janusian thinking—“actively conceiving two or more opposite or antithetical ideas, concepts, or images simultaneously,” according to the author’s definition—is proposed as a specific thought process that operates in the act of creation.
People also say ‘Janus-like’.
Posted by: David Corfield on January 20, 2007 11:58 AM | Permalink | Reply to this
### Re: Terminology
David wrote:
How about ‘Janusian objects’ after the Roman God Janus?
Oddly enough, I thought of that as I was writing my previous comment, but dismissed it as too fanciful. I think it’s a more accurate word than any of the others, though it’s a bit unusual.
Posted by: Tom Leinster on January 20, 2007 12:21 PM | Permalink | Reply to this
### Re: Terminology
Mathematicians seem more conservative than physicists when it comes to naming. But Alan Weinstein’s explanation for his ‘hopfish algebras’, see Friday 21 October entry from my old blog shows some humour is possible.
Posted by: David Corfield on January 20, 2007 12:32 PM | Permalink | Reply to this
### Re: Terminology
David wrote:
Mathematicians seem more conservative than physicists when it comes to naming.
Really? What about amoebas, train tracks, child’s drawings, shtukas, and bubble trees? Not to mention perversity and cacti.
Posted by: John Baez on January 21, 2007 2:45 AM | Permalink | Reply to this
### Re: Terminology
Hmm. I guess things have moved on since the times when people wouldn’t stray far from terms like ‘normal’, ‘regular’ and ‘weak’.
I’d still like to hear of Janusian objects sitting in a pair of bicategories if such a thing exists.
Posted by: David Corfield on January 21, 2007 5:03 PM | Permalink | Reply to this
### Re: Terminology
I’ve never seen the words ‘Janusian’ or ‘Janus-like’; what I’ve seen is ‘Janus-faced’. This is the one that gets the most Google hits, and it has the advantage of suggesting the word ‘two-faced’, to help out people who don’t know their Roman mythology.
Unfortunately, ‘Janus-faced’, like ‘two-faced’, has the connotation of ‘deceitful’. But, I guess ‘schizophrenic’ is also negative. I’d suggest ‘ambidextrous’, but that’s already taken.
Posted by: John Baez on January 21, 2007 2:31 AM | Permalink | Reply to this
### Re: Terminology
How about ‘Janusian objects’ after the Roman God Janus?
Ooh, I like that too! And you can justify it to those conservative mathematicians using Rothenberg’s interpretation, since even conservative mathematicians love to think of themselves as creative.
Posted by: Toby Bartels on January 21, 2007 4:13 AM | Permalink | Reply to this
### Re: Terminology
Janusian thinking–“actively conceiving two or more opposite or antithetical ideas, concepts, or images simultaneously,” according to the author’s definition–is proposed as a specific thought process that operates in the act of creation.
If there is some residual sense of “oppositeness” to “Janusian” or “Janus-faced”, as in the figure of Janus facing in opposite directions, then it doesn’t seem to me all that accurate for describing objects formerly known as schizophrenic.
“Schizophrenic”, literally “split-minded”, was perhaps ill-conceived for the reason Tom gives, but perhaps if we just keep the “schizo” and change the “phrenic” to something more accurate? I thought of “schizomorphic”, which already sounds mathematical, but there may be better options. (“Schizomorphic” already has various technical meanings, including an Aristasian one, but I think these could safely be ignored.)
Or, how about “ambimorphic”? I think I like that even more.
Posted by: Todd Trimble on January 21, 2007 9:50 AM | Permalink | Reply to this
### Re: Terminology
I feel a little silly: “Janusian” is certainly appropriate (thinking of the object as a target object in a category and in its opposite). I was somehow getting diverted by a different sense of “opposite” – sorry!
I’m not prepared to do a complete about-face just yet – for some reason, “Janusian” sounds to me just a little precious, or like a word Gene Roddenberry might have used. I may be in the minority though.
Posted by: Todd Trimble on January 21, 2007 2:47 PM | Permalink | Reply to this
### Re: Terminology
I agree that something doesn’t quite sit right with me about the term. On the other hand, if we end up deciding mythology is fair game I’ll work on finding something about tangles to name after Shelob.
Posted by: John Armstrong on January 21, 2007 4:05 PM | Permalink | Reply to this
### Janice Ian, musician; Re: Terminology
“Janusian” was what I thought I heard when, in the 1960s, I heard the name of the singer-songwriter Janice Ian. See her home page
www.janisian.com/
Then I heard and liked her song “at seventeen” – which I already knew was a Fermat Prime. Finally, I got to meet her, at a World Science Fiction convention, as an anthology of stories was written based on her lyrics.
So I have no problem with the term. Is cerberus-headed the phrase from triality? Or tridental?
Posted by: Jonathan Vos Post on January 23, 2007 7:41 AM | Permalink | Reply to this
### Re: Terminology
The reference to Janus in a similar (but informal) context goes back to Weyl in his 1932 paper “Topology and Abstract Algebra as Two Roads of Mathematical Comprehension” (1995 English translation available here: http://www.jstor.org/stable/2975040). There he speaks of the reals as “a Janus head with two oppositely directed faces”, referring to it being the domain of algebraic operations but also a continuous manifold, and to the interplay between the two structures.
Posted by: rsb on August 3, 2013 2:53 AM | Permalink | Reply to this
### Re: Terminology
The reference to Janus in a similar (but informal) context goes back to Weyl in his 1932 paper “Topology and Abstract Algebra as Two Roads of Mathematical Comprehension” (1995 English translation available here: http://www.jstor.org/stable/2975040). There he speaks of the reals as “a Janus head with two oppositely directed faces”, referring to it being the domain of algebraic operations but also a continuous manifold, and to the interplay between the two structures.
Posted by: rsb on August 3, 2013 2:53 AM | Permalink | Reply to this
### Re: Terminology
According to the OED, the word “schizophrenia” entered the English language in 1910, as an invented name for the mental illness. This illness has nothing to do with “split personalities”. Unfortunately, well-educated people at the time had a better knowledge of Greek than of psychiatry, and so the literal meaning “split mind” subsequently crept into the language. The OED cites TS Eliot as misusing the term in 1933 (Use of Poetry and Use of Criticism). So while I agree that we should now undo this ignorance, I wanted to point out that Simmons etc do have admirable company in their mistake.
Posted by: Sam on October 30, 2010 9:07 AM | Permalink | Reply to this
Post a New Comment |
# Questions & Answers of Engineering Mechanics
#### Topics of Engineering Mechanics 51 Question(s) | Weightage 03 (Marks)
Question No. 16
A rigid ball of weight 100 N is suspended with the help of a string. The ball is pulled by a horizontal force F such that the string makes an angle of $30^\circ$ with the vertical. The magnitude of force F (in N) is __________
Question No. 17
A point mass M is released from rest and slides down a spherical bowl (of radius R) from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is
Question No. 40
A block of mass m rests on an inclined plane and is attached by a string to the wall as shown in the figure. The coefficient of static friction between the plane and the block is 0.25. The string can withstand a maximum force of 20 N. The maximum value of the mass (m) for which the string will not break and the block will be in static equilibrium is ____________ kg.
Take $\cos\theta=0.8$ and $\sin\theta=0.6$.
Acceleration due to gravity g = 10 m/$s^2$
Question No. 41
A two-member truss PQR is supporting a load W. The axial forces in members PQ and QR are respectively
Question No. 116
A point mass having mass M is moving with a velocity V at an angle $\theta$to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is
Question No. 140
A mass of 2000 kg is currently being lowered at a velocity of 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is $150\mathrm{kg}-\mathrm m^2$. On applying the brake, the mass is brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is __________
Question No. 216
A force F is acting on a bent bar which is clamped at one end as shown in the figure.
The CORRECT free body diagram is
Question No. 219
A rigid link PQ is undergoing plane motion as shown in the figure ($\style{font-family:'Times New Roman'}{\left(V_{P\;}and\;V_Q\;are\;non-zero\right)}.$\style{font-family:'Times New Roman'}{V_{QP}}$is the relative velocity of point Q with respect to point P. Which one of the following is TRUE? ##### Show Answer Question No. 240 An inextensible massless string goes over a frictionless pulley. Two weights of 100 N and 200 N are attached to the two ends of the string. The weights are released from rest, and start moving due to gravity. The tension in the string (in N) is __________ ##### Show Answer Question No. 241 A circular disc of radius 100 mm and mass 1 kg, initially at rest at position A, rolls without slipping down a curved path as shown in figure. The speed $v$ of the disc when it reaches position B is _________ m/s. Acceleration due to gravity g = 10 m/s2. ##### Show Answer Question No. 242 A rigid rod (AB) of length $L=\sqrt{2}$ m is undergoing translational as well as rotational motion in the x-y plane (see the figure). The point A has the velocity$V_{1\;}=\widehat i+2\widehat j$m/s. The end B is constrained to move only along the x direction. The magnitude of the velocity V2 (in m/s) at the end B is __________ ##### Show Answer Question No. 23 A wheel of radius r rolls without slipping on a horizontal surface shown below. If The velocity of point P is 10 m/s in the horizontal direction, the magnitude of velocity of point Q (in m/s) is ________ ##### Show Answer Question No. 26 Two identical trusses support a load of 100 N as shown in the figure. The length of each truss is 1.0m; cross-sectional area is 200 mm2, young’s modulus E = 200GPa. The force in the truss AB (in N) is ______. ##### Show Answer Question No. 49 For the turss shown in figure, the magnitude of the force in member PR and the support reaction at R are respectively ##### Show Answer Question No. 50 A ball of mass 0.1 kg, initially at rest, is dropped from height of 1 m. Ball hits the ground and bounces off the ground. Upon impact with the ground, the velocity reduces by 20% . The height (in m) to which the ball will rise is _____. ##### Show Answer Question No. 116 A small ball of mass 1 kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be____. ##### Show Answer Question No. 140 The initial velocity of an object is 40 m/s. The acceleration a of the object is given by the following expression: a = – 0.1 v Where v is the instantaneous velocity of the object. The velocity of the object after 3 seconds will be____. ##### Show Answer Question No. 142 For the truss shown in figure , the magnitude of the force (in kN) in the member SR is ##### Show Answer Question No. 228 A weight of 500 N is supported by two metallic ropes as shown in the figure. The values of tensions T1 and T2 are respectively ##### Show Answer Question No. 236 The value of moment of inertia of the section shown in the figure about the axis-XX is ##### Show Answer Question No. 237 Figure shows a wheel rotating about O2. Two points A and B located along the radius of wheel have speeds of 80 m/s and 140 m/s respectively. The distance between the points A and B is 300 mm. The diameter of the wheel (in mm) is__________ ##### Show Answer Question No. 255 A bullet spins as the shot is fired from a gun. For this purpose, two helical slots as shown in the figure are cut in the barrel. Projections A and B on the bullet engage in each of the slots. Helical slots are such that one turn of helix is completed over a distance of 0.5 m. If velocity of bullet when it exits the barrel is 20 m/s, its spinning speed in rad/s is _______. ##### Show Answer Question No. 19 A circular object of radius r rolls without slipping on a horizontal level floor with center having velocity V. The velocity at the point of contact between the object and floor is ##### Show Answer Question No. 41 A block R of mass 100 kg is placed on a block S of mass 150 kg as shown in the figure.Block R is tied to the wall by a massless and inextensible string PQ.If the coefficient of static friction for all surfaces is 0.4, the minimum force F (in n kN) needed to move the block S is ##### Show Answer Question No. 45 A block weighing 200 N is in contact with a level plane whose coefficients of static and kinetic friction are 0.4 and 0.2, respectively. The block is acted upon by a horizontal force (in newton) P=10t, where t denotes the time in seconds. The velocity (in m/s) of the block attained after 10 seconds is _______ ##### Show Answer Question No. 117 A two member truss ABC is shown in the figure. The force (in kN) transmitted in member AB is _______ ##### Show Answer Question No. 146 A truck accelerates up a 10° incline with a crate of 100 kg. Value of static coefficient of friction between the crate and the truck surface is 0.3. The maximum value of acceleration (in m/s2) of the truck such that the crate does not slide down is _______ ##### Show Answer Question No. 219 A mass m1 of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with a stationary mass m2 of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is ##### Show Answer Question No. 241 A body of mass (M) 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is _______ ##### Show Answer Question No. 244 An annular disc has a mass m, inner radius R and outer radius 2R. The disc rolls on a flat surface without slipping. If the velocity of the center of mass is v, the kinetic energy of the disc is ##### Show Answer Question No. 316 In a statically determinate plane truss, the number of joints (j) and the number of members (m) are related by ##### Show Answer Question No. 320 A point mass is executing simple harmonic motion with an amplitude of 10 mm and frequency of 4 Hz. The maximum acceleration (m/s2) of the mass is _______ ##### Show Answer Question No. 342 For the truss shown in the figure, the forces F1 and F2 are 9 kN and 3 kN, respectively. The force (in kN) in the member QS is ##### Show Answer Question No. 346 A frame is subjected to a load P as shown in the figure. The frame has a constant flexural rigidity EI. The effect of axial load is neglected. The deflection at point A due to the applied load P is ##### Show Answer Question No. 347 A wardrobe (mass 100 kg, height 4 m, width 2 m, depth 1 m), symmetric about the Y-Y axis, stands on a rough level floor as shown in the figure. A force P is applied at mid-height on the wardrobe so as to tip it about point Q without slipping. What are the minimum values of the force (in newton) and the static coefficient of friction μ between the floor and the wardrobe, respectively? ##### Show Answer Question No. 350 A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in newton) required to maintain equilibrium of the ladder is _______ ##### Show Answer Question No. 33 A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external force F as shown in the figure below. The force F is suddenly removed. At the instant of force removal, the magnitude of vertical reaction developed at the support is ##### Show Answer Question No. 44 A force of 400 N is applied to the brake drum of 0.5 m diameter in a band-brake system as shown in the figure, where the wrapping angle is 180°. If the coefficient of friction between the drum and the band is 0.25, the braking torque applied, in N.m. is ##### Show Answer Question No. 48 Two steel truss members, AC and BC, each having cross sectional area of 100 mm2, are subjected to a horizontal force F as shown in figure. All the joints are hinged. If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is ##### Show Answer Question No. 49 Two steel truss members, AC and BC, each having cross sectional area of 100 mm2, are subjected to a horizontal force F as shown in figure. All the joints are hinged. The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is ##### Show Answer Question No. 22 The coefficient of restitution of a perfectly plastic impact is ##### Show Answer Question No. 32 A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force Fx (in N) exerted by the rubber band as a function of distance x (in m) is given by Fx = 300x2. If the stone is displaced by 0.1m from the un-stretched position (x = 0) of the rubber band, the energy stored in the rubber band is ##### Show Answer Question No. 43 A 1 kg block is resting on a surface with coefficient of friction μ = 0.1. A force of 0.8 N is applied to the block as shown in figure. The friction force is ##### Show Answer Question No. 10 A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is µ = 0.2.A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right? ##### Show Answer Question No. 41 A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of 2L/3 from the hinge so that the rod swings to the right. The reaction at the hinge is ##### Show Answer Question No. 9 A straight rod of length L(t), hinged at one end and freely extensible at the other end, rotates through an angle about the hinge. At time t, L(t) =1 m,$\dot L\left(t\right)$= 1 m/s, $\theta \left(t\right)=\frac{\mathrm{\pi }}{4}$ rad and$\dot\theta\left(t\right)\$= 1 rad/s. The magnitude of the velocity at the other end of the rod is
Question No. 10
A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 kg. The mass of the counterweight, for static balance, is
Question No. 33
A circular disk of radius R rolls without slipping at a velocity v. The magnitude of the velocity at point P (see figure) is
Question No. 34
Consider a truss PQR loaded at P with a force F as shown in the figure
The tension in the member QR is
A block of mass M is released from point P on a rough inclined plane with inclination angle $\theta$, shown in the figure below. The coefficient of friction is $\mu$. If $\mu <\mathrm{tan}\theta$, then the time taken by the block to reach another point Q on the inclined plane, where PQ = s, is |
# The exponential form of complex number
• December 21st 2012, 02:23 AM
mous99
The exponential form of complex number
z = (-i) ^1/3
z^3 = -i
let z = re ^i ϴ
z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]
pls help me to understand how to find "3ϴ"?
what is the formula to find 3ϴ??
• December 21st 2012, 03:10 AM
chiro
Re: The exponential form of complex number
Hey mous99.
For this problem do you know how to use De-Moivre's formula for the angle? (Hint: its in the form of (ϴ + 2kπ)/n where n = 3 k = 0,1,2) [In this case theta = [3/2]*π]
• December 21st 2012, 03:56 AM
Deveno
Re: The exponential form of complex number
behold:
$e^{i (\theta + \phi )} = \cos(\theta + \phi ) + i \sin(\theta + \phi )$
$= \cos(\theta)\cos(\phi ) - \sin(\theta)\sin(\phi ) + i (\sin(\theta)\cos(\phi ) + \cos(\theta)\sin(\phi ))$
$= (\cos(\theta) + i \sin(\theta))(\cos(\phi ) + i \sin(\phi )) = e^{i \theta}e^{i \phi }$
so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles).
• December 21st 2012, 06:03 AM
mous99
Re: The exponential form of complex number
i cant see the invalid equation.
can u show me again?
thanks...
• December 21st 2012, 11:41 AM
HallsofIvy
Re: The exponential form of complex number
Unfortunately, Latex does not seem to be working today. I have removed the Latex tags.
Quote:
Originally Posted by Deveno
behold:
e^{i (\theta + \phi )} = \cos(\theta + \phi ) + i \sin(\theta + \phi ) = \cos(\theta)\cos(\phi ) - \sin(\theta)\sin(\phi ) + i (\sin(\theta)\cos(\phi ) + \cos(\theta)\sin(\phi ))
= (\cos(\theta) + i \sin(\theta))(\cos(\phi ) + i \sin(\phi )) = e^{i \theta}e^{i \phi }
so, in order to add (resp. divide) two complex numbers on the unit circle you simply add (resp. subtract) their arguments (angles).
Quote:
The exponential form of complex number
z = (-i) ^1/3
z^3 = -i
let z = re ^i ϴ
z^3 = (r^3)(e^i3ϴ) - i = e ^i[(3π/2) + 2nπ]
pls help me to understand how to find "3ϴ"?
what is the formula to find 3ϴ??
Just what you wrote: 3ϴ= [(3π/2) + 2nπ]
Now, solve for ϴ. |
# viewSpace to objectSpace
## Recommended Posts
mokaschitta 124
Hi, I am currently doing some rayCasting on the GPU using GLSL and I once again stumbled across a problem. Right now I am doing the raycasting in viewSpace, down the negative -z axis. I am raycasting 3D Metaballs which are accelarated using spatialHashing via texture lookups. My hashFunction simply takes the current Position and gives me all metaballs in that bucket. Anyways now since I started doing the raycasting in viewSpace I need to convert the position back to object/ModelViewSpace to find the right bucket. Am I right that I would have to multiply the current position with the inverse of the ModelViewMatrix? Thanks!
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jpetrie 13102
Quote:
Am I right that I would have to multiply the current position with the inverse of the ModelViewMatrix?
Yes.
##### Share on other sites
mokaschitta 124
I am kind of scared that this could ruin the performance (since I would have to make multiple matrix multiplications per fragment).- Maybe I should try raycasting in object space directly.
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mokaschitta 124
Just in case anybody wants to know. I do the raycasting in objectSpace now which saves me alot of matrix Multiplications. Basically I reconstruct the nearplane in objects space each time the camera changes like this:
// get the current modelview matrix glGetFloatv(GL_MODELVIEW_MATRIX , modelview); //right & upvector of the current camera setup Vec3f Right(modelview[0], modelview[4], modelview[8]); Vec3f Up(modelview[1], modelview[5], modelview[9]); Vec3f dir = at-cpos; //camera dir dir.normalize(); Vec3f nPos = cpos+dir*near; //find center point on the nearplane Vec3f ntl,ntr,nbl,nbr; //find corners of the nearplane ntl = nPos+(Up*halfHeight)-(Right*halfWidth); ntr = nPos+(Up*halfHeight)+(Right*halfWidth); nbl = nPos-(Up*halfHeight)-(Right*halfWidth); nbr = nPos-(Up*halfHeight)+(Right*halfWidth); //halfWidth and halfHeight as set in glFrustum
then I basically just draw a quad with the four calculated corners (which should be fullscreen since its on the nearplane) and pass the cameraPosition to my shader. The ray direction can then simply be calculated like this:
varying vec3 msPos; //vertex position of the quad in objectSpaceuniform vec3 camPos;void main(){ vec3 rayOrigin = msPos; vec3 rayDir = normalize(rayOrigin-camPos);}
alternatively you can render the nearplane quad to a texture and do the same thing with a texture lookUp.
It took me quite some time to figgure this out so I hope this will help somebody :) |
# Kiss
HD Video
3.06 min
2017
Kiss was made in collaboration with scientists at Nokia Bell Labs in which I used their Schlieren imaging machinery. Schlieren imaging is commonly used in fluid dynamics, as it can show hidden air currents on a television monitor that would usually be invisible to the naked eye. In Kiss the Schlieren apperatus is used to capture a person blowing a kiss. When we blow someone a kiss we pretend that we are actually blowing someting physical, as if a kiss is a real object. With this video we can actually see the kiss traveling through the air from the persons hand. Although it lasts only a split second, we can still see the kiss leave the person’s hand and continue on its way towards the its lucky recciever. |
## tisdag 21 juni 2016
### New Quantum Mechanics 3: Why?
Modern physics is being based on (i) relativity theory and (ii) quantum mechanics, both viewed to be correct beyond any conceivable doubt, but nevertheless (unfortunately) being incompatible. The result is a modern physics based on shaky grounds of contradictory theories from which anything can emerge, and so has done in the form of string theory and multiversa beyond thinkable experimental verification.
The basic model of quantum mechanics is Schrödinger's equation as a linear equation in a wave function depending on $3N$ spatial dimensions for an atom with $N$ electrons. Schrödinger's equation is an ad hoc model arrived at by a purely formal extension of classical mechanics without direct physical meaning and rationale. Schrödinger's equation is thus viewed as being given by God with the job of physical interpretation being left to humanity in endless quarrels. In this sense quantum mechanics is rather religion than science and the present state of physics maybe a fully logical result.
Experimental support for Schrödinger's equation is in incontestable form only available in the case of Hydrogen with $N=1$, since for larger $N$ the multidimensionality prevents both analytical and computational solution. The message of books on quantum mechanics that solutions of Schrödinger's equation always (have to) agree with observations, rather reflect a belief that a God-given equation cannot be wrong, than actual human experience.
But if we as scientists do not welcome the idea of an equation given by God beyond human comprehension, then we may find motivation to search for an alternative atomic model which is computable and thus is possible to compare with physical experiment. This is my motivation anyway.
And God said:
And then there were Atoms!
### New Quantum Mechanics 2: Computational Results
I have now tested the atomic model for an atom with $N$ electrons of the previous post formulated as a classical free boundary problem in $N$ single-electron charge densities with non-overlapping supports filling 3d space with joint charge density as a sum of electron densities being continuously differentiable across inter-electron boundaries.
I have computed in spherical symmetry on an increasing sequence of radii dividing 3d space into a sequence of shells filled by collections of electrons smeared into spherically symmetric shell charge distribution. The electron-electron repulsive energy is computed with a reduction factor of $\frac{n-1}{n}$ for the electrons in a shell with $n$ electrons to account for lack of self repulsion.
Below is a typical result for Xenon with 54 electrons organised in shells with 2, 8, 18, 18 and 8 electrons with ground state energy -7413 to be compared with -7232 measured and with the energy distribution in the 5 shells displayed in the order of total energy, kinetic energy, kernel potential energy and inter-electron energy. Here the blue curve represents electron charge density, green is kernel potential and red is inter-electron potential. The inter-shell boundaries are adaptively computed so as to represent a preset 2-8-18-18-8 configuration in iterative relaxation towards a ground state of minimal energy.
In general computed ground state energies agree with measured energies within a few percent for all atoms up to Radon with 86 electrons.
The computations indicate that it may well be possible to build an atomic model based on non-overlapping electronic charge densities as a classical continuum mechanical model with electrons keeping individuality by occupying different regions of space, which agrees reasonably well with observations. The model is an $N$-species free boundary problem in three space dimensions and as such is readily computable for any number of $N$ for both ground states, excited states and dynamic transitions between states.
We recall the the standard model in the form of Schrödinger's equation for a wave function depending on $3N$ space dimensions, is computationally demanding already for $N=2$ and completely beyond reach for larger $N$. As a result the full $3N$-dimensional Schrödinger equation is always replaced by some radically reduced model such as Hartree-Fock with optimization over a "clever choice" of a few "atomic orbitals", or Thomas-Fermi and Density Functional Theory with different forms of electron densities.
The present model is an electron density model, which as a free boundary problem with electric individuality is different from Thomas-Fermi and DFT.
We further recall that the standard Schrödinger equation is an ad hoc model with only formal justification as a physical model, in particular concerning the kinetic energy and the time dependence, and as such should perhaps better not be taken as a given ready-made model which is perfect and as such canonical (as is the standard view).
Since this standard model is uncomputable, it is impossible to show that the results from the model agree with observations, and thus claims of perfection made in books on quantum mechanics rather represent an ad hoc preconceived idea of unquestionable ultimate perfection than true experience.
## onsdag 1 juni 2016
### New Quantum Mechanics 1 as Classical Free Boundary Problem
Let me (as a continuation of the sequence of posts on Finite Element Quantum Mechanics 1-5) present an alternative formulation of the eigenvalue problem for Schrödinger's equation for an atom with $N$ electrons starting from an Ansatz for the wave function
• $\psi (x) = \sum_{j=1}^N\psi_j(x)$ (1)
as a sum of $N$ electronic real-valued wave functions $\psi_j(x)$, depending on a common 3d space coordinate $x\in R^3$ with non-overlapping spatial supports $\Omega_1$,...,$\Omega_N$, filling 3d space, satisfying
• $H\psi = E\psi$ in $R^3$, (2)
where $E$ is an eigenvalue of the (normalised) Hamiltonian $H$ given by
• $H(x) = -\frac{1}{2}\Delta - \frac{N}{\vert x\vert}+\sum_{k\neq j}V_k(x)$ for $x\in\Omega_j$,
where $V_k(x)$ is the potential corresponding to electron $k$ defined by
• $V_k(x)=\int\frac{\psi_k^2(y)}{2\vert x-y\vert}dy$, for $x\in R^3$,
and the wave functions are normalised to correspond to unit charge of each electron:
• $\int_{\Omega_j}\psi_j^2(x) dx=1$ for $j=1,..,N$.
One can view (2) as a formulation of the eigenvalue problem for Schrödinger's equation, starting from an Ansatz for the total wave function as a sum of electronic wave function according to (1), as a classical free boundary problem in $R^3$, where the electron configuration is represented by a partition of $R^3$ into non-overlapping domains representing the supports of the electronic wave functions $\psi_j$ and the total wave function $\psi$ is continuously differentiable.
Defining $\rho_j = \psi_j^2$, we have
• $\psi\Delta\psi = \frac{1}{2}\Delta\rho-\frac{1}{4\rho}\vert\nabla\rho\vert^2$,
and thus (2) upon multiplication by $\psi$ takes the form
• $-\frac{1}{4}\Delta\rho+\frac{1}{8\rho}\vert\nabla\rho\vert^2-\frac{N\rho}{\vert x\vert}+V\rho = E\rho$ in $R^3$, (3)
where
• $\rho_j\ge 0$, $support(\rho_j)=\Omega_j$ and $\rho_j=0$ else,
• $\int_{\Omega_j}\rho_jdx =1$,
• $\rho =\sum_j\rho_j$,
• $V\rho=\sum_{k\neq j}V_k\rho_j$ in $\Omega_j$,
• $\Delta V_j=2\pi\rho_j$ in $R^3$.
The model (3) (or equivalently (2)) is computable as a system in 3d and will be tested against observations. In particular the ground state of smallest eigenvalue/energy E is computable by parabolic relaxation of (3) in $\rho$. Continuity of $\psi$ then corresponds to continuity of $\rho$.
We can view the formulation (3) in the same way as that explored for gravitation, with the potential $V_j$ primordial and the electronic density $\rho_j$ defined by $\rho_j =\frac{1}{2\pi}\Delta V_j$ as a derived quantity, with in particular total electron-electron repulsion energy given by the neat formula
• $\frac{1}{2\pi}\sum_{k\neq j}\int V_k\Delta V_jdx=-\frac{1}{2\pi}\sum_{k\neq j}\int\nabla V_k\cdot\nabla V_jdx$
in terms of potentials with an analogous expression for the kernel-electron attraction energy.
For the choice of free boundary condition see later post.
### Many Big Bangs: Universe Bigger Than You Think
Astronomer Royal Lord Rees has made a statement:
• There may have been more than one Big Bang, the Astronomer Royal has said and claims the world could be on the brink of a revolution as profound as Copernicus discovering the Earth revolved around the Sun.
• Many people suspect that our Big Bang was not the only one, but there’s a whole ensemble of Big Bangs, a whole archipelago of Big Bangs.
• The theory is still highly controversial, but Lord Rees said he would ‘bet his dog’ on the theory being true.
This fits with the view I have presented in posts on a new view on gravitation, dark matter and dark energy, with gravitational potential $\phi$ viewed as primordial from which matter density $\rho$, which may be both positive and negative, is generated by
• $\rho = \Delta\phi$,
through local action in space of the Laplacian $\Delta$. In this model a Big Bang corresponds to a small local fluctuation of $\phi$ around zero, which generates an much bigger fluctuation of matter density by the action of the Laplacian.
In this model substantial matter may be generated locally from small fluctuations of gravitational potential opening the possibility of an endless number of Big Bangs seemingly created out of nothing.
You can test the model in the app Dark Energy on App Store. |
# Optica Hecht Cuarta Edicion Espanol
Optica Hecht Cuarta Edicion Espanol
References
Category:Optical illusions
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Category:Optical illusions involving geometric figuresQ:
Importance of Vector Space as Quotient of Euclidean Space
This is an excerpt from my textbook about Vector Spaces:
The idea of vector spaces is the idea of an abstract space in which our scalar products is compatible with a quotient operation, $A \oplus B$ where $A$ and $B$ are two different (real or complex) vector spaces. That means that for all points $x \in A, y \in B,$ and a scalar $a$, one has that the “dot product” $a(x+y)$ makes sense.
Given the vector space $A$ and the nonzero vector $x \in A$, the space $A/x$ is indeed a space, in which we can make the quotient operation, and it is a vector space with respect to the usual scalar product.
I am a little confused by the excerpt. Is it telling me that the abstract space, for all points $x \in A$ and a scalar $a$, we have that $a(x+y)$ makes sense? That is, if one identifies the vector space with itself (in the same way we have that the vector space $Z_n$ can be identified with itself in the rational field $\mathbb{Q}$ with the usual scalar product), then for all points $x \in A$ and a scalar $a$, $a(x+y)$ makes sense?
Why do we need to take the quotient of the first vector space to obtain the second vector space?
A:
We choose an additive $+$ and a scalar product on $A$ so that we can make sense of the scalar product $a(x+y)$ where $x$ is some fixed vector of $A$ and $y$ is any vector of $A$.
The vector space $A$ is an abritrary vector space and could be an $n$-dimensional vector space.
The function $a\colon A\to\mathbb{K}$ defined by $a(x+y):=a(x)\cdot a(y)$ where $x$ is a fixed vector and $y$ is a vector in $A$,
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optica hecht cuarta edicion espanol:.. Historical vii. | Academic libraries at a glance | Dc ontario government libraries and data centersCatalytic oxidation of methyl tert-butyl ether with different catalysts.
Methyl tert-butyl ether (MTBE) is a common gasoline component. In gasoline, MTBE is used mainly as a gasoline oxygenate to enhance octane levels. Based on the differences in selectivity, productivity and cost between catalysts, in this work, four commercially-available catalysts were used to catalyze the oxidation of MTBE to tert-butanol (TBA) and carbon oxides (CO2 and CO). The catalysts were HZSM-5, NH2-HZSM-5, SAPO-11 and SAPO-14. After the initial step, the coke deposition was determined by the atomic absorption spectroscopy and potassium permanganate combustion methods. The results indicated that the activity of catalyst used in the second reaction was higher than that of the first reaction. This indicated that the catalyst had better selectivity for TBA. After the reaction, the CO2 and CO concentrations were determined by a gas chromatograph. The coke deposition and the specific activity of the reaction mixture were increased when NH2-HZSM-5 was used. The results suggest that NH2-HZSM-5 has a better ability to maintain the catalytic activity during the reaction of MTBE to TBA.Q:
How do we force a bijective function into a set of functions?
For context, I’m just trying to fill the gaps in my knowledge of function theory.
If I have a function $f$
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then
if [ “$(tail -c 1 ‘ file_content else read -p ‘–>’ file_content2 fi else echo “Sorry but file does not contain ‘abc'” fi See the updated code below for the OP’s updated question. I’ve added the ‘3’ to the grep as I assume you are looking to see if something other than a line starting with abc occurs in the file, and then check if there’s anything after the first line that begins with.jpg. if [[ “$1” == “ABC” ]]
then
if grep -q.jpg |
# Show sequence converges?
1. Sep 24, 2009
1. The problem statement, all variables and given/known data
Show that the sequence P_n = [(n+1)/n, [(-1)^n]/n] converges.
2. Relevant equations
A sequence p_n converges to a point p if and only if every neighborhood about p contains all the terms p_n for sufficiently large indices n; to any neighborhood U about p, there corresponds an index N such that p_n exists in U whenever n > or = N.
3. The attempt at a solution
The sequence converges to the point p = (1 0). Let U be the open ball around p with radius r>0. Need to show that |p_n - p|< r for all n > or = N i.e. ....
|((n+1)/n, ((-1)^n)/n) - (1, 0)| < r
sqrt of ((n+1)/n - 1)^2 + (((-1)^n)/n)^2 < r
Now I assume the n and N come into play, but I don't know how.
2. Sep 24, 2009
### snipez90
Simply further. You should get something like 2/n^2 under the square root. |
# I need help finishing a gcd proof
• Sep 25th 2012, 07:39 PM
bskcase98
I need help finishing a gcd proof
the question was to prove:
if m and n are positive integers and gcd(m, n) = d, then gcd(2n - 1, 2m - 1) = 2d - 1.
my approach was the following:
given gdc(m,n) = d, the d|m and d|n also m = dp and n = dq for some integers p and q.
so, (2n - 1, 2m - 1) = (2dq - 1, 2dp - 1)
2dq - 1 = 2d - 1(2d(q-1) + 2d(q-2) +...+ 2d + 1) = (2d - 1)(s) for some integer s.
2dp - 1 = 2d - 1(2d(p-1) + 2d(p-2) +...+ 2d + 1) = (2d - 1)(t) for some integer t.
so (2d - 1) is a divisor of (2n - 1, 2m - 1), however I do not know how to show that it is the gcd(2n - 1, 2m - 1) or how to show that
gcd(s,t) = 1.
my instructor suggested proof by induction showing true for m + n < s, then true for m + n = s. I can not see how to start the proof
using his suggestion. any help would be appreciated.
• Sep 25th 2012, 08:18 PM
chiro
Re: I need help finishing a gcd proof
Hey bskcase98.
If something is a greatest divisor (let alone a greatest common divisor), then it means that all divisors must be less than that divisor. In other words if you have n and the greatest divisor of n is a then if n = ab then b <= a. From this can you show that this holds?
• Sep 25th 2012, 10:53 PM
johnsomeone
Re: I need help finishing a gcd proof
You've proven that, if $d = gcd(a,b)$, then $(2^d - 1)$ divides the $gcd((2^a - 1), (2^b-1))$.
If you could prove that $gcd((2^a - 1), (2^b-1))$ divides $(2^d - 1)$, then you'd be done, since they'd then be equal.
Recall in a previous thread of yours called "need help with an Eucledian algorthim and gcd involving factorials and large exponets", I gave a long derivation? In there I wrote:
"Proposition: If d divides $(b^y - 1)$ and d divides $(b^y - 1)$, then d divides $(b^{gcd(x, y)} - 1)$. Rather than work out a proof, I'll run through it step by step for this problem."
Well, now that proposition is exactly what you need. Switching the labelling back to the current problem, you'd be done if you could prove:
Lemma: If $c$ divides $(2^a - 1)$ and $c$ divides $(2^b - 1)$, then $c$ divides $(2^{gcd(a, b)} - 1)$.
If you can prove that Lemma, then, since $c$ = $gcd((2^a - 1), (2^b-1))$ divides both $(2^a - 1)$ and $(2^b - 1)$, it would follow that
$c = gcd((2^a - 1), (2^b-1))$ divides $(2^{gcd(a, b)} - 1)$.
And since you already have shown that $(2^{gcd(a,b)} - 1)$ divides the $gcd((2^a - 1), (2^b-1))$, you'd then be done.
That insane calculation I did in that other problem lays out - perhaps - the way for you to prove the above Lemma.
By the way, that insane calculation is, I believe, called Euclid's Algorithm for finding the gcd.
Just a thought. Good luck. |
Quora is one of my favorite time-waster websites, and somehow, I signed myself up to get a semi-weekly email from them that always seems to draw me into reading all the way down the email. Yesterday, I came across this question:
Since we’ve been studying energy in my send year Matter and Interactions course, and just finished studying centripetal forces in the previous chapter, this seemed like a wonderful problem for our last class of 2017.
We quickly went to estimating the things we knew to solve this. We’ve talked about the solar constant before and we knew that at the radius of the earth, every square meter receives 1400W of light energy on average. Knowing that it takes light 8 minutes to reach the Earth from the Sun, we quickly calculated the total power of the earth to be around 10^26W, and so we knew that the sun was losing 10^26 J of mass energy every second.
We knew that this energy was coming from fusion and that the sun was losing rest energy. We found the energy using the idea
$\Delta E =\Delta m c^2$
$\Delta m = 1.1 \times 10^9 kg$
This was a pretty huge mass, but we weren’t as worried when we realized the mass of the sun is ~10^31 kg.
We then wondered how long the sun had been around, and figured a good estimate would be slightly longer than the earth, or about 4 billion years. Since we only care about the how the orbit of the earth might have changed, we wondered how much the mass of the sun changed while the earth was around. Googling 4 billion years in seconds gave us 10^17s, so the sun has lost about 10^26 kg of mass or around 100 Earths of mass.
Though this turns out to be only about 0.01% of the Sun’s mass, we still wondered if this would have a noticeable effect on the Earth’s orbit, and it was at this point that we realized computational modeling could give us an answer. We’ve already written a model of the earth-sun which includes code like this:
dt = 86400 #make time step 1 day while t < 4e9 * dt: #update forces Fg = (-Gm_sunm=_earth/r**2)*norm(r) #update momentum earth.p = earth.p+Fg*dt #update position earth.pos = earth.pos + earth.p/earth.m*dt
And we realized that we could add a single line about the gravitational force calculation that read: m_sun = m_sun -1.1e9 *dt
to account for the changing mass of the sun. We then thought we could write a program that plotted two earths—one feeling a force from a constant mass sun, and one feeling a force from a changing mass sun, and see how they departed.
All of this was great until we realized we’d set a timestep of a single day, and a wile loop that would need to run for 4billion years, and around this time, the class was almost over, but we thought of a few problems:
• We can’t increase the size of the time step by much because our approximation that the final momentum is equal to the average momentum is conditioned upon the idea that the net force is constant over the time interval
• Our errors from the time steps accumulate as time goes forward in in our program, so we weren’t even sure that a 1-day long time step would work for a 4 billion year long calculation
The beauty of this was we saw that this would be a nearly impossible problem to solve in closed form—integrating a force that depends on both position and changing mass seems daunting to me, but it’s totally do-able (in theory) computationally. We also all agreed that the answer is likely to be there is no discernable change in the orbit.
I’m sure there is some algorithm or method that would allow you to use a larger time step, or somehow quickly compute these trajectories, but I must admit I don’t know what it is off the top of my head, and would welcome ideas from my readers.
tags:
This is part of a three post series on the hiring process in independent schools from multiple perspectives. This post gives my advice to teaching candidates as someone who evaluated applicants to for our physics openings last year.
To learn about how we conducted our search for physics teaching candidates last year, read my post, Some thoughts on conducting a physics teacher search.
To learn about the perspective of a candidate applying for jobs at independent and boarding schools, see Megan’s post, How to get hired in an independent school
Now that we’ve completed our job search and hired 3 teachers, I thought I would share a few pieces of advice for job applicants.
## Consider independent schools
If you’ve thought about becoming a teacher, you might think you need to complete a degree in education, be certified or hold an advanced degree in physics in order to teach high school physics. While all of these things can be helpful, none of them are necessary to teach in an independent school, and we routinely hire teachers who have none of these qualifications. Almost none of my colleagues are certified to teach, few hold education degrees, though most do hold advanced degrees (many earn them while teaching at my school).
My school hires recent graduates—we’ve often hired those who think that they might want to teach for a few years before applying to graduate school. We also hire experienced candidates who’ve worked years in an industry and who aren’t sure that they and it hires people who have worked in industry for years and are ready to make a career change. Just about the only prerequisite I would say is common to our faculty and successful candidates is a deep love of their discipline, a passion for learning and an ability to connect with students.
And of course, if you’re an experienced teacher in a public or private school, we’d love to consider you for a teaching position as well. So what’s the next step?
## Think carefully about using a placement service
There are a number of teacher placement services that specialize in placing teachers in independent schools. Nearly every independent school uses these services for the majority of their hiring. Using a placement service is pretty simple, especially for physics teachers scarce and in demand. You fill out an online application, submit a personal statement, resume and a transcript if you are a recent grad, and will usually complete a short interview with a placement associate. After that, the placement service will begin sending you referrals of schools that they’ve sent your file to, according to your school and geographic preferences. You then contact the schools that interest you (or not, if they indicate such), and begin the job interview process at each of those schools. Along the way, you might also go to a hiring fair, where hundreds of schools and thousands of candidates gather in large hotel ballrooms for a series of 30-minute interviews that can feel like speed dating gone wild.
While a placement service can do a lot of the grunt work for you, I’d encourage you to seriously consider whether or not you need to use a placement service at all. Here are some reasons you might want to consider doing a search as a free agent.
• Placement services aren’t gatekeepers: You don’t need a placement service to find out about openings—most schools post them on their website, and the NAIS Job board (which you can even subscribe to via RSS). Also, no school gets so many applications that they would pass over a well-qualified application that came unsolicited—many times, these applications stand out more because they don’t come through the placement service.
• You get to customize your application: to a large degree, applications from placement services all look the same. When I browse through the website, I see a list of brief bios that are all formatted the same, and when I pull up the full file, I get cover sheet template that is the same for every candidate I read. I actually have to dig to find the unique documents that you have control over—your personal statement and resume. But if you simply email your application to us, those documents are the very first things we see.
• You are a free agent: If a school hires you, then the school must pay 15% of your starting salary to the placement service. Though I have no evidence of this, I’ve got to think that a person making a hiring decision choosing between two equally qualified candidates would be much more inclined to take the one who doesn’t come at a 15% premium, and heck, you might even be able to negotiate a signing bonus.
• Your blog is your application: If you’re reading this blog, there’s a good chance that you’re already part of the MTB0S/PhTBoS. You’ve got 10, 100, or 1000 followers, and you’re actively having conversations about teaching online. Sure, you could have a placement service send out your materials schools for you, or, you could save everyone a lot of trouble and send a letter and a link to your blog to schools that interest you, and if your letter crossed my desk, you’d be fast-tracked for a phone interview. There just aren’t that many candidates out there who are thinking about math/physics teaching with the level of depth, consistency, and engagement that is common on the MTBoS.
### If you do use a placement service
Ok, so you still want to use a placement service? That’s great. There’s nothing wrong with that. But I do ask that you do one thing. Please make sure your file is complete before the service begins to send out referrals. I can’t tell you how many partial files I read this year—I’d read a great resume, a wonderful reference or two, and then I’d look for a personal statement only to find it’s not there. Incomplete files give a poor impression to reviews, and I’m sure some just pass on them and never look back. Usually, I’d set a reminder to come back in a week to see if the file is complete, and a disappointing number of times, it wasn’t.
Also, a surprising number of times, it seemed like the documents in candidates files were first faxed to a working coal mine and back before being added to a file. I’m not sure how this happens in 2017, but I’d encourage you to make sure you give the placement service pristine PDF copies of all of your materials to prevent this from happening. It’s your responsibility to make sure your file is the best reflection of you.
Remeber also that you are likely just one row in a big web table of applicants, or one file in a stack of documents, each with the same format. You should think carefully about how the documents you have control over—your resume and your personal statement stand out and reflect who you are. At the moment I’m reading your file, I’m often looking for something, anything that will make me think it’s worth half an hour to offer you an interview. Why not take me out of the confines of the placement service’s website or off the page and direct me to a website that shares some of your work. I’d love to read a comment you wrote for a student, an interesting class activity you crafted, or just about anything else that shows me you are going out of your way to share who you are.
### Don’t forget the power of networking
Just like any job search, having a contact with someone at the school can obviously be a huge help. But don’t think that means you need to have some sort of deep connection. If you’ve commented on this blog, we’ve chatted at an AAPT meeting, or you’ve just emailed me out of the blue at some point to ask a teaching question, we’re connected. The MTBoS/PhTBoS really is one big family, and we seek to help each other as much as we can, so don’t be afraid to reach out.
### Your personal statement is the key
This also applies to your resume—bring the detail. Since we aren’t dealing with hundreds of applicants, I don’t mind long resumes. If you need a 3rd or 4th page to really describe everything that’s involved with coaching your robotics team, take it.
### Prepare for every interview
If you we offer you a phone interview, that means we found something interesting about you in your application and want to learn more. Like your personal statement, we are looking for you to tell us a story through this interview—don’t be afraid to reinterpret our questions to take us to the things you most want to talk about and give us the details that we can latch on to in follow-up questions. One of the things I love most about interviewing candidates is that I usually find myself coming away with new approaches and ideas for my own teaching, but for this to happen, you have to be ready to push past surface level responses to our questions.
We are also expecting that you are going make an effort to learn something about us. The easiest place to show us that you’ve done some preparation is when we turn the last half of the interview over to you and ask you what questions you have for us. Just about the worst thing, you can say at this point, and in any interview is “I think we’ve covered them all.” This is a great chance for you to ask us the questions that will help you to determine if our school is right for you, and I can’t tell you how great it was to hear a candidate say “yes, I have a whole list here.”
### Don’t forget the invisible interview
Any good advice about interviewing will tell you that your interview begins the moment you step through the door of the building or office—how you treat the receptionist is often as important as how you treat the CEO. This is doubly or triply true at a school. At my boarding school, we arrange for most teaching candidates to fly the closest airport to us, which is about 45 minutes away. Candidates get picked up by our school driver, a wonderful man who drives members of the community to and from the airport. If you engaged the school driver in a thoughtful conversation, and he happened to run into the headmaster, who is an occasional passenger to the airport, I’m sure it would be a significant point in your favor. Likewise, if you were rude or dismissive to him, or anyone you encounter during your visit, that could easily find its way back to a member of the search team and hurt your candidacy.
Since interview visits to our school are all day affairs, we usually host candidates overnight, and it’s not uncommon to arrange some sort of informal social gathering or dinner for the candidate the evening before. When I visited 20 years ago, I remember the headmaster saying he was going to drop me off at a small gathering of a few faculty at a colleague’s house. Being the naive 21-year old I was, I remember thinking how cool it was that faculty got together for wine and cheese socials during the middle of the week and that they were willing to invite a stranger like me. But in actuality, this was an informal part of the interview process too. No, participants didn’t fill out some sort of evaluation card rating my strengths and weaknesses afterward, but I am sure that the impressions I made with attendees at that event did play into the many many factors that were considered in making the decision to hire me.
Remember, we aren’t just looking to hire a teacher to teach a few classes, we are hiring a neighbor, and if we have brought you to campus, we are all pulling for you and hoping that you will have a great visit.
My advice, use this to your advantage you’re going to meet a lot of people and get a chance to learn from perspectives about our community—use that opportunity to its fullest.
This is part of a three post series on the hiring process in independent schools from multiple perspectives. This post discusses how we ran a search for a physics teacher last year.
To read advice I’d give to candidates looking for STEM jobs in independent schools, see my post, Advice for a physics job seeker from someone who just helped to hire three.
To learn about the perspective of a candidate applying for jobs at independent and boarding schools, see Megan’s post, How to get hired in an independent school
This year we had an opening for a physics teacher, and we ended up hiring a great one. In this post, I wanted to share some of the things we learned.
It’s always a challenge to find a physics teachers—there just aren’t that many of them. It’s even more challenging to find a physics teacher at a boarding school where we need our teachers to also coach and do dorm duty. I know this is a truism even at non-boarding schools, and for other STEM subjects, like math.
## Expanding the pool
We started this search knowing we wanted to begin with a large and diverse pool of candidates. In the past, we’ve simply posted a job description on our website and initiated a search with one of the major independent school search firms. Unsurprisingly, this approach doesn’t lead to the most diverse applicant pool—there just aren’t that many physics teachers out there who are familiar with the boarding/independent school world looking for jobs. Just as an example, one placement service we worked with sent us 20 applications for our physics opening, one of whom was a candidate of color and only two of whom were women. Compare this to the 84 candidates the placement service provided for a French opening. It seems that placement services have just as many challenges as we do in finding qualified candidates for physics openings. In order to get the diverse applicant pool we wanted, we needed to be more active in our search.
We worked to expand or pool in 3 ways. First, we posted the job on the NAIS Job Board. This is a simple and obvious first step, but one we’ve neglected in the past. It’s free to post ads on this site, and the job board serves as a clearinghouse for most jobs in the independent school world, so it’s very widely read. It easily yielded as many applicants as the placement service we routinely use.
We also posted to relevant listservs, most notably the Modeling Physics Listserv hosted by modelinginstruction.org. Even if you aren’t 100% into modeling, this is a great place to put your job in front of a great group of teachers who are deeply committed to physics, professional development, and student inquiry. You do have to be a member of this group in order to post, but if you have an opening that you would like to be shared, I would be happy to post it for you on this listserv. Other listservs worth considering might be your local AAPT subsection listserv, and the PHYS-L listserv for physics teachers.
By far the best thing we did in this job search was sending emails to the department chairs of about 80 physics programs around the country. Sadly, there’s no mailing list for this, so a colleague and I spent an hour of so googling “physics department chair for X University” and filling in a large spreadsheet, which I’ve posted here to save you a bit of work should you choose to do the same. In building this list, we made an explicit effort to target women’s colleges, historically black colleges and colleges with PhysTEC programs focused on physics teacher education. Our email included short 1 page flyer that could be posted on a bulletin board (do schools even do this anymore?), along with a description of our school and opening, and a statement that we would be happy to talk to anyone interested in talking about Physics Teaching, regardless of whether they were interested in our school or not. This letter turned out to be one of our best sources of potential applicants. Though these new applicants lacked experience, many demonstrated a deep commitment to physics education through their undergraduate careers and that rivaled many of the experienced applicants we saw. Three of our five finalists who came to campus contacted us first through this outreach, and the diversity candidates that came to us through this outreach was far greater than any other application source in our search.
Three of our five finalists who came to campus contacted us first through this outreach, and the diversity candidates that came to us through this outreach was far greater than any other application source in our search.
## Managing the Workflow
We had 3 people participating on the initial screening of applicants, and we knew we needed a way to manage the workflow of reviewing and contacting candidates. It’s easy to get buried in email about a job search, so we turned to Trello, a great to-do manager that specializes in group collaboration. Using Trello, we were able to set up lists for todos, as well as lists of candidates, complete with individual checklists listing each step of the review process.
As much as possible, we wanted to make sure we were keeping candidates informed of their status as we proceeded in our search, keeping in mind that most searches can drag on for much longer than you might first imagine. One thing that helped us tremendously in this effort was to pre-draft a set of correspondence templates for each stage of the search, from when an application is received, to the point where finalists are invited for a visit to campus. We tried to make sure we followed up at each of the following stages, and whenever more than a few weeks of time passed following our last contact with a candidate if only to say that we are still considering the application and appreciate the candidate’s patience.
### Have a low bar for phone screens
Our efforts worked, and we got a strong pool of diverse applicants. Because of this, we felt we should try to phone interview as many candidates as possible, which led to us doing about 20 phone interviews for this search. Though I’m not sure this approach would scale to a search that had hundreds of applicants, it turned out to be the right decision for us, as we met some great physics teachers, had wonderful conversations about teaching physics and were able to even more clearly identify our most standout candidates. For a number of the candidates that didn’t move forward in the search, I believe we established professional connections that will last into the future.
In order to make our phone interviews fair, we also tried as much as possible to work from a uniform set of questions. We conducted our interviews using skype/google hangouts (and usually had to switch to the other when we had audio problems with our first attempt) and always made sure we had two people on each call so that we could be taking notes while the other was talking. We drafted specific questions for candidates coming directly from college or without teaching experience. Having the question document open during the interview helped tremendously to make sure we were covering roughly the same ground with each interview.
The decision to move forward after a phone interview and invite a candidate for an on-campus interview can be a bit unpredictable and drawn out at a boarding school, given all the various needs the school has in terms of coaching, residential life, housing and other considerations. This is where it became especially important to stay in touch with candidates as the search progressed. But once a candidate did make it to the on-campus interview stage, the headmaster’s office took over, scheduling the day and arranging meetings with students, and teachers and administrators from all aspects of the life of the school. It’s also customary to ask teachers to do a mock lesson, but we’ve found it better to instead ask candidates to participate as a co-teacher in one of our classes. Most often, our students are working in small groups on problems as part of a modeling cycle, and so it’s very instructive to have a second teacher in the room to circulate and push students’ thinking, and it’s very useful to us to see how the teacher actually interacts with students in a small group setting, as this represents the dominant mode of learning for most of our classes. One other thing we do is try to make sure that we provided each visiting candidate with a complete copy of his/her schedule before arriving on campus. We also wrote a clear explanation of what we are looking for when the candidate visits our classes, to make explicit that this was an evaluative part of the interview process, and we didn’t simply want them to passively observe our class. Before arriving on campus we sent them the problems/lesson we were working on, and told them we would like them to interact and observe the students in the problem-solving process, and then followed up later in the day to see what observations they had about the work they saw. Often, this turned out to be a critical differentiator in evaluating candidates. Stronger candidates made sharper and more incisive observations of the students in the class.
In the end, we had three wonderful candidates visit the school, none of whom came through one of the major search firms. We ended up hiring our first choice—a candidate who found us through the NAIS job board posting.
## Stay conscious of your unconscious bias
Unconscious bias can be a significant obstacle in an effective search. The unconscious biases we all have to implicitly favor those candidates that are most similar to us have the potential to derail even the most well-intentioned efforts to seek out a diverse pool.I know that if I let it, it would be easy to for me to fill our applicant pool with geeky, Duke basketball loving white guys like me. I would be most comfortable in those interviews, and as a result, I’d likely make those interviewees feel more comfortable, and probably feel a stronger “gut feeling” that this white, male, candidate is going to be a great fit for our department or program. Unconscious bias is how our brain processes information and how we work as social animals, and we have to work hard at every step of the search to make sure we aren’t allowing our own biases to interfere with the integrity of the search and the needs of the program. To learn more about how to do this, I strongly encourage you to check out the great resources Google’s HR department has created in their re:Work program on Unconscious Bias at Work. You’ll find incredible materials here focused on how to unbias your workplace, with a particular emphasis on the hiring process. Google has even created a step by step workbook and presentation you can adapt to do training at your own school. The one hour video Google re:Work produced on unconscious bias is one of the most informative and practical presentations on this subject that I’ve seen.
One takeaway I’d personally like to try from this is to try to adopt more of the practice of structured interviews in our phone screens. Although we did develop a set of common questions, we weren’t as careful in thinking beforehand about what excellent, average and poor responses to our questions looked like, and I think this would have been a very helpful exercise, given the research on how ineffective unstructured interviews can be. At the same time, we have to allow for personalization of the interview process adapt to the candidate’s experiences and background.
For those of us who work in privileged boarding and independent schools, I think we should also spend a fair amount of time considering our privilege as individuals familiar with and knowledgeable about the boarding school world. We must recognize that the candidates who are most familiar with the independent and boarding school world are going to naturally appear far more fluid and capable in interviews and informal settings throughout the interview process. This initial familiarity, however, may not be an indicator that their performance or adjustment to our community expectations will be superior to other candidates who are less familiar with the boarding school milieu and may, in fact, pick up our culture and habits fairly quickly. Nearly 20 years ago, this was certainly me. Though I was fortunate to attend a private college, I attended public school all my life, was the first in my family to go to college, and spent much of my interview at the school I work at now in awe of the incredible contrasts between this school and my own educational upbringing. I’m thankful that those that interviewed me were willing to see beyond all of the ways in which I might not have registered as a “good fit” for the culture – not because I wasn’t capable but because it was so foreign to me. Those who interviewed me gave me a chance learn.
## Further ideas
The lesson I’ve learned from this is that good schools should always looking for great teachers. We need to always be reaching out make connections with other teachers. Mainly, this is good practice—staying connected with professionals in your field keeps you abreast of the latest practices in the field, and also allows your best ideas to make it into the world. But making these connections also makes it much easier to find teachers when you do have an opening.
Reaching out directly to physics departments at colleges and universities turned out to be the biggest bang for our buck in terms of bringing a pool of qualified candidates from a range of backgrounds that wouldn’t have otherwise considered our opening. It has made me think that we could do even more outreach with our neighboring colleges and universities and invite any seniors who are considering teaching to spend a day visiting our school to get a better sense of what independent school teaching is like.
Our school asks each class/discipline to develop specific honor expectations for our classes. My physics colleagues and I put together this document, and I welcome your feedback to make it even better.
Honor Expectations in Physics
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Earlier in the summer, I wrote about paradigm lab assessments as a way to improve student understanding of the arguments we made in board meetings.
Summer is over, and it’s time to get to work cranking out these assessments. I’ve written a draft of the first assessment I’d like to do with my Honors Physics class, and I would love your feedback to make this even better.
CVPM Paradigm Assessment (Google Doc—feel free to edit and comment)
Some questions from me:
• This will be the first paradigm assessment for my students—am I testing too much here?
• I don’t want this to feel like a lab report. Would it be better to break this up into a quiz or some sort of form on Canvas?
I’ll admit, I can become infatuated with edTech from time to time. I also envision that it will have a much more powerful influence on my classroom than it ever seems to have. Despite this, my classroom is still fairly un-techy; I’m inconsistent at best when it comes to using our LMS, Canvas, and on any given you’re most likely to find my students solving physics problems in small groups using pencil and paper.
Still, I do dream of an app that will be a true game changer in my classroom, and I think the app of my dreams would look something like this:
• Quickly record and tag observations about students and groups—I want to be able to walk around the room, see a cool thing a student is doing, and write a note about it and tag it to the student to refer back to later when I’m writing comments. I’d also like to have the choice to make these comments private. It’d be even better if I could add photos, short video recordings, and audio notes. Double bonus if I could tag more than just the student, for instance #insightfulquestion.
• Let students submit photos and screencasets for feedback—It’s become a very common method of getting help in my class for students to send me a quick photo of their work for feedback. I can quickly look at the photo and write a few questions to push their thinking, or offer a hint. Right now, those emails pile up in my inbox, and I don’t have a nice way of organizing them or quickly searching for them. Putting all of these photos/screencasts in an app and then being able to organize and tag them would be a godsend.
• Let students create multimedia portfolios of their work and get feedback from me—I’d love for students to be able to create portfolios of their work—they could add photos, video and audio recordings and tag their work with various tags like #goallessproblem or #explainedmistake. I’d then be able to go through the portfolio of a student and give feedback, both as text, and ideally as written annotation. Bonus if the student can respond to my feedback and we can have a bit of a dialogue.
And of course, I’d want all of this to work nicely on multiple platforms, including my iPhone, student phones of every flavor of iOS and Android and laptops (with extra bonus for instantly de-flipping photos snapped using the webcam).
As best I can tell, this app doesn’t exist as a single app, but I have found 3 contenders that each do part of this. They are:
• Class Dojo—the observation app. Class dojo makes it super easy to record feedback about individual students or groups. If I never invite my students to have their own accounts, then my notes notes would be private, but if I do set my students up with accounts, it seems like students would see any notes I write about them. Class Dojo is fast—it’s really easy to group students and quickly leave feedback. My heisitation is that Class Dogo does seem to be the the app embodiement of Skinnerism, and I don’t want to create a points economy in my class is fed by Class Dojo like I’ve read about in so many elementary school classrooms
• Seesaw—an incredible digital portfolio. Seesaw really seems to do some great stuff for allowing you and students to quickly add multimedia to a digitial portfolio, but the ability to quickly add a note about a student leaves a lot to be desired—it’s very clunky and slow. But I’ve heard from a number of high school teachers that really like this app.
• Flipgrid—the new video sharing wonder. Flipgrid seems like a great way for students to share video. Many of its features seem to overlap with Seesaw, but Flipgrid’s presentation is unique, and I’ve seen a few links on Twitter of people using it as a tool for having people leave video messages when they miss a teacher or professor for office hours which seems cool (but not something I need to do). Flipgrid seems like it could be my photo/screencast submission app, but I’m not sure it does photos, which would be a dealbreaker, I’m afraid.
There’s a crazy part of me that thinks it wouldn’t be so bad to try to use all of these apps this year. But the sane part of me knows this is a really bad idea. Here’s where I’m reaching out to you—are there things these apps can do that I’m missing? Is there a way to get to 80% of what I want to do with just one of these apps?
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More details that make Twitter Math Camp so great:
• Thank you notes: Have you ever had a conference where you were literally provided with thank you notes and encouraged to write a note thanking anyone who made the conference useful for you? Have you ever seen a beach bag full of handwritten thank you notes? Go to TMC18, and you’ll see both of these.
• Simply amazing volunteers, and not wasting their time: Lots of conferences have volunteers to help make things run smoothly, but I’ve often noticed that those volunteers end up overtaxed and really struggle to participate fully in the conference itself. Somehow, TMC has voluneteers—someone stopped by the first few minutes of every session I was at to make sure the AV was working properly, but at the same time, I noticed that the volunteers were able to participate every bit as much as the anyone else.
• Flex sessions: Have you ever had a great idea at a conference and then just wanted to be able to talk about it with 10 other people? Or have you just wanted to do some yoga in the middle of the conference? All of this is possible at TMC. Flex sessions are 1 hour long open sessions that any can offer on the 3rd day of TMC, and you can propose one as late as a few hours before these sessions start. So great.
I also truly love the idea of having 2.5-hour morning sessions that extend for 3 days. This really lets you get to know a group of people in a way that seldom happens at a 3-day conference. Our closing circle at Elizabeth’s session was truly special. Elizabeth shared a really open look into how she organizes her classroom and truly delivered on her description to provide an immersive experience and a real window into her teaching.
TMC is dedicated to welcoming others and constantly re-examining itself to make sure it is being as inclusive as possible. Really, I’ve never been part of a more self-reflective community that is so committed to sharing, connecting and welcoming new people to the joy of math and math teaching.
Perhaps the best example of this was the flex session I sat in on with Tina Cardone discussing the TMC application process. Tina is one of the leaders of the registration process, and she shared a ton of statistics with the group about the TMC attendees. The thought that Tina and the rest of the TMC board put into planning this process—they’ve thought of so much and made so many efforts to diversify TMC. A few things I learned from attending this session is that nearly 110 of the TMC slots are filled by presenters. The remaining 90 slots were filled by a lottery and wait list, and only about 94 people completed the lottery application—and ultimately everyone who completed the lottery was able to gain attendance to TMC. This surprised me—I had submitted two session proposals, both of which were rejected (rightly so, when I compare my ideas to the sessions I saw), and after my sessions were rejected, I was really worried about my odds in the lottery.
The challenges of creating a 200 person conference that continues to allow all the people who know and love this community to attend while still remaining open to newcomers is one that I think might be right up there with P vs NP. There were a number of ideas for how we might be able to increase access and awareness of TMC, but I think this is something that our community will be struggling with for some time to come. I did have one big idea about how we might reach pre-service teachers and others who just love math, and I’m going to blog about that in another post.
With this post, my time at TMC must come to an end, as I’ve got to drive back to Delaware and start getting ready for school. I’m sad to miss the last day and the big reveal for TMC18, but I’m leaving knowing that this is a vibrant organization with an amazing future ahead of it. |
# More Binomial Prob
• Apr 17th 2010, 06:49 PM
dwsmith
More Binomial Prob
A fertility clinic claims to have a procedure for selecting the gender of a child. Among nine randomly selected couples who wished to have girls, seven got the intended result. Test a hypothesis to determine if the procedure does better than chance.
The answer is .0898; however, I obtained $9C7*.5^7*.5^2=.0703125$.
What is the issue here?
• Apr 17th 2010, 06:52 PM
Debsta
Quote:
Originally Posted by dwsmith
A fertility clinic claims to have a procedure for selecting the gender of a child. Among nine randomly selected couples who wished to have girls, seven got the intended result. Test a hypothesis to determine if the procedure does better than chance.
The answer is .0898; however, I obtained $9C7*.5^7*.5^2=.0703125$.
What is the issue here?
The answer to what is 0.0898??
• Apr 17th 2010, 07:00 PM
dwsmith
• Apr 17th 2010, 07:05 PM
Debsta
Quote:
Originally Posted by dwsmith
There is no "question" - you are testing a hypothesis.
• Apr 17th 2010, 07:07 PM
dwsmith
Yes, I understand that. The probability of having a boy and girl is 50%.
So I used the binomial procedure.
• Apr 17th 2010, 07:14 PM
Debsta
Quote:
Originally Posted by dwsmith
Yes, I understand that. The probability of having a boy and girl is 50%.
So I used the binomial procedure.
OK I get where the 0.07 came from just not sure where you got the 0.08 figure from
• Apr 17th 2010, 07:16 PM
dwsmith
• Apr 17th 2010, 07:18 PM
Debsta
Quote:
Originally Posted by dwsmith
Is the question that you posted word for word from your book. If it is then their "answer" doesn't make a lot of sense????
• Apr 17th 2010, 07:19 PM
dwsmith
Word for word my friend.
• Apr 17th 2010, 07:23 PM
Debsta
Quote:
Originally Posted by dwsmith
Word for word my friend.
How can the "answer" to "Test a hypothesis to determine if the procedure does better than chance" be a number???
Suggestion: Toss the book.(Thinking)
• Apr 17th 2010, 07:24 PM
dwsmith
Trust me, I know the book is garbage. |
# Simple probability question
A person from group A has 20% chance of having some characteristic
A person from group B has 30% chance of having the same characteristic
How can I calculate the probability of a person belonging to both groups having the given characteristic?
edit
Since a number of people pointed out that it's impossible to know, I'll change my question.
Let's assume that there are 10 different groups. Do I need to know the probabilities for each possible combination, or can I infere at least some probabilities?
edit
I've added a solution witch I think is plausible for some cases.
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Given this information, you can't do this. – Peter Flom Sep 27 '12 at 20:31
What additional information do I need? If you want some context, think disease risk assesment – sabof Sep 27 '12 at 20:32
I think the only that will help is observations of people from both groups. There's no theoretical way of calculating the interaction effect - the answer could be anything between 30% and 100%. – Peter Ellis Sep 27 '12 at 20:35
Essentially, you'd need to know what you are asking. From what you have, there is no way to know if the answer should be something between 20 and 30, or something greater than 30, or even, possibly something less than 20. Risks can be additive, but they needn't be: There can be (and often is) an interaction. E.g. Radon exposure is related to lung cancer. Smoking is too. But the effects of radon are greater for smokers than nonsmokers. But it can go the other way, too. – Peter Flom Sep 27 '12 at 20:36
@Glen_b no, that's a mistake from me, the interaction could be anything down to zero. – Peter Ellis Sep 28 '12 at 11:59
PeterFlom is right that the information you have is not enough to answer the question. However when you ask what do I need to know I could say that if a divine spirit told you P(AUB)=.6 then since P(A)=.2 and P(B)=.3 then the desired answer .1. It comes from the well known formula in probability
For any two event A and B P(AUB)=P(A)+P(B)-P(A∩B). So P(A∩B)=P(A)+P(B)-P(AUB)
So you only need to know P(AUB). This also makes it clear why you can't solve with the information at hand You do need to know P(AUB)
I am adding to my answer because the OP mentioned that his problem is considering information where A=[set of radiologists in the sample space] and B=[set of members if the sample space with cancer] and he wants to understand what is the probability that a member of the cancer set that also is a radiologist will have cancer. He states that he thinks that the probability that a cancer patient who is a radiologist would have a probability of 0.1 for having cancer. He thinks that is too low and presumably even 0.2 might seem too low as well. I have 2 responses to that.
1. It seems that the answer to that question is really asking for P[B|A]
P[B|A]=P{A∩B)/P(A). This does not have the same bounds as P(A∩B).
For this problem P(B|A) is not known either since we do not know P(A∩B). In this case P[B|A]=P(A∩B)/0.2 =5 P(A∩B). So P[B|A] has a lower bound of 0 and an upper bound of 5(.2)=1! So P[B|A] ** can be any probability regardless of what P(A) and P(B) are!**
2. Why is a value less than 0.2 plausible?
This is just a theory, but given no information about the smoker's profession the probability he/she has cancer is 0.2. Now radiologist understand the dangers of cancer better than most smokers who are not radiologists. So the radiologists that smoke might tend to be light smokers. Now light smokers are less likely to have cancer because smokers with cancer predominantly have lung cancer and light smokers are less likely to have lung cancer than the moderate or heavy smokers.
Initially it was not clear that the OP wanted P(B|A) to me. After explaining the problem I think that it is because he is asking for the probability of cancer when you know the individual is a radiologist. .
-
I think what I need is something else. It makes little sense for the answer to be less than .3 in my context. I'm looking for models suitable for risk assessment, and an example maximally similar to mine. I'm out of my water to be more specific. – sabof Sep 27 '12 at 22:35
This result is a correct formula for any two events. What models are you talking about. There seems to be details about this problem that you haven't told us about. With the information you gave us P(AUB) can have a range of values. I said that you can't know P(AUB) and 0.6 was just a hypothetical value. If you know P(A)=0.2 and P(B)=0.3 then 0.0<=P(A∩B)<=0.2 and 0.3<=P(AUB)<=0.5. If the answer can't be in the interval [0. 0.2] then either P(A) is not equal to 0.2 or P(B) is not 0.3 or they both are different from what you specified. – Michael Chernick Sep 28 '12 at 2:33
There is no other possibility without violating the laws of probability! – Michael Chernick Sep 28 '12 at 2:33
That does not seem weird to me. That makes perfect sense to me. The smoking radiologist form a subset of both the set of radiologists and the smokers. So P(A∩B) has to be no bigger than the minimum of P(A) and P(B). – Michael Chernick Sep 28 '12 at 8:58
-1 for deducing from $P(A) = 0.2$, $P(B) = 0.3$, and the claim of a divine spirit that $P(A\cup B) =0.6$ that $P(A\cap B)=0.1$ instead of questioning said allegedly heavenly claim, and for not bothering to correct the error even when it has been pointed out. – Dilip Sarwate Dec 27 '12 at 23:14 |
# infinity divided by constant
So, if we take the difference of two infinities we have a couple of possibilities. Start at the smaller of the two and list, in increasing order, all the integers that come after that. Again, $$a$$ must not be negative infinity to avoid some potentially serious difficulties. In the case of our example this would yield the new number. If you move into complex numbers for instance things can and do change. Then can double type really do in the current 64-bit cyber world or maybe even 1024-bit one in the future? With infinity you have the following. $$a < 0$$) to a really, really large positive number and stay really, really large and positive. With infinity this is not true. $$a < 0$$) from a really, really large negative number will still be a really, really large negative number. A number over zero or infinity over zero, the answer is infinity. values are all in doubles, the output will be, ‘as expected’, Infinity. However, when the code is being executed, a DivideByZeroException with an explanation like ‘Attempted to divide by zero’ would occur at run time. Depending on the relative size of the two integers it might take a very, very long time to list all the integers between them and there isn’t really a purpose to doing it. For other types such as integer and decimal, they cannot hold infinity either positive or negative in the C# language so far. Hence, it must not be possible to list out all First, I am going to define this axiom (assumption) that infinity divided by infinity is equal to one: ∞. Bad? The biggest problem we think is that the infinity concept only applies to the double and float values. With addition, multiplication and the first sets of division we worked this wasn’t an issue. Also, please note that I’m not trying to give a precise proof of anything here. Subtracting a negative number (i.e. Again, we avoided a quotient of two infinities of the same type since, again depending upon the context, there might still be ambiguities about its value. A number over infinity, the answer is zero. Otherwise, serious problems would come along. Infinity simply isn’t a number and because there are different kinds of infinity it generally doesn’t behave as a number does. But this contradicts the initial assumption that we could list out all the numbers in the interval $$\left(0,1\right)$$. Because we could list all these integers between two randomly chosen integers we say that the integers are countably infinite. Notice that this number is in the interval $$\left(0,1\right)$$ and also notice that given how we choose the digits of the number this number will not be equal to the first number in our list, $${x_1}$$, because the first digit of each is guaranteed to not be the same. Division of a number by infinity is somewhat intuitive, but there are a couple of subtleties that you need to be aware of. If it is, there are some serious issues that we need to deal with as we’ll see in a bit. a spider crawling mainly in the .NET of Revit and Navisworks, RevitNetAddinWizard & NavisworksNetAddinWizard, Revit .NET API: FilterRule, ElementParameterFilter and FilteredElementCollector.WherePasses(), Revit API 2012: Extensible Storage – Manage Sub Schema/Entity. —. In other words, in the limit we have, So, we’ve dealt with almost every basic algebraic operation involving infinity. This is not correct of course but may help with the discussion in this section. Section 7-7 : Types of Infinity. Eventually we will reach the larger of the two integers that you picked. Mark Ryan has taught pre-algebra through calculus for more than 25 years. What we’ve got to remember here is that there are really, really large numbers and then there are really, really, really large numbers.
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## The Annals of Probability
### Sojourns and Extremes of Gaussian Processes
Simeon M. Berman
#### Abstract
Let $X(t), 0 \leqq t \leqq 1$, be a real Gaussian process with mean 0 and continuous sample functions. For $u > 0$, form the process $u(X(t) - u)$. In this paper two related problems are studied. (i) Let $G$ be a nonnegative measurable function, and put $L = \int^1_0 G(u(X(t) - u)) dt$. For certain classes of processes $X$ and functions $G$, we find, for $u \rightarrow \infty$, the limiting conditional distribution of $L$ given that it is positive. (ii) For the same class of processes $X$, we find the asymptotic form of $P(\max_{\lbrack 0,1 \rbrack} X(t) > u)$ for $u \rightarrow \infty$. Finally, these results are extended to the process with the "moving barrier," $X(t) - f(t)$, where $f$ is a continuous function.
#### Article information
Source
Ann. Probab., Volume 2, Number 6 (1974), 999-1026.
Dates
First available in Project Euclid: 19 April 2007
https://projecteuclid.org/euclid.aop/1176996495
Digital Object Identifier
doi:10.1214/aop/1176996495
Mathematical Reviews number (MathSciNet)
MR372976
Zentralblatt MATH identifier
0298.60026
JSTOR |
# Tree species diversity distribution
Estimating tree diversity distribution patterns using the r.series.diversity addon for GRASS GIS
### Intro
In this post I’ll use the r.series.diversity addon for GRASS GIS to compute the tree species diversity distribution patterns in the contiguous USA, based on tree distribution data from . With the addon, you can compute a number of diversity indices, including the Renyi entropy index and a number of specialized cases of the Renyi enthropy, viz.the species richness, the Shannon index, the Shannon based effective number of species (ENS), the Simpson index (inverse and gini variants), and Pielou’s eveness . Install the addon first (see here how).
### The data
published raster maps with the estimated live tree basal area (square feet/acre) for 324 tree species in the United States. The raster maps, at a resolution of 250 meter and representing the time period 2000-2009, are based on MODIS imagery, raster data describing relevant environmental parameters, and field plot data of tree species basal area. The methods are described in - and . We’ll use the basal area as proxy for the species abundances.
### Data import
You can download the data from this website. There are three zip files, each containing approximately 80 raster layers (img format), one per species. The map projection is Albers Conical Equal Area (datum NAD83). First step is to download the data and unzip the rasters from the three zip files in one folder.
Working with data in GRASS GIS differs from what you may be used to in e.g. QGIS. So for readers less familiar with GRASS GIS, I would recommend to first read the introduction to GRASS GIS here. Here you are also introduced to another important concept, the region.
The next step is to create a new location and mapset. Below the steps are oulined in more details. Skip these steps if you are already familiar with how to create a new location and mapset.
Steps to create a new mapset
1. Open GRASS and select the option to create a new location
1. Enter a name and optionally a description. In the example, I named the new location NAD83CA, which is the location in which I will keep all source data in Albers Conical Equal Area (datum NAD83) projection.
1. Select the method to create a new location. In this case, we want to read the projection and datum terms from the georeferenced file.
1. In the next window, you can select one of the raster images you downloaded and unzipped.
1. In the next window you are presented with a summary of the new location you are about to create. If it all looks good, click finish.
2. Now you are asked if you want to import the data layer that you used to create the location. If you select yes, the data layer will be imported in the PERMANENT mapset. The default region extent and resolution will be set based on the imported data layer. Selected ‘no.’ You’ll import the layer together with the others in one go in the next steps.
3. If you selected ‘Create user mapset’ in step 2, you are now asked for a name for the new mapset.
After creating the new location and mapset, you can continue with importing the data layers in your newly created mapset. The files have an .img extension. We can check with the function gdalinfo what kind of raster file this is.
Check the raster file type
Run the following on the command line.
# Go to the directory with your raster files
cd Desktop/treesUSA
# Get information about the layer
gdalinfo s10.img
This will give you a lot of information, but what we are after is the file type. You can find this near the top of the output; it is a HFA/Erdas Imagine Image (.img).
Now that you know the type of the raster layers, you can import them using the r.import function.
Importing all raster layers in a folder
You can use the function r.import to import all the layers in a folder in one go. Go to Menu > File > Import raster data > Simplified raster import with reprojection. Don’t worry about the reprojection bit. The coordinate reference system (crs) of the mapset is equal to that of the raster layers (obviously, we used one of the raster layer to define the crs of the mapset), so no reprojection is carried out.
See the numbers in the image above: (1) Select as source type Directory and (2) select the directory with the raster layers. (3) Select Erdas Imagine Images (.img) as format and make sure the extension is set to img. After selecting the directory and raster format, a list will appear with all raster layers (4). Select all layers, and deselect Add imported layers into layer tree (5). Optionally, go to Import settings (6) to change some settings (e.g., if you have some RAM to spare, increase the maximum memory to be used).
Before we compute the diversity indici, we need to set the computational region and define a mask. The region defines the extent and resolution that will be used for all raster computations. It can be set with the g.region function.
Set the region
The region should match the raster layers we just imported. To do so, open one of them and right-click on the layer name. In the context menu, select ‘Set computation region from selected map’
Alternatively, you can use the g.region function on the command line or console.
g.region raster=s10
The tree distribution data is for the continental part of the USA, excluding Alaska. So we want to mask out all other areas. To be able to do so, first download the USA boundaries, e.g., from the GADM website and import the data. The vector layer can subsequently be used to define the mask.
To download a geopackage with the vector data, use this link. After download, extract the geopackage. Note that the the data is in latlon (wgs84). Easiest is to use the v.import function, which reprojects the data on the fly. Make sure to select the layer with the national boundary gadm36_USA_0.
You can also run this from the command line or console.
v.import input=gadm36_USA_gpkg/gadm36_USA.gpkg layer=gadm36_USA_0
Next, we can create a mask with the r.mask function. This function essentially converts the vector layer to a raster layer, with the special name MASK. You can display this layer like any other layer.
# Set the mask
r.mask vector=gadm36_USA_0
Remember that all raster operations use the extent and resolution set by the region. This is true for the MASK layer as well. This means that only the areas within the bounds of the region are included. E.g. Alaska falls outside the region.
### Compute diversity
Now we are finally ready to compute the tree species diversity, using the r.series.diversity addon for GRASS GIS. This can be done in two steps on the command line.
Get a list of input layers
You can create a list with the names of the raster layers using the g.list function. The names of our raster layers all start with a s. The following command will create a list with all names, and assign these to the variable MAPS.
# List all raster layers with species
MAPS=g.list type=raster pattern=s* separator=comma mapset=trees
The variable MAPS can now be used as input in the r.series.diversity function.
Use this as input for the r.series.diversity function.
Compute diversity indici
The flags indicate which diversity indici to calculate. In the code below, these are (r) Renyi enthropy index, (s) the richness index, (h), the shannon index, (p) the reversed Simpson index, (g) the Gini-Simpson index,(e) the Pielou’s evenness index, and (n) the Shannon effective number of species.
# Compute the diversity layers
r.series.diversity -s -h -p -g -e input=\$MAPS output=TreeDiversity
Note that with this dataset, the computation will take a really long time. Long enough for you to forget about it for the rest of the day, hopefully to have it finished next morning. So yes, the code definitely needs some optimization (any suggestion as to how to accomplish this are most welcome).
### The results
The output are maps with patterns of tree diversity across contiguous United States. In Figure 9 the maps with the species richness and the Shannon-wiener diversity index are shown. The maps show that in the mideastern states like West Virginia, Kentucky and Tennessee, tree diversity is relatively high, while tree diversity in the states on the west coast is clearly lower.
The observed patterns correspond fairly well with the map of the tree diversity by . This is not surprising. Both the map by and the species distribution maps by are based on the Forest Inventory and Analysis (FIA) field plot data from the U.S. Forest Service.
The tree diversity map by shows a different pattern, with a clear diversity hotspot in the southeastern United States. The use of different sources of information (and underlying methodologies and assumptions) might be one reason for the differences. Jenkins et al. used the range maps of tree species by , digitized by the .
In addition, the diversity map by Jenkins et al. is based on the range maps of 641 species, out of the more than 700 different native and naturalized tree species in the United States . The dataset contains a much smaller subset of 324 species. If these are e.g., the most common species, this could result in an underestimation of the species richness in regions with more rare or endemic species. And indeed, these maps show that the southeastern United States has a high number of endemic tree species. Another point to reiterate is that the basal area was used as proxy indicator for tree abundance.
I have uploaded the data to Open Science Framework (OSF) data sharing site from the Center for Open Science. If you use the data, please cite as follows: Van Breugel, P. (2020, August 3). Tree diversity distribution patterns in contiguous USA. https://doi.org/10.17605/OSF.IO/UC6DA.
### References
Breugel, Paulo van. 2016. “R.series.diversity - a GRASS GIS Addon to Compute Diversity Indici over Input Layers.” https://trac.osgeo.org/grass/browser/grass-addons/grass7/raster/r.series.diversity.
GRASS Development Team. 2020. Geographic Resources Analysis Support System (GRASS GIS) Software. USA: Open Source Geospatial Foundation. https://grass.osgeo.org.
Heath, Linda S., Sarah M. Anderson, Marla R. Emery, Jeffrey A. Hicke, Jeremy Littell, Alan Lucier, Jeffrey G. Masek, et al. 2015. “Indicators of Climate Impacts for Forests: Recommendations for the US National Climate Assessment Indicators System.” NRS-GTR-155. Newtown Square, PA: U.S. Department of Agriculture, Forest Service, Northern Research Station. https://doi.org/10.2737/NRS-GTR-155.
Jenkins, Clinton N., Kyle S. Van Houtan, Stuart L. Pimm, and Joseph O. Sexton. 2015. US Protected Lands Mismatch Biodiversity Priorities.” Proceedings of the National Academy of Sciences 112 (16): 5081–86. https://doi.org/10.1073/pnas.1418034112.
Legendre, P., and L. Legendre. 1998. Numerical Ecology. Second English edition. Developments in Environmental Modelling 20. Amsterdam: Elsevier.
Little, Elbert L. Jr. 1971. Atlas of United States Trees: Volume 1: Conifers and Important Hardwoods. First Edition edition. United States Government Printing Office,, Washington:
Little, Elbert Luther. 1979. Checklist of United States Trees (Native and Naturalized). D.C. : Forest Service, U.S. Department of Agriculture.
US Geological Survey. 1999. “Digital Representation of Tree Species Range Maps from "Atlas of United States Trees" by Elbert L. Little, Jr.” Geosciences; Environmental Change Science Center, USGS. http://esp.cr.usgs.gov/data/atlas/little/.
Wilson, B. Tyler, Andrew J. Lister, and Rachel I. Riemann. 2012. “A Nearest-Neighbor Imputation Approach to Mapping Tree Species over Large Areas Using Forest Inventory Plots and Moderate Resolution Raster Data.” Forest Ecology and Management 271 (May): 182–98. https://doi.org/10.1016/j.foreco.2012.02.002.
Wilson, Barry Tyler, Andrew J Lister, R I Riemann, Douglas M Griffith, and M Douglas. 2013. “Live Tree Species Basal Area of the Contiguous United States (2000-2009).” Newtown Square, PA: USDA Forest Service, Rocky Mountain Research Station. http://dx.doi.org/10.2737/RDS-2013-0013.
##### Paulo van Breugel
###### Lecturer & researcher
My interests range from biodiversity and ecology to spatial data analysis. I am also what one could describe as a lifelong learner; I enjoy to learn new things and widen my horizon, both professionally and personally. |
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
Analytic Geometry: (lesson 3 of 3)
## Hyperbola
### Definitions:
1. A hyperbola is the locus of points where the difference in the distance to two fixed points (called the foci) is constant.
2. A hyperbola is the set of all points (x, y) in the plane the difference of whose distances from two fixed points is some constant. The two fixed points are called the foci.
A hyperbola comprises two disconnected curves called its arms or branches which separate the foci.
Hyperbola can have a vertical or horizontal orientation.
### Hyperbola centered in the origin
Standard equation of a hyperbola centered at the origin (horizontal orientation)
Example 1:
Standard equation of a hyperbola centered at the origin (vertical orientation)
Example 2:
### Foci
The foci for a horizontal oriented hyperbola are given by
The foci for a vertical oriented hyperbola are given by
### Asymptote:
Asymptotes of a horizontal oriented hyperbola are determined by
Asymptotes of a vertically oriented hyperbola are determined by
### Eccentricity:
The eccentricity is given by
Example 3:
Consider the equation
Find: a, b, foci, asymptotes, and eccentricity.
Foci:
Asymptotes:
Eccentricity:
Picture:
### Hyperbola centered in (u,v):
Horizontal oriented hyperbola centered at (u, v)
Vertical oriented hyperbola centered at (u, v)
Foci:
The foci for a horizontal oriented hyperbola centered at (u, v):
The foci for a vertical oriented hyperbola centered at (u, v):
Asymptote:
Asymptotes of a horizontal oriented hyperbola are determined by
Asymptotes of a vertically oriented hyperbola are determined by
Eccentricity:
The eccentricity is given by
### Parametric Equations:
Horizontal oriented hyperbola:
Vertical oriented hyperbola: |
# Satellite Question
1. Apr 1, 2004
### angelical_kitten
I am a first-year physics/chemistry major in First-term Calculus-based Physics. Normally, I have no trouble with my physics assignments, but this problem...man, i have no idea even where to start, even after analyzing it for two hours straight! Any help would be appreciated!
5. You are an astronaut in the space shuttle pursuing a satellite in need of repair. You find yourself in a circular orbit of the same radius as the satellite, but 20 km behind it.
(a) How long will it take to reach the satellite if you reduce your orbital radius by 0.9 km?
km
2. Apr 1, 2004
For circular orbit,
$$a_\textrm{centripetal} = \frac{v^2}{r}$$
correct? Now, this centripetal acceleration must be provided by gravity (what else is going to produce it?), so acceleration due to gravity is
$$a_\textrm{gravity} = G\frac{M_\textrm{Earth}}{r^2}$$
So centripetal acceleration and acceleration due to gravity will be equal. Now we have an equation relating velocity and radius. Can you take it from here?
The second problem works the same way, just backwards.
3. Apr 1, 2004
### angelical_kitten
Oh, I forgot! I know that this relationship needs to be used somehow, but I still can't seem to see how:
(mv^2)/r = Gm*(m of earth)/(r^2)
4. Apr 1, 2004
### angelical_kitten
don't i need to know the original radius? or does that not really matter?
5. Apr 1, 2004
It seems to me that you do, but everybody's working me so hard that I'm running around like a madman and I could very well be wrong. Is the initial orbital radius given? Do they want you to solve it symbolically?
See where you can take that equation.
6. Apr 1, 2004
### angelical_kitten
The initial orbital radius isn't given, but they still expect me to come up with a numerical answer...lol, soon I'm probably just going to start guessing, hope I come up with the right answer!
7. Apr 2, 2004
*clip*
This all was misleading, so forget it.
Last edited: Apr 2, 2004
8. Apr 2, 2004
All right. Here we go.
$$\frac{1}{v_i^2} - \frac{1}{v_f^2} = \frac{r_0 - r_0 + .9 km}{GM}$$
Take it from there.
9. Apr 3, 2004
### angelical_kitten
ok...i figured out what i was supposed to do...
$$\frac{mv^2}{r} = \frac{GmM}{r^2}$$
m cancels out, leaving:
$$\frac{v^2}{r} = \frac{GM}{r^2}$$
solving for v gives:
$$v = \sqrt{GM}*r^{-1/2}$$
taking the derivative gives us $$dv$$ in terms of $$r$$ and $$dr$$:
$$dv = \sqrt{GM}*r^{-3/2}*dr$$
Now, I guess I was supposed to assume that the satellite was a low-orbiting body, thus the orbital radius is nearly equal to the radius of the earth.
Thus, $$r = 6.38 * 10^3 km$$
$$M = 5.97 * 10^{24} kg$$
$$G = 6.67 * 10^{-11} Nm^2kg^{-2}$$
$$dr = 0.9 km$$
Solving gives:
$$dv = 2.02 m/s$$
Because $$dr$$ is so small in comparison with the orbital radius, the effect it has on the rest of the solution is negligible.
Because all aspects of the two orbiting bodies are effectively equal except for the velocities, we can now use:
$$t = \frac{d}{v}$$
$$d = 20 km$$
$$v = 2.02 m/s$$
From this, we obtain $$t = 9.90 hrs$$
Part B uses these equations, but in reverse.
10. Apr 3, 2004
$$v_f = v_i + \Delta v$$ |
# Why do banks have capital requirements on deposits?
In article from Reuters: Big U.S. banks hunger for loans, capital relief as deposits pile up, we read:
Combined with rules that require more capital for bigger balance sheets, that makes deposits more expensive to hold, instead of profitable.
If I understand correctly, it means that the banks has to keep capital against the deposit. On first sight, it seem quiet strange to me: since deposits are basically cash, and not a risk loan that the bank should be required to hold capital. So why is that?
• think about what happens when the person you have given loan to defaults? The capital requirement against the deposit would help you return the money to your depositor. – nimbus3000 Feb 4 at 11:17
• @nimbus3000, Thanks. but doesn't a loan also require capital? I mean there are already capital requirement for loans, why should we add also to deposits (which is basically the bank receiving cash) – d_e Feb 4 at 11:22
• Filing this as evidence for the idea that interest rates need to be negative: if they were lower, there would be fewer deposits and less of what the article calls "excess liquidity". – pjc50 Feb 5 at 9:38
• @pjc50 filing that as evidence you want to abolish cash – user253751 Feb 5 at 14:10
When someone deposits money at the bank, it immediately appears on the balance sheet as both, an Asset and a Liability: on the liability side, it will sit as something along the lines of "deposit owned to customers", and on the Asset side as "cash" (this is just regular "double entry accounting").
If the bank then lends part of this deposit as a loan or mortgage, it will generate some income on this deposit. The deposit will still be sitting on the balance sheet as a Liability ("deposit owned to customers"), but on the Asset side the "cash" item will have turned into "Mortgage due from customers" or "Loan due from customers".
The bank's balance sheet grows the same second when the deposit enters the bank and the bank will need to allocate capital against the size of the balance sheet. The difference is that when the deposit sits on the Asset side just as "cash" it generates no income for the bank, whilst the bank has to pay interest to the customer + allocate capital. But when the bank manages to "transform" the asset from cash into a loan or a mortgage, it will generate some income (and it will thereby justify allocating capital against it, i.e. it will put the capital "to work").
Ps: that's why the headline of the article you reference is:
"Big U.S. banks hunger for loans, capital relief as deposits pile up"
I.e. they feel the need to turn that "cash" item on the asset side of their balance sheet into a "loan" item, but the demand for loans is not there right now (so they also plead for at least some capital relief on those deposits, at least until loan demand picks up again).
• Thanks it is much clearer now. I understand the banks motivation. only one thing I seem to be missing. "will need to allocate capital against the size of the balance sheet." : As far as I know capital allocations does not depend on the value of the balance sheet per se, but depends on the assets in the balance. i.e. safe loan will require less capital than riskier one. I fail to understand why to impose capital requirements on "cash" asset? – d_e Feb 4 at 11:43
• @d_e: good point, capital requirement depends on Risk-Weighted assets, and these in turn depend on the type of asset. I am not 100% sure if "cash" attracts any capital charge, but even if it's small, it is "annoying" for the bank because it will reduce the "return on capital" that the bank will generate. Also, banks worry about their leverage ratio (i.e. Assets / Capital), and this goes up even if capital requirement doesn't. – Jan Stuller Feb 4 at 11:46
• @d_e: I edited your question, because I am myself curious if anyone knows the answer to whether cash attracts capital charge. Pls feel free to un-mark my answer so that the question attracts more responses. – Jan Stuller Feb 4 at 11:49
• "cash" is not really valid (if not physical). All digital cash is ultimately traced back to the central bank authority. So "digital cash asset" would be converted to another asset, e.g. "central bank deposit", "government debt", "mortage", "corporate debt" etc. The first of those has a risk weight of 0%, and likely as does 2nd. That potentially leaves capital (from a capital ratio) perspective unimpacted by a deposit. – Attack68 Feb 4 at 22:38
• @d_e yes they are forced to do that. typically what they do though is deposit the cash with another bank who does have access to the FED, or they buy a T-Bill, or they transact a reverse repo. – Attack68 Feb 5 at 16:59
This is an answer from European perspective.
As @JanStuller has explained a retail cash deposit results in two balance sheet entries.
In the simplest form:
(Liability) Customer Demand Deposit, e.g. €100
(Asset) Central Bank Deposit, e.g. €100
### Capital Requirements
The asset can be transformed to other forms but according to Capital Requirements Regulation Article 114 a cash deposit with the ECB has a risk weight of 0%. Therefore no additional capital is required to be held against this customer deposit on this measure.
$$\text{new ratio} = \frac{\text{old capital}}{\text{old risk weighted assets + 0% x €100}}$$
### Liquidity Coverage Ratio
Article 412 essentially states that liquid assets must cover outflows minus inflows over the next 30 days. Again the asset form is important but here 100% of the central bank deposit is counted, but the liability is only counted from 5% to 10%, as retail deposits are generally considered sticky Article 421 so there is a 'run-off' rate. This impacts the ratio favourably.
$$\text{new ratio} = \frac{\text{old HQLA + 100% x €100}}{\text{old net outflows + 95% x €100}}$$
### Net Stable Funding Ratio
This basically says that the available stable funding (ASF) exceeds the required stable funding. Basel states that demand deposits qualify for 90-95% ASF while central bank deposits amount to no RSF, so here again the ratio is impacted favourably.
$$\text{new ratio} = \frac{\text{old ASF + 95% x €100}}{\text{old RSF outflows + 0% x €100}}$$
### Leverage Ratio
This is a banks Tier 1 capital divided by total exposure. The exposure of a central bank deposit is zero (I believe) Article 11. Therefore in this scenario the ratio is unchanged.
$$\text{new ratio} = \frac{\text{old capital}}{\text{old risk exposure + 0% x €100}}$$
### Net Interest Income
The profit for the bank here is the interest rate differential between the central bank deposit and the retail deposit account. In Europe this is problematic since the central bank rate is lower than 0%.
### Other factors
When that asset is not a central bank deposit it clearly impacts the above calculations. On top of that there may well be additional capital charges for market risk, i.e. if you buy a long dated treasury bond instead of T-bill.
• That's a great answer. +1 – Jan Stuller Feb 5 at 7:32
Afaik and as Jan Stuller already mentioned, banks have to meet requirements to the Leverage Ratio, which gets mandatory with CRR II in 2021. For simplicity, most banks will have to meet a minimum of 3%.
Important note: riskweights do not play a role here.
That's why the leverage ratio is decreasing when customers put their money to the banks, leading to banks will need to increase the numerator (=capital) to increase their Leverage Ratio. (But since demand for loans is low and cash or "riskless" bonds don't earn money, how should banks increase their capital...)
• "Important note: riskweights do not play a role here." => do I understand your post correctly that cash would attract zero risk weights and therefore zero capital consumption, but it is only the leverage ratio that drives the need to increase capital? – Jan Stuller Feb 4 at 13:39
• Indeed, LR is an indicator that doesn't take riskweights into account. – simzoor Feb 4 at 13:50
• Understood. But just to be sure: are you absolutely certain that cash doesn't attract any Risk Weighted Assets? (intuitively it shouldn't, because it is the "safest" asset... so it'd make sense) – Jan Stuller Feb 4 at 13:54
• To be a bit more clear: I'm assuming cash as the money that banks have on their account of the related central bank for payments (ECB, Bundesbank, Banque de France etc.). It is a bit more tricky with "cash equivalents" (e.g. nostro accounts with other "normal" banks won't get a RW of 0%, but 20%). eba.europa.eu/single-rule-book-qa/-/qna/view/publicId/2015_2001 – simzoor Feb 4 at 14:23
• Excellent, thank you!! – Jan Stuller Feb 4 at 14:34
Another problem is the bank's GSIB score. There are about 10 components that go into the GSIB score calculation, but one of them is (bank equity)/(balance sheet).
Having a poor GSIB score means the bank will have to a surcharge assessment on it's balances.
So in other words, if a bank just takes in more deposits - and does nothing with them - the leverage ratio will not change. But the GSIB score increases. Of course, a rational person would say, "who cares?" as this is just cash vs balances. But the rules are the rules and they will now need even more capital set aside.
It's important to understand that the GSIB scores are a step function. So, a bank will have so far that it can go until hit hits a new tier and then all balances are subject to a higher assesment.
• +1 from me for the GSIB score (of which I was unaware). But I don't think the following statement of yours is correct: So in other words, if a bank just takes in more deposits - and does nothing with them - the leverage ration will not change. The leverage ratio will increase if a bank takes in more deposits, because its balance sheet grows, but its capital (i.e. equity) does not grow. You cannot count "cash" from the Asset part of a balance sheet into the Equity part of a balance sheet. Equity only grows via retained earnings or rights issues. – Jan Stuller Feb 4 at 15:57
• There are quite a few ratios under Basel. I'm not an expert on the terminology. I think just a balance that sits in the Fed vs the customer deposit is OK. The problem is the capital ratio. (Not enough bank equity vs the balance sheet). I know in the money market research that I get sometimes it's all about the GSIB limit for the new balances. That worry dwarfs worries about other items. (Because of the step function nature as well) – JoshK Feb 4 at 16:01
Apologies if this answer is a bit off the beaten track.
The reason capital requirements are imposed on deposits is historical. In the United States they were introduced by Alexander Hamilton (1755-1804) who was knowledgeable about British banking practices.
In the 18th century people understood that banking can be dangerous. The failure of Law's Bank in France in 1720 is an extreme example of what can go wrong with completely unregulated banking. It seemed sensible that, to reassure depositors, there should be a requirement that at least X% of the amount you deposit should be set aside in some kind of "reserve".
People such as Hamilton understood that the need for such reserve requirements is not only prudential (for the benefit of depositors) but even more importantly for macro-financial reasons (for the stability of the banking system, to avoid an infinite expansion of the money supply).
The reason deposit reserve requirements exist today (where they do exist) is mainly traditional. As other answers show, nowadays there are many other tools used to regulate the banking system (LCR, leverage ratios, etc. etc.). But reserve requirements were the standard way for over 200 years.
• Great insights, didn't know many of these facts! Learned something new. Just to stress a point: deposit reserves are a separate measure to capital allocation (I am sure you know, just wanted to highlight it). Even if % of deposits needs to be "reserved" as cash and cannot be re-loaned, these reserves will still sit on a balance sheet as an "Asset", they won't increase the Capital part of the balance sheet. – Jan Stuller Feb 4 at 20:15
• There has been more bank runs in regulated than irregulated markets. Compare Scotland (irregulated) with England (regulated) during 18th-19th century or US (regulated) with Canada (irregulated) during the same period. – d-b Feb 5 at 8:59
Bank reserves are based on liabilities because that is the money people expect to be able to get back whenever they want it.
If I put money into a brokerage account and direct my bank to "Buy GameStop!" with it and the GameStop tanks, well that is too bad for me.
If I put my money into a savings account (loan my money to the bank) and my bank buys GameStop and GameStop tanks, I can still get my money back, and too bad for the bank.
So the bank needs to have some money liquid (in reserve) so that they still have money for me even if their investments do poorly (they don't keep the whole amount in a vault, but a percentage).
On top of that, the article seems to specifically be talking about the globally significant banking institution (GSIB) surcharge. The GSIB surcharge was created after the 2008 financial crisis based on the idea that some banks have become so interconnected and large that if they failed it would be catastrophic for the entire financial system.
One of the calculations that goes into GSIB is "assets under trust"--basically a complicated way of saying how large the bank is. Larger banks are required to be more careful (keep more money in reserve) than smaller banks.
• there is a technical difference between 'own funds = T1 + T2 capital' and 'reserve requirements'. It might be easy to meet reserve requirements by borrowing money in capital markets and posting a necessary amount to the central bank, but own funds cannot be as engineered therefore regulation is structured around capital ratio and leverage ratio. – Attack68 Feb 4 at 23:37 |
# Tim Sullivan
### Welcome!
I am Associate Professor in Predictive Modelling in the Mathematics Institute and School of Engineering at the University of Warwick. I am also a Turing Fellow at the Alan Turing Institute. I have wide interests in uncertainty quantification the broad sense, understood as the meeting point of numerical analysis, applied probability and statistics, and scientific computation. On this site you will find information about how to contact me, my research, publications, and teaching activities.
### Testing whether a learning procedure is calibrated in JMLR
The article “Testing whether a learning procedure is calibrated” by Jon Cockayne, Matthew Graham, Chris Oates, Onur Teymur, and myself has just appeared in its final form in the Journal of Machine Learning Research. This article is part of our research on the theoretical foundations of probabilistic numerics and uncertainty quantification, as we seek to explore what it means for the uncertainty associated to a computational result to be “well calibrated”.
J. Cockayne, M. M. Graham, C. J. Oates, T. J. Sullivan, and O. Teymur. “Testing whether a learning procedure is calibrated.” Journal of Machine Learning Research 23(203):1–36, 2022. https://jmlr.org/papers/volume23/21-1065/21-1065.pdf
Abstract. A learning procedure takes as input a dataset and performs inference for the parameters $$\theta$$ of a model that is assumed to have given rise to the dataset. Here we consider learning procedures whose output is a probability distribution, representing uncertainty about $$\theta$$ after seeing the dataset. Bayesian inference is a prime example of such a procedure, but one can also construct other learning procedures that return distributional output. This paper studies conditions for a learning procedure to be considered calibrated, in the sense that the true data-generating parameters are plausible as samples from its distributional output. A learning procedure whose inferences and predictions are systematically over- or under-confident will fail to be calibrated. On the other hand, a learning procedure that is calibrated need not be statistically efficient. A hypothesis-testing framework is developed in order to assess, using simulation, whether a learning procedure is calibrated. Several vignettes are presented to illustrate different aspects of the framework.
Published on Friday 5 August 2022 at 14:50 UTC #publication #prob-num #cockayne #graham #oates #teymur
### Randomised integration for deterministic operator differential equations in Calcolo
The article “Randomised one-step time integration methods for deterministic operator differential equations” by Han Cheng Lie, Martin Stahn, and myself has just appeared in its final form in Calcolo. In this paper, we extend the analysis of Conrad et al. (2016) and Lie et al. (2019) to the case of evolutionary systems in Banach spaces or even Gel′fand triples, this being the right setting for many evolutionary partial differential equations.
H. C. Lie, M. Stahn, and T. J. Sullivan. “Randomised one-step time integration methods for deterministic operator differential equations.” Calcolo 59(1):13, 33pp., 2022. doi:10.1007/s10092-022-00457-6
Abstract. Uncertainty quantification plays an important role in applications that involve simulating ensembles of trajectories of dynamical systems. Conrad et al. (Stat. Comput., 2017) proposed randomisation of deterministic time integration methods as a strategy for quantifying uncertainty due to time discretisation. We consider this strategy for systems that are described by deterministic, possibly non-autonomous operator differential equations defined on a Banach space or a Gel′fand triple. We prove pathwise and expected error bounds on the random trajectories, given an assumption on the local truncation error of the underlying deterministic time integration and an assumption that the absolute moments of the random variables decay with the time step. Our analysis shows that the error analysis for differential equations in finite-dimensional Euclidean space carries over to infinite-dimensional settings.
Published on Friday 25 February 2022 at 17:00 UTC #publication #prob-num #lie #stahn
### GParareal: A time-parallel ODE solver using Gaussian process emulation
Kamran Pentland, Massimiliano Tamborrino, James Buchanan, Lynton Appel and I have just uploaded a preprint of our latest article, “GParareal: A time-parallel ODE solver using Gaussian process emulation”, to the arXiv. In this paper, we show how a Gaussian process emulator for the difference between coarse/cheap and fine/expensive solvers for a dynamical system can be used to enable rapid and accurate solution of that dynamical system in a way that is parallel in time. This approach extends the now-classical Parareal algorithm in a probabilistic way that allows for efficient use of both runtime and legacy data gathered about the coarse and fine solvers, which may be a critical performance advantage for complex dynamical systems for which the fine solver is too expensive to run in series over the full time domain.
Abstract. Sequential numerical methods for integrating initial value problems (IVPs) can be prohibitively expensive when high numerical accuracy is required over the entire interval of integration. One remedy is to integrate in a parallel fashion, “predicting” the solution serially using a cheap (coarse) solver and “correcting” these values using an expensive (fine) solver that runs in parallel on a number of temporal subintervals. In this work, we propose a time-parallel algorithm (GParareal) that solves IVPs by modelling the correction term, i.e. the difference between fine and coarse solutions, using a Gaussian process emulator. This approach compares favourably with the classic parareal algorithm and we demonstrate, on a number of IVPs, that GParareal can converge in fewer iterations than parareal, leading to an increase in parallel speed-up. GParareal also manages to locate solutions to certain IVPs where parareal fails and has the additional advantage of being able to use archives of legacy solutions, e.g. solutions from prior runs of the IVP for different initial conditions, to further accelerate convergence of the method - something that existing time-parallel methods do not do.
Published on Tuesday 1 February 2022 at 12:00 UTC #preprint #prob-num #pentland #tamborrino #buchanan #appel
### Γ-convergence of Onsager-Machlup functionals in Inverse Problems
The articles “Γ-convergence of Onsager–Machlup functionals” (“I. With applications to maximum a posteriori estimation in Bayesian inverse problems” and “II. Infinite product measures on Banach spaces”) by Birzhan Ayanbayev, Ilja Klebanov, Han Cheng Lie, and myself have just appeared in their final form in the journal Inverse Problems.
The purpose of this work is to address a long-standing issue in the Bayesian approach to inverse problems, namely the joint stability of a Bayesian posterior and its modes (MAP estimators) when the prior, likelihood, and data are perturbed or approximated. We show that the correct way to approach this problem is to interpret MAP estimators as global weak modes in the sense of Helin and Burger (2015), which can be identified as the global minimisers of the Onsager–Machlup functional of the posterior distribution, and hence to provide a convergence theory for MAP estimators in terms of Γ-convergence of these Onsager–Machlup functionals. It turns out that posterior Γ-convergence can be assessed in a relatively straightforward manner in terms of prior Γ-convergence and continuous convergence of potentials (negative log-likelihoods). Over the two parts of the paper, we carry out this programme both in generality and for specific priors that are commonly used in Bayesian inverse problems, namely Gaussian and Besov priors (Lassas et al., 2009; Dashti et al., 2012).
B. Ayanbayev, I. Klebanov, H. C. Lie, and T. J. Sullivan. “Γ-convergence of Onsager–Machlup functionals: I. With applications to maximum a posteriori estimation in Bayesian inverse problems.” Inverse Problems 38(2):025005, 32pp., 2022. doi:10.1088/1361-6420/ac3f81
Abstract (Part I). The Bayesian solution to a statistical inverse problem can be summarised by a mode of the posterior distribution, i.e. a MAP estimator. The MAP estimator essentially coincides with the (regularised) variational solution to the inverse problem, seen as minimisation of the Onsager–Machlup functional of the posterior measure. An open problem in the stability analysis of inverse problems is to establish a relationship between the convergence properties of solutions obtained by the variational approach and by the Bayesian approach. To address this problem, we propose a general convergence theory for modes that is based on the Γ-convergence of Onsager–Machlup functionals, and apply this theory to Bayesian inverse problems with Gaussian and edge-preserving Besov priors. Part II of this paper considers more general prior distributions.
B. Ayanbayev, I. Klebanov, H. C. Lie, and T. J. Sullivan. “Γ-convergence of Onsager–Machlup functionals: II. Infinite product measures on Banach spaces.” Inverse Problems 38(2):025006, 35pp., 2022. doi:10.1088/1361-6420/ac3f82
Abstract (Part II). We derive Onsager–Machlup functionals for countable product measures on weighted $$\ell^{p}$$ subspaces of the sequence space $$\mathbb{R}^\mathbb{N}$$. Each measure in the product is a shifted and scaled copy of a reference probability measure on $$\mathbb{R}$$ that admits a sufficiently regular Lebesgue density. We study the equicoercivity and Γ-convergence of sequences of Onsager–Machlup functionals associated to convergent sequences of measures within this class. We use these results to establish analogous results for probability measures on separable Banach or Hilbert spaces, including Gaussian, Cauchy, and Besov measures with summability parameter $$1 \leq p \leq 2$$. Together with Part I of this paper, this provides a basis for analysis of the convergence of maximum a posteriori estimators in Bayesian inverse problems and most likely paths in transition path theory.
Published on Wednesday 5 January 2022 at 12:00 UTC #publication #inverse-problems #modes #map-estimators #ayanbayev #klebanov #lie
### Dimension-independent MCMC on spheres
Han Cheng Lie, Daniel Rudolf, Björn Sprungk and I have just uploaded a preprint of our latest article, “Dimension-independent Markov chain Monte Carlo on the sphere”, to the arXiv. In this paper, motivated by problems such as Bayesian binary classification over continuous spaces, for which the parameter space is naturally an infinite-dimensional sphere of functions, we consider MCMC methods for inference on spheres of Hilbert spaces. In particular, we construct MCMC methods that have dimension-independent performance in terms of their acceptance probability, spectral gap, etc.; we also show how more naive approaches may lack basic properties such as Markovianity and reversibility; how how even sophisticated geometric MCMC approaches can still suffer from the curse of dimension.
Abstract. We consider Bayesian analysis on high-dimensional spheres with angular central Gaussian priors. These priors model antipodally-symmetric directional data, are easily defined in Hilbert spaces and occur, for instance, in Bayesian binary classification and level set inversion. In this paper we derive efficient Markov chain Monte Carlo methods for approximate sampling of posteriors with respect to these priors. Our approaches rely on lifting the sampling problem to the ambient Hilbert space and exploit existing dimension-independent samplers in linear spaces. By a push-forward Markov kernel construction we then obtain Markov chains on the sphere, which inherit reversibility and spectral gap properties from samplers in linear spaces. Moreover, our proposed algorithms show dimension-independent efficiency in numerical experiments.
Published on Wednesday 22 December 2021 at 12:00 UTC #preprint #mcmc #lie #rudolf #sprungk |
# What is a combination of capacitors?
Jun 15, 2018
See answer below
#### Explanation:
Given: What is a combination of capacitors?
A circuit can contain more than one capacitor. The capacitors can be grouped in series or in parallel.
Series grouping: $\frac{1}{C} _ \left(e q\right) = \frac{1}{C} _ 1 + \frac{1}{C} _ 2 + \ldots$
Parallel grouping: ${C}_{e q} = {C}_{1} + {C}_{2} + \ldots$ |
## Labor Economics
7. U.S. Trucking pays its drivers $40,000 per year, while American Trucking pays its drivers$38,000 per year. For both firms, truck drivers average 240,000 miles per year. For simplicity, assume both firms require driving 60 million miles each year. Truck driving jobs are the same regardless of which firm one works for, except that U.S. Trucking gives each of its trucks a safety inspection ever 50,000 miles while American Trucking gives each of its trucks a safety inspection every 36,000 miles. This difference in safety inspection rates results in a different rate of fatal accidents between the two companies. In particular, one driver for U.S Trucking dies in an accident every 12 million miles while one driver for American Trucking dies in an accident every 15 million miles. What is the value of a trucker’s life implied by the compensating differential between the two firms? |
# What product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?
I want to do Wurtz reaction of 1-bromo-3-chlorocyclobutane:
Wikipedia suggests to do an intramolecular reaction. Instead, I remember reading somewhere that bromine is more reactive and hence we should do an intermolecular reaction, and attach two carbons to where $\ce{Br}$ was attached.
So, what product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane? |
SMS scnews item created by Stephan Tillmann at Thu 9 Oct 2014 1020
Type: Seminar
Distribution: World
Expiry: 8 Jan 2015
Calendar1: 29 Oct 2014 1100-1200
CalLoc1: Carslaw 535A
Auth: [email protected] (assumed)
# Ergodic theorems for amenable groups
### Sasha Fish (Sydney)
GTA Seminar - Wednesday, 29 October, 11:00-12:00 in Carlaw 535A
## Ergodic theorems for amenable groups
We will talk on the validity of the mean ergodic theorem along left Følner sequences in a countable amenable group G. Although the weak ergodic theorem always holds along any left Følner sequence in G, we will provide examples where the mean ergodic theorem fails in quite dramatic ways. On the other hand, if G does not admit any ICC quotients, e.g. if G is virtually nilpotent, then we will prove that the mean ergodic theorem does indeed hold along any left Følner sequence. Based on the joint work with M. Bjorklund (Chalmers). |
An interview question in the wild, the Category Problem asks to model a graph of categories of goods in SQL. The sample solution involves a couple of recursive SQL queries that traverse the edges of the model graph. The sample solution works in PostgreSQL 9.6. This post includes a reference implementation in Ruby.
Problem
The problem asks you to implement in SQL the following rules.
1. There are products. Each product has a name.
2. There are categories. Each category has a name.
3. Each product may belong to one or more categories.
4. Each category may have subcategories.
5. Categories that have products do not have subcategories.
6. Each category has at most one parent.
7. You may ask a product for its corresponding categories.
8. You may ask a category for its corresponding products.
Consider the following diagram of an example state consisting of products and categories.
In the diagram, a box corresponds to a product and a circle to a category. Arrows start from a parent category and end in a child category. This is relation subcategory. Dotted lines connect categories with their corresponding products. This is relation belongs-to.
Your database must forbid violations of the rules. Consider the following examples. By rule 5, it must not be possible to add products to category Food. By rule 6, it must not be possible for Fast Food to be subcategory of Food and Home Delivered Goods.
The categories of Sandwich are Gas Station Food and Food. The categories of Pizza are Fast Food and Food.
The only product of category Fast Food is Pizza. The products of category Food are Sandwich and Pizza.
Solution
The sample solution is the following.
The sample solution consists of two parts. The first part is the construction of the database schema required by rules 1 to 6. The second part is the construction of two recursive queries, one for rule 7 and one for rule 8. Section Reference implementation in Ruby includes a reference implementation in Ruby that you can use to understand the semantics of the sample solution.
Database schema
The following is an explanation of the construction of the database schema.
1. Each product has a name.
We create a table of products.
Example usage.
Gives.
2. Each category has a name.
We create a table of categories.
Example usage.
Gives.
3. Each product may belong to one or more categories.
Meaning, each product may or may not belong to one or more categories. We assume that categories may correspond to multiple products. Given these reasons, relation belongs-to is a many-to-many relation. The following associative table (a.k.a. cross-reference table) corresponds to relation belongs-to.
Example usage.
Gives.
4. Each category may have subcategories.
AND
6. Each category has at most one parent.
We address rules 4 and 6 in one step. These rules mean that relation subcategory is a one-to-many relation. We implement relation subcategory by appending an additional attribute to table categories.
Example usage.
Gives.
5. Categories that have products do not have subcategories.
We implement this rule by functions if_no_sub_c and if_no_prod.
We constraint relation belongs-to by function if_no_sub_c.
Example usage.
Gives.
We constraint relation subcategory by function if_no_prod.
Example usage.
Gives.
Recursive queries
The following is an explanation of the two recursive queries.
7. You may ask a product for its corresponding categories.
This rule asks for those categories that are ancestors of a given product in relation subcategory. We compute the ancestor categories by function categories.
For a given product, the computation starts from the categories that contain the product. Those categories are given by the following SELECT statement.
The SELECT statement is the base case of a recursive WITH clause. The recursive step consists in selecting those categories that are parent of some category given by a previous recursive call. The the recursive step corresponds to the following SELECT statement.
The result of the recursive WITH clause is projected onto the name before returning by the following SELECT statement.
Example usage.
Gives.
8. You may ask a category for its corresponding products.
This rule asks that we collect the products of categories that are descendant of given category in relation subcategory. The following function products implements the rule.
Example usage.
Gives.
Reference implementation in Ruby
The reference implementation is the following.
Example usage.
Gives.
1. Each product has a name.
We create a class of products. Example usage. Gives. 2. Each category has a name.
We create a class of categories. Example usage. Gives. 3. Each product may belong to one or more categories.
Given a category, we record the products that belong to the category in array Category#@products. We append products to a category by means of method Category#add_product. Example usage. Gives. 4. Each category may have subcategories.
Given a category, we record the children categories in array Category#@children. We append subcategories by means of method Category#add_category. Example usage. Gives. 5. Categories that have products do not have subcategories.
This is a constraint on relations _belongs-to_ and _subcategory_. We constraint each relation by constraining corresponding methods Category#add_product and Category#add_category. Example usage. Gives. 6. Each category has at most one parent.
This is a constraint on relation _subcategory_. We apply the constraint by constraining method Category#add_category. We create instance variable Category#@parent to label each category with its parent. Example usage. Gives. 7. You may ask a product for its corresponding categories.
This rule asks for those categories that are ancestors of a given product in relation _subcategory_. We compute the ancestor categories by method Product#categories. For a given product, the computation starts from the categories that contain the product. Those categories are recorded in array Product#@parent_categories. The array is populated by method Category#add_product. Example usage. Gives. 8. You may ask a category for its corresponding products.
This rule asks that we collect the products of categories that are descendant of a given category in relation _subcategory_. We do so by traversing relation _subcategory_ from the given category all the way down to leaf descendants. We do the traversal and collect corresponding products by method Category#all_products. Example usage. Gives.
Other uses of WITH clauses
• Apply several data modifications in the same query (section 7.8.2).
• Apply recursive self-references in a data-modifying query (section 7.8.2). |
Canadian Mathematical Society www.cms.math.ca
location: Publications → journals → CMB
Abstract view
# Genericity of certain classes of unitary and self-adjoint operators
Published:1998-06-01
Printed: Jun 1998
• J. R. Choksi
• M. G. Nadkarni
Format: HTML LaTeX MathJax PDF PostScript
## Abstract
In a paper [1], published in 1990, in a (somewhat inaccessible) conference proceedings, the authors had shown that for the unitary operators on a separable Hilbert space, endowed with the strong operator topology, those with singular, continuous, simple spectrum, with full support, form a dense $G_\delta$. A similar theorem for bounded self-adjoint operators with a given norm bound (omitting simplicity) was recently given by Barry Simon [2], [3], with a totally different proof. In this note we show that a slight modification of our argument, combined with the Cayley transform, gives a proof of Simon's result, with simplicity of the spectrum added.
MSC Classifications: 47B15 - Hermitian and normal operators (spectral measures, functional calculus, etc.)
© Canadian Mathematical Society, 2014 : https://cms.math.ca/ |
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