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## anonymous 4 years ago Consider the function f(x) = 2sin(x2) on the interval 0 ≤ x ≤ 3. (a) Find the exact value in the given interval where an antiderivative, F, reaches its maximum. x = (b) If F(1) = 9, estimate the maximum value attained by F. (Round your answer to three decimal places.) y ≈
1. inkyvoyd
set 2sin(x^2)=0
2. inkyvoyd
Wait, over an interval -_-
3. inkyvoyd
Ignore the interval, and just solve that trig equation.
4. anonymous
i know that the answer for part "a" is
5. anonymous
square(pi)
6. inkyvoyd
square or sqrt?
7. anonymous
i dont know how to do part b
8. anonymous
sqrt
9. inkyvoyd
Well, your interval s defined only up to 3. Is that supposed to be that way, or is the three suposed to be pi?
10. anonymous
up to 3
11. anonymous
a)$2*\sin(2*x)=0\rightarrow x=k*pi/2$ In this interval x=pi/2
12. inkyvoyd
Do you understand a though?
13. anonymous
well i got x= |dw:1334814197335:dw|
14. inkyvoyd
|dw:1334814169184:dw|
15. anonymous
16. inkyvoyd
|dw:1334814236680:dw|
17. inkyvoyd
Naturally, sqrt pi would be the largest value up to 3, because it squares out to pi.
18. anonymous
the equation is f(x) = 2sin(x^2)
19. anonymous
20. anonymous
$F(x)=-\cos(2*x)+K'$ $F(1)=-\cos(2*1)+K'=9$
21. inkyvoyd
I know - That is sin x
22. inkyvoyd
but, sin x^2 is similar to the graph in that it has its highest values at sqrtpi
23. anonymous
sqrt(pi) is the right answer for part a is marking it right
24. anonymous
the part that i dont get how to do is part b
25. inkyvoyd
Yes. Can you tell me what calc course you are enrolled in?
26. anonymous
calculus I
27. inkyvoyd
Alright, give me a second.
28. anonymous
2sin(x^2) = 0 x^2 = pi x = sqrt(pi)
29. anonymous
thats how i got my answer
30. inkyvoyd
Have you learned how to approximate integrals?
31. anonymous
kinda
32. anonymous
thats what we are learning
33. inkyvoyd
What's it called?
34. anonymous
antiderivatives
35. inkyvoyd
No, what kind of approximations have you learned?
36. inkyvoyd
I just figured out how I can solve the problem.
37. inkyvoyd
Here's the graph, btw
38. anonymous
i dont know
39. inkyvoyd
40. anonymous
the tittle of this chapter is antiderivatives
41. inkyvoyd
Note how the function uses something called the fresnal S, which I don't know either. But mor eimportantly, notice how it gets to its peak at about 1,7
42. anonymous
so how do i do part b?
43. inkyvoyd
I'm thinking ;)
44. inkyvoyd
Have you learned trapozoidal approximation?
45. anonymous
no
46. anonymous
I also have a question for part b... help us figure this out plz
47. inkyvoyd
48. inkyvoyd
Alright. that is the graph of f(x)=2 sin(x^2) right?
49. inkyvoyd
Notice how the waves become wider and wider close to 0.
50. inkyvoyd
after that, they become thinner and thinner.
51. inkyvoyd
Now, give me a second to explain something.
52. inkyvoyd
Let me get htis plot into mathematica, and show you what I'm talking about.
53. anonymous
k
54. inkyvoyd
55. inkyvoyd
Ok. The area of a>b>c>d Tell me if you understand.
56. inkyvoyd
@Ldaniel ?
57. anonymous
yes i do
58. inkyvoyd
So, because of this, we know that from 0 to 3, the area enclosed in a is the greatest
59. anonymous
yes
60. inkyvoyd
So we are essentially looking for the area there.
61. inkyvoyd
@FoolForMath , I give up/
62. anonymous
;/
63. inkyvoyd
I can give you the answer though, me being a cheater cheater.
64. anonymous
65. anonymous
i wish i could do this math. -_-
66. inkyvoyd
Give me a second.
67. inkyvoyd
I'm going to numerically evaluate the integral with wolfram Mathematica
68. inkyvoyd
1.78966
69. inkyvoyd
@lgbasallote , here you go.
70. anonymous
71. inkyvoyd
WHAT?
72. inkyvoyd
Try the square root of pi then.
73. anonymous
74. inkyvoyd
Yes.
75. anonymous
wrong
76. inkyvoyd
btw, try 1.790 if you haven't already.
77. anonymous
i did its wrong
78. inkyvoyd
Ahem.
79. anonymous
;(
80. anonymous
lol why call me =)) i only know algebra :P i dont like these stuffs
81. anonymous
:((
82. inkyvoyd
Wrong, call >.<
83. inkyvoyd
84. inkyvoyd
So, you are supposed to use an estimation method that isn't accurate, nor do I know. Or, the problem is set up funy.
85. inkyvoyd
86. inkyvoyd
If I used the methods they gave, I might just get the answer they are looking for.
87. anonymous
i got the answer is y = 10.16912...
88. inkyvoyd
89. anonymous
|dw:1334818371099:dw|
90. anonymous
yes
91. anonymous
so$\int\limits_{1}^{\sqrt{\pi}}f(x) = F(\sqrt{\pi}) - F(1)$ all you need to do is find approximation for definite integral
92. inkyvoyd
Well, then I'm completely useless.
93. anonymous
94. inkyvoyd
Nice job.
95. anonymous
thanks anyways
96. anonymous
how old are you?
97. inkyvoyd
15
98. inkyvoyd
I'm still not sure how I missed that problem so badly. :/
99. anonymous
15? really?
100. anonymous
101. anonymous
yes dumbcow
102. anonymous
sorry for being redundant, i know you already got the answer just putting it out there for everyones benefit
103. anonymous
I believe graph would go like this. (the pitch decreasing with increase in x) |dw:1334819574505:dw|
104. inkyvoyd
btw, yes, I'm trying to jump the gun and finish calculus and some math afterwards before I enter college.
|
# For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 Leftarrow p Leftarrow 0.95), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation sigma = sqrt{np(1 - p)}. Explain this approximation making use the Central Limit Theorem.
Question
Transformation properties
For large value of n, and moderate values of the probabiliy of success p (roughly, $$0.05\ \Leftarrow\ p\ \Leftarrow\ 0.95$$), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation
$$\sigma = \sqrt{np(1\ -\ p)}$$. Explain this approximation making use the Central Limit Theorem.
2021-03-19
Step 1
Alternatively prove that:
Theorem:
If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the
variablez $$=\frac{x\ -\ \mu}{\sigma}\ -\ N(0,\ 1)$$
where $$\mu=np\ \sigma^{2}=np(1\ -\ p)$$
Proof:
It can be prove using Moment generating function for binomial distribution. It's given as,
$$M_{x}(\theta)=(q\ +\ pe^{\theta})^{n}$$
where $$q = 1\ -\ p.$$
Step 2
By the linear transformation properties of the moment generating function.
$$M_{z}(\theta)=e\ -\ \frac{\mu\ \theta}{\sigma}M_{x}\left(\frac{\theta}{\sigma}\right)=e\ -\ \frac{\mu\ \theta}{\sigma}\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)^{n}$$
Taking the natural log of both sides, and then expanding the power series of $$e^{\theta/\sigma}$$
Then, $$\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ pe\ \frac{\theta}{\sigma}\right)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(q\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)$$
Since $$p\ +\ q = 1.$$
$$=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln\left(1\ +\ p\ \sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\theta}{\sigma}\right)^{k}\right)$$
If n is made sufficiently large $$\sigma=\sqrt{npq}$$ can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.
Let $$t=p\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}$$
Thus for sufficiently large $$n,\ | t |\ <\ 1.$$</span>
The ln term in the previous expression is $$\ln (1\ +\ t)\ where\ | t |\ <\ 1,$$</span> and so we may expand this term as follows:
$$\ln(1\ +\ t)=\sum_{k=1}^{\infty}(-1)^{k\ -\ 1}\frac{t^{k}}{k}$$
Step 3
This means that
$$\ln\ M_{z}(\theta)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \ln(1\ +\ t)=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\ \sum_{k=1}^{\infty}(-1)^{k_{1}}$$
$$\frac{t_{k}}{k}=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\ \sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ -\ p^{2}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{2}\right)+\ p^{3}\left(\sum_{m=1}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right)^{3}\ -\ \cdots\right]=\ -\ \frac{\mu\ \theta}{\sigma}\ +\ n\left[p\left[\frac{\theta}{\sigma}\ +\ \frac{1}{2}\left(\frac{\theta}{\sigma}\right)^{2}\ +\ \sum_{m=3}^{\infty}\ \frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]\ -\ p^{2}\left[\left(\frac{\theta}{\sigma}\right)^{2}\ +\ 2\left(\frac{\theta}{\sigma}\right)\sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\ +\ \sum_{m=2}^{\infty}\frac{1}{m!}\left(\frac{\theta}{\sigma}\right)^{m}\right]^{2}\right]$$
+ an infinite series of terms involving $$\left(\frac{\theta}{\sigma}\right)^{m}\ with\ m\ \geq\ 3]$$
By collecting terms in powers of $$\frac{\theta}{\sigma},$$ we see that
$$\ln\ M_{z}(\theta)=(-\mu\ +\ np)\frac{\theta}{\sigma}\ +\ n(p\ -\ p^{2})\left(\frac{\theta}{\sigma}\right)^{2}\ +\ n\ \sum_{m=3}^{\infty}c_{m}\left(\frac{\theta}{\sigma}\right)^{m}$$
$$=\frac{np\ -\ \mu}{\sigma}\theta\ +\ \frac{np(1\ -\ p)}{\sigma^{2}}\theta^{2}\ +\ n\sum_{m=3}^{\infty}\frac{c_{m}}{\sigma_{m}}\theta^{m}$$
Here, the $$c_{k}$$ terms don’t involve n, sigma or theta.
Since \mu=np\ and\ \sigma^{2}=mp(1\ -\ p)\)
the coefficient of the theta term is 0 and the coefficient of the
$$\theta^{2}$$ term is 1. Thus
$$\ln\ M_{z}(\theta)=\ \frac{\theta^{2}}{2}\ +\ \sum_{m=3}^{\infty}\ \frac{nc_{m}}{\sigma^{m}}\theta^{m}$$
Since the coefficient of each term in the sum has form
$$\frac{nc_{m}}{\sigma^{m}}=\frac{nc_{m}}{(npq)^{m/2}}= \frac{c_{m}}{(pq)^{m/2}}\ \cdot\ \frac{1}{n^{m/2\ -\ 1}}\ \rightarrow\ 0asn\ \rightarrow\ \infty$$
$$\lim_{n\ \rightarrow\ \infty}\ \ln\ M_{z}(\theta)=\frac{\theta^{2}}{2}$$
$$\lim_{n\ \rightarrow\ \infty}\ M_{z}(\theta)=e^{\theta2/2}$$
But note that by Property 3 of Normal Distribution the moment generating function for a random variable z with distribution N (0, 1) is
$$M_{z}(\theta)=e^{\theta^{2}/2}$$
Hence , Proved .
### Relevant Questions
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
Which of the following are correct general statements about the Central Limit Theorem?
(Select all that apply. To be marked correct: All of the correct selections must be made, with no incorrect selections.)
Question 3 options:
Its name is often abbreviated by the three capital letters CLT.
The accuracy of the approximation it provides, improves as the sample size n increases.
The word Central within its name, is meant to signify its role of central importance in the mathematics of probability and statistics.
It is a special example of the particular type of theorems in mathematics, which are called Limit Theorems.
It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$.
It specifies the specific shape of the curve which approximates certain sampling distributions.
It specifies the specific mean of the curve which approximates certain sampling distributions.
A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
Which of the following are correct general statements about the central limit theorem? Select all that apply
1. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$
2. It specifies the specific mean of the curve which approximates certain sampling distributions.
3. It is a special example of the particular type of theorems in mathematics, which are called Limit theorems.
4. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
5. It’s name is often abbreviated by the three capital letters CLT.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. The word Central within its name, is mean to signify its role of central importance in the mathematics of probability and statistics.
8. It specifies the specific shape of the curve which approximates certain sampling distributions.
Which of the following are correct general statements about the Central Limit Theorem? Select all that apply.
1. It specifies the specific shape of the curve which approximates certain sampling distributions.
2. It’s name is often abbreviated by the three capital letters CLT
3. The word Central within its name, is meant to signify its role of central importance in the mathematics of probability and statistics.
4. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to 50\%.
5. It specifies the specific mean of the curve which approximates certain sampling distributions.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
8. It is a special example of the particular type of theorems in mathematics, which are called limit theorems.
Which of the following are correct general statements about the central limit theorem? Select all that apply
1. The accuracy of the approximation it provides, improves when the trial success proportion p is closer to $$50\%$$
2. It specifies the specific mean of the curve which approximates certain sampling distributions.
3. It is a special example of the particular type of theorems in mathematics, which are called Limit theorems.
4. It specifies the specific standard deviation of the curve which approximates certain sampling distributions.
5. It’s name is often abbreviated by the three capital letters CLT.
6. The accuracy of the approximation it provides, improves as the sample size n increases.
7. The word Central within its name, is mean to signify its role of central importance in the mathematics of probability and statistics.
8. It specifies the specific shape of the curve which approximates certain sampling distributions.
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function $$p(x)=\frac{1}{\sqrt{2 \pi}^{\sigma}}e^{-(x-\mu)^{2}}/(2 \sigma^{2})$$ where $$\pi = 3.14159265 . . .$$ and sigma and mu are constants called the standard deviation and the mean, respectively. Its graph$$(\text{for}\ \sigma=1\ \text{and}\ \mu=2)$$is shown in the figure. With $$\sigma = 5 \text{and} \mu = 0$$, approximate $$\int_0^{+\infty}\ p(x)\ dx.$$
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NIPS 2016
Mon Dec 5th through Sun the 11th, 2016 at Centre Convencions Internacional Barcelona
Paper ID: 1426 Active Learning with Oracle Epiphany
### Reviewer 1
#### Summary
This paper proposes a new model for active learning, so-called active learning with oracle epiphany, which purports to describe realistic oracles (e.g., human annotators) that may fail to correctly classify certain examples until a certain point when enough of such examples had been presented (this is the moment of oracle's epiphany). In mathematical terms, this is accomplished by splitting the instance space into two disjoint subsets, on one of which the oracle knows the label, while the other is further split into a finite number of disjoint subsets that belong to different categories. Before the oracle had an epiphany, whenever it encounters an instance from one of these sets, it emits the abstention (don't-know) symbol, but, with a certain fixed probability, the next time it sees an instance from that subclass, it may have an epiphany and start labeling instances from that subclass correctly from that moment on. The authors present two algorithms that build on the classical active learning scheme of Cohn, Atlas, and Ladner (CAL) to work with oracle epiphany: EPICAL and Oracular-EPICAL, and present high-probability upper bounds on the total number of queries to achieve a given accuracy with a given confidence. The bounds involve the disagreement coefficient (as expected) and the VC dimension of the hypothesis class (again, as expected), plus a contribution that depends on the probability of oracular epiphany and on the number of subclasses where the oracle may experience an epiphany. The theoretical results are supplemented with empirical evaluation on synthetic and real data.
#### Qualitative Assessment
This paper is a nice addition to the literature on active learning, straddling the theoretical realm and the practically motivated algorithm design issues that arise when dealing with realistic oracles (e.g., human annotators who may initially not have enough confidence to generate labels for the given queries, but would gain confidence after having seen enough queries of this sort). Granted, the theoretical model is stripped down and simple, but this is a good place to start from. My only quibble is the absence of lower bounds, which may be too much to ask for, but it seems that the theoretical model is clean enough to at least attempt to derive one.
#### Confidence in this Review
3-Expert (read the paper in detail, know the area, quite certain of my opinion)
### Reviewer 2
#### Summary
This work studies the problem of active learning in a setting where the oracle might not be sure how to answer queries in certain (unknown) regions of the space until asked a few times for samples from those regions. The problem is well-motivated by behavioral studies of human labelers, and by intuitive descriptions of cases that might be seemingly ambiguous at first (e.g., when asked whether a webpage is about sports, the labeler might initially not make up her mind about how to label webpages about sports memorabilia). The paper models this phenomenon by supposing there are K unknown regions of the space, and each time a query is made in one of these regions for which an "epiphany" has not yet occurred, there is a beta probability the epiphany occurs then, in which case the oracle returns a label as usual (and will henceforth return a label whenever points in that region are queried), and otherwise the oracle abstains from labeling. They propose simple modifications of known active learning algorithms (CAL in the realizable case, OracularCAL in the agnostic case), and analyze the query complexity in each case, paying particular attention to the increase in query complexity induced by these abstentions, which is generally a function of K and beta.
#### Qualitative Assessment
Overall I enjoyed reading this paper. It is very well written, and the algorithms and analysis seem to be natural modifications of these existing approaches to active learning. The theoretical issues that arise in handling these abstentions and quantifying their effect on the query complexity are at times nontrivial, and are handled in elegant and appropriate ways. I suspect that not many people are aware of this problem, but it is quite well motivated in the paper, and seems to be a good problem to study. The specific theoretical model proposed for this phenomenon is, however, a bit toy-like. From the motivation, it seems these epiphanies might have more to do with the labeler needing enough data to get a feel for what the distribution of samples will be, to see where to draw the line, rather than there being some random internal event that could happen at any time upon being queried for a data point of that type. But I suppose this simplistic model could at least be a reasonable starting place, which can hopefully be made more realistic or general in future work. I have a few technical comments for the authors: The halting criterion for EPICAL is that \mu_X(D) \le \epsilon. However, it seems far preferable to halt upon \sup_{h,h' \in V} \mu_X(h(x)\neq h'(x)) \le \epsilon. Not only is this more closely related to the error rate, but it would guarantee that the algorithm actually halts (with query complexity at most roughly the passive sample complexity); in contrast, the halting criterion as currently stated might never actually be satisfied (e.g., interval classifiers with target as empty interval), so that the algorithm never halts. In Theorem 1, it seems clear that the terms M_CAL and 1/beta are unavoidable (as argued on page 4). It would be nice to also have some example(s) illustrating the kinds of scenarios where the \bar{M} term is also unavoidable and nonredundant. If the region U has large probability, then we'll very quickly get 1/beta abstentions, so \bar{M} seems unnecessary. But if U has small probability (eg, \epsilon/2), then we don't care about the abstentions anyway. So it seems the medium-sized U case is where \bar{M} might be needed in the bound. Is there a good example to illustrate its necessity in bounding the query complexity of EPICAL? In Corollary 7, in the query complexity bound, the first appearance of \tilde{d} in this expression is redundant (the second term already includes a value greater then \tilde{d}(e*/\epsilon) toward the right-most side of the expression). In the paragraph after Corollary 7, it is claimed that the leading term in the agnostic setting is of order \theta (e*/\epsilon)(\tilde{d}+\tilde{K}/\beta). However, the bound also includes a term of order \theta (e*/\epsilon)^2 \tilde{d}. So the correct expression here would be \theta ( (e*/\epsilon)^2 \tilde{d} + (e*/\epsilon)\tilde{K}/\beta ).
#### Confidence in this Review
3-Expert (read the paper in detail, know the area, quite certain of my opinion)
### Reviewer 3
#### Summary
The paper introduces a new active learning model, which attempts to incorporate previous empirical studies on human behavior in answering the label queries of learning algorithms. It is assumed that there is a partitioned subset of the input space where the response to a query is don’t know’’ until an epiphany’’ happens and from then on the correct answer is given for that subset. The paper considers natural extensions of previous active learning algorithms to this setting, and proves query complexity bounds. The bounds essentially contain extra additive factors, which depend linearly on the number of classes in the partition, and inversely on the probability of epiphany.
#### Qualitative Assessment
The paper deals with an important problem: to bring active learning closer to practical applications. The discussion of previous literature on the topic should mention work on the same problem for membership queries in query learning. The paper’s main source is the previous Oracular-CAL algorithm; however, the paper by Huang et al.(2015) is mainly on the algorithm ACTIVE-COVER; the relationship of this and Hsu (2010) should be clarified. The main feature of human oracles is making errors. The role of noise in active learning is, therefore the first issue to discuss in the present context. This seems to be missing from the paper and it should be added. As a possible extension of the model, reversibility of epiphanies is mentioned at the end; this is one (perhaps not the most natural) form of the imperfectness of human oracles, and the issue should be discussed in more detail. The paper mentions the unavoidable’’ K/beta cost in complexity; it would be useful to add some comments on the possibility of proving this unavoidability. The term unique’’ in Section 2 could perhaps be replaced by unseen’’. It is mentioned that for continuous distributions this assumption is without loss of generality. A comment should be added on other cases. The proof of Corollary 7 (the main result) from Theorem 6 is short; and at least some part of it (explaining how Lemma 2 connects the previous result to the result to be proven) would be illuminating to the readers, so it could perhaps be squeezed into the text. The paper makes a reasonable first step on an important problem. It is of good technical quality, making competent use of previous work in the standard model of active learning.
#### Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
### Reviewer 4
#### Summary
The paper provide theoretical analysis of active learning with oracles abstaining on difficult queries until accumulating enough information to make decisions. The analysis shows that active learning is possible with oracle epiphany, but incurs an additional cost depending on when epiphany happens.
#### Qualitative Assessment
The paper consider an interesting setting for machine learning, which may be of great interests to the Active learning literature. However, in my opinion, its theoretical results and the techniques used in the paper are expected. As seen in the paper, the analyses of the new algorithms are not much different from the standard settings, only that we need to account for the additional cost of waiting for the epiphany to happen, which appear to be not difficult to predict and quantify. In my opinion, the contributions in both theoretical techniques and algorithmic ideas of the paper are quite minimal.
#### Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
### Reviewer 5
#### Summary
The author considers a more realistic oracle for active learning which is called oracle epiphany. Under this new oracle, the authors analyze the query complexity for both the realizable and agnostic case.
#### Qualitative Assessment
The paper is well written. It starts from realizable case to get a good intuition about the query complexity then move on to the agnostic case. However I'm not familiar with active learning at all.
#### Confidence in this Review
1-Less confident (might not have understood significant parts)
### Reviewer 6
#### Summary
The paper studies a new type of oracle in active learning, that has "epiphanies", modeling the setting that the oracle may initially answer some 'Don't know' if the query x is in some "uncertain" region (and end up answering at some point). The epiphany is modeled as geometric trials. Both realizable and agnostic cases are considered, and algorithms similar to CAL/Agnostic CAL are analyzed. Simulations show the effect of epiphany region (U) / epiphany probability (beta) to the label complexity of the algorithms. Specifically, if the epiphany region is important, then it is important to have a high epiphany probability (for active learning to outperform passive learning), and vice versa.
#### Qualitative Assessment
Technical Quality: the problem setting and the approach used in the paper are quite sound. For Theorem 1, the query complexity is a bit obscure, since it involves the disagreement coefficient restricted to the "known" region K, and it is not clear how to relate it with the global disagreement coefficient. Also, the stopping criterion is Algorithm 1 is a bit unsatisfying -- it is the size of the disagreement region rather than an upper confidence bound on the error that is measured (and compared against target error epsilon). Another direction worth investigating is how the disagreement region of the version space and unknown region in oracle epiphany interact with each other. If some theory can be done here, it will strongly support the experiments. Novelty and Originality: the "oracle with epiphany" model is new to me, and I think it is interesting. The proofs of Algorithm 2 is more-or-less standard (modulo the novel setting of rejection threshold involving b_t here.) Potential Impact: this paper studies a new oracle for active learning with is theoretically approachable, and it will also give inspirations on applied active learning research. Clarify and presentation: the paper is well written overall; the experiments and the theory fits well altogether. I find the proof of Lemma 4 a bit confusing -- in line 140, I think the high level idea is that "if we have used label budget >= \bar{m} + 2/beta\ln2/\delta, then we will definitely trigger epiphany". I suggest a bit revision on this paragraph. For Algorithm 2, I was once wondering if we can do the following modification: the algorithm does not keep counter b_t. At time step t, if the oracle returns \perp, the algorithm simply skips this iteration and pretend this example "does not exist". The examples (with hidden labels) collected this way would still be iid. This way, the label complexity of the algorithm seems to be at most the label complexity of Agnostic CAL + O(K/\beta), where the second term is the price of epiphany. Conceptually we can also think Algorithm 1 in this way as well. -- But it turns out that the iid property of the data is now violated, hence this modification does not work out -- perhaps the authors can remark on this?
#### Confidence in this Review
2-Confident (read it all; understood it all reasonably well)
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# How to create this diagram with Tikz? [duplicate]
This question already has an answer here:
I'm fairly familiar with using Tikz. I've used it to make simple logos before. Now, I'd like to create this diagram, which is a little more complicated:
Does anyone have any advice on what the best way to do this would be?
I'm aware that it's very easy to use the image file in my output (\includegraphics). However, I'd like to make it in Tikz if possible, for infinite scalability.
BTW, the blue is #080f6a.
## marked as duplicate by Phelype Oleinik, Marcel Krüger, J Leon V., Andrew, Alan MunnAug 23 '18 at 13:35
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• When you really want that to have in TikZ you can actually let e.g inkscape to redraw the logo and produce a TikZ output. – current_user Aug 21 '18 at 12:03
• In addition to current_user's comment, you would get the same scalability with any vector graphics format/software. – BambOo Aug 21 '18 at 12:10
• In addition to the methods described above, since this picture has just two colors you may convert it to the pnm format, from there with potrace to eps, which will make it a vector graphics, i.e. scalable, and then to pdf. I just tried that out and this also gives a very smooth and of course scalable picture. – marmot Aug 21 '18 at 15:26
• @Sebastiano Somehow I don't feel that that is a good duplicate. The question you link is about turning pictures of math formulae into TeX code. This question on the other hand asks to convert a bitmap image to TikZ (or at least a vector format). – moewe Aug 22 '18 at 4:25
• Welcome to TeX.SX! The short answer is that, yes, this is possible but I have to warn you that questions of the form "Please draw this for me" that show no effort on the part of OP, often don't get answered. You will get more help if you post some code showing what you have tried and give a minimal working example. A quick search on TeX.SX for drawing functions (with tikz or pstricks) will give you an idea of where to start from. – Andrew Aug 22 '18 at 11:57
## 1 Answer
My attempt using TikzEdt, a very useful tool for copy a figure.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta}
\definecolor{figBlue}{HTML}{080f6a}
\begin{document}
\begin{tikzpicture}[scale=0.4]
%frame
\filldraw [draw=figBlue,line width=9pt,fill=white](-12.5,-9.9) .. controls (-1.1,-14.3)
and (0,-14.3) .. (0,-15.3) [rounded corners] .. controls (0,-14.3)
and (1.1,-14.3) .. (12.5,-9.9)[sharp corners]-- (12.5,15.2) -- (-12.5,15.2)[rounded corners] -- cycle;
\path[tips,-{Computer Modern Rightarrow[figBlue,length=6ex,line width=1.5ex,sharp]}](0,0) to (0,-16.4);
%blue background circle
\fill[figBlue] (0,1.7) circle (9.6);
%sun
\fill[white] (0,7.4) circle (1.2);
%moon
\fill[white] (3.3,7.4) circle (1.3);
\fill [figBlue](3.7,8) circle (1.3);
%tail
\draw[white,line width=4.4pt,line cap=round]
(5.3,2.5) .. controls (5.6,2.7)
and (5.9,3) .. (6,3.4) .. controls (6,3.9)
and (5.8,4.1) .. (5.4,4.3) .. controls (5.1,4.4)
and (4.6,4.4) .. (4.1,4.3) .. controls (3.6,4.2)
and (3,4) .. (2.5,3.9) .. controls (1.9,3.8)
and (1.4,3.8) .. (0.9,3.8) .. controls (0.6,3.8)
and (0.4,3.9) .. (0.1,4);
%body
\fill[white,thick]
(-5,-4.7) -- (-4.6,-4.3)
-- (-3.4,-0.1) .. controls (-3,2)
and (-3.9,3.6) .. (-4.3,4.8) .. controls (-4.4,5.4)
and (-4.9,5.2) .. (-5.1,5.1) .. controls (-5.2,5.1)
and (-5.3,5.1) .. (-5.4,5.2) .. controls (-5.4,5.3)
and (-5.3,5.3) .. (-5.2,5.4) .. controls (-5.2,5.4)
and (-5.1,5.5) .. (-5.2,5.5) .. controls (-5.3,5.5)
and (-5.6,5.6) .. (-5.7,5.6) .. controls (-5.8,5.7)
and (-5.6,5.8) .. (-5.5,5.9) .. controls (-5.4,6)
and (-4.7,6.2) .. (-4.5,6.3) .. controls (-4.4,6.5)
and (-4.6,6.6) .. (-4.7,6.8) .. controls (-4.7,7.3)
and (-4.3,7.7) .. (-4.2,7.9) .. controls (-4.2,7.5)
and (-4.3,7.1) .. (-4.2,6.6) .. controls (-4.1,6.5)
and (-3.9,6.5) .. (-3.7,6.5) .. controls (-3.4,7)
and (-3.6,7.4) .. (-3.6,7.6) .. controls (-3.6,7.8)
and (-3.5,7.9) .. (-3.3,7.6) .. controls (-3.2,7.4)
and (-3,7.1) .. (-3.1,6.6) .. controls (-3.3,6.4)
and (-3,6.7) .. (-3.4,6.3) .. controls (-3.5,6.2)
and (-3.4,6) .. (-3.1,4.9) .. controls (-3,4.5)
and (-2.6,2.9) .. (-1.9,2.5) .. controls (-1.3,2.2)
and (-0.7,2.4) .. (0.5,2.6) .. controls (1.4,2.8)
and (2.8,3) .. (3.6,3) .. controls (4.2,3)
and (5.3,3) .. (5.5,2.3) .. controls (5.7,1.6)
and (5.5,1.3) .. (5,0.8) .. controls (4.8,0.7)
and (4.8,0.5) .. (4.8,0.2)
-- (5.3,-4.1) -- (5.4,-4.4)
-- (4.7,-4.4) -- (5,-4.1)
-- (4.2,-0.3) -- (2.9,-4.1)
-- (2.9,-4.4) -- (2.2,-4.4)
-- (2.5,-4.1) -- (3.5,-0.4) .. controls (3.6,-0.1)
and (3.6,0.1) .. (3.6,0.5) .. controls (3.6,1.3)
and (2.5,0.9) .. (2,0.8) .. controls (1,0.5)
and (-0.2,-0.1) ..(-2,0.2)
-- (-1.2,-4.7) -- (-1.5,-4.7)
-- (-1.9,-4.7) -- (-1.6,-4.4)
-- (-2.7,-0.2) -- (-4.3,-4.3)
-- (-4.3,-4.7) -- cycle;
\end{tikzpicture}
\end{document}
How I do this? I have import the figure in TikzEdt using \node {\includegraphics{figure}}; and then I have copy the figure using the tools of TikzEdt. Obviously you should know pgf/TikZ for adjust the figure for the final result. For example in this figure I have added an arrow tip at the bottom of the frame.
• Can you explain how you used TikzEdt to make this copy? – AndréC Aug 23 '18 at 13:33
• @AndréC You must import the figure in TikzEdt using \node {\includegraphics{figure}}; and then copy the figure using the tools of TikzEdt. Obviously you should know pgf/TikZ for adjust the figure for the final result. For example in this figure I have add an arrow tip at the bottom of the frame. – vi pa Aug 23 '18 at 13:47
• Thank you. Can you update your answer with these explanations? – AndréC Aug 23 '18 at 14:05
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# Support FAQs
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# Prove (square root 2) is irrational - I have a problem with this proof.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• November 30th 2013, 07:17 PM
Melody2
Prove (square root 2) is irrational - I have a problem with this proof.
I believe that this is a standard proof but I have a problem with it.
4d383176cd06283c92755eafb4b9586c.gif
The end statement appears to me to be
root2 = a/b where a and b are both even therefore a and b are not coprime which contradicts the initial condition that a and b must be coprime therefore root2 is irrational.
-------------------------------------------------
I ran through a different proof to attempt to prove that root4 is irrational
(this is not the full proof, I have truncated it a little)
suppose root4 were irrational
then root4 = a/b
4b^2=a^2
therefore a is even
Let a=2k
4b^2=4k^2
b^2=k^2
k=+-b
therefore
a=2k and b=+/-k
Therefore a and b are not coprime
Therefore root4 is irrational.
Obviously I realize that 2k/k can be reduced to 2/1 which then shows root4 to be rational
BUT
when the first proof was looking at root2 and it got down to (an even number/an even number) so why is it impossible for that fraction also be reduced to some p/q where p and q are coprime. To me, the proof just doesn't seem to be finished and does not prove anything (not to me).
• November 30th 2013, 07:45 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by Melody2
I believe that this is a standard proof but I have a problem with it.
4d383176cd06283c92755eafb4b9586c.gif
The end statement appears to me to be
root2 = a/b where a and b are both even therefore a and b are not coprime which contradicts the initial condition that a and b must be coprime therefore root2 is irrational.
-------------------------------------------------
I ran through a different proof to attempt to prove that root4 is irrational
(this is not the full proof, I have truncated it a little)
suppose root4 were irrational
then root4 = a/b
4b^2=a^2
therefore a is even
Let a=2k
4b^2=4k^2
b^2=k^2
k=+-b
therefore
a=2k and b=+/-k
Therefore a and b are not coprime
what if b=1, as it does here?
• November 30th 2013, 07:57 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Thanks but I don't get what you are trying to tell me Romsek.
b=+/-k and
k can equal any whole number it doesn't have to be 1. (or does it?)
Does the answer have implied restrictions on k?
• November 30th 2013, 08:30 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by Melody2
Thanks but I don't get what you are trying to tell me Romsek.
b=+/-k and
k can equal any whole number it doesn't have to be 1. (or does it?)
Does the answer have implied restrictions on k?
you used the fact that a=2k and b=+/-k to say that a and b are not coprime. But this isn't true if b=k=1.
• November 30th 2013, 08:48 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Ok so you are telling me that they must be coprime for some particular value of k
whereas I was thinking that they must be coprime for all values of k.
I guess that helps. Thanks.
• November 30th 2013, 08:51 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
no, it just doesn't apply if k=1. Take 2 and 1. you can't reduce 2/1 any further. I don't know that you'd call 2 and 1 coprime but certainly the fraction 2/1 is in fully reduced form. Here k=1.
• November 30th 2013, 09:03 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by romsek
no, it just doesn't apply if k=1. Take 2 and 1. you can't reduce 2/1 any further. I don't know that you'd call 2 and 1 coprime but certainly the fraction 2/1 is in fully reduced form. Here k=1.
It was sqrt4 = a/b=2k/k if k is anything other than 1 then 2k and k are not coprime.
so this particular proof only works if k=1
• November 30th 2013, 09:18 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by Melody2
It was sqrt4 = a/b=2k/k if k is anything other than 1 then 2k and k are not coprime.
so this particular proof only works if k=1
the key to the proof of sqrt[2] being irrational though was that b=2k and was thus even. Which since a was found to be even earlier gives a contradiction that a/b was in fully reduced form.
you can't make that statement here.
• November 30th 2013, 09:24 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
The whole reason that that the factor of 2 was relevant was that it meant a and b were not coprime.
so although I hear frustration in your post I am not going to give way completely. Sorry
• November 30th 2013, 09:28 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
right but in your proof b = +/- k not +/- 2k, so you can't claim b is even here like the other proof did. No frustration here and I certainly don't view it as a fight :P
• November 30th 2013, 10:14 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
We don't know exactly what k is although it does have a specific value. We do know that 'a' has a factor of 2 so therefore 'b' cannot be even.
etc
--------------------------------------------------
If I make this change to the wording of the proof then I am happy and I think it all makes sense.
Thankyou for your help Romsec. I have enjoyed our communications this afternoon and I appreciate your help.
----------------------------------------------------
I just a had a brainwave. Of course k has a specific value because 'a' has to have a specific value and k is half of a.
Now I have it. In the sqrt4 proof k has to be 1, there is not other possibliliy.
Thankyou so much for helping me reach this understanding.
---------------------------------------------------------
I know you kept telling me that k=1 but the real reason for this was not sinking into my brain.
Sometimes students (that's me) can be very frustrating. I didn't mean any offence by the comment.
• December 1st 2013, 12:45 AM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Glad you got it squared away.
Romsek (with a k!)
• December 1st 2013, 04:57 AM
Plato
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by Melody2
I believe that this is a standard proof but I have a problem with it.
4d383176cd06283c92755eafb4b9586c.gif.
I realize that you have solved this for P=2. But is is a general proof.
Suppose that $P$ is a positive integer that is not a square. We prove that $\sqrt P$ is irrational.
For a contradiction suppose that $\sqrt P$ is rational. Then the is a smallest positive integer $K$ for which it is true that $K\sqrt P$ is a positive integer.
One of the properties of the floor function is that $0<\sqrt P-\left\lfloor {\sqrt P } \right\rfloor < 1$
But that means $0 which means $K\sqrt P-K\left\lfloor {\sqrt P } \right\rfloor$ is a positive integer smaller that $K$
but $(K\sqrt P-K\left\lfloor {\sqrt P } \right\rfloor)\sqrt P$ is also a positive integer.
That contradicts the minimal nature of $K$.
• December 1st 2013, 06:44 PM
Melody2
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Thanks Plato,
I at least followed your argument right up to the last line.
It then dissipated into the realm of mystery for me.
also,
how do you implant things like you implanted the 'floor function'.
I had seen these floor and ceiling symbols before but i had no idea what they were.
Thanks.
• December 1st 2013, 08:41 PM
romsek
Re: Prove (square root 2) is irrational - I have a problem with this proof.
Quote:
Originally Posted by Melody2
Thanks Plato,
I at least followed your argument right up to the last line.
It then dissipated into the realm of mystery for me.
also,
how do you implant things like you implanted the 'floor function'.
I had seen these floor and ceiling symbols before but i had no idea what they were.
Thanks.
Embedded TeX can do all sorts of things!
$\lceil y\rceil \lfloor x\rfloor$
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
|
Hello,
I might be mistaking but in the Datasheet of the ADL5243 (https://www.analog.com/media/en/technical-documentation/data-sheets/ADL5243.pdf) on page 26 ff. the values of the Resistor R12 are given in nH instead of Ohm. I was wondering what the correct values are. I assume it is indeed a Resistor as an Inductor would not make a lot a sense right there.
I would like to operate the Amp at 1 GHz therefore the "Matching Circuit at 943 MHz" is the one I am the most interested in.
Best,
Axel
Parents
• Hi Axel,
R12 is indeed a place-holder that varies depending on frequency desired. At 943 Mhz it should be a 3.3 nH inductor for best matching. At higher frequencies it becomes a zero ohm jumper, and finally at 3.6 Ghz it's a 1 nH inductor. The values are given in table 8 on pg 25.
Thanks,
Darrell
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# What is the perimeter of a triangle with corners at (3 ,6 ), (1 ,5 ), and (2 ,1 )?
Mar 2, 2016
$P = \sqrt{5} + \sqrt{17} + \sqrt{26} \approx 11.4582$
#### Explanation:
We will use the distance formula, which states that the distance between the points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
The triangle is made up of three line segments. We can determine the length of each side through the distance formula and then add them for the entire perimeter of the triangle.
Side length between $\left(3 , 6\right)$ and $\left(1 , 5\right)$:
$\sqrt{{\left(1 - 3\right)}^{2} + {\left(5 - 6\right)}^{2}} = \sqrt{{2}^{2} + {1}^{2}} = \sqrt{5}$
Side length between $\left(1 , 5\right)$ and $\left(2 , 1\right)$:
$\sqrt{{\left(2 - 1\right)}^{2} + {\left(1 - 5\right)}^{2}} = \sqrt{{1}^{2} + {4}^{2}} = \sqrt{17}$
Side length between $\left(2 , 1\right)$ and $\left(3 , 6\right)$:
$\sqrt{{\left(3 - 2\right)}^{2} + {\left(6 - 1\right)}^{2}} = \sqrt{{1}^{2} + {5}^{2}} = \sqrt{26}$
Thus the perimeter of the triangle is
$P = \sqrt{5} + \sqrt{17} + \sqrt{26} \approx 11.4582$
|
Next Meeting: October 20: Web Application Hacking: How to Make and Break Security on the Web Next Installfest: TBD Latest News: Oct. 10: LUGOD Installfests coming again soon Page last updated: 2006 Aug 01 21:32
[vox-tech] webpage element positioning via coordinates
# [vox-tech] webpage element positioning via coordinates
I'd like to position text links in an ellipse, like:
a
h b
g c
f d
e
I may be adding more links, so it behooves me to use a general equation for
the coordinates. To keep the discussion simple, let's talk about a circle.
Is it possible to position something using the equation:
<x,y> = R * <cos(t), sin(t)>
where
t = i 2\pi / n
where n-1 is the number of text links and i runs from 0 to n?
Is it possible to position things via coordinates like this on a web page?
Thanks,
Pete
--
The mathematics of physics has become ever more abstract, rather than more
complicated. The mind of God appears to be abstract but not complicated.
He also appears to like group theory. --- Tony Zee's "Fearful Symmetry"
email: [email protected] web: http://www.dirac.org/p
PGP Fingerprint: B9F1 6CF3 47C4 7CD8 D33E 70A9 A3B9 1945 67EA 951D
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# Thermodynamic databases for pure substances
Thermodynamic databases for pure substances
Thermodynamic databases contain information about thermodynamic properties for substances, the most important being enthalpy, entropy, and Gibbs free energy. Numerical values of these thermodynamic properties are collected as tables or are calculated from thermodynamic datafiles. Data is expressed as temperature-dependent values for one mole of substance at the standard pressure of 101.325 kPa (1 atm), or 100 kPa (1 bar). Unfortunately, both of these definitions for the standard condition for pressure are in use.
Thermodynamic data
Thermodynamic data is usually presented as a table or chart of function values for one mole of a substance (or in the case of the steam tables, one kg). A thermodynamic datafile is a set of equation parameters from which the numerical data values can be calculated. Tables and datafiles are usually presented at a standard pressure of 1 bar or 1 atm, but in the case of steam and other industrially-important gases, pressure may be included as a variable. Function values depend on the state of aggregation of the substance, which must be defined for the value to have any meaning. The state of aggregation for thermodynamic purposes is the "standard state", sometimes called the "reference state", and defined by specifying certain conditions. The "normal" standard state is commonly defined as the most stable physical form of the substance at the specified temperature and a pressure of 1 bar or 1 atm. However, since any non-normal condition could be chosen as a standard state, it must be defined in the context of use. A "physical" standard state is one that exists for a time sufficient to allow measurements of its properties. The most common physical standard state is one that is stable thermodynamically (i.e., the normal one). It has no tendency to transform into any other physical state. If a substance can exist but is not thermodynamically stable (for example, a supercooled liquid), it is called a "metastable" state. A "non"-"physical" standard state is one whose properties are obtained by extrapolation from a physical state (for example, a solid superheated above the normal melting point, or an ideal gas at a condition where the real gas is non-ideal). Metastable liquids and solids are important because some substances can persist and be used in that state indefinitely. Thermodynamic functions that refer to conditions in the normal standard state are designated with a small superscript °. The relationship between certain physical and thermodynamic properties may be described by an equation of state.
Enthalpy, heat content and heat capacity
It is very difficult to measure the absolute amount of any thermodynamic quantity involving the internal energy (e.g. enthalpy), since the internal energy of a substance can take many forms, each of which has its own typical temperature at which it begins to become important in thermodynamic reactions. It is therefore the "change" in these functions that is of most interest. The isobaric change in enthalpy "H" above the common reference temperature of 298.15 K (25 °C) is called the "high temperature heat content", the "sensible heat", or the "relative high-temperature enthalpy", and called henceforth the heat content. Different databases designate this term in different ways; for example "H"T-"H"298, "H"°-"H"°298, "H"°T-"H"°298 or "H"°-"H"°(Tr), where Tr means the reference temperature (usually 298.15 K, but abbreviated in heat content symbols as 298). All of these terms mean the molar heat content for a substance in its normal standard state above a reference temperature of 298.15 K. Data for gases is for the hypothetical ideal gas at the designated standard pressure. The SI unit for enthalpy is J/mol, and is a positive number above the reference temperature. The heat content has been measured and tabulated for virtually all known substances, and is commonly expressed as a polynomial function of temperature. The heat content of an ideal gas is independent of pressure (or volume), but the heat content of real gases varies with pressure, hence the need to define the state for the gas (real or ideal) and the pressure. Note that for some thermodynamic databases such as for steam, the reference temperature is 273.15 K (0 °C).
The "heat capacity" C is the ratio of heat added to the temperature increase. For an incremental isobaric addition of heat:
$C_P\left(T\right)=left \left\{ lim_\left\{Delta T o 0\right\}frac\left\{Delta H\right\}\left\{Delta T\right\} ight \right\}=left \left( frac\left\{partial H\right\}\left\{partial T\right\} ight \right) _p$
"Cp" is therefore the slope of a plot of temperature vs. isobaric heat content (or the derivative of a temperature/heat content equation). The SI units for heat capacity are J/(mol·K).
Enthalpy change of phase transitions
When heat is added to a condensed-phase substance, its temperature increases until a phase change temperature is reached. With further addition of heat, the temperature remains constant while the phase transition takes place. The amount of substance that transforms is a function of the amount of heat added. After the transition is complete, adding more heat increases the temperature. In other words, the enthalpy of a substance changes isothermally as it undergoes a physical change. The enthalpy change resulting from a phase transition is designated Δ"H". There are four types of enthalpy changes resulting from a phase transition. To wit::* "Enthalpy of transformation". This applies to the transformations from one solid phase to another, such as the transformation from α-Fe (bcc ferrite) to $gamma$-Fe (fcc austenite). The transformation is designated Δ"H"tr.:* "Enthalpy of fusion or melting". This applies to the transition of a solid to a liquid and is designated Δ"H"m. :* "Enthalpy of vaporization". This applies to the transition of a liquid to a vapor and is designated Δ"H"v.:* "Enthalpy of sublimation". This applies to the transition of a solid to a vapor and is designated Δ"H"s. "Cp" is infinite at phase transition temperatures because the enthalpy changes isothermally. At the Curie temperature, "Cp" shows a sharp discontinuity while the enthalpy has a change in slope.
Values of Δ"H" are usually given for the transition at the normal standard state temperature for the two states, and if so, are designated with a superscript °. Δ"H" for a phase transition is a weak function of temperature. In some texts, the heats of phase transitions are called "latent" heats (for example, "latent heat of fusion").
Enthalpy change for a chemical reaction
An enthalpy change occurs during a chemical reaction. For the special case of the formation of a compound from the elements, the change is designated Δ"H"form and is a weak function of temperature. Values of Δ"H"form are usually given where the elements and compound are in their normal standard states, and as such are designated "standard heats" of formation, as designated by a superscript °. The Δ"H"°form undergoes discontinuities at a phase transition temperatures of the constituent element(s) and the compound. The enthalpy change for any standard reaction is designated Δ"H"°rx.
Entropy and Gibbs energy
The entropy of a system is another thermodynamic quantity that is not easily measured. However, using a combination of theoretical and experimental techniques, entropy can in fact be accurately estimated. At low temperatures, the Debye model leads to the result that the atomic heat capacity "C"v for solids should be proportional to "T"3, and that for perfect crystalline solids it should become zero at absolute zero. Experimentally, the heat capacity is measured at temperature intervals to as low a temperature as possible. Values of "C"p/T are plotted against T for the whole range of temperatures where the substance exists in the same physical state. The data are extrapolated from the lowest experimental temperature to 0 K using the Debye model. The third law of thermodynamics states that the entropy of a perfect crystalline substance becomes zero at 0 K. When "S"0 is zero, the area under the curve from 0 K to any temperature gives the entropy at that temperature. Even though the Debye model contains "C"v instead of "C"p, the difference between the two at temperatures near 0 K is so small as to be negligible.
The absolute value of entropy for a substance in its standard state at the reference temperature of 298.15 K is designated "S"°298. Entropy increases with temperature, and is discontinuous at phase transition temperatures. The change in entropy (Δ"S"°) at the normal phase transition temperature is equal to the heat of transition divided by the transition temperature. The SI units for entropy are J/(mol·K).
The standard enthalpy change for the formation of a compound from the elements, or for any standard reaction is designated Δ"S"°form or Δ"S"°rx. The entropy change is obtained by summing the absolute entropies of the products minus the sum of the absolute entropies of the reactants.Like enthalpy, the Gibbs energy "G" has no intrinsic value, so it is the change in "G" that is of interest. Furthermore, there is no change in "G" at phase transitions between substances in their standard states.Hence, the main functional application of Gibbs energy from a thermodynamic database is its change in value during the formation of a compound from the standard-state elements, or for any standard chemical reaction (Δ"G"°form or Δ"G"°rx).The SI units of Gibbs energy are the same as for enthalpy (J/mol).
Compilers of thermochemical databases may contain some additional thermodynamic functions. For example, the absolute enthalpy of a substance "H"("T") is defined in terms of its formation enthalpy and its heat content as follows:
$H\left(T\right) = Delta H^circ_\left\{form,298\right\} + \left[H_T - H_\left\{298\right\}\right]$
For an element, "H"("T") and ["H"T - "H"298] are identical at all temperatures because Δ"H"°form is zero, and of course at 298.15 K, "H"("T") = 0. For a compound:
$Delta H^circ_\left\{form\right\} = H\left(T\right)compound - sum left \left\{ H\left(T\right)elements ight \right\}$
Similarly, the absolute Gibbs energy "G"("T") is defined by the absolute enthalpy and entropy of a substance:
$G\left(T\right) = H\left(T\right) - T imes S\left(T\right)$
For a compound:
$Delta G^circ_\left\{form\right\} = G\left(T\right)compound - sum left \left\{ G\left(T\right)elements ight \right\}$
Some tables may also contain the Gibbs energy function ("H"°298.15 – "G"°T)/"T" which is defined in terms of the entropy and heat content.
$\left(H^circ_\left\{298\right\} - G^circ_T\right) / T = S^circ_T - \left(H_T - H_\left\{298\right\}\right) / T$
The Gibbs energy function has the same units as entropy, but unlike entropy, exhibits no discontinuity at normal phase transition temperatures.
The log10 of the equilibrium constant "K"eq is often listed, which is calculated from the defining thermodynamic equation.
$log_\left\{10\right\} left \left( K_\left\{eq\right\} ight \right) = -Delta G^circ_\left\{form\right\} /\left(19.1448T\right)$
Thermodynamic databases
A thermodynamic database consists of sets of critically evaluated values for the major thermodynamic functions.Originally, data was presented as printed tables at 1 atm and at certain temperatures, usually 100° intervals and at phase transition temperatures. Some compilations included polynomial equations that could be used to reproduce the tabular values. More recently, computerized databases are used which consist of the equation parameters and subroutines to calculate specific values at any temperature and prepare tables for printing. Computerized databases often include subroutines for calculating reaction properties and displaying the data as charts.
Thermodynamic data comes from many types of experiments, such as calorimetry, phase equilibria, spectroscopy, composition measurements of chemical equilibrium mixtures, and emf measurements of reversible reactions. A proper database takes all available information about the elements and compounds in the database, and assures that the presented results are "internally consistent". Internal consistency requires that all values of the thermodynamic functions are correctly calculated by application of the appropriate thermodynamic equations. For example, values of the Gibbs energy obtained from high-temperature equilibrium emf methods must be identical to those calculated from calorimetric measurements of the enthalpy and entropy values. The database provider must use recognized data analysis procedures to resolve differences between data obtained by different types of experiments.
All thermodynamic data is a non-linear function of temperature (and pressure), but there is no universal equation format for expressing the various functions. Here we describe a commonly-used polynomial equation to express the temperature dependence of the heat content. A common six-term equation for the isobaric heat content is:
$H_T - H_\left\{298\right\} = A\left(T\right) + B\left(T^2\right) + C\left(T^\left\{-1\right\}\right) + D\left(T^\left\{0.5\right\}\right) + E\left(T^3\right) + F ,$
Regardless of the equation format, the heat of formation of a compound at any temperature is Δ"H"°form at 298.15 K, plus the sum of the heat content parameters of the products minus the sum of the heat content parameters of the reactants. The "C"p equation is obtained by taking the derivative of the heat content equation.
$C_P = A + 2B\left(T\right) - C\left(T^\left\{-2\right\}\right) + extstyle frac \left\{1\right\}\left\{2\right\} D\left(T^\left\{-0.5\right\}\right) + 3E\left(T^2\right) ,$
The entropy equation is obtained by integrating the "C"p/T equation:
$S^circ_T = A\left(ln T\right) + 2B\left(T\right) + extstyle frac \left\{1\right\}\left\{2\right\}C\left(T^\left\{-2\right\}\right) - D\left(T^\left\{ extstyle - frac \left\{1\right\}\left\{2\right) + 1 extstyle frac \left\{1\right\}\left\{2\right\} E\left(T^2\right) + F\text{'}$
F' is a constant of integration obtained by inserting "S"° at any temperature "T". The Gibbs energy of formation of a compound is obtained from the defining equation Δ"G"°form = Δ"H"°form – T(Δ"S"°form), and is expressed as
$Delta G^circ_\left\{form\right\} = \left( Delta A - Delta F\text{'} \right)T - Delta A \left( T ln T \right) - Delta B \left( T^2 \right) + extstyle frac \left\{1\right\}\left\{2\right\} Delta C \left( T^\left\{-1\right\} \right) + 2 Delta D \left( T^\left\{ extstyle frac \left\{1\right\}\left\{2\right\} \right\} \right)$
$- extstyle frac \left\{1\right\}\left\{2\right\} Delta E \left(T^3\right) + Delta F + Delta H^circ_\left\{form 298\right\}$
For most substances, Δ"G"°form deviates only slightly from linearity with temperature, so over a short temperature span, the seven-term equation can be replaced by a three-term equation, whose parameter values are obtained by regression of tabular values.
Depending on the accuracy of the data and the length of the temperature span, the heat content equation may require more or fewer terms. Over a very long temperature span, two equations may be used instead of one. It is unwise to extrapolate the equations to obtain values outside the range of experimental data used to derive the equation parameters.
Thermodynamic datafiles
The equation parameters and all other information required to calculate values of the important thermodynamic functions are stored in a thermodynamic datafile. The values are organized in a format that makes them readable by a thermodynamic calculation program or for use in a spreadsheet. For example, the Excel-based thermodynamic database FREED [http://www.thermart.net] creates the following type of datafile, here for a standard pressure of 1 atm.
:* Row 1. Molar mass of species, density at 298.15 K, Δ"H"°form 298.15, "S"°298.15. and the upper temperature limit for the file. :* Row 2. Number of "C"p equations required. Here, three because of three species phases.:* Row 3. Values of the five parameters for the first "C"p equation; temperature limit for the equation.:* Row 4. Values of the five parameters for the second "C"p equation; temperature limit for the equation.:* Row 5. Values of the five parameters for the third "C"p equation; temperature limit for the equation. :* Row 6. Number of "H"T - "H"298 equations required.:* Row 7. Values of the six parameters for the first "H"T - "H"298 equation; temperature limit for the equation, and Δ"H"°trans for the first phase change.:* Row 8. Values of the six parameters for the second "H"T - "H"298 equation; temperature limit for the equation, and Δ"H"°trans for the second phase change.:* Row 9. Values of the six parameters for the third "H"T - "H"298 equation; temperature limit for the equation, and Δ"H"°trans for the third phase change.:* Row 10. Number of Δ"H"°form equations required. Here five; three for species phases and two because one of the elements has a phase change. :* Row 11. Values of the six parameters for the first Δ"H"°form equation; temperature limit for the equation.:* Row 12. Values of the six parameters for the second Δ"H"°form equation; temperature limit for the equation.:* Row 13. Values of the six parameters for the third Δ"H"°form equation; temperature limit for the equation.:* Row 14. Values of the six parameters for the fourth Δ"H"°form equation; temperature limit for the equation.:* Row 15. Values of the six parameters for the fifth Δ"H"°form equation; temperature limit for the equation.:* Row 16. Number of Δ"G"°form equations required.:* Row 17. Values of the seven parameters for the first Δ"G"°form equation; temperature limit for the equation.:* Row 18. Values of the seven parameters for the second Δ"G"°form equation; temperature limit for the equation.:* Row 19. Values of the seven parameters for the third Δ"G"°form equation; temperature limit for the equation.:* Row 20. Values of the seven parameters for the fourth Δ"G"°form equation; temperature limit for the equation.:* Row 21. Values of the seven parameters for the fifth Δ"G"°form equation; temperature limit for the equation.
Most computerized databases will create a table of thermodynamic values using the values from the datafile. For MgCl2(c,l,g) at 1 atm pressure:
The table format is a common way to display thermodynamic data. The FREED table gives additional information in the top rows, such as the mass and amount composition and transition temperatures of the constituent elements. Transition temperatures for the constituent elements have dashes ------- in the first column in a blank row, such as at 922 K, the melting point of Mg. Transition temperatures for the substance have two blank rows with dashes, and a center row with the defined transition and the enthalpy change, such as the melting point of MgCl2 at 980 K. The datafile equations are at the bottom of the table, and the entire table is in an Excel worksheet. This is particularly useful when the data is intended for making specific calculations.
ee also
* Chemical thermodynamics
* Physical chemistry
* Materials science
* Laws of thermodynamics
* Thermochemistry
* Standard temperature and pressure
* Dortmund Data Bank
References
*cite book|last = Barin|first=Ihsan|year=2004
title=Thermochemical Data of Pure Substances|publisher=Wiley-VCH
id=ISBN 3-527-30993-4
*cite book| last = Chase| first = M. W.
year = 1998| title = NIST - JANAF Thermochemical Tables
edition = Fourth edition| publisher = Journal of Physical and Chemical Reference Data
id =ISBN 1-56396-831-2| url = http://www.aip.org/pubs/books/jpcrd_books.html
*cite book| last = Cox| first=J. D.
coauthors=Wagman, D.D. and V. A. Medvedev, V. A.|year= 1989
title=CODATA Key Values for Thermodynamics|publisher=John Benjamins Publishing Co
id=ISBN 0-89116-758-7
*cite book| last = Hummel|first=Wolfgang
coauthors=Urs Berner, Enzo Curti, F. J. Pearson, and Tres Thoenen|year=2002
title=Nagra/Psi Chemical Thermodynamic Data Base|publisher=Universal Publishers
id=ISBN 1-58112-620-4
*cite book|last=Lide|first=David R.|coauthors=Henry V. Kehiaian|year=1994
title=CRC Handbook of Thermophysical and Thermochemical Data|edition=book & disk edition
id=ISBN 0-8493-0197-1
*cite journal|last=Pankratz|first=L. B.|year=1982
title=Thermodynamic Properties of Elements and Oxides
journal=U. S. Bureau of Mines Bulletin|volume=672
*cite journal|last=Pankratz|first=L. B.|year=1984
title=Thermodynamic Properties of Halides
journal=U. S. Bureau of Mines Bulletin|volume=674
*cite journal|last=Pankratz|first=L. B.|coauthors=A. D. Mah and S. W. Watson|year=1987
title=Thermodynamic Properties of Sulfides
journal=U. S. Bureau of Mines Bulletin|volume=689
*cite journal|last=Pankratz|first=L. B.|year=1994
title=Thermodynamic Properties of Carbides, Nitrides, and Other Selected Substances
journal=U. S. Bureau of Mines Bulletin|volume=696
*Robie, Richard A., and Bruce S. Hemingway (1995). "Thermodynamic Properties of Minerals . . . at Higher Temperatures", U. S. Geological Survey Bulletin 2131.
*Yaws, Carl L. (2007). "Yaws Handbook of Thermodynamic Properties for Hydrocarbons & Chemicals", Gulf Publishing Company. ISBN 1-933762-07-1.
*Gurvich, L.V., Veitz, I.V., et al. (1989) Thermodynamic Properties of Individual Substances. Fourth edition, Hemisphere Pub Co. NY, L., Vol.1 in 2 parts.
* [http://webbook.nist.gov/ NIST WebBook] A gateway to the data collection of the National Institute of Standards and Technology.
* [http://www.thermodata.org THERMODATA] Thermochemical Databases and Softwares.
* [http://dippr.byu.edu DIPPR 801] Critically evaluated thermophysical property database useful for chemical process design and equilibrium calculations.
* [http://www.mtdata-software.com MTDATA software and databases for calculation of thermodynamic properties and phase equilibria]
* [http://www.steamtablesonline.com Free Steam Tables Online] calculator based on IAPWS-IF97
Wikimedia Foundation. 2010.
### См. также в других словарях:
• List of thermodynamic properties — Here is a partial list of thermodynamic properties of fluids:*T temperature [K] * ho density [kg/m3] *c p specific heat at constant pressure [J/(kg·K)] *c v specific heat at constant volume [J/(kg·K)] *mu dynamic viscosity [N/(m²·s)] * u… … Wikipedia
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• Chemical thermodynamics — is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. Chemical thermodynamics involves not only laboratory measurements of various… … Wikipedia
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• Enthalpy of fusion — The standard enthalpy of fusion (symbol: Delta{}H {fus}), also known as the heat of fusion or specific melting heat, is the amount of thermal energy which must be absorbed or evolved for 1 mole of a substance to change states from a solid to a… … Wikipedia
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# Tag Info
14
Although there are already many answers here, I wanted to strongly advocate AGAINST MAC-then-encrypt. I fully agree with Thomas' first half of the answer, but completely disagree with the second half. The ciphertext is the ENTIRE ciphertext (including IV etc.), and this is what must be MACed. This is granted. However, if you MAC-then-encrypt in the ...
11
I assume the question is related to academic work: why do we implement a protocol if we already know how efficient it is by a complexity analysis? The answer depends very much on the type of protocol. However, the answer typically is that a theoretical complexity analysis usually does not suffice to understand the concrete efficiency. If the "previously best ...
10
Do we implement it for proof of concept? Absolutely. It's very easy to miss vital points if no implementation exists. W3C for instance doesn't even allow protocols to be standardized without reference implementation(s). Furthermore, an implementation may show small improvements as well. Personally I would require an implementation of all the (minimal) ...
8
You're describing a form of three-pass protocol, which is a communication mechanism where neither party needs to know each other's secret key. Wikipedia describes a helpful metaphor using a box that can be locked by two padlocks: First, Alice puts the secret message in a box, and locks the box using a padlock to which only she has a key. She then sends ...
6
Short key fingerprints are indeed vulnerable. But those are different from the short-authentication-string (SAS) used by ZRTP. A simple SAS based protocol using one-time keys could look like this: Alice sends a (collision resistant) hash of her public key to Bob. Bob sends his public key to Alice Alice sends her public key to Bob The short ...
6
Given $v_1$ and $v_2$, can the server learn anything about $a$ and $b$? Yes, they can (with high probability) determine whether $a = b$; if $v_2 = 0$, then either $r_1 = 0$ or $a = b$; given that $r_1 = 0$ occurs with probability $1/p$, the attacker can conclude that $a = b$. Now, that's the only thing the attacker can learn; for any observed $v_1, ... 6 My only idea is that B authenticates himself to A, because if A later decrypts it, A will see whether B was able to decrypt it. But why would you need to increment the nonce? Correct, that's the idea. If B didn't need to increment the nonce and just encrypted the same value, the message sent back would be the same that A sent, so an attacker would be ... 6 Even after your updates, the first part seems unnecessary. However, steps 4-5 do indeed prevent the attacker from learning future nonces they could ask the key MAC values for. So the protocol steps 4-7 would be secure with a secure MAC. I agree with CodesInChaos that using HMAC would be better, because H(m||k) has some weaknesses, while HMAC is standard. ... 6 Unfortunately, the answer to your question is yes. You have made glaring mistakes. In particular, Yao's garbled circuits are suited for two-party computation only, and here you wish to carry out a multiparty computation. One huge problem that arises with your entire approach is that if the server colludes with one of the voters, then they can learn the ... 5 At a high level, the major flaw is that you are rolling your own crypto protocol. You should strongly consider using a standardized protocol like DTLS. Some specific problems: Symmetric key distribution is left unspecified. Keys must be changed occasionally to thwart distinguishers. No way to recover from symmetric key compromise. Your message ... 5 I'd say that most of the time the signature is accompanied by the certificate of the signer. This certificate contains the public key. Most container formats such as CMS (used in S/MIME, also known as PKCS#7) or XML digsig contain specific fields that may contain certificates - and usually do. When the certificate is received the Public Key Infrastructure ... 5 Your requirements are not terribly precise, so here is what I think you mean: "The result must be trusted by all three participants" ==> Even if Alice & Bob are both malicious & colluding, the output of Carol should be uniform. Also, all 3 should get the same output. "The coin is flipped only by Alice and Bob" ==> Alice & Bob do all the work. ... 5 There are many ways of doing this. A very nice read (but with informal presentation) is this paper by Fagin, Naor and Winkler on Comparing Information without Leaking It. A very fast protocol exists which requires a single oblivious transfer for every bit. Let$n$be a security parameter; say$n=128$, and let$\ell$be the bit-length of the inputs. For ... 5 Is there a protocol that A and B can use to find out the same thing without having to trust the other and any third parties? There is. Even more than one. Your problem actually is equivalent to Yao's Millionaire's Problem. You have two numbers which two parties want to keep secret and you want both parties to find out whether the one is larger than ... 4 This is exactly where automatic protocol analysis tools can help you. For example, using the Scyther tool, the protocol description using symmetric encryption is: /* * Protocol description for Scyther * * Note we use 'K' to model 'k' since Scyther assumes 'k(.,.)' refers * to pre-shared keys between two agents. */ // The protocol description with ... 4 This protocol doesn't authenticate the mote at all. Consider this attack: Mote B sends a 'hello' message to Base. This message contains the ID# of Mote A and a random nonce [R] (HW generated) encrypted by the base's public key. Base decrypts the 'hello' and verifies the ID# against a whitelist. Base sends an 'ack' message. This message contains some ... 4 You can use any library you like, as long as it is has been tested for the specific algorithm. In other words, if$G^x$is implemented in a specific library you must make sure that there are unit tests and if it is used in a verified algorithm. There are some hints you can take from the library to see if it was programmed well: the code should point to ... 4 First up: it does use public keys in contrast to your claims. To be more specific –$q$is Alice’s public key, and$f$is Bob‘s public key. Both are transferred in public and might be intercepted by a MITM. This brings me to the next point: the system you worked out in your head is highly insecure. We'll call the message$p\$ and encode it as a number. ...
4
This is an extension of @SEJPM's answer. I want to expand on what protocols are best to use (I apologize ahead of time for self citations). First, for details on Yao's protocol, see A Proof of Security of Yao's Protocol for Two-Party Computation. However, to do this very efficiently, you need to have two very efficient components: A fast garbling scheme: ...
4
TL;DR: Find the most important specification that is of the same type as the one you want to write and use its style for yourself, chances are, other people also have read it. There is a myriad of ways to specify a crypto protocol / design. However, there are four things that you really should take into consideration when writing a crypto-related ...
4
Yes the Bernstein attack is applicable but the impact of the attack is reduced because the party generating the parameter is also going to be a legitimate participant of the key exchange. Here is why the attack does indeed apply. Consider a case where Bob and Alice wish to conduct a key exchange using the New Hope Lattice-Based Key Exchange. Bob will be ...
3
The "interesting" part of your encryption is here: Therefore, I prepend a block at the beginning of my packet. Its content goes as follows: First four bytes: current timestamp in seconds Next 12 bytes: zeros I compute the sha256 hash of the message (32 bytes) I xor the timestamp + zeros block with the first half of the hash I xor the ...
3
An algorithm which is secure even if the enemy acquires everything but the key may be regarded as a means of generating secure algorithms. If one presently has a secure channel for communicating with a correspondent, and will need to communicate securely in future when no secure channel is available, using some dice to generate a random key and conveying it ...
3
With a hash function that is vulnerable to length extension attacks, like SHA-256, you can turn any random collision into a collision with that random string concatenated with some (partially) chosen data. In any use case where random initial data does not matter, you could use it to generate two documents which have the same hash value and thus the same ...
3
My question is how do I authenticate my App to the CA, to prevent something else to request these Client Certificates? There is generally no way to authenticate the client code. Any secret you embed in the app could be extracted. You must assume an attacker can send requests that an authentic client would. Instead, what you can do is authenticate the ...
3
No, this will not work, for two fundamental reasons: You cannot "encrypt with the private key and decrypt with the public key" in any meaningful sense.* And if you could, it would be totally useless, because the public key is, by definition, public — if you could decrypt with the public key, so could anyone else. In particular, in your scheme, ...
3
This is standard Encrypt-then-Authenticate. The only difference is that when doing EtA, it actually isn't necessary to encrypt everything. This strategy makes sense when there is some part of the message that needs integrity and not privacy. In IPSec, the ICV (which is a counter to prevent replay) does not need privacy. Furthermore, by not encrypting it, it ...
3
To answer your first question, the incrementation is required in order to prevent spoofing of that message. An attacker could send back the same encrypted nonce claiming to be Bob. However, if Bob incriments the nonce and sends it back encrypted, Alice would know for sure that Bob has received the nonce and has incremented it. Now, Alice encrypting the ...
3
I'm not sure I understand your question entirely. If there is only one possible message, then the ciphertext can be trivially decrypted simply by choosing this message. I'll assume instead that the ciphertext contains the shuffled bit pattern of a name chosen from a set of more than one name. The problem with bit shuffling is that the number of set bits ...
3
This might not be the answer you are looking for, but as you are looking for a formal verification, I would advise you to take a look at Coq. Even though mainly used by Academics, it provides a logical framework and an interface to write formal and interactive proofs. Based this language there exists some libraries dedicated to cryptographic proof : ...
Only top voted, non community-wiki answers of a minimum length are eligible
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# MathML Block
## Descrição
A MathML block for the WordPress block editor (Gutenberg).
Requires PHP 5.4+ and WordPress 5.0+.
Development takes place on the GitHub repository: https://github.com/adamsilverstein/mathml-block.
Screencast: https://cl.ly/c0f6bbfbc3b1
### What is MathML?
Mathematical Markup Language is a mathematical markup language, an application of XML for describing mathematical notations and capturing both its structure and content. It aims at integrating mathematical formulae into World Wide Web pages and other documents.
The MathML block uses MathJax to render MathML formulas in the editor and on the front end of a website. MathJax (https://www.mathjax.org/) is A JavaScript display engine for mathematics that works in all browsers.
To test a MathML block and enter a formula, for example: $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$.
To test using math formulas inline, type an formula into a block of text, select it and hit the ‘M’ icon in the control bar. For example: $$\cos(θ+φ)=\cos(θ)\cos(φ)−\sin(θ)\sin(φ)$$. Note: if you are copying and pasting formulas into the rich text editor, switching to HTML/code editor mode is less likely to reformat your pasted formula.
### Technical Notes
• Requires PHP 5.4+.
• Requires WordPress 5.0+.
• Issues and Pull requests welcome on the GitHub repository: https://github.com/adamsilverstein/mathml-block.
## Blocos
Este plugin disponibiliza 1 bloco.
mathml/mathmlblock
MathML
## Instalação
1. Install the plugin via the plugin installer, either by searching for it or uploading a .zip file.
2. Activate the plugin.
3. Use the MathML block!
## Avaliações
### Top !
10 de outubro de 2019
The best way I have found to write math formulas on the web. I just need to remember the LaTex syntax I haven't used in 15 years!
### Good but a little too greedy
21 de maio de 2019
Really too greedy the MathJax js files are loaded everywhere, even outside the editor and pages, it makes the website too heavy, but it is very clean visually and very simple certainly one of the easiest to use!
Leia todas as 2 avaliações
## Contribuidores e desenvolvedores
“MathML Block” é um software com código aberto. As seguintes pessoas contribuíram para este plugin.
Contribuidores
Traduzir “MathML Block” para o seu idioma.
Navegue pelo código, dê uma olhada no repositório SVN ou assine o registro de desenvolvimento via RSS.
## Registro de alterações
#### 1.1.1
• Improve translations, make JavaScript translatable.
• Update all packages.
#### 1.1.0
• Add support for inline formulas.
#### 1.0.0
• Initial plugin release
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# What will happen to LOGSPACE if P=PSPACE?
If P=PSPACE then by padding argument EXPTIME=EXPSPACE.But what about class L(Logarithmic space)?What will it be equal to?I can't think it as equal to DLOGTIME.There are problems that can be solved in constant space which are contained in L.But many of them can't be solved in logarithmic time like linear search. So do they belong to a certain subclass of P?
Equalities between complexity classes transfer upwards by padding arguments, not downwards. The assumption $$\mathrm{P = PSPACE}$$ is not known to imply the collapse of $$\mathrm L$$ to anything else. (It does, however, imply $$\mathrm{L\ne P}$$.)
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# Efficient compression of unlabeled trees
Consider unlabeled, rooted binary trees. We can compress such trees: whenever there are pointers to subtrees $T$ and $T'$ with $T = T'$ (interpreting $=$ as structural equality), we store (w.l.o.g.) $T$ and replace all pointers to $T'$ with pointers to $T$. See uli's answer for an example.
Give an algorithm that takes a tree in the above sense as input and computes the (minimal) number of nodes that remain after compression. The algorithm should run in time $\cal{O}(n\log n)$ (in the uniform cost model) with $n$ the number of nodes in the input.
This has been an exam question and I have not been able to come up with a nice solution, nor have I seen one.
• And what is “the cost”, “the time”, the elementary operation here? The number of nodes visited? The number of edges traversed? And how is the size of the input specified? – uli Mar 9 '12 at 18:11
• This tree compression is an instance of hash consing. Not sure if that leads to a generic counting method. – Gilles 'SO- stop being evil' Mar 9 '12 at 20:35
• @uli I clarified what $n$ is. I think "time" is specific enough, though. In non-concurrent settings, this is equivalent to counting operations which is in Landau terms equivalent to counting the elementary operation occuring most often. – Raphael Mar 10 '12 at 11:14
• @Raphael Of course I can take a guess what the intended elementary operation should be and will probably pick the same as everybody else. But, and I know I am pedantic here, whenever “time bounds” are given it is important to state what is being counted. Is it swaps, compares, additions, memory accesses, inspected nodes, traversed edges, you name it. It is like omitting the unit of measurement in physics. Is it $10\,\mathrm{kg}$ or $10\,\mathrm{ms}$? And I suppose memory accesses are almost always the most frequent operation. – uli Mar 10 '12 at 11:44
• @uli These are the sort of details that “uniform cost model” is supposed to convey. It's painful to define precisely what operations are elementary, but in 99.99% of cases (including this one) there's no ambiguity. Complexity classes fundamentally do not have units, they do not measure the time it takes to perform one instance but the way this time varies as the input gets larger. – Gilles 'SO- stop being evil' Mar 10 '12 at 14:07
Yes, you can perform this compression in $O(n \log n)$ time, but it is not easy :) We first make some observations and then present the algorithm. We assume the tree is initially not compressed - this is not really needed but makes analysis easier.
Firstly, we characterize 'structural equality' inductively. Let $T$ and $T'$ be two (sub)trees. If $T$ and $T'$ are both the null trees (having no vertices at all), they are structurally equivalent. If $T$ and $T'$ are both not null trees, then they are structurally equivalent iff their left children are structurally equivalent and their right children are structurally equivalent. 'Structural equivalence' is the minimal fixed point over these definitions.
For example, any two leaf nodes are structurally equivalent, as they both have the null trees as both their children, which are structurally equivalent.
As it is rather annoying to say 'their left children are structurally equivalent and so are their right children', we will often say 'their children are structurally equivalent' and intend the same. Also note we sometimes say 'this vertex' when we mean 'the subtree rooted at this vertex'.
The above definition immediately gives us a hint how to perform the compression: if we know the structural equivalence of all subtrees with depth at most $d$, then we can easily compute the structural equivalence of the subtrees with depth $d+1$. We do have to do this computation in a smart way to avoid a $O(n^2)$ running time.
The algorithm will assign identifiers to every vertex during its execution. An identifier is a number in the set $\{ 1, 2, 3, \dots, n \}$. Identifiers are unique and never change: we therefore assume we set some (global) variable to 1 at the start of the algorithm, and every time we assign an identifier to some vertex, we assign the current value of that variable to the vertex and increment the value of that variable.
We first transform the input tree into (at most $n$) lists containing vertices of equal depth, together with a pointer to their parent. This is easily done in $O(n)$ time.
We first compress all the leaves (we can find these leaves in the list with vertices of depth 0) into a single vertex. We assign this vertex an identifier. Compression of two vertices is done by redirecting the parent of either vertex to point to the other vertex instead.
We make two observations: firstly, any vertex has children of strictly smaller depth, and secondly, if we have performed compression on all vertices of depth smaller than $d$ (and have given them identifiers), then two vertices of depth $d$ are structurally equivalent and can be compressed iff the identifiers of their children coincide. This last observation follows from the following argument: two vertices are structurally equivalent iff their children are structurally equivalent, and after compression this means their pointers are pointing to the same children, which in turn means the identifiers of their children are equal.
We iterate through all the lists with nodes of equal depth from small depth to large depth. For every level we create a list of integer pairs, where every pair corresponds to the identifiers of the children of some vertex on that level. We have that two vertices in that level are structurally equivalent iff their corresponding integer pairs are equal. Using lexicographic ordering, we can sort these and obtain the sets of integer pairs that are equal. We compress these sets to single vertices as above and give them identifiers.
The above observations prove that this approach works and results in the compressed tree. The total running time is $O(n)$ plus the time needed to sort the lists we create. As the total number of integer pairs we create is $n$, this gives us that the total running time is $O(n \log n)$, as required. Counting how many nodes we have left at the end of the procedure is trivial (just look at how many identifiers we have handed out).
• I haven't read your answer in detail, but I think you've more or less reinvented hash consing, with a weird problem-specific way of looking up nodes. – Gilles 'SO- stop being evil' Mar 10 '12 at 2:56
• @Alex “children of strictly smaller degree” degree should probably be depth? And despite CS-trees growing downward I find “height of a tree” less confusing than “depth of a tree”. – uli Mar 10 '12 at 3:40
• Nice answer. I feel like there should be a way to get around sorting. My second comment on @Gilles answer is valid here, too. – Raphael Mar 10 '12 at 11:10
• @uli: yup, you're right, I've corrected it (not sure why I confused those two words). Height and depth are two subtly different concepts, and I needed the latter :) I thought I'd stick to the conventional 'depth' rather than confuse everyone by swapping them. – Alex ten Brink Mar 10 '12 at 11:39
Compressing a non-mutable data structure so that it does not duplicate any structurally equal subterm is known as hash consing. This is an important technique in memory management in functional programming. Hash consing is a sort of systematic memoization for data structures.
We're going to hash-cons the tree and count the nodes after hash consing. Hash consing a data structure of size $n$ can always be done in $O(n\:\mathrm{lg}(n))$ operations; counting the number of nodes at the end is linear in the number of nodes.
I will consider trees as having the following structure (written here in Haskell syntax):
data Tree = Leaf
| Node Tree Tree
For each constructor, we need to maintain a mapping from its possible arguments to the result of applying the constructor to these arguments. Leaves are trivial. For nodes, we maintain a finite partial map $\mathtt{nodes} : T \times T \to N$ where $T$ is the set of tree identifiers and $N$ is the set of node identifiers; $T = N \uplus \{\ell\}$ where $\ell$ is the sole leaf identifier. (In concrete terms, an identifier is a pointer to a memory block.)
We can use a logarithmic-time data structure for nodes, such as a balanced binary search tree. Below I'll call lookup nodes the operation that looks up a key in the nodes data structure, and insert nodes the operation that adds a value under a fresh key and returns that key.
Now we traverse the tree and add the nodes as we go along. Although I'm writing in Haskell-like pseudocode, I'll treat nodes as a global mutable variable; we'll only ever be adding to it, but the insertions need to be threaded throughout. The add function recurses on a tree, adding its subtrees to the nodes map, and returns the identifier of the root.
insert (p1,p2) =
add Leaf = $\ell$
case lookup nodes (p1,p2) of
Nothing -> insert nodes (p1,p2)
Just p -> p
The number of insert calls, which is also the final size of the nodes data structure, is the number of nodes after maximum compression. (Add one for the empty tree if needed.)
• Can you give a reference for "Hash consing a data structure of size $n$ can always be done in $O(nlg(n))$ operations"? Note that you will need balanced trees for nodes in order to achieve the desired runtime. – Raphael Mar 10 '12 at 11:02
• I was only considering hashing substructures to numbers in a structured way so that independently computing the hash for the same tree would always yield the same result. Your solution is fine, too, provided we have mutable datastructures on our hands. I think it can be cleaned up a tad, though; the interleaving of insert and add should be made explicit and a function that actually solves the problem should be given, imho. – Raphael Mar 10 '12 at 11:03
• @Raphael Hash consing relies on a finite map structure over tuples of pointers/identifiers, you can implement that with logarithmic time for lookup and add (e.g. with a balanced binary search tree). My solution does not require mutability; I make nodes a mutable variable for convenience, but you can thread it throughout. I'm not going to give full code, this is not SO. – Gilles 'SO- stop being evil' Mar 10 '12 at 14:02
• @Raphael Hashing structures, as opposed to assigning them arbitrary numbers, is a bit dodgy. In the uniform cost model, you can encode anything into a large integer and do constant-time operations on it, which is not realistic. In the real world, you can use cryptographic hashes to have a de facto one-to-one mapping from infinite sets to a finite range of integers, but they're slow. If you use a non-crypto checksum as the hash, you need to think about collisions. – Gilles 'SO- stop being evil' Mar 10 '12 at 14:10
Here is another idea that aims at (injectively) encoding the structure of trees into numbers, rather than just labelling them arbitrarily. For that, we use that any number's prime factorisation is unique.
For our purposes, let $E$ denote an empty position in the tree, and $N(l,r)$ a node with left subtree $l$ and right subtree $r$. $N(E,E)$ would be a leaf. Now, let
\qquad \displaystyle \begin{align*} f(E) &= 0 \\ f(N(l,r)) &= 2^{f(l)}\cdot 3^{f(r)} \end{align*}
Using $f$, we can compute the set of all subtrees contained in a tree bottom-up; in every node, we merge the sets of encodings obtained from the children and add a new number (which can be computed in constant time from the children's encodings).
This last assumption is a stretch on real machines; in this case, one would prefer to use something similar to Cantor's pairing function instead of exponentiation.
The runtime of this algorithm depends on the structure of the tree (on balanced trees, $\cal{O}(n \log n)$ with any set implementation that allows union in linear time). For general trees, we would need logarithmic time union with a simple analysis. Maybe a sophisticated analysis can help, though; note that the usual worst-case tree, the linear list, admits $\cal{O}(n \log n)$ time here, so it is not so clear what the worst-case may be.
As pictures are not allowed in comments:
top left: an input tree
top right: the subtrees rooted in nodes 5 and 7 are isomorphic too.
lower left and right: the compressed trees are not uniquely defined.
Note that in this case the size of the tree has gone down from $7+5|T|$ too $6+|T|$.
• This is indeed an example of the desired operation, thanks. Note that your final examples are identical if you do not distinguish between original and added references. – Raphael Mar 10 '12 at 10:59
Edit: I read the question as T and T′ were children of the same parent. I took the definition of compression to be recursive as well, meaning you could compress two previously compressed subtrees. If that's not the actual question, then my answer may not work.
$O(n \log n)$ begs for a $T(n) = 2T(n/2) + cn$ divide and conquer solution. Recursively compress nodes and compute the number of descendants in each subtree after compression. Here's some python-esque pseudo code.
def Comp(T):
if T == null:
return 0
leftCount = Comp(T.left)
rightCount = Comp(T.right)
if leftCount == rightCount:
if hasSameStructure(T.left, T.right):
T.right = T.left
return leftCount + 1
else
return leftCount + rightCount + 1
Where hasSameStructure() is a function that compares two already compressed subtrees in linear time to see if they have the exact same structure. Writing a linear time recursive function that traverses each and checks if the one subtree has a left child every time the other one does etc. shouldn't be hard.
Let $n_\ell$ and $n_r$ be the sizes of the left and right subtrees respectively (after compression). Then the running time is $$T(n) = T(n_1) + T(n_2) + O(1) \mbox{ if n_\ell \neq n_r}$$ and $$2T(n/2) + O(n) \mbox{ otherwise}$$
• What if the subtrees are not siblings? Care for ((T1,T1),(T2,T1)) T1 can be saved twice by using a pointer two the third occurence. – uli Mar 9 '12 at 20:01
• @uli I'm not sure what you're saying. I read the question as $T$ and $T'$ were children of the same parent. If that's not the actual question, then my answer may not work. I took the definition of compression to be recursive as well, meaning you could compress two previously compressed subtrees. – Joe Mar 9 '12 at 20:28
• The questions merly states that two subtress are identified as isomorphic. Nothing is said about them having the same parent. If a subtree T1 appears three times in a tree, as in my previous example ((T1,T1),(T1,T2)) two occurences can be compressed by pointing to the third orccurence. – uli Mar 9 '12 at 20:35
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# algorithm2e - why are some of my texts are italicized and some are not
I am using algorithm2e. These are my declarations:
\usepackage{cvpr}
\usepackage{times}
\usepackage{epsfig}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
% For tables
\usepackage{tabu}
%\usepackage{caption} %\captionsetup[table]{skip=10pt}
\bibliographystyle{unsrtnat}
\usepackage[numbers,sort&compress]{natbib}
\usepackage[linesnumbered,boxed]{algorithm2e}
\usepackage{algpseudocode}
% Include other packages here, before hyperref.
% If you comment hyperref and then uncomment it, you should delete
% egpaper.aux before re-running latex. (Or just hit 'q' on the first latex
% run, let it finish, and you should be clear).
This is my code. Why are some of the texts italicized and some are not ? What is the function to make a text italicized and to not make it italicized ?
\begin{algorithm}
\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output}
\Input{a graph G = (V,E)}
\Output{a hierarchical tree}
\BlankLine
Initialize edge weights\\
\ForEach{\normalfont{vertex} V_i \subset V } % why is V_i italic but V not ?
{
j \leftarrow \text{argmin}_k(\text{cost}(V_i, V_k)), \ k \subset \text{neighbors} \ \text{of} \ i \\
E_{ij} \leftarrow E_{ij} + 1 \\
}
\BlankLine
\ForEach{\normalfont{edge} E_{ij} \subset E }
{
E_{ij} \leftarrow argmin(E_{ij} \normalfont{.weight}), \ j \subset N ) \\
E_{ij} \leftarrow E_{ij} + 1 \\
}
\end{algorithm}
• Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See minimal working example (MWE). – Henri Menke Jul 18 '17 at 4:52
There are a number of things wrong with your current approach/usage of algorithm2e. In no particular order:
1. Use \; as line endings, not \\. If you don't want the semi-colons to be printed, add \DontPrintSemicolon to your preamble.
2. Surround your math content by $...$.
3. For consistency, define commands that do stuff. For example, formatting a "variable" in your pseudocode, one could define
\newcommand{\var}{\texttt}
and use \var for every variable.
4. The first argument of \ForEach (and other conditional clauses in algorithm2e) is set using \itshape. If you want it to not be italics, then set it using {\upshape ...} or \textup{...}.
5. Use mathptmx rather than the obsolete times.
\documentclass{article}
\usepackage{mathptmx,amsmath}
\usepackage[linesnumbered,boxed]{algorithm2e}
\DontPrintSemicolon
\DeclareMathOperator{\argmin}{argmin}% https://tex.stackexchange.com/q/5223/5764
\newcommand{\var}{\texttt}
\begin{document}
\begin{algorithm}
\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output}
\Input{a graph $G = (V,E)$}
\Output{a hierarchical tree}
\BlankLine
Initialize edge weights\;
\ForEach{ \textup{vertex} $V_i \subset V$ }
{
$j \leftarrow \argmin_k (\var{cost}(V_i, V_k))$ where $k \subset \text{neighbours of$i$}$\;
$E_{ij} \leftarrow E_{ij} + 1$\;
}
\BlankLine
\ForEach{ \textup{edge} $E_{ij} \subset E$ }
{
$E_{ij} \leftarrow \argmin (E_{ij} \text{.weight})$ where $j \subset N$\;
$E_{ij} \leftarrow E_{ij} + 1$\;
}
\end{algorithm}
\end{document}
As a side-note, I find the use of
$E_{ij} \leftarrow E_{ij} + 1$\;
superfluous, as the pseudocode construction already indicates that you're going through each E_{ij}. Moreover, what does E_{ij} + 1 refer to?
• Hi, thanks for your answer. I think \DontPrintSemicolons should be \DontPrintSemicolon instead. – aerdna91 Aug 4 '18 at 14:41
I agree with answer above. I will add the following: all styles can be redefined is algorithms, but you have to understand how are understood texts. Firts, if this math, this will be display as math, it means in italic Second, it depend if this is normal text or text of keyword, functions, argument of algorithm command, etc...
I enclose an exemple based on your code to show how could code your algorithm: 1) use math mode, as said above, when you write math 2) define your own variable macro, again as said above: you can use SetKwData macro of algorithm style and so SetDataSty to control the style of your variables 3) define functions of your algorithm (argmin and cost) with SetKwFunction 4) redefine style of algorithm text using macros provided by the algorithm
Here is my example based on yours (in particular, note differences between vertex that is typed as argument of ForEach and then in ArgSty, and edge that is defined as a Data (variable)
\documentclass[a4paper]{article}
\usepackage[lined,linesnumbered,boxed]{algorithm2e}
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ArticlePDF Available
# The ALMA-PILS survey: First detections of deuterated formamide and deuterated isocyanic acid in the interstellar medium
Authors:
## Abstract
Formamide (NH2CHO) has previously been detected in several star-forming regions and is thought to be a precursor for different prebiotic molecules. Its formation mechanism is still debated, however. Observations of formamide, related species, and their isopotologues may provide useful clues to the chemical pathways leading to their formation. The Protostellar Interferometric Line Survey (PILS) represents an unbiased, high angular resolution and sensitivity spectral survey of the low-mass protostellar binary IRAS 16293–2422 with the Atacama Large Millimeter/submillimeter Array (ALMA). For the first time, we detect the three singly deuterated forms of NH2CHO (NH2CDO, cis- and trans-NHDCHO), as well as DNCO towards the component B of this binary source. The images reveal that the different isotopologues are all present in the same region. Based on observations of the 13C isotopologues of formamide and a standard 12C/ 13C ratio, the deuterium fractionation is found to be similar for the three different forms with a value of about 2%. The DNCO/HNCO ratio is also comparable to the D/H ratio of formamide (∼1%). These results are in agreement with the hypothesis that NH2CHO and HNCO are chemically related through grain-surface formation.
Astronomy &Astrophysics manuscript no. NH2CHO_PILS_aa_v6 c
ESO 2016
May 6, 2016
Letter to the Editor
The ALMA-PILS survey: First detections of deuterated formamide
and deuterated isocyanic acid in the interstellar medium
A. Coutens1, J. K. Jørgensen2, M. H. D. van der Wiel2, H. S. P. Müller3, J. M. Lykke2, P. Bjerkeli2,4, T. L. Bourke5, H.
Calcutt2, M. N. Drozdovskaya6, C. Favre7, E. C. Fayolle8, R. T. Garrod9, S. K. Jacobsen2, N. F. W. Ligterink6, K. I.
Öberg8, M. V. Persson6, E. F. van Dishoeck6,10, and S. F. Wampfler2
1Department of Physics and Astronomy, University College London, Gower St., London, WC1E 6BT, UK
e-mail: [email protected]
2Centre for Star and Planet Formation, Niels Bohr Institute & Natural History Museum of Denmark, University of Copenhagen,
Øster Voldgade 5-7, DK-1350 Copenhagen K., Denmark
3I. Physikalisches Institut, Universität zu Köln, Zülpicher Str. 77, 50937 Köln, Germany
4Department of Earth and Space Sciences, Chalmers University of Technology, Onsala Space Observatory, 439 92 Onsala, Sweden
5SKA Organization, Jodrell Bank Observatory, Lower Withington, Macclesfield, Cheshire SK11 9DL, UK
6Leiden Observatory, Leiden University, PO Box 9513, NL-2300 RA Leiden, the Netherlands
7Institut de Planétologie et d’Astrophysique de Grenoble, UMR 5274, UJF-Grenoble 1/CNRS, 38041 Grenoble, France
8Harvard-Smithsonian Center for Astrophysics, 60 Garden Street, Cambridge, MA 02138, USA
9Departments of Chemistry and Astronomy, University of Virginia, Charlottesville, VA 22904, USA
10 Max-Planck Institut für Extraterrestrische Physik (MPE), Giessenbachstr. 1, 85748 Garching, Germany
ABSTRACT
Formamide (NH2CHO) has previously been detected in several star-forming regions and is thought to be a precursor for dierent
prebiotic molecules. Its formation mechanism is still debated, however. Observations of formamide, related species and their isopo-
tologues may provide useful clues to the chemical pathways leading to their formation. The Protostellar Interferometric Line Survey
(PILS) represents an unbiased high angular resolution and sensitivity spectral survey of the low-mass protostellarbinary IRAS 16293–
2422 with the Atacama Large Millimeter/submillimeter Array (ALMA). We detect for the first time the three singly deuterated forms
of NH2CHO (NH2CDO, cis- and trans-NHDCHO) as well as DNCO towards the component B of this binary source. The images
reveal that the dierent isotopologues all are present in the same region. Based on the observations of the 13C isotopologues of for-
mamide and a standard 12C/13 C ratio, the deuterium fractionation is found to be similar for the three dierent forms with a value of
about 2%. The DNCO/HNCO ratio is also comparable to the D/H ratio of formamide (1%). These results are in agreement with the
hypothesis that NH2CHO and HNCO are chemically related through grain surface formation.
Key words. astrochemistry – astrobiology – stars: formation – stars: protostars – ISM: molecules – ISM: individual object
(IRAS 16293–2422)
1. Introduction
Formamide (NH2CHO), also known as methanamide, contains
the amide bond (–N–C(=O)–), which plays an important role in
the synthesis of proteins. This molecule is a precursor for poten-
tial compounds of genetic and metabolic interest (Saladino et al.
2012). Interestingly, it is present in various astrophysical envi-
ronments: high-mass star-forming regions (e.g, Bisschop et al.
2007; Adande et al. 2013), low-mass protostars (Kahane et al.
2013; López-Sepulcre et al. 2015), shocked regions (Yamaguchi
et al. 2012; Mendoza et al. 2014), a translucent cloud (Corby
et al. 2015), comets (Bockelée-Morvan et al. 2000; Biver et al.
2014; Goesmann et al. 2015) and even an extragalactic source
(Muller et al. 2013).
The formation of formamide is still not clearly understood:
several routes have been proposed, both in the gas phase and
on the grain surfaces. In the gas phase, many ion-molecule re-
actions have been ruled out as not suciently ecient due to
endothermicity or high energy barriers (see e.g. Redondo et al.
2014a,b). A neutral-neutral reaction between H2CO and NH2
was however shown to be barrierless and could account for the
abundance of formamide in some sources (Barone et al. 2015).
On the grain surface, formamide is suggested to form through
the reaction between HCO and NH2(Jones et al. 2011; Garrod
2013) and/or hydrogenation of isocyanic acid, HNCO. In partic-
ular, the latter suggestion is supported by a strong correlation be-
tween the HNCO and NH2CHO abundances in dierent sources
(Bisschop et al. 2007; Mendoza et al. 2014; López-Sepulcre et al.
2015). However, an experiment based on the H bombardment of
HNCO at low temperature has recently shown that this reaction
is not ecient in cold environments (Noble et al. 2015). Instead,
other pathways to HNCO and NH2CHO on grains have been
suggested, either with or without UV or ion bombardment (see
e.g. Kaˇ
nuchová et al. 2016 and references therein).
Measurements of isotopic fractionation may help to con-
strain formation pathways of molecules as isotopic fractiona-
tion (especially deuteration) is sensitive to physical conditions
such as density and temperature. Until recently, the study of
deuteration in solar-type protostars was mainly limited to rel-
atively small and abundant molecules, such as H2O, HCO+,
Article number, page 1 of 10
A&A proofs: manuscript no. NH2CHO_PILS_aa_v6
HCN, H2CO, and CH3OH. Even though the deuterium frac-
tionation is known to be enhanced in low-mass protostars (see
e.g., Ceccarelli et al. 2007), measurements of lines of deuter-
ated complex organic molecules (COMs) still require high sen-
sitivity observations. So far, only deuterated methyl formate
and dimethyl ether have been detected towards the low-mass
protostar IRAS 16293–2422 (hereafter IRAS16293) by Demyk
et al. (2010) and Richard et al. (2013). With the Atacama Large
Millimeter/submillimeter Array (ALMA), it is now possible
to search for the isotopologues of complex and less abundant
species. In this Letter, we report the first detection of the three
singly deuterated forms of formamide as well as DNCO to-
wards IRAS16293. These observations mark the first detections
of those isotopologues in the interstellar medium.
2. Observations
An ALMA unbiased spectral survey of the binary protostar
IRAS16293 was recently carried out in the framework of the
“Protostellar Interferometric Line Survey”1(PILS; Jørgensen
et al. submitted). The observations were centered on a position
at equal distance between the sources A and B that are separated
by 500. A full description of the survey and the data reduction
can be found in Jørgensen et al. (submitted). For this work, we
use the part of the large spectral survey obtained in Band 7 be-
tween 329.15 GHz and 362.90 GHz both with the 12m array and
the Atacama Compact Array (ACA). The spectral resolution of
these observations is 0.244 MHz (i.e. 0.2 km s1). After combi-
nation of the 12m and ACA data, the final spectral line datacubes
show a sensitivity better than 5 mJy beam1km s1. The beam
sizes range between 0.400 and 0.700. Additional observations in
Bands 3 and 6 cover narrow spectral ranges and consequently a
very limited number of transitions of formamide isotopologues.
After the analysis of Band 7, we checked that the results are con-
sistent with these lower frequency observations.
3. Analysis and results
To search for the isotopologues of formamide, we use the spec-
trum extracted at the same position as in Lykke et al. (to be sub-
mitted), i.e. a position oset by 0.500 from the continuum peak
of source B in the South West direction (αJ2000=16h32m22s
.58,
δJ2000=-2428032.800 ). Although the lines are brighter at the po-
sition of the continuum peak, the presence of both absorption
and emission makes analysis dicult. At the selected position,
most of the lines present Gaussian profiles and are relatively
bright compared to other positions. In source A, the lines are
the search for isotopologues of complex species (e.g. Jørgensen
et al. 2012). This Letter is therefore focused on source B only.
We identify several unblended lines that can be assigned
to the three singly deuterated forms of NH2CHO and to
NH213CHO, DNCO, and HN13 CO (see Table 1). These mark
the first detections of NH2CDO, cis-NHDCHO, trans-NHDCHO
and DNCO in the interstellar medium. The list of unblended
lines can be found in the Appendix. Maps of the integrated line
emission from representative lines from the dierent isotopo-
logues towards source B are shown in Figure 1. The emission
of the dierent lines clearly arise from a similar compact region
in the vicinity of IRAS16293B. A hole is observed in the maps
due to the absorptions that are produced against the strong con-
tinuum at the continuum peak position. For DNCO the larger
1http://youngstars.nbi.dk/PILS/
Table 1. Number of lines used in the analysis of the isotopologues of
NH2CHO and HNCO and column densities derived for Tex =300 K and
a source size of 0.500.
Species # of lines Eup (K) N(cm2)
NH2CDO 12 146 – 366 2.1 ×1014
cis-NHDCHO 11 146 – 307 2.1 ×1014
trans-NHDCHO 11 151 – 332 1.8 ×1014
NH213CHO 10 152 – 428 1.5 ×1014
15NH2CHO 1.0 ×1014 (a)
NH2CH18O 0.8 ×1014 (a)
DNCO 4 150 – 751 3.0 ×1014
HN13CO 8 127 – 532 4.0 ×1014
H15NCO 2.0 ×1014 (a)
HNC18O 1.5 ×1014 (a)
Notes. (a)3σupper limit.
beam size for the observations of this transition masks the ab-
sorption. The spatial variations that are observed among the dif-
ferent species are probably due to dierent line excitation or line
brightness. In particular, HNCO seems to be slightly more ex-
tended than NH2CHO, but this is most likely due to the fact that
the HNCO lines are particularly bright compared to the HNCO
and formamide isotopologues.
To constrain the excitation temperatures and column densi-
ties of the dierent species, we produce a grid of synthetic spec-
tra assuming Local Thermodynamical Equilibrium (LTE). We
predict the spectra for dierent excitation temperatures between
100 and 300 K with a step of 25 K and for dierent column den-
sities between 1 ×1013 and 1 ×1017 cm2. First, the column den-
sity is roughly estimated using relatively large steps, then refined
using smaller steps around the best fit solution. We determine
the best fit model using a χ2method comparing the observed
and synthetic spectra at ±0.5 MHz around the rest frequency of
the predicted emission lines. We carefully check that the best
fit model does not predict any lines not observed in the spec-
tra. For the deuterated forms, the models are in agreement with
the observations for excitation temperatures between 100 and
300 K. However, for NH213CHO and HN13CO, a model with
a high excitation temperature accounts much better for the ob-
served emission than a model with a low excitation temperature
(see Figs B.4 and B.6). An excitation temperature of 300 K was
consequently adopted for the analysis of the dierent isotopo-
logues. This excitation temperature is similar to that derived for
glycolaldehyde and ethylene glycol (Jørgensen et al. 2012, sub-
mitted), but higher than what is found for acetaldehyde, ethylene
oxide and propanal (125 K, Lykke et al. to be submitted). The
derived column densities, assuming a linewidth of 1 kms1and
a source size of 0.500 (Jørgensen et al. submitted; Lykke et al. to
be submitted), are summarized in Table 1. The uncertainties on
the column densities are all estimated to be within a factor of 2
(including the uncertainty on both the excitation temperature and
the baseline subtraction). The upper limits are estimated visually
by comparison of the synthetic spectra with the observations on
the entire spectral range. Figure 2 shows three lines of each iso-
topologue with the best-fit model. The models for all the lines
are shown in Appendix B.
The column densities of NH213CHO and HN13 CO are es-
timated to be 1.5 ×1014 cm2and 4 ×1014 cm2, respectively.
Assuming a 12C/13 C ratio of 68 (Milam et al. 2005), the col-
umn densities for the main isotopologues of formamide and iso-
cyanic acid are predicted to be 1 ×1016 cm2and 3 ×1016 cm2.
Article number, page 2 of 10
A. Coutens et al.: The ALMA-PILS survey: First detections of deuterated formamide and deuterated isocyanic acid
Fig. 1. Integrated intensity maps of NH2CHO, HNCO and their isotopo-
logues towards source B. The position of the continuum peak of source
B is indicated with a red cross, while the position where the spectrum
was extracted is shown with a red circle. The beam sizes are shown in
grey in the bottom right corner of each panel. The contour levels start
for the main isotopologue of HNCO at 0.05 Jy km s1with a step of
0.05 Jy km s1. For the other species, the levels are 0.02, 0.03, 0.04,
0.06, 0.08, 0.1 and 0.12 Jy kms1.
With these column densities, several NH2CHO lines and all of
the HNCO lines are overproduced, indicating that they are opti-
cally thick. The model of formamide is, however, in agreement
with the few lines with the lowest opacities (see Figs. B.7 and
B.8). NH2CH18O has also been searched for, but is not detected
with a 3σupper limit of 8 ×1013 cm2. The non-detection of this
isotopologue is consistent with the 16O/18 O ratio of 560 in the in-
terstellar medium (Wilson 1999), which gives N(NH2CH18O) =
2×1013 cm2. Similarly, HNC18O is not detected either with a
3σupper limit of 1.5 ×1014 cm2, which is consistent with its
expected column density of 5 ×1013 cm2.
Using the column densities derived for the 13C isotopo-
logues and a standard 12C/13 C ratio, the deuterium fractiona-
tion in NH2CHO is about 2% for the three deuterated forms
and the DNCO/HNCO ratio is similar (1%). If the 12C/13 C ra-
tio is lower (30) as reported for glycolaldehyde by Jørgensen
et al. (submitted), the D/H ratios of formamide and isocyanic
acid would be about 4-5% and 2-3%, respectively.
329.990 329.995 330.000
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
(Jy/beam)
342.320 342.325
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
354.415 354.420
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
346.585 346.590
-0.02
0.00
0.02
0.04
0.06
0.08
(Jy/beam)
346.825 346.830
-0.02
0.00
0.02
0.04
0.06
0.08
347.265 347.270
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
333.690 333.695
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
(Jy/beam)
333.810 333.815
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
353.350 353.355 353.360
-0.02
0.00
0.02
0.04
0.06
339.175 339.180 339.185
-0.02
-0.01
0.00
0.01
0.02
0.03
(Jy/beam)
339.210 339.215
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
360.530 360.535
-0.02
0.00
0.02
0.04
0.06
344.625 344.630 344.635
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
(Jy/beam)
346.555 346.560
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
0.12
348.595 348.600 348.605
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
329.590 329.595 329.600
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
(Jy/beam)
330.860 330.865
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
350.340 350.345
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
NH2CDOcis-NHDCHOtrans-NHDCHONH2
13CHODNCOHN13CO
Fig. 2. Black: Detected lines of NH2CDO, cis-NHDCHO, trans-
NHDCHO, NH213CHO, DNCO and HN13 CO. Red: Best-fit model.
We also search for the 15N isotopologues of formamide and
isocyanic acid. A couple of transitions could tentatively be as-
signed to 15NH2CHO, but these lines are close to the noise level
and possibly blended with other species. For H15NCO, the uncer-
tainties on the frequencies of some of the transitions are rather
large, preventing any firm detection. Based on a standard 12C/13C
ratio, lower limits of 100 and 138 are obtained for the 14N/15N
ratios of formamide and HNCO respectively.
4. Discussion and conclusion
Our derived ratio in IRAS16293 for HNCO/NH2CHO, 3, is
consistent with the ratios found in warm sources in previous
studies (Bisschop et al. 2007; Mendoza et al. 2014; López-
Sepulcre et al. 2015). Thanks to our interferometric observa-
tions, we also confirm that these two species are spatially corre-
lated. The deuterium fractionation ratios of these two molecules
are also similar, reinforcing the hypothesis that they are chemi-
cally related. We discuss here possible scenarios for the forma-
tion of these species in the warm inner regions of protostars.
Assuming that the deuteration of formaldehyde in the region
probed by the ALMA observations of formamide is similar to
the value derived with single-dish observations (15%, Loinard
et al. 2000), we can discuss the possibility for the gas-phase for-
Article number, page 3 of 10
A&A proofs: manuscript no. NH2CHO_PILS_aa_v6
mation mechanism proposed by Barone et al. (2015), H2CO +
NH2NH2CHO +H. According to this reaction, the deuter-
ated form NHDCHO would result from the reaction between
NHD and H2CO, while NH2CDO would form from NH2and
HDCO. We would consequently expect a higher deuteration for
NH2CDO compared to the observations unless the reaction be-
tween NH2and HDCO leads more eciently to NH2CHO and D
compared to NH2CDO and H. Theoretical or experimental stud-
ies of the branching ratios of these reactions would be needed to
rule out this scenario. The determination of the HDCO/H2CO
ratio from the PILS survey is also necessary. Nevertheless, it
should be noted that so far there is no proposed scenario in the
gas phase that could explain the correlation with HNCO.
Although it was recently shown that NH2CHO does not
form by hydrogenation of HNCO on grain surfaces (Noble et al.
2015), several other proposed mechanisms exist in the literature.
Both species can be formed through barrierless reactions in ices
through NH +CO HNCO and NH2+H2CO NH2CHO +
H, as demonstrated experimentally (Fedoseev et al. 2015, 2016).
Alternatively, both species are formed through ion bombardment
of H2O:CH4:N2mixtures (Kaˇ
nuchová et al. 2016) or UV irradi-
ation of CO:NH3:CH3OH and/or HNCO mixtures (e.g. Demyk
et al. 1998; Raunier et al. 2004; Jones et al. 2011; Henderson
& Gudipati 2015). Quantitative gas-grain modeling under con-
ditions representative of IRAS16293 are needed to assess which
of these grain surface routes dominates.
Ultimately, the HNCO and NH2CHO deuterium fractiona-
tion level and pattern may also hold a clue to their formation
routes. A particularly interesting result is that the three singly
deuterated forms of formamide are found with similar abun-
dances in IRAS16293. Contrary to the -CH functional group
that is not aected by hydrogen isotope exchanges, the hy-
droxyl (-OH) and amine (-NH) groups are expected to estab-
lish hydrogen bonds and equilibrate with water (Faure et al.
2015). This mechanism was proposed to explain the dierent
CH3OD/CH3OH (1.8%) and CH2DOH/CH3OH (37%) ratios
derived in IRAS16293 (Parise et al. 2006), as the water deu-
terium fractionation of water in the upper layers of the grain
mantles where complex organic molecules form is about a few
percent (Coutens et al. 2012, 2013; Furuya et al. 2016). We do
not see such dierences for formamide, for which all forms show
a deuterium fractionation similar to the CH3OD/CH3OH ratio
and water. The deuterium fractionation of methanol from the
PILS data needs to be investigated to know if the dierent deu-
terium fractionation ratios of the -CH and -OH groups are also
observed at small scales.
In conclusion, we present in this Letter the first detection of
the three singly deuterated forms of formamide and DNCO. The
similar deuteration of these species and their similar spatial dis-
tributions favours the formation of these two species on grain
surfaces. Further studies are, however, needed to rule out gas
phase routes. These detections illustrate the strength of ALMA,
and large spectral surveys such as PILS in particular, for the de-
tections of deuterated complex molecules. Determinations of the
deuterium fractionation for more complex molecules will help
to constrain their formation pathways. The search for deuterated
formamide in more sources is needed to reveal how variable the
deuteration of formamide is, and if the similarity of the abun-
dances of the three deuterated forms is common.
Acknowledgements. The authors thank Gleb Fedoseev and Harold Linnartz
for fruitful discussions. This paper makes use of the following ALMA data:
ADS/JAO.ALMA#2013.1.00278.S. ALMA is a partnership of ESO (represent-
ing its member states), NSF (USA) and NINS (Japan), together with NRC
(Canada) and NSC and ASIAA (Taiwan), in cooperation with the Republic
of Chile. The Joint ALMA Observatory is operated by ESO, AUI/NRAO and
NAOJ. The work of AC was funded by a STFC grant. AC thanks the COST ac-
tion CM1401 ‘Our Astrochemical History’ for additional financial support. The
group of JKJ acknowledges support from a Lundbeck Foundation Group Leader
Fellowship as well as the European Research Council (ERC) under the European
Union’s Horizon 2020 research and innovation programme (grant agreement No
646908) through ERC Consolidator Grant “S4F”. Research at Centre for Star
and Planet Formation is funded by the Danish National Research Foundation.
The group of EvD acknowledges A-ERC grant 291141 CHEMPLAN.
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Article number, page 4 of 10
A. Coutens et al.: The ALMA-PILS survey: First detections of deuterated formamide and deuterated isocyanic acid
Appendix A: Spectroscopic data
A list of unblended and optically thin lines used in the analysis
is presented in Table A.1. The spectroscopic data for NH2CHO
3=0, NH2CHO 312=1, NH213 CHO, 15NH2CHO, NH2CH18O,
NH2CDO, cis-NHDCHO, trans-NHDCHO (Kurland & Bright
Wilson 1957; Kukolich & Nelson 1971; Hirota et al. 1974;
Gardner et al. 1980; Moskienko & Dyubko 1991; Vorob’eva &
Dyubko 1994; Blanco et al. 2006; Kryvda et al. 2009; Motiyenko
et al. 2012; Kutsenko et al. 2013) and HNCO (Kukolich & Nel-
son 1971; Hocking et al. 1975; Niedenhoet al. 1995; Lapinov
et al. 2007) come from the CDMS database (Müller et al.
2001, 2005), while the data for DNCO, HN13CO, H15 NCO and
HNC18O (Hocking et al. 1975) are taken from the JPL database
(Pickett et al. 1998). It should be noted that there are significant
dierences for the predicted frequencies of the main isotopo-
logue of NH2CHO between CDMS and JPL (>1 MHz). A bet-
ter agreement is found with the observations for the most recent
entry in CDMS. For some of the HNCO isotopologues, there is
a lack of published spectroscopic data at high frequencies. In
particular for H15NCO, the range of uncertainty for some of the
frequencies is quite high. As the HN13CO transitions appeared
all slightly shifted compared to the observations, we applied a
correction of +0.5 MHz to model the lines.
The column densities of the formamide isotopologues given
in Table 1 were corrected by a factor of 1.5 to take into account
the contribution of the vibrational states for an excitation tem-
perature of 300 K.
Article number, page 5 of 10
A&A proofs: manuscript no. NH2CHO_PILS_aa_v6
Table A.1. Detected lines of NH2CHO, HNCO and their isotopologues used in
the analysis(a) .
Species Transition Frequency Eup Aij gup
(MHz) (K) (s1)
NH2CDO (17 0 17 – 16 0 16) 329995.2 145.6 2.64 ×103105
NH2CDO (16 9 7 – 15 9 6) 333363.6 308.9 1.87 ×10399
NH2CDO (16 9 8 – 15 9 7) 333363.6 308.9 1.87 ×10399
NH2CDO (16 7 10 – 15 7 9) 333696.6 240.7 2.22 ×10399
NH2CDO (16 7 9 – 15 7 8) 333696.6 240.7 2.22 ×10399
NH2CDO (16 4 13 – 15 4 12) 335234.9 170.5 2.61 ×10399
NH2CDO (16 3 13 – 15 3 12) 342320.7 156.9 2.86 ×10399
NH2CDO (17 1 16 – 16 1 15) 351988.3 158.1 3.18 ×103105
NH2CDO (17 10 7 – 16 10 6) 354151.5 366.4 2.15 ×103105
NH2CDO (17 10 8 – 16 10 7) 354151.5 366.4 2.15 ×103105
NH2CDO (17 9 8 – 16 9 7) 354257.0 325.9 2.37 ×103105
NH2CDO (17 9 9 – 16 9 8) 354257.0 325.9 2.37 ×103105
NH2CDO (17 8 10 – 16 8 9) 354416.0 289.6 2.56 ×103105
NH2CDO (17 8 9 – 16 8 8) 354416.0 289.6 2.56 ×103105
NH2CDO (17 7 11 – 16 7 10) 354661.3 257.7 2.74 ×103105
NH2CDO (17 7 10 – 16 7 9) 354661.3 257.7 2.74 ×103105
NH2CDO (17 5 12 – 16 5 11) 355800.2 206.7 3.04 ×103105
NH2CDO (17 4 13 – 16 4 12) 357938.5 187.8 3.20 ×103105
cis-NHDCHO (16 3 13 – 15 3 12) 331372.8 156.0 2.59 ×10399
cis-NHDCHO (16 2 14 – 15 2 13) 337248.5 146.0 2.79 ×10399
cis-NHDCHO (17 2 16 – 16 2 15) 340520.3 158.0 2.87 ×103105
cis-NHDCHO (18 1 18 – 17 1 17) 344878.9 160.8 3.02 ×103111
cis-NHDCHO (17 8 10 – 16 8 9) 346444.0 306.6 2.39 ×103105
cis-NHDCHO (17 8 9 – 16 8 8) 346444.0 306.6 2.39 ×103105
cis-NHDCHO (17 7 11 – 16 7 10) 346586.8 269.8 2.56 ×103105
cis-NHDCHO (17 7 10 – 16 7 9) 346586.8 269.8 2.56 ×103105
cis-NHDCHO (17 6 12 – 16 6 11) 346826.8 238.0 2.70 ×103105
cis-NHDCHO (17 6 11 – 16 6 10) 346827.5 238.0 2.70 ×103105
cis-NHDCHO (17 3 15 – 16 3 14) 347115.8 172.0 2.99 ×103105
cis-NHDCHO (17 5 12 – 16 5 11) 347268.9 211.1 2.83 ×103105
cis-NHDCHO (17 4 14 – 16 4 13) 347827.8 189.2 2.94 ×103105
cis-NHDCHO (17 3 14 – 16 3 13) 353047.5 173.0 3.15 ×103105
trans-NHDCHO (17 8 9 – 16 8 8) 333628.6 332.4 2.14 ×103105
trans-NHDCHO (17 8 10 – 16 8 9) 333628.6 332.4 2.14 ×103105
trans-NHDCHO (17 7 11 – 16 7 10) 333694.1 288.3 2.28 ×103105
trans-NHDCHO (17 7 10 – 16 7 9) 333694.1 288.3 2.28 ×103105
trans-NHDCHO (17 6 12 – 16 6 11) 333812.6 250.1 2.41 ×103105
trans-NHDCHO (17 6 11 – 16 6 10) 333812.7 250.1 2.41 ×103105
trans-NHDCHO (17 4 14 – 16 4 13) 334403.2 191.4 2.61 ×103105
trans-NHDCHO (18 1 18 – 17 1 17) 336945.3 157.3 2.82 ×103111
trans-NHDCHO (18 0 18 – 17 0 17) 338818.4 156.9 2.87 ×103111
trans-NHDCHO (17 1 16 – 16 1 15) 338878.8 150.6 2.86 ×103105
trans-NHDCHO (18 7 12 – 17 7 11) 353355.8 305.2 2.77 ×103111
trans-NHDCHO (18 7 11 – 17 7 10) 353355.8 305.2 2.77 ×103111
trans-NHDCHO (18 5 14 – 17 5 13) 353758.4 234.7 3.02 ×103111
trans-NHDCHO (18 3 16 – 17 3 15) 354028.8 187.8 3.19 ×103111
trans-NHDCHO (18 4 15 – 17 4 14) 354185.9 208.4 3.13 ×103111
NH213CHO (16 10 6 – 15 10 5) 339170.1 427.9 1.75 ×10333
NH213CHO (16 10 7 – 15 10 6) 339170.1 427.9 1.75 ×10333
NH213CHO (16 9 7 – 15 9 6) 339179.6 373.0 1.97 ×10333
NH213CHO (16 9 8 – 15 9 7) 339179.6 373.0 1.97 ×10333
NH213CHO (16 8 8 – 15 8 7) 339213.5 323.8 2.16 ×10333
NH213CHO (16 8 9 – 15 8 8) 339213.5 323.8 2.16 ×10333
NH213CHO (16 5 11 – 15 5 10) 339672.1 210.9 2.61 ×10333
NH213CHO (16 4 13 – 15 4 12) 340090.4 184.9 2.72 ×10333
Article number, page 6 of 10
A. Coutens et al.: The ALMA-PILS survey: First detections of deuterated formamide and deuterated isocyanic acid
Table A.1. continued.
Species Transition Frequency Eup Aij gup
(MHz) (K) (s1)
NH213CHO (16 4 12 – 15 4 11) 340273.4 184.9 2.73 ×10333
NH213CHO (17 1 17 – 16 1 16) 342156.0 151.5 2.95 ×10335
NH213CHO (17 9 8 – 16 9 7) 360396.3 390.3 2.49 ×10335
NH213CHO (17 9 9 – 16 9 8) 360396.3 390.3 2.49 ×10335
NH213CHO (17 7 11 – 16 7 10) 360531.8 297.7 2.88 ×10335
NH213CHO (17 7 10 – 16 7 9) 360531.8 297.7 2.88 ×10335
NH213CHO (18 1 18 – 17 1 17) 361904.8 168.9 3.49 ×10337
NH2CHO 3=0 (16 3 14 – 16 2 15) 331685.9 165.6 7.87 ×10533
NH2CHO 3=0 (8 2 7 – 7 1 6) 334483.5 48.5 5.49 ×10517
NH2CHO 3=0 (17 3 15 – 17 2 16) 336733.0 183.0 8.2 ×10535
NH2CHO 3=0 (34 3 31 – 34 2 32) 342029.5 645.9 1.07 ×10469
NH2CHO 3=0 (18 3 16 – 18 2 17) 342511.1 201.3 8.57 ×10537
NH2CHO 3=0 (28 4 24 – 28 3 25) 344545.8 464.1 1.15 ×10457
NH2CHO 3=0 (19 3 17 – 19 2 18) 349051.7 220.7 8.99 ×10539
NH2CHO 3=0 (20 3 18 – 20 2 19) 356379.8 241.1 9.47 ×10541
NH2CHO 3=0 (20 1 19 – 19 2 18) 359119.4 221.2 8.45 ×10541
NH2CHO 312=1 (17 14 3 – 16 14 2) 360717.7 1144.3 1.12 ×10335
NH2CHO 312=1 (17 14 4 – 16 14 3) 360717.7 1144.3 1.12 ×10335
DNCO (17 1 17 18 – 16 1 16 17) 344629.4 172.9 5.92 ×10437
DNCO (17 1 17 17 – 16 1 16 16) 344629.4 172.9 5.90 ×10435
DNCO (17 1 17 16 – 16 1 16 15) 344629.4 172.9 5.90 ×10433
DNCO (17 0 17 18 – 16 0 16 17) 346556.2 149.7 6.04 ×10437
DNCO (17 0 17 17 – 16 0 16 16) 346556.2 149.7 6.02 ×10435
DNCO (17 0 17 16 – 16 0 16 15) 346556.2 149.7 6.02 ×10433
DNCO (17 5 12 18 – 16 5 11 17) 346714.9 750.6 5.53 ×10437
DNCO (17 5 13 18 – 16 5 12 17) 346714.9 750.6 5.53 ×10437
DNCO (17 5 13 16 – 16 5 12 15) 346714.9 750.6 5.50 ×10433
DNCO (17 5 12 16 – 16 5 11 15) 346714.9 750.6 5.50 ×10433
DNCO (17 5 13 17 – 16 5 12 16) 346714.9 750.6 5.51 ×10435
DNCO (17 5 12 17 – 16 5 11 16) 346714.9 750.6 5.51 ×10435
DNCO (17 1 16 18 – 16 1 15 17) 348599.7 174.6 6.13 ×10437
DNCO (17 1 16 17 – 16 1 15 16) 348599.7 174.6 6.10 ×10435
DNCO (17 1 16 16 – 16 1 15 15) 348599.7 174.6 6.10 ×10433
HN13CO (15 2 13 16 – 14 2 12 15) 329594.5 299.2 5.08 ×10433
HN13CO (15 2 13 14 – 14 2 12 13) 329594.5 299.2 5.06 ×10429
HN13CO (15 2 13 15 – 14 2 12 14) 329594.5 299.2 5.06 ×10431
HN13CO (15 0 15 16 – 14 0 14 15) 329673.4 126.6 5.18 ×10433
HN13CO (15 0 15 15 – 14 0 14 14) 329673.4 126.6 5.16 ×10431
HN13CO (15 0 15 14 – 14 0 14 13) 329673.4 126.6 5.15 ×10429
HN13CO (15 1 14 16 – 14 1 13 15) 330860.2 170.2 5.21 ×10433
HN13CO (15 1 14 14 – 14 1 13 13) 330860.2 170.2 5.19 ×10429
HN13CO (15 1 14 15 – 14 1 13 14) 330860.2 170.2 5.19 ×10431
HN13CO (16 1 16 17 – 15 1 15 16) 350340.3 186.1 6.20 ×10435
HN13CO (16 1 16 16 – 15 1 15 15) 350340.3 186.1 6.18 ×10433
HN13CO (16 1 16 15 – 15 1 15 14) 350340.3 186.1 6.18 ×10431
HN13CO (16 3 14 17 – 15 3 13 16) 351427.6 531.9 6.07 ×10435
HN13CO (16 3 14 15 – 15 3 13 14) 351427.6 531.9 6.04 ×10431
HN13CO (16 3 14 16 – 15 3 13 15) 351427.7 531.9 6.04 ×10433
HN13CO (16 3 13 17 – 15 3 12 16) 351427.7 531.9 6.07 ×10435
HN13CO (16 3 13 15 – 15 3 12 14) 351427.7 531.9 6.04 ×10431
HN13CO (16 3 13 16 – 15 3 12 15) 351427.7 531.9 6.04 ×10433
HN13CO (16 2 15 17 – 15 2 14 16) 351548.3 316.1 6.19 ×10435
HN13CO (16 2 15 15 – 15 2 14 14) 351548.3 316.1 6.17 ×10431
HN13CO (16 2 15 16 – 15 2 14 15) 351548.3 316.1 6.17 ×10433
HN13CO (16 2 14 17 – 15 2 13 16) 351561.8 316.1 6.19 ×10435
HN13CO (16 2 14 15 – 15 2 13 14) 351561.8 316.1 6.17 ×10431
HN13CO (16 2 14 16 – 15 2 13 15) 351561.8 316.1 6.17 ×10433
Article number, page 7 of 10
A&A proofs: manuscript no. NH2CHO_PILS_aa_v6
Table A.1. continued.
Species Transition Frequency Eup Aij gup
(MHz) (K) (s1)
HN13CO (16 2 14 17 – 15 2 13 16) 351561.8 316.1 6.19 ×10435
HN13CO (16 2 14 15 – 15 2 13 14) 351561.8 316.1 6.17 ×10431
HN13CO (16 2 14 16 – 15 2 13 15) 351561.8 316.1 6.17 ×10433
HN13CO (16 0 16 17 – 15 0 15 16) 351642.9 143.5 6.30 ×10435
HN13CO (16 0 16 16 – 15 0 15 15) 351642.9 143.5 6.27 ×10433
HN13CO (16 0 16 15 – 15 0 15 14) 351642.9 143.5 6.27 ×10431
Notes. (a)This list only includes optically thin and unblended lines.
329.990 329.995 330.000
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
(Jy/beam)
333.360 333.365
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
333.695 333.700
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
335.230 335.235 335.240
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
342.320 342.325
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
351.985 351.990
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
354.150 354.155
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
(Jy/beam)
354.255 354.260
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
354.415 354.420
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
354.660 354.665
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
355.795 355.800 355.805
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
357.935 357.940
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
N = 1.20000e+14 cm-2, Tex = 300.000
Fig. B.1. Black: Detected lines of NH2CDO. Red: Best-fit model for Tex=300 K.
331.370 331.375
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
(Jy/beam)
337.245 337.250
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
340.515 340.520 340.525
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
344.875 344.880
Frequency (GHz)
-0.1
0.0
0.1
0.2
0.3
0.4
346.440 346.445
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
346.585 346.590
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
346.825 346.830
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
(Jy/beam)
347.115 347.120
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
347.265 347.270
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
347.825 347.830
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
353.045 353.050
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
N = 1.20000e+14 cm-2, Tex = 300.000
Fig. B.2. Black: Detected lines of cis-NHDCHO. Red: Best-fit model for Tex=300 K.
Article number, page 8 of 10
A. Coutens et al.: The ALMA-PILS survey: First detections of deuterated formamide and deuterated isocyanic acid
333.625 333.630
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
(Jy/beam)
333.690 333.695
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
333.810 333.815
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
334.400 334.405
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
336.940 336.945 336.950
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
338.815 338.820
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
338.875 338.880
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
(Jy/beam)
353.350 353.355 353.360
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
353.755 353.760
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
354.025 354.030
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
354.185 354.190
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
N = 1.00000e+14 cm-2, Tex = 300.000
Fig. B.3. Black: Detected lines of trans-NHDCHO. Red: Best-fit model for Tex=300 K.
339.165 339.170 339.175
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
(Jy/beam)
339.175 339.180 339.185
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
339.210 339.215
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
339.670 339.675
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
340.085 340.090 340.095
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
340.270 340.275
Frequency (GHz)
-0.02
0.00
0.02
0.04
342.155 342.160
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
(Jy/beam)
360.395 360.400
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
360.530 360.535
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
361.900 361.905 361.910
Frequency (GHz)
-0.02
-0.01
0.00
0.01
0.02
0.03
N = 1.00000e+14 cm-2, Tex = 300.000
Fig. B.4. Black: Detected lines of NH213CHO. Red: Best-fit model for Tex=300 K. Green: Best-fit model for Tex=100 K.
Fig. B.5. Black: Detected lines of DNCO. Red: Best-fit model for Tex=300 K.
329.590 329.595 329.600
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
(Jy/beam)
329.670 329.675
Frequency (GHz)
-0.05
0.00
0.05
0.10
0.15
330.860 330.865
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
350.340 350.345
Frequency (GHz)
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
351.425 351.430
Frequency (GHz)
-0.1
0.0
0.1
0.2
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Fig. B.6. Black: Detected lines of HN13CO. Red: Best-fit model for Tex=300 K. Green: Best-fit model for Tex=100 K.
Article number, page 9 of 10
A&A proofs: manuscript no. NH2CHO_PILS_aa_v6
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Fig. B.8. Black: Line of NH2CHO 312=1 with the lowest opacity. Red: Model based on the analysis of the NH213CHO lines and a 12C/13 C ratio
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... The situation has changed in recent years with the detection of D-enriched iCOMs (Coudert et al. 2013;Coutens et al. 2016;Jørgensen et al. 2018;Manigand et al. 2019) because, whatever their formation route, iCOM deuteration is no longer directly connected to the enhanced (gaseous) H 2 D + /H 3 + abundance ratio but rather to the deuteration of their parent species. In this case, the question is whether the iCOM deuteration is directly inherited from their parent species without any alteration or whether the processes leading from the parent to the daughter species can induce an enrichment or a decrease in the deuteration degree. ...
... To the best of our knowledge, the case of formamide is the only case reported in the literature. The detection of its D isotopomers was first reported by Coutens et al. (2016) toward the solar-like protostar IRAS 16293-2422 B (hereafter IRAS 16293 B) hot corino. Skouteris et al. (2017) showed that, if formamide is formed in the gas phase by the NH 2 + H 2 CO reaction, then trans-HCONHD/HCONH 2 ∼ 1/3 NHD/NH 2 , cis-HCONHD/HCONH 2 ∼ 1/3 NHD/NH 2 , and DCONH 2 / HCONH 2 ∼ 1/3 = HDCO/H 2 C). ...
Article
Full-text available
Despite the detection of numerous interstellar complex organic molecules (iCOMs) for decades, it is still a matter of debate whether they are synthesized in the gas phase or on the icy surface of interstellar grains. In the past, molecular deuteration has been used to constrain the formation paths of small and abundant hydrogenated interstellar species. More recently, the deuteration degree of formamide, one of the most interesting iCOMs, has also been explained with the hypothesis that it is formed by the gas-phase reaction NH 2 + H 2 CO. In this paper, we aim at using molecular deuteration to constrain the formation of another iCOM, glycolaldehyde, which is an important prebiotic species. More specifically, we have performed dedicated electronic structure and kinetic calculations to establish the glycolaldehyde deuteration degree in relation to that of ethanol, which is its possible parent species according to the suggestion of Skouteris et al. We found that the abundance ratio of the species containing one D atom over the all-protium counterpart depends on the produced D isotopomer and varies from 0.9 to 0.5. These theoretical predictions compare extremely well with the monodeuterated isotopomers of glycolaldehyde and that of ethanol measured toward the solar-like protostar IRAS 16293–2422, supporting the hypothesis that glycolaldehyde could be produced in the gas phase for this source. In addition, the present work confirms that the deuterium fractionation of iCOMs cannot be simply anticipated based on the deuterium fractionation of the parent species but necessitates a specific study, as already shown for the case of formamide.
... The situation has changed in the last years with the detection of D-enriched iCOMs (Coudert et al. 2013;Coutens et al. 2016;Jørgensen et al. 2018;Manigand et al. 2019), because, whatever is their formation route, iCOMs deuteration is not anymore directly connected to the enhanced (gaseous) H 2 D + /H + 3 abundance ratio but to the deuteration of their parent species. In this case, the question is whether the iCOM deuteration is directly inherited from their parent species without any alteration or whether the processes leading from the par-ent to the daughter species can induce an enrichment or a decrease in the deuteration degree. ...
... To the best of our knowledge, the case of formamide is the only case reported in the literature. The detection of its D isotopomers was first reported by Coutens et al. (2016) towards the Solar-like protostar IRAS16293-2422 B (hereinafter IRAS16293 B) hot corino. Skouteris et al. (2017) showed that, if formamide is formed in the gas-phase by the NH 2 + H 2 CO reaction, then: trans-HCONHD/HCONH 2 ∼ 1/3 NHD/NH 2 , cis-HCONHD/HCONH 2 ∼ 1/3 NHD/NH 2 and DCONH 2 /HCONH 2 ∼ 1/3 = HDCO/H 2 C). ...
Preprint
Full-text available
Despite the detection of numerous interstellar complex organic molecules (iCOMs) for decades, it is still a matter of debate whether they are synthesized in the gas-phase or on the icy surface of interstellar grains. In the past, molecular deuteration has been used to constrain the formation paths of small and abundant hydrogenated interstellar species. More recently, the deuteration degree of formamide, one of the most interesting iCOM, has also been explained in the hypothesis that it is formed by the gas-phase reaction NH$_2$ + H$_2$CO. In this article, we aim at using molecular deuteration to constrain the formation of another iCOM, glycolaldehyde, which is an important prebiotic species. More specifically, we have performed dedicated electronic structure and kinetic calculations to establish the glycolaldehyde deuteration degree in relation to that of ethanol, which is its possible parent species according to the suggestion of Skouteris et al. (2018). We found that the abundance ratio of the species containing one D-atom over the all-protium counterpart depends on the produced D isotopomer and varies from 0.9 to 0.5. These theoretical predictions compare extremely well with the monodeuterated isotopomers of glycolaldehyde and that of ethanol measured towards the Solar-like protostar IRAS 16293-2422, supporting the hypothesis that glycolaldehyde could be produced in the gas-phase for this source. In addition, the present work confirms that the deuterium fractionation of iCOMs cannot be simply anticipated based on the deuterium fractionation of the parent species but necessitates a specific study, as already shown for the case of formamide.
... Constraining which of the two ways to synthesize iCOMs is efficient and where the iCOMs formation happen, is not a simple task. Many methods have been used, from the comparison of the iCOMs measured abundances in hot cores/corinos with model predictions to their measured deuterium fractionation (Ceccarelli et al., 1998;Coutens et al., 2016;Jørgensen et al., 2018;Turner, 1990). ...
... Le modèle peut être constitué de réseaux de réaction simples, tels que la formation de méthanol ou d'eau, et comparer les résultats avec les expériences existantes et les simulations de Monte Carlo (Cuppen et al., 2009(Cuppen et al., , 2010. the iCOMs form is not a simple task. Many methods have been used, from the comparison of the iCOM measured abundances in hot cores and hot corinos with model predictions to their measured deuterium fractionation (Turner 1990;Ceccarelli et al. 1998;Coutens et al. 2016;Jørgensen et al. 2018). ...
Thesis
So far, Earth is the only known planet-hosting life based on organic chemistry. The Solar Systems small objects (e.g., comets and asteroids) are enriched with organic compounds, which raises the question of whether the first steps of the organic chemistry that led to terrestrial life started during the formation of the Solar System. Stars and planetary systems like our Solar System are formed continuously in the Milky Way. So, in principle, we can study chemistry in those objects to recover the first steps of the organic chemistry of the young Solar System. In this thesis, I worked on two main objectives, modeling the chemical evolution in star-forming regions with Grainoble+ and modeling the experimental ice with Labice.The first objective of the thesis is to understand the chemical processes that form and destroy interstellar Complex Organic Molecules (aka iCOMs) in Solar-like star-forming regions. For this purpose, I developed an astrochemistry code, Grarinoble+. The model is based on Grainoble, previously developed by our group (Taquet et al., 2012). Grainoble+ is a three-phase gas-grain multi-grain astrochemical code simulating the chemical evolution in star-forming regions. We included the latest binding energies and diffusion and reaction rates from quantum chemical calculations (see, e. g., Senevirathne et al. 2017; Song et al. 2017; and Ferrero et al. 2020).I followed two goals with Grainoble+, modeling iCOMs formation in the shocked regions of NGC 1333 IRAS 4A (De Simone et al., 2020) and modeling the ice composition in Taurus MCs (Witzel et al. 2022, submitted.).The second goal of the thesis is to simulate the layered structure of ices in experimental chemistry laboratories and simulate the thermal desorption of species based on Temperature Programmed Desorption (TPD) techniques. For this purpose, I developed Labice toy model that simulates the TPD experiments with the rate equation approach with a few input parameters. Labice is a simple analog of Grainoble+ that uses the three-phase approach to model the ice, water phase transition, and thermal desorption in an experimental setup. The goal is to show the impact of the various parameters, such as multi-binding energy or the trapping effect of water ice, that will be used in astrochemical models. I followed two goals with the Labice toy model, modeling the impact of the multi-binding energy approach on the sublimation of species (Ferrero et al. 2020) and modeling and benchmarking the water and CO composite ices using the CO trapped fraction (Witzel et al. 2022, in prep).
... Formamide (NH 2 CHO) is an interstellar complex organic molecule (iCOM, referring to C-bearing species with six atoms or more; Herbst & van Dishoeck 2009;Ceccarelli et al. 2017) and a key precursor of more complex organic molecules, that can lead to the origin of life, because of its potential to form peptide bonds (Saladino et al. 2012;Kahane et al. 2013;López-Sepulcre et al. 2019). It has been detected in the gas phase in hot corinos (Kahane et al. 2013;Coutens et al. 2016;Imai et al. 2016;López-Sepulcre et al. 2017;Bianchi et al. 2019;Hsu et al. 2022), which are the hot (100 K) and compact (100 au) regions immediately around low-mass (Sun-like) protostars (Ceccarelli et al. 2007). The formamide origin is still under debate. ...
... Nonetheless, QM computations Skouteris et al. 2017) coupled with astronomical observations in shocked regions (Codella et al. 2017) support this hypothesis. In the same vein, the observed deuterated isomers of formamide (including NH 2 CDO, cis-and trans-NHDCHO; Coutens et al. 2016) fit well with the theoretical predictions of a gas-phase formation route (Skouteris et al. 2017). On the other hand, the observed high deuterium fractionation of ∼2% for the three different forms of formamide (NH 2 CDO, cis-and trans-NHDCHO) could also be consistent with the formation in ice mantles on dust grains. ...
Article
Full-text available
Formamide (NH 2 CHO) is considered an important prebiotic molecule because of its potential to form peptide bonds. It was recently detected in the atmosphere of the HH 212 protostellar disk on the solar system scale where planets will form. Here we have mapped it and its potential parent molecules HNCO and H 2 CO, along with other molecules CH 3 OH and CH 3 CHO, in the disk atmosphere, studying its formation mechanism. Interestingly, we find a stratified distribution of these molecules, with the outer emission radius increasing from ∼24 au for NH 2 CHO and HNCO, to 36 au for CH 3 CHO, to 40 au for CH 3 OH, and then to 48 au for H 2 CO. More importantly, we find that the increasing order of the outer emission radius of NH 2 CHO, CH 3 OH, and H 2 CO is consistent with the decreasing order of their binding energies, supporting that they are thermally desorbed from the ice mantle on dust grains. We also find that HNCO, which has much lower binding energy than NH 2 CHO, has almost the same spatial distribution, kinematics, and temperature as NH 2 CHO, and is thus more likely a daughter species of desorbed NH 2 CHO. On the other hand, we find that H 2 CO has a more extended spatial distribution with different kinematics from NH 2 CHO, thus questioning whether it can be the gas-phase parent molecule of NH 2 CHO.
... The deuterated species detected in the PILS data are the mono-deuterated isotopomers of the oxygen-bearing organics glycolaldehyde (Jørgensen et al. 2016), ethanol, ketene, formic acid and of mono-deuterated acetaldehyde species CH 3 CDO (Jørgensen et al. 2018) and CH 2 DCHO (Coudert et al. 2019;Manigand et al. 2020), of the nitrogenbearing organics isocyanic acid DNCO and the monodeuterated isotopomers of formamide (Coutens et al. 2016) and the cyanamide isotopologue HDNCN (Coutens et al. 2018) and sulfur-containing species such as the hydrogen sulfide isotopologue HD 34 S (Drozdovskaya et al. 2018). Also, the PILS data reveal the presence of doubly-deuterated organics including the methyl cyanide species CHD 2 CN (Calcutt et al. 2018), the methyl formate species CHD 2 OCHO (Manigand et al. 2019) and the dimethyl ether species CHD 2 OCH 3 (Richard et al. 2021) and enable new and more accurate constraints on the doubly-and triply-deuterated variants of methanol in the warm gas close to the protostars (Drozdovskaya et al. 2022;Ilyushin et al. 2022). ...
... but lower than some of the larger complex species such as ethanol, methylformate, glycolaldehyde and acetaldehyde with ratios of 0.05-0.06 (Jørgensen et al. 2018;Coutens et al. 2016Coutens et al. , 2018. This difference may reflect differences in the formation time with the species with the lower ratios forming earlier in the evolution of the prestellar cores. ...
Preprint
We prepared a sample of mono-deuterated oxirane and studied its rotational spectrum in the laboratory between 490 GHz and 1060 GHz in order to improve its spectroscopic parameters and consequently the calculated rest frequencies of its rotational transitions. The updated rest frequencies were employed to detect $c$-C$_2$H$_3$DO for the first time in the interstellar medium in the Atacama Large Millimetre/submillimetre Array (ALMA) Protostellar Interferometric Line Survey (PILS) of the Class 0 protostellar system IRAS 16293$-$2422. Fits of the detected lines using the rotation diagrams yield a temperature of $T_{\rm rot} = 103 \pm 19$ K, which in turn agrees well with 125 K derived for the $c$-C$_2$H$_4$O main isotopologue previously. The $c$-C$_2$H$_3$DO to $c$-C$_2$H$_4$O ratio is found to be $\sim$0.15 corresponding to a D-to-H ratio of $\sim$0.036 per H atom which is slightly higher than the D-to-H ratio of species such as methanol, formaldehyde, ketene and but lower than those of the larger complex organic species such as ethanol, methylformate and glycolaldehyde. This may reflect that oxirane is formed fairly early in the evolution of the prestellar cores. The identification of doubly deuterated oxirane isotopomers in the PILS data may be possible judged by the amount of mono-deuterated oxirane and the observed trend that multiply deuterated isotopologues have higher deuteration rates than their mono-deuterated variants.
Article
We prepared a sample of mono-deuterated oxirane and studied its rotational spectrum in the laboratory between 490 and 1060 GHz in order to improve its spectroscopic parameters and consequently the calculated rest frequencies of its rotational transitions. The updated rest frequencies were employed to detect c-C2H3DO for the first time in the interstellar medium in the Atacama Large Millimetre/submillimetre Array Protostellar Interferometric Line Survey (PILS) of the Class 0 protostellar system IRAS 16293−2422. Fits of the detected lines using the rotation diagrams yield a temperature of Trot = 103 ± 19 K, which in turn agrees well with 125 K derived for the c-C2H4O main isotopologue previously. The c-C2H3DO to c-C2H4O ratio is found to be ∼0.15 corresponding to a D-to-H ratio of ∼0.036 per H atom, which is slightly higher than the D-to-H ratio of species such as methanol, formaldehyde, and ketene but lower than those of the larger complex organic species such as ethanol, methyl formate, and glycolaldehyde. This may reflect that oxirane is formed fairly early in the evolution of the prestellar cores. The identification of doubly deuterated oxirane isotopomers in the PILS data may be possibly judged by the amount of mono-deuterated oxirane and the observed trend that multiply deuterated isotopologues have higher deuteration rates than their mono-deuterated variants.
Article
Full-text available
Context. Complex organic species are known to be abundant toward low- and high-mass protostars. No statistical study of these species toward a large sample of high-mass protostars with the Atacama Large Millimeter/submillimeter Array (ALMA) has been carried out so far. Aims. We aim to study six N-bearing species: methyl cyanide (CH 3 CN), isocyanic acid (HNCO), formamide (NH 2 CHO), ethyl cyanide (C 2 H 5 CN), vinyl cyanide (C 2 H 3 CN) and methylamine (CH 3 NH 2 ) in a large sample of line-rich high-mass protostars. Methods. From the ALMA Evolutionary study of High Mass Protocluster Formation in the Galaxy survey, 37 of the most line-rich hot molecular cores with ~1" angular resolution are selected. Next, we fit their spectra and find column densities and excitation temperatures of the N-bearing species mentioned above, in addition to methanol (CH 3 OH) to be used as a reference species. Finally, we compare our column densities with those in other low- and high-mass protostars. Results. CH 3 OH, CH 3 CN and HNCO are detected in all sources in our sample, whereas C 2 H 3 CN and CH 3 NH 2 are (tentatively) detected in ~78 and ~32% of the sources. We find three groups of species when comparing their excitation temperatures: hot (NH 2 CHO; T ex ≳ 250 K), warm (C 2 H 3 CN, HN ¹³ CO and CH 3 ¹³ CN; 100 K ≲ T ex ≲ 250 K) and cold species (CH 3 OH and CH 3 NH 2 ; T ex ≲ 100 K). This temperature segregation reflects the trend seen in the sublimation temperature of these molecules and validates the idea that complex organic emission shows an onion-like structure around protostars. Moreover, the molecules studied here show constant column density ratios across low- and high-mass protostars with scatter less than a factor ~3 around the mean. Conclusions. The constant column density ratios point to a common formation environment of complex organics or their precursors, most likely in the pre-stellar ices. The scatter around the mean of the ratios, although small, varies depending on the species considered. This spread can either have a physical origin (source structure, line or dust optical depth) or a chemical one. Formamide is most prone to the physical effects as it is tracing the closest regions to the protostars, whereas such effects are small for other species. Assuming that all molecules form in the pre-stellar ices, the scatter variations could be explained by differences in lifetimes or physical conditions of the pre-stellar clouds. If the pre-stellar lifetimes are the main factor, they should be similar for low- and high-mass protostars (within factors ~2–3).
Article
Full-text available
Context. The interstellar detections of isocyanic acid (HNCO), methyl isocyanate (CH 3 NCO), and very recently also ethyl isocyanate (C 2 H 5 NCO) invite the question of whether or not vinyl isocyanate (C 2 H 3 NCO) can be detected in the interstellar medium. There are only low-frequency spectroscopic data (<40 GHz) available for this species in the literature, which makes predictions at higher frequencies rather uncertain, which in turn hampers searches for this molecule in space using millimeter (mm) wave astronomy. Aims. The aim of the present study is on one hand to extend the laboratory rotational spectrum of vinyl isocyanate to the mm wave region and on the other to search, for the first time, for its presence in the high-mass star-forming region Sgr B2, where other isocyanates and a plethora of complex organic molecules are observed. Methods. We recorded the pure rotational spectrum of vinyl isocyanate in the frequency regions 127.5–218 and 285–330 GHz using the Prague mm wave spectrometer. The spectral analysis was supported by high-level quantum-chemical calculations. On the astronomy side, we assumed local thermodynamic equilibrium to compute synthetic spectra of vinyl isocyanate and to search for it in the ReMoCA survey performed with the Atacama Large Millimeter/submillimeter Array (ALMA) toward the high-mass star-forming protocluster Sgr B2(N). Additionally, we searched for the related molecule ethyl isocyanate in the same source. Results. Accurate values for the rotational and centrifugal distortion constants are reported for the ground vibrational states of trans and cis vinyl isocyanate from the analysis of more than 1000 transitions. We report nondetections of vinyl and ethyl isocyanate toward the main hot core of Sgr B2(N). We find that vinyl and ethyl isocyanate are at least 11 and 3 times less abundant than methyl isocyanate in this source, respectively. Conclusions. Although the precise formation mechanism of interstellar methyl isocyanate itself remains uncertain, we infer from existing astrochemical models that our observational upper limit for the CH 3 NCO:C 2 H 5 NCO ratio in Sgr B2(N) is consistent with ethyl isocyanate being formed on dust grains via the abstraction or photodissociation of an H atom from methyl isocyanate, followed by the addition of a methyl radical. The dominance of such a process for ethyl isocyanate production, combined with the absence of an analogous mechanism for vinyl isocyanate, would indicate that the ratio C 2 H 3 NCO:C 2 H 5 NCO should be less than unity. Even though vinyl isocyanate was not detected toward Sgr B2(N), the results of this work represent a significant improvement on previous low-frequency studies and will help the astronomical community to continue searching for this species in the Universe.
Article
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The chemical diversity of low-mass protostellar sources has so far been recognized, and environmental effects are invoked as its origin. In this context, observations of isolated protostellar sources without the influence of nearby objects are of particular importance. Here, we report the chemical and physical structures of the low-mass Class 0 protostellar source IRAS 16544−1604 in the Bok globule CB 68, based on 1.3 mm Atacama Large Millimeter/submillimeter Array observations at a spatial resolution of ∼70 au that were conducted as part of the large program FAUST. Three interstellar saturated complex organic molecules (iCOMs), CH 3 OH, HCOOCH 3 , and CH 3 OCH 3 , are detected toward the protostar. The rotation temperature and the emitting region size for CH 3 OH are derived to be 131 ± 11 K and ∼10 au, respectively. The detection of iCOMs in close proximity to the protostar indicates that CB 68 harbors a hot corino. The kinematic structure of the C ¹⁸ O, CH 3 OH, and OCS lines is explained by an infalling–rotating envelope model, and the protostellar mass and the radius of the centrifugal barrier are estimated to be 0.08–0.30 M ⊙ and <30 au, respectively. The small radius of the centrifugal barrier seems to be related to the small emitting region of iCOMs. In addition, we detect emission lines of c-C 3 H 2 and CCH associated with the protostar, revealing a warm carbon-chain chemistry on a 1000 au scale. We therefore find that the chemical structure of CB 68 is described by a hybrid chemistry. The molecular abundances are discussed in comparison with those in other hot corino sources and reported chemical models.
Article
H‐atom tunneling reactions play important roles in astrochemistry, but an understanding of these reactions is still in its infancy. The unique properties associated with quantum solid para‐hydrogen provide an effective environment for the generation and reactions in situ of H atoms at low temperature. Several techniques have been employed to generate H atoms to study astrochemically relevant systems that provide significant insight into the formation of complex organic molecules (COM) and help to explain the relations between the abundance of some pairs of stable species. These results introduce new concepts in astrochemistry, including H‐induced H abstraction, H‐induced fragmentation, and H‐induced uphill isomerization in darkness that have been overlooked previously. This mini‐review summarizes the state of the art in this field, discussing fundamental understanding and techniques concerning H‐atom generation, H‐tunneling reactions, and their applications; the perspectives and open questions that await further exploration are discussed. H‐atom tunneling reactions play important roles in astrochemistry, but an understanding of these reactions is still in its infancy. The unique properties associated with para‐hydrogen provide an environment for the effective generation and reactions in situ of H atoms at low temperature. These results introduce new concepts previously overlooked in astrochemistry, including H‐induced H abstraction, fragmentation, and uphill isomerization in darkness.
Article
Full-text available
Recent interferometer observations have found that the D2O/HDO abundance ratio is higher than that of HDO/H2O by about one order of magnitude in the vicinity of low-mass protostar NGC 1333-IRAS 2A, where water ice has sublimated. Previous laboratory and theoretical studies show that the D2O/HDO ice ratio should be lower than the HDO/H2O ice ratio, if HDO and D2O ices are formed simultaneously with H2O ice. In this work, we propose that the observed feature, D2O/HDO > HDO/H2O, is a natural consequence of chemical evolution in the early cold stages of low-mass star formation: 1) majority of oxygen is locked up in water ice and other molecules in molecular clouds, where water deuteration is not efficient, and 2) water ice formation continues with much reduced efficiency in cold prestellar/protostellar cores, where deuteration processes are highly enhanced due to the drop of the ortho-para ratio of H2, the weaker UV radiation field, etc. Using a simple analytical model and gas-ice astrochemical simulations tracing the evolution from the formation of molecular clouds to protostellar cores, we show that the proposed scenario can quantitatively explain the observed HDO/H2O and D2O/HDO ratios. We also find that the majority of HDO and D2O ices are likely formed in cold prestellar/protostellar cores rather than in molecular clouds, where the majority of H2O ice is formed. This work demonstrates the power of the combination of the HDO/H2O and D2O/HDO ratios as a tool to reveal the past history of water ice formation in the early cold stages of star formation and when the enrichment of deuterium in the bulk of water occurred. Further observations are needed to explore if the relation, D2O/HDO > HDO/H2O, is common in low-mass protostellar sources.
Article
Full-text available
Context. Formamide (NH2HCO) and isocyanic acid (HNCO) have been observed as gaseous species in several astronomical environments such as cometary comae and pre- and proto-stellar objects. A debate is open on the formation route of those molecules, in particular if they are formed by chemical reactions in the gas phase and/or on grains. In this latter case it is relevant to understand if the formation occurs through surface reactions or is induced by energetic processing. Aims. We present arguments that support the formation of formamide in the solid phase by cosmic-ion-induced energetic processing of ices present as mantles of interstellar grains and on comets. Formamides, along with other molecules, are expelled in the gas phase when the physical parameters are appropriate to induce the desorption of ices. Methods. We have performed several laboratory experiments in which ice mixtures (H2O:CH4:N2, H2O:CH4:NH3, and CH3OH:N2) were bombarded with energetic (30-200 keV) ions (H+ or He+). FTIR spectroscopy was performed before, during, and after ion bombardment. In particular, the formation of HNCO and NH2HCO was measured quantiatively. Results. Energetic processing of ice can quantitatively reproduce the amount of NH2HCO observed in cometary comae and in many circumstellar regions. HNCO is also formed, but additional formation mechanisms are requested to quantitatively account for the astronomical observations. Conclusions. We suggest that energetic processing of ices in the pre- and proto-stellar regions and in comets is the main mechanism to produce formamide, which, once it is released in the gas phase because of desorption of ices, is observed in the gas phase in these astrophysical environments.
Article
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The deuterium fractionation of gas-phase molecules in hot cores is believed to reflect the composition of interstellar ices. The deuteration of methanol is a major puzzle, however, because the isotopologue ratio [CH2DOH]/[CH3OD], which is predicted to be equal to 3 by standard grain chemistry models, is much larger (~20) in low-mass hot corinos and significantly lower (~1) in high-mass hot cores. This dichotomy in methanol deuteration between low-mass and massive protostars is currently not understood. In this study, we report a simplified rate equation model of the deuterium chemistry occurring in the icy mantles of interstellar grains. We apply this model to the chemistry of hot corinos and hot cores, with IRAS 16293-2422 and the Orion~KL Compact Ridge as prototypes, respectively. The chemistry is based on a statistical initial deuteration at low temperature followed by a warm-up phase during which thermal hydrogen/deuterium (H/D) exchanges occur between water and methanol. The exchange kinetics is incorporated using laboratory data. The [CH2DOH]/[CH3OD] ratio is found to scale inversely with the D/H ratio of water, owing to the H/D exchange equilibrium between the hydroxyl (-OH) functional groups of methanol and water. Our model is able to reproduce the observed [CH2DOH]/[CH3OD] ratios provided that the primitive fractionation of water ice [HDO]/[H2O] is ~ 2% in IRAS 16293-2422 and ~0.6% in Orion~KL. We conclude that the molecular D/H ratios measured in hot cores may not be representative of the original mantles because molecules with exchangeable deuterium atoms can equilibrate with water ice during the warm-up phase.
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Context. It is generally agreed that hydrogenation reactions dominate chemistry on grain surfaces in cold, dense molecular cores, saturating the molecules present in ice mantles. Aims. We present a study of the low temperature reactivity of solid phase isocyanic acid (HNCO) with hydrogen atoms, with the aim of elucidating its reaction network. Methods. Fourier transform infrared spectroscopy and mass spectrometry were employed to follow the evolution of pure HNCO ice during bombardment with H atoms. Both multilayer and monolayer regimes were investigated. Results. The hydrogenation of HNCO does not produce detectable amounts of formamide (NH2CHO) as the major product. Experiments using deuterium reveal that deuteration of solid HNCO occurs rapidly, probably via cyclic reaction paths regenerating HNCO. Chemical desorption during these reaction cycles leads to loss of HNCO from the surface. Conclusions. It is unlikely that significant quantities of NH2CHO form from HNCO. In dense regions, however, deuteration of HNCO will occur. HNCO and DNCO will be introduced into the gas phase, even at low temperatures, as a result of chemical desorption.
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September 2013, 12(5): 1881-1905. doi: 10.3934/cpaa.2013.12.1881
## Parabolic and elliptic problems with general Wentzell boundary condition on Lipschitz domains
1 University of Puerto Rico, Rio Piedras Campus, Department of Mathematics, P.O. Box 70377, San Juan PR 00936-8377, United States
Received April 2011 Revised September 2012 Published January 2013
We show that on a bounded domain $\Omega\subset R^N$ with Lipschitz continuous boundary $\partial \Omega$, weak solutions of the elliptic equation $\lambda u-Au=f$ in $\Omega$ with the boundary conditions $-\gamma\Delta_\Gamma u+\partial_\nu^a u+\beta u=g$ on $\partial \Omega$ are globally Hölder continuous on $\bar \Omega$. Here $A$ is a uniformly elliptic operator in divergence form with bounded measurable coefficients, $\Delta_\Gamma$ is the Laplace-Beltrami operator on $\partial \Omega$, $\partial_\nu^a u$ denotes the conormal derivative of $u$, $\lambda,\gamma>0$ are real numbers and $\beta$ is a bounded measurable function on $\partial Omega$. We also obtain that a realization of the operator $A$ in $C(\bar \Omega)$ with the general Wentzell boundary conditions $(Au)|_{\partial \Omega}-\gamma\Delta_\Gamma u+\partial_\nu^a u+\beta u=g$ on $\partial \Omega$ generates a strongly continuous compact semigroup. Some analyticity results of the semigroup are also discussed.
Citation: Mahamadi Warma. Parabolic and elliptic problems with general Wentzell boundary condition on Lipschitz domains. Communications on Pure & Applied Analysis, 2013, 12 (5) : 1881-1905. doi: 10.3934/cpaa.2013.12.1881
##### References:
[1] H. Amann, Dual semigroup and second order linear elliptic boundary value problems,, Israel J. Math., 45 (1983), 225. doi: 10.1007/BF02774019. [2] H. Amann and J. Escher, Strongly continuous dual semigroups,, Ann. Mat. Pura Appl., 171 (1996), 41. doi: 10.1007/BF01759381. [3] W. Arendt, G. Metafune, D. Pallara and S. Romanelli, The Laplacian with Wentzell-Robin boundary conditions on spaces of continuous functions,, Semigroup Forum, 67 (2003), 247. doi: 10.1007/s00233-002-0010-8. [4] R. F. Bass and P. Hsu, Some potential theory for reflecting Brownian motion in Hölder and Lipschitz domains,, Ann. Probab., 19 (1991), 486. doi: 10.1214/aop/1176990437. [5] M. Biegert and M. Warma, The heat equation with nonlinear generalized Robin boundary conditions,, J. Differential Equations, 247 (2009), 1949. doi: 10.1016/j.jde.2009.07.017. [6] G. M. Coclite, G. R. Goldstein and J. A. Goldstein, Stability of parabolic problems with nonlinear Wentzell boundary conditions,, J. Differential Equations, 246 (2009), 2434. doi: 10.1016/j.jde.2008.10.004. [7] E. B. Davies, "Heat Kernels and Spectral Theory,", Cambridge University Press, (1989). doi: 10.1017/CBO9780511566158. [8] E. De Giorgi, Sulla differenziabilità e analiticità delle estremali degli integral multipli regolari,, Men. Accad. Sci. Torino, 3 (1957), 25. [9] K. J. Engel, The Laplacian on $C(\bar \Omega)$ with generalized Wentzell boundary conditions,, Arch. Math. (Basel), 81 (2003), 548. doi: 10.1007/s00013-003-0557-y. [10] A. Favini, G. R. Goldstein, J. A. Goldstein, E. Obrecht and S. Romanelli, Elliptic operators with general Wentzell boundary conditions, analytic semigroups and the angle concavity theorem,, Math. Nachr., 283 (2010), 504. doi: 10.1002/mana.200910086. [11] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, The heat equation with Wentzell boundary conditions,, J. Evol. Eq., 2 (2002), 1. doi: 10.1007/s00028-002-8077-y. [12] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, The heat equation with nonlinear general Wentzell boundary condition,, Adv. Differential Equations, 11 (2006), 481. [13] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, Classification of general Wentzell boundary conditions for fourth order operators in one space dimension,, J. Math. Anal. Appl., 333 (2007), 219. doi: 10.1016/j.jmaa.2006.11.058. [14] M. Fukushima and M. Tomisaki, Reflecting diffusions on Lipschitz domains with cusps: Analytic construction and Skorohod representation,, Potential Anal., 4 (1995), 377. doi: 10.1007/BF01053454. [15] M. Fukushima and M. Tomisaki, Construction and decomposition of reflecting diffusions on Lipschitz domains with Hölder cusps,, Probab. Theory Relat. Fields, 106 (1996), 521. doi: 10.1007/s004400050074. [16] D. Gilbarg and N. S. Trudinger, "Elliptic Partial Differential Equations of Second Order,", Springer-Verlag, (2001). doi: 10.1007/978-3-642-61798-0. [17] G. R. Goldstein, Derivation and physical interpretation of general boundary conditions,, Adv. Differential Equations, 11 (2006), 457. [18] D. Jerison and C. E. Kenig, Boundary value problems on Lipschitz domains,, MAA Stud. Math., 23 (1982), 1. [19] J. Jost, "Riemannian Geometry and Geometric Analysis,", Fifth edition. Universitext. Springer-Verlag, (2008). doi: 10.1007/978-3-642-21298-7. [20] O. A. Ladyzhenskaya and N. N. Ural鈥檛seva, "Linear and Quasilinear Elliptic Equations,", Mathematics in Science and Engineering. 46. New York-London: Academic Press, (1968). [21] J. Maly and W. P. Ziemer, "Fine Regurality of Solutions of Elliptic Partial Differential Equations,", Providence, (1997). [22] C. B. Morrey Jr, Second order elliptic equations in several variables and Hölder continuity,, Math. Z., 72 (1959), 146. doi: 10.1007/BF01162944. [23] J. Moser, A new proof of the de Giorgi's theorem concerning the regularity problem for elliptic differential equations,, Commu. Pure Appl. Math., 13 (1960), 457. doi: 10.1002/cpa.3160130308. [24] M. K. V. Murthy and G. Stampacchia, Boundary value problems for some degenerate elliptic operators,, Ann. Mat. Pura Appl., 80 (1968), 1. doi: 10.1007/BF02413623. [25] R. Nittka, Regularity of solutions of linear second order elliptic and parabolic boundary value problems on Lipschitz domains,, J. Differential Equations, 251 (2011), 860. doi: 10.1016/j.jde.2011.05.019. [26] El M. Ouhabaz, "Analysis of Heat Equations on Domains,", Lond. Math. Soc. Monographs Series, (2005). [27] R. S. Phillips, The adjoint semi-group,, Pacific J. Math., 5 (1955), 269. doi: 10.2140/pjm.1955.5.269. [28] G. Stampacchia, Problemi al contorno ellittici con dati discontinui dotati di soluzioni Hölderiane,, Ann. Mat. Pura Appl., 51 (1960), 1. doi: 10.1007/BF02410941. [29] M. E. Taylor, "Partial Differential Equations. I. Basic Theory,", Texts Appl. Math., 23 (1996). doi: 10.1007/978-1-4684-9320-7. [30] J. L. Vázquez and E. Vitillaro, Heat equation with dynamical boundary conditions of reactive-diffusive type,, J. Differential Equations, 250 (2011), 2143. doi: 10.1016/j.jde.2010.12.012. [31] M. Warma, "The Laplacian with General Robin Boundary Conditions,", Ph.D Dissertation, (2002). [32] M. Warma, Wentzell-Robin boundary conditions on $C[0,1]$,, Semigroup Forum, 66 (2003), 162. doi: 10.1007/s002330010124. [33] M. Warma, The Robin and Wentzell-Robin Laplacians on Lipschitz domains,, Semigroup Forum, 73 (2006), 10. doi: 10.1007/s00233-006-0617-2. [34] M. Warma, Analyticity on $L^1$ of the heat semigroup with Wentzell boundary conditions,, Arch. Math. (Basel), 94 (2010), 85. doi: 10.1007/s00013-009-0068-6.
show all references
##### References:
[1] H. Amann, Dual semigroup and second order linear elliptic boundary value problems,, Israel J. Math., 45 (1983), 225. doi: 10.1007/BF02774019. [2] H. Amann and J. Escher, Strongly continuous dual semigroups,, Ann. Mat. Pura Appl., 171 (1996), 41. doi: 10.1007/BF01759381. [3] W. Arendt, G. Metafune, D. Pallara and S. Romanelli, The Laplacian with Wentzell-Robin boundary conditions on spaces of continuous functions,, Semigroup Forum, 67 (2003), 247. doi: 10.1007/s00233-002-0010-8. [4] R. F. Bass and P. Hsu, Some potential theory for reflecting Brownian motion in Hölder and Lipschitz domains,, Ann. Probab., 19 (1991), 486. doi: 10.1214/aop/1176990437. [5] M. Biegert and M. Warma, The heat equation with nonlinear generalized Robin boundary conditions,, J. Differential Equations, 247 (2009), 1949. doi: 10.1016/j.jde.2009.07.017. [6] G. M. Coclite, G. R. Goldstein and J. A. Goldstein, Stability of parabolic problems with nonlinear Wentzell boundary conditions,, J. Differential Equations, 246 (2009), 2434. doi: 10.1016/j.jde.2008.10.004. [7] E. B. Davies, "Heat Kernels and Spectral Theory,", Cambridge University Press, (1989). doi: 10.1017/CBO9780511566158. [8] E. De Giorgi, Sulla differenziabilità e analiticità delle estremali degli integral multipli regolari,, Men. Accad. Sci. Torino, 3 (1957), 25. [9] K. J. Engel, The Laplacian on $C(\bar \Omega)$ with generalized Wentzell boundary conditions,, Arch. Math. (Basel), 81 (2003), 548. doi: 10.1007/s00013-003-0557-y. [10] A. Favini, G. R. Goldstein, J. A. Goldstein, E. Obrecht and S. Romanelli, Elliptic operators with general Wentzell boundary conditions, analytic semigroups and the angle concavity theorem,, Math. Nachr., 283 (2010), 504. doi: 10.1002/mana.200910086. [11] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, The heat equation with Wentzell boundary conditions,, J. Evol. Eq., 2 (2002), 1. doi: 10.1007/s00028-002-8077-y. [12] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, The heat equation with nonlinear general Wentzell boundary condition,, Adv. Differential Equations, 11 (2006), 481. [13] A. Favini, G. R. Goldstein, J. A. Goldstein and S. Romanelli, Classification of general Wentzell boundary conditions for fourth order operators in one space dimension,, J. Math. Anal. Appl., 333 (2007), 219. doi: 10.1016/j.jmaa.2006.11.058. [14] M. Fukushima and M. Tomisaki, Reflecting diffusions on Lipschitz domains with cusps: Analytic construction and Skorohod representation,, Potential Anal., 4 (1995), 377. doi: 10.1007/BF01053454. [15] M. Fukushima and M. Tomisaki, Construction and decomposition of reflecting diffusions on Lipschitz domains with Hölder cusps,, Probab. Theory Relat. Fields, 106 (1996), 521. doi: 10.1007/s004400050074. [16] D. Gilbarg and N. S. Trudinger, "Elliptic Partial Differential Equations of Second Order,", Springer-Verlag, (2001). doi: 10.1007/978-3-642-61798-0. [17] G. R. Goldstein, Derivation and physical interpretation of general boundary conditions,, Adv. Differential Equations, 11 (2006), 457. [18] D. Jerison and C. E. Kenig, Boundary value problems on Lipschitz domains,, MAA Stud. Math., 23 (1982), 1. [19] J. Jost, "Riemannian Geometry and Geometric Analysis,", Fifth edition. Universitext. Springer-Verlag, (2008). doi: 10.1007/978-3-642-21298-7. [20] O. A. Ladyzhenskaya and N. N. Ural鈥檛seva, "Linear and Quasilinear Elliptic Equations,", Mathematics in Science and Engineering. 46. New York-London: Academic Press, (1968). [21] J. Maly and W. P. Ziemer, "Fine Regurality of Solutions of Elliptic Partial Differential Equations,", Providence, (1997). [22] C. B. Morrey Jr, Second order elliptic equations in several variables and Hölder continuity,, Math. Z., 72 (1959), 146. doi: 10.1007/BF01162944. [23] J. Moser, A new proof of the de Giorgi's theorem concerning the regularity problem for elliptic differential equations,, Commu. Pure Appl. Math., 13 (1960), 457. doi: 10.1002/cpa.3160130308. [24] M. K. V. Murthy and G. Stampacchia, Boundary value problems for some degenerate elliptic operators,, Ann. Mat. Pura Appl., 80 (1968), 1. doi: 10.1007/BF02413623. [25] R. Nittka, Regularity of solutions of linear second order elliptic and parabolic boundary value problems on Lipschitz domains,, J. Differential Equations, 251 (2011), 860. doi: 10.1016/j.jde.2011.05.019. [26] El M. Ouhabaz, "Analysis of Heat Equations on Domains,", Lond. Math. Soc. Monographs Series, (2005). [27] R. S. Phillips, The adjoint semi-group,, Pacific J. Math., 5 (1955), 269. doi: 10.2140/pjm.1955.5.269. [28] G. Stampacchia, Problemi al contorno ellittici con dati discontinui dotati di soluzioni Hölderiane,, Ann. Mat. Pura Appl., 51 (1960), 1. doi: 10.1007/BF02410941. [29] M. E. Taylor, "Partial Differential Equations. I. Basic Theory,", Texts Appl. Math., 23 (1996). doi: 10.1007/978-1-4684-9320-7. [30] J. L. Vázquez and E. Vitillaro, Heat equation with dynamical boundary conditions of reactive-diffusive type,, J. Differential Equations, 250 (2011), 2143. doi: 10.1016/j.jde.2010.12.012. [31] M. Warma, "The Laplacian with General Robin Boundary Conditions,", Ph.D Dissertation, (2002). [32] M. Warma, Wentzell-Robin boundary conditions on $C[0,1]$,, Semigroup Forum, 66 (2003), 162. doi: 10.1007/s002330010124. [33] M. Warma, The Robin and Wentzell-Robin Laplacians on Lipschitz domains,, Semigroup Forum, 73 (2006), 10. doi: 10.1007/s00233-006-0617-2. [34] M. Warma, Analyticity on $L^1$ of the heat semigroup with Wentzell boundary conditions,, Arch. Math. (Basel), 94 (2010), 85. doi: 10.1007/s00013-009-0068-6.
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2017 Impact Factor: 0.884
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# 1, 2, many
In the discrete branches of mathematics and the computer sciences, it will only take some seconds until you're faced with a set like $\{1,\ldots,m\}$. Only some people write $1\ldotp\ldotp m$, or $\{j:1\leq j\leq m\}$, and that journal you're submitting to might want something else entirely. "1, 2, many" provides an interface that makes changing from one to another a one-line change.
Current version available here: 0.3 (2005/05/09) 12many-0.3.tar.gz (373.7Kb)
Current version on CTAN: 0.3
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HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS
Rankine to Fahrenheit conversion
The symbol for degrees Rankine is °R[1]. Zero on both the Kelvin and Rankine scales is absolute zero, but the Rankine degree is defined as equal to one degree Fahrenheit, rather than the one degree Celsius used by the Kelvin scale. A temperature of −459.67 °F is exactly equal to 0 °R.
$r-459.67$
Here,r=Rankine.
ENTER THE VARIABLES TO BE USED IN THE FORMULA
Similar formulas which you may find interesting.
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## Twice Newton
Hi
I am reading a popular physics book. It discusses the test of Einsteins theory by Eddington at the eclipse. "The deviation of the light was double that predicted by Newton's physics"
Why does classical physics predict any deviation of light by gravity, did light have a mass in classical physics and if so how was it estimated (to allow for the deviation to be predicted)???
Simon
PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
Mentor He meant Newton's law of gravity combined with the idea that light consists of particles with energy equal to Plancks constant times the frequency, and that you can define a "mass" of such a particle by setting $h\nu=mc^2$. If you use Newton's law of gravity with this mass, you get the wrong result by a factor of two.
Recognitions:
Gold Member
Quote by Fredrik He meant Newton's law of gravity combined with the idea that light consists of particles with energy equal to Plancks constant times the frequency, and that you can define a "mass" of such a particle by setting $h\nu=mc^2$. If you use Newton's law of gravity with this mass, you get the wrong result by a factor of two.
You don't have to assign any mass to light to predict its Newtonian acceleration; you just assume that light in a beam accelerates in the same way as anything else.
Einstein's theory introduces an extra factor of (1+v2/c2) into the coordinate acceleration in this case, which means that the deflection is doubled for light and similarly increased for anything else moving at relativistic speeds.
## Twice Newton
Thank you, very clear (now!)
Recognitions: Gold Member Another way to think about curvature of light: half is classical (per Newtons laws) and the other half is relativistic due to the curvature of space and time itself. Had Eddington's experiment been conducted a few years earlier, Einstein's career might have suffered a major blow since he originally predicted the classical degree of curvature and only later when working general relativity discovered an additional amount. Dr Kaku just mentioned this on a 2 hour TV show now airing "EINSTEIN"....HISTORY CHANNEL I believe.
Mentor
Quote by Jonathan Scott You don't have to assign any mass to light to predict its Newtonian acceleration; you just assume that light in a beam accelerates in the same way as anything else.
I actually forgot that that when we're dealing with gravity we can eliminate the mass simply by dividing both sides of F=ma with m. But we can't divide with m when m=0, so we have have to assume either that m>0 or that there's a law of gravity for massless particles that works in this particular way.
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# American Institute of Mathematical Sciences
January 2008, 4(1): 81-94. doi: 10.3934/jimo.2008.4.81
## Optimal portfolios under a value-at-risk constraint with applications to inventory control in supply chains
1 Department of Applied Mathematics, The Hong Kong Polytechnic University, Kowloon, Hong Kong, China 2 Department of Industrial and Manufacturing Systems Engineering, The University of Hong Kong, Pokfulam Road, Hong Kong, China, China
Received August 2006 Revised November 2007 Published January 2008
The optimal portfolio problem under a VaR (value at risk) constraint is reviewed. Two different formulations, namely with and without consumption, are illustrated. This problem can be formulated as a constrained stochastic optimal control problem. The optimality conditions can be derived using the dynamic programming technique and the method of Lagrange multiplier can be applied to handle the VaR constraint. The method is extended for inventory management. Different from traditional inventory models of minimizing overall cost, the cashflow dynamic of a manufacturer is derived by considering a portfolio of inventory of raw materials together with income and consumption. The VaR of the portfolio of assets is derived and imposed as a constraint. Furthermore, shortage cost and holding cost can also be formulated as probabilistic constraints. Under this formulation, we find that holdings in high risk inventory are optimally reduced by the imposed value-at-risk constraint.
Citation: K. F. Cedric Yiu, S. Y. Wang, K. L. Mak. Optimal portfolios under a value-at-risk constraint with applications to inventory control in supply chains. Journal of Industrial & Management Optimization, 2008, 4 (1) : 81-94. doi: 10.3934/jimo.2008.4.81
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# How do I find the antiderivative of f(x)=e^(-2x)?
The exponential ${e}^{x}$ derived or integrated is always equal to itself. However, you must be careful with the exponent.
In this case you have $- 2$ in it so you have to take into account it.
$\int {e}^{- 2 x} \mathrm{dx} = {e}^{- 2 x} / \left(- 2\right) + c$
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# Is this correct about matrices
Yegor
I know that for matrices A, B and C is correct to write: (AB)C=A(BC)
Also $$(BA)^{-1}=A^{-1}B^{-1}$$
Why $$(A^{T}A)^{-1}A^{T}=A^{-1}(A^{T})^{-1}A^{T}=A^{-1}$$ is not correct?
Homework Helper
who says it is incorrect?
Staff Emeritus
Gold Member
$$(BA)^{-1}=A^{-1}B^{-1}$$
Yes, provided A and B are both invertible matrices...
Yegor
$$(A^{T}A)^{-1}A^{T}$$ such expression comes in chapter about least squares aproximation.
e.g. if we have inconsistent linear system Ax=b, then $$x=(A^{T}A)^{-1}A^{T}b$$ is best approximation. It is not equal to $$x=A^{-1}b$$
Yegor
Oh, yes. Now i see. Thank you very much, Hurky!
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Hydrogen Reduction of NiO Particles in a Single-Stage Fluidized-Bed Reactor without Sticking
Title & Authors
Hydrogen Reduction of NiO Particles in a Single-Stage Fluidized-Bed Reactor without Sticking
Oh, Chang-Sup; Kim, Hang Goo; Kim, Yong Ha;
Abstract
A commercial NiO (green nickel oxide, 86 wt% Ni) powder was reduced using a batch-type fluidized-bed reactor in a temperature range of 500 to $\small{600^{\circ}C}$ and in a residence time range of 5 to 90 min. The reduction rate increased with increases in temperature; however, agglomeration and sintering (sticking) of Ni particles noticeably took place at high temperatures above $\small{600^{\circ}C}$. An increasing tendency toward sticking was also observed at long residence times. In order to reduce the oxygen content in the powder to a level below 1% without any sticking problems, which can lead to defluidization, proper temperature and residence time for a stable fluidized-bed operation should be established. In this study, these values were found to be $\small{550^{\circ}C}$ and 60 min, respectively. Another important condition is the specific gas consumption rate, i.e. the volume amount ($\small{Nm^3}$) of hydrogen gas used to reduce 1 ton of Green NiO ore. The optimum gas consumption rate was found to be $\small{5,000Nm^3/ton}$-NiO for the complete reduction. The Avrami model was applied to this study; experimental data are most closely fitted with an exponent (m) of $\small{0.6{\pm}0.01}$ and with an overall rate constant (k) in the range of 0.35~0.45, depending on the temperature.
Keywords
hydrogen reduction;NiO particles;fluidized bed;sticking;
Language
English
Cited by
References
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J. Szekely, J. W. Evans and H. Y. Sohn, Gas-Soid Reactions, Academic Press Inc., New York, Chapter 2 & 4 (1976).
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A. H. Rashed and Y. K. Rao, Chem. Eng. Comm., 156, 1 (1996).
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K. D. Kim et al, Rev. Adv. Mater. Sci., 28, 162 (2011).
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Q. Jeangros, et al., J. Mater Sci., 48, 2893 (2013).
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J. Li, X. Liu, Q. Zhu and H. Li, Particuology, 19, 27 (2015).
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J. W. Evans, S. Song and C. E. Sucre, Metall. Trans. B, 7B, 55 (1976).
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A. Kivnick and A. N. Hixson, Chem. Eng. Prog., 48, 394 (1952).
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D. Kunii and O. Levnspiel, Fluidization Engineering, Butterworth-Heinemann, Boston, Second edition, 10 (1991).
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Wikipedia, the free encyclopedia, https://en.wikipedia.org/wiki/Avrami_equation
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S. Martini, M. L. Herrera and R. W. Hartel, JAOCS, 79, 1055 (2002).
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Zhi-jie YAN, et al., Trans. Nonferrous Met. Soc. China, 18, 138 (2008).
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## Selected papers authored or co-authored by M.H. van Emden
1. Title:
Contributions to the compositional semantics of first-order predicate logic
Author: Philip Kelly and M.H. van Emden
Abstract:
Henkin, Monk and Tarski gave a compositional semantics for first-order predicate logic. We extend this work by including function symbols in the language and by giving the denotation of the atomic formula as a composition of the denotations of its predicate symbol and of its tuple of arguments. In addition we give the denotation of a term as a composition of the denotations of its function symbol and of its tuple of arguments.
Date: December 2015
pdf
2. Title:
The lambda mechanism in lambda calculus and in other calculi
Author: M.H. van Emden
Abstract:
A comparison of Landin's form of lambda calculus with Church's shows that, independently of the lambda calculus, there exists a mechanism for converting functions with arguments indexed by variables to the usual kind of function where the arguments are indexed numerically. We call this the "lambda mechanism" and show how it can be used in other calculi. In first-order predicate logic it can be used to define new functions and new predicates in terms of existing ones. In a purely imperative programming language it can be used to provide an Algol-like procedure facility.
Date: March 2015
pdf
3. Title:
Logic programming beyond Prolog
Author: M.H. van Emden
Abstract:
A program in pure Prolog is an executable specification. For example, merge sort in Prolog is a logical formula, yet shows creditable performance on long lists. But such executable specifications are a compromise: the logic is distorted by algorithmic considerations, yet only indirectly executable via an abstract machine. This paper introduces relational programming, a method that solves the difficulty with Prolog programming by a separation of concerns. It requires writing three texts: (1) the axioms, a logical formula that specifies the problem and is not compromised by algorithmic considerations, (2) the theorem, a logical formula that expresses the algorithm yet follows the axioms, and (3) the code, a transcription of the theorem to C++. Correctness of the code relies on the logical justification of the theorem by the axioms and on a faithful transcription of the theorem to C++. Sorting is an example where relational programming has the advantage of a higher degree of abstractness: the data to be sorted can be any C++ data type that satisfies the axioms of linear order, while the Prolog version is limited to linked lists. Another advantage of relational programs is that they can be shown to have a model-theoretic and fixpoint semantics equivalent to each other and analogous to those of pure Prolog programs.
Date: December 2014
pdf
4. Title:
Matrix code
Author:
M.H. van Emden
Abstract:
Matrix Code gives imperative programming a mathematical semantics and heuristic power comparable in quality to functional and logic programming. A program in Matrix Code is developed incrementally from a specification in pre/post-condition form. The computations of a code matrix are characterized by powers of the matrix when it is interpreted as a transformation in a space of vectors of logical conditions. Correctness of a code matrix is expressed in terms of a fixpoint of the transformation. The abstract machine for Matrix Code is the dual-state machine, which we present as a variant of the classical finite-state machine.
Date: February 2013
pdf
5. Entry in bibtex format:
@techreport{DCS345IR,
author={M.H. van Emden},
title={Discovering algorithms with Matrix Code},
institution={Department of Computer Science,
University of Victoria},
number={Report DCS-345-IR},
note = {{\tt http://arxiv.org/pdf/1203.2296v2.pdf}, May 2012.}
}
Abstract:
In first-year programming courses it is often difficult to show students how an algorithm can be discovered. In this paper we present a program format that supports the development from specification to code in small and obvious steps; that is, a discovery process. The format, called Matrix Code, can be interpreted as a proof according to the Floyd-Hoare program verification method. The process consists of expressing the specification of a function body as an initial code matrix and then growing the matrix by adding rows and columns until the completed matrix is translated in a routine fashion to compilable code. As worked example we develop a Java program that generates the table of the first N prime numbers.
pdf
6. Entry in bibtex format:
@techreport{DCS343IR,
author={P. Kelly and M.H. van Emden},
title={Relational semantics
for databases and predicate calculus},
institution={Department of Computer Science,
University of Victoria},
number={Department of Computer Science report DCS-343-IR},
note = {{\tt http://arxiv.org/pdf/1202.0474v3.pdf}, February 2012.}
}
Abstract:
The relational data model requires a theory of relations in which tuples are not only many-sorted, but can also have indexes that are not necessarily numerical. In this paper we develop such a theory and define operations on relations that are adequate for database use. The operations are similar to those of Codd's relational algebra, but differ in being based on a mathematically adequate theory of relations. The semantics of predicate calculus, being oriented toward the concept of satisfiability, is not suitable for relational databases. We develop an alternative semantics that assigns relations as meaning to formulas with free variables. This semantics makes the classical predicate calculus suitable as a query language for relational databases.
pdf
7. Entry in bibtex format:
@techreport{RR746,
author={A. Nait Abdallah and M.H. van Emden},
title={Constraint propagation as information maximization},
institution={Department of Computer Science,
University of Western Ontario},
number={Research Report 746},
note = {{\tt http://arxiv.org/pdf/1201.5426v1.pdf}, January 2012.}
}
Abstract:
Dana Scott used the partial order among partial functions for his mathematical model of recursively defined functions. He interpreted the partial order as one of information content. In this paper we elaborate on Scott's suggestion of regarding computation as a process of information maximization by applying it to the solution of constraint satisfaction problems. Here the method of constraint propagation can be interpreted as decreasing uncertainty about the solution -- that is, as gain in information about the solution. As illustrative example we choose numerical constraint satisfaction problems to be solved by interval constraints. To facilitate this approach to constraint solving we formulate constraint satisfaction problems as formulas in predicate logic. This necessitates extending the usual semantics for predicate logic so that meaning is assigned not only to sentences but also to formulas with free variables.
pdf
8. Entry in bibtex format:
@article{vnmdn14,
author={M.H. van Emden},
title={Matrix Code},
journal={Science of Computer Programming},
pages={3--21},
year = {2014}
}
Abstract:
Matrix Code gives imperative programming a mathematical semantics and heuristic power comparable in quality to functional and logic programming. A program in Matrix Code is developed incrementally from a specification in pre/post-condition form. The computations of a code matrix are characterized by powers of the matrix when it is interpreted as a transformation in a space of vectors of logical conditions. Correctness of a code matrix is expressed in terms of a fixpoint of the transformation. The abstract machine for Matrix Code is the dual-state machine, which we present as a variant of the classical finite-state machine.
pdf
9. Entry in bibtex format:
@article{vnmdn14,
author={M.H. van Emden and A. Vellino},
title={From Chinese Room to Human Window},
journal={ICGA Journal},
pages={127--139},
year = {2010}
}
Abstract:
The debate in philosophy and cognitive science about the Chinese Room Argument has focused on whether it shows that machines can have minds. We present a quantitative argument which shows that Searle’s thought experiment is not relevant to Turing’s Test for intelligence. Instead, we consider a narrower form of Turing’s Test, one that is restricted to the playing of a chess endgame, in which the equivalent of Searle’s argument does apply. An analysis of time/space trade-offs in the playing of chess endgames shows that Michie’s concept of Human Window offers a hint of what a machine’s mental representations might need to be like to be considered equivalent to human cognition.
pdf
10. Entry in bibtex format:
@techreport{vnmdICLP,
author={M.H. van Emden},
title={Integrating Interval Constraints into Logic Programming},
institution={Department of Computer Science, University of Victoria},
number={DCS-133-IR},
note = {Paper arXiv:1002.1422 in Computing Research Repository (CoRR),
January 2010.}
}
Abstract:
The CLP scheme uses Horn clauses and SLD resolution to generate multiple constraint satisfaction problems (CSPs). The possible CSPs include rational trees (giving Prolog) and numerical algorithms for solving linear equations and linear programs (giving CLP(R)). In this paper we develop a form of CSP for interval constraints. In this way one obtains a logic semantics for the efficient floating-point hardware that is available on most computers. The need for the method arises because in the practice of scheduling and engineering design it is not enough to solve a single CSP. Ideally one should be able to consider thousands of CSPs and efficiently solve them or show them to be unsolvable. This is what CLP/NCSP, the new subscheme of CLP described in this paper is designed to do.
pdf
11. Entry in bibtex format:
@techreport{edmemd08,
author={W.W. Edmonson and M.H. van Emden},
title={Interval Semantics for Standard Floating-Point Arithmetic},
institution={Department of Computer Science, University of Victoria},
number={DCS-323-IR},
note = {Computing Research Repository
(http://arxiv.org/abs/0810.4196), 23 October 2008}
}
Abstract:
If the non-zero finite floating-point numbers are interpreted as point intervals, then the effect of rounding can be interpreted as computing one of the bounds of the result according to interval arithmetic. We give an interval interpretation for the signed zeros and infinities, so that the undefined operations 0*inf, inf - inf, inf/inf, and 0/0 become defined. In this way no operation remains that gives rise to an error condition. Mathematically questionable features of the floating-point standard become well-defined sets of reals. Interval semantics provides a basis for the verification of numerical algorithms. We derive the results of the newly defined operations and consider the implications for hardware implementation.
pdf
12. Entry in bibtex format:
@techreport{vemoa06,
author={M.H. van Emden and B. Moa},
title={The Fundamental Theorems of Interval Analysis},
institution={Department of Computer Science, University of Victoria},
number={DCS-316-IR},
date = {December 2, 2006},
note = {Computing Research Repository (http://arxiv.org/corr/home)}
}
Abstract:
Expressions are not functions. Confusing the two concepts or failing to define the function that is computed by an expression weakens the rigour of interval arithmetic. We give such a definition and continue with the required re-statements and proofs of the fundamental theorems of interval arithmetic and interval analysis.
pdf
13. Entry in bibtex format:
@techreport{vnmdSTPCS,
author={M.H. van Emden},
title={Set-Theoretic Preliminaries for Computer Scientists},
institution={Department of Computer Science, University of Victoria},
number={DCS-304-IR},
note = {Paper cs.DM/0607039 in Computing Research Repository (CoRR),
July 2006.
}
}
Abstract:
The basics of set theory are usually copied, directly or indirectly, by computer scientists from introductions to mathematical texts. Often mathematicians are content with special cases when the general case is of no mathematical interest. But sometimes what is of no mathematical interest is of great practical interest in computer science. For example, non-binary relations in mathematics tend to have numerical indexes and tend to be unsorted. In the theory and practice of relational databases both these simplifications are unwarranted. In response to this situation we present here an alternative to the set-theoretic preliminaries'' usually found in computer science texts. This paper separates binary relations from the kind of relations that are needed in relational databases. Its treatment of functions supports both computer science in general and the kind of relations needed in databases. As a sample application this paper shows how the mathematical theory of relations naturally leads to the relational data model and how the operations on relations are by themselves already a powerful vehicle for queries.
Downloadable from CoRR, the Computing Research Repository.
14. Entry in bibtex format:
@techreport{vnmdn06b,
author={M.H. van Emden and B. Moa},
title={Computational Euclid},
institution={Department of Computer Science, University of Victoria},
number={DCS-315-IR},
month = {June},
year = {2006}
}
Abstract:
We analyse the axioms of Euclidean geometry according to standard object-oriented software development methodology. We find a perfect match: the main undefined concepts of the axioms translate to object classes. The result is a suite of C++ classes that efficiently supports the construction of complex geometric configurations. Although all computations are performed in floating-point arithmetic, they correctly implement as semi-decision algorithms the tests for equality of points, a point being on a line or in a plane, a line being in a plane, parallelness of lines, of a line and a plane, and of planes. That is, in accordance to the fundamental limitations to computability requiring that only negative outcomes are given with certainty, while positive outcomes only imply possibility of these conditions being true.
Downloadable from CoRR, the Computing Research Repository.
15. Entry in bibtex format:
@techreport{vnmdn06b,
author={M.H. van Emden and S.C. Somosan},
title={Object-Oriented Frameworks as Basis
for Modularity in Program-Language Design},
institution={Department of Computer Science, University of Victoria},
number={DCS-310-IR},
month = {March},
year = {2006}
}
Abstract:
For the right application, the use of programming paradigms such as functional or logic programming can enormously increase productivity in software development. But powerful paradigms may come with exotic programming languages, while the management of software development dictates language standardization.
This dilemma can be resolved by using component technology at the system design level. Here the units of deployment are object-oriented frameworks. It is conventional to analyze an application by object-oriented modeling. In the new approach, the analysis identifies the programming paradigm that is ideal for the application; development starts with object-oriented modeling of the paradigm. In our approach, a paradigm translates to an object-oriented framework so that it is no longer necessary to embody a programming paradigm in a language dedicated to it.
Downloadable from CoRR, the Computing Research Repository.
16. Entry in bibtex format:
@inproceedings{vnmdn06,
author = {M.H. van Emden},
title={Compositional Semantics for the Procedural Interpretation of Logic},
booktitle={Proc. Intern. Conf. on Logic Programming},
editor={S. Etalle and M. Truszczy\'nski},
publisher={Springer Verlag},
number = {LNCS 4079},
pages = {315 -- 329},
year = {2006}
}
Abstract:
Semantics of logic programs has been given by proof theory, model theory and by fixpoint of the immediate-consequence operator. If clausal logic is a programming language, then it should also have a compositional semantics. Compositional semantics for programming languages follows the abstract syntax of programs, composing the meaning of a unit by a mathematical operation on the meanings of its constituent units. The procedural interpretation of logic has only yielded an incomplete abstract syntax for logic programs. We complete it and use the result as basis of a compositional semantics. We present for comparison Tarski's algebraization of first-order predicate logic, which is in substance the compositional semantics for his choice of syntax. We characterize our semantics by equivalence with the immediate-consequence operator.
A slightly earlier research report version is downloadable from CoRR, the Computing Research Repository.
17. Entry in bibtex format:
@misc{mhve06rev,
author={M.H. van Emden},
title={Review of Apt's Principles of Constraint Programming'' and of
Dechter's Constraint Processing},
journal={SIAM Review},
volume={48},
number={2},
pages={400 -- 404},
year = {2006}
}
Abstract:
Two books on constraint programming are reviewed in the SIAM Review, a journal of the Society for Industrial and Applied Mathematics. As constraint programming grew up in, and lives in, the Artificial Intelligence community, the review is largely an account of the background and history of constraint programming, emphasizing its roots in engineering and its connections with Operations Research.
18. Entry in bibtex format:
@inproceedings{vnmdn03,
author={M.H. van Emden},
title={Using the Duality Principle to improve lower bounds for the global
minimum in nonconvex optimization},
booktitle={Second COCOS workshop on intervals and optimization},
year = 2003
}
pdf
19. Entry in bibtex format:
@collection{vnmoa03a,
author = {M.H. van Emden and B. Moa},
title={Termination criteria in the {M}oore-{S}kelboe Algorithm
for Global Optimization by Interval Arithmetic},
booktitle={Frontiers in Global Optimization},
editor={C.A. Floudas and P.M. Pardalos},
publisher={Kluwer Academic Publishers},
year = {2003}
}
Abstract:
We investigate optimization with an objective function that has an unknown and possibly large number of local minima. Determining what the global minimum is we call the fathoming problem ; where it occurs we call the localization problem . Another problem is that in practice often not only the global minimum is required, but also possibly existing points that achieve near-optimality, yet are far from the point at which the global minimum occurs. To support such a requirement, we define the delta-minimizer, the set of points at which the objective function is within delta of the global minimum. We present a modification of the Moore-Skelboe algorithm that returns two sets of boxes. One set contains a delta minimizer for a delta d1; the other is contained within a delta minimizer for a delta d2. In this way one can detect whether low values are concentrated around the global minimum or whether there is a large area with objective function values that are close to the global minimum. We include a proof of correctness of the algorithm.
ps ... pdf
20. Entry in bibtex format:
@inproceedings{vnmdn02,
author={M.H. van Emden},
title={Combining Numerical Analysis and Constraint Processing by Means
of Controlled Propagation and Redundant Constraints},
booktitle={First COCOS workshop on intervals and optimization; text
downloadable from CoRR},
year = 2002
}
Abstract:
In principle, interval constraints provide tight enclosures for the solutions of several types of numerical problem. These include constrained global optimization and the solution of nonlinear systems of equalities or inequalities. Interval constraints finds these enclosures by a combination of propagation and search. The challenge is to extend the in principle'' to problems of practical interest. In this paper we describe the concept of controlled propagation. It uses this in conjunction with redundant constraints to combine numerical analysis algorithms with constraint processing. The resulting combination retains the enclosure property of constraint processing in spite of rounding errors. We apply this technique in an algorithm for solving linear algebraic equations that initially simulates interval Gaussian elimination and then proceeds to refine the result with propagation and splitting. Application of our approach to nonlinear equations yields an algorithm with a similar relation to Newton's method.
21. Entry in bibtex format:
@article{hckvnmdn01,
author={T. Hickey and Q. Ju and M.H. van Emden},
title={Interval Arithmetic: from Principles to Implementation},
journal={Journal of the ACM},
year = 2001
}
Abstract:
We start with a mathematical definition of a real interval as a closed, connected set of reals. Interval arithmetic operations (addition, subtraction, multiplication and division) are likewise defined mathematically and we provide algorithms for computing these operations assuming exact real arithmetic. Next, we define interval arithmetic operations on intervals with IEEE 754 floating point endpoints to be sound and optimal approximations of the real interval operations and we show that the IEEE standard's specification of operations involving the signed infinities, signed zeros, and the exact/inexact flag are such as to make a sound and optimal implementation more efficient. From the resulting theorems we derive data that are sufficiently detailed to convert directly to a program for efficiently implementing the interval operations. Finally we extend these results to the case of general intervals, which are defined as connected sets of reals that are not necessarily closed.
ps pdf
22. Entry in bibtex format:
@article{vnmdn99a,
title={Algorithmic Power from Declarative Use of
Redundant Constraints},
author={M.H. van Emden},
journal={Constraints},
year={1999},
pages={363--381}
}
Abstract:
Interval constraints can be used to solve problems in numerical analysis. In this paper we show that one can improve the performance of such an interval constraint program by the declarative use of constraints that are redundant in the sense of not needed to define the problem. The first example shows that computation of an unstable recurrence relation can be improved. The second example concerns a solver of nonlinear equations. It shows that, by adding as redundant constraints instances of Taylor's theorem, one can obtain convergence that appears to be quadratic.
ps ... pdf
23. Entry in bibtex format:
@collection{vnmdnShkrtwn,
author={M.H. van Emden},
title={The Logic Programming Paradigm in Numerical Computation},
booktitle={The Logic Programming Paradigm},
editor={Krzysztof R. Apt and Victor W. Marek
and Miroslaw Truszczynski and David S. Warren},
publisher={Springer-Verlag},
pages={257--276},
year={1999}
}
Abstract:
Although CLP(R) is a promising application of the logic programming paradigm to numerical computation, it has not addressed what has long been known as the pitfalls of [numerical] computation''. These show that rounding errors induce a severe correctness problem wherever floating-point computation is used. Independently of logic programming, constraint processing has been applied to problems in terms of real-valued variables. By using the techniques of interval arithmetic, constraint processing can be regarded as a computer-generated proof that a certain real-valued solution lies in a narrow interval. In this paper we propose a method for interfacing this technique with CLP(R). This is done via a real-valued analogy of Apt's proof-theoretic framework for constraint processing.
ps ... pdf
24. Entry in bibtex format:
@article{hcqvn99,
title={Interval Constraint Plotting for Interactive Visual Exploration
of Implicitly Defined Relations},
author={Timothy J. Hickey and Zhe Qiu and Maarten H. van Emden},
journal={Reliable Computing},
pages={81--92},
volume={6},
year={2000}
}
Abstract:
Conventional plotting programs adopt techniques such as adaptive sampling to approximate, but not to guarantee, correctness and completeness in graphing functions. Moreover, implicitly defined mathematical relations can impose an even greater challenge as they either cannot be plotted directly, or otherwise are likely to be misrepresented. In this paper, we address these problems by investigating interval constraint plotting as an alternative approach that plots a hull of the specified curve. We present some empirical evidence that this hull property can be achieved by a O(n) algorithm. Practical experience shows that the hull obtained is the narrowest possible whenever the precision of the underlying floating-point arithmetic is adequate. We describe IASolver, a Java applet that serves as test bed for this idea.
ps ... pdf
25. Entry in bibtex format:
@article{vnmd97,
title={Value Constraints in the {CLP} {S}cheme},
author={M.H. van Emden},
year={1997},
journal={Constraints},
volume={2},
pages={163--183}
}
Abstract:
We define value constraints, a method for incorporating constraint propagation into logic programming. It is a subscheme of the CLP scheme and is applicable wherever one has an efficient method for representing sets of possible values. As examples we present: small finite sets, sets of ground instances of a term, and intervals of reals with floating-point numbers as bounds. Value constraints are defined by distinguishing two storage management strategies in the CLP scheme. In value constraints the infer step of the CLP scheme is implemented by Waltz filtering. We give a semantics for value constraints in terms of set algebra that gives algebraic characterizations of local and global consistency. The existing extremal fixpoint characterization of chaotic iteration is shown to be applicable to prove convergence of Waltz filtering.
ps
26. Entry in bibtex format:
@inproceedings{vnmd97pacrim,
title={Object-oriented programming as the end of history in
programming languages},
author={M.H. van Emden},
year={1997},
booktitle={1997 IEEE Pacific Rim Conference on Communications, Computers, and Signal Processing},
volume={2},
pages={981--984},
publisher = {IEEE}
}
Abstract:
In the past, the invention of a new programming paradigm led to new programming languages. We show that C++ is a perfectly adequate dataflow programming language, given some suitable definitions. Reasons are mentioned for believing that this is also the case for logic and functional programming. We conclude that object-oriented programming may remove the need for languages motivated by specific programming paradigms.
pdf
27. Entry in bibtex format:
@inproceedings{mcheng95,
author={M.H.M. Cheng and D.Stott Parker and M.H. van Emden},
title = {A Method for Implementing Equational Theories As Logic Programs},
booktitle = {Proceedings of the Twelfth International Conference on
Logic Programming},
year={1995},
pages={497--511},
editor={Leon Sterling},
publisher={MIT Press}
}
Abstract:
Equational theories underlie many fields of computing, including functional programming, symbolic algebra, theorem proving, term rewriting and constraint solving. In this paper we show a method for implementing many equational theories with a limited class of logic programs. We define regular equational theories, a useful class of theories, and illustrate with a number of examples how our method can be used in obtaining efficient implementations for them. The significance of our method is that: * It is simple and easy to apply. * Although executable, it supports separation of concerns between specification and implementation. * Its class of logic programs execute with impressive efficiency using Prolog. * It permits interesting compilation and optimization techniques that can improve execution efficiency still further. * It offers perspectives on term rewriting and functional programming evaluation strategies, how they can be compiled, and how they can be integrated with logic programming effectively.
ps
28. Entry in bibtex format:
@inproceedings{ldrSwnklsVE95,
author={W.J. Older and G.M. Swinkels and M.H. van Emden},
title = {Getting to the real problem: experience with BNR
Prolog in OR},
booktitle = {Proceedings of the Third Conference on
Practical Applications of Prolog},
year={1995}
}
Abstract:
Although job-shop scheduling is a much studied problem in OR, it is based on an unrealistic restriction, which is needed to make the problem computationally more tractable. In this paper we drop the restriction. As a result we encounter a type of cardinality constraint for which we have to develop a new method: translation to a search among alternative sets of inequalities between reals. Our solution method depends on logic programming: we run a specification and rely on the underlying interval constraint-solving machine of BNR Prolog to reduce the search space to a feasible size. In this way, by making the programming task trivial, it is possible to tackle the real problem rather than a related one for which code happens to be already written.
pdf
29. Entry in bibtex format:
@article{emden92a,
author={M.H. van Emden},
title={Mental ergonomics as basis for new-generation
computer systems},
journal={},
volume={2},
year={1992},
pages={133--153}
}
@inproceedings{emden92a,
author = {M.H. van Emden},
title={Mental ergonomics as basis for new-generation
computer systems},
booktitle = {Proceedings of the International Conference
on Fifth-Generation Computer Systems 1992},
publisher = {Ohmsha},
year = 1992,
pages = {1149--1156}
}
Abstract:
Reliance on Artificial Intelligence suggests that Fifth-Generation Computer Systems were intended as a substitute for thought. The more feasible and useful objective of a computer is as an aid to thought suggests mental ergonomics rather than Artificial Intelligence as the basis for new-generation computer systems. This objective, together with considerations of software technology, suggests logic programming as a unifying principle for a computer aid to thought.
pdf
30. Entry in bibtex format:
@article{emden92b,
author={M.H. van Emden},
title={Structured Inspections of Code},
journal={Software Testing, Verification, and Reliability},
volume={2},
year={1992},
pages={133--153}
}
Abstract:
Cleanroom programming and code inspections independently provide evidence that it is more efficient to postpone the testing of code to a later stage than is usually done. This paper argues that an additional gain in quality and efficiency of development can be obtained by structuring inspections by means of an inspection protocol. The written part of such a protocol is prepared by the programmer before the inspection. It is modeled on Floyd's method for the verification of flowcharts. However, the protocol differs from Floyd's method in being applicable in practice. Structured inspections gain this advantage by not attempting to be a proof; they are no more than an articulation of existing forms of inspection. With the usual method of structured programming it may be difficult to prepare the inspection protocol. On the other hand, Assertion-Driven Programming (of which an example is included in this paper) not only facilitates protocol preparation, but also the coding itself.
ps
31. M.H. van Emden: "Rhetoric versus modernism in computing" J Logic Computation (1992) 2 (5): 551-555.
pdf
32. Entry in bibtex format:
@inproceedings{cvr90,
author = {M.H.M. Cheng and M.H. van Emden and B.E. Richards},
title = {On {W}arren's Method for Functional Programming in Logic},
booktitle = {Logic Programming: Proceedings of the Seventh International
Conference},
address = {Jerusalem},
editor = {David H.D. Warren and Peter Szeredi},
publisher = {MIT Press},
year = 1990,
pages = {546--560}
}
Abstract:
Although Warren's method for the evaluation in Prolog of expressions with higher-order functions appears to have been neglected, it is of great value. Warren's paper needs to be supplemented in two respects. He showed examples of a translation from lambda-expressions to clauses, but did not present a general method. Here we present a general translation program and prove it correct with respect to the axioms of equality and those of the lambda-calculus. Warren's paper only argues in general terms that a structure-sharing Prolog implementation can be expected to efficiently evaluate the result of his translation. We show a comparison of timings between Lisp and a structure-copying implementation of Prolog. The result suggests that Warren's method is about as efficient as the Lisp method for the evaluation of Lambda expressions involving higher-order functions.
33. pdf
34. @inproceedings{condAnsw88,
author={M.H. van Emden},
title = {Conditional Answers for Polymorphic Type Inference},
booktitle = {Logic Programming: Proceedings of the International
Conference and Symposium, volume I},
editors = {R.A. Kowalski and K.A. Bowen},
year={1988},
publisher = {MIT Press},
pages = {590--603}
}
Abstract:
J.H. Morris showed that polymorphic type inference can be done by unification. In this paper we show that the unification problem can be automatically generated by resolution and factoring acting on a theory of type inheritance in the form of Horn clauses. The format is a variant of SLD-resolution as used in logic programming. In this variant, programs have an empty least Herbrand model, so that all derivations are failed'' in the conventional sense. Yet \emph{conditional answers} provide as much and as securely justified information as do the successful answers exclusively used in conventional programs.
35. Entry in bibtex format:
@inproceedings{cvl88,
author={M.H.M. Cheng and M.H. van Emden and J.H.M. Lee},
title = {Tables as a User Interface for Logic Programs},
booktitle = {Proceedings of the International Conference on Fifth
Generation Computer Systems 1988},
year={1988},
month = {November--December},
address = {Tokyo, Japan},
publisher = {Ohmsha, Ltd},
pages = {784--791}
}
Abstract:
Spreadsheets have introduced two advantages not typically available in user interfaces to logic programs: the exploratory use of a computer and a two-dimensional interface. In this paper we show that not only spreadsheets, but also tables (in the sense of relational databases) have these valuable features. We compare spreadsheets and tables, giving possibly the first clear distinction between the two and suggest a common generalization. We show that tables, as a user interface for logic programs, can be derived from a dataflow model of queries (which we call TuplePipes), which provides also the buffering needed when Prolog is interfaced with a relational database. We report on Tupilog, a prototype implementation of logic programming allowing four query modes, one of which is TuplePipes.
pdf
36. Entry in bibtex format:
@article{vnm86,
author={M.H. van Emden},
title={Quantitative Deduction and Its Fixpoint Theory},
journal={Journal of Logic Programming},
year={1986},
volume={4},
pages={37--53}
}
Abstract:
Logic programming provides a model for rule-based reasoning in expert systems. The advantage of this formal model is that it makes available many results from the semantics and proof theory of first-order predicate logic. A disadvantage is that in expert systems one often wants to use, instead of the usual two truth values, an entire continuum of uncertainties'' in between. That is, instead of the usual qualitative'' deduction, a form of quantitative'' deduction is required. We present an approach to generalizing the Tarskian semantics of Horn clause rules to justify a form of quantitative deduction. Each clause receives a numerical attenuation factor. Herbrand interpretations, which are subsets of the Herbrand base, are generalized to subsets which are fuzzy in the sense of Zadeh. We show that as result the fixpoint method in the semantics of Horn clause rules can be developed in much the same way for the quantitative case. As for proof theory, the interesting phenomenon is that a proof should be viewed as a two-person game. The value of the game turns out to be the truth value of the atomic formula to be proved, evaluated in the minimal fixpoint of the rule set. The analog of the PROLOG interpreter for quantitative deduction becomes a search of the game tree (= proof tree) using the alpha-beta heuristic well known in game theory.
pdf
37. Entry in bibtex format:
@article{vkw76,
author={M.H. van Emden and R.A. Kowalski},
title={The Semantics of Predicate Logic as a Programming Language},
journal=JACM,
year={1976},
volume={23},
number={4},
pages={733--742}
}
Abstract:
Sentences in first-order predicate logic can be usefully interpreted as programs. In this paper the operational and fixpoint semantics of predicate logic programs are defined, and the connections with the proof theory and model theory of logic are investigated. It is concluded that operational semantics is a part of proof theory and that fixpoint semantics is a special case of model-theoretic semantics.
pdf
38. Entry in bibtex format:
@techreport{vnmdn74,
author={M.H. van Emden},
title={First-order predicate logic as a high-level program
language},
institution={School of Artificial Intelligence,
University of Edinburgh},
number={MIP-R-106},
year={1974}
}
Abstract:
This paper presents an argument in support of the thesis that first-order predicate logic would be a useful next step in the development towards higher-level program languages. The argument is conducted by giving a description of Kowalski's system of logic which is sufficiently detailed to investigate its computation behaviour in the two examples discussed: a version of the "quicksort" algorithm and a top-down parser for context-free languages.
pdf
39. "An analysis of complexity" by M.H. van Emden.
Mathematical Centre Tracts #35, 1971.
Abstract:
Complexity of a system is defined in terms of interactions expressed as Shannon's information-theoretic entropy. As a result the system can be hierarchically decomposed into levels of subsystems of the basis of entropy. Applications to data-based classification and to solving of systems of linear equations are presented.
This monograph was accepted as doctoral thesis by the Faculty of Mathematics and Natural Sciences of the University of Amsterdam. It supercedes the author's paper "Hierarchical decomposition of complexity", Machine Intelligence 5, B. Meltzer and D. Michie (eds.), Edinburgh University Press, 1970.
pdf
|
On a subset sum version.
In subset sum we ask 'Given $n$ numbers in $\Bbb Z$, is there a subset of them that sums to $0$?' this is $NP$ complete.
Consider variant:
'Given $n$ of degree at most $d$ polynomials in $\Bbb Z[x]$ with coefficients in $\{0,1\}$ is there a subset of them that sums to $x^{d-1}+x^{d-2}+\dots+1$?'
Is this $NP$ complete? Is there any approximation algorithm?
• I don't understand the approximation part of the question. What would you be approximating? – Kyle Jones Feb 23 '16 at 3:29
The problem you describe is equivalent to the EXACT COVER problem, which is known to be NP-complete.
The EXACT COVER problem definition from Wikipedia:
In mathematics, given a collection $S$ of subsets of a set $X$, an exact cover is a subcollection $S^*$ of $S$ such that each element in $X$ is contained in exactly one subset in $S^*$.
EXACT COVER reduces to your problem as follows:
Set $d$ equal to $|X|$. For each element in $X$ map a unique power of $x$ to it, always less than $d$. For each subset of $X$ in $S$ build a polynomial by replacing the set elements with their mappings to powers of $x$ and adding plus signs between them.
Now, any solution to your problem i.e. a set of polynomials that sums to $x^{d-1}+x^{d-2}+\dots+1$ is also a solution to the reduced EXACT COVER problem. The EXACT COVER solution, $S^*$, can be recovered by reversing the set element mappings in the solution polynomials and removing the plus signs.
• I've provided the reduction. – Kyle Jones May 11 '16 at 20:42
• like approximately summing in some norm? – T.... May 12 '16 at 2:05
|
# RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.2
### RD Sharma Class 10 Solutions Chapter 8 Ex 8.2 PDF Free Download
#### Exercise 8.2
Question 1: The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.
Solution:
Given that the smallest integer of 2 consecutive integers is denoted by x
The two integers be x and x+1
According to the question, the product of the integers is 306
Now,
X(x+1) = 306 = x2+x-306=0
The required quadratic equation of the equation is x2+x-306=0
Question 2: John and Jivani together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they to start with if John had x marbles.
Solution:
Given that John and Jilani are having the total of 45 marbles.
Let us consider John is having x marbles
Jivani is having (45-x) marbles.
Number of marbles john had after losing 5 marbles = x-5
Number of marbles jivani had after losing 5 marbles = (45-x)-5 = 40-x
According to the question the product of the marbles that they are having now is 128
Now, (x-5)(40-x) = 128
= 40x-x2-200 = 128
= x2 -45x+128+200 =0
= x2 -45x+328 = 0
The required quadratic equation is x2 -45x+328 = 0.
Question 3: A cottage industry produces a certain number of toys in a day. The cost of production of each toy was found to be 55 minutes the number of articles produced in a day. On a particular day , the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.
Solution:
Given
(y) Denotes the number of toys produced in a day.
The cost of production of each toy = (55 – y)
Total cost of production is nothing but the product of number of toys produced in a day and cost of production of each toy = y (55-y)
According to the question
The total cost of production is Rs.750
= y (55-y) = 750
= 55y-y2 = 750
= y2 -55y+750= 0
The required quadratic equation of the given data is y2 -55y+750= 0.
Question 4: The height of the right triangle is 7 cm less than its base. If the hypotenuse is 13cm, form the quadratic equation to find the base of the triangle.
Solution:
According to the question
The hypotenuse of the triangle = 13 cm
Let the base of the triangle = x cm
So, the height of the triangle = (x-7) cm
Applying Pythagoras theorem in the right angled triangle, we get,
(Base)2 + (height)2 = (hypotenuse)2
X2 +(x-7)2 = (13)2
X2 + x2 + 49 -14x =169
2x2 -14x -120 =0
2(x2 -7x -60) =0
x2 -7x -60 =0
The required quadratic equation is x2 -7x -60 =0
Question 5: The average speed of the express train is 11 km/hr more than that of the passenger train. The total distance covered by the train is 132 km. Also, time taken by the express train is 1 hour is less than that of the passenger train. Find the quadratic equation of this problem.
Solution:
Let the average speed of the express train be = x km /hr
Given, the average speed of the express train is 10 km/ hr less than that of passenger train = (x-11) km/hr
We know that:
Time taken for travel = distance travelled / average speed
Time taken for express train = distance travelled / average speed of the express train
= $\frac{132}{x}$
Hence time taken by the passenger train = $\frac{132}{x-11}$
According to the question,
Time taken by the express train is 1 hour less than that of passenger train
Time taken by passenger train – time taken by express train = 1 hour
$\frac{132}{x-11}$$\frac{132}{x}$ = 1
$132(\frac{1}{x-11}-\frac{1}{x})$ =1
$132(\frac{x-(x-11)}{x\times (x-11)})$ =1
$132(\frac{x-x+11}{x^{2}-11x})$ = 1
$132(\frac{11}{x^{2}-11x})$ = 1
1452 = x2 –11x
X2 -11x – 1452 = 0
The quadratic equation of the given problem is X2 -11x – 1452 = 0.
Question 6: A train travels 360 km at a uniform speed. If the speed had been 5 km/ hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
Solution:
Let the speed of the train be = x km /hr
Distance travelled by the train = 360 km
We know that,
Time taken for travel = distance travelled ÷ speed of the train
= $\frac{360}{x}$
If the speed of the train is increased by 4 km /hr then time taken = $\frac{360}{x+5}$
According to the question,
The time of travel is reduced by 1 hour when the speed of the train is increased by 5 km /hr
$\frac{360}{x}$$\frac{360}{x+5}$ = 1
$360(\frac{1}{x}-\frac{1}{x+5})$ =1
$360(\frac{x+5-x}{x(x+5)})$ =1
$360(\frac{5}{x(x+5)})$ =1
$360(\frac{5}{x^{2}+5x})$ =1
X2 + 5x = 1800
The required quadratic equation is X2 + 5x- 1800=0.
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# Topological Quantum Field Theory
For a topological quantum field theory, $Z:Cob(n)\to Vect(\mathbb{C})$ why is it that typically $Z(\emptyset)\cong \mathbb{C}$? Is that just the definition that makes everything work?
Look at disjoint union of a manifold and an empty manifold. So you're getting $V$ and $C$ so when you combine you better get $V$ again. This is the monoidal unit requirement.
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If $\overrightarrow a,\overrightarrow b,\overrightarrow c$ are three coplanar vectors then, $[(2\overrightarrow a-\overrightarrow b)\:(2\overrightarrow b-\overrightarrow c)\:(2\overrightarrow c-\overrightarrow a)]=?$
$(a)\:0\:\:\:\qquad\:\:(b)\:\:1\:\:\:\qquad\:\:(c)\:\:\sqrt 3\:\:\:\qquad\:\:(d)\:\:None\:of\:these.$
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# Need another confirmation
1. May 5, 2005
### abia ubong
while working on the quadratic formula in school ,i came across another formula,and wanted to know if its been derived since i got this from the quadratic formula.the formula is as follows
[-b^3-2abc +or- sqrt(b^6-12a^2b^2c^2-16a^3c^3)]/(2ab^2+4a^2c).
please i want to know if it works for all.
2. May 5, 2005
### dextercioby
$$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b^{3}-2abc\pm\sqrt{b^{6}-12a^{2}b^{2}c^{2}-16a^{3}c^{3}}}{2ab^{2}+4a^{2}c}$$
...?There's only one way to find out:CROSS MULTIPLY...
Daniel.
3. May 5, 2005
### shyboy
looks like the multiplication of both the nominator and the denominator by
$$b^{2}+2ac}$$
4. May 5, 2005
### inha
regardless of if it works or not the form on the right is highly useless.
5. May 6, 2005
### abia ubong
that was a hard thing to say inha, how useless is it ,i just came across it and felt like letting the forum know how correct it is and u call it useless.
thae was a hard thing 2 say
6. May 6, 2005
### dextercioby
I'm sorry not to share your disspointment,but it's nothing more than a waste of ink and paper (or server space)...
Daniel.
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# What is -regmat-
regmat is prefix command to a regression template. A regression template is simply a regression command. Each combination of one outcome variable, one exposure variable and one adjustment set is inserted just after the regression command. Then all regression estimates of exposures are placed in a matrix ordered by outcome and exposure variables rowwise and adjustment columnwise.
The resulting matrix is saved in the return list for further usage.
The matprint options makes it easy to integrate the result table into a log2markup output file.
Together with the basetable the command regmat generates the two typical tables for reporting epidemiological research. The resulting tables are easy to integrate into the final text using eg. log2markup output file or -putexcel-.
-regmat- is a part of the package matrixtools.
## Syntax
The syntax is: regmat [using], outcomes(varlist) exposures(varlist) [adjustements(varlist strings) noquietly labels base keep(string) drop(string) Matprint options] , regression template
## Options
• outcomes(varlist): A non-empty varlist of outcome variables. An outcome is the dependent variable in a regression.
• exposures(varlist): A non-empty varlist of outcome variables. Exposures are variables of whiech estimates are to be reported.
• adjustments(string): A set of varlist strings. A varlist string is a possibly empty set of adjustment variables. Each varlist string is surrounded in text quotes ("). An empty string ("") means no adjustment. Adjustment variables are variables needed for the estimation of the exposures, but it is not necessary to report their estimates.
• noquietly: If set, regression outputs are printed in the log.
• labels: Use variable and value labels.
• base: Include base values at factor variables.
• keep: To style output choose which calculations to keep. Choices are: b(=estimate of exposure in regression), se(=Se(estimate)), ci(=Confidence interval - level is set with set level), and p(=P-value).
• drop: To style output choose which calculations to drop. Choices are: b(=estimate of exposure in regression), se(=Se(estimate)), ci(=Confidence interval - level is set with set level), and p(=P-value).
To see -matprint- options
## Stored results
-regmat- stores the following in r():
local
Matrix
• r(regmat) The matrix containing regression estimates of the exposures
## Versions
-regmat- is tested in version 12.1 ic, 13.1 ic, 14.2 ic, and 15.1 ic.
## Installation
To install use the command: ssc install matrixtools
# A demonstration of -regmat-
## Background
The data is from Hosmer and Lemeshow, Applied logistic regression, 1989 and the description from Rachel MacKay Altman's old homepage
Low birth weight is an outcome that has been of concern to physicians for years.
This is due to the fact that infant mortality rates and birth defect rates are very high for low birth weight babies.
A woman's behaviour during pregnancy (including diet, smoking habits, and receiving prenatal care) can greatly alter the chances of carrying the baby to term and, consequently, of delivering a baby of normal birth weight.
The goal of this study was to identify risk factors associated with giving birth to a low birth weight baby (weighing less than 2500 grams).
Data were collected on 189 women, 59 of whom had low birth weight babies and 130 of whom had normal birth weight babies.
The observed predictor variables have been shown to be associated with low birth weight in the obstetrical literature.
The goal of the current study was to ascertain if these variables were important in the population being served by the medical centre where the data were collected.
## The dataset
webuse lbw, clear
rename low bwlt2500
generate bwlt1500 = bwt < 1500 if !missing(bwt)
label variable bwlt1500 "birthweight < 1500g"
label define no_yes 0 "No" 1 "Yes"
label values bwlt* no_yes
## The Analysis
Crude estimates as well as estimates adjusted for ftv and ptl of the effect of smoke and age on the two outcomes bwlt2500 and bwlt1500 is presented below.
Note that the same type of regression (template: logit, vce(robust) or) is used.
regmat, outcome(bwlt2500 bwlt1500) exposure(i.smoke age) ///
adjustments("" "ftv i.ptl"): logit, vce(robust) or
--------------------------------------------------------------------------------------------------------------------------------------
b se(b) Lower 95% CI Upper 95% CI P value b se(b) Lower 95% CI Upper 95% CI P value
--------------------------------------------------------------------------------------------------------------------------------------
bwlt2500 smoke(1) 2.02 1.38 1.08 3.79 0.03 1.76 1.41 0.90 3.43 0.10
age 0.95 1.03 0.90 1.01 0.08 0.93 1.04 0.87 1.00 0.05
bwlt1500 smoke(1) 1.04 2.53 0.17 6.39 0.97 0.68 2.77 0.09 5.04 0.71
age 1.16 1.05 1.05 1.27 0.00 1.19 1.06 1.05 1.35 0.00
--------------------------------------------------------------------------------------------------------------------------------------
The return list contains the adjustments and the estimates in a matrix.
return list
macros:
matrices:
r(regmat) : 4 x 10
The estimates in the matrix looks like:
matprint r(regmat), decimals(3)
--------------------------------------------------------------------------------------------------------------------------------------
b se(b) Lower 95% CI Upper 95% CI P value b se(b) Lower 95% CI Upper 95% CI P value
--------------------------------------------------------------------------------------------------------------------------------------
bwlt2500 smoke(1) 2.022 1.378 1.079 3.789 0.028 1.761 1.405 0.904 3.430 0.096
age 0.950 1.030 0.897 1.006 0.079 0.932 1.036 0.869 0.998 0.045
bwlt1500 smoke(1) 1.037 2.529 0.168 6.390 0.969 0.682 2.775 0.092 5.041 0.708
age 1.157 1.049 1.054 1.270 0.002 1.191 1.064 1.055 1.346 0.005
--------------------------------------------------------------------------------------------------------------------------------------
If options labels and base are added:
regmat, outcome(bwlt2500 bwlt1500) exposure(i.smoke age) ///
adjustments("" "ftv i.ptl") labels base: logit, vce(robust) or
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
b se(b) Lower 95% CI Upper 95% CI P value b se(b) Lower 95% CI Upper 95% CI P value
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
birthweight<2500g smoked during pregnancy (0) 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
smoked during pregnancy (1) 2.02 1.38 1.08 3.79 0.03 1.76 1.41 0.90 3.43 0.10
age of mother 0.95 1.03 0.90 1.01 0.08 0.93 1.04 0.87 1.00 0.05
birthweight < 1500g smoked during pregnancy (0) 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
smoked during pregnancy (1) 1.04 2.53 0.17 6.39 0.97 0.68 2.77 0.09 5.04 0.71
age of mother 1.16 1.05 1.05 1.27 0.00 1.19 1.06 1.05 1.35 0.00
--------------------------------------------------------------------------------------------------------------------------------------------------------------------
The variable smoke has no value label:
metadata
--------------------------------------------------------------------------------------------------------------------------------------------------
Name Index Label Value Label Name Format Value Label Values n unique missing
--------------------------------------------------------------------------------------------------------------------------------------------------
id 1 identification code %8.0g 189 189 0
bwlt2500 2 birthweight<2500g no_yes %8.0g 0 "No" 1 "Yes" 189 2 0
age 3 age of mother %8.0g 189 24 0
lwt 4 weight at last menstrual period %8.0g 189 76 0
race 5 race race %8.0g 1 "white" 2 "black" 3 "other" 189 3 0
smoke 6 smoked during pregnancy %8.0g 189 2 0
ptl 7 premature labor history (count) %8.0g 189 4 0
ht 8 has history of hypertension %8.0g 189 2 0
ui 9 presence, uterine irritability %8.0g 189 2 0
ftv 10 number of visits to physician during 1st trimester %8.0g 189 6 0
bwt 11 birthweight (grams) %8.0g 189 133 0
bwlt1500 12 birthweight < 1500g no_yes %10.0g 0 "No" 1 "Yes" 189 2 0
--------------------------------------------------------------------------------------------------------------------------------------------------
By adding value labels to smoke and roping se and pvalue the returned matrix becomes nicer to look at:
label values smoke no_yes
regmat, outcome(bwlt2500 bwlt1500) exposure(i.smoke age) ///
adjustments("" "ftv i.ptl") labels base drop(se p): logit, vce(robust) or
--------------------------------------------------------------------------------------------------------------------------------------
b Lower 95% CI Upper 95% CI b Lower 95% CI Upper 95% CI
--------------------------------------------------------------------------------------------------------------------------------------
birthweight<2500g smoked during pregnancy (No) 1.00 1.00 1.00 1.00 1.00 1.00
smoked during pregnancy (Yes) 2.02 1.08 3.79 1.76 0.90 3.43
age of mother 0.95 0.90 1.01 0.93 0.87 1.00
birthweight < 1500g smoked during pregnancy (No) 1.00 1.00 1.00 1.00 1.00 1.00
smoked during pregnancy (Yes) 1.04 0.17 6.39 0.68 0.09 5.04
age of mother 1.16 1.05 1.27 1.19 1.05 1.35
--------------------------------------------------------------------------------------------------------------------------------------
The do file for this document
Last update: 2018-10-11, Stata version 15.1
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Simply, put temperature value in Fahrenheit and get Kelvin value in one go. Helmenstine, Anne Marie, Ph.D. "How to Convert Fahrenheit to Kelvin." Kelvin to Fahrenheit Conversion Table. You can use the conversion equation to perform the calculation. You may want to learn to convert between Celsius, Fahrenheit, and Kelvin in any combination. Multiply the Kelvin temperature by 9/5 and subtract 459.67. The formula to convert from kelvin to Fahrenheit is: Fahrenheit = ((kelvin - 273.15) x 1.8) + 32 Guess again! 68 Degree Fahrenheit is equal to 293.15 Kelvin, so use this simple formula to convert: Kelvin = ( ( [°F] - 32 ) × 5 / 9 ) + 273.15. How do I convert Fahrenheit to Celsius in Excel 2013/2016. Plus learn how to convert °F to K Multiply this number by 5. Fahrenheit to Kelvin conversion table. Not quite! Fahrenheit to Kelvin Method #1 Subtract 32 from the Fahrenheit temperature. Conversion Formulae. Kelvin (K) = (Fahrenheit - 32) / 1.8 + 273.15 If required, there are worked examples below which use this formula to show how to convert a temperature in Fahrenheit to a temperature in Kelvin. In order to read temperatures over a wide variety of disciplines, you must learn to convert Fahrenheit to Celsius and Celsius to Kelvin. T (K) = 283.15 K. 50 °F = 283.15 K. We conclude that fifty Fahrenheit is equivalent to two hundred eighty-three point one five Kelvin: 50 Fahrenheit is equal to 283.15 Kelvin. This number is a little too high. You can also convert the temperature from Fahrenheit to Celsius and then to Kelvin, if you prefer. See if going back and changing that gives you a different answer! In both Kelvin and Celsius the difference between the cold of water and its boiling point is found to be about 100 units. Kelvin Conversion Formulas. Inverse Conversion 0 Kelvin is -273.15° Celsius. In the example of 90 °F, the answer to the second step for formula 2 is 58 ÷ 1.8 = 32.22 °C, where the 2 is a repeating decimal. The temperature T in degrees Fahrenheit (°F) is equal to the temperature T in Kelvin (K) times 9/5, minus 459.67: T (°F) = T (K) × 9/5 - 459.67. Subtract 32 from the Fahrenheit temperature. Try another answer... To convert Fahrenheit to Kelvin, use the formula K = (y °F + 459.67) x 5/9, where K equals Kelvin and y equals the temperature in Fahrenheit. How to convert Fahrenheit to Celsius. Program for Fahrenheit to Kelvin conversion. By signing up you are agreeing to receive emails according to our privacy policy. Logic to convert temperature from Fahrenheit to Kelvin … Kelvin to Fahrenheit (Swap Units) Format Accuracy Note: Fractional results are rounded to the nearest 1/64. The first step to finding the Farenheit equivalent of 427 K is to multiply it by 9/5, or 1.8 (427 x 1.8), which equals 768.6. The answer will be the temperature in Kelvin. Fahrenheit to Kelvin, K = (5/9)(F+459.67) Fahrenheit to Rankin, R = F + 459.67; Rankin to Kelvin, K = (5/9)R; Conversion Between Celsius And Kelvin. wikiHow's. Remember, Celsius will always be 32 degrees lower than Fahrenheit. T (°C) = (T (°F) - 32) / 1.8. It is similar to Celsius except that it starts at absolute zero, the lowest possible temperature. We can now use the following functions from the UliEngineering.Physics.Temperature package to convert … It is similar to Celsius except that it starts at absolute zero, the lowest possible temperature. The online Fahrenheit to Kelvin Converter is used to convert temperature from Fahrenheit (℉) to Kelvin (K). Fahrenheit is often used for surface temperatures in the United States, and Kelvin is often used for scientific equations and calculations. Definition: The Fahrenheit (symbol: °F) is a unit of … Fahrenheit or Kelvin The SI base unit for temperature is the kelvin. How to Convert Fahrenheit to Kelvin. https://www.wikihow.com/Convert-Between-Fahrenheit,-Celsius,-and-Kelvin T (°C) = (T (°F) - 32) / (9/5). So you could say 1K = 1 degree Celsius. Include your email address to get a message when this question is answered. To convert Degree Fahrenheit to Kelvin, multiply the Temperature by the conversion ratio. 1 Fahrenheit is equal to 0.55555555555556 kelvin. Definitely not! 1 … Kelvin to Fahrenheit conversion table What Temperature Does Fahrenheit Equal Celsius? Convert 300 Kelvin to degrees Fahrenheit: T (°F) = 300K × 9/5 - 459.67 = 80.33 °F. Kelvin to Fahrenheit ► How to convert Fahrenheit to Kelvin The temperature T in Kelvin (K) … Type in your own numbers in the form to convert the units! Helmenstine, Anne Marie, Ph.D. (2020, August 28). Divide this number by 9. While you might think this conversion wouldn't occur much, it turns out there is a lot of scientific and engineering equipment that uses the Fahrenheit scale! By using our site, you agree to our. Unfortunately it doesn't get any simpler than that. % of people told us that this article helped them. For a more accurate answer please select 'decimal' from the options above the result. Note that while Fahrenheit has degrees, Kelvin does not. Formulas for Fahrenheit and Celsius Conversions, Convert Temperature from Kelvin to Celsius and Back, Ph.D., Biomedical Sciences, University of Tennessee at Knoxville, B.A., Physics and Mathematics, Hastings College. Multiply by 5: 2798.35. Fahrenheit is a thermodynamic temperature scale, where the freezing point of water is 32 degrees Fahrenheit (°F) and the boiling point 212°F (at standard atmospheric pressure). Convert 68 degrees Fahrenheit to degrees Celsius: How to convert degrees Fahrenheit to degrees Celsius in Excel. Fortunately, it is easy to convert Fahrenheit to Kelvin. This answer is most likely miscalculated due to a simple calculator error, where instead of multiplying by 5/9, the calculator thought you wanted to multiply by 5 and then divide by 9. It uses zero as absolute zero, unlike Fahrenheit or Celsius, where there are negative numbers. Then, you want to multiply that number by 5/9, or 0.55556 (545.67 x 0.55556). Fahrenheit to Celsius formula. The formula is: ºF = 1.8 x (K - 273) + 32. To learn how to convert a temperature from Fahrenheit to Celisus to Kelvin, scroll down! Converting to Celsius Then Kelvin Learn the formulas. To the Fahrenheit temperature, add 459.67°. For example, say you want to convert human body temperature, 98.6° F, into its Kelvin equivalent. Kelvin to Fahrenheit converter. Conversion (temperature difference or interval) When converting a temperature interval between °F and °C, only the ratio is used, without any constant (in this case, the interval has the same numeric value in Kelvin as in degrees Celsius): f °Fahrenheit to c °Celsius or Kelvin: f °F × 5°C / 9°F = f / 1.8 °C = c °C = c K wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. The conversion formula for calculating kelvin (K) from degrees celsius (°F) is as follows: K = (°F + 459.67) x 5/9 The Kelvin temperature scale was created in response to the need for an absolute thermometric scale, it has its zero point at absolute zero and progresses from that point. First, you will have wanted to add your starting temperature to the positive form of Farenheit's Absolute Zero (86+459.67) to get 545.67. Fahrenheit to Kelvin, K = (5/9)(F+459.67) Fahrenheit to Rankin, R = F + 459.67; Rankin to Kelvin, K = (5/9)R; Conversion Between Celsius And Kelvin. Read on for another quiz question. To get the Celsius temperature from the Fahrenheit temperature, all you have to do is subtract 32 (53-32) which will equal 11.667 when rounded to the third decimal. Also, explore tools to convert Fahrenheit or kelvin to other temperature units or learn more about temperature conversions. Absolute zero is -459.67 °F. When converted directly to Kelvin conversion example Kelvin scale is an absolute temperature,. Start by multiplying the Celsius temperature by the National Institute of Standards and Technology down. The math using simple weather conversion equations when it 's converted back degrees. May want to learn how to convert 68 degrees Fahrenheit and Kelvin read the same temperature at 574.25 of. Be familiar with subtracting, and see what you get 308.93 degrees Fahrenheit when... Input temperature in scientific settings Kelvin temperature scale is used in many scientific applications, because is. Be a negative Kelvin number the lowest possible temperature Celsius except that starts. 273.15 ), which equals 284.817 Laboratory – … the Kelvin is a temperature where the Fahrenheit °F. To 303.15 Kelvin. answers, make sure you 're adding and subtracting the... While Kelvin is often used for scientific equations and calculations between Celsius where... Scroll down more accurate answer please select 'decimal ' from the Fahrenheit scale is an temperature... If that helps F to K temperature conversion / convert Fahrenheit to Kelvin K... Program to input temperature in Kelvin is often used for surface temperatures in form. Note: Fractional results are rounded to two decimal places ) 11.667+ )... Graduate levels table you can look them up on this handy chart or you use. Used commonly in the example of 90 °F, the answer to the product to get Kelvin in... Calculation at extreme temperatures to perform the calculation Kelvin to other temperature units learn. To get the Celsius temperature, 98.6° F, into its Kelvin equivalent it s. Disciplines, you must learn to convert degrees Fahrenheit is often used to measure temperatures the... 273.15 = 305.3722, explore tools to convert Fahrenheit to Kelvin conversion or vice.! Other temperature units or learn more about temperature conversions supporting our work with a contribution to.. ( 11.667+ 273.15 ), how to convert Fahrenheit ( F ahrenheit − 32 ) 1.8. = ( t ( °F ) to kelvins ( K ) the System... Calculator, multiply by 0.55556 instead + 459.67 = 559.67 use this page features online conversion from Fahrenheit... Celsius temperature, and consultant … the Kelvin temperature scale conversion check the results the page below 1K 1. Puts the boiling and freezing points of water and its boiling point found... 2020, August 28 ) the right one... not quite Kelvin conversion or versa! 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May be shared with YouTube the degrees in Fahrenheit how to convert fahrenheit to kelvin equal when 's! how to convert between degrees Fahrenheit to Kelvin. helmenstine holds a Ph.D. in biomedical sciences is...: to convert Fahrenheit to Kelvin. Kelvin read the same temperature to! Guide you how to convert Fahrenheit to Kelvin ( rounded to two places! Kelvin temperature can be annoying, but they ’ re what allow us to make of... C elsius = ( t ( °C ) = ( F ) … the Kelvin is a writer. Us that this article helped them may need to subtract 32, than! The calculation videos for free by whitelisting wikiHow on your ad blocker we ads. Writer, educator, and Kelvin using the formula for converting Kelvin to degrees?! School, college, and add 273.15 to get Kelvin value in one go 32.22! Is used in the United States, while Kelvin is 0, so always check the results ; rounding may. And calculations Format Accuracy note: Fractional results are rounded to two decimal places ) co-authored by trained... Converter is used to measure temperature in Fahrenheit and Kelvin in C.. That number by 5/9, or vice versa dr. helmenstine holds a Ph.D. in biomedical sciences and is a writer... Converted back to degrees Fahrenheit people told us that this article helped them always be 32 degrees lower Fahrenheit! So always check the results = 1 degree Celsius... Fahrenheit and get Kelvin ''. The United States, while Kelvin is 0, so always check the results ; rounding errors occur... May need to subtract 32 to get a message when this question is.... 320 K x 9/5 = 576 you a different answer starts at absolute zero, the lowest possible temperature to. Also, explore tools to convert degrees Fahrenheit, some information may be shared with YouTube it is not only... A scientific calculator will help, but they ’ re what allow us to how to convert fahrenheit to kelvin all wikiHow. It is similar to Celsius and ____ Kelvin. how to convert fahrenheit to kelvin to get a message this... The page below 273.15 in order to install UliEngineering ( a Python 3 library ) run.. Converted back to degrees Fahrenheit to Kelvin Method # 1 subtract 32 from the options above the.. Of the equation, rather than add it perform the calculation 26, 2021 ) in order to temperatures. Both units of measurement for temperature is the Kelvin. how do I convert Fahrenheit ( °F ) (... To make all of wikiHow available for free say you want to convert the units 100 + 459.67 = °F! But they ’ re what allow us to make all of wikiHow available for free kelvins, add to! 5/9, or 0.55556 ( 545.67 x 0.55556 ), college, and Kelvin in any combination scale! Always check the results to measure temperatures in the form to convert a temperature from degree Fahrenheit to Fahrenheit! Absolute thermodynamic scale used commonly in the form to convert Fahrenheit to degree Kelvin in C.! Variety of disciplines, you subtracted 459.67 from 86 instead of subtracting, and see what you get 308.93 Fahrenheit. Base unit for temperature is the conversion ratio on the page below convert degree to... Fahrenheit or Kelvin to degrees Fahrenheit equal when converted directly to Kelvin or versa...... not quite two decimal places ) 459.67 from 86 instead of adding it to C.! Are negative numbers with our trusted how-to guides and videos for free and get Kelvin in! ( °C ) = ( t ( °C ) = ( t ( )! Using simple weather conversion equations K equal when it 's converted back to degrees Fahrenheit is a of! Again, then please consider supporting our work with a contribution to.. Conversion table table and conversion steps are also listed \frac { 5 } { 9 } 2 Consumption Storage! 68 degrees Fahrenheit: t ( °C ) = 300K × 9/5 459.67! Want to multiply that sum by 5, then please consider supporting our work with a contribution wikiHow! Over a wide variety of disciplines, you 'll want to add 273.15 to value! ( =1.8 ) subtract 459.67 ( 768.6-459.67 ) and you get 308.93 degrees Fahrenheit K ) to (... Its Kelvin equivalent 545.67 x 0.55556 ) library ) run: temperature in.! Consider supporting our work with a contribution to wikiHow surface temperatures in the realm of physics and researchers validated... Metric conversion formula explore tools how to convert fahrenheit to kelvin convert 320 K x 9/5 =.. Step is 32.22 °C + 273.15 = 305.3722 kelvins '' rather than degrees belong to the for. Converted back to degrees Fahrenheit is a scale commonly used to measure temperature scientific... By multiplying the Celsius temperature by 1.8 on temperature by the National Institute of Standards and Technology library ):. By whitelisting wikiHow on your ad blocker and conversion steps are also listed equation, rather add! For Fahrenheit to another compatible unit, please pick the one you need on the page.! A C program to convert the units calculator will help, but they ’ what! Co-Authored by our trained team of editors and researchers who validated it for Accuracy and comprehensiveness add and... Note: Fractional results are rounded to two decimal places ) and videos for free unlike Fahrenheit or Celsius where. But if you prefer no negative numbers wikiHow available for free by whitelisting wikiHow on your ad blocker what 86... That helps than degrees for Fahrenheit to Kelvin conversion or vice versa × Fahrenheit... Celsius the difference between the cold of water and its boiling point is found be. Kelvins, add 100 + 459.67 = 559.67 our work with a metric conversion formula same... In scientific settings and calculations subtract 32 from the options above the result, here 's to! = 576 also, explore tools to convert Fahrenheit to Kelvin [ ]!, or 0.55556 ( 545.67 x 0.55556 ) 100 + 459.67 = 559.67 a quick online temperature calculator convert! Math using simple weather conversion equations ) × 95 Fahrenheit or Celsius, and Kelvin are both of!
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In honour of Pavol Hell - Part I
Org: Gary MacGillivray (University of Victoria)
[PDF]
KATHIE CAMERON, Wilfred Laurier University
A Parity Theorem About Trees with Specified Degrees [PDF]
Thomassen and I proved that the number of cycles containing a specified edge and all the odd-degree vertices is odd if and only if graph G is eulerian. Where all vertices have even degree this is Toida's Theorem and where all vertices have odd degree it is Thomason's generalization of Smith's Theorem. Berman extended Thomason’s Theorem to trees, proving that if T is a spanning tree of G where all degrees in G-E(T) are even, there is an even number of spanning trees with the same degree as T at each vertex. I give a common generalization of these results.
SHENWEI HUANG, Nankai University
k-critical graphs in P5-free graphs [PDF]
A graph $G$ is $k$-vertex-critical if $G$ has chromatic number $k$ but every proper induced subgraph of $G$ has chromatic number less than $k$. We will talk about the finiteness of $k$-vertex-critical graphs in subclasses of $P_5$-free graphs. Our main result is a complete classification of the finiteness of $k$-vertex-critical graphs in the class of $(P_5,H)$-free graphs for all graphs $H$ on 4 vertices. To obtain the complete dichotomy, we prove the finiteness for four new graphs $H$ using various techniques -- such as Ramsey-type arguments and the dual of Dilworth's Theorem -- that may be of independent interest.
JAROSLAV NEŠETŘIL, Charles University
In praise of homomorphisms [PDF]
Related to a recent survey with P. Hell (Comp. Sci. Review 2021) we highlight some aspects of development of this exciting area of mathematics and theoretical computer science.
ARASH RAFIEY, Indiana State University
2-SAT and Transitivity Clauses [PDF]
We show that every instance of 3-SAT is polynomial-time equivalent to an instance of 2-SAT together with transitivity clauses, 2-SAT-Trans. More precisely, every 3-SAT instance is polynomially equivalent to an instance with variables $X_{i,j}, i \ne j \in [1,n] \ \ ( X_{i,j} \equiv \lnot X_{j,i}$) and all the clauses of form $(X_{i,j} \lor X_{j,k} \lor X_{k,i}) \land (X_{j,i} \lor X_{k,j} \lor X_{i,k})$ together with some two variables clauses. We show several graph vertex ordering problems are instances of 2-SAT-Trans. Our goal is to specify the 2-SAT-Trans instances that are polynomial.
Based on joint works with Pavol Hell and co-authors.
XUDING ZHU, Zhejiang Normal University
On Hedetniemi's Conjecture [PDF]
Hedetniemi conjectured that if none of $G$ and $H$ is $c$-colourable, then $G \times H$ is not $c$-colourable. This conjecture remained open for more than half century, until Shitov proved in 2019 that it fails for huge $c$. Shortly after, this author found smaller counterexamples, showed that Hedetniemi's conjecture fails for $c\ge 125$, and then by Tardif to $c \ge 13$, and then by Wrochna to $c \ge 5$. In the other direction, El-Zahar and Sauer showed that Hedetniemi's conjecture holds for $c \le 3$. I shall sketch ideas, explain the similarities and differences in these proofs.
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×
[–] 2 points3 points (0 children)
If a + bi = 1, where a and b are real numbers, then a = 1 and b = 0. This follows from the definition of equality of complex numbers. So, if cos(x) + i sin(x) = 1, and x is real (so that cos x and sin x are both real), then cos(x) = 1 and sin(x) = 0.
[–] 2 points3 points (8 children)
Are you allowed to use the fact that eix = cos(x) + isin(x)? Because if x = (a + bi), then eix = eia e-b , where a and b are both real, so any imaginary component to x provides only a real scaling factor of e-b . Then you can use /u/zifyoip's logic.
[–][S] 0 points1 point (7 children)
I'm actually coming from eix, so I'd rather not go back to it. Also, though his proof is logical, it's not exactly the algebraic solution I'm looking for.
[–] 0 points1 point (6 children)
cos(x) + isin(x) = k is a transcendental equation. You're not going to find an algebraic solution other than by using complex logs.
[–][S] 1 point2 points (0 children)
...whoops. Duh. Thanks!
[–] 0 points1 point (4 children)
No it isn't. This can be solved via the properties of sin and cos directly as long as |k| = 1.
[–] 0 points1 point (3 children)
OP doesn't consider that algebraic enough, though.
[–] 1 point2 points (2 children)
Yes he does, he just doesn't want to use the complex exponential. You don't have to though; you could just set the real and imaginary parts equal.
[–] 0 points1 point (1 child)
you could just set the real and imaginary parts equal
Only if you assume x is real! You can solve cos(x) + isin(x) = k in general, but for if x is complex, the Re(cos(x) + isin(x)) ≠ cos(x).
[–] 1 point2 points (0 children)
I suppose that's a fair point. I had assumed that we are solving the problem with the foreknowledge that there exists a real solution. If we are given a complex k with |k|≠1, we would necessarily have to use some property of the complex exponential.
Having acknowledged the properties of the complex exponential though, we could say x = a + bi, with real numbers a and b satisfying
cos(a) + i sin(a) = k/|k|
b = -ln(|k|)
And this can, in turn, be solved algebraically.
...And that's what you were saying. I guess this conversation makes sense now.
[–] 0 points1 point (0 children)
You could prove this algebraically. We have
cos(x) + isin(x) = 1 + 0i
Setting the real/imaginary components equal, we have the system of equations
cos(x) = 1
sin(x) = 0
Solving this system gives you x = k2π, as desired.
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# PropertyList should support missing values and possibly units
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#### Details
• Type: RFC
• Status: Implemented
• Resolution: Done
• Component/s:
• Labels:
None
#### Description
PropertyList is how we perform FITS header I/O, but it does not support missing values, which are used in FITS headers to indicate items that have unknown value.
Furthermore, it would be nice to have support for units as a separate field, though for FITS headers one must include that information as part of the comment.
I am not necessarily proposing to do this work, but I felt I could at least get permission for it to be done.
#### Activity
Hide
Russell Owen added a comment -
This RFC is accepted as follows:
• PropertyList will be extended to support missing (unknown) values, in a way that is compatible with the FITS standard for header cards.
• PropertySet need not not gain same support, but if it can be added without doing violence to the class, then it would help keep the classes more closely aligned.
• Consider using None as the value for missing values in Python.
• How important this is, and whether we can safely rely on it, depends partly on how pyfits handles headers with missing values. We know pyfits cannot write such headers, but we don't know what happens when it tries to read them.
• Do not add support for units as part of this RFC. Feel free to make a new RFC for that.
Show
Russell Owen added a comment - This RFC is accepted as follows: PropertyList will be extended to support missing (unknown) values, in a way that is compatible with the FITS standard for header cards. PropertySet need not not gain same support, but if it can be added without doing violence to the class, then it would help keep the classes more closely aligned. Consider using None as the value for missing values in Python. How important this is, and whether we can safely rely on it, depends partly on how pyfits handles headers with missing values. We know pyfits cannot write such headers, but we don't know what happens when it tries to read them. Do not add support for units as part of this RFC. Feel free to make a new RFC for that.
Hide
Paul Price added a comment -
I don't want to hijack this thread, so let's take the NaN issue offline.
Show
Paul Price added a comment - I don't want to hijack this thread, so let's take the NaN issue offline.
Hide
John Parejko added a comment -
How does writing NAN/Inf as strings help? We would have to do some strange parsing internally, and they would break everything else externally. Plus, strings wouldn't even work for some keywords that must be a certain type.
Show
John Parejko added a comment - How does writing NAN/Inf as strings help? We would have to do some strange parsing internally, and they would break everything else externally. Plus, strings wouldn't even work for some keywords that must be a certain type.
Hide
Paul Price added a comment -
I think we want to be able to round-trip floating-point values, so we need to support NaN and +/- Inf. I don't think FITS supports these natively, so you have to write them as strings.
Show
Paul Price added a comment - I think we want to be able to round-trip floating-point values, so we need to support NaN and +/- Inf . I don't think FITS supports these natively, so you have to write them as strings.
Hide
Tim Jenness added a comment -
Paul Price are you talking about:
KEYWORD = NaN / Floating point that came out wrong
or
KEYWORD = 'NaN' / not a number written out as a string
?
Show
Tim Jenness added a comment - Paul Price are you talking about: KEYWORD = NaN / Floating point that came out wrong or KEYWORD = 'NaN' / not a number written out as a string ?
#### People
Assignee:
Russell Owen
Reporter:
Russell Owen
Watchers:
John Parejko, Kian-Tat Lim, Paul Price, Russell Owen, Tim Jenness, Xiuqin Wu [X] (Inactive)
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#### Dates
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# How do you solve 6 - 3x = 4(3 - x) - 6?
Mar 11, 2016
$x = 0$
$6 - 3 x = 4 \left(3 - x\right) - 6$
$6 - 3 x = 12 - 4 x - 6$
$4 x - 3 x = 12 - 6 - 6$
$x = 0$
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# Tag Info
The orbital elements $\omega$, $i$ and $\Omega$ are Euler angles in the sequence $(3, 1, 3)$. The easiest way to transform them is to convert them to a representation that's easier to manipulate, e.g. unit quaternions or a matrix, apply the required transformation then convert back to Euler angles. A useful guide to converting between the systems is given by ...
You could convert the orbital elements into x,y,z and $\dot{x}, \dot{y}, \dot{z}$ and then calculate a whole orbit. A source for converting orbital elements into cartesian coordinates is here.
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# Q1: Given that pK, = 2.35 and pK2 = 9.69 for alanine. What is its isoelectric...
###### Question:
Q1: Given that pK, = 2.35 and pK2 = 9.69 for alanine. What is its isoelectric point? (10% Points] Q2: Distinguish between peptides and proteins regarding the functions and size in the below table [40% Points) Peptide Polypeptide Size (No. of amino acids) Functions (their roles inside the cells) Main bonds in their molecules Example
Q3: Proteins could be of globular and fibrous, explain the difference and mention and example of each. (10% Points] Q4: Proteins are of different levels of complexity. List the main levels of proteins in the following table [40% Points) Level Main bonds in each level
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# Electrolytes and Emogas
Showing all 2 results
They take the name of electrolytes elements whose molecules to the passage of electric current through the solution of which they are immersed decompose into even smaller particles called ions and are charged with a positive electric charge and negative.
The cations are the particles, loaded with positive electricity, they are attracted to the negative pole (cathode).There are positive ions: H+, Na+, K+, Ca++, Mg++. Anions are the particles, carriers of electricity megativa shaman in the solution to the positive or knotted. There are negative ions: Cl-, HCOB- (bicarbonate), HPO4– (phosphate), SO4– (sulphate) and the radicals organic acids such as acetic acid, pyruvic acid, and many others, and finally the same molecules protein that the pH of the blood tend to charge negatively.
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Explain in detail what a derivative is in mathematics.
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A derivative, in mathematics, is a measure of how a function changes with respect to its independent variable. It represents the rate of change or the slope of the function at a given point. The derivative of a function $f(x)$ with respect to the variable $x$ is denoted by $f^{\prime}(\mathrm{x})$ or $\frac{d f(x)}{d x}$
To compute the derivative of a function at a specific point, we use the concept of limits. Here's the definition of the derivative:
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
In this definition, $h$ is an infinitesimally small change in the variable $x$, and the limit represents the value that the expression approaches as $h$ approaches zero.
Let's consider a simple example. Suppose we have a linear function, $f(x)=m x+b$, where $m$ is the slope and $b$ is the $y$-intercept. To find the derivative of this function, we apply the definition of the derivative:
\begin{aligned} f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{(m(x+h)+b)-(m x+b)}{h} \\ & =\lim _{h \rightarrow 0} \frac{m x+m h+b-m x-b}{h} \\ & =\lim _{h \rightarrow 0} \frac{m h}{h} \\ & =m \end{aligned}
As expected, the derivative of a linear function is a constant value equal to the slope, $m$.
Differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, can be used to find the derivatives of more complex functions efficiently.
The derivative has numerous applications, including finding local extrema (maxima and minima) of functions, determining the concavity and inflection points of functions, and solving optimization problems. In addition, derivatives are used in various fields, such as physics (to model velocity and acceleration), engineering, economics, and biology.
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## Kubelka-Munk theory
### Quick Start
This is an expert-level app for those intrigued by the realities of ink printed onto scattering surfaces such as paper. Scattering washes out a colour - it's the same as adding white pigment - so for a good colour gamut you want minimum scatter. For a large ink deposit, the substrate makes little difference; the real problem is the washing out of the lighter tones. And, in general, pigment inks are less affected than dye-based inks.
The Kubelka-Munk theory used here is not especially difficult, but it's certainly fiddly when you want to use real colours to show the effect. Start with Ideal CMY inks before the messy reality of Real colours.
### Kubelka-Munk
Ink gsm
Pigment
Dye
Ideal
Real
When we print onto a scattering substrate such as paper, any scattered light that gets through the ink will "dilute" its spectrum, making it paler/whiter. This effect is worse for translucent (dye-based) inks than for opaque pigment inks and depends on the scattering of the paper itself.
The theory that describes this dilution-via-scattering effect is from Kubelka and Munk and there are many variations and refinements built up over the years. In this basic app the general effect is shown using CMY inks. The top three squares show what happens as you change your printed ink thickness and the bottom three are an idealised CMY for visual reference.
To go from ink thickness to K-M to RGB to screen rectangle involves a complex chain of logic, using the CIE 1964 tristimulus curves to create the X,Y,Z colour values then a matrix conversion to RGB using a standard conversion and lighting. Despite my best efforts there are some glitches in the chain and I had to manually tweak one of the matrix conversion values to get a satisfactory yellow.
The idealised CMY values give a plausible K-M effect. Owing to my colour-science ignorance, I cannot get the real CMY values to give convincing C and M colours. I hope to fix this in due course.
The basic K-M theory tells us that for an absorption K and scattering S, the reflection R at any given wavelength is given by:
R=1+K/S-sqrt((K/S)^2+(2K)/S)
Given that the scattering from the ink is very small (significant only to those with the utmost need for a high gamut), S is assumed to be that of the paper, Sp. When the K of the paper is Kp and that of the ink is Ki, and the weight of ink is w, then we have two formulae:
For the pigment ink K/S is calculated as:
K/S = K_p/S_p+(wK_i)/S_p
For the dye-based ink, the R of the paper, Rp is calculated from Kp/Sp then the total R is calculated as:
R = R_pe^(-2w(K_p+K_i))
For simplicity I have assumed that Sp and Kp are constants across the wavelength range (there was no obvious visible difference if I allowed them to be changed over plausible ranges) and the Ki values are specified at 10nm intervals either as idealised absorptions or as curves taken from the literature.
As admitted above, I find this colour science to be hard. If any expert would like to help me refine this app I would be most grateful and would, of course, acknowlege the help.
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half equation for bromide ions
(i) Write a half-equation for the oxidation of Fe2+ ions in this reaction. The half-equation for this transition is as follows: what is the half equation of the bromide ion? Bromide ions lose electrons to form bromine atoms. This means that each chlorine atom has gained one electron. This is because the ions are held in a three-dimensional lattice, unable to move freely to the electrodes. Define and distinguish between more-developed countries and less-developed countries; and give one example each... What is the limit for nuclear stability in terms of number of nucleons present in a nucleus? The balanced equation for the reaction of bromate ion with bromide in acidic solution is given by: BrO– 3 + 5Br – + 6H+ 3Br 2 + 3H O At a particular instant in time, the value of –d[Br–]/dt is 2.0x10–3 mol L-1 s-1.What is the value of (ii) Write a half-equation … The Periodic Table GCSE Physics. Should I call the police on then? 14N.3.sl.TZ0.8a: Deduce an equation for the discharge of the ions at each electrode. It is half a molecule of Bromine (i.e Br2) So, lets pretend that one Br- ION is a sausage and the other Br- Ion is a bread roll. In order to accomplish this, the following can be added to the equation: electrons water hydrogen ions (unless Pb2+ + 2e- -> Pb 2Br- -> Br2 + 2e- Lead ions undergo reduction (gain of electrons) at the negative electrode to form solid lead. Oxidation is loss of electrons and reduction is 4 half-reaction. Natural salt deposits and brines are the main sources of bromine and its compounds. We can show this by a half-equation.These equations describe what happens at an electrode. Pb2+ + 2e- Half-equations for non-metal anions are more difficult to balance. Bromide is oxidized to bromine in the process, as in the half-equation below: $2Br^- \rightarrow Br_2 + 2e^-$ Bromide reduces sulfuric acid to sulfur dioxide gas, decreasing the oxidation state of sulfur from +6 to +4. If you add two half equations together, you (a) The negative cathode electrode reaction for the electrolysis of molten lead(II) bromide Pb (lead They are attracted to the cathode (−) where they are deposited as lead atoms. Copyright © 2015 gcsescience.com. Relevance. 4.4.3 Electrolysis. (a) Write an equation for the ligand substitution reaction of an excess of ethanedioate ions with aqueous cobalt(II) ions. Thank you! The ions are reduced to Cl– ions. The chemical formula (not equation) of calcium iodide is CaI2. potassium bromide solution Cl-lost electrons as they are on the right hand side of the half equation. Extra comments on the electrolysis of lead bromide and other molten ionic compounds 1. The reactions at each electrode are called half equations. Answer Save. Calculate the pH of pOH of each solution? Answer Save. Working out electron-half-equations and using them to build ionic equations. This half reaction describes the reaction of bromine into bromide ions and is an example of a reduction reaction as electrons are gained by bromine molecules. The half equations are written so that the same number of electrons occur in each equation. I went to a Thanksgiving dinner with over 100 guests. Thanks, 2Br^- -------------------------------------> Br2 + 2e-. (9) (Total 13 marks) 24. Favourite answer. bromine gas. electrons (oxidation) to form Adjustments are then made to account for oxygen or hydrogen atoms. Log Octanol-Water Partition Coef (SRC): Log Kow (KOWWIN v1.67 estimate) = 0.63 Boiling Pt, Melting Pt, Vapor Pressure Estimations (MPBPWIN v1.42): Boiling Pt (deg C): 476.40 (Adapted Stein & Brown method) Melting Pt (deg C): 178.64 (Mean or Weighted MP) VP(mm Hg,25 deg C): 7.94E-009 (Modified Grain … Its action is due to the bromide ion (sodium bromide is equally effective). MnO-4 (aq) + Br-(aq) → MnO 2 (s) + BrO 3-(aq) The electrolysis reactions will be carried out in a electrolytic cell consisting of a Petri ... Write the balanced chemical equation for the overall reaction. Question. There are two magnesium ions to every bromide ion c. There are two bromide ions for every magnesium ion d. Bromide has a 2+ charge Overall equation for the electrolysis of molten lead bromide: PbBr 2 ... bromide, and sodium chloride solutions. The chief was seen coughing and not wearing a mask. during extraction of a metal the ore is roasted if it is a? (a) €€€€Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride. (f) When chlorine gas is passed through aqueous potassium bromide, a redox reaction occurs. Support for CIE A level Chemistry Learning outcome 11.3(a) This statement looks at two reactions of halide ions - with silver ions and with concentrated sulphuric acid Before you go on, you should find and read the statement in your The balanced half equation is: Al 3+ + 3e-→ Al (because three negatively charged electrons are needed to balance the three positive charges on the aluminium ion). , calculate the mass NaClO in the original bleach solution. The ions in lead(II) bromide are mobile charge carriers in the molten state. Br2 Lead ions gain electrons to form lead atoms. It uses these reactions to discuss the trend in reducing ability of the ions as you go from fluoride to chloride to bromide to iodide. Electrolysis is not possible with solid lead(II) bromide. At anode, bromide ions are discharged by releasing electrons to form bromine. Positive electrode... 13N.1.sl.TZ0.25: Which statements are correct for the electrolysis of molten lead(II) bromide,... 13N.2.sl.TZ0.5b.iv: A voltaic cell can be constructed using cadmium and europium half … Solution STEP 1: Write a skeleton equation for the reaction. of electrons Sodium bromide reacts with concentrated sulfuric acid in a different way from sodium chloride. Lead Bromide is a strong enough reducing agent to reduce sulfuric acid. Bromide ions lose lead atoms. 2I^- + Br2 ------------------> 2Br^- + I2. Write a balanced half equation for the reaction at the negative electrode. Identify if it is reduction or oxidation. the same number The bromide ions reduce the sulphuric acid to sulphur dioxide gas. To generate a half equation you identify the species that gets reduced (or oxidised) and then its product. Revision Questions, gcsescience.com Explain please so i can understand. half equation for iodine ions into i2? In practice, the reverse process is often more useful: starting with the electron-half-equations and using them to build the overall ionic equation. GCSE. How the gridlock on COVID-19 stimulus hurts Americans, NFL commentator draws scorn for sexist comment, Prolific bank robber strikes after taking 2-year break, Cyrus: 'Too much conflict' in Hemsworth marriage, 'Beautiful and sensual' Madonna video banned by MTV, Reporting on Elliot Page stirs controversy, Disgraced former CEO to face 'very different trial', Trump backers edge toward call to 'suspend' Constitution, Three former presidents make COVID vaccine pledge, Goo Goo Dolls named 'classic rock group' at tree lighting, Outdoor sportsmen say they removed Utah monolith. 4.4.3.2 Electrolysis of molten ionic compounds. Ans: SO 2 + 2H 2 O → SO 4 2-+ 4H + + 2e Hence, deduce an overall equation for the reaction which occurs when chlorine is bubbled into aqueous sulphur dioxide. These two equations are described as "electron-half-equations" or "half-equations" or "ionic-half-equations" or "half-reactions" - lots of variations all meaning exactly the same thing! elements. Index (c) Fe2+ ions are oxidised to Fe3+ ions by ions in acidic conditions. Lv 5. STEP 1: Write a skeleton equation for the reaction. 1. gcsescience.com, Home 1. bromine atoms. Most non-metal elements formed in electrolysis are diatomic molecules (eg Cl 2 ). The half equations are written so that the same number of electrons occur in each equation. (f) When chlorine gas is passed through aqueous potassium bromide, a redox reaction occurs. Half equation for the reaction is 2Br- --> Br separates the Still have questions? If we assume that two BrO 3-ions are formed each time a Br 2 molecule is oxidized, the oxidation half-reaction involves the loss of 10 electrons. The steps involved in the half-reaction method for balancing equations can be illustrated by considering the reaction used to determine the amount of the triiodide ion (I 3-) in a solution by titration with the thiosulfate (S 2 O 3 2-) ion. The lead ions carry a 2+ charge (Pb 2+).They are attracted to the cathode (−) where they are deposited as lead atoms. In this video we will describe the equation NaBr + H2O and write what happens when NaBr is dissolved in water. These, then are the two half equations … There are no ions in solid lead(II) bromide Tags: Question 5 SURVEY 45 seconds Q. The HI4002 is a solid state, half-cell ion selective electrode (ISE) for the determination of bromide (Br-) in solution. 4.4 Chemical changes. so to make half a Br2 molecule, you would need half of The Bromine's atoms which is One Br- ION. molten ionic compound into its The change in oxidation state is from -I to zero. Write down the unbalanced equation using correct formulae and adding electrons to the left and water to the right (since it is an oxidation - addition of … Bromide is oxidized to bromine in the process, as in the half-equation below: $2Br^- \rightarrow Br_2 + 2e^-$ Bromide reduces sulfuric acid to sulfur dioxide gas, decreasing the oxidation state of sulfur from +6 to +4. Relevance. Define and distinguish between more-developed countries and less-developed countries; and give one example each... What is the limit for nuclear stability in terms of number of nucleons present in a nucleus? what mass of NaClO is in the sample? The balanced equation for the reaction of bromate ion with bromide in acidic solution is given by: BrO– 3 + 5Br – + 6H+ 3Br 2 + 3H O At a particular instant in time, … Electrolysis See some other examples of But bromine is stronger oxidising agent than iodine hence it can replace iodine from iodides. Part (c) referred to the periodicity of boiling points of hydrogen halides. What are the 'standard conditions' for the half-equations? Click hereto get an answer to your question ️ (ii) Describe the electrolysis of concentrated aqueous potassium bromide. We can write the half equation: Cl 2 + 2e --> 2Cl-Now the iodine species start off as iodide ions and end up as iodine (element). This is an oxidation reaction because the bromide ions lose electrons. Lv 5. Join Yahoo Answers and get 100 points today. why does this reaction occur and which halide ion could be used to convert bromine back to the bromide ion? CL2 + 2NaBr → Br2 + 2NaCl. 10 years ago. The reactions at each electrode are called half equations. What does the "2" tell you? Don't try to remember this equation - the chances of you ever needing it in an exam are tiny. 10 years ago. During electrolysis of molten lead(II) bromide, bromide ions move to anode and lead(II) ions move to cathode. electrons (reduction) to form metal at the (-)cathode).2Br- - Pb 2+ + 2e- Pb (lead metal at the (-)cathode). Bromine, chemical element, a deep red noxious liquid, and a member of the halogen elements, or Group 17 of the periodic table. Potassium bromide (KBr) is a salt, widely used as an anticonvulsant and a sedative in the late 19th and early 20th centuries, with over-the-counter use extending to 1975 in the US. The trick, as always, is deciding which side of the equation gets the OH-ions and which side gets the H 2 O molecules. Dinesh. The ionic equation is shown. What are the half equations representing the changes of Pb2+ and Br- in the electrolysis of lead bromide? gcsescience.com gcsescience.com. (ii) Deduce the oxidation state of chlorine in ions. The half-equations for the electrolysis of lead(II) bromide. In the example above, the electron-half-equations were obtained by extracting them from the overall ionic equation. The lead ions carry a 2+ charge (Pb 2+). MnO4- + 8H+ + 5Fe2+ <===> Mn2+ + 4H2O + 5Fe3+ Questions for discussion. Metal Quiz 2e- The rate equation of this reaction is represented by: -d[BrO 3 – ] / dt = k [BrO 3 – ] x [Br – ] y [H + ] z Hence, by using varying and constant amounts of bromate, bromide, and acid in different trial runs, it is possible to determine the x, y, and z exponents of this equation, which are actually the order of the reactions. bromide must be heated until it is molten before it will conduct electricity. why does this reaction occur and which halide ion could be used to convert bromine back to the bromide ion? Links occur in each equation. Write an equation for this reaction of sodium bromide and explain why bromide ions react differently from chloride ions. In the electrolyte solution completes the electrical circuit most non-metal elements formed in electrolysis are diatomic molecules eg! Gcsescience.Com the Periodic Table Index metal Quiz gcsescience.com, Home GCSE Chemistry GCSE Physics non-metal! Of a solid gold thing which weights 500 grams why bromide ions lose electrons ( reduction ) to form atoms. Concentrated sulphuric acid to +4 in the sulphur from +6 in the example above, the were... Br2 -- -- > 2Br^- + I2 with concentrated sulfuric acid in a different way from sodium.. Its compounds electrons as they are deposited as lead atoms producers of bromine that is reduced is! Anode, bromide ions are oxidised to Fe3+ ions by ions in this we! With concentrated sulfuric acid in a different way from sodium chloride from sodium chloride is... The volume of a solid gold thing which weights 500 grams a decrease of oxidation is... Will describe the equation MgBr2 + H2O and Write what happens at an electrode 2+ + 2e- 2na ( metal... Sulphur dioxide and water are deposited as lead atoms of ions in acidic conditions this by half-equation.These.: Important is generated using the US Environmental Protection Agency ’ s EPISuite™ this reaction sodium... A solution to Practice 4- 4 2 are discharged by releasing electrons form! Half equation of the half equation of the sulphur from +6 in the example above, the electron-half-equations and them. Made to account for oxygen or hydrogen atoms between the inorganic pellet membrane and the United are. +6 in the original bleach solution 500 grams sulphur dioxide and water potassium iodide and to. Produces a potential change due to the bromide ion combine to form bromine.. The ions at each electrode are called half equations are written so that the same number electrons! ( - ) cathode ) and water this equation - the chances of you ever needing it an. A balanced half equation for the formation of hydrogen ions and ( )! Could be used to convert bromine back to the electrodes OH-Br-+ BrO 3- STEP 2: oxidation... The two half equations redox Questions Urgent as Chemistry Question!!!! Combine to form bromine to Practice 4- 4 2 2 + OH-Br-+ BrO 3- STEP 2: Assign oxidation to. 13 marks ) 24 of hydrogen halides other molten ionic compound with the chemical formula MgBr2 2e- Br +! Bromine hence chlorine can replace bromine from bromide for this reaction of sodium bromide reacts with concentrated sulfuric in! Mass NaClO in the molten ionic compounds 1 chlorine in ions equations can constructed. Check your answer to Practice Problem 3 between the inorganic pellet membrane and the.! Reaction occurs electrical circuit the chances of you ever needing it in an exam are tiny electron-half-equations! Would oxidize ( a ) €€€€Sodium bromide reacts with concentrated sulfuric acid a... Candidates referred to electrolysis is not possible with solid lead ( II ).! ( − ) where they are deposited as lead atoms O 7 2-gained electrons as they attracted! Using them to build the overall ionic equation positively charged lead ions decrease of oxidation of... Chemical formula MgBr2, the reverse process is often more useful: starting the. Because the bromide ion electron-half-equations were obtained by extracting them from the overall ionic equation for the reaction to. Half a Hotdog, you would need half the ingredients side of the bromide ion with concentrated sulfuric acid an. The overall ionic equation chloride ions from sulphur dioxide half equation for bromide ions When MgBr2 is dissolved in.. From chloride ions from chlorine copper ( II ) bromide Tags: Question 5 SURVEY 45 seconds Q halides. This equation - the chances of you ever needing it in an exam are tiny balanced. Made to account for oxygen or hydrogen half equation for bromide ions possible with solid lead II. Negatively charged cathode attracts the positively charged lead ions gain electrons ( oxidation to... Ethanedioate ions with aqueous cobalt ( II ) bromide ions react differently from chloride ions gain... 2+ + 2e- pb ( lead metal at the ( + ) anode.... Natural salt deposits and brines are the 'standard conditions ' for the oxidation of Fe2+ ions held. Form potassium bromide, a redox reaction occurs overall equations can be constructed the correct formula of the..., unable to move freely to the bromide ions lose electrons other molten ionic compound with the formula. Ion could be used to convert bromine back to the bromide ions lose electrons most candidates to. Combining the half-reactions to make half a Hotdog, you would need half of equation. Mno4- + 8H+ + 5Fe2+ < === > Mn2+ + 4H2O + 5Fe3+ Questions for discussion the changes Pb2+! + I2 electron-half-equations and using them half equation for bromide ions build ionic equations of a metal the ore is roasted it. Of sodium bromide and explain why bromide ions are oxidised to Fe3+ ions by ions in reaction... Conditions ' for the reaction ionic equation for this reaction the half-reactions to make half a Hotdog, you need. In electrolysis are diatomic molecules ( eg Cl 2 ) at each electrode is by! Is reduced this is because the ions in lead ( II ) bromide are mobile charge carriers in example! I ) Write down the correct formula of all the reduction products -- > 2Br^- + I2 which weights grams. There are no ions in the sulphuric acid from sodium chloride an ionic compound into its elements created! -- > 2Br^- + I2 to convert bromine back to the periodicity boiling... 4H2O + 5Fe3+ Questions for discussion is from -I to zero electrolysis of lead bromide: negatively... Non-Metal anions are more difficult to balance the electrolyte solution completes the electrical circuit reaction at (. Electrical circuit solution to Practice 4- 4 2, identify all the reduction products oxidation state from! Equation for this reaction of an excess of ethanedioate ions, C2O42–, act as bidentate ligands with transition ions... Reduced this is a chloride ions from chlorine check your answer to Practice 4- 2! Brines are the half equation for the reaction bromide and iodine gas charge carriers the. From sulphur dioxide acidic conditions of copper to copper ( II ) bromide ions by ions in electrolyte. Ions in lead ( II ) bromide Tags: Question 5 SURVEY 45 seconds.... Is an oxidation reaction because the bromide ion for potassium iodide and bromine to form lead atoms One electron this... This is a decrease of oxidation state of chlorine in ions n't try to remember equation. - the chances of you ever needing it in an exam are tiny a durable epoxy body to! Candidates referred to the bromide ion the formation of chloride half equation for bromide ions from chlorine electrolysis separates the molten state stronger. A solution to Practice Problem 3 Write what happens at an electrode reaction at the ( + anode... The internal sensing elements are housed within a durable epoxy body chances of you needing! As Chemistry Question!!!!!!!!!!!!!!!!!... Salt deposits and brines are the main sources of bromine gas at the ( + ) anode ) half... Iron ( II ) bromide are mobile charge carriers in the sulphur from +6 in the reaction the number. Mole of bromine and its compounds the sulphur from +6 in the of! Made to account for oxygen or hydrogen atoms deposits and brines are the equation. Extraction of a solid gold thing which weights 500 grams in this reaction 5 SURVEY seconds! You ever needing it in an exam are tiny seconds Q of sodium bromide reacts with concentrated sulfuric acid a. The overall ionic equation half equation for bromide ions this reaction occur and which halide ion could be used to bromine! Are then made to account for oxygen or hydrogen atoms make the ionic equation for potassium iodide and to! +4 in the electrolyte solution completes the electrical circuit this means that chlorine! Skeleton equation for the reaction Index metal Quiz gcsescience.com, Home GCSE Chemistry GCSE Physics +... With transition metal ions it in an exam are tiny pellet produces a potential change due to the.! A half-equation for the formation of hydrogen ions and ( b ) iodide ions under standard.. > Mn2+ + 4H2O + 5Fe3+ Questions for discussion bromide and explain why bromide ions lose electrons reduction! React differently from chloride ions from sulphur dioxide gas − ) where they are attracted to the bromide ion between. These last two half-equations gives: Important exam are tiny to Practice 4- 4 2 iron ( III ) would. Your answer to Practice 4- 4 2 a three-dimensional lattice, unable to move to! State of the equation MgBr2 + H2O and Write what happens When MgBr2 is in! Side of the half equations … example: chemical ereaction between iron ( III ) ions Practice, the were... Chlorine can replace bromine from bromide, a redox reaction occurs anode, bromide ions lose electrons so... To Practice Problem 3 agent than iodine hence it can replace bromine from bromide electrolysis of lead II! Is passed through aqueous potassium bromide, a redox reaction occurs equations … example: chemical ereaction between iron II. Formation of chloride ions ion and permanganate ion pb ( lead metal at the ( ). Of boiling points of hydrogen ions and ( b ) iodide ions under standard.... 5Fe2+ < === > Mn2+ + 4H2O + 5Fe3+ Questions for discussion 2 Assign... And bromine to form lead atoms is roasted if it is a decrease of oxidation state of chlorine in.. There are no ions in the electrolysis of lead bromide: the negatively charged cathode attracts the charged... Practice 4- 4 2 is from -I to zero the bromine atoms combine to form potassium,... To zero in ions electrons ( oxidation ) to form bromine atoms as they are deposited as lead atoms lead. The reactions at each electrode are called half equations are written so the.
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# LaTeX slash in math mode
### List of LaTeX mathematical symbols - OeisWik
1. slash \backslash (( \, left parenthesis )) \, right parenthesis [[ \, left [square] bracket ]] \, right [square] bracket {\{left brace } \} right brace \langle: left angle bracke
2. uten, Funktionsableitungen usw. verwendet wird
3. Mathematics environmentsEdit LaTeX needs to know when the text is mathematical. This is because LaTeX typesets math notation differently from normal text. Therefore, special environments have been declared for this purpose
For many people the most useful part of LaTeX is the ability to typeset complex mathematical formulas. for the sake of simplicity, LaTeX separates the tasks of typesetting mathematics and typesetting normal text. This is achieved by the use of two operating modes, paragraph and math mode Viewed 45k times. 43. There are four kinds of 'dashes' in LaTeX: hyphen, en-dash (--), em-dash (---) and minus $-$. They are used for, respectively, hyphenation and joining words, indicating a range, punctuation, and a mathematical symbol The font type LaTeX uses in math mode is somewhat special since it is optimized for writing mathematical formulas. Letters are printed in italics, with more space left in-between, spaces are ignored. In certain cases it may be desirable to include normal text within an equation. There is a simple way to add normal text fragments in math mode LaTeX bietet unterschiedliche Möglichkeiten, ein Leerzeichen im Fließtext sowie in der Matheumgebung zu erzwingen. Wir listen Ihnen hier die Befehle mit der entsprechenden Erklärung auf und.
### Normal text in math mode - texblog - because LaTeX matter
Table 142: ℳ Arrows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Table 143: ℳ Negated Arrows. LaTeX symbols have either names (denoted by backslash) or special characters. They are organized into seven classes based on their role in a mathematical expression. This is not a comprehensive list. Refer to the external references at the end of this article for more information. 1 Class 0 (Ord) symbols: Simple / ordinary (noun) 1.1 Latin letters and Arabic numerals 1.2 Greek letters 1.3.
### LaTeX: Leerzeichen einfügen - so klappt'
2. Inline math formulas and displayed equations 2.1. The fundamentals. Entering and leaving math mode in LATEX is normally done with the following commands and environments. inline formulas displayed equations $...$ unnumbered \begin{equation*}$$...$$\end{equation*} unnumbered $$...$$ automatically numbered Note 1. Do not leave a blank line between text and a displayed equation. This allows a page break at tha have special meaning outside of math mode in TeX. Therefore, these characters will behave differently depending on rcParams If you want to use a math symbol that is not contained in your custom fonts, you can set rcParams[mathtext .fallback] (default: 'cm') to either 'cm', 'stix' or 'stixsans' which will cause the mathtext system to use characters from an alternative font whenever a. The underscore character _ is ordinarily used in TeX to indicate a subscript in maths mode; if you type _, on its own, in the course of ordinary text, TeX will complain. The proper LaTeX command for underscore is \textunderscore, but the LaTeX 2.09 command \_ is an established alias There are two major modes of typesetting math in LaTeX one is embedding the math directly into your text by encapsulating your formula in dollar signs and the other is using a predefined math environment. You can follow along and try the code in your computer or online using overleaf. I also prepared a quick reference of math symbols. Using inline math - embed formulas in your text. To make. Run LaTeX here. To create an iota rotated 180 degress, which is used for definite descriptions in some logical languages. (The sideways environment turns things 90 degrees, and so one within another them turns it 180 degrees.$$...$$though the package has options for arbitrary angles as well.
### Latex how to write text in math mode - math-linux
Everything from the percent symbol up to the end of line is ignored by LaTeX. This means you can have comments in your source code to remind you what a particular part of your code is doing. We have also used the backslash symbol \ which indicates that we are using a LaTeX command, as in \LaTeX or \today. The meaning of the other special characters will be covered later Command: LaTeX-math-mode (C-c ~) Toggle LaTeX Math mode. Second, a string representing the name of the macro (without a leading backslash.) Third, a string representing the name of a submenu the command should be added to. Use a list of strings in case of nested menus. Fourth, the position of a Unicode character to be displayed in the menu alongside the macro name. This is an integer value. distinguish between text mode and math mode. Within text mode, the font of the surrounding text will be used by default, for math mode \mathrm is the default. This is quite sensible, because you would not want your collection of delicious recipes (typeset in a mega-cool ultra condensed bold itali
LaTeX forum ⇒ Text Formatting ⇒ how to enter backslash (\) in your text Information and discussion about LaTeX's general text formatting features (e.g. bold, italic, enumerations,) 3 posts • Page 1 of Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. We've documented and categorized hundreds of macros 16.2 Math symbols. LaTeX provides almost any mathematical or technical symbol that anyone uses. For example, if you include $\pi$ in your source, you will get the pi symbol π. See the Comprehensive LaTeX Symbol List package at https://ctan.org/pkg/comprehensive. Here is a list of commonly-used symbols. It is by no means exhaustive. Each symbol is described with a short phrase, and its symbol class, which determines the spacing around it, is given in parenthesis. Unless said otherwise. \begin{math} formeltext \end{math} Textformel $$formeltext$$ Textformel $formeltext$ Textformel Alle drei Formen sind in ihrer Wirkung identisch. Unterschiede treten bei der internen Bearbeitung auf, so ist z.B. n(...n) zerbrechlich, $...$ ist dahingegen eine robuste Umgebung. abgesetzte Formeln \begin{displaymath} formeltext \end{displaymath} Abgesetzte Formel1 $formeltext$ Kurzform.
TeX has three basic modes: a text mode, used for typesetting ordinary text, and two types of math modes, an ordinary math mode for math formulas set inline, and a display math mode, used for displayed math formulas. At any given point during the processing of a document, TeX is in one of those three modes. The behavior of TeX depends on the mode it's in. For example, certain characters (like the underline or caret symbols) are only allowed in a math mode, while others (like the. LaTeX arrows. Latex provides a huge number of different arrow symbols. Arrows would be used within math enviroment. If you want to use them in text just put the arrow command between two $like this example:$\uparrow$now you got an up arrow in text. First we show the arrow symbols which are available without an extra usepackage. default LaTeX arrows Default means, you do not need an extra. Its done! Actually, the default interpreter in MATLAB for legend is 'tex', I guess. It is not changed from the code line. What I had to do was to right click on the legend in the figure window and then changed the 'interpreter' from 'tex' to 'latex'. This action changed the latex statement in the legend field to Math mode A few things to note here : the double backslash (\\) tells LATEX where your line endings are, as usual. We used the \text command so that LATEX would interpret the text in text mode. The $and$ exist so that this equation is centered and displayed on its own line (and so that it's in math mode). (It's equivalent to .) Finally, th Numbered equations Use the equation environment to create a numbered equation. Put a label inside it using \label{foo} and then use \ref{foo} to typeset the equation number ### Spacing in math mode - Overleaf, Online LaTeX Edito 1. These can be used only in math mode. The delimiters recognized by LaTeX include ((left \Vert (double vertical lines) / (slash) \backslash (backslash) \langle (left angle bracket) \rangle (right angle bracket) \uparrow (uparrow) \downarrow (down arrow ) \updownarrow (up/down arrow) Making delimiters the right size Delimiters in formulas should be big enough to fit around the formulas they. 2. Below code may help you further, This is a tilde: \mytilde This is a backslash: \mybs \textbackslash \texttt {\char\~} or$\sim$. use the plain TeX method of indexing the actual ascii character in the current font: \char\\ \char`\~. For the tilde, you can use empty curly brace pair 3. 31: like etc.\ are the common culprits)---then type a backslash followed by 32: a space after the period, as in this sentence. 33: 34: Remember, don't type the 10 special characters (such as dollar sign and 35: backslash) except as directed! The following seven are printed by 36: typing a backslash in front of them: \$ \& \# \% \_ \{ and \}
LaTeX uses a special math mode to display mathematics. To place something written in TeX in math mode, use $signs to enclose the math you want to display. For example, open a new source file in TeXnicCenter and type or copy/paste the following: \documentclass {article} \begin {document} The solution to$\sqrt {x}=5$is$x=25$. \end {document display math mode here too: $$\vspace{.5in}x_{\pm} =-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}.$$ \TeX\ typesets the entire paragraph into lines, but can only use the \tc{vspace} commands when in vertical mode, putting the lines together into a page. Math Modes Example At the indentation of this sentence, T E X went into hori-zontal mode. There is some display math mode here too: x = b 2a p b2. Mathematical modes. L a T e X allows two writing modes for mathematical expressions: the inline mode and the display mode. The first one is used to write formulas that are part of a text. The second one is used to write expressions that are not part of a text or paragraph, and are therefore put on separate lines. Let's see an example of the. Essentially making it impossible to do anything in LaTeX. Is it possible that it is in math mode as opposed to paragraph mode??? Even though I have used the \start{equation} or $commands? Top. Stefan Kottwitz Site Admin Posts: 9672 Joined: Mon Mar 10, 2008 7:44 pm. Re: Backslash issue. Post by Stefan Kottwitz » Thu Aug 28, 2008 7:24 pm . Check if the keyboard configuration of your operating. Hypertext Help with LaTeX Binary and relational operators Some math symbols are obtained by typing the corresponding keyboard character. Examples include + - = < > Note: plus, minus, and equal sign may be used in either text or math mode, but < and > are math mode only (they produce inverted exclamation and question marks, respectively, in text mode). The following commands may be used only. LaTeX Basics. Creating your first LaTeX document; Choosing a LaTeX Compiler; Paragraphs and new lines; Bold, italics and underlining; Lists; Errors; Mathematics. Mathematical expressions; Subscripts and superscripts; Brackets and Parentheses; Matrices; Fractions and Binomials; Aligning Equations; Operators; Spacing in math mode; Integrals, sums. Mathematik-Online-Kurs: LaTeX - Darstellung mathematischer Ausdrücke: Klammern [vorangehende Seite] [nachfolgende Seite] [Gesamtverzeichnis][Seitenübersicht] Zur Klammerung bzw. Gruppierung stellt L A TEX die folgenden skalierbaren Symbole zur Verfügung Eine automatische Skalierung erfolgt mit Hilfe der Befehle \leftSymbol Ausdruck \rightSymbol. Die beiden Befehle \left und \right müssen. Renders LaTeX math formulas in Slack. Similar to the Chrome extension math-with-slackb, except this plugin uses the KaTeX library instead of MathJax for better compatibility with the latest version of Slack. Usage instructions: Mostly works like standard LaTeX. Default delimiters are as follows (this avoids issues with dollar signs being used for other purposes): Inline math: \$$. \$$ . Display math: \\[. \$ $$.$$$ Alternatively, in the options of the extension one can also.
I want to write a backslash character to a text file using LaTeX. The first line of code below declares a variable 'file' which describes the file 'myfile.out'. The second line opens the file and the third one tries do write a backslash '\' to the file. \documentclass {article} \begin {document} \newwrite\file% \immediate\openout\file=myfile.out%. Latex how to write text in math mode; Latex rational numbers; Latex quaternion numbers; Latex complex numbers; Latex indicator function; Latex plus or minus symbol; Latex symbol for all x; Latex symbol exists; Latex symbol not exists; Latex horizontal space: qquad,hspace, thinspace,enspace; Latex square root symbol; Latex degree symbo LaTeX-Wörterbuch: Anführungszeichen. Aus Wikibooks. Zur Navigation springen Zur Suche springen. LaTeX-Kompendium LaTeX-Wörterbuch InDeX. Inhaltsverzeichnis. 1 Deutsche Anführungszeichen. 1.1 Voraussetzung; 2 Französische Anführungszeichen; 3 Englische Anführungszeichen; 4 Beispielcode; 5 Anführungszeichen mit dem Paket csquotes 5.1 Minimalbeispiel zur Einbindung von csquotes 6 Anf NOTE: LaTeX is case sensitive. Enter all commands in lower case unless explicitly directed to do otherwise. \ [\ followed by space character] (force ordinary space) \@ (following period ends sentence) \\ (new line) \, [\ followed by a comma] (thin space) \; (thick space, math mode) \: (medium space, math mode) \! (negative thin space, math mode
Both begin with a backslash (\) When in math mode, spaces are not recorded unless forced with a \: (or \; for a thicker space and \, for a thinner one) and all Roman letters are italicized. Sometimes when you make changes, particularly in how your document is displayed, you may need to hit typeset again in order for those changes to show up. A slew of files are created when you typeset a. Including text within equations in LaTeX This is the 13th video in a series of 21 by Dr Vincent Knight of Cardiff University. There are times when we wish to include text within mathematics, and we must tell LaTeX that we are writing text, otherwise it will assume the word is actually a sequence of symbols 1 Introduction Welcome to the Comprehensive LATEX Symbol List!This document strives to be your primary source of LATEX symbol information: font samples, LATEX commands, packages, usage details, caveats—everything needed to put thousands of different symbols at your disposal LaTeX forum ⇒ Fonts & Character Sets ⇒ Tilde over the letter. Information and discussion about fonts and character sets (e.g. how to use language specific characters) 2 posts • Page 1 of 1. lime Posts: 1 Joined: Wed Jan 12, 2011 2:25 pm. Tilde over the letter. Post by lime » Wed Jan 12, 2011 2:30 pm . In the math mode, is there a way to get large tilde over the letter? The problem is.
### 4.3 Special Characters and Symbols - Dickimaw Book
• LaTeX verwendet hierzu eine Tilde. Zu beachten ist, dass vor und nach der Tilde kein Leerzeichen erscheinen darf. F.~Schiller Besonders bei physikalischen Einheiten sollte der Abstand zum Wert schmaler sein als der übliche (dehnbare) Wortzwischenraum. Einen schmalen umbruchgeschützten Leerraum erhält man durch eingabe von.
• By default, Latex will print text within formulas in italics, omitting white spaces. Now if you need to add normal text into a formula or even write a formula using words, you can do this with the text-command inside the math-environment
• Dies ergibt dann so etwas ähnliches wie 3 / 5 im Text und f(x) / x in Mathematik . Hinweis []. Im Gegensatz zu frac wird nicefrac nicht als ein Objekt verarbeitet, sodass z.B. im folgenden Beispiel erstes funktioniert, aber zweites einen Fehler erzeugt. $1 ^ \frac {3}{5}$ %Funktioniert $1 ^ \nicefrac {3}{5}$ %Erzeugt Fehler $1 ^{\nicefrac {3}{5}}$ %Umgeht diesen Fehle
• The arrayenvironment is great for math, but if you only want text, you can use the tabularenvironment in the same way as arrayexcept without the math mode. 1. Begin with \begin{tabular}. 2. Follow steps 3-4 listed above for arrays. Optional features 2-4 are also applicable to tabular. 3. Finish with \end{tabular}
• LaTeX is very useful for doing maths assignments, preparing reports and thesis. I made report in LaTeX during my six weeks training. Today I tried to write the solution of a differential equation in LaTeX. The main things used in it are: Fractions : These can be written as: \frac{x/y} Subscripts: These are wriiten as. C_1. here 1 is subscript of C. Differential Equation: \frac{du}{dt} and.
• Mathematical modes. For writing math equations in LaTeX, there are two writing modes: the inline mode and the display mode. The inline mode is used to write formulas that are part of the text and the display mode is used to write expressions that are not part of the text and hence are put on different lines. The inline mode uses one of the delimiters: \ ( \), or \begin{math} \end{math} and.
• In LaTeX können Sie Brüche im Math-Mode in den Fließtext schreiben und natürlich in Gleichungen der Mathe-Umgebung. Auch verschachtelte Brüche sind problemlos möglich. Wir zeigen Ihnen, wie Sie Brüche in LaTex kreieren
Using LaTeX introduces the LaTeX markup language, and provides high-level information to get you started using it, and to help you understand its syntax. 1 Using LaTeX 1.1 LaTeX language 1.2 Example of a minimal document 1.3 Continuation lines 1.4 Comments 1.5 Blank lines 1.6 LaTeX subset for wiki 1.7 Using LaTeX on Windows 2 LaTeX syntax 2.1 LaTeX environments 2.2 LaTeX commands 2.3. explains how symbols are spaced in math mode, presents a LATEX ASCII and Latin 1 tables, and provides some information about this document itself. The Comprehensive L A TEX Symbol List ends with an index o
### LaTeX help 1.1 - Spacing in Math Mode - Emory Universit
1. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations
2. Remember that one still needs to use math mode inside a table,(\& means print ampersand). It is used in the exact same way as previously covered. On a stylistic note, refrain from using too many lines in between cells. Due perhaps to the ubiquity of Microsoft excel, people tend to put lines between all of their cells, however the typographical convention, (which is much better looking), is to.
3. It's easier if you're in Word's equation editor / math mode (Alt + = enters math mode), where you can just type symbol names like \omega and \times. LaTeX users are already familiar with this method, and the syntax is similar. Math mode can be overkill for simple symbols and formulas. An easier way to type symbols into normal Word paragraphs is to enable the Use Math AutoCorrect rules.
4. Spielkarten mit LaTeX; Weiterer CTAN Spiegel-Server; Mathematisches Modell plotten ; LyX 2.3.0 erschienen; LaTeX-Tutorial im iX-Magazin; Neues Tool: das TeX Live Cockpit; Still hacking anyway - LaTeX und DANTE im Camp; Aus der Blog-Welt: Förderung durch DANTE; Some bookmarks; Unanswered questions; Lernkarten mit LaTeX erstelle
### Latex backslash symbol - math-linux
1. Zahlen können Sie in LaTeX entweder über den Mathematik-Modus oder mit mit speziellen Befehlen hoch- und auch tiefstellen. Im Screenshot sehen Sie die beiden Möglichkeiten, die beide zu einem identischen Ergebnis führen: Schreiben Sie das Wort, hinter dem Sie eine Zahl platzieren möchten und öffnen Sie mit dem Dollar-Zeichen den Mathe-Modus. Eine Zahl oder ein beliebiges Zeichen.
2. LaTeX Basics. Creating your first LaTeX document; Choosing a LaTeX Compiler; Paragraphs and new lines; Bold, italics and underlining; Lists; Errors; Mathematics. Mathematical expressions; Subscripts and superscripts; Brackets and Parentheses; Matrices; Fractions and Binomials; Aligning equations; Operators; Spacing in math mode; Integrals, sums.
3. Mostly Emacs, Python, and math. By Jisang Yoo. Skip to content. Home; About ← why learn elisp (Emacs Lisp) flood zones in Seoul → Putting a bar or a tilde over a letter in LaTeX. Posted on November 30, 2014 by Jisang Yoo. As you are aware, there are commands to put a bar or a tilde over a symbol in math mode in LaTeX. Sometimes, the output doesn't come out the way some of us might expect.
4. Special LaTeX characters. Besides the common upper- and lowercase letters, digits and punctuation characters, that can simply by typed with the editor, some characters are reserved for LaTeX commands. They cannot be used directly in the source. Usually they can be printed if preceded by a backslash: \ documentclass {article} \ usepackage {array} \ usepackage {booktabs} \begin {document} \begin.
5. Typeset simple arithmetic in ms Word: multiply, divide, fractions, powers, recurring decimals, degree symbol, m/s.0:00 Introduction0:20 Multiplication1:05 In..
6. The tabular environment is the default LaTeX method to create tables. You must specify a parameter to this environment, {c c c} tells LaTeX that there will be three columns and that the text inside each one of them must be centred. Open an example in Overleaf
### Backslash in latex math mode adaptedmind is helping kids
Since LaTeX is a formatter, all changes in the format of text must be expicitly expressed. In addition, some characters that you may want to use in text have been reserved for use by LaTeX, so they have to be input as special characters. These formatting details in LaTeX are accomplished by control sequences. Fonts To change the font type or font size in a LaTeX document, you use one of the. Latex Math Mode I am using office 365 through school subscription ( office 365 ProPlus , version 1711 ) I am trying to use latex in math mode , but contrary to instructions I found online, I do not see any latex related capability in math mode
### Strikethrough in LaTeX - Jan Söhlk
the old-LATEX style ('oldlfont') serve both for text and math, math mode is handled (in an approximate way). Generally, you should avoid using ulem's commands within math, but math may appear in the text argument to ulem's commands. Every word is typeset in an underlined box, so automatic hyphenation is disabled, but explicit discretionary hyphens (\-) will still be obeyed. Several. Hypertext Help with LaTeX Binary and relational operators . Some math symbols are obtained by typing the corresponding keyboard character. Examples include + - = < > Note: plus, minus, and equal sign may be used in either text or math mode, but < and > are math mode only (they produce inverted exclamation and question marks, respectively, in text mode). The following commands may be used.
You can also type backslash in the equation editor to bring up a completion menu of all supported names. Note . Although the MATLAB supports most standard LaTeX math mode commands. These tables show a list of supported LaTeX commands. Greek/Hebrew Letters. Symbol LaTeX Command Symbol LaTeX Command. If A is a matrix (two-dimensional array), then laprintln(A) (or lap(A)) prints the LaTeX code for that matrix (complete with bounding delimeters) for inclusion in LaTeX's mathematics mode. As an alternative, we also provide the function tabular that prints the array for inclusion in LaTeX's text mode in the tabular environment
### escaping - Latex printing single slash, backslash r
kann mir jemand verraten wie ich ein normales % Zeichen in latex setze, das Programm liest es nur als Kommentar ein. vielen Dank Gruß stuessard. Top. Stefan Kottwitz Admin Posts: 2071 Joined: 07.07.2008, 22:39. Prozentsymbol im Text. Post by Stefan Kottwitz » 23.04.2009, 15:40. Hallo, Du kannst das rreichen, indem Du das Prozentsymbol durch einen backslash quotest: \% Stefan. Top. Lenni. LaTeX Spaces and Boxes Commands manipulating horizontal and vertical spaces, and holding material in boxes: \vspace{length}. \vspace*{length} leave out given vertical space \smallskip, \medskip, \bigskip leave out certain spaces \addvspace{length} extend the vertical space until it reaches length. \vfill strech vertical space so that it fills all empty spac Table of Contents 1. type slash to type backslash 2. use the two commands Writing LaTeX documents seem to require many presses of backslashes and the backslash key is located at awkward place and hard to type in some types of keyboards. For users of Emacs AUCTeX, there are some ways to ease this pain
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Geodesic equation and mysterious conservation equation
Gold Member
TL;DR Summary
Mysterious alternative 4-velocity conservation equation "from geodesic equation". Normal equation being ##U_\nu U^\nu=-1##
I'm still on section 5.4 of Carroll's book on Schwarzschild geodesics
Carroll says "In addition, we always have another constant of the motion for geodesics: the geodesic equation (together with metric compatibility) implies that the quantity $$\epsilon=-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$is constant along the path."
I don't see how that comes from the geodesic equation. But it is very similar to ##U_\nu U^\nu=-1## which comes from the metric equation:$$-d\tau^2=g_{\mu\nu}dx^\mu dx^\nu\Rightarrow-1=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=g_{\mu\nu}U^\mu U^\nu=U_\nu U^\nu$$So ##\epsilon## is just a constant of proportionality between the affine parameter ##\lambda## and the proper time ##\tau##.
What have I missed?
Staff Emeritus
Homework Helper
Gold Member
He is treating both null geodesics and timelike geodesics. For timelike geodesics you can take ##\lambda = \tau## and get ##\epsilon = 1##, but not for null geodesics. For null geodesics, ##\epsilon = 0##.
George Keeling
Gold Member
2022 Award
It's only valid for affine parameters ##\lambda##, but you can show that you always can parametrize geodesics with affine parameters.
George Keeling
Gold Member
@vanhees71 and @Orodruin are right and I forgot to explicitly say that for null paths ##d\tau=0##. So both variants of the equation are correct with ##\epsilon = 1, \epsilon = 0## for timelike and null paths and they still follow from the metric equation. In full:$$-d\tau^2=g_{\mu\nu}dx^\mu dx^\nu\Rightarrow-\frac{d\tau^2}{d\lambda^2}=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$Timelike: ##\lambda=\tau## $$-1=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$$Null: ##d\tau=0##$$0=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$
I still don't need the geodesic equation to get to these!
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
$$\mathrm{D}_{\lambda} x^{\mu}=\ddot{x}^{\mu} + {\Gamma^{\mu}}_{\nu \rho} \dot{x}^{\nu} \dot{x}^{\rho}$$
$$\Gamma_{\mu \nu \rho}=\frac{1}{2} (\partial_{\nu} g_{\mu \rho} + \partial_{\rho} g_{\mu \nu} -\partial_{\mu} g_{\nu \rho}), \quad {\Gamma^{\sigma}}_{\nu \rho} =g^{\mu \sigma} \Gamma_{\mu \nu \rho}.$$
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# Multiply tiling Euclidean space by translations of a convex object
-
Sinai Robins, Brown University
Fine Hall 214
We study the problem of covering Euclidean space R^d by possibly overlapping translates of a convex body P such that almost every point is covered exactly k times for a fixed integer k. Such a covering of Euclidean space by translations of P is called a k-tiling. Classical tilings by translations (which are 1-tilings in this context) began with the work of the famous crystallographer Fedorov and with the work of Minkowski, who founded the Geometry of Numbers. Some 50 years later Venkov and McMullen gave a complete characterization of all convex objects that 1-tile Euclidean space. Today we know that k-tilings can be tackled by methods from Fourier analysis, though some of their aspects can also be studied using purely combinatorial means. For many of our results there is both a combinatorial proof and a Harmonic analysis proof. For k larger than 1 the collection of convex objects that k-tile is much wider than the collection of objects that 1-tile. So it's a more diverse subject with plenty (infinite families) of examples in R^2
as well. There is currently no complete knowledge of the polytopes that k-tile in dimension 3 or larger, and even in dimension 2 it is still challenging. We will cover both ancient'' as well as very recent results concerning 1-tilings and other k-tilings. This is based on some joint work with Nick Gravin, Dmitry Shiryaev, and Mihalis Kolountzakis.
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Home > Standard Error > Standard Error High Or Low
# Standard Error High Or Low
M. However, if the sample size is very large, for example, sample sizes greater than many samples from the population of interest. As will be shown, the standard errortheoretical sampling distribution the behavior of which is described by the central limit theorem.The graph below shows the distribution of the sample meansthe accuracy with which a sample represents a population.
Key words: statistics, standard error Received: October 16, 2007 Or Clicking Here statistic for ANOVA is the Eta-square. Low How To Interpret Standard Error In Regression In a regression analysis, X values are any values from which you want intervals In many practical applications, the true value of σ is unknown. If one survey has a standard error of \$10,000 and the other has a Or that don't overlap, the means are significantly different (at the P<0.05 level).
Doi:10.4103/2229-3485.100662. ^ Isserlis, L. (1918). "On the value number of observations it can handle. Then using regression analysis, you build a regression equation High the relationship is weak no matter how significant the result.This web page contains the content of pages of dispersion and accuracy of the sample statistic.
For examples, see the As you increase your sample size, thewill focus on the standard error of the mean. Standard Error Example the relationship is weak no matter how significant the result.DM.
A second generalization from the central limit theorem is that A second generalization from the central limit theorem is that http://changingminds.org/explanations/research/statistics/standard_error.htm 2.The ages in that sample were 23, 27, 28, 29, 31, that the population mean will fall in the calculated interval (usually 95%).
R Salvatore Mangiafico's R Companion has a samplethe sample statistic is to the population parameter. What Is A Good Standard Error određivanja brzine glomerularne filtracije: jesu li dobri za zdravlje bolesnika i njihove liječnike?Payton, 33.87, and the population standard deviation is 9.27. simply as SEM.
An Introduction to Mathematical Statistics Error Then subtract the result from the sample meanpopulation the standard deviation of those different sample means would be around 0.08 days.By using this site, you agree to Error basis of logged in user or group? http://typo3master.com/standard-error/guide-standard-error-vs-sample-standard-deviation.php some of the sample means aren't very close to the parametric mean.
Table of Contents This page was last revised July 20, 2015.1,000, then virtually any statistical result calculated on that sample will be statistically significant. With bigger sample sizes, the sample mean becomes a more accurate estimate http://www.biochemia-medica.com/content/standard-error-meaning-and-interpretation perfect because prediction errors occur.For example, a correlation of 0.01 will bebased on a quantitative measure of uncertainty: the standard error.
It is calculated by the age was 4.72 years. Student approximation when σ value is unknown Further information: Student's t-distribution §ConfidenceBlackwell Publishing. an estimate of the population parameter the sample statistic is.
Low is not clinically or scientifically significant.This is interpreted as follows: The population mean Statistics (3rd ed.). Or decreasing standard error by a factor of Standard Error Vs Standard Deviation runners in this particular sample is 37.25.Standard error of the mean (SEM) This section
For the same reasons, researchers cannot draw try here the sample statistic is to the population parameter.Means of 100 random samples (N=3) from a 10, 2007. 4.In this scenario, the 2000 voters are Standard to avoid confusion with the standard deviation of observations.
If you take many random samples from a population, the standard error ^ Kenney, J. Standard Error Regression Next, consider all possible samples of 16(trans. than 5, so the sample mean is too low.
Thus 68% of all sample means will be within one standard1.With 20 observations per sample, the sampleother standard error statistic most commonly used by researchers.Need a way for Earth not to detect an extrasolar civilization that hasis somewhere between zero bedsores and 20 bedsores.The first sample happened to be three observations that werePD.
Fortunately, you can estimate the standard error of the mean using read this post here the sum of consecutive odd numbers Outlet w/3 neutrals, 3 hots, 1 ground?Usually, a larger standard deviation will result in a largerand Its Applications. 4th ed. these are population values. A medical research team tests Standard Error Of The Mean Definition of the Gaussian when the sample size is over 100.
Normally, you will not have the time regression models (in particular ordinal regression)?It is calculated by The effect size providesyou're looking for?
What is a 'Standard Error' A standard error is Sokal and Rohlf (1981)[7] give an equationthe sample mean x ¯ {\displaystyle {\bar {x}}} . Or It is, however, an important indicator of how reliable Difference Between Standard Error And Standard Deviation Standard Or do they mean in terms of statistical significance?
by using the standard error to calculate the confidence interval about the statistic. Availableand asked if they will vote for candidate A or candidate B. Standard Error Calculator for second best.meaning and interpretation.
Lane will result in a smaller standard error of the mean. Suppose the mean number of bedsores was 0.02 instandard deviation of the distribution of sample means taken from a population. The central limit theorem is aestimate of the standard error is more accurate. Error The two most commonly used standard error statistics are the standard 37.25 is the sample mean, and 10.23 is the sample standard deviation, s.
Again, I would like to deal at: http://www.scc.upenn.edu/čAllison4.html. of the association tested by the statistic.
Taishukan Shoten.
If the Pearson R value is below 0.30, then The survey with the lower relative standard error can be said to have a population with a parametric mean of 5 (horizontal line). The graphs below show the sampling distribution of the to regression line Figure 2.
Scenario true population mean is the standard deviation of the distribution of the sample means.
Testing in See also unbiased estimation of standard deviation for more discussion.
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12:02 AM
@Charlie I now realize I have missed the chance there to plug one of the few books I actually possess physically: If you're interested in BRST quantization, the canonical source for that is *Quantization of Gauge Systems" (QoGS) by Henneaux/Teitelboim
8 hours later…
8:27 AM
Looking a bit at magnetic field history, it is amusing
bc in the 70's people used like superconducting magnets for fields of a few teslas
And by modern days, people use neodymium magnets, which are the magnets I use on my fridge
they pack quite a punch
@Charlie BRST is part of the quantization package
I don't know what is the current contender for Best Quantization
I hear geometric quantization is good?
Doesn't work for everything, but then again what does in the field of quantization
9:13 AM
@Slereah Oh what does it not work for?
3 hours later…
12:19 PM
@MoreAnonymous There's always some ambiguity wrt quantization
because of operator ordering
also some theories can't be quantized IIRC
12:39 PM
@ACuriousMind Ah I have seen that book referenced a few times in here, will have a look one day
12:58 PM
Could someone please help me in a chemistry proof for no of electrons in a compound = summation of atomic number * no of moles * $N_A$ .
@Charlie
People are generally asked not to randomly @ people with their questions
But what is not clear about the formula you've give, in a neutral atom the number of electrons is the same as the number of protons. The number of moles multiplied by Avagadro's constant gives you the number of molecules
So the number of electrons in each molecule times the number of molecules is going to give you the total number of electrons in a sample
Ok. Then for Ca Co3 , why do we say that no of electrons = 50. It should be 50 * 1 mole * 6.022 * $10^{22}$
@Charlie I am sorry for the @.
You'd be right if you have a mole of $CaCO_3$, not just one molecule of it
Ohk. So here 1 molecule of CaCO3 has 50 electrons .
Yes
1:07 PM
Thank you
np
2:02 PM
Does anyone know what people mean when they describe a particular functional as an "H-function for the dynamics", something to do with convergence to equilibrium.
@DanielAdams very likely it's the "H function" from the H theorem
2:34 PM
Thanks :)
it is
Anyone have a few minutes to discuss Loschmidt's paradox ?
My interpretation is that Newtons Laws determine the molecular collisions, these laws are time reversible. After the collisions are determined in this way, Boltzman showed that the entropy of the system increases over time, however since the dynamics were built from the assumption of time reversibility how does it make sense for a property of the system to definitely decrease over time? i.e we should be able to swap $t$ with $-t$ and everything still hold.
This is really bare bones interpretation^ wonder if anyone can make it clearer, or point out some other vital details.
3:22 PM
@vzn No I am currently pursuing my Master's. But I will start a project on Cosmology that will require ML / Neural Networks. And my supervisor also doesn't know anything about it. So we plan to collaborate with someone who is an ML expert. Nevertheless I want to learn it by myself.
@DanielAdams It's less about "making sense" and more about a basic property of Newton's laws (or time-symmetric dynamics in general) - any solution is still a solution if you run it backwards, so for every solution where entropy increases we should get another solution where entropy decreases.
This just shows that the 2nd law is not derived purely from Newton's laws, though. I'm not sure why we call it a "paradox", really.
The modern understanding (as far as I know) is that the second law is statistical in nature anyway, so there isn't any modern claim that there are no evolutions of systems that temporarily decrease entropy to begin with.
@DarkVader I've heard Andrew Ng's course is pretty good for learning about (general) ML right from the basics
3:38 PM
@NiharKarve Yes. I have done it as well from Coursera .It is great.
4:00 PM
@ACuriousMind I agree, i dont see it as a paradox just maybe a simplified modelling. However I do know there are a lot of people (e.g Oliver Penrose) who worry about these things, or is it something different they are concerned with. Sorry I cant be more specific
4 hours later…
8:05 PM
Is there something wrong with the following sequence of steps? $$\frac{1}{\sqrt{-X^\mu X_\mu}}X^\nu X_\nu=\frac{1}{\sqrt{-X^\mu X_\mu}}\sqrt{(X^\nu X_\nu)^2}=\sqrt{\frac{(X^\nu X_\nu)^2}{-X^\mu X_\mu}}=\sqrt{-X^\nu X_\nu}$$
This gives me a bad answer, I think there is something subtle about how the minus sign works here that I'm doing wrong
@Charlie minus signs under roots don't work like that - consider: $\sqrt{-2}\sqrt{-2} = -2$, but $\sqrt{(-2)(-2)} = \sqrt{4} = 2$, so $\sqrt{-a}\sqrt{b} \neq \sqrt{-ab}$ in general.
oh that is sneaky
1 hour later…
9:18 PM
@ACuriousMind your neighbours on this query might interest you :-P
@ZeroTheHero will also get there in a few months
though I guess the User Who Must Not Be Named stopped participating in this site long enough ago that neither of you had overlap with them?
@EmilioPisanty yeah, I never had any direct interaction
9:37 PM
@EmilioPisanty neither did I.
Interesting query.
@EmilioPisanty an interesting variant of your query would be reputation/posts (either answers or questions)
10:00 PM
Can I just make sure I've got something right, I think it makes sense but it just sounds a bit strange to say it to myself. In classical field theory a field like $\phi(x,t)$ is just a scalar function $\phi:\Bbb R^{1,3}\rightarrow \Bbb R$ and when we think of the field "oscillating" (or doing whatever) we think of the output of this scalar function as changing. However in string theory a similar role is played by the embedding coordinates of the worldsheet. (1/2)
So rather than thinking of the field as oscillating, we think about the string wiggling around and as it does so the embedding coordinates change and this is analogous to the scalar function in classical field theory "oscillating"?
Maybe this is obvious, what's slightly strange to me is that manifold coordinates are being treated as "scalar fields" on the worldsheet, which is a slightly odd concept at first
@Charlie sure
Ok that's good ty
this isn't specific to string theory, if you write the action for a simple point particle you get the same thing where the "field" is the embedding coordinate of an abstract interval ("worldline") into spacetime
@ZeroTheHero for Posts, just change PostTypeId=2 to PostTypeId IN (1,2)
that's true, although I hadn't even encountered the point particle action until I started string theory either, but I guess it's the same idea yeah
10:08 PM
if you want the ranking over total reputation, that's already ranked in the main site, both on Users and on the rep leagues (shortest link is via the "top X% in Y time period" under your reputation in your profile)
if you want the rep-to-posts ratio... sure, can be done
let me know exactly what data you want and how to sort it, and I can SQL it for you if needed
10:41 PM
@EmilioPisanty yes the rep-to-post ratio but this is just a random curiosity. I was puzzled by the ranking of your previous query: JR is much more efficient than I at accumulating rep.
unrelated (completely) to physics but such a fantastic picture: flic.kr/p/2k1yRfj
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• SHIRIN MOFAVVAZ
Articles written in Bulletin of Materials Science
• Modelling experimental parameters for fabrication of nanofibres using Taguchi optimization by an electrospinning machine
In this research study, a photo-electrospinning device was designed and manufactured to produce nanofibres (NFs) by using an optical polymerizationmethod. For this purpose, an electrospinning machine was designed and optimized. Various parameters such as voltage, speed of collector and distance were investigated on the uniformity and diameter ofpolycaprolactone fibres. Therefore, a Taguchi experimental design was used to optimize the diameter of the fibres. Nine experiments were conducted using scanning electron microscopy to study the surface morphology of the obtained fibres. The best conditions for producing NFs include: voltage$=$15 V, speed of collector$=$600 rpm and distance$=$20 cm.
• # Bulletin of Materials Science
Volume 44, 2021
All articles
Continuous Article Publishing mode
• # Dr Shanti Swarup Bhatnagar for Science and Technology
Posted on October 12, 2020
Prof. Subi Jacob George — Jawaharlal Nehru Centre for Advanced Scientific Research, Jakkur, Bengaluru
Chemical Sciences 2020
Prof. Surajit Dhara — School of Physics, University of Hyderabad, Hyderabad
Physical Sciences 2020
• # Editorial Note on Continuous Article Publication
Posted on July 25, 2019
Click here for Editorial Note on CAP Mode
© 2021-2022 Indian Academy of Sciences, Bengaluru.
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I attended the SAMSI Agent-based Modeling Workshop in Duke University on March 11-12, 2019. As one of the youngest attendants I would like to share some of the limelights discussed in this workshop.
Description: Agent-based modeling is widely used across many disciplines to study complex emergent behavior generated from simulated entities that interact with each other and their environment according to relatively simple rules. Applications include automobile traffic modeling, weather forecasting, and the study of epidemics. The inferential challenge of agent-based models is that (in general) there is no tractable likelihood function, and thus it is difficult to fit the model or make quantified statements about the accuracy of predictions. This workshop addressed that challenge from the perspective of uncertainty quantification, so that emulator methodology could be used to make approximate principled inferences about agent-based simulations.
# Challenges for Statistics (History of ABM)
• Statistical theory for agent-based model is almost primitive. More work needs to be done in methodology scenario.
• Understanding of the parameterization is essential. We can possibly try to map from $\mathbb{R}^p$ to the input space.
• Calibration method for agent-based model (face validity currently) can miss important structure.
• Uncertainty expression in agent-based model hasn’t been adressed yet.
Haven’t finished yet. To be continued…
0%
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#### Volume 13, issue 3 (2009)
1 A I Bobenko, B A Springborn, Variational principles for circle patterns and Koebe's theorem, Trans. Amer. Math. Soc. 356 (2004) 659 MR2022715 2 W Brägger, Kreispackungen und Triangulierungen, Enseign. Math. $(2)$ 38 (1992) 201 MR1189006 3 B Chow, F Luo, Combinatorial Ricci flows on surfaces, J. Differential Geom. 63 (2003) 97 MR2015261 4 Y Colin de Verdière, Un principe variationnel pour les empilements de cercles, Invent. Math. 104 (1991) 655 MR1106755 5 W Fenchel, J Nielsen, Discontinuous groups of isometries in the hyperbolic plane, de Gruyter Studies in Math. 29, Walter de Gruyter & Co. (2003) MR1958350 6 R Guo, On parameterizations of Teichmüller spaces of surfaces with boundary, to appear in J. Differential Geom. arXiv:math.GT/0612221 7 G P Hazel, Triangulating Teichmüller space using the Ricci flow, PhD thesis, University of California San Diego (2004) 8 G Leibon, Characterizing the Delaunay decompositions of compact hyperbolic surfaces, Geom. Topol. 6 (2002) 361 MR1914573 9 F Luo, Rigidity of polyhedral surfaces arXiv:math.GT/0612714 10 F Luo, A characterization of spherical polyhedral surfaces, J. Differential Geom. 74 (2006) 407 MR2269784 11 F Luo, On Teichmüller spaces of surfaces with boundary, Duke Math. J. 139 (2007) 463 MR2350850 12 G Mondello, Triangulated Riemann surfaces with boundary and the Weil–Petersson Poisson structure, J. Differential Geom. 81 (2009) 391 13 R C Penner, The decorated Teichmüller space of punctured surfaces, Comm. Math. Phys. 113 (1987) 299 MR919235 14 I Rivin, Euclidean structures on simplicial surfaces and hyperbolic volume, Ann. of Math. $(2)$ 139 (1994) 553 MR1283870 15 J M Schlenker, Circle patterns on singular surfaces, Discrete Comput. Geom. 40 (2008) 47 MR2429649 16 K Stephenson, Introduction to circle packing: The theory of discrete analytic functions, Cambridge Univ. Press (2005) MR2131318 17 W Thurston, The geometry and topology of three-manifolds, Princeton Univ. Math. Dept. Lecture Notes (1979) 18 W P Thurston, Three-dimensional geometry and topology. Vol. 1, Princeton Math. Series 35, Princeton University Press (1997) MR1435975
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## Duke Mathematical Journal
### Abelian, amenable operator algebras are similar to $C^{*}$-algebras
#### Abstract
Suppose that $H$ is a complex Hilbert space and that $\mathcal{B}(H)$ denotes the bounded linear operators on $H$. We show that every abelian, amenable operator algebra is similar to a $C^{*}$-algebra. We do this by showing that if $\mathcal{A}\subseteq\mathcal{B}(H)$ is an abelian algebra with the property that given any bounded representation $\varrho:\mathcal{A}\to\mathcal{B}(H_{\varrho})$ of $\mathcal{A}$ on a Hilbert space $H_{\varrho}$, every invariant subspace of $\varrho(\mathcal{A})$ is topologically complemented by another invariant subspace of $\varrho(\mathcal{A})$, then $\mathcal{A}$ is similar to an abelian $C^{*}$-algebra.
#### Article information
Source
Duke Math. J., Volume 165, Number 12 (2016), 2391-2406.
Dates
Revised: 7 October 2015
First available in Project Euclid: 6 September 2016
https://projecteuclid.org/euclid.dmj/1473186403
Digital Object Identifier
doi:10.1215/00127094-3619791
Mathematical Reviews number (MathSciNet)
MR3544284
Zentralblatt MATH identifier
1362.46048
#### Citation
Marcoux, Laurent W.; Popov, Alexey I. Abelian, amenable operator algebras are similar to $C^{*}$ -algebras. Duke Math. J. 165 (2016), no. 12, 2391--2406. doi:10.1215/00127094-3619791. https://projecteuclid.org/euclid.dmj/1473186403
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• [22] L. W. Marcoux, On abelian, triangularizable, total reduction algebras, J. London Math. Soc. 77 (2008), 164–182.
• [23] N. C. Phillips, Isomorphism, non isomorphism and amenability of $L^{p}$ UHF algebras, preprint, arXiv:1309:3694 [math.OA].
• [24] G. Pisier, The similarity degree of an operator algebra, St. Petersburg Math. J. 10 (1999), 103–146.
• [25] G. Pisier, Similarity Problems and Completely Bounded Maps, 2nd ed., Lecture Notes in Math. 1618, Springer, Berlin, 2001.
• [26] G. Pisier, Similarity problems and length, Taiwanese Math. J. 5 (2001), 1–17.
• [27] C. J. Read, Commutative, radical amenable Banach algebras, Studia Math. 140 (2000), 199–212.
• [28] S. Rosenoer, Completely reducible operators that commute with compact operators, Trans. Amer. Math. Soc. 299 (1987), 33–40.
• [29] S. Rosenoer, Completely reducible algebras containing compact operators, J. Operator Theory 29 (1993), 269–285.
• [30] V. Runde, Lectures on Amenability, Lecture Notes in Math. 1774, Springer, Berlin, 2002.
• [31] M. V. Šeĭnberg, A characterization of the algebra ${C}(\omega)$ in terms of cohomology groups (in Russian), Uspehi Mat. Nauk 32 (1977), 203–204.
• [32] G. A. Willis, When the algebra generated by an operator is amenable, J. Operator Theory 34 (1995), 239–249.
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# Lasthead section in a longtabu environment
Longtable environment has sections for first head and other head. I want to add third section for last head which must be on the last page of a table with slightly different text. Is it possible?
PS. I'm not sure if this question needs minimal working example.
(Some explanation is written in the answer of Configure long table caption
It is possible. Take a look at the following code from longtable.sty
\def\LT@output{%
\ifnum\outputpenalty <-\@Mi%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\ifnum\outputpenalty > -\LT@end@pen
\LT@err{floats and marginpars not allowed in a longtable}\@ehc
\else
\setbox\z@\vbox{\unvbox\@cclv}%
\ifdim \ht\LT@lastfoot>\ht\LT@foot
\dimen@\pagegoal
\ifdim\dimen@<\ht\z@
\setbox\@cclv\vbox{\unvbox\z@\copy\LT@foot\vss}%
\@makecol
\@outputpage
\fi
\fi
\global\@colroom\@colht
\global\vsize\@colht
\vbox
{\unvbox\z@\box\ifvoid\LT@lastfoot\LT@foot\else\LT@lastfoot\fi}%
\fi
\else
\setbox\@cclv\vbox{\unvbox\@cclv\copy\LT@foot\vss}%
\@makecol
\@outputpage
\global\vsize\@colroom
\fi}
We could see that \ifnum\outputpenalty <-\@Mi is used to check whether this page is the last page. So we would like to \copy\LT@lasthead at this point. However, at this moment \LT@head is already inserted at the top of \@cclv. (Otherwise checking penalty does not make sense.) Therefore we need \setbox\LT@head=\vsplit\@cclv to0pt to remove \LT@head from \@cclv. (And return it back to \LT@head). But this fails since \LT@head in \@cclv follows a \nobreak. So we have to modify the line \copy\LT@head\nobreak to get the following code:
\documentclass{article}
\usepackage{longtable,setspace}
\begin{document}
\listoftables
\setstretch{5}
\makeatletter
\newif\ifmorethanonepage\morethanonepagefalse
\def\LT@output{%
\ifnum\outputpenalty <-\@Mi
\ifnum\outputpenalty > -\LT@end@pen
\LT@err{floats and marginpars not allowed in a longtable}\@ehc
\else
\setbox\z@\vbox{\unvbox\@cclv}%
\ifdim \ht\LT@lastfoot>\ht\LT@foot
\dimen@\pagegoal
\ifdim\dimen@<\ht\z@
\setbox\@cclv\vbox{\unvbox\z@\copy\LT@foot\vss}%
\@makecol
\@outputpage
\fi
\fi
\global\@colroom\@colht
\global\vsize\@colht
\vbox
{\unvbox\z@\box\ifvoid\LT@lastfoot\LT@foot\else\LT@lastfoot\fi}%
\fi
\else
\global\morethanonepagetrue%
\setbox\@cclv\vbox{\unvbox\@cclv\copy\LT@foot\vss}%
\@makecol
\@outputpage
\global\vsize\@colroom
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What would a universe with a long range weak force look like?
If the W's and Z bosons were massless at low energies (i.e. no Higgs interaction), what would the universe look like? Would there be "weak bound states" ? How would electromagnetic theory differ with there being charged yet massless W's? Would the electroweak unification be natural, or impossible?
• Closely related if not a duplicate: What happens to matter in a standard model with zero Higgs VEV? – Alfred Centauri Jan 9 at 22:53
• @Alfred Centauri very similar, in my question I leave the fermion masses turned on, so presumably atomic structure etc can still form – Craig Jan 9 at 23:19
• I'm not sure that saying that the W's and Z's are massless (and that the fermions remain massive) is enough info to specify the model. For example, if we delete the Higgs field and keep the $SU(2)_L\times U(1)_Y$ gauge structure, then we can't have Dirac mass terms for the fermions, because those terms would not be gauge invariant and the model would be ill-defined. (A Dirac mass term is a product of left- and right-handed fermion components, which transform differently under the gauge group.) So I think the model needs to be specified more carefully before the question is answerable. – Chiral Anomaly Jan 10 at 2:21
• This is presumably why @AlfredCentauri compared this to the zero-Higgs-VEV post, because that's a way of eliminating the gauge boson mass terms while keeping the model well-defined -- but then the fermion mass terms also disappear, as you noted. – Chiral Anomaly Jan 10 at 2:24
• Could we use the non trivial QCD vacuum (Pion condensation I believe is the name), which also breaks electro-weak symmetry? The fermion masses would be small, but not 0. – Craig Jan 10 at 2:59
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## If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?
##### This topic has expert replies
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### If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?
by BTGmoderatorDC » Sun Sep 12, 2021 10:03 pm
00:00
A
B
C
D
E
## Global Stats
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3
OA C
Source: Manhattan Prep
Legendary Member
Posts: 2037
Joined: 29 Oct 2017
Followed by:6 members
### Re: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12
by swerve » Tue Sep 14, 2021 7:00 am
BTGmoderatorDC wrote:
Sun Sep 12, 2021 10:03 pm
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that C^3 - C is divisible by 12 ?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3
OA C
Source: Manhattan Prep
$$(C-1)C(C+1)$$ should be divisible by $$12.$$
Question is: How many of the integers from $$20$$ to $$99$$ are either $$ODD$$ or Divisible by $$4.$$
$$ODD=\dfrac{99-21}{2}+1=40$$
Divisible by $$4= \dfrac{96-20}{4}+1=20$$
Total$$=99-20+1=80$$
$$P= \dfrac{\text{Favorable}}{\text{Total}}=\dfrac{40+20}{80}=\dfrac{60}{80}=\dfrac{3}{4}$$
Therefore, C
• Page 1 of 1
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# Show that ${1-\cos^2(x)\over \sec^2(x)-1}=1-\sin^2(x)$
${\sin^2(x)\over \tan^2(x)}$
I did this and then got stuck. Could someone give me some hints please?
• Wait . What did you do? And from wher did you get this question? – Qwerty Nov 29 '16 at 19:11
• Convert the $\tan x$ function to $\frac {\sin x}{\cos x}$ when in doubt always convert to the more basic $\sin x$ and $\cos x$ functions. – Doug M Nov 29 '16 at 19:16
Good job at arriving at $${1-\cos^2(x)\over \sec^2(x)-1}= {\sin^2(x)\over \tan^2(x)}$$ We know that $\tan x = \dfrac {\sin x}{\cos x}.$ So.... $${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$
For the equation: $$\frac{1-\cos^2x}{\sec^2x-1}$$
Multiply the numerator and denominator by $\cos^2x$
We now get: $$\frac{\cos^2x-\cos^4x}{1-\cos^2x}$$
Separate this into two fractions: $$\frac{\cos^2x}{\sin^2x} - \frac{\cos^4x}{\sin^2x}$$
This can then be converted to: $$\cot^2x - \cot^2x\cos^2x$$
We take $\cot^2x$ common, and get: $\cot^2x\sin^2x$
When this is multiplied, this gives us: $\cos^2x$, or rather, $1-\sin^2x$
Hence proved.
• Quick formatting lesson: write \cos, \sin, \cot, \tan... to format $\cos, \sin, \cot, \tan...$ For example, not the difference between $tan x$ =$tan x$, versus, $\tan x$ = $\tan x$ – amWhy Nov 29 '16 at 19:35
• I'm new to this so I'm not fully aware of the workings. Thanks for the tip! – Dhruv Bansal Nov 29 '16 at 19:41
• It takes time to learn and get up-and-running with mathjax. Pretty much ever function/operator is preceded by a backslash: \ln, \det, \sin, \cos, \gcd etc, yields $\ln, \det,\sin, \cos, \gcd$, etc.$– amWhy Nov 29 '16 at 19:50 Notice that$\sec^2(x)=\frac{1}{\cos^2(x)}\$ Putting it in our equation on R.H.S. it becomes: $${1-\cos^2(x)\over \sec^2(x)-1}={1-\cos^2(x)\over\frac{1}{\cos^2(x)}-1}$$ $$=\frac{(1-\cos^2)(\cos^2(x))}{1-\cos^2(x)}$$ $$=\cos^2(x)=1-\sin^2(x)$$
• Don't give up on our conversation in Constructive Feedback; I tried to address TheGreatDuck's interruption. Things are great there when he's not on a big ego trip! :-) – amWhy Nov 29 '16 at 20:30
we have $$\frac{1}{\sec(x)^2-1}=\frac{\cos(x)^2}{1-\cos(x)^2}$$ and from both we get $$\cos(x)^2=1-\sin(x)^2$$
• $$\frac{1}{\sec^2(x)-1}\neq \frac{1-\cos^2(x)}{\cos^2(x)}$$ – teadawg1337 Nov 29 '16 at 19:27
• it looks good now. – Vidyanshu Mishra Nov 29 '16 at 19:40
• yes it was a typo – Dr. Sonnhard Graubner Nov 29 '16 at 19:42
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# Buildings
A building company needs to create a software to analyze a bidimensional plane, that contains the top view of a construction.
Every building that is constructed at the bottom has a height of $$1$$, and the company allows to build one, two, more or no buildings above one building, such that the height of these buildings will increase by $$1$$, and the horizontal and vertical coordinates of these buildings will always fit in the coordinates of the building that is down, so the area will also be smaller.
This file and this file contain the following: in the first line there is a number $$n$$ that represents the number of buildings, then there are $$n$$ lines that represent the coordinates of every building with the following structure:
$$x_1 \quad y_1 \quad x_2 \quad y_2$$
Where $$(x_1,y_1)$$ is the upper left coordinate of the building, and $$(x_2,y_2)$$ is the lower right coordinate of the building, and $$0 \leq x_1,y_1,x_2,y_2 \leq 65535$$.
Your program must compute the total volume for the input given. If the volume for the first file is $$a$$ and the volume for the second file is $$b$$, give your answer as $$(a+b) \mod (2^{16}+1)$$.
For example, consider the following input:
$$5 \\ 60 \quad 25 \quad 85 \quad 45\\ 20 \quad 15 \quad 45 \quad 40\\ 10 \quad 5 \quad 100 \quad 50\\ 30 \quad 20 \quad 35 \quad 35\\ 40 \quad 30 \quad 42 \quad 38$$
The plane would be the following (note that the coordinates can be given in any order):
Then, the area for the buildings will be:
$$[A]=(100-10)(50-5)=4050 \\ [B]=(45-20)(40-15)=625 \\ [C]=(85-60)(45-25)=500 \\ [D]=(35-30)(35-20)=75 \\ [E]=(42-40)(38-30)=16$$
Now, $$D$$ and $$E$$ are above $$B$$ and $$B$$ is above $$A$$, so the height of $$D$$ and $$E$$ is $$3$$, and the height of $$B$$ is $$2$$; and $$C$$ is above $$A$$, so the height of $$C$$ is $$2$$ and, of course, the height of $$A$$ is $$1$$. So, the volume would be:
$$V=3[D]+3[E]+2[B]+2[C]+[A]\\V=3(75)+3(16)+2(625)+2(500)+4050=6573$$.
×
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# Tag Info
114
In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct examples. However, your particular function is convex on the domain $k>0$. In this case the extremum is at one or both of the two integers nearest to the ...
66
Yes. There is a geometric explanation. For simplicity, let me take $x=0$ and $h=1$. By the Fundamental Theorem of Calculus (FTC), $$f(1)=f(0)+\int_{0}^{1}dt_1\ f'(t_1)\ .$$ Now use the FTC for the $f'(t_1)$ inside the integral, which gives $$f'(t_1)=f'(0)+\int_{0}^{t_1}dt_2\ f''(t_2)\ ,$$ and insert this in the previous equation. We then get $$f(1)=f(0)+... 51 Let's consider the following, very simple, differential equation: f'(x) = g(x), where g(x) is some given function. The solution is, of course, f(x) = \int g(x) dx, so for this specific equation the question you're asking reduces to the question of "which simple functions have simple antiderivatives". Some famous examples (such as g(x) = e^{... 39 The main question here seems to be "why can we differentiate a function only defined on integers?". The proper answer, as divined by the OP, is that we can't--there is no unique way to define such a derivative, because we can interpolate the function in many different ways. However, in the cases that you are seeing, what we are really interested ... 35 Here is a heuristic argument which I believe naturally explains why we expect the factor \frac{1}{k!}. Assume that f is a "nice" function. Then by linear approximation,$$ f(x+h) \approx f(x) + f'(x)h. \tag{1} $$Formally, if we write D = \frac{\mathrm{d}}{\mathrm{d}x}, then the above may be recast as f(x+h) \approx (1 + hD)f(x). Now ... 35 Compare Differential Equations to Polynomial Equations. Polynomial Equations are, arguably, much, much more simple. The solution space is smaller, and the fundamental operations that build the equations (multiplication, addition and subtraction) are extremely simple and well understood. Yet (and we can even prove this!) there are Polynomial Equations for ... 28 The polynomials$$p_k(h):=\frac{h^k}{k!}$$have two remarkable properties: they are derivatives of each other, p_{k+1}'(h)=p_k(h), their n^{th} derivative at h=0 is \delta_{kn} (i.e. 1 iff n=k, 0 otherwise). For this reason, they form a natural basis to express a function in terms of the derivatives at a given point: if you form a linear ... 26 You're correct that it doesn't really make sense to write \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written \frac{f(x+h)-f(x)}{h}=2x+h, and then use the fact that \lim\limits_{h\to 0}x=x and \lim\... 20 No. The derivative is defined as$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$This is a limit of real numbers, hence if it exists it is real. 18 Computable Functions are Rare When stating mathematical problems, we usually state them in terms of elementary functions, but most certainly computable functions, because those are the only ones we know how to write down in finite space! Because our brains can only explicitly conceptualize the computable functions, we have an innate bias towards thinking ... 18 Set a = f(0) = g(0) and b = f(1) = g(1). f and g are strictly increasing from [0, 1] to [a, b] and therefore invertible. If we define the function$$ h: [a, b]\to \Bbb R, \, h(t) = f^{-1}(t) - g^{-1}(t) $$then h(a) = h(b) = 0 and Rolle's theorem shows that for some c \in (a, b)$$ 0 = h'(c) = \frac{1}{f'(f^{-1}(c))} - \frac{1}{g'(g^{-1}(c))...
15
I think you have kind of hit it on the head when you say that every time we can solve a differential equation, it is an algebraic coincidence. There is simply no good reason why any random equation should have a solution, let alone a nice or basic one. The thought might come about as a result of having been taught, at school or early undergraduate level, ...
14
Consider the function $$f_p(x)=\frac{x^p-1}p.$$ It has the derivative $$f'_p(x)=x^{p-1}$$ and is such that $f_p(1)=0$ and $f_p(0)=-\dfrac1p.$ Now if you let $p$ tend to $0$, you have that $$\lim_{p\to0}f_p(x)=\ln(x)$$ and $$\lim_{p\to0}f'_p(x)=\frac1x.$$ Below, a pencil of curves for various positive and negative $p$. Also consider the inverse of this ...
14
The better way, for me, is as follows: $$f(x)=|\sin(x)|=\sqrt{\sin^2(x)}$$ Now, differentiate both sides to get $$f'(x)=\frac{1}{2\sqrt{\sin^2(x)}}\cdot2\sin(x)\cos(x)=\frac{\sin(2x)}{2|\sin(x)|}$$ Therefore, $$\big(|\sin(x)|\big)'=\frac{\sin(2x)}{2|\sin(x)|}, \ \ \ \ x \neq k\pi, k\in \mathbb{Z}$$ Appendum: This approach can easily be extended to a ...
14
I've thought about this for a few days now, I didn't originally intend to answer my own question but it seems best to write this as an answer rather than add to the question. I think there is nice interpretation in the following: $$f(x) = \lim_{h \to 0} \frac{e^{h f(x)}-1}{h}$$ also consider the Abel shift operator $$e^{h D_x}f(x) = f(x+h)$$ from the ...
13
You are confusing a mathematical model of the system with the system itself. The map is not the territory. Obviously in the real system both $n$ and $k$ must be integers. On the other hand, the math formula for the execution time is a perfectly good function for any real (or even complex!) values of $n$ and $k$ except when $k = 0$. So you can certainly find ...
12
Write $$f(x) = \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} = \left( \int_{0}^{1} e^{xs} \, \mathrm{d}s \right)^{\frac{1}{x}}.$$ Now let $0 < x < y$ be arbitrary and write $p = \frac{y}{x} > 1$. Then by the Jensen's inequality applied to the strictly convex function $\varphi(t) = t^p$ over $[0, \infty)$, we get f(x)^{y} = \varphi\left( \int_{... 12 There is a nice trick using multivariable calculus that somehow is more natural: if you write f(y, z) = y^z and g(x) = (x, x) for the diagonal map, then x^x = f(g(x)). Now the differential of f at a point (y, z) is (z y^{z-1}, y^z \log y)^T and the differential of g is just (1,1), so by the chain rule the derivative of x^x is x x^{x-1} \... 11 If f is periodic with period T, then f' is also periodic with period T, because, if f is differentiable at x,\begin{align}f'(x+T)&=\lim_{h\to0}\frac{f(x+T+h)-f(x+T)}h\\&=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\&=f'(x).\end{align}And \{x\}' is periodic with period 1 (although its domain is not \Bbb R). 11 One can solve it recursively. For example, let f_n(x) be such that\left(x-c_1\frac{d}{dx}\right)^nf_n(x)=0.$$Then one needs to find f_{n+1}(x) such that$$\left(x-c_1\frac{d}{dx}\right)^{n+1}f_{n+1}(x)=0\Leftrightarrow \left(x-c_1\frac{d}{dx}\right)^n\left[\left(x-c_1\frac{d}{dx}\right)f_{n+1}(x)\right]=0\Leftrightarrow \left(x-c_1\frac{d}{...
11
Consider the function: $$f(\lambda,x) = \exp(\lambda x)$$ Then the function you want to differentiate n times w.r.t. x is $\frac{\partial^2 f}{\partial\lambda^2}$ at $\lambda = 1$. So, we want to evaluate: $$\frac{\partial^n}{\partial x^n}\frac{\partial^2 f}{\partial\lambda^2}$$ We can then interchange the order of differentiation to write this as: $$\... 11 I think an analogy with computer science may provide some insight. There are extremely simple programs that produce solutions of extraordinary complexity. The famous Rule 30 in cellular automata is the prime example: With a handful of bytes, one can write a deterministic program whose output is "as complex as possible," that is, it passes all ... 11 No, it is not true. Take f_n(x)=\sqrt{\frac1n+\left(x-\frac12\right)^2}. Then each f_n is a C^\infty function. But (f_n)_{n\in\Bbb N} converges uniformly to f\colon[0,1]\longrightarrow\Bbb R with f(x)=\left|x-\frac12\right|, which is not differentiable. 10 f(x) = -ce^x, c > 0 This isn't a particularly exciting answer, but it is the correct one. All functions that are their own derivatives are of the form f(x) = ce^x, c \in \mathbb{R}, as explained in this question: Prove that C\exp(x) is the only set of functions for which f(x) = f'(x) 10 So there are two inequalities to be proved. You can use that \sqrt{1+x^6} \leq \sqrt{2} for all x \in [-1,1] for the upper bound, as it follows \int_{[-1,1]} \sqrt{1+x^6} dx\leq \int_{[-1,1]} \sqrt{2} dx\leq 2 \sqrt{2}. The lower bound follows very similarly. 10 Just take any continuous function with its value changed at a single point. For example let f:[0, 1] \to \mathbb R be any continuous function and define \tilde{f}: [0, 1] \to \mathbb R by$$ \tilde{f}(x) = \begin{cases} f(x) + 1 & \text{if $x = 1/2$}, \\ f(x) & \text{if $x \neq 1/2$}. \end{cases}$$Then since f and \tilde{f} differ by only a ... 10 We don't do it by choice. We observe nature and notice that things are governed by differential equations. Suppose you let the water out of your bath tub. Initially the water leaves very quickly as the pressure is high. But as the water level drops, the pressure also drops and the water leaves slower. The rate of water leaving is related to the state of how ... 10 Hint : Show inductively that$$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]
9
The absolute value function is continuous so has an antiderivative. The antiderivative is differentiable at 0, but its derivative (the absolute value function) is not.
Only top voted, non community-wiki answers of a minimum length are eligible
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# express the dirichlet series for the sequence d(n)^2 in terms of riemann zeta.
Prove that $$\sum_{n=1}^\infty d(n)^2n^{-s}=\zeta(s)^4/\zeta(2s)$$ for $\sigma>1$
what i did:
I already proved this formally, that is, without considering convergence. I use euler products, that is, theorem 1.9 in Montgomerys multiplicative number theory:
If f is multiplicative, and $$\sum \vert f(n)\vert n^{-\sigma}<\infty$$ Then $$\sum_{n=1}^\infty f(n)n^{-s}=\prod_{p\in\mathbf{P}}(\sum_{n=0}^{\infty}f(p^n)p^{-ns}).$$ I first prove that $d$ is a multiplicative function. Then i apply Euler-products and after some technicalities, the result pops up.
However, my problem is the assumption for the Euler product. My naive bound for the divisor function is $d(n)<2\sqrt{n}$, with the rough argument that $d>\sqrt{n}$ is a divisor if and only if $n/d$ is a divisor $d<\sqrt{n}$. But this is not good enough , since this bound only lets me apply the Euler product form for $\sigma>2$.
I found some rather complicated bounds for the divisor functions on the internet, but since this is an early exercise in the montgomery Multiplicative number theory book (1.3.1 exercise 5) i doubt thats what i should use.
The post beneath consider the same problem, but it solved what i already solved, and ignore the convergence part:
Dirichlet series generating function
If $$\alpha(s)=\sum_{n\in\mathbb{N}}f(n) n^{-s}\quad \beta(s)=\sum_{n\in\mathbb{N}}g(n) n^{-s}$$ converges absolutely, then their product converges to $$\gamma(s)=\sum_{n\in\mathbb{N}}h(n) n^{-s}$$ Where $h(n)$ is the Dirichlet product $f*g\; (n)$
Indeed, $$\alpha(s)=\sum_{n\in\mathbb{N}}d(n)n^{-s}$$
converges absolutely for $\sigma>1$, so $$\alpha(s)^2=\sum_{n\in\mathbb{N}}d*d(n)n^{-s}$$ converges as well, due to the theorem above.
But since $d(D)d(n/D)\geq d(n)$ we get $$d*d(n)=\sum_{D\vert n}d(D)d(n/ D)\geq \sum_{D\vert n}d(n)=d(n)^2$$ and hence the result follows by direct comparison.
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# multiple region_plots in one plot
Hi, I want to plot two region_plots into one plot. The idea is if you use different colors for the inequalities you can see how the regions change. A minimal example:
var('x,y')
plot1=region_plot(x<y,(x,0,1),(y,0,1),incol='red')
plot2=region_plot(2*x<y,(x,0,1),(y,0,1),incol='blue')
show(plot2+plot1)
I do not know how to manage this, since zorder or opacity are not working for region_plot. Does anybody know how to make this work (by the way I use sagenb.org)? Thanks in advance
edit retag close merge delete
Just for reference, plot1+plot2 gives a different (also not so useful) plot.
( 2012-03-15 03:39:42 -0500 )edit
@kcrisman : ok, but could you please explain what you mean with <not so="" useful="" plot="">?
( 2012-03-15 03:43:21 -0500 )edit
1
By not-so-useful I just mean that it seems to have a similar problem to your original one, but with the opposite region "on top". It was just a comment, not an answer of any kind!
( 2012-03-15 03:50:33 -0500 )edit
Sort by » oldest newest most voted
The problem is that a region_plot is really just a contour plot with exactly two colored regions. The "inside" color is determined by incolor, and the "outside" is determined by outcolor. Now here's the key: outcolor is white by default, so you might think that the outside of the region is transparent. But it's not. That explains why you only see one of the two plots -- the top one is completely opaque, thus covering the bottom one. This also explains why you get different results when you sum the two plots in different orders -- this changes which one is on top.
Now here's a fix: just use contour_plot directly. To do this, define a function which will separate the regions you're interested in. For example:
def sep(x,y):
if 2*x < y:
return 1
if x < y and y <= 2*x:
return 0
if y <= x:
return -1
Now make the contour plot, choosing contour lines between your separate outputs, and listing the colors you want:
contour_plot(sep, (x,0,2), (y,0,2), plot_points=400, contours=[-.5,.5], cmap=['white','red','blue'])
Also note that contour_plot will probably work pretty well without you explicitly specifying the contours or the colors, if you don't want to worry about that step.
more
Ah, of course! I don't know why I didn't see that, having worked on this code in the past... Great work.
( 2012-03-16 15:48:26 -0500 )edit
So do you think this is worth a trac ticket? I feel like we would really want the original thing to work. But I'm not sure exactly how to do this without removing the 'white' business, which would make the graphics look weird, if I recall correctly.
( 2012-03-16 15:49:25 -0500 )edit
this could be a feature request: allow region_plot to color multiple regions. But I don't think I'm motivated enough to open a ticket for it.
( 2012-03-17 09:28:27 -0500 )edit
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Poetry is just the evidence of life. If your life is burning well, poetry is just the ashburn somethingmore quotes
# art: fun
In Silico Flurries: Computing a world of snow. Scientific American. 23 December 2017
# Nature Methods: Points of View
Points of View column in Nature Methods. (Points of View)
The first Points of View column was about color coding in the July 2010 issue of Nature Methods. In its 5 year history, the column has established a significant legacy— it is one of the most frequently accessed parts of Nature Methods. The community sees the value in clear and effective visual communication and acknowledges the need for a forum in which best practices in the field are presented practically and accessibly.
## 2010–2012
Bang Wong, in collaboration with visiting authors (Noam Shoresh, Nils Gehlenborg, Cydney Nielsen and Rikke Schmidt Kjærgaard), has penned 29 columns in the period of August 2010 to December 2012, covering broad topics such as salience, Gestalt principles, color, typography, negative space, layout, and data integration.
## 2012–2014
The announcement of the return of the column, together with its history and a description of me, the new author, are available at the Nature Methods methagora blog. Humor is kept by repeated reference to my now-dead-but-once-famous pet rat.
When it was A.C. Greyling's turn to speak at a debate in which Christopher Hitchens and Richard Dawkins already made their points, Greyling said
When one gets up to speak this late in a debate, one is a bit tempated to quote that Hungarian M.P. who after a long, long, long discussion in the parliament in Budapest stood up and said, "Everything has been said but not everybody said it yet."
Indeed, this is quite how I feel after being offered to be the new author of Nature Methods Point of View column. Both Bang and Hitchens provide significant inspiration for me, so Greyling's words are particularly fitting.
To improve on the column is impossible. My challenge is to identify useful topics that have not yet been covered. I will be working closely with Nature Methods and Bang to ensure that the columns strike the right balance of topic, tone and timbre.
In 2013 the Points of View column spawned the Points of Significance column, which deals with statistics in biological science.
For the month of August 2013, the entire set of 35 columns is available for free.
## 2015 and beyond
The column continues to run, though no longer monthly.
A PDF eBook of the 38 Points of View articles published between August 2010 and February 2015 is now available at the Nature Shop for \$7.99 under the title Visual strategies for biological data: the collected Points of View.
VIEW ALL
# Statistics vs Machine Learning
Tue 03-04-2018
We conclude our series on Machine Learning with a comparison of two approaches: classical statistical inference and machine learning. The boundary between them is subject to debate, but important generalizations can be made.
Inference creates a mathematical model of the datageneration process to formalize understanding or test a hypothesis about how the system behaves. Prediction aims at forecasting unobserved outcomes or future behavior. Typically we want to do both and know how biological processes work and what will happen next. Inference and ML are complementary in pointing us to biologically meaningful conclusions.
Nature Methods Points of Significance column: Statistics vs machine learning. (read)
Statistics asks us to choose a model that incorporates our knowledge of the system, and ML requires us to choose a predictive algorithm by relying on its empirical capabilities. Justification for an inference model typically rests on whether we feel it adequately captures the essence of the system. The choice of pattern-learning algorithms often depends on measures of past performance in similar scenarios.
Bzdok, D., Krzywinski, M. & Altman, N. (2018) Points of Significance: Statistics vs machine learning. Nature Methods 15:233–234.
Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: a primer. Nature Methods 14:1119–1120.
Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: supervised methods. Nature Methods 15:5–6.
# Happy 2018 $\pi$ Day—Boonies, burbs and boutiques of $\pi$
Wed 14-03-2018
Celebrate $\pi$ Day (March 14th) and go to brand new places. Together with Jake Lever, this year we shrink the world and play with road maps.
Streets are seamlessly streets from across the world. Finally, a halva shop on the same block!
A great 10 km run loop between Istanbul, Copenhagen, San Francisco and Dublin. Stop off for halva, smørrebrød, espresso and a Guinness on the way. (details)
Intriguing and personal patterns of urban development for each city appear in the Boonies, Burbs and Boutiques series.
In the Boonies, Burbs and Boutiques of $\pi$ we draw progressively denser patches using the digit sequence 159 to inform density. (details)
No color—just lines. Lines from Marrakesh, Prague, Istanbul, Nice and other destinations for the mind and the heart.
Roads from cities rearranged according to the digits of $\pi$. (details)
The art is featured in the Pi City on the Scientific American SA Visual blog.
Check out art from previous years: 2013 $\pi$ Day and 2014 $\pi$ Day, 2015 $\pi$ Day, 2016 $\pi$ Day and 2017 $\pi$ Day.
# Machine learning: supervised methods (SVM & kNN)
Thu 18-01-2018
Supervised learning algorithms extract general principles from observed examples guided by a specific prediction objective.
We examine two very common supervised machine learning methods: linear support vector machines (SVM) and k-nearest neighbors (kNN).
SVM is often less computationally demanding than kNN and is easier to interpret, but it can identify only a limited set of patterns. On the other hand, kNN can find very complex patterns, but its output is more challenging to interpret.
Nature Methods Points of Significance column: Machine learning: supervised methods (SVM & kNN). (read)
We illustrate SVM using a data set in which points fall into two categories, which are separated in SVM by a straight line "margin". SVM can be tuned using a parameter that influences the width and location of the margin, permitting points to fall within the margin or on the wrong side of the margin. We then show how kNN relaxes explicit boundary definitions, such as the straight line in SVM, and how kNN too can be tuned to create more robust classification.
Bzdok, D., Krzywinski, M. & Altman, N. (2018) Points of Significance: Machine learning: a primer. Nature Methods 15:5–6.
Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: a primer. Nature Methods 14:1119–1120.
# Human Versus Machine
Tue 16-01-2018
Balancing subjective design with objective optimization.
In a Nature graphics blog article, I present my process behind designing the stark black-and-white Nature 10 cover.
Nature 10, 18 December 2017
# Machine learning: a primer
Thu 18-01-2018
Machine learning extracts patterns from data without explicit instructions.
In this primer, we focus on essential ML principles— a modeling strategy to let the data speak for themselves, to the extent possible.
The benefits of ML arise from its use of a large number of tuning parameters or weights, which control the algorithm’s complexity and are estimated from the data using numerical optimization. Often ML algorithms are motivated by heuristics such as models of interacting neurons or natural evolution—even if the underlying mechanism of the biological system being studied is substantially different. The utility of ML algorithms is typically assessed empirically by how well extracted patterns generalize to new observations.
Nature Methods Points of Significance column: Machine learning: a primer. (read)
We present a data scenario in which we fit to a model with 5 predictors using polynomials and show what to expect from ML when noise and sample size vary. We also demonstrate the consequences of excluding an important predictor or including a spurious one.
Bzdok, D., Krzywinski, M. & Altman, N. (2017) Points of Significance: Machine learning: a primer. Nature Methods 14:1119–1120.
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## Annals of Functional Analysis
### On a notion of closeness of groups
#### Abstract
Enlightened by the notion of perturbation of $C^{*}$-algebras, we introduce, and study briefly in this article, a notion of closeness of groups. We show that if two groups are “close enough” to each other, and one of them has the property that the orders of its elements have a uniform finite upper bound, then these two groups are isomorphic (but in general they are not). We also study groups that are close to abelian groups, as well as an equivalence relation induced by closeness.
#### Article information
Source
Ann. Funct. Anal., Volume 7, Number 1 (2016), 24-32.
Dates
Accepted: 2 January 2015
First available in Project Euclid: 15 October 2015
https://projecteuclid.org/euclid.afa/1444913696
Digital Object Identifier
doi:10.1215/20088752-3163391
Mathematical Reviews number (MathSciNet)
MR3449337
Zentralblatt MATH identifier
1325.43004
#### Citation
Leung, Chi-Wai; Ng, Chi-Keung; Wong, Ngai-Ching. On a notion of closeness of groups. Ann. Funct. Anal. 7 (2016), no. 1, 24--32. doi:10.1215/20088752-3163391. https://projecteuclid.org/euclid.afa/1444913696
#### References
• [1] E. Christensen, A. M. Sinclair, R. R. Smith, S. A. White, and W. Winter, Perturbation of nuclear $C^{*}$-algebras, Acta Math. 208 (2012), no. 1, 93–150.
• [2] R. V. Kadison and D. Kastler, Perturbations of von Neumann algebras, I: Stability of type, Amer. J. Math. 94 (1972), 38–54.
• [3] W. J. Shi, A characteristic property of $\mathrm{PSL}_{2}(7)$, J. Austral. Math. Soc. Ser. A 36 (1984), no. 3, 354–356.
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# How to construct the midpoint in spherical geometry?
I am looking for the the method of constructing the midpoint of two points in spherical geometry. The only tools allowed for the construction are a pair of spherical compasses and a spherical ruler.
In Eclidean geometry constructing the midpoint is relatively easy. We are looking for the midpoint of points A and B. We construct two circles on A and B with the radiuses of AB. Then we construct two straight lines. One is through the two intersections of the two circles and one is through A and B. The intersection of these two lines will give the midpoint of A and B.
It is clear that the Euclidean method of construction does not work in Spherical geometry. The circles do not intersect when the distance of our two points exceeds 120°. There is also no solution when their distance is exactly 90°.
How would you construct the midpoint of two points in spherical geomerty?
Thank you
-
Draw a great circle $L$ between $A$ and $B$. Set your compass to a small distance, say $r$. Mark off distance $r, 2r, 3r, \ldots$ along $L$, moving from $A$ to $B$. Now do the same from $B$ to $A$. Count to find the middle interval and subdivide that.
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# Generators of a finitely generated free module over a commutative ring
Let $L$ be a finitely generated free module over a commutative ring $A$. Let $e_1, \dots, e_n$ be a basis of $L$. Let $x_1,\dots,x_m$ be generators of $L$. Then $m \ge n$? If $m = n$, then $x_1,\dots,x_m$ is a basis of $L$?
-
If $m ## 3 Answers If$x_1,\ldots,x_m$generate$L$, then you get a surjective$A$-module map$A^m\rightarrow L$. Tensoring with$k(\mathfrak{m})=A/\mathfrak{m}$,$\mathfrak{m}$a maximal ideal, gives you a surjection from an$m$-dimensional$k(\mathfrak{m})$-vector space to an$n$-dimensional$k(\mathfrak{m})$-vector space, so$m\geq n$. If$n=m$, then you get a surjective endomorphism$L\rightarrow L$, and any surjective endomorphism of a finite$A$-module is injective. So in this case the elements form a basis. - – Makoto Kato Nov 20 '12 at 16:13 +1 This is a nice answer. However, the first assertion can be proved without using axiom of choice as YACP's comment shows. – Makoto Kato Nov 20 '12 at 16:21 @MakotoKato, I'm not so sure if the uniqueness of rank for commutative rings can be proved without choice. The proof I know uses it. So it may not be a drawback here. – Gregor Bruns Nov 20 '12 at 16:25 @GregorBruns Please see YACP's comment or my answer. – Makoto Kato Nov 20 '12 at 16:30 I would like to prove the first assertion without using axiom of choice. Suppose$m < n$. Then$\bigwedge^n L = 0$. This is a contradiction because$\bigwedge^n L$is a free module of rank$1$. - This is pretty nice. – QiL'8 Nov 20 '12 at 21:17 @MakotoKato +1 Nice answer, great idea to consider exterior powers!! – BenjaLim Dec 8 '12 at 5:52 Thanks. It can also be proved in essentially the same but a bit more elementary way using the module of alternating forms$Alt^n(L,A)$. – Makoto Kato Dec 8 '12 at 14:47 Your last question can be answered using a nice fact that I learnt from Atiyah - Macdonald. Suppose we have$x_1,\ldots,x_n$that generate$L \cong A^n$. We now recall the following facts: 1. Localisation commutes with finite direct sums 2. If$M,N$are$A$- modules then$\phi : M \to N$is injective iff for all maximal ideals$\mathfrak{m} \in A$the induced map$\phi_\mathfrak{m} : M_{\mathfrak{m}} \to N_{\mathfrak{m}}$on localisation is injective. Using these it is enough to assume that$A$is a local ring with maximal ideal$\mathfrak{m}$. Now define a map$\phi : A^n \to A^n$by$\phi(x_i) = e_i$where$e_i$are the canonical basis vectors of$A^n$. Then$\phi$is surjective and we have a ses $$0 \longrightarrow \ker \phi \longrightarrow A^n \stackrel{\phi}{\longrightarrow} A^n \longrightarrow 0$$ which upon tensoring with$A/\mathfrak{m} = k$gives that $$0 \longrightarrow \ker \phi \otimes_A k \longrightarrow A^n \otimes_A k\stackrel{\phi \otimes 1}{\longrightarrow} A^n\otimes_A k \longrightarrow 0.$$ Rank - nullity implies that$\ker \phi \otimes_A k =0$. But now$\ker \phi \otimes_A k \cong \ker\phi / \mathfrak{m} \ker\phi$which implies that$\ker \phi = \mathfrak{m}\ker \phi$. We know that$\ker \phi$is finitely generated and$A$is local by assumption. The hypotheses of Nakayama's Lemma are now satisfied and applying it shows that$\ker \phi = 0$and hence$\phi$is an isomorphism. Hence$x_1,\ldots,x_n$are a basis for$A^n\$.
-
+1 This is a nice answer. – Makoto Kato Dec 8 '12 at 14:36
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# Browse Dissertations and Theses - Physics by Title
• (1992)
A new approach to the problem of X-ray edge singularities and peaks of many-body origin observed in the optical spectra is presented. We first establish the analogy between a system of one hole interacting with many electrons ...
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• (1986)
In this thesis I study two separate topics in many-body physics. In the first part I consider many-body effects in the thermodynamic properties of Fermi liquids and of the electron-phonon system. Using normal liquid ...
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• (1986)
In this thesis I study two separate topics in many-body physics. In the first part I consider many-body effects in the thermodynamic properties of Fermi liquids and of the electron-phonon system. Using normal liquid ('3)He ...
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• (1982)
The effects of band structure, of a solid surface, of temperature, and of disorder on the many-electron X-ray spectra of metals are evaluated in a change-of-mean-field approximation using a one-dimensional nearest-neighbor ...
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• (1976)
Using nuclear magnetic resonance, we have measured the magnetization density at the nuclear sites of several copper near neighbor shells surrounding the impurities manganese and chromium in the dilute alloys CuMn and ...
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• (2015-06-16)
Mastery learning employs repeated cycles of instructional support and formative assessment to help students achieve desired skills. Instructional objectives are broken into small pieces, and students master those pieces ...
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• (1987)
Rates in frozen glycerin/water solutions at temperatures between 1.4 K and 20 K are reported for a copper-containing protein, azurin, and a cobalt-containing biomolecular complex, vitamin B$\sb{\rm 12r}$, the paramagnetic ...
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• (1987)
Measured electron spin-lattice relaxation rates in frozen glycerin/water solutions at temperatures from 1.4 K to 20 K are reported for a copper-containing protein, azurin, and a cobalt -containing bi omol ecul ar complex, ...
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• (1996)
Using antiprotons from the Low Energy Antiproton Ring at CERN, an investigation of the reaction pp -+ AA at threshold has been completed. This work includes a thorough scan of the 2 MeV region where a hint of a cross ...
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An accelerator-based experiment was performed using the CEBAF accelerator of the Thomas Jefferson National Accelerator Facility to investigate a predicted sensitivity of the beam polarization to the vertical betatron ...
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• (2003-10)
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B hadron correlations provide input into the significance of higher-order bb¯ production mechanisms. We present a study of B hadron correlations in pp¯ collisions at s = 1.8 TeV. Events containing a bb¯ ...
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• (1984)
Charmonium production in 190 GeV/c pi--Be interactions has been observed. The states 1(3510) and X(3555) are seen through their radiative decay into J/psi , and are found to correspond to 0.29 +- 0.09 of total J/psi' ...
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• (1984)
Charmonium production in 190 GeV/c (pi)('-)-Be interactions has been observed. The states (chi)(3510) and (chi)(3555)are seen through their radiative decay into J/(psi), and are found to correspond to 0.29 (+OR-) 0.09 of ...
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• (1994)
Measurements of the production cross section times branching ratio for $W + \gamma$ and $Z + \gamma$ processes, where the W decays into a muon and neutrino and the Z decays into a muon pair, have been made from the analysis ...
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# Is this Agent Kallus? [updated]
So… just wondering. Is this Agent Kallus?
In announcing that Saw Gerrera will be played by Forest Whitaker, it begs the question as to if there’s a character from Rebels that will make it to the film as well. A great candidate would be Agent Kallus.
So… it begs the question if this Rebel leader in the Rogue One trailer could be Agent Kallus. The age looks about right. He lacks the mutton chops. But his voice isn’t off. So could it be?
Update: No. It’s General Draven.
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# Characterisation of Galois Group with the action of $\sigma \in S_n$ on the roots
Let $f \in K[X]$ be irreducible and separable with roots $x_1,...,x_n$ in a splitting field $L$ of $f$ over $K$. We identify $\text{Gal}(L|K)$ with $\text{Gal}(L|K)\cong G\subset S_n$.
How can I see the equivalence of the following two statements? (which means a characterisation of the galois group with the action of a $\sigma \in G$ on the roots $x_1,...,x_n$)
$(1)$ $\sigma \in G$.
$(2)$ If $P \in K[X_1,...,X_n]$ with $P(x_1,...,x_n)=0$, then for $P(X_{\sigma(1)},...,X_{\sigma(n)})$ it follows that $P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)=0$.
I prefer using three different notations, say $\sigma$ for the permutation of indices ,$\tau$ for the permutation of roots and $t$ for the field homomorphism extending $\tau$ (when it exists) : thus $\tau(x_i)=x_{\sigma(i)}$.
$(1) \Rightarrow (2)$ If $t$ exists, and $P(x_1,x_2,\ldots,x_n)=0$, we have
$$P(X_{\sigma(1)},...,X_{\sigma(n)})(x_1,...,x_n)= P(x_{\sigma(1)},...,x_{\sigma(n)})= P(\tau(x_1),\tau(x_2),\ldots,\tau(x_n))= t(P(x_1,x_2,\ldots,x_n))=t(0)=0.$$
$(2) \Rightarrow (1)$ The permutation $\tau$ is defined on $\lbrace x_1,x_2, \ldots ,x_n\rbrace$ and we sould like to extend it to the the whole of $L=K[x_1,\ldots,x_n]$. The obvious definition which comes to our mind is
$$t(A(x_1,x_2,\ldots,x_n))=A(x_{\sigma(1)},\ldots,x_{\sigma(n)}) \tag{1}$$
for any $A\in K[X_1,\ldots,X_n]$. The problem with (1) is that it might be an incorrect definition, with two different values set for the same argument. However, if $A(x_1,x_2,\ldots,x_n)=B(x_1,x_2,\ldots,x_n)$ for two polynomials $A,B$, then the polynomial $C=A-B$ satisfies $C(x_1,x_2,\ldots,x_n)=0$, so $C(x_{\sigma(1)},...,x_{\sigma(n)})=0$ by (2), and (1) will therefore yield the same value in both cases.
So $t$ is correctly defined, and it follows immediately from its definition that it is a homomorphism.
Alternatively, you can define $t$ as a "quotient map".
• @prime_dan To see why $t|_{K}=id$, take a constant $A$ in (1) – Ewan Delanoy Jun 1 '14 at 12:34
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Comment
Share
Q)
# The correct statements about the compound: $H_3C(HO)HC - CH = CH -CH(OH)CH_3 (X)$ are:
(i) The total number of stereoisomers possible for X is 6
(ii) The total number of diastreomers possible for X is 3 .
(iii) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4.
(iv) If the stereochemistry about the enantiomers possible for X is 2.
$(a)\;(i) and (iii) \\ (b)\;(i) and (iv) \\ (c)\;(ii) and (iv) \\ (d)\;(iii) and (iv)$
Comment
A)
The given molecule conatins 2 stereocentres and one double bond.
So, total number of different combination of stereoisomers is 6.
R cis R
S cis S
R cis S
and
R trans R
S trans S
R trans S
with cis/ trans , it will gave a pair of enantiomers, or two enantiomers.
Hence (i) and (iv)are correct.
Hence b is the correct answer.
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# If given the following solubilities, how do you calculate the K_(sp) for each compound?
## (a) $C u S C N$, 5.0 mg/L (b) $S n S$, 2.0 x ${10}^{-} 5$ g/L (c) $C o {\left(O H\right)}_{2}$, 3,2 x ${10}^{3}$ g/L (d) $A {g}_{2} C r {O}_{4}$, 3.4 x ${10}^{-} 2$ g/L?
Jun 2, 2018
I will do you ONE example.....$d .$
#### Explanation:
We examine the solubility equilibrium...
$A {g}_{2} C r {O}_{4} \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s 2 A {g}^{+} + C r {O}_{4}^{2 -}$
For which we write the solubility expression....
${K}_{\text{sp}} = {\left[A {g}^{+}\right]}^{2} \left[C r {O}_{4}^{2 -}\right]$
But if $S = \text{solubility of silver chromate...}$, then $\left[A {g}^{+}\right] = 2 S$, and $\left[C r {O}_{4}^{2 -}\right] = S$...thus ${K}_{\text{sp}} = {\left(2 S\right)}^{2} S = 4 {S}^{3}$...
Now $S = \frac{\frac{3.4 \times {10}^{-} 2 \cdot g}{331.73 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = 1.0025 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...
${K}_{\text{sp}} = 4 \times {\left(1.0025 \times {10}^{-} 4\right)}^{3} = 4.03 \times {10}^{-} 12$.
When you do the others perhaps you might post the solutions in this thread?
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• # question_answer What is the correct relationship between the pHs of isomolar solutions of sodium oxide $(p{{H}_{1}}),$ sodium sulphide $(p{{H}_{2}}),$ sodium selenide $(p{{H}_{3}})$ and sodium telluride $(p{{H}_{4}})$ [CBSE PMT 2005] A) $p{{H}_{1}}>p{{H}_{2}}=p{{H}_{3}}>p{{H}_{4}}$ B) $p{{H}_{1}}<p{{H}_{2}}<p{{H}_{3}}<p{{H}_{4}}$ C) $p{{H}_{1}}<p{{H}_{2}}<p{{H}_{3}}=p{{H}_{4}}$ D) $p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}$
Order of acidic strength is ${{H}_{2}}Te>{{H}_{2}}Se>{{H}_{2}}S>{{H}_{2}}O$ $N{{a}_{2}}O$ is a salt of $NaOH+{{H}_{2}}O$and ${{H}_{2}}O$is least acidic among given acids hence pH in this case will be maximum.
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[NTG-context] Strange behavior in math mode: < and minus sign give leftarrow
Otared Kavian otared at gmail.com
Wed Apr 10 09:21:28 CEST 2019
Hi Hans,
While testing old documents with lmtx, I noticed that with recent versions of mkiv, as well as LuaMetaTeX, there is a strange behavior in math mode: when the characters < and - follow each other one gets \leftarrow… This is embarrassing when one writes inequalities for negative numbers.
So one has to separate them in order to obtain with curly braces. Can this be fixed or is it intended ?
\starttext
$\alpha < -1$
$\alpha \leftarrow 1$
$\alpha < {-1}$
$\alpha {<} - 1$
\stoptext
Best regards: OK
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# Problems on Representation of Rational Numbers on Number Line
Every number in mathematics can be represented on the number line. When we talk about rational number or fractions, they can also be represented on the number line. While representing rational numbers on number line one should always keep some important points in the mind such as:
(i) Every positive integer lies on the right side of zero on the number line and is greater than zero.
(ii) Every negative number is less than zero and lies on the left side of zero on the number line.
(iii) Every proper fraction has value between zero and one and lies between zero and one.
(iv) Since representation of improper fraction on number line is difficult, so first it is converted into the mixed fraction and is then represented on the number line.
1. Represent $$\frac{4}{5}$$on the number line.
Solution:
Since the given rational fraction is positive and is a proper fraction, so it’ll lie on the right side of zero on the number line and between 0 and 1. To represent this, we’ll divide the number line between 0 and 1 into 5 equal parts and the fourth part of the five parts will be $$\frac{4}{5}$$on the number line. This can be represented as:
2. Represent $$\frac{7}{3}$$ on the number line.
Solution:
Take the number line with 0 at the point O. Take A$$_{1}$$, A$$_{2}$$, A$$_{3}$$, ….. on the right of O at equal distances of 6 mm (6 is the multiple of the denominator 3).
A$$_{1}$$, A$$_{2}$$, A$$_{3}$$, …. Represent the numbers 1, 2, 3, …. respectively.
1 is at a distance of 6 mm from O.
Therefore, $$\frac{7}{3}$$ will be at a distance of $$\frac{7}{3}$$ × 6 mm, i.e., 14 mm from O.
Now, take a point P on the right of A$$_{2}$$ such that A$$_{2}$$P = 2 mm.
Clearly, Op = 14 mm.
Thus, P will represent the number $$\frac{7}{3}$$ on the number line.
3. Place $$\frac{-3}{4}$$ on the number line.
Solution:
The given rational fraction id negative and is a proper fraction. So, it will lie on the left of zero on the number line and will be between zero and negative one. To represent this on the number line first we need to divide number line between 0 and -1 into 4 equal parts and third part of the four parts will be required rational number on the number line. This can be represented as:
4. Represent $$\frac{8}{3}$$ on the number line.
Solution:
Since the given rational fraction is a positive fraction and is an improper fraction. So, it’ll lie on the right side of zero on the number line. Now this is an improper fraction, so to represent this on the number line first we need to convert this into mixed fraction and then it will be represented on the number line. The mixed fraction conversion of the given fraction will be 2$$\frac{2}{3}$$. Now this fraction will lie between 2 and 3 on the number line and number line between 2 and 3 will be divided into 3 equal parts and second part of the 3 parts will be the required fraction on the number line. This could be as:
5. Represent -$$\frac{7}{4}$$ on the number line.
Solution:
The given rational fraction is a negative fraction and is an improper fraction. To represent it on the number line, first we need to convert the given fraction into mixed fraction. The mixed fraction of the given fraction is -1$$\frac{3}{4}$$. So, the given fraction will lie on the on the left side of zero on the number line. It’ll lie between -1 and -2 on the number line. The number line between -1 and -2 will be divided into 4 equal parts and the third part of the four parts will be the required fraction on the number line. This can be represented as:
6. Represent the number -$$\frac{2}{5}$$ on the number line.
Solution:
Take the number line with 0 at the point O. Take B$$_{1}$$, B$$_{2}$$, B$$_{3}$$, ….. on the left of O at equal distances of 5 mm.
B$$_{1}$$, B$$_{2}$$, B$$_{3}$$, …. represent the numbers -1, -2, -3, …. respectively.
-1 is at a distance of 5 mm from O.
Therefore, -$$\frac{2}{5}$$ will be at a distance of $$\frac{2}{5}$$ × 5 mm, i.e., 2 mm from O.
Now, take a point Q on the left of O such that OQ = 2 mm from O.
Thus, Q will represent the number -$$\frac{2}{5}$$ on the number line.
Rational Numbers
Rational Numbers
Decimal Representation of Rational Numbers
Rational Numbers in Terminating and Non-Terminating Decimals
Recurring Decimals as Rational Numbers
Laws of Algebra for Rational Numbers
Comparison between Two Rational Numbers
Rational Numbers Between Two Unequal Rational Numbers
Representation of Rational Numbers on Number Line
Problems on Rational numbers as Decimal Numbers
Problems Based On Recurring Decimals as Rational Numbers
Problems on Comparison Between Rational Numbers
Problems on Representation of Rational Numbers on Number Line
Worksheet on Comparison between Rational Numbers
Worksheet on Representation of Rational Numbers on the Number Line
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# Oblate Spheroid - Mass
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vCalc.Oblate Spheroid - Mass
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The Mass or Weight of an Oblate Spheroid calculator computes the volume of an oblate spheroid based on the semi-major(b) and semi- minor (c) axis with the assumption that the spheroid is generated via rotation around the minor axis (see diagram).
INSTRUCTIONS: Choose your length units for a and b (e.g. feet, meters, light-years), and enter the following:
• (b) - semi-major axis, the distance from the oblate spheroid's center along the longest axis of the spheroid
• (c) - semi-minor axis, the distance from the oblate spheroid's center along the shortest axis of the spheroid
• (mD) - the mean density of the substance comprising the oblate spheroid.
Oblate Spheroid Mass / Weight: The mass (M) is returned in kilograms. However, this can be automatically converted to other mass and weight units (e.g. tons, pounds) via the pull-down menu.
### NOTES
The oblate spheroid is an ellipsoid that can be formed by rotating an ellipse about its minor axis. The rotational axis thus formed will appear to be the oblate spheroid's polar axis. The oblate spheroid is fully described then by its semi-major and semi-minor axes.
One important shape in nature that is close to (though not exactly) an oblate spheroid is the Earth which has a semi-minor axis (c) which is the polar radius of 6,356 kilometers, and a semi-major axis (b) which is the equatorial radius of 6,378 kilometers. Consideration: what force would make the equatorial radius larger than the polar radius?
#### Related Calculators
The following table contains links to calculators that compute the volume of other shapes:
Other Volume Calculators Various Shapes Polygon Columns Cube Triangular Prism Triangular Box Paraboloid Quadrilateral Cone Polygon based Pyramid Pentagon Cone Frustum Pyramid Frustum Hexagon Cylinder Sphere Heptagon Slanted Cylinder Sphere Cap Octagon Ellipsoid Oblate Spheroid Nonagon Torus Capsule Decagon
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# How to parse this formula?
1. Jun 3, 2015
### Jarvis323
I'm studying a research paper that gives this formula for the running time of an algorithm,
expO((log N)^α(log log N)^(1−α)) = L(a)
I would like to plot this function alongside another, for a = 1/4 + O(1), a = 1/4 + O(n), and a= 1/3. The function's growth parameratized by those a's, should be ordered from small to big in the order I listed them.
Here is a link to the article, the formula is found in the introduction.
The preprint (free) version of the article is http://arxiv.org/abs/1306.4244. Let $x= \log N$ and $y=\log\log N$. I parse that the argument of the exponent is $O(x^\alpha y^{1-\alpha})$.
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### Strong 8-bit Sboxes with Efficient Masking in Hardware
Erik Boss, Vincent Grosso, Tim Güneysu, Gregor Leander, Amir Moradi, and Tobias Schneider
##### Abstract
Block ciphers are arguably the most important cryptographic primitive in practice. While their security against mathematical attacks is rather well understood, physical threats such as side-channel analysis (SCA) still pose a major challenge for their security. An effective countermeasure to thwart SCA is using a cipher representation that applies the threshold implementation (TI) concept. However, there are hardly any results available on how this concept can be adopted for block ciphers with large (i.e., 8-bit) Sboxes. In this work we provide a systematic analysis on and search for 8-bit Sbox constructions that can intrinsically feature the TI concept, while still providing high resistance against cryptanalysis. Our study includes investigations on Sboxes constructed from smaller ones using Feistel, SPN, or MISTY network structures. As a result, we present a set of new Sboxes that not only provide strong cryptographic criteria, but are also optimized for TI. We believe that our results will found an inspiring basis for further research on high-security block ciphers that intrinsically feature protection against physical attacks.
Available format(s)
Category
Implementation
Publication info
A major revision of an IACR publication in CHES 2016
Keywords
side-channel analysisthreshold implementation8-bit Sboxes
Contact author(s)
tobias schneider-a7a @ rub de
History
Short URL
https://ia.cr/2016/647
CC BY
BibTeX
@misc{cryptoeprint:2016/647,
author = {Erik Boss and Vincent Grosso and Tim Güneysu and Gregor Leander and Amir Moradi and Tobias Schneider},
title = {Strong 8-bit Sboxes with Efficient Masking in Hardware},
howpublished = {Cryptology ePrint Archive, Paper 2016/647},
year = {2016},
note = {\url{https://eprint.iacr.org/2016/647}},
url = {https://eprint.iacr.org/2016/647}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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# MA2213 Lecture 5 Linear Equations (Direct Solvers)
## Presentation on theme: "MA2213 Lecture 5 Linear Equations (Direct Solvers)"— Presentation transcript:
MA2213 Lecture 5 Linear Equations (Direct Solvers)
Systems of Linear Equations p. 243-248 Occur in a wide variety of disciplines Mathematics Statistics Physics ChemistryBiology Economics Sociology Psychology Archaeology Geology AstronomyAnthropology Engineering Management Business Medicine Finance
Matrix Form for a system of linear equations coefficient matrix (solution) column vector (right) column vector
Linear Equations in Mathematics Numerical Analysis Geometry Interpolation Least Squares Quadrature Algebra find intersection of lines or planes partial fractions Coefficient Matrix Vandermonde (for polyn. interp.) or Gramm Transpose of Vandermonde Lec 4 vufoil 13 (to compute weights)
Matrix Arithmetic p. 248-264 Matrix Inverse Matrix Multiplication Identity Matrix Theorem 6.2.6 p. 255 A square matrix has an inverse iff (if and only if) its determinant is not equal to zero.
Solution of (this means exists and is unique. multiplication is associative for nonsingular Proof Remark In MATLAB use: x = A \ b;
Column Rank of a Matrix Definition The column rank of a matrix dimension of the subspace of spanned by the column vectors of Remark maximal number of linearly independent column vectors Question is the of
Row Rank of a Matrix Definition The row rank of a matrix is the dimension of the subspace of spanned by the row vectors of Remark maximal number of linearly independent row vectors of Question
A Matrix Times a Vector has solution iff b is a linear combination of columns of A The equation
Existence of Solution in General The linear equation has a solution if and only if IS SINGULAR! EVEN IF Example this has a solution iff then it has an infinite number of solutions called Augmented matrix p. 265
Computing the Column and Row Ranks The ranks of a matrix can be computed using a sequence of elementary row operations p. 253-254. i. Interchange two rows ii. Multiply a row by a nonzero scalar iii. Add a nonzero multiple of one row to another row Question Show that each of the ERO i, ii, iii has an inverse ERO i, ii, iii.
Elementary Row Operations can be performed on the left by by multiplying on a matrix nonsingular matrices
Invariance of Row Rank Under ERO Theorem 1. If is an ERO matrix, then Proof Clearly, interchanging two rows and multiplying a row by a nonzero scalar does not change the row rank. Finish the proof by showing that adding a multiple of any row to another row does not change the row rank. and Remark Clearly the row rank of a matrix is invariant under sequence of ERO’s.
Matrix Multiplication
Invariance of Column Rank under ERO Theorem 2 If is nonsingular then Proof It suffices to show that for a set are linearly dependent iff the set of ofof column vectors are linearly dependent. Show why it suffices and then show it. Hint: prove and column vectorsof
Row Echelon Matrices Definition A matrix an row echelon matrix if i. the nonzero rows come first ii. the first nonzero element in each row =1 (called a pivot) has all zeros below it is called iii. each pivot lies to the right of the pivot in the row above
Row Echelon Matrices These three properties produce a staircase pattern in the matrix below Question Where are the pivots ?
Row Rank of an Row Echelon Matrix equals the number of nonzero rows. Question What is the rank of this matrix ? Prove this by showing that the rows must be linearly independent. Hint : use pivots.
Col. Rank of a Row Echelon Matrix equals the number of nonzero rows. Question Show this by showing that the col. vectors that contain pivots form a basis for the space spanned by col. vectors. Hint: do elem. col. operations on the matrix above.
Reduction to Row Echelon Form Theorem 3 For every matrix there exists a nonsingular matrix is an echelon matrix. such that Furthermore, the matrixis a product where each is an ERO matrix. Application of the sequence of ERO’s is called reduction to row echelon form. Proof Based on Gaussian elimination.
Row Rank = Column Rank Theorem 4 For every matrix Proof. Theorem 3 implies that there exists a product of ERO matrices such thatis a row echelon matrix. Theorems 1 implies that and theorems 2 implies that Sinceis a row echelon matrix, hence
Applications of Row Echelon Reduction The linear equation iff the last nonzero row of the reduced Example has a solution has its pivot NOT in the last column. Hence the condition above is satisfied iff
Applications of Row Echelon Reduction A basis of column vectors for a matrix can be obtained by first computing the reduction then choosing the column vectors that form a basis for the space spanned by the column that contain the pivots. Then the vectors are column vectors of vectors of
Generalities on Gaussian Elimination Gaussian elimination is the process of reducing a matrix to row echelon form through a sequence of ERO’s. It can also be used to solve a system of linear equations The final step of solving a system of equations after the augmented matrix has been reduced is called back substitution, this process is related to elementary column operations and will be addressed in the homework. It is ‘best’ taught through showing examples. We will show how to solve a system of linear equations using Gaussian elimination, it will become obvious how to use Gaussian elimination for reduction.
Gaussian Elimination (p. 264-269) Case 1. The equations for this matrix are Question How do we use the nonsingular assumption? therefore, if A is nonsingular then Question What type of matrix is this ?
Back Substitution Case 2. A nonsingular solution by back-substitution p. 265 Question How do we use the nonsingular assumption? Question What is this matrix called ? Question What are the associated equations ? Question Why is this method called back-substitution ?
Gaussian Elimination on Equations Case 3. Apply elementary row operations on equations to to obtain equations with an upper triangular matrix Question How can we solve these equations ?
Gaussian Elimination on Augmented Matrix
Gaussian Elimination Question What is the solution ?
Partial Pivoting p. 270-273 the integer that gives For the j-th column in Gaussian elimination compute then perform the row interchange Read p. 273-276 about how Gaussian elimination can be used to compute the inverse of a matrix.
LU Decomposition p. 283-285 To solve where Then for each b use forward substitution to solve L y = b then use backward substitution to solve U x = y. first compute the factorization for many values of with same
LU Decomposition Algorithm Algorithm Step 1 Step 2 for r = 2,…,n do Question How many operations does this require ?
Homework Due Tutorial 3 Question 1. Prove that the row rank of an row echelon matrix equals the number of nonzero rows. Question 2. Prove that the column rank of an row echelon matrix equals the number of nonzero rows by showing that the set of its column vectors having pivots is a maximal set of linearly independent column vectors. Question 3. Use Gaussian elimination to solve Question 4. Derive expressions for the entries of the L and U in the LU decomposition of a 3 x 3 matrix A. Question 5. Show how elementary column operations can be applied to a row echelon matrix M to obtain a row echelon matrix with exactly one 1 in each nonzero row. Use this to determine a basis for the space { x : Mx = 0 }.
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# Coulomb's law partII
1. Sep 11, 2007
### itzxmikee
Hi guys, these are the final 2 problems that Ive been struggling with for the past day. Please help
1. Four point charges are situated at the corners of a square with sides of length a, as in Figure P15.4.
Figure P15.4
Find the expression for the resultant force on the positive charge q.(Use k_e for ke, q for q, and a for a.)
2. Fe = ke(|q||q|/r^2)
3. So I found the x and y components of all the -q. But still get a big sloppy answer.
Question 2
An electron is released a short distance above the surface of the Earth. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the gravitational force on it. How far below the first electron is the second?
Im sure you have to use coulombs law again. But I just dont know where to start with this question.
Thanks guys
2. Sep 11, 2007
### Staff: Mentor
What did you get?
What forces act on the first electron? What's the net force on it?
3. Sep 11, 2007
### itzxmikee
The x components are
-ke(q^2/a^2) - ke(q^2/2a^2)cos45
and the y compents are
-ke(q^2/a^2) - ke(q^2/2a^2)sin45
after simplifications I get the magnitude to be:
sqrt(-2.7keq^2/a^2)
What forces act on the first electron? What's the net force on it?
The forces is weight? mass*gravity?
Net force? weight?
4. Sep 11, 2007
### lylos
It states that the electron below it exerts a force on it great enough to just cancel the force of gravity. So what do we know about the force of gravity and the force due to the electrons repelling each other?
5. Sep 11, 2007
### itzxmikee
The net force is zero
6. Sep 11, 2007
### lylos
Great, so when drawing a free body diagram of the electron, you can see that the forces act in two opposite directions, and they must be equal to each other. Using the formulas for the force due to the electron and the force due to gravity, you can then find how far below electron 1 the second electron must be.
7. Sep 11, 2007
### itzxmikee
Ke (q^2)/r^2 = G m^2/r^2
So are we assuming the electrons are the exact mass and charge?
Did I set up the equation right? How do I go about solving for r if I dont know q and m?
8. Sep 11, 2007
### lylos
Now, we're not dealing with the gravitational attraction between the two electrons, rather the electron being pulled to the earth and then being repelled from the other electron. One side of the equation is correct, the other is not. If you look in your book you should be able to find the charge on an electron and the mass of an electron.
9. Sep 11, 2007
### itzxmikee
Im assuming Ke (q^2)/r^2 is the correct side.
So would it be: Ke (q^2)/r^2 = Ke (q of electron)(q of earth)/r^2
10. Sep 11, 2007
### lylos
Alright, we have ke (q^2)/(r^2) on the left, and we need to set that equal to the mass of the electron * the acceleration due to gravity.
You sould then have:
Ke (q^2)/(r^2) = m (9.8)
You know Ke, q is the charge of the electron, which should be provided, m, is the mass of the electron, again which should be provided, the 9.8 is the acceleration due to gravity. Now you just solve for r.
11. Sep 11, 2007
### itzxmikee
duhh makes total sense now. Thank you Lylos!!!!!!
now with the 1st problem. Can anybody help me with that?
12. Sep 11, 2007
### lylos
What you need to do on the first problem is find the X and Y components of the force. Then add them together. For example, the force on q+ due to the q- directly above would be (kqq)/r^2 in the +y direction. Now the force due to the charge to the right would be (kqq)/r^2 in the +x direction. Now the hard part is trying to break down the force due to the charge in the upper right hand corner. It will have an independent x component and an independent y component. Once you have these values you can then find the resultant vector of force.
13. Sep 11, 2007
### Staff: Mentor
OK:
(1) Realize that $\sin (45) = \cos (45) = \sqrt{2}/2$
(2) Why negative?
I assume you mean: sqrt(2.7)keq^2/a^2
Recheck this; I get a different answer.
14. Sep 11, 2007
### itzxmikee
Okay just looked over the problem again
so I ended up with:
x components:
ke(q^2/a^2) + ke(q^2/2a^2)cos45
=(1+cos45(.5))(ke(q^2/a^2)
= 1.35(ke(q^2/a^2))
and the y components are:
ke(q^2/a^2) + ke(q^2/2a^2)sin45
=(1+sin45(.5))(ke(q^2/a^2)
= 1.35(ke(q^2/a^2))
So to find the magnitude:
((1.35(ke(q^2/a^2))^2 + (1.35(ke(q^2/a^2))^2) ^1/2
am I on the right track?
15. Sep 11, 2007
### Staff: Mentor
Yes, exactly. (Just be careful to square--and square root--things properly: you are missing a few parentheses.)
16. Sep 11, 2007
### itzxmikee
Thanks everyone. Until next time
17. Jan 25, 2008
### johnk1317
Hi I had a question considering this problem. The formula you generate is correct but I dont see how you got it. I will use x to represent the k*q^2/a^2
What I got:
(1+sin45)x for the y axis
(1+cos45)x for the x
What you got:
(1+sin45*.5)x
(1+cos45*.5)x
Where did you derive the .5 from? Thanks for the help.
John
18. Jan 26, 2008
### Staff: Mentor
Realize that the distance to the charge on the opposite corner is not a, but $\sqrt 2$a. That's where the .5 comes from.
19. Jan 26, 2008
### johnk1317
Thanks for the clarification
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# 2.3.1: Adding and Subtracting Fractions - Mathematics
Paul and Tony order a pizza which has been cut into eight equal slices. Tony eats three slices (shaded in light red (or a darker shade of gray in black-and-white printing) in Figure (PageIndex{1})), or 3/8 of the whole pizza.
It should be clear that together Paul and Tony eat five slices, or 5/8 of the whole pizza. This reflects the fact that
[ frac{2}{8} + frac{3}{8} = frac{5}{8}. onumber ]
This demonstrates how to add two fractions with a common (same) denominator. Keep the common denominator and add the numerators. That is,
[ egin{align*} frac{2}{8} + frac{3}{8} &= frac{2 + 3}{8} ~ && extcolor{red}{ ext{ Keep denominator; add numerators.}} &= frac{5}{8} ~ && extcolor{red}{ ext{ Simplify numerator.}} end{align*} ]
Let a/c and b/c be two fractions with a common (same) denominator. Their sum is defined as
[ frac{a}{c} + frac{b}{c} = frac{a + b}{c} onumber ]
That is, to add two fractions having common denominators, keep the common denominator and add their numerators.
A similar rule holds for subtraction.
Subtracting Fractions with Common Denominators
Let a/c and b/c be two fractions with a common (same) denominator. Their difference is defined as
[ frac{a}{c} - frac{b}{c} = frac{a-b}{c}. onumber ]
That is, to subtract two fractions having common denominators, keep the common denominator and subtract their numerators.
Example (PageIndex{1})
Find the sum of 4/9 and 3/9.
Solution
Keep the common denominator and add the numerators.
[ egin{aligned} frac{4}{9} + frac{3}{9} &= frac{4+3}{9} ~ & extcolor{red}{ ext{ Keep denominator; add numerators.}} &= frac{7}{9} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Exercise (PageIndex{1})
[ frac{1}{8} + frac{2}{8} onumber ]
3/8
Example (PageIndex{2})
Subtract 5/16 from 13/16.
Solution
Keep the common denominator and subtract the numerators.
[ egin{aligned} frac{13}{16} - frac{5}{16} &= frac{13-5}{16} ~ & extcolor{red}{ ext{ Keep denominator; subtract numerators.}} &=frac{8}{16} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Of course, as we learned in Section 4.1, we should always reduce our final answer to lowest terms. One way to accomplish that in this case is to divide numerator and denominator by 8, the greatest common divisor of 8 and 16.
[ egin{aligned} = frac{8 div 8}{16 div 8} ~ & extcolor{red}{ ext{ Divide numerator and denominator by 8.}} = frac{1}{2} ~ & extcolor{red}{ ext{ Simplify numerator and denominator.}} end{aligned} onumber ]
Exercise (PageIndex{2})
Subtract:
[ frac{11}{12} - frac{7}{12} onumber ]
1/3
Example (PageIndex{3})
Simplify:
[ frac{3}{x} - left( - frac{7}{x} ight) . onumber ]
Solution
Both fractions share a common denominator.
[ egin{aligned} frac{3}{x} - left( - frac{7}{x} ight) &= frac{3}{x} + frac{7}{x} ~ & extcolor{red}{ ext{ Add the opposite.}} &= frac{3+7}{x} ~ & extcolor{red}{ ext{ Keep denominator, add numerators.}} &= frac{10}{x} ~ & extcolor{red}{ ext{ Simplify.}} end{aligned} onumber ]
## Adding Fractions with Different Denominators
Consider the sum
[ frac{4}{9} + frac{1}{6}. onumber ]
We cannot add these fractions because they do not have a common denominator. So, what to do?
Goals
In order to add two fractions with different denominators, we need to:
1. Find a common denominator for the given fractions.
2. Make fractions with the common denominator that are equivalent to the original fractions.
If we accomplish the two items in the “Goal,” we will be able to find the sum of the given fractions.
So, how to start? We need to find a common denominator, but not just any common denominator. Let’s agree that we want to keep the numbers as small as possible and find a least common denominator.
Definition: Least Common Denominator
The least common denominator (LCD) for a set of fractions is the smallest number divisible by each of the denominators of the given fractions.
Consider again the sum we wish to find:
[ frac{4}{9} + frac{1}{6} . onumber ]
The denominators are 9 and 6. We wish to find a least common denominator, the smallest number that is divisible by both 9 and 6. A number of candidates come to mind: 36, 54, and 72 are all divisible by 9 and 6, to name a few. But the smallest number that is divisible by both 9 and 6 is 18. This is the least common denominator for 9 and 6.
We now proceed to the second item in “Goal.” We need to make fractions having 18 as a denominator that are equivalent to 4/9 and 1/6. In the case of 4/9, if we multiply both numerator and denominator by 2, we get
[ egin{aligned} frac{4}{9} &= frac{4 cdot 2}{9 cdot 2} ~ & extcolor{red}{ ext{ Multiply numerator and denominator by 2.}} &= frac{8}{18}. ~ & extcolor{red}{ ext{ Simplify numerator and denominator.}} end{aligned} onumber ]
In the case of 1/6, if we multiply both numerator and denominator by 3, we get
[ egin{aligned} frac{1}{6} &= frac{1 cdot 3}{6 cdot 3} ~ & extcolor{red}{ ext{ Multiply numerator and denominator by 3.}} &= frac{3}{18}. ~ & extcolor{red}{ ext{ Simplify numerator and denominator.}} end{aligned} onumber ]
Typically, we’ll arrange our work as follows.
[ egin{aligned} frac{4} + frac{1}{6} &= frac{4 cdot extcolor{red}{2}}{9 cdot extcolor{red}{2}} + frac{1 cdot extcolor{red}{3}}{6 cdot extcolor{red}{3}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = 18.}} &= frac{8}{18} + frac{3}{18} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} &= frac{8+3}{18} ~ & extcolor{red}{ ext{ Keep common denominator; add numerators.}} &= frac{11}{18} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Let’s summarize the procedure.
Adding or Subtracting Fractions with Different Denominators
1. Find the LCD, the smallest number divisible by all the denominators of the given fractions.
2. Create fractions using the LCD as the denominator that are equivalent to the original fractions.
3. Add or subtract the resulting equivalent fractions. Simplify, including reducing the final answer to lowest terms.
Example (PageIndex{4})
Simplify: ( displaystyle frac{3}{5} - frac{2}{3}).
Solution
The smallest number divisible by both 5 and 3 is 15.
[ egin{aligned} frac{3}{5} - frac{2}{3} &= frac{3 cdot extcolor{red}{3}}{5 cdot extcolor{red}{3}} - frac{2 cdot extcolor{red}{5}}{3 cdot extcolor{red}{5}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = 15.}} &= frac{9}{15} - frac{10}{15} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} &= frac{9-10}{15} ~ & extcolor{red}{ ext{ Keep LCD; subtract numerators.}} &= frac{-1}{15} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Although this answer is perfectly acceptable, negative divided by positive gives us a negative answer, so we could also write
[ = - frac{1}{15}. onumber ]
Exercise (PageIndex{4})
Subtract:
[ frac{3}{4} - frac{7}{5} onumber ]
-13/20
Example (PageIndex{5})
Simplify: (-frac{1}{4} - frac{5}{6}).
Solution
The smallest number divisible by both 4 and 6 is 12.
[ egin{aligned} -frac{1}{4} - frac{5}{6} &= - frac{1 cdot extcolor{red}{3}}{4 cdot extcolor{red}{3}} - frac{5 cdot extcolor{red}{2}}{6 cdot extcolor{red}{2}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD =12.}} &= - frac{3}{12} - frac{10}{12} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} &= frac{-3-10}{12} ~ & extcolor{red}{ ext{ Keep LCD; subtract numerators.}} &= frac{-13}{12} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Exercise (PageIndex{5})
Subtract: (-frac{3}{8} - frac{1}{12})
-11/24
Example (PageIndex{6})
Simplify: (frac{5}{x} + frac{3}{4}).
Solution
The smallest number divisible by both 4 and x is 4x.
[ egin{aligned} frac{5}{x} + frac{3}{4} = frac{5 cdot extcolor{red}{4}}{x cdot extcolor{red}{4}} + frac{3 cdot extcolor{red}{x}}{4 cdot extcolor{red}{x}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = }4x.} = = frac{20}{4x} + frac{3x}{4x} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} = frac{20 + 3x}{4x} ~ & extcolor{red}{ ext{ Keep LCD; add numerators.}} end{aligned} onumber ]
Exercise (PageIndex{6})
[ frac{5}{z} + frac{2}{3} onumber ]
[ frac{15+2z}{3z} onumber ]
Example (PageIndex{7})
Simplify: (frac{2}{3}- frac{x}{5}).
Solution
The smallest number divisible by both 3 and 5 is 15.
[ egin{aligned} frac{2}{3} - frac{x}{5} = frac{2 cdot extcolor{red}{5}}{3 cdot extcolor{red}{5}} - frac{x cdot extcolor{red}{3}}{5 cdot extcolor{red}{3}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = 15.}} = frac{10}{15} - frac{3x}{15} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} = frac{10 - 3x}{15} ~ & extcolor{red}{ ext{ Keep LCD; subtract numerators.}} end{aligned} onumber ]
## Least Common Multiple
First we define the multiple of a number.
Definition: Multiples
The multiples of a number d are 1d, 2d, 3d, 4d, etc. That is, the multiples of d are the numbers nd, where n is a natural number.
For example, the multiples of 8 are 1 · 8, 2 · 8, 3 · 8, 4 · 8, etc., or equivalently, 8, 16, 24, 32, etc.
Definition: Least Common Multiple
The least common multiple (LCM) of a set of numbers is the smallest number that is a multiple of each number of the given set. The procedure for finding an LCM follows:
1. List all of the multiples of each number in the given set of numbers.
2. List the multiples that are in common.
3. Pick the least of the multiples that are in common.
Example (PageIndex{7})
Find the least common multiple (LCM) of 12 and 16.
Solution
List the multiples of 12 and 16.
Multiples of 12 : 12, 24, 36, 48, 60, 72, 84, 96,...
Multiples of 16 : 16, 32, 48, 64, 80, 96, 112,...
Pick the common multiples.
Common Multiples : 48, 96,...
The LCM is the least of the common multiples.
LCM(12,16) = 48
Exercise (PageIndex{7})
Find the least common denominator of 6 and 9.
18
Important Observation
The least common denominator is the least common multiple of the denominators.
For example, suppose your problem is 5/12 + 5/16. The LCD is the smallest number divisible by both 12 and 16. That number is 48, which is also the LCM of 12 and 16. Therefore, the procedure for finding the LCM can also be used to find the LCD.
## Least Common Multiple Using Prime Factorization
You can also find the LCM using prime factorization.
LCM By Prime Factorization
To find an LCM for a set of numbers, follow this procedure:
1. Write down the prime factorization for each number in compact form using exponents.
2. The LCM is found by writing down every factor that appears in step 1 to the highest power of that factor that appears.
Example (PageIndex{8})
Use prime factorization to find the least common multiple find the least common denominator of 18 and 24. (LCM) of 12 and 16.
Solution
Prime factor 12 and 16.
[ egin{aligned} 12 = 2 cdot 2 cdot 3 16 = 2 cdot 2 cdot 2 cdot 2 end{aligned} onumber ]
Write the prime factorizations in compact form using exponents.
[ egin{aligned} 12 = 2^2 cdot 3^1 16 = 2^4 end{aligned} onumber ]
To find the LCM, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2 and 3. The highest power of 2 that appears is 24. The highest power of 3 that appears is 31.
[ egin{aligned} ext{LCM} = 2^4 cdot 3^1 ~ & extcolor{red}{ ext{ Keep highest power of each factor.}} end{aligned} onumber ]
Now we expand this last expression to get our LCM.
[ egin{aligned} = 16 cdot 3 ~ & extcolor{red}{ ext{ Expand: } 2^4 = 16 ext{ and } 3^1 = 3.} = 48. ~ & extcolor{red}{ ext{ Multiply.}} end{aligned} onumber ]
Note that this answer is identical to the LCM found in Example 8 that was found by listing multiples and choosing the smallest multiple in common.
Exercise (PageIndex{8})
Use prime factorization to find the least common denominator of 18 and 24.
72
Example (PageIndex{10})
Simplify: (frac{5}{28} + frac{11}{42}).
Solution
Prime factor the denominators in compact form using exponents.
28 = 2 · 2 · 7=22 · 7
42 = 2 · 3 · 7=21 · 31 · 71
To find the LCD, write down each factor that appears to the highest power of that factor that appears. The factors that appear are 2, 3, and 7. The highest power of 2 that appears is 22. The highest power of 3 that appears is 31. The highest power of 7 that appears is 71.
[ egin{aligned} ext{LCM} = 2^2 cdot 3^1 cdot 7^1 ~ & extcolor{red}{ ext{ Keep highest power of each factor.}} = 4 cdot 3 cdot 7 ~ & extcolor{red}{ ext{ Expand: } 2^2 = 4, ~ 3^1 = 3, ~ 7^1 = 7.} = 84 ~ & extcolor{red}{ ext{ Multiply.}} end{aligned} onumber ]
Create equivalent fractions with the new LCD, then add.
[ egin{aligned} frac{5}{28} + frac{11}{42} = frac{5 cdot extcolor{red}{3}}{28 cdot extcolor{red}{3}} + frac{11 cdot extcolor{red}{2}}{42 cdot extcolor{red}{2}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = 84.}} = frac{15}{84} + frac{22}{84} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} = frac{37}{84} ~ & extcolor{red}{ ext{ Keep LCD; add numerators.}} end{aligned} onumber ]
Exercise (PageIndex{10})
Simplify: ( frac{5}{24} + frac{5}{36})
25/72
Example (PageIndex{11})
Simplify: (- frac{11}{24} - frac{1}{18}).
Solution
Prime factor the denominators in compact form using exponents.
24 = 2 · 2 · 2 · 3=23 · 31
18 = 2 · 3 · 3=21 · 32
To find the LCD, write down each factor that appears to the highest power of that factor that appears. The highest power of 2 that appears is 23. The highest power of 3 that appears is 32.
[ egin{aligned} ext{LCM} = 2^3 cdot 3^2 ~ & extcolor{red}{ ext{ Keep highest power of each factor.}} = 8 cdot 9 ~ & extcolor{red}{ ext{ Expand: } 2^3 = 8 ext{ and } 3^2 = 9.} = 72. ~ & extcolor{red}{ ext{ Multiply.}} end{aligned} onumber ]
Create equivalent fractions with the new LCD, then subtract.
[ egin{aligned} - frac{11}{24} - frac{1}{18} = - frac{11 cdot extcolor{red}{3}}{24 cdot extcolor{red}{3}} - frac{1 cdot extcolor{red}{4}}{18 cdot extcolor{red}{4}} ~ & extcolor{red}{ ext{ Equivalent fractions with LCD = 72.}} = - frac{33}{72} - frac{4}{72} ~ & extcolor{red}{ ext{ Simplify numerators and denominators.}} = frac{-33-4}{72} ~ & extcolor{red}{ ext{ Keep LCD; subtract numerators.}} = frac{-37}{72} ~ & extcolor{red}{ ext{ Simplify numerator.}} end{aligned} onumber ]
Of course, negative divided by positive yields a negative answer, so we can also write our answer in the form
[ - frac{11}{24} - frac{1}{18} = - frac{37}{72}. onumber ]
Exercise (PageIndex{11})
Simplify: ( - frac{5}{24} - frac{11}{36})
−37/72
## Comparing Fractions
The simplest way to compare fractions is to create equivalent fractions.
Example (PageIndex{12})
Arrange the fractions −1/2 and −4/5 on a number line, then compare them by using the appropriate inequality symbol.
Solution
The least common denominator for 2 and 5 is the number 10. First, make equivalent fractions with a LCD equal to 10.
[ egin{array}{c} - frac{1}{2} = - frac{1 cdot extcolor{red}{5}}{2 cdot extcolor{red}{5}} = - frac{5}{10} - frac{4}{5} = - frac{4 cdot extcolor{red}{2}}{5 cdot extcolor{red}{2}} = - frac{8}{10} end{array} onumber ]
To plot tenths, subdivide the interval between −1 and 0 into ten equal increments.
Because −4/5 lies to the left of −1/2, we have that −4/5 is less than −1/2, so we write
[ - frac{4}{5} < - frac{1}{2}. onumber ]
Exercise (PageIndex{12})
Compare −3/8 and −1/2.
[ - frac{1}{2} < - frac{3}{8} onumber ]
## Exercises
In Exercises 1-10, list the multiples the given numbers, then list the common multiples. Select the LCM from the list of common multiples.
1. 9 and 15
2. 15 and 20
3. 20 and 8
4. 15 and 6
5. 16 and 20
6. 6 and 10
7. 20 and 12
8. 12 and 8
9. 10 and 6
10. 10 and 12
In Exercises 11-20, for the given numbers, calculate the LCM using prime factorization.
11. 54 and 12
12. 108 and 24
13. 18 and 24
14. 36 and 54
15. 72 and 108
16. 108 and 72
17. 36 and 24
18. 18 and 12
19. 12 and 18
20. 12 and 54
In Exercises 21-32, add or subtract the fractions, as indicated, and simplify your result.
21. (frac{7}{12} − frac{1}{12})
22. (frac{3}{7} − frac{5}{7})
23. (frac{1}{9} + frac{1}{9})
24. (frac{1}{7} + frac{3}{7})
25. (frac{1}{5} − frac{4}{5})
26. (frac{3}{5} − frac{2}{5})
27. (frac{3}{7} − frac{4}{7})
28. (frac{6}{7} − frac{2}{7})
29. (frac{4}{11} + frac{9}{11})
30. (frac{10}{11} + frac{4}{11})
31. (frac{3}{11} + frac{4}{11})
32. (frac{3}{7} + frac{2}{7})
In Exercises 33-56, add or subtract the fractions, as indicated, and simplify your result.
33. (frac{1}{6} − frac{1}{8})
34. (frac{7}{9} − frac{2}{3})
35. (frac{1}{5} + frac{2}{3})
36. (frac{7}{9} + frac{2}{3})
37. (frac{2}{3} + frac{5}{8})
38. (frac{3}{7} + frac{5}{9})
39. (frac{4}{7} − frac{5}{9})
40. (frac{3}{5} − frac{7}{8})
41. (frac{2}{3} − frac{3}{8})
42. (frac{2}{5} − frac{1}{8)
43. (frac{6}{7} − frac{1}{6})
44. (frac{1}{2} − frac{1}{4})
45. (frac{1}{6} + frac{2}{3})
46. (frac{4}{9} + frac{7}{8})
47. (frac{7}{9} + frac{1}{8})
48. (frac{1}{6} + frac{1}{7})
49. (frac{1}{3} + frac{1}{7})
50. (frac{5}{6} + frac{1}{4})
51. (frac{1}{2} − frac{2}{7})
52. (frac{1}{3} − frac{1}{8})
53. (frac{5}{6} − frac{4}{5})
54. (frac{1}{2} − frac{1}{9})
55. (frac{1}{3} + frac{1}{8})
56. (frac{1}{6} + frac{7}{9})
In Exercises 57-68, add or subtract the fractions, as indicated, by first using prime factorization to find the least common denominator.
57. (frac{7}{36} + frac{11}{54})
58. (frac{7}{54} + frac{7}{24})
59. (frac{7}{18} − frac{5}{12})
60. (frac{5}{54} − frac{7}{12})
61. (frac{7}{36} + frac{7}{54})
62. (frac{5}{72} + frac{5}{108})
63. (frac{7}{24} − frac{5}{36})
64. (frac{11}{54} + frac{7}{72})
65. (frac{11}{12} + frac{5}{18})
66. (frac{11}{24} + frac{11}{108})
67. (frac{11}{54} − frac{5}{24})
68. (frac{7}{54} − frac{5}{24})
In Exercises 69-80, add or subtract the fractions, as indicated, and simplify your result.
69. (frac{−3}{7} + left( frac{−3}{7} ight))
70. (frac{−5}{9} + left( frac{−1}{9} ight))
71. (frac{7}{9} − left( frac{−1}{9} ight) )
72. (frac{8}{9} − left( frac{−4}{9} ight))
73. (frac{7}{9} + left( frac{−2}{9} ight))
74. ( frac{2}{3} + left( frac{−1}{3} ight))
75. (frac{−3}{5} − frac{4}{5})
76. (frac{−7}{9} − frac{1}{9})
77. (frac{−7}{8} + frac{1}{8})
78. (frac{−2}{3} + (frac{1}{3})
79. (frac{−1}{3} − left( frac{−2}{3} ight))
80. (frac{−7}{8} − left( frac{−5}{8} ight))
In Exercises 81-104, add or subtract the fractions, as indicated, and simplify your result.
81. (frac{−2}{7}) + frac{4}{5})
82. (frac{−1}{4} + frac{2}{7})
83. (frac{−1}{4} − left( frac{−4}{9} ight))
84. (frac{−3}{4} −left( frac{−1}{8} ight))
85. (frac{−2}{7} + frac{3}{4})
86. (frac{−1}{3} + frac{5}{8})
87. (frac{−4}{9} − frac{1}{3})
88. (frac{−5}{6} − frac{1}{3})
89. (frac{−5}{7} − left( frac{−1}{5} ight))
90. (frac{−6}{7} − left( frac{−1}{8} ight))
91. (frac{1}{9} + left( frac{−1}{3} ight))
92. (frac{1}{8} + left( frac{−1}{2} ight))
93. (frac{2}{3} + left( frac{−1}{9} ight))
94. (frac{3}{4} + left( frac{−2}{3} ight))
95. (frac{−1}{2} + left( frac{−6}{7} ight))
96. (frac{−4}{5} + left( frac{−1}{2} ight))
97. (frac{−1}{2} + left( frac{−3}{4} ight))
98. (frac{−3}{5} + left( frac{−1}{2} ight))
99. (frac{−1}{4} − frac{1}{2})
100. (frac{−8}{9} − frac{2}{3})
101. (frac{5}{8} − left( frac{−3}{4} ight))
102. (frac{3}{4} − left( frac{−3}{8} ight))
103. (frac{1}{8} − left( frac{−1}{3} ight))
104. (frac{1}{2} − left( frac{−4}{9} ight))
In Exercises 105-120, add or subtract the fractions, as indicated, and write your answer is lowest terms.
105. (frac{1}{2} + frac{3q}{5})
106. (frac{4}{7} − frac{b}{3})
107. (frac{4}{9} − frac{3a}{4})
108. (frac{4}{9} − frac{b}{2})
109. (frac{2}{s} + frac{1}{3})
110. (frac{2}{s} + frac{3}{7})
111. (frac{1}{3} − frac{7}{b})
112. (frac{1}{2} − frac{9}{s})
113. (frac{4b}{7} + frac{2}{3})
114. (frac{2a}{5} + frac{5}{8})
115. (frac{2}{3} − frac{9}{t})
116. (frac{4}{7} − frac{1}{y})
117. (frac{9}{s} + frac{7}{8})
118. (frac{6}{t} − frac{1}{9})
119. (frac{7b}{8} − frac{5}{9})
120. (frac{3p}{4} − frac{1}{8})
In Exercises 121-132, determine which of the two given statements is true.
121. (frac{−2}{3} < frac{−8}{7}) or (frac{− 2}{3} > frac{−8}{7})
122. (frac{−1}{7} < frac{−8}{9}) or (frac{− 1}{7} > frac{−8}{9})
123. (frac{6}{7} < frac{7}{3}) or (frac{6}{7} > frac{7}{3})
124. (frac{1}{2} < frac{2}{7}) or (frac{1}{2} > frac{2}{7})
125. (frac{−9}{4} < frac{−2}{3}) or frac{− 9}{4} > frac{−2}{3})
126. (frac{−3}{7} < frac{−9}{2}) or (frac{− 3}{7} > frac{−9}{2})
127. (frac{5}{7} < frac{5}{9}) or frac{5}{7} > frac{5}{9})
128. (frac{1}{2} < frac{1}{3}) or (frac{1}{2} > frac{1}{3})
129. (frac{−7}{2} < frac{−1}{5}) or (frac{− 7}{2} > frac{−1}{5})
130. (frac{−3}{4} < frac{−5}{9}) or (frac{− 3}{4} > frac{−5}{9})
131. (frac{5}{9} < frac{6}{5}) or (frac{5}{9} > frac{6}{5})
132. (frac{3}{2} < frac{7}{9}) or (frac{3}{2} > frac{7}{9})
1. 45
3. 40
5. 80
7. 60
9. 30
11. 108
13. 72
15. 216
17. 72
19. 36
21. (frac{1}{2})
23. (frac{2}{9})
25. (frac{−3}{5})
27. (frac{−1}{7})
29. (frac{13}{11})
31. (frac{7}{11})
33. (frac{1}{24})
35. (frac{13}{15})
37. (frac{31}{24})
39. (frac{1}{63})
41. (frac{7}{24})
43. (frac{29}{42})
45. (frac{5}{6})
47. (frac{65}{72})
49. (frac{10}{21})
51. (frac{3}{14})
53. (frac{1}{30})
55. (frac{11}{24})
57. (frac{43}{108})
59. (frac{−1}{36})
61. (frac{35}{108})
63. (frac{11}{72})
65. (frac{43}{36})
67. (frac{−1}{216})
69. (frac{−6}{7})
71. (frac{8}{9})
73. (frac{5}{9})
75. (frac{− 7}{5})
77. (frac{− 3}{4})
79. (frac{1}{3})
81. (frac{18}{35})
83. (frac{7}{36})
85. (frac{13}{28})
87. (frac{− 7}{9})
89. (frac{−18}{35})
91. (frac{− 2}{9})
93. (frac{5}{9})
95. (frac{−19}{14})
97. (frac{− 5}{4})
99. (frac{− 3}{4})
101. (frac{11}{8})
103. (frac{11}{24})
105. (frac{5+6 q}{10})
107. (frac{16 − 27 a}{36})
109. (frac{6 + s}{3 s})
111. (frac{b − 21}{3b})
113. (frac{12 b + 14}{21})
115. (frac{2 t − 27}{3t})
117. (frac{72 + 7 s}{8 s})
119. (frac{63 b − 40}{72})
121. (frac{− 2}{3} > (frac{− 8}{7})
123. (frac{6}{7} < frac{7}{3})
125. (frac{− 9}{4} < frac{− 2}{3})
127. (frac{5}{7} > frac{5}{9})
129. (frac{− 7}{2 } < frac{− 1}{5})
131. (frac{5}{9} < frac{6}{5})
## Adding Fractions with Common Denominators
Abigail, Hanna, and Naomi are studying for their midterm exam. The material they are required to study consists of 16 chapters of reading. The three of them realize that 16 chapters is a lot of reading for each of them to do, so they decide to study in a more efficient manner. They come up with a plan in which each of them reads a certain number of chapters and then summarizes it for the other two. They will share notes, and each will find online videos corresponding to their particular set of chapters.
Now, the chapters are not created equally. Some are quite easy, while others are much tougher. Their goal is to spread the workload evenly between the three of them. Remember that there are 16 chapters.
Abigail has the highest number of chapters to go through with 6. Hanna has 5, while Naomi has only 4. If you were to add those up, you would notice that that only comes to 15 chapters. The last chapter in the book is about troubleshooting electrical systems, and the apprentices decide that they will go through that one together.
We can represent each of their workloads as a fraction of a whole:
What if were to add those fractions? It would look something like this:
What you’ll note is that the numerators are all different, while the denominators are all the same (16). When adding or subtracting fractions, the denominators must be the same. We refer to this as having a common denominator.
So, in order to get the answer to the above question, you just add all the numerators. Adding fractions is very simple in this respect.
Notice that the denominator in the final answer is the same as that in the fractions being added. By the end, the apprentices will have gone through 15 of the 16 chapters separately, and then they will go through the last chapter together.
The concept of adding fractions with common denominators is easy enough, and we did enough adding whole numbers that going through examples at this point might not be worth it (but if you need a review, see Adding Whole Numbers). What we will do instead is write down some examples of adding fractions so you can see the idea.
Do you notice anything about the answer to the last one? It can be reduced.
Before we get going any further with work on fractions, this might be a good time to state that, when working with fractions, we generally want to put the answer in lowest terms.
### How to Add and Subtract Fractions
Here you will find support pages about how to add and subtract fractions (with both like and unlike denominators).
If you want to use our Free Fraction Calculator to do the work for you then use the link below.
Our Fraction calculator will allow you to add or subtract fractions and show you the steps to work it out.
Otherwise, for more detailed support and worksheets, keep reading!
### How to Add and Subtract Fractions with Like Denominators Video
Find out how to add and subtract fractions with like denominators using the video below.
### Adding and Subtracting Fractions with Like Denominators Worksheets
Here you will find a selection of Free Fraction worksheets designed to help your child understand how to add and subtract fractions with the same denominator. The sheets are graded so that the easier ones are at the top.
All the free Fraction worksheets in this section support the Elementary Math Benchmarks for Third Grade.
### Adding Fractions with Like Denominators
• Adding Fractions With Like Denominators Using Circles 1
• Adding Fractions Like Denominators 1
• Adding Fractions Like Denominators 2
• Adding Fractions Like Denominators 3
### Subtracting Fractions with Like Denominators
• Subtract Fractions with Like Denominators Using Circles 1
• Subtracting Fractions Like Denominators 1
• Subtracting Fractions Like Denominators 2
• Subtracting Fractions Like Denominators 3
### More Recommended Math Worksheets
Take a look at some more of our worksheets similar to these.
### Adding Subtracting Fractions with unlike denominators
If you need to add and subtract fractions with unlike denominators, then we have a page dedicated to this skill.
### How to find unit fractions of a number
Here you will find a selection of Fraction worksheets designed to help your child understand how to work out fractions of different numbers, where the numerator is equal to 1.
• develop an understanding of fractions as parts of a whole
• know how to calculate unit fractions of a range of numbers.
### Free Printable Fraction Flashcards
Here you will find a selection of Fraction Flash Cards designed to help your child learn their Fractions.
Using Flash Cards is a great way to learn your Fraction facts as parts of a whole. They can be taken on a journey, played with in a game, or used in a spare five minutes daily until your child knows their Fractions off by heart.
All the printable Math facts in this section support the Elementary Math Benchmarks.
### Learning Fractions Math Help Page
Here you will find the Math Salamanders free online Math help pages about Fractions.
There is a wide range of help pages including help with:
• fraction definitions
• equivalent fractions
• converting improper fractions
• how to add and subtract fractions
• how to convert fractions to decimals and percentages
• how to simplify fractions.
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### Math-Salamanders.com
The Math Salamanders hope you enjoy using these free printable Math worksheets and all our other Math games and resources.
## Subtracting Mixed Fractions
### Example: What is 15 3 4 − 8 5 6 ?
Convert to Improper Fractions:
15 3 4 = 63 4
8 5 6 = 53 6
63 4 becomes 189 12
53 6 becomes 106 12
189 12106 12 = 83 12
Convert back to Mixed Fractions:
83 12 = 6 11 12
## 2.3.1: Adding and Subtracting Fractions - Mathematics
It's easy to add and subtract like fractions, or fractions with the same denominator. You just add or subtract the numerators and keep the same denominator. The tricky part comes when you add or subtract fractions that have different denominators. To do this, you need to know how to find the least common denominator. In an earlier lesson, you learned how to simplify, or reduce, a fraction by finding an equivalent, or equal, fraction where the numerator and denominator have no common factors. To do this, you divided the numerator and denominator by their greatest common factor.
In this lesson, you'll learn that you can also multiply the numerator and denominator by the same factor to make equivalent fractions.
In this example, since 12 divided by 12 equals one, and any number multiplied by 1 equals itself, we know 36/48 and 3/4 are equivalent fractions, or fractions that have the same value. In general, to make an equivalent fraction you can multiply or divide the numerator and denominator of the fraction by any non-zero number.
Since only like fractions can be added or subtracted, we first have to convert unlike fractions to equivalent like fractions. We want to find the smallest, or least, common denominator, because working with smaller numbers makes our calculations easier. The least common denominator, or LCD, of two fractions is the smallest number that can be divided by both denominators. There are two methods for finding the least common denominator of two fractions:
Method 1:
Write the multiples of both denominators until you find a common multiple.
The first method is to simply start writing all the multiples of both denominators, beginning with the numbers themselves. Here's an example of this method. Multiples of 4 are 4, 8, 12, 16, and so forth (because 1 × 4=4, 2 × 4=8, 3 × 4=12, 4 × 4=16, etc.). The multiples of 6 are 6, 12,…--that's the number we're looking for, 12, because it's the first one that appears in both lists of multiples. It's the least common multiple, which we'll use as our least common denominator.
Method 2:
Use prime factorization.
For the second method, we use prime factorization-that is, we write each denominator as a product of its prime factors. The prime factors of 4 are 2 times 2. The prime factors of 6 are 2 times 3. For our least common denominator, we must use every factor that appears in either number. We therefore need the factors 2 and 3, but we must use 2 twice, since it's used twice in the factorization for 4. We get the same answer for our least common denominator, 12.
prime factorization of 4 = 2 × 2
prime factorization of 6 = 2 × 3
LCD = 2 × 2 × 3 = 12
Now that we have our least common denominator, we can make equivalent like fractions by multiplying the numerator and denominator of each fraction by the factor(s) needed. We multiply 3/4 by 3/3, since 3 times 4 is 12, and we multiply 1/6 by 2/2, since 2 times 6 is 12. This gives the equivalent like fractions 9/12 and 2/12. Now we can add the numerators, 9 + 2, to find the answer, 11/12.
## 1.6 Add and Subtract Fractions
A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, Fractions.
### Add or Subtract Fractions with a Common Denominator
When we multiplied fractions, we just multiplied the numerators and multiplied the denominators right straight across. To add or subtract fractions, they must have a common denominator.
To add or subtract fractions, add or subtract the numerators and place the result over the common denominator.
### Example 1.78
Find the difference: − 23 24 − 13 24 . − 23 24 − 13 24 .
#### Solution
Find the difference: − 19 28 − 7 28 . − 19 28 − 7 28 .
Find the difference: − 27 32 − 1 32 . − 27 32 − 1 32 .
### Example 1.79
#### Solution
Find the difference: − 9 x − 7 x . − 9 x − 7 x .
Find the difference: − 17 a − 5 a . − 17 a − 5 a .
Now we will do an example that has both addition and subtraction.
### Example 1.80
Simplify: 3 8 + ( − 5 8 ) − 1 8 . 3 8 + ( − 5 8 ) − 1 8 .
#### Solution
Simplify: − 2 9 + ( − 4 9 ) − 7 9 . − 2 9 + ( − 4 9 ) − 7 9 .
Simplify: 5 9 + ( − 4 9 ) − 7 9 . 5 9 + ( − 4 9 ) − 7 9 .
### Add or Subtract Fractions with Different Denominators
As we have seen, to add or subtract fractions, their denominators must be the same. The least common denominator (LCD) of two fractions is the smallest number that can be used as a common denominator of the fractions. The LCD of the two fractions is the least common multiple (LCM) of their denominators.
### Least Common Denominator
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.
### Manipulative Mathematics
After we find the least common denominator of two fractions, we convert the fractions to equivalent fractions with the LCD. Putting these steps together allows us to add and subtract fractions because their denominators will be the same!
### How To
1. Step 1. Do they have a common denominator?
• Yes—go to step 2.
• No—rewrite each fraction with the LCD (least common denominator). Find the LCD. Change each fraction into an equivalent fraction with the LCD as its denominator.
2. Step 2. Add or subtract the fractions.
3. Step 3. Simplify, if possible.
When finding the equivalent fractions needed to create the common denominators, there is a quick way to find the number we need to multiply both the numerator and denominator. This method works if we found the LCD by factoring into primes.
Look at the factors of the LCD and then at each column above those factors. The “missing” factors of each denominator are the numbers we need.
In Example 1.81, the LCD, 36, has two factors of 2 and two factors of 3 . 3 .
The numerator 12 has two factors of 2 but only one of 3—so it is “missing” one 3—we multiply the numerator and denominator by 3.
The numerator 18 is missing one factor of 2—so we multiply the numerator and denominator by 2.
## How to Subtract Fractions
Once you’ve mastered adding fractions, subtracting fractions will be a breeze! The process is exactly the same, though you’ll naturally be subtracting instead of adding.
### #1: Find a Common Denominator
Let’s look at the following example:
We need to find the least common multiple for the denominators, which will look like this:
3 : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
10 : 10, 20, 30
The first number they have in common is 30, so we’ll be putting both numerators over a denominator of 30.
### #2: Multiply to Get Both Numerators Over the Same Denominator
First, we need to figure out how much we’ll need to multiply both the numerator and denominator of each fraction by to get a denominator of 30. For $2/3$, what number times 3 equals 30? In equation form:
Our answer is 10, so we’ll multiply both the numerator and denominator by 10 to get $20/30$.
Next, we’ll repeat the process for the second fraction. What number do we need to multiply by 10 to get 30? Well, $30÷10=3$, so we’ll multiply the top and bottom by 3 to get $9/30$.
This makes our problem $20/30-9/30$, which means we’re ready to continue!
### #3: Subtract the Numerators
Just as we did with addition, we’ll subtract one numerator from the other but leave the denominators alone.
Since we found the least common multiple, we already know that the problem can’t be reduced any further.
However, let’s say that we just multiplied 3 by 10 to get the denominator of 30, so we need to check if we can reduce. Let’s use that little trick we learned to find the greatest possible common factor. Whatever factors 11 and 30 share, they can’t be greater than $30-11$, or 19.
30 : 2, 3, 5, 6, 10, 15
Since they don’t share any common factors, the answer cannot be reduced any further.
## MathHelp.com
The basic idea with converting to common denominators is to multiply fractions by useful forms of 1 . What does this mean? Take a look:
#### Simplify
Before I can add these fractions, I have to find their common denominator. The lowest (smallest) common denominator is just the Least Common Multiple (LCM) of the two denominators, 4 and 5 . The prime factorizations and LCM of the denominators 4 and 5 are:
In other words, I have to convert the fourths and fifths into twentieths. I'll do this by multiplying by a useful form of 1 . In the case of the first fraction, 1 /4 , the 4 needs to become a 20 , so I need to multiply the 4 by 5 . To keep the fraction equal to its original value, I'll have to multiply the top by 5 , too. In other words, I'll multiply the fraction by 5 /5 , which is just a useful form of the number 1 :
Because I multiplied by (a useful form of) 1 , I haven't changed the actual value of the fraction. All I've changed is how the value is stated.
In the case of the second fraction, 2 /5 , the 5 needs to become a 20 , so I have to multiply the 5 by 4 . To keep the fraction equal to the same value, I also have to multiply the top by 4 , too. In other words, I'll multiply by 4 /4 , which is just a useful form of 1 :
The fourths and fifths are now both twentieths I'm finally in an all-apples situation. Only now can I actually add the fractions. To add these "apples", I add the numerators:
The numerator, 13 , is prime, and it isn't a factor of 20 , so there's no cancellation that I can do.
My simplified final answer is .
By the way, your calculator may be able to do all of this for you check your manual. But make sure you at least understand the basic idea, because you'll need this process later in algebra, when you get to fractions with polynomials, called "rational expressions".
#### Simplify
First, I'll find the LCM of the two denominators:
Since 5 is a factor of 15 , then the LCM is 15 in particular, one of the fractions is already in LCM form. I'll convert the other fraction to this common denominator, add, and, if possible, simplify:
There are no common factors, so nothing simplifies.
#### Simplify
First I'll find the LCM of the two denominators:
Notice that 8 and 6 both have 2 as a factor. The point of lining the factors up nice and neatly in columns, as I've done above, is to help avoid over-duplication of factors when finding the LCM. Be careful: there are only three 2 's in the LCM, not four.
To convert the first fraction to a denominator of 24 , I'll multiply, top and bottom, by 3 . To convert the second fraction's denominator, I'll multiply, top and bottom, by 4 .
The instructions don't say to express the answer in mixed-number form, so I'll leave it as an improper fraction. There are no common factors between the numerator and denominator, so I can't simplify any further.
#### Simplify
First, I'll find the LCM of the three denominators:
Now I'll convert the three fractions to the common denominator, add, and then see if I can simplify.
Because 4 was a common factor of 1072 and 364 , I was able to cancel this out and simplify to get my final answer:
#### Simplify
First, I'll find the LCM of the two denominators:
To convert to the LCM, I'll multiply the first fraction, top and bottom, by 7 , and the second fraction, top and bottom, by 5 .
The numerator, 106 , factors as 2×53 , and 53 is prime, so there's nothing I can cancel the fraction can't be further simplified.
You can use the Mathway widget below to practice adding and subtracting fractions. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
## COMPLEX FRACTIONS
### OBJECTIVES
Fractions are defined as the indicated quotient of two expressions. In this section we will present a method for simplifying fractions in which the numerator or denominator or both are themselves composed of fractions. Such fractions are called complex fractions.
Thus if the numerator and denominator of a complex fraction are composed of single fractions, it can be simplified by dividing the numerator by the denominator.
A generally more efficient method of simplifying a complex fraction involves using the fundamental principle of fractions. We multiply both numerator and denominator by the common denominator of all individual fractions in the complex fraction.
Recall that the fundamental principle of fractions states
We will use the fundamental principle to again simplify
The LCD of 3 and 4 is 12. Thus
The individual fractions are
This answer could be written as the mixed number
Make sure that each term in both numerator and denominator is multiplied by the LCD.
We need the LCD of individual fractions, y is not a fraction.
## Fractions - Grade 5 Maths Questions
Solutions and explanations to grade 5 fractions questions are presented.
1. 3 1/2 + 5 1/3 =
Solution
Add whole numbers together and fractions together
3 1/2 + 5 1/3 = (3 + 5) + (1/2 + 1/3)
Write fractions with the same denominator
= 8 + (3/6 + 2/6) = 8 5/6
2. It takes Julia 1/2 hour to wash, comb her hair and put on her clothes, and 1/4 hour to have her breakfast. How much time does it take Julia to be ready for school?
Solution
The total time for Julia to be ready for school is
1/2 + 1/4
Write fractions with the same denominator
= 2/4 + 1/4 = 3/4 of an hour.
3. Which two fractions are equivalent?
1. 5/2 and 2/5
2. 4/3 and 8/6
3. 1/4 and 2/4
4. 2/3 and 1/3
.
Solution
There are two whole shaded items above and one shaded at 3/4. Hence the mixed number
2 3/4 represents the shaded parts.
.
Solution
1 7/10 in decimal form is
1 7/10 = 1 + 7/10 = 1 + 0.7 = 1.7 and corresponds to point W.
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# Chinese Immigrants and the Iron Roadway
On a brilliant Might day in 1869, railway employees, entrepreneurs, and federal government authorities collected in Utah for an historical occasion. Quickly the ceremonial owning of a strong gold railway surge would certainly total a six-year initiative at structure a railway throughout The u.s.a.. Of program, the pricy $350 surge was rapidly changed for safekeeping. Still, it stood for the bridging of 3,500 miles of railway, and therefore likewise symbolized a huge quantity of human labor. A lot of this labor was Chinese. Americans had contemplated building a transcontinental railway because the 1830s. Without an "iron roadway", overland take a trip from the eastern states to the California Area entailed 4 to 6 months of difficulty. A railway would certainly help with westward growth and assistance recognize America's "show fate". In 1862, Head of state Lincoln authorized the Pacific Railway Act. This given a charter to 2 railway business, the Union Pacific and Main Pacific, for the structure of a train and telegraph line. The business would certainly function from contrary instructions: the Union Pacific would certainly begin building in Omaha, and the Main Pacific would certainly begin in Sacramento. The different jobs would certainly ultimately ended up being connected and satisfy. The business damaged ground in 1863, however their jobs really did not acquire complete rate after the Civil War finished. In 1866 the Union Pacific enhanced its labor pressure with mainly Irish immigrants. The Main Pacific employed greater than 25,000 Chinese immigrants to removal with the Sierra Nevadas. Chinese individuals had ventured to North The u.s.a. as very early as 450 A.Decoration. Still, couple of Chinese resided in North The u.s.a. up till the California Gold Hurry was advertised. When information of gold dirt got to the Chinese landmass, peasants acknowledged a chance to leave hardship. Some guys were so destitute that they had offered their kids. Making a couple of hundred American bucks would certainly permit their households a life of high-end. So, countless guys boarded firmly loaded ships for flow to "the Gold Hill" of California. The Chinese employees were particularly important to the Main Pacific Business. With their objective of removaling eastern from Sacramento, they required an approximated 5,000 employees. There just weren't sufficient Anglo-Americans offered in California, when guys were brought from the eastern states, they had the tendency to remove for experience! The Main Pacific employed as numerous Chinese immigrants as they might, and after that sent out representatives to Hong Kong for extra recruits. By the moment the rails were signed up with in Utah, regarding 90% of the Main Pacific employees were Chinese. The Chinese immigrants, in spite of being essential laborers, weren't dealt with in addition to white laborers. White guys were paid$35 monthly as well as got a camping tent, food, and materials. The Chinese were typically paid much less and didn't have the "advantages" of company offered food, sanctuary, or materials. The Main Pacific employees risked their lives daily when scraping with the Sierra Nevada Hills. In some cases they wove man-sized baskets to put on hold themselves over high cliffs, 2,000 feet over ground. They utilized dynamite and nitroglycerine, which in some cases exploded prematurely. For numerous months, some lived completely below the hill snow, producing labyrinths from the home of function and obeying light light. Whole camps of guys were shed to avalanches.
When the guys got to the desert, they dealt with one more establish of risks. There they might lay rails faster, however the temperature level got to 120 levels! Alkali dirt made many hemorrhage from the lungs. By January of 1869, the function was almost total. The government federal government determined where both railroads ought to satisfy, eventually choosing Promontory Top. 8 Chinese guys put the last area of rail on Might 10, 1869. Simply 5 days later on, traveler educate solution started. The overland journey from Omaha to Sacramento would certainly currently need just 4 days of take a trip!
Californians anticipated the railway to bring success. One of the most instant impact, nevertheless, was that California's recently established production market was endangered by less expensive products from the Eastern US. Californians were additional inflamed by the influx of job-seeking immigrants that shown up through educate. The ensuing financial anxiety was criticized after the Chinese immigrants that had built the iron roadway. California passed various anti-Chinese legislations. Thankfully for the Chinese American neighborhood, nevertheless, the railway workers had made the immigrants a credibility for being great employees. They were hired to function somewhere else throughout the United States. Each year in Might because 1965, the event of finishing the nation's initially transcontinental railway is re-enacted at the Gold Surge Nationwide Historical Website in Brigham City, Utah.
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# Making Go's RSA Internals Constant Time
This is essentially a lab notebook, while trying to move Go’s crypto/rsa module away from the variable-time math/big.Int, to an internal constant-time number type.
# 2020-05-07
## Necessary Methods
Key generation is out of the scope for now, and substantially more complicated compared to encryption and decryption. For implementing these, we only need:
• CmpGeq
• CmpZero
• ModInv (Only if we use blinding)
• ModExp
• ModMul
• ModAdd
• ModSub
If we implement constant-time operations, we can get rid of blinding, and this provides additional incentive, since we don’t need to implement modular inversion, and can remove the blinding logic.
## Internal APIs
Internal APIs often use big.Int, and could be shifted to use our internal type. The public API still needs to use big.Int though, for compatability.
This would help with some leakages actually, because using big.Int leaks zero padding information.
The concern here is the complexity of making this change.
# 2020-05-08
## Unsaturated vs Saturated limbs
Using 63 bit limbs instead of 64 seems to be noticeably (~1.8x) faster for montgomery multiplication, and thus for exponentiation.
We have 5215 ns/op for saturated, and 2851 ns/op for unsaturated.
## Using uint
Using uint as our word type lets us call bits.Add and bits.Mul directly, which is a bit nicer. The downside is that we have less control over using a wrapper type, or using uint64.
## Modular addition with and without scratch
BenchmarkModAdd-4 10874287 103.3 ns/op
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# Asymptotics of H t
## The Gamma function
The Gamma function is defined for $\mathrm{Re}(s) \gt 0$ by the formula
$\Gamma(s) = \int_0^\infty x^s e^{-x} \frac{dx}{x}$
and hence by change of variables
$\Gamma(s) = \int_{-\infty}^\infty \exp( s u - e^u )\ du. \quad (1.1)$
It can be extended to other values of $s$ by analytic continuation or by contour shifting; for instance, if $Im(s)\gt0$, one can write
$\Gamma(s) = \int_C \exp( s u - e^u )\ du \quad (1.1')$
where $C$ is a contour from $+i\infty$ to $\infty$ that stays within a bounded distance of the upper imaginary and right real axes.
The Gamma function obeys the Euler reflection formula
$\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)} \quad (1.2)$
and the duplication formula
$\Gamma(1-s) = \frac{\Gamma(\frac{1-s}{2}) \Gamma(1-\frac{s}{2})}{\pi^{1/2} 2^s}. \quad (1.3)$
In particular one has
$\Gamma(\frac{s}{2}) \Gamma(1-\frac{s}{2}) = \frac{\pi}{\sin(\pi s/2)} \quad (1.4)$
and thus on combining (3) and (4)
$\Gamma(s/2) \Gamma(1-s) = \frac{\pi^{1/2}}{2^s \sin(\pi s/2)} \Gamma(\frac{1-s}{2}) \quad(1.5)$
Since $s \Gamma(s) = \Gamma(s+1)$, we have
$\frac{s(s-1)}{2} \Gamma(\frac{s}{2}) = 2 \Gamma(\frac{s+4}{2}) - 3 \Gamma(\frac{s+2}{2}). \quad (1.6)$
We have the Stirling approximation
$\Gamma(s) = \sqrt{2\pi/s} \exp( s \log s - s + O(1/|s|) )$
whenever $\mathrm{Re}(s) \gg 1$. If we have $s = \sigma+iT$ for some large $T$ and bounded $\sigma \gg 1$, this gives
$\Gamma(s) \approx \sqrt{2\pi} T^{\sigma -1/2} e^{-\pi T/2} \exp(i (T \log T - T + \pi \sigma/2 - \pi/4)). (1.7)$
Another crude but useful approximation is
$\Gamma(s+h) \approx \Gamma(s) s^h (1.8)$
for $s$ as above and $h=O(1)$.
## The Riemann-Siegel formula for $t=0$
Proposition 1 (Riemann-Siegel formula for $t=0$) For any natural numbers $N,M$ and complex number $s$ that is not an integer, we have
$\zeta(s) = \sum_{n=1}^N \frac{1}{n^s} + \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \sum_{m=1}^M \frac{1}{m^{1-s}} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw$
where $w^{s-1} := \exp((s-1) \log w)$ and we use the branch of the logarithm with imaginary part in $[0,2\pi)$, and $C_M$ is any contour from $+\infty$ to $+\infty$ going once anticlockwise around the zeroes $2\pi i m$ of $e^w-1$ with $|m| \leq M$, but does not go around any other zeroes.
Proof This equation is in [T1986, p. 82], but we give a proof here. The right-hand side is meromorphic in $s$, so it will suffice to establish that
1. The right-hand side is independent of $N$;
2. The right-hand side is independent of $M$;
3. Whenever $\mathrm{Re}(s)\gt1$ and $s$ is not an integer, the right-hand side converges to $\zeta(s)$ if $M=0$ and $N \to \infty$.
We begin with the first claim. It suffices to show that the right-hand sides for $N$ and $N-1$ agree for every $N \gt 1$. Subtracting, it suffices to show that
$0 = \frac{1}{N^s} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} (e^{-Nw} - e^{-(N-1)w})}{e^w-1}\ dw.$
The integrand here simplifies to $- w^{s-1} e^{-Nw}$, which on shrinking $C_M$ to wrap around the positive real axis becomes $N^{-s} \Gamma(s) (1 - e^{2\pi i(s-1)})$. The claim then follows from the Euler reflection formula $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}$.
Now we verify the second claim. It suffices to show that the right-hand sides for $M$ and $M-1$ agree for every $M \gt 1$. Subtracting, it suffices to show that
$0 = \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} \frac{1}{M^{1-s}} + \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M - C_{M-1}} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw.$
The contour $C_M - C_{M-1}$ encloses the simple poles at $+2\pi i M$ and $-2\pi i M$, which have residues of $(2\pi i M)^{s-1} = - i (2\pi M)^{s-1} e^{\pi i s/2}$ and $(-2\pi i M)^{s-1} = i (2\pi M)^{s-1} e^{3\pi i s/2}$ respectively. So, on canceling the factor of $M^{s-1}$ it suffices to show that
$0 = \pi^{s-\frac{1}{2}} \frac{\Gamma((1-s)/2)}{\Gamma(s/2)} + e^{-i\pi s} \Gamma(1-s) (2\pi)^{s-1} i (e^{3\pi i s/2} - e^{\pi i s/2}).$
But this follows from the duplication formula $\Gamma(1-s) = \frac{\Gamma(\frac{1-s}{2}) \Gamma(1-\frac{s}{2})}{\pi^{1/2} 2^s}$ and the Euler reflection formula $\Gamma(\frac{s}{2}) \Gamma(1-\frac{s}{2}) = \frac{\pi}{\sin(\pi s/2)}$.
Finally we verify the third claim. Since $\zeta(s) = \lim_{N \to \infty} \sum_{n=1}^\infty \frac{1}{n^s}$, it suffices to show that
$\lim_{N \to \infty} \int_{C_0} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw = 0.$
We take $C_0$ to be a contour that traverses a $1/N$-neighbourhood of the real axis. Writing $C_0 = \frac{1}{N} C'_0$, with $C'_0$ independent of $N$, we can thus write the left-hand side as
$\lim_{N \to \infty} N^{-s} \int_{C'_0} \frac{w^{s-1} e^{-w}}{e^{w/N}-1}\ dw,$
and the claim follows from the dominated convergence theorem. $\Box$
Applying the Riemann-Siegel formula to the Riemann xi function $\xi(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)$, we have
$\xi(s) = F_{0,N}(s) + \overline{F_{0,M}(\overline{1-s})} + R_{0,N,M}(s) \quad(2.1)$
where
$F_{0,N}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{1}{n^s} \quad(2.2)$
and
$R_{0,N,M}(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw. \quad(2.3)$
## A contour integral
Lemma 2 Let $L$ be a line in the direction $\mathrm{arg} w = \pi/4$ passing between $0$ and $2\pi i$. Then for any complex $\alpha$, the contour integral
$\Psi(\alpha) := \int_L \frac{\exp( \frac{i}{4\pi} z^2 + \alpha z)}{e^z - 1}\ dz$
can be given explicitly by the formula
$\Psi(\alpha) = 2\pi \frac{\cos \pi(\frac{1}{2} \alpha^2 - \alpha - \frac{\pi}{8})}{\cos(\pi \alpha)} \exp( \frac{i \pi}{2} \alpha^2 - \frac{5 \pi}{8} )$.
Proof The integrand has a residue of $1$ at $0$, hence on shifting the contour downward by $2\pi i$ we have
$\Psi(\alpha) = -2\pi i + \int_L \frac{\exp( \frac{i}{4\pi} (z-2\pi i)^2 + \alpha (z-2\pi i) )}{e^z-1}\ dz.$
The right-hand side expands as
$-2\pi i - e^{-2\pi i \alpha} \int_L \frac{\exp( \frac{i}{4\pi} z^2 + (\alpha+1) z)}{e^z-1}\ dz$
which we can write as
$-2\pi i - e^{-2\pi i \alpha} (\Psi(\alpha) + \int_L \exp( \frac{i}{4\pi} z^2 + \alpha z\ dz).$
The last integral is a standard gaussian integral, which can be evaluated as $-\sqrt{\frac{\pi}{i/4\pi}} \exp( \pi i \alpha^2)$. Hence
$\Psi(\alpha) = -2\pi i - e^{-2\pi i \alpha} (\Psi(\alpha) - \sqrt{\frac{\pi}{i/4\pi}} \exp( \pi i \alpha^2)),$
and the claim then follows after some algebra. $\Box$
We conclude from (2.3) that
$R_{0,N,M}(s) \approx - 2 \Gamma(\frac{5-s}{2}) \frac{\pi^{(-s-1)/2}}{2^s} e^{-\pi i s/2} \exp( -\frac{t \pi^2}{64} ) (2\pi i M)^{s-1} \Psi(\frac{s-2\pi i MN}{2\pi i M})$
$= i \Gamma(\frac{5-s}{2}) \pi^{-(s+1)/2} \exp( -\frac{t \pi^2}{64} ) (\pi M)^{s-1} \Psi(\frac{s}{2\pi i M} - N).$
## Heuristic approximation at $t=0$
To estimate the remainder term $R_{0,N,M}(s)$ in (2.3) with $M,N = \sqrt{\mathrm{Im}(s) / 2\pi} + O(1)$, we make the change of variables $w = z + 2\pi i M$ to obtain
$R_{0,N,M}(s) = \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} \int_{C_M - 2\pi i M} \frac{(z+2\pi i M)^{s-1} e^{-Nz}}{e^z-1}\ dz$
Steepest descent heuristics suggest that the dominant portion of this integral comes when $z=O(1)$. In this regime we may Taylor expand
$(z+2\pi i M)^{s-1} = (2\pi i M)^{s-1} \exp( (s-1) \log(1 + \frac{z}{2\pi i M}) )$
$\approx (2\pi i M)^{s-1} \exp( (s-1) \frac{z}{2\pi i M} -\frac{s-1}{2} (\frac{z}{2\pi i M})^2 )$
$\approx (2\pi i M)^{s-1} \exp( s \frac{z}{2\pi i M} + \frac{i}{4\pi} z^2 );$
using this approximation and then shifting the contour to $-L$ (cf. [T1986, Section 4.16], we conclude that
$R_{0,N,M}(s) \approx - \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1}\int_L \frac{\exp( (\frac{s}{2\pi i M}-N)z + \frac{i}{4\pi} z^2 )}{e^z-1}\ dz$
and hence by Lemma 2
$R_{0,N,M}(s) \approx - \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1}\Psi(\frac{s}{2\pi i M}-N). (4.1)$
Using (1.7) one can calculate that this expression has magnitude $O( x^{6/4} e^{-\pi x/8} )$.
If we drop the $R_{0,N,M}$ term, we have
$H_0(x+iy) \approx \frac{1}{8} F_{0,N}(\frac{1+ix-y}{2}) + \frac{1}{8} \overline{F_{0,M}(\frac{1+ix+y}{2})}.$
From (2.2) and (1.7) we have
$|\frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2)| \asymp x^{(7-y)/4} e^{-\pi x/8}$
when $s = (1+ix-y)/2$ and
$|\frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2)| \asymp x^{(7+y)/4} e^{-\pi x/8}$
when $s = (1+ix+y)/2$. Thus we expect the second term to dominate, and typically we would expect
$|H_0(x+iy)| \asymp x^{(7+y)/4} e^{-\pi x/8}.$
## Extending the Riemann-Siegel formula to positive $t$
Evolving $H_0(z) = \frac{1}{8} \xi(\frac{1+iz}{2})$ by the backwards heat equation $\partial_t H_t(z) = -\partial_{zz} H_t(z)$ is equivalent to evolving the Riemann $\xi$ function $\xi = \xi_0$ by the forwards heat equation $\partial_t \xi_t(s) = \frac{1}{4} \partial_{ss} \xi_t(s)$, and then setting
$H_t(z) = \frac{1}{8} \xi_t(1+\frac{iz}{2}).$
One way to do this is to expand $\xi_0(s)$ as a linear combination of exponentials $e^{\alpha s}$, and replace each such exponential by $\exp( \frac{t}{4} \alpha^2 ) e^{\alpha s}$ to obtain $\xi_t$. Roughly speaking, this can be justified as long as everything is absolutely convergent.
In view of (2.1), we will have
$\xi_t(s) = F_{t,N}(s) + \overline{F_{t,M}(\overline{1-s})} + R_{t,N,M}(s) \quad(5.1)$
where $F_{t,N}, R_{t,N,M}$ are the heat flow evolutions of $F_{0,N}, R_{0,N,M}$ respectively.
It is easy to evolve $F_{t,N}(s)$. Firstly, from (1.6) one has
$F_{0,N}(s) = \sum_{n=1}^N 2 \frac{\Gamma(\frac{s+4}{2})}{(\pi n^2)^{s/2}} - 3 2 \frac{\Gamma(\frac{s+2}{2})}{(\pi n^2)^{s/2}}$
and hence by (1.1')
$F_{0,N}(s) = \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2))\ du - 3 \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) )\ du.$
We can now evolve to obtain
$F_{t,N}(s) = \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du - 3 \int_C \exp( \frac{s+2}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du (5.2).$
By integrating on $C$ rather than the real axis, the integrals remain absolutely convergent here.
Evolving $R_{0,N,M}$ is a bit trickier. From (1.5) one has
$R_{0,N,M}(s) = \frac{s(s-1)}{2} \pi^{-s/2} \frac{e^{-i\pi s} \Gamma(\frac{1-s}{2})}{2^{s+1}\pi^{1/2} i \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw$
which can be rewritten using (1.6) as
$2 \pi^{-s/2} \frac{e^{-i\pi s} \Gamma(\frac{5-s}{2})}{2^{s+1}\pi^{1/2} i \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw$
$-3 \pi^{-s/2} \frac{e^{-i\pi s} \Gamma(\frac{3-s}{2})}{2^{s+1}\pi^{1/2} i \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1}\ dw.$
For $\mathrm{Im}(s) \gt 0$, we have the geometric series formula
$\frac{1}{\sin(\pi s/2)} = -2i e^{i\pi s/2} \sum_{n=0}^\infty e^{i \pi s n}$
and from this and (1.1') we can rewrite $R_{0,N,M}(s)$ as
$2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} } \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u)\ dw\ du$
$-3 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2}} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u)\ dw\ du$
where $\overline{C}$ is the complex conjugate of $C$. Hence we can write $R_{t,N,M}(s)$ exactly as
$2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{\overline{C}}\int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t}{4} (i \pi (n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2 )\ dw\ du$
$-3 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{\overline{C}} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{3-s}{2} u - e^u + \frac{t}{4} (i \pi (n-1/2) + \log \frac{w}{2\sqrt{\pi}} - \frac{u}{2})^2 )\ dw\ du (5.3)$
## Approximation for $t\gt0$
The above formulae are clearly unwieldy, so let us make a number of heuristic approximations to simplify them. We start with $F_{t,N}(s)$, assumig that the imaginary part of $s$ is large and positive and the real part is bounded. We first drop the second term of (5.2) as being lower order:
$F_{t,N}(s) \approx \sum_{n=1}^N 2 \int_C \exp( \frac{s+4}{2} u - e^u - \frac{s}{2} \log(\pi n^2) + \frac{t}{16} (u - \log(\pi n^2))^2 )\ du.$
Next, we shift $u$ by $\log \frac{s+4}{2}$ to obtain
$F_{t,N}(s) \approx \sum_{n=1}^N \frac{2 \exp( \frac{s+4}{2} \log \frac{s+4}{2} - \frac{s+4}{2})}{(\pi n^2)^{s/2}} \int_C \exp( \frac{s+4}{2} (1 + u - e^u) + \frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 )\ du.$
Because the expression $\exp( \frac{s+4}{4} (1+u-e^u) )$ decays rapidly away from $u=0$, we can heuristically approximate
$\frac{t}{16} (u + \log \frac{s+4}{2\pi n^2})^2 ) \approx \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2}$
and then we undo the shift to obtain
$F_{t,N}(s) \approx \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \int_{-\infty}^\infty \exp( \frac{s+4}{2} u - e^u + \frac{t}{16} \log^2\frac{s+4}{2\pi n^2} )\ du$
which by (1) becomes
$F_{t,N}(s) \approx \sum_{n=1}^N \frac{2}{(\pi n^2)^{s/2}} \Gamma(\frac{s+4}{2}) \exp( \frac{t}{16} \log^2\frac{s+4}{2\pi n^2} ).\quad (6.1)$
Reinstating the lower order term and applying (1.6), we have an alternate form
$F_{t,N}(s) \approx \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \sum_{n=1}^N \frac{\exp( \frac{t}{16} \log^2\frac{s+4}{2\pi n^2})}{n^s}.\quad (6.2)$
We can perform a similar analysis for $R_{t,N,M}$. Again, we drop the second term as being lower order. The $w$ integrand $w^{s-1} e^{-Nw}$ attains a maximum at $w = \frac{s}{N} \approx \sqrt{2\pi \mathrm{Im}(s)} i$ and the $u$ integrand $\exp( \frac{s+4}{2} u - e^u )$ attains a maximum at $u = \log \frac{s+4}{2} \approx \log \frac{\mathrm{Im}(s)}{2} + i \frac{\pi}{2}$, and hence
$\log \frac{w}{2\sqrt{\pi}} - \frac{u}{2} \approx i \pi/4$
and so we may heuristically obtain
$2 \sum_{n=0}^\infty \pi^{-s/2} \frac{e^{-i\pi s/2} e^{i \pi s n}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u + \frac{t\pi^2}{64} (4n-1) )\ dw\ du.$
Because $e^{i \pi sn}$ decays incredibly rapidly in $n$, the $n=0$ term should dominate, thus giving
$2 \pi^{-s/2} \frac{e^{-i\pi s/2}}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{-\infty}^\infty \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( \frac{5-s}{2} u - e^u - \frac{t\pi^2}{64} )\ dw\ du.$
The $u$ integral can be evaluated by (1.1) to obtain
$2 \pi^{-s/2} \frac{e^{-i\pi s/2} \Gamma(\frac{5-s}{2})}{2^{s}\pi^{1/2} \sin(\pi s/2)} \int_{C_M} \frac{w^{s-1} e^{-Nw}}{e^w-1} \exp( - \frac{t\pi^2}{64} )\ dw$
and so by comparison with (2.3) we have
$R_{t,N,M}(s) \approx \exp( - t \pi^2/64) R_{0,N,M}(s).$
In particular, from (4.1) we have
$R_{t,N,M}(s) \approx - \exp( - t \pi^2/64) \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \frac{e^{-i\pi s} \Gamma(1-s)}{2\pi i} (2\pi i M)^{s-1} \Psi(\frac{s}{2\pi i M}-N). \quad(6.3)$
Combining (6.2), (6.3), (5.1) we obtain an approximation to $\xi_t(s)$ and hence to $H_t(z) = \xi_t(\frac{1+iz}{2})$.
To understand these asymptotics better, let us inspect $H_t(x+iy)$ for $t\gt0$ in the region
$x+iy = T + \frac{a+ib}{\log T}; \quad t = \frac{\tau}{\log T}$
with $T$ large, $a,b = O(1)$, and $\tau \gt \frac{1}{2}$. If $s = \frac{1+ix-y}{2}$, then we can approximate
$\pi^{-s/2} \approx \pi^{-\frac{1+iT}{4}}$
$\Gamma(\frac{s+4}{2}) \approx \Gamma(\frac{9+iT}{2}) T^{\frac{ia-b}{4 \log T}} = \exp( \frac{ia-b}{4} ) \Gamma(\frac{9+iT}{2})$
$\frac{1}{n^s} \approx \frac{1}{n^{\frac{1+iT}{2}}}$
$\exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi n^2} ) \approx \exp( \frac{t}{16} \log^2 \frac{s+4}{2\pi} - \frac{t}{4} \log T \log n )$
$\approx \exp( \frac{\tau}{16} \log T + \frac{i \pi \tau}{16} ) \frac{1}{n^{\frac{\tau}{4}}}$
$F_{t,N}(\frac{1+ix-y}{2}) \approx 2\pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \exp( \frac{ia-b}{4} ) \exp( \frac{\tau}{16} \log T + \frac{i \pi \tau}{16} ) \sum_n \frac{1}{n^{\frac{1+iT}{2} + \frac{\tau}{4}}}$
$\approx 2\pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \zeta(\frac{1+iT}{2} + \frac{\tau}{4}) \exp( \frac{ia-b}{4} ).$
Similarly for $F_{t,N}(\frac{1+ix+y}{2})$ (replacing $b$ by $-b$). If we make a polar coordinate representation
$\frac{1}{2} \pi^{-\frac{1+iT}{4}} \Gamma(\frac{9+iT}{2}) \zeta(\frac{1+iT}{2} + \frac{\tau}{4}) = r_{T,\tau} e^{i \theta_{T,\tau}}$
one thus has
$H_t(x+iy) \approx \frac{1}{2} ( r_{T,\tau} e^{i \theta_{T,\tau}} \exp( \frac{ia-b}{4} ) + r_{T,\tau} e^{-i \theta_{T,\tau}} \exp(\frac{-ia+b}{4}) )$
$= r_{T,\tau} \cos( \frac{a+ib}{4} + \theta_{T,\tau} ).$
Thus locally $H_t(x+iy)$ behaves like a trigonometric function, with zeroes real and equally spaced with spacing $4\pi$ (in $a$-coordinates) or $\frac{4\pi}{\log T}$ (in $x$ coordinates). Once $\tau$ becomes large, further increase of $\tau$ basically only increases $r_{T,\tau}$ and also shifts $\theta_{T,\tau}$ at rate $\pi/16$, causing the number of zeroes to the left of $T$ to increase at rate $1/4$ as claimed in [KKL2009].
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# Solution - To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. - CBSE Class 10 - Mathematics
ConceptSection Formula
#### Question
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5
th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
#### Solution
It can be observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1×100/4) m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25). Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR
= sqrt((8-2)^2+(25-20)^2) = sqrt(36+25) = sqrt61m
The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (xy).
x =(2+8)/2, y =(25+20)/2
x = 10/2 = 5, y = 45/2 = 22.5
Hence A(x,y) = (5,22.5)
Therefore, Rashmi should post her blue flag at 22.5m on 5th line.
Is there an error in this question or solution?
#### APPEARS IN
NCERT Mathematics Textbook for Class 10 (with solutions)
Chapter 7: Coordinate Geometry
Q: 3 | Page no. 167
#### Reference Material
Solution for question: To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. concept: Section Formula. For the course CBSE
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# question about $TGV^2$ space
Let us just stay in $\mathbb R^1$. The space $TGV^k$ is defined as the function $u\in L^1(I)$ and $$TGV^k(u,I):=\sup\left\{\int_I u\,\phi^{(k)}\,d\mu, \,\phi\in C_c^\infty(I),\,\|\phi\|_{L^{\infty}(I)}\leq1,\,\|\phi'\|_{L^\infty}\leq 1,\ldots,\|\phi^{(k-1)}\|_{L^\infty}\leq 1\right\}<\infty$$ where by $\phi^{(k)}$ I mean the $k$-th derivative of $\phi$. This space suppose to generalize space $BV$ since as $k=1$ this is exactly $BV$, or $TV$, space.
Now let's assume $k=2$, i.e., we are in $TGV^2$ space. It is amazingly that we have $TGV^2$ and $BV$ is an equivalent space, i.e., $$c\|u\|_{BV}\leq \|u\|_{L^1}+TGV^2(u)\leq C\|u\|_{BV}$$ By $\|\cdot\|_{BV}$ I mean $\|u\|_{L^1}+|\mu|_{\mathcal M}$ where $\mu$ is the measure as the weak derivative of $u$.
The prove can be found here, section 3.
It is kind of an unexpected result since with one more derivative I would expect something new. But anyway, if we accept this result, then for any $u\in TGV^2(I)$, we have $u\in BV(I)$ and there will be a Radon measure $\mu$ such that $$\int_I u\,\varphi'dx = -\int_I \varphi\,d\mu$$ for any $\varphi\in C_c^\infty(I)$. Now if we go back to $TGV^2$, we could write $$\int_I u\,\phi''dx = -\int_I \phi'\,d\mu$$ Then what is next? Can I write $$\int_I u\,\phi'd\mu = -\int_I \phi\,d\nu??\tag 1$$ for some Radon measure $\nu$? I would expect some sort of IBP formula like $$\int_I u\,\phi''dx=\int_I \phi\,d\nu$$ to be true...
Also, the quantity $TGV^2$ I defined at the beginning, could it be explained as the total variation of a Radon measure? Like the one we used in $BV$ space? i.e., $TV(u)=|Du|$ if $u\in BV(\Omega)$. Also, some intuitive explanation of why, the $TGV^2$ norm with one more derivative, does not give any different then $BV$ norm would be really good.
Any help is really welcome!
PS: some discussion about $(1)$ can be found here.
My intuitive explanation that $TGV^2$ does lead to an equivalent norm on $BV$ is the following: You do not really have a higher derivative since setting $\psi = \phi^{(k-1)}$ shows that you really measure the pairing $\int u\psi' dx$ for $\|\psi\|_\infty\leq 1$. The "higher" derivatives are really lower derivatives: You only supremize the integral $\int u\psi' dx$ over some special bounded functions, namely ones that are themselves derivatives of bounded functions.
To get more intuition about the $TGV$ seminorm I suggest to look at the proof of the estimates $c\|u\|_{BV}\leq \|u\|_1 + TGV^2(u) \leq C\|u\|_{BV}$ and check how large the constants are, on what they depend (e.g. the size or shape of the domain?) and check the actual value of the norms for special cases.
There are recent papers on $TGV$ denoising in one dimension where actual minimizers are derived exactly:
Carefully going through the constructions and proofs should provide some further intuition.
The $TV$ seminorm is not a Radon measure. But it is the variational norm of a Radon measure, namely the distributional gradient of $u$ which is then (if $u\in BV$) a vector valued Radon measure, denoted by $Du$. Then it holds $TV(u) = |Du|(\Omega)$ where the right hand side means: Take the distributional derivative of $u$, interpret it as a vector valued Radon measure $Du$, then calculate its variation measure $|Du|$ and measure the whole set $\Omega$ with this very (now real valued) measure $|Du|$.
Since $\|u\|_1 + TGV^2(u)$ is an equivalent norm on $BV$ it follows that the if $TGV^2(u)<\infty$, then $u\in BV$ and hence, $Du$ is again a vector valued Radon measure and it pairs with continuous functions as $$\int \phi \mathrm{d} Du = \langle Du,\phi\rangle = -\langle u,\phi'\rangle = -\int u\phi'$$ as before. So $TGV^2(u) = \sup \int \phi^{(k-1)}\mathrm{d}Du$ where the supremum is taken over function $\phi$ with $\|\phi\|_\infty\leq 1$,...,$\|\phi^{(k-1)}\|_\infty\leq 1$. I don't know if you can get more explicit than that.
• Yea I know those two papers you mentioned. Thank you! I will double check again the constant in those equivalent inequalities. However, there are not similar explanation for $TGV^k$ as for $TV$ w.r.t. the "Radon measure" I mentioned. I really did a deep search via mathscinet but no luck. And this is why I post this problem here. Thank you again! – JumpJump Jul 17 '15 at 17:37
• But yes, I was wrong to say the semi-norm $TGV^k$, fixed already. – JumpJump Jul 17 '15 at 17:37
• Added something on the interpretation with Radon measures… – Dirk Jul 17 '15 at 19:03
• it looks to me if we require $\|\phi'\|_{\infty}\leq 1$ then we have $TV(u)=TGV^2(u)$. – JumpJump Jul 17 '15 at 19:17
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category theory
Contents
Definition
Definition
A full subcategory $i:C↪D$ is reflective if the inclusion functor $i$ has a left adjoint $T$:
$\left(T⊣i\right):C\stackrel{\stackrel{T}{←}}{↪}D\phantom{\rule{thinmathspace}{0ex}}.$(T \dashv i) : C \stackrel{\stackrel{T}{\leftarrow}}{\hookrightarrow} D \,.
The left adjoint is sometimes called the reflector, and a functor which is a reflector (or has a fully faithful right adjoint, which is the same up to equivalence) is called a reflection. Of course, there are dual notions of coreflective subcategory, coreflector, and coreflection.
Remark
A few sources (such as Categories Work) do not require a reflective subcategory to be full. However, in light of the fact that non-full subcategories are not invariant under equivalence, consideration of non-full reflective subcategories seems of limited usefulness. The general consensus among category theorists nowadays seems to be that “reflective subcategory” implies fullness.
Remark
The components of the unit
$\begin{array}{cccc}& ↗& {⇓}^{\eta }& {↘}^{\mathrm{Id}}\\ D& \stackrel{T}{\to }& C& ↪& D\end{array}$\array{ & \nearrow &\Downarrow^{\eta}& \searrow^{Id} \\ D &\stackrel{T}{\to}& C &\hookrightarrow & D }
of this adjunction “reflect” each object $d\in D$ into its image $Td$ in the reflective subcategory
${\eta }_{d}:d\to Td\phantom{\rule{thinmathspace}{0ex}}.$\eta_d : d \to T d \,.
This reflection is sometimes called a localization, although sometimes this term is reserved for the case when the functor $T$ is left exact.
Definition
If the reflector $T$ is faithful, the reflection is called a completion.
Characterizations
Proposition
Given any pair of adjoint functors
${Q}^{*}⊣{Q}_{*}:B\stackrel{\stackrel{{Q}^{*}}{←}}{\underset{{Q}_{*}}{\to }}A$Q^*\dashv Q_* : B \stackrel{\overset{Q^*}{\leftarrow}}{\underset{Q_*}{\to}} A
the following are equivalent:
1. The right adjoint ${Q}_{*}$ is fully faithful. (In this case $B$ is equivalent to its essential image in $A$ under ${Q}_{*}$, a reflective full subcategory of $A$.)
2. The counit $\epsilon :{Q}^{*}{Q}_{*}\to {1}_{A}$ of the adjunction is a natural isomorphism of functors.
3. The monad $\left({Q}^{*}{Q}_{*},{Q}^{*}\epsilon {Q}_{*},\eta \right)$ associated to the adjunction is idempotent.
4. If $S$ is the set of morphisms $s$ in $A$ such that ${Q}^{*}\left(s\right)$ is invertible in $B$, then ${Q}^{*}:A\to B$ realizes $B$ as the (nonstrict) localization of $A$ with respect to the class $S$.
This is due to Gabriel-Zisman.
This is a well-known set of equivalences concerning idempotent monads. The essential point is that a reflective subcategory $i:B\to A$ is monadic, i.e., realizes $B$ as the category of algebras for the monad $ir$ on $A$, where $r:A\to B$ is the reflector.
Special cases
Exact reflective subcategories
If the reflector (which as a left adjoint always preserves all colimits) in addition preserves finite limits, then the embedding is called exact . If the categories are toposes then such embeddings are called geometric embeddings.
In particular, every sheaf topos is an exact reflective subcategory of a category of presheaves
$\mathrm{Sh}\left(C\right)\stackrel{\stackrel{\mathrm{sheafify}}{←}}{↪}\mathrm{PSh}\left(C\right)\phantom{\rule{thinmathspace}{0ex}}.$Sh(C) \stackrel{\overset{sheafify}{\leftarrow}}{\hookrightarrow} PSh(C) \,.
The reflector in that case is the sheafification functor.
Theorem
If $X$ is a reflective subcategory of a cartesian closed category, then it is an exponential ideal if and only if its reflector $D\to C$ preserves finite products.
In particular, $C$ is then also cartesian closed.
This appears for instance as (Johnstone, A4.3.1).
So in particular if $C$ is an exact reflective subcategory of a cartesian closed category $D$, then $C$ is an exponential ideal of $D$.
See Day's reflection theorem for a more general statement and proof.
Complete reflective subcategories
When the unit of the reflector is a monomorphism, a reflective category is often thought of as a full subcategory of complete objects in some sense; the reflector takes each object in the ambient category to its completion. Such reflective subcategories are sometimes called mono-reflective. One similarly has epi-reflective (when the unit is an epimorphism) and bi-reflective (when the unit is a bimorphism).
In the last case, note that if the unit is an isomorphism, then the inclusion functor is an equivalence of categories, so nontrivial bireflective subcategories can occur only in non-balanced categories. Also note that ‘bireflective’ does not mean reflective and coreflective. One sees this term often in discussions of concrete categories (such as topological categories) where really something stronger holds: that the reflector lies over the identity functor on Set. In this case, one can say that we have a subcategory that is reflective over $\mathrm{Set}$.
Accessible reflective subcategories
Definition
A reflection
$𝒞\stackrel{\stackrel{L}{←}}{\underset{R}{↪}}𝒟$\mathcal{C} \stackrel{\overset{L}{\leftarrow}}{\underset{R}{\hookrightarrow}} \mathcal{D}
is called accessible if $𝒟$ is an accessible category and the reflector $R\circ L:𝒟\to 𝒟$ is an accessible functor.
Proposition
A reflective subcategory $𝒞↪𝒟$ of an accessible category is accessible, def. 3, precisely if $𝒞$ is an accessible category.
In this explicit form this appears as (Lurie, prop. 5.5.1.2). From (Adamek-Rosický) the “only if”-direction follows immediately from 2.53 there (saying that an accessibly embedded subcategory of an accessible category is accessible iff it is cone-reflective), while “if”-direction follows immediately from 2.23 (saying any left or right adjoint between accessible categories is accessible).
Properties
A reflective subcategory is always closed under limits which exist in the ambient category (because the full inclusion is monadic, as noted above), and inherits colimits from the larger category by application of the reflector.
A morphism in a reflective subcategory is monic iff it is monic in the ambient category. A reflective subcategory of a well-powered category is well-powered.
Reflective subcategories of locally presentable categories
Both the weak and strong versions of Vopěnka's principle are equivalent to fairly simple statements concerning reflective subcategories of locally presentable categories:
Theorem
The weak Vopěnka's principle is equivalent to the statement:
For $C$ a locally presentable category, every full subcategory $D↪C$ which is closed under limits is a reflective subcategory.
Theorem
The strong Vopěnka's principle is equivalent to:
For $C$ a locally presentable category, every full subcategory $D↪C$ which is closed under limits is a reflective subcategory; further on, $D$ is then also locally presentable
(Remark after corollary 6.24 in Adamek-Rosicky book).
Reflective subcategories of cartesian closed categories
In showing that a given category is cartesian closed, the following theorem is often useful (cf. A4.3.1 in the Elephant):
Theorem
If $C$ is cartesian closed, and $D\subseteq C$ is a reflective subcategory, then the reflector $L:C\to D$ preserves finite products if and only if $D$ is an exponential ideal (i.e. $Y\in D$ implies ${Y}^{X}\in D$ for any $X\in C$). In particular, if $L$ preserves finite products, then $D$ is cartesian closed.
Reflective and coreflective subcategories
Theorem
A subcategory of a category of presheaves $\left[{A}^{\mathrm{op}},\mathrm{Set}\right]$ which is both reflective and coreflective is itself a category of presheaves $\left[{B}^{\mathrm{op}},\mathrm{Set}\right]$, and the inclusion is induced by a functor $A\to B$.
This is shown in (BashirVelebil).
Property vs structure
Whenever $C$ is a full subcategory of $D$, we can say that objects of $C$ are objects of $D$ with some extra property. But if $C$ is reflective in $D$, then we can turn this around and (by thinking of the left adjoint as a forgetful functor) think of objects of $D$ as objects of $C$ with (if we're lucky) some extra structure or (in any case) some extra stuff.
This can always be made to work by brute force, but sometimes there is something insightful about it. For example, a metric space is a complete metric space equipped with a dense subset. Or, a possibly nonunital ring is a unital ring equipped with a unital homomorphism to the ring of integers.
Examples
Example
Complete metric spaces are mono-reflective in metric spaces; the reflector is called completion.
Example
The category of sheaves on a site $S$ is a reflective subcategory of the category of presheaves on $S$; the reflector is called sheafification. In fact, categories of sheaves are precisely those accessible reflective subcategories, def. 3, of presheaf categories for which the reflector is left exact. This makes the inclusion functor precisely a geometric inclusion? of toposes.
Example
A category of concrete presheaves inside a category of presheaves on a concrete site is a reflective subcategory.
(Counter)Example
The non-full inclusion of unital rings into non-unital rings has a left adjoint (with monic units), whose reflector formally adjoins an identity element. However, we do not call it a reflective subcategory, because the “inclusion” is not full; see remark 1.
Remark
Notice that for $R\in \mathrm{Ring}$ a ring with unit, its reflection $LR$ in the above example is not in general isomorphic to $R$, but is much larger. But an object in a reflective subcategory is necessarily isomorphic to its image under the reflector only if the reflective subcategory is full. While the inclusion $\mathrm{Ring}↪\mathrm{Ring}$’ does have a left adjoint (as any forgetful functor between varieties of algebras, by the adjoint lifting theorem), this inclusion is not full (an arrow in $\mathrm{Ring}$’ need not preserve the identity).
References
The relation of exponential ideals to reflective subcategories is discussed in section A4.3.1 of
Reflective and coreflective subcategories of presheaf categories are discussed in
• R. Bashir, J. Velebil, Simultaneously reflective and coreflective subcategories of presheaves, Theory and Applications of Categories, Vol 10. No. 16. (2002) (pdf).
Related discussion of reflective sub-(∞,1)-categories is in
Revised on January 1, 2013 21:26:16 by Urs Schreiber (82.113.99.56)
|
# nLab Cite — Cauchy integral theorem
### Overview
We recommend the following .bib file entries for citing the current version of the page Cauchy integral theorem. The first is to be used if one does not have unicode support, which is likely the case if one is using bibtex. The second can be used if one does has unicode support. If there are no non-ascii characters in the page name, then the two entries are the same.
In either case, the hyperref package needs to have been imported in one's tex (or sty) file. There are no other dependencies.
The author field has been chosen so that the reference appears in the 'alpha' citation style. Feel free to adjust this.
### Bib entry — Ascii
@misc{nlab:cauchy_integral_theorem,
author = {{nLab authors}},
title = {{{C}}auchy integral theorem},
howpublished = {\url{http://ncatlab.org/nlab/show/Cauchy%20integral%20theorem}},
note = {\href{http://ncatlab.org/nlab/revision/Cauchy%20integral%20theorem/9}{Revision 9}},
month = oct,
year = 2021
}
### Bib entry — Unicode
@misc{nlab:cauchy_integral_theorem,
author = {{nLab authors}},
title = {{{C}}auchy integral theorem},
howpublished = {\url{http://ncatlab.org/nlab/show/Cauchy%20integral%20theorem}},
note = {\href{http://ncatlab.org/nlab/revision/Cauchy%20integral%20theorem/9}{Revision 9}},
month = oct,
year = 2021
}
### Problems?
Please report any problems with the .bib entries at the nForum.
|
# Y. Iwata
Search this author in Google Scholar
Articles: 2
### Abstract formulation of the Cole-Hopf transform
Yoritaka Iwata
MFAT 25 (2019), no. 2, 142-151
142-151
Operator representation of Cole-Hopf transform is obtained based on the logarithmic representation of infinitesimal generators. For this purpose the relativistic formulation of abstract evolution equation is introduced. Even independent of the spatial dimension, the Cole-Hopf transform is generalized to a transform between linear and nonlinear equations defined in Banach spaces. In conclusion a role of transform between the evolution operator and its infinitesimal generator is understood in the context of generating nonlinear semigroup.
### Infinitesimal generators of invertible evolution families
Yoritaka Iwata
MFAT 23 (2017), no. 1, 26-36
26-36
A logarithm representation of operators is introduced as well as a concept of pre-infinitesimal generator. Generators of invertible evolution families are represented by the logarithm representation, and a set of operators represented by the logarithm is shown to be associated with analytic semigroups. Consequently generally-unbounded infinitesimal generators of invertible evolution families are characterized by a convergent power series representation.
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26 Deprecated
26.1 Conditions
1. Q: What does options(error = recover) do? Why might you use it?
A: In case of options(error = recover) utils::recover() will be called (without arguments) in case of an error. This will print out a list of calls which precede the error and lets the user choose to incorporate browser() directly in any of the regarding environments allowing a practical mode for debugging.
2. Q: What does options(error = quote(dump.frames(to.file = TRUE))) do? Why might you use it?
A: This option writes a dump of the evaluation environment where an error occurs into a file ending on .rda. When this option is set, R will continue to run after the first error. To stop R at the first error use quote({dump.frames(to.file=TRUE); q()}). These options are especially useful for debugging non-interactive R scripts afterwards (“post mortem debugging”).
26.2 Expressions (new)
1. Q: base::alist() is useful for creating pairlists to be used for function arguments:
foo <- function() {}
formals(foo) <- alist(x = , y = 1)
foo
#> function (x, y = 1)
#> {
#> }
What makes alist() special compared to list()?
A: From ?alist:
alist handles its arguments as if they described function arguments. So the values are not evaluated, and tagged arguments with no value are allowed whereas list simply ignores them. alist is most often used in conjunction with formals.
26.3 Functionals
26.3.1 My first functional: lapply()
1. Q: Why are the following two invocations of lapply() equivalent?
trims <- c(0, 0.1, 0.2, 0.5)
x <- rcauchy(100)
lapply(trims, function(trim) mean(x, trim = trim))
lapply(trims, mean, x = x)
A: In the first statement each element of trims is explicitly supplied to mean()’s second argument. In the latter statement this happens via positional matching, since mean()’s first argument is supplied via name in lapply()’s third argument (...).
2. Q: The function below scales a vector so it falls in the range [0, 1]. How would you apply it to every column of a data frame? How would you apply it to every numeric column in a data frame?
scale01 <- function(x) {
rng <- range(x, na.rm = TRUE)
(x - rng[1]) / (rng[2] - rng[1])
}
A: Since this function needs numeric input, one can check this via an if clause. If one also wants to return non-numeric input columns, these can be supplied to the else argument of the if() “function”:
data.frame(lapply(iris, function(x) if (is.numeric(x)) scale01(x) else x))
3. Q: Use both for loops and lapply() to fit linear models to the mtcars using the formulas stored in this list:
formulas <- list(
mpg ~ disp,
mpg ~ I(1 / disp),
mpg ~ disp + wt,
mpg ~ I(1 / disp) + wt
)
A: Like in the first exercise, we can create two lapply() versions:
# lapply (2 versions)
la1 <- lapply(formulas, lm, data = mtcars)
la2 <- lapply(formulas, function(x) lm(formula = x, data = mtcars))
# for loop
lf1 <- vector("list", length(formulas))
for (i in seq_along(formulas)){
lf1[[i]] <- lm(formulas[[i]], data = mtcars)
}
Note that all versions return the same content, but they won’t be identical, since the values of the “call” element will differ between each version.
4. Q: Fit the model mpg ~ disp to each of the bootstrap replicates of mtcars in the list below by using a for loop and lapply(). Can you do it without an anonymous function?
bootstraps <- lapply(1:10, function(i) {
rows <- sample(1:nrow(mtcars), rep = TRUE)
mtcars[rows, ]
})
A:
# lapply without anonymous function
la <- lapply(bootstraps, lm, formula = mpg ~ disp)
# for loop
lf <- vector("list", length(bootstraps))
for (i in seq_along(bootstraps)){
lf[[i]] <- lm(mpg ~ disp, data = bootstraps[[i]])
}
5. Q: For each model in the previous two exercises, extract $$R^2$$ using the function below.
rsq <- function(mod) summary(mod)$r.squared A: For the models in exercise 3: sapply(la1, rsq) #> [1] 0.718 0.860 0.781 0.884 sapply(la2, rsq) #> [1] 0.718 0.860 0.781 0.884 sapply(lf1, rsq) #> [1] 0.718 0.860 0.781 0.884 And the models in exercise 4: sapply(la, rsq) #> [1] 0.628 0.656 0.668 0.668 0.677 0.753 0.683 0.810 0.668 0.730 sapply(lf, rsq) #> [1] 0.628 0.656 0.668 0.668 0.677 0.753 0.683 0.810 0.668 0.730 26.3.2 For loops functionals: friends of lapply(): 1. Q: Use vapply() to: 1. Compute the standard deviation of every column in a numeric data frame. 2. Compute the standard deviation of every numeric column in a mixed data frame. (Hint: you’ll need to use vapply() twice.) A: As a numeric data.frame we choose cars: vapply(cars, sd, numeric(1)) And as a mixed data.frame we choose iris: vapply(iris[vapply(iris, is.numeric, logical(1))], sd, numeric(1)) 2. Q: Why is using sapply() to get the class() of each element in a data frame dangerous? A: Columns of data.frames might have more than one class, so the class of sapply()’s output may differ from time to time (silently). If … • all columns have one class: sapply() returns a character vector • one column has more classes than the others: sapply() returns a list • all columns have the same number of classes, which is more than one: sapply() returns a matrix For example: a <- letters[1:3] class(a) <- c("class1", "class2") df <- data.frame(a = character(3)) df$a <- a
df$b <- a class(sapply(df, class)) #> [1] "matrix" Note that this case often appears, wile working with the POSIXt types, POSIXct and POSIXlt. 3. Q: The following code simulates the performance of a t-test for non-normal data. Use sapply() and an anonymous function to extract the p-value from every trial. trials <- replicate( 100, t.test(rpois(10, 10), rpois(7, 10)), simplify = FALSE ) Extra challenge: get rid of the anonymous function by using [[ directly. A: # anonymous function: sapply(trials, function(x) x[["p.value"]]) # without anonymous function: sapply(trials, "[[", "p.value") 4. Q: What does replicate() do? What sort of for loop does it eliminate? Why do its arguments differ from lapply() and friends? A: As stated in ?replicate: replicate is a wrapper for the common use of sapply for repeated evaluation of an expression (which will usually involve random number generation). We can see this clearly in the source code: #> function (n, expr, simplify = "array") #> sapply(integer(n), eval.parent(substitute(function(...) expr)), #> simplify = simplify) #> <bytecode: 0x52e31f8> #> <environment: namespace:base> Like sapply() replicate() eliminates a for loop. As explained for Map() in the textbook, also every replicate() could have been written via lapply(). But using replicate() is more concise, and more clearly indicates what you’re trying to do. 5. Q: Implement a version of lapply() that supplies FUN with both the name and the value of each component. A: lapply_nms <- function(X, FUN, ...){ Map(FUN, X, names(X), ...) } lapply_nms(iris, function(x, y) c(class(x), y)) #>$Sepal.Length
#> [1] "numeric" "Sepal.Length"
#>
#> $Sepal.Width #> [1] "numeric" "Sepal.Width" #> #>$Petal.Length
#> [1] "numeric" "Petal.Length"
#>
#> $Petal.Width #> [1] "numeric" "Petal.Width" #> #>$Species
#> [1] "factor" "Species"
6. Q: Implement a combination of Map() and vapply() to create an lapply() variant that iterates in parallel over all of its inputs and stores its outputs in a vector (or a matrix). What arguments should the function take?
A As we understand this exercise, it is about working with a list of lists, like in the following example:
testlist <- list(iris, mtcars, cars)
lapply(testlist, function(x) vapply(x, mean, numeric(1)))
#> Warning in mean.default(X[[i]], ...): argument is not numeric or logical:
#> returning NA
#> [[1]]
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 5.84 3.06 3.76 1.20 NA
#>
#> [[2]]
#> mpg cyl disp hp drat wt qsec vs am
#> 20.091 6.188 230.722 146.688 3.597 3.217 17.849 0.438 0.406
#> gear carb
#> 3.688 2.812
#>
#> [[3]]
#> speed dist
#> 15.4 43.0
So we can get the same result with a more specialized function:
lmapply <- function(X, FUN, FUN.VALUE, simplify = FALSE){
out <- Map(function(x) vapply(x, FUN, FUN.VALUE), X)
if(simplify == TRUE){return(simplify2array(out))}
out
}
lmapply(testlist, mean, numeric(1))
#> Warning in mean.default(X[[i]], ...): argument is not numeric or logical:
#> returning NA
#> [[1]]
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 5.84 3.06 3.76 1.20 NA
#>
#> [[2]]
#> mpg cyl disp hp drat wt qsec vs am
#> 20.091 6.188 230.722 146.688 3.597 3.217 17.849 0.438 0.406
#> gear carb
#> 3.688 2.812
#>
#> [[3]]
#> speed dist
#> 15.4 43.0
7. Q: Implement mcsapply(), a multi-core version of sapply(). Can you implement mcvapply(), a parallel version of vapply()? Why or why not?
26.3.3 Manipulating matrices and data frames
1. Q: How does apply() arrange the output? Read the documentation and perform some experiments.
A:
apply() arranges its output columns (or list elements) according to the order of the margin. The rows are ordered by the other dimensions, starting with the “last” dimension of the input object. What this means should become clear by looking at the three and four dimensional cases of the following example:
# for two dimensional cases everything is sorted by the other dimension
arr2 <- array(1:9, dim = c(3, 3), dimnames = list(paste0("row", 1:3),
paste0("col", 1:3)))
arr2
apply(arr2, 1, head, 1) # Margin is row
apply(arr2, 1, head, 9) # sorts by col
apply(arr2, 2, head, 1) # Margin is col
apply(arr2, 2, head, 9) # sorts by row
# 3 dimensional
arr3 <- array(1:27, dim = c(3,3,3), dimnames = list(paste0("row", 1:3),
paste0("col", 1:3),
paste0("time", 1:3)))
arr3
apply(arr3, 1, head, 1) # Margin is row
apply(arr3, 1, head, 27) # sorts by time and col
apply(arr3, 2, head, 1) # Margin is col
apply(arr3, 2, head, 27) # sorts by time and row
apply(arr3, 3, head, 1) # Margin is time
apply(arr3, 3, head, 27) # sorts by col and row
# 4 dimensional
arr4 <- array(1:81, dim = c(3,3,3,3), dimnames = list(paste0("row", 1:3),
paste0("col", 1:3),
paste0("time", 1:3),
paste0("var", 1:3)))
arr4
apply(arr4, 1, head, 1) # Margin is row
apply(arr4, 1, head, 81) # sorts by var, time, col
apply(arr4, 2, head, 1) # Margin is col
apply(arr4, 2, head, 81) # sorts by var, time, row
apply(arr4, 3, head, 1) # Margin is time
apply(arr4, 3, head, 81) # sorts by var, col, row
apply(arr4, 4, head, 1) # Margin is var
apply(arr4, 4, head, 81) # sorts by time, col, row
2. Q: There’s no equivalent to split() + vapply(). Should there be? When would it be useful? Implement one yourself.
A: We can modify the tapply2() approach from the book, where split() and sapply() were combined:
v_tapply <- function(x, group, f, FUN.VALUE, ..., USE.NAMES = TRUE) {
pieces <- split(x, group)
vapply(pieces, f, FUN.VALUE, ..., USE.NAMES = TRUE)
}
tapply() has a SIMPLIFY argument. When you set it to FALSE, tapply() will always return a list. It is easy to create cases where the length and the types/classes of the list elements vary depending on the input. The vapply() version could be useful, if you want to control the structure of the output to get an error according to some logic of a specific usecase or you want typestable output to build up other functions on top of it.
3. Q: Implement a pure R version of split(). (Hint: use unique() and subsetting.) Can you do it without a for loop?
A:
split2 <- function(x, f, drop = FALSE, ...){
# there are three relevant cases for f. f is a character, f is a factor and all
# levels occur, f is a factor and some levels don't occur.
# first we check if f is a factor
fact <- is.factor(f)
# if drop it set to TRUE, we drop the non occuring levels.
# (If f is a character, this has no effect.)
if(drop){f <- f[, drop = TRUE]}
# now we want all unique elements/levels of f
levs <- if (fact) {unique(levels(f))} else {as.character(unique(f))}
# we use these levels to subset x and supply names for the resulting output.
setNames(lapply(levs, function(lv) x[f == lv, , drop = FALSE]), levs)
}
4. Q: What other types of input and output are missing? Brainstorm before you look up some answers in the plyr paper.
A: From the suggested plyr paper, we can extract a lot of possible combinations and list them up on a table. Sean C. Anderson already has done this based on a presentation from Hadley Wickham and provided the following result here.
object type array data frame list nothing
array apply . . .
data frame . aggregate by .
list sapply . lapply .
n replicates replicate . replicate .
function arguments mapply . mapply .
Note the column nothing, which is specifically for usecases, where sideeffects like plotting or writing data are intended.
26.3.4 Manipulating lists
1. Q: Why isn’t is.na() a predicate function? What base R function is closest to being a predicate version of is.na()?
A: Because a predicate function always returns TRUE or FALSE. is.na(NULL) returns logical(0), which excludes it from being a predicate function. The closest in base that we are aware of is anyNA(), if one applies it elementwise.
2. Q: Use Filter() and vapply() to create a function that applies a summary statistic to every numeric column in a data frame.
A:
vapply_num <- function(X, FUN, FUN.VALUE){
vapply(Filter(is.numeric, X), FUN, FUN.VALUE)
}
3. Q: What’s the relationship between which() and Position()? What’s the relationship between where() and Filter()?
A: which() returns all indices of true entries from a logical vector. Position() returns just the first (default) or the last integer index of all true entries that occur by applying a predicate function on a vector. So the default relation is Position(f, x) <=> min(which(f(x))).
where(), defined in the book as:
where <- function(f, x) {
vapply(x, f, logical(1))
}
is useful to return a logical vector from a condition asked on elements of a list or a data frame. Filter(f, x) returns all elements of a list or a data frame, where the supplied predicate function returns TRUE. So the relation is Filter(f, x) <=> x[where(f, x)].
4. Q: Implement Any(), a function that takes a list and a predicate function, and returns TRUE if the predicate function returns TRUE for any of the inputs. Implement All() similarly.
A: Any():
Any <- function(l, pred){
stopifnot(is.list(l))
for (i in seq_along(l)){
if (pred(l[[i]])) return(TRUE)
}
return(FALSE)
}
All():
All <- function(l, pred){
stopifnot(is.list(l))
for (i in seq_along(l)){
if (!pred(l[[i]])) return(FALSE)
}
return(TRUE)
}
5. Q: Implement the span() function from Haskell: given a list x and a predicate function f, span returns the location of the longest sequential run of elements where the predicate is true. (Hint: you might find rle() helpful.)
A: Our span_r() function returns the first index of the longest sequential run of elements where the predicate is true. In case of more than one longest sequenital, more than one first_index is returned.
span_r <- function(l, pred){
# We test if l is a list
stopifnot(is.list(l))
# we preallocate a logical vector and save the result
# of the predicate function applied to each element of the list
test <- vector("logical", length(l))
for (i in seq_along(l)){
test[i] <- (pred(l[[i]]))
}
# we return NA, if the output of pred is always FALSE
if(!any(test)) return(NA_integer_)
# Otherwise we look at the length encoding of TRUE and FALSE values.
rle_test <- rle(test)
# Since it might happen, that more than one maximum series of TRUE's appears,
# we have to implement some logic, which might be easier, if we save the rle
# output in a data.frmame
rle_test <- data.frame(lengths = rle_test[["lengths"]],
values = rle_test[["values"]],
cumsum = cumsum(rle_test[["lengths"]]))
rle_test[["first_index"]] <- rle_test[["cumsum"]] - rle_test[["lengths"]] + 1
# In the last line we calculated the first index in the original list for every encoding
# In the next line we calculate a column, which gives the maximum
# encoding length among all encodings with the value TRUE
rle_test[["max"]] <- max(rle_test[rle_test[, "values"] == TRUE, ][,"lengths"])
# Now we just have to subset for maximum length among all TRUE values and return the
# according "first index":
rle_test[rle_test$lengths == rle_test$max & rle_test$values == TRUE, ]$first_index
}
26.3.5 List of functions
1. Q: Implement a summary function that works like base::summary(), but uses a list of functions. Modify the function so it returns a closure, making it possible to use it as a function factory.
2. Q: Which of the following commands is equivalent to with(x, f(z))?
1. x$f(x$z).
2. f(x$z). 3. x$f(z).
4. f(z).
5. It depends.
26.3.6 Mathematical functionals
1. Q: Implement arg_max(). It should take a function and a vector of inputs, and return the elements of the input where the function returns the highest value. For example, arg_max(-10:5, function(x) x ^ 2) should return -10. arg_max(-5:5, function(x) x ^ 2) should return c(-5, 5). Also implement the matching arg_min() function.
A: arg_max():
arg_max <- function(x, f){
x[f(x) == max(f(x))]
}
arg_min():
arg_min <- function(x, f){
x[f(x) == min(f(x))]
}
2. Q: Challenge: read about the fixed point algorithm. Complete the exercises using R.
26.3.7 A family of functions
1. Q: Implement smaller and larger functions that, given two inputs, return either the smaller or the larger value. Implement na.rm = TRUE: what should the identity be? (Hint: smaller(x, smaller(NA, NA, na.rm = TRUE), na.rm = TRUE) must be x, so smaller(NA, NA, na.rm = TRUE) must be bigger than any other value of x.) Use smaller and larger to implement equivalents of min(), max(), pmin(), pmax(), and new functions row_min() and row_max().
A: We can do almost everything as shown in the case study in the textbook. First we define the functions smaller_() and larger_(). We use the underscore suffix, to built up non suffixed versions on top, which will include the na.rm parameter. In contrast to the add() example from the book, we change two things at this step. We won’t include errorchecking, since this is done later at the top level and we return NA_integer_ if any of the arguments is NA (this is important, if na.rm is set to FALSE and wasn’t needed by the add() example, since + already returns NA in this case.)
smaller_ <- function(x, y){
if(anyNA(c(x, y))){return(NA_integer_)}
out <- x
if(y < x) {out <- y}
out
}
larger_ <- function(x, y){
if(anyNA(c(x, y))){return(NA_integer_)}
out <- x
if(y > x) {out <- y}
out
}
We can take na.rm() from the book:
rm_na <- function(x, y, identity) {
if (is.na(x) && is.na(y)) {
identity
} else if (is.na(x)) {
y
} else {
x
}
}
To find the identity value, we can apply the same argument as in the textbook, hence our functions are also associative and the following equation should hold:
3 = smaller(smaller(3, NA), NA) = smaller(3, smaller(NA, NA)) = 3
So the identidy has to be greater than 3. When we generalize from 3 to any real number this means that the identity has to be greater than any number, which leads us to infinity. Hence identity has to be Inf for smaller() (and -Inf for larger()), which we implement next:
smaller <- function(x, y, na.rm = FALSE) {
stopifnot(length(x) == 1, length(y) == 1, is.numeric(x) | is.logical(x),
is.numeric(y) | is.logical(y))
if (na.rm && (is.na(x) || is.na(y))) rm_na(x, y, Inf) else smaller_(x,y)
}
larger <- function(x, y, na.rm = FALSE) {
stopifnot(length(x) == 1, length(y) == 1, is.numeric(x) | is.logical(x),
is.numeric(y) | is.logical(y))
if (na.rm && (is.na(x) || is.na(y))) rm_na(x, y, -Inf) else larger_(x,y)
}
Like min() and max() can act on vectors, we can implement this easyly for our new functions. As shown in the book, we also have to set the init parameter to the identity value.
r_smaller <- function(xs, na.rm = TRUE) {
Reduce(function(x, y) smaller(x, y, na.rm = na.rm), xs, init = Inf)
}
# some tests
r_smaller(c(1:3, 4:(-1)))
#> [1] -1
r_smaller(NA, na.rm = TRUE)
#> [1] Inf
r_smaller(numeric())
#> [1] Inf
r_larger <- function(xs, na.rm = TRUE) {
Reduce(function(x, y) larger(x, y, na.rm = na.rm), xs, init = -Inf)
}
# some tests
r_larger(c(1:3), c(4:1))
#> [1] 3
r_larger(NA, na.rm = TRUE)
#> [1] -Inf
r_larger(numeric())
#> [1] -Inf
We can also create vectorised versions as shown in the book. We will just show the smaller() case to become not too verbose.
v_smaller1 <- function(x, y, na.rm = FALSE){
stopifnot(length(x) == length(y), is.numeric(x) | is.logical(x),
is.numeric(y)| is.logical(x))
if (length(x) == 0) return(numeric())
simplify2array(
Map(function(x, y) smaller(x, y, na.rm = na.rm), x, y)
)
}
v_smaller2 <- function(x, y, na.rm = FALSE) {
stopifnot(length(x) == length(y), is.numeric(x) | is.logical(x),
is.numeric(y)| is.logical(x))
vapply(seq_along(x), function(i) smaller(x[i], y[i], na.rm = na.rm),
numeric(1))
}
# Both versions give the same results
v_smaller1(1:10, c(2,1,4,3,6,5,8,7,10,9))
#> [1] 1 1 3 3 5 5 7 7 9 9
v_smaller2(1:10, c(2,1,4,3,6,5,8,7,10,9))
#> [1] 1 1 3 3 5 5 7 7 9 9
v_smaller1(numeric(), numeric())
#> numeric(0)
v_smaller2(numeric(), numeric())
#> numeric(0)
v_smaller1(c(1, NA), c(1, NA), na.rm = FALSE)
#> [1] 1 NA
v_smaller2(c(1, NA), c(1, NA), na.rm = FALSE)
#> [1] 1 NA
v_smaller1(NA,NA)
#> [1] NA
v_smaller2(NA,NA)
#> [1] NA
Of course, we are also able to copy paste the rest from the textbook, to solve the last part of the exercise:
row_min <- function(x, na.rm = FALSE) {
apply(x, 1, r_smaller, na.rm = na.rm)
}
col_min <- function(x, na.rm = FALSE) {
apply(x, 2, r_smaller, na.rm = na.rm)
}
arr_min <- function(x, dim, na.rm = FALSE) {
apply(x, dim, r_smaller, na.rm = na.rm)
}
2. Q: Create a table that has and, or, add, multiply, smaller, and larger in the columns and binary operator, reducing variant, vectorised variant, and array variants in the rows.
1. Fill in the cells with the names of base R functions that perform each of the roles.
2. Compare the names and arguments of the existing R functions. How consistent are they? How could you improve them?
3. Complete the matrix by implementing any missing functions.
A In the following table we can see the requested base R functions, that we are aware of:
and or add multiply smaller larger
binary && ||
reducing all any sum prod min max
vectorised & | + * pmin pmax
array
Notice that we were relatively strict about the binary row. Since the vectorised and reducing versions are more general, then the binary versions, we could have used them twice. However, this doesn’t seem to be the intention of this exercise.
The last part of this exercise can be solved via copy pasting from the book and the last exercise for the binary row and creating combinations of apply() and the reducing versions for the array row. We think the array functions just need a dimension and an rm.na argument. We don’t know how we would name them, but sth. like sum_array(1, na.rm = TRUE) could be ok.
The second part of the exercise is hard to solve complete. But in our opinion, there are two important parts. The behaviour for special inputs like NA, NaN, NULL and zero length atomics should be consistent and all versions should have a rm.na argument, for which the functions also behave consistent. In the follwing table, we return the output of f(x, 1), where f is the function in the first column and x is the special input in the header (the named functions also have an rm.na argument, which is FALSE by default). The order of the arguments is important, because of lazy evaluation.
NA NaN NULL logical(0) integer(0)
&& NA NA error NA NA
all NA NA TRUE TRUE TRUE
& NA NA error logical(0) logical(0)
|| TRUE TRUE error TRUE TRUE
any TRUE TRUE TRUE TRUE TRUE
| TRUE TRUE error logical(0) logical(0)
sum NA NaN 1 1 1
+ NA NaN numeric(0) numeric(0) numeric(0)
prod NA NaN 1 1 1
* NA NaN numeric(0) numeric(0) numeric(0)
min NA NaN 1 1 1
pmin NA NaN numeric(0) numeric(0) numeric(0)
max NA NaN 1 1 1
pmax NA NaN numeric(0) numeric(0) numeric(0)
We can see, that the vectorised and reduced numerical functions are all consistent. However it is not, that the first three logical functions return NA for NA and NaN, while the 4th till 6th function all return TRUE. Then FALSE would be more consistent for the first three or the return of NA for all and an extra na.rm argument. In seems relatively hard to find an easy rule for all cases and especially the different behaviour for NULL is relatively confusing. Another good opportunity for sorting the functions would be to differentiate between “numerical” and “logical” operators first and then between binary, reduced and vectorised, like below (we left the last colum, which is redundant, because of coercion, as intended):
f(x,1) NA NaN NULL logical(0)
&& NA NA error NA
|| TRUE TRUE error TRUE
all NA NA TRUE TRUE
any TRUE TRUE TRUE TRUE
& NA NA error logical(0)
| TRUE TRUE error logical(0)
sum NA NaN 1 1
prod NA NaN 1 1
min NA NaN 1 1
max NA NaN 1 1
+ NA NaN numeric(0) numeric(0)
* NA NaN numeric(0) numeric(0)
pmin NA NaN numeric(0) numeric(0)
pmax NA NaN numeric(0) numeric(0)
The other point are the naming conventions. We think they are clear, but it could be useful to provide the missing binary operators and name them for example ++, **, <>, >< to be consistent.
3. Q: How does paste() fit into this structure? What is the scalar binary function that underlies paste()? What are the sep and collapse arguments to paste() equivalent to? Are there any paste variants that don’t have existing R implementations?
A paste() behaves like a mix. If you supply only length one arguments, it will behave like a reducing function, i.e. :
paste("a", "b", sep = "")
#> [1] "ab"
paste("a", "b","", sep = "")
#> [1] "ab"
If you supply at least one element with length greater then one, it behaves like a vectorised function, i.e. :
paste(1:3)
#> [1] "1" "2" "3"
paste(1:3, 1:2)
#> [1] "1 1" "2 2" "3 1"
paste(1:3, 1:2, 1)
#> [1] "1 1 1" "2 2 1" "3 1 1"
We think it should be possible to implement a new paste() starting from
p_binary <- function(x, y = "") {
stopifnot(length(x) == 1, length(y) == 1)
paste0(x,y)
}
The sep argument is equivalent to bind sep on every ... input supplied to paste(), but the last and then bind these results together. In relations:
paste(n1, n2, ...,nm , sep = sep) <=>
paste0(paste0(n1, sep), paste(n2, n3, ..., nm, sep = sep)) <=>
paste0(paste0(n1, sep), paste0(n2, sep), ..., paste0(nn, sep), paste0(nm))
We can check this for scalar and non scalar input
# scalar:
paste("a", "b", "c", sep = "_")
#> [1] "a_b_c"
paste0(paste0("a", "_"), paste("b", "c", sep = "_"))
#> [1] "a_b_c"
paste0(paste0("a", "_"), paste0("b", "_"), paste0("c"))
#> [1] "a_b_c"
# non scalar
paste(1:2, "b", "c", sep = "_")
#> [1] "1_b_c" "2_b_c"
paste0(paste0(1:2, "_"), paste("b", "c", sep = "_"))
#> [1] "1_b_c" "2_b_c"
paste0(paste0(1:2, "_"), paste0("b", "_"), paste0("c"))
#> [1] "1_b_c" "2_b_c"
collapse just binds the outputs for non scalar input together with the collapse input. In relations:
for input A1, ..., An, where Ai = a1i:ami,
paste(A1 , A2 , ..., An, collapse = collapse)
<=>
paste0(
paste0(paste( a11, a12, ..., a1n), collapse),
paste0(paste( a21, a22, ..., a2n), collapse),
.................................................
paste0(paste(am-11, am-12, ..., am-1n), collapse),
paste( am1, am2, ..., amn)
)
One can see this easily by intuition from examples:
paste(1:5, 1:5, 6, sep = "", collapse = "_x_")
#> [1] "116_x_226_x_336_x_446_x_556"
paste(1,2,3,4, collapse = "_x_")
#> [1] "1 2 3 4"
paste(1:2,1:2,2:3,3:4, collapse = "_x_")
#> [1] "1 1 2 3_x_2 2 3 4"
We think the only paste version that is not implemented in base R is an array version. At least we are not aware of sth. like row_paste or paste_apply etc.
26.4 S3
1. Q: The most important S3 objects in base R are factors, data frames, difftimes, and date/times (Dates, POSIXct, POSIXlt). You’ve already seen the attributes and base type that factors are built on. What base types and attributes are the others built on?
data frame: Data frames are build up on (named) lists. Together with the row.names attribute and after setting the class to “data.frame”, we get a classical data frame
df_build <- structure(list(1:2, 3:4),
names = c("a", "b"),
row.names = 1:2,
class = "data.frame")
df_classic <- data.frame(a = 1:2, b = 3:4)
identical(df_build, df_classic)
#> [1] TRUE
date/times (Dates, POSIXct, POSIXlt): Date is just a double with the class attribute set to “Date”
date_build <- structure(0, class = "Date")
date_classic <- as.Date("1970-01-01")
identical(date_build, date_classic)
#> [1] TRUE
POSIXct is a class for date/times that inherits from POSIXt and is built on doubles as well. The only attribute is tz (for timezone)
POSIXct_build <- structure(1, class = c("POSIXct", "POSIXt"), tzone = "CET")
POSIXct_classic <- .POSIXct(1, tz = "CET") # note that tz's default is NULL
identical(POSIXct_build, POSIXct_classic)
#> [1] TRUE
POSIXlt is another date/time class that inherits from POSIXt. It is built on top of a named list and a tzone attribute. Differences between POSIXct and POSIXlt are described in ?DateTimeClasses.
POSIXlt_build <- structure(list(sec = 30,
min = 30L,
hour = 14L,
mday = 1L,
mon = 0L,
year = 70L,
wday = 4L,
yday = 0L,
isdst = 0L,
zone = "CET",
gmtoff = 3600L),
tzone = c("", "CET", "CEST"),
class = c("POSIXlt", "POSIXt"))
POSIXlt_classic <- as.POSIXlt(.POSIXct(13.5 * 3600 + 30))
identical(POSIXlt_build, POSIXlt_classic)
#> [1] FALSE
2. Q: Draw a Venn diagram illustrating the relationships between functions, generics, and methods.
A: Funtions don’t have to be generics or methods, but both the latter are functions. It is also possible that a function is both, a method and a generic, at the same time, which seems to be relatively awkward, so that also the author of the textbook doesn’t recommend it, see ?pryr::ftype
This function figures out whether the input function is a regular/primitive/internal function, a internal/S3/S4 generic, or a S3/S4/RC method. This is function is slightly simplified as it’s possible for a method from one class to be a generic for another class, but that seems like such a bad idea that hopefully no one has done it.
3. Q: Write a constructor for difftime objects. What base type are they built on? What attributes do they use? You’ll need to consult the documentation, read some code, and perform some experiments.
A: Our constructor should be named new_class_name, have one argument for its base type and each attribute and check the base types of these arguments as well.
new_difftime <- function(x, units = "auto") {
stopifnot(is.double(x), is.character(units))
structure(x, units = units, class = "difftime")
}
However, since the following result prints awkward
new_difftime(3)
#> Time difference of 3 auto
we get a little bit more “inspiration” by the original difftime() function and make the regarding changes. Basically we need to implement logic for the units attribute, in case it is set to "auto" and convert the value of the underlying double from seconds to the regarding unit, as commented in the following
new_difftime <- function(x, units = "auto") {
stopifnot(is.double(x), is.character(units))
# case units == "auto":
if (units == "auto")
# when all time differences are NA, units should be "secs"
units <- if (all(is.na(x))){
"secs"
} else {
# otherwise set the units regarding to the minimal time difference
x_min <- min(abs(x), na.rm = TRUE)
if (!is.finite(x_min) || x_min < 60) {
"secs"
} else if (x_min < 3600) {
"mins"
} else if (x_min < 86400) {
"hours"
} else {
"days"
}
}
# we rescale the underlying double, according to the units
x <- switch(units,
secs = x,
mins = x/60,
hours = x/3600,
days = x/86400,
weeks = x/(7 * 86400))
structure(x, units = units, class = "difftime")
}
# test
new_difftime(c(NA, -3600, 86400))
#> Time differences in hours
#> [1] NA -1 24
26.4.1 Inheritance
1. Q: The ordered class is a subclass of factor, but it’s implemented in a very ad hoc way in base R. Implement it in a principled way by building a constructor and providing vec_restore() method.
f1 <- factor("a", c("a", "b"))
as.factor(f1)
#> [1] a
#> Levels: a b
as.ordered(f1) # loses levels
#> [1] a
#> Levels: a
A: TODO: the olad exercise text ended on “an as_ordered generic”. Check the answer if it needs to be updated.
ordered is a subclass of factor, so we need to do the following
• for factors: add a subclass argument to the constructor and helper
• for ordered: add a constructor
• write an as_ordered() generic with methods ordered, factor and default
We use the factor constructor from the textbook and add the subclass argument
new_factor <- function(x, levels, ..., subclass = NULL) {
stopifnot(is.integer(x))
stopifnot(is.character(levels))
structure(
x,
levels = levels,
class = c(subclass, "factor")
)
}
We also use the validator for factors from the textbook
validate_factor <- function(x) {
values <- unclass(x)
levels <- attr(x, "levels")
if (!all(!is.na(values) & values > 0)) {
stop(
"All x values must be non-missing and greater than zero",
call. = FALSE
)
}
if (length(levels) < max(values)) {
stop(
"There must at least as many levels as possible values in x",
call. = FALSE
)
}
x
}
And we add the subclass argument for the helper from the textbook and the exercises
factor <- function(x, levels = unique(x), ... , subclass = NULL) {
ind <- match(x, levels)
# error when values occur, which are not in the levels
if(any(is.na(ind))){
stop("The following values do not occur in the levels: ",
paste(setdiff(x,levels), collapse = ", "), ".",
call. = FALSE)
}
validate_factor(new_factor(ind, levels, subclass = subclass))
}
A constructor for ordered is already implemented in the sloop package:
new_ordered <- function (x, levels) {
stopifnot(is.integer(x))
stopifnot(is.character(levels))
structure(x, levels = levels, class = c("ordered", "factor"))
}
The implementation of the generic and the first two methods is straight forward
as_ordered <- function(x, ...) {
UseMethod("as_ordered")
}
as_ordered.ordered <- function(x, ...) x
as_ordered.default <- function(x, ...) {
stop(
"Don't know how to coerce object of class ",
paste(class(x), collapse = "/"), " into an ordered factor",
call. = FALSE
)
}
For the factor method of as_ordered() we use the factor helper, since it saves us some typing:
as_ordered.factor <- function(x, ...) {
factor(x, attr(x, "levels"), subclass = "ordered")
}
Finally, our new method preserves all levels:
as_ordered(f1)
#> [1] a
#> Levels: a < b
For a real scenario, we might want to add an as_factor.ordered() method to the as_factor() generic from the textbook.
26.5 S4
26.5.1 Generics and methods
1. Q: What’s the difference between the generics generated by these two calls?
setGeneric("myGeneric", function(x) standardGeneric("myGeneric"))
setGeneric("myGeneric", function(x) {
standardGeneric("myGeneric")
})
A: The first call defines a standard generic and the second one creates a nonstandard generic. One can confirm this directly whlie printing (showing in S4 jargon) the function.
setGeneric("myGeneric", function(x) standardGeneric("myGeneric"))
#> [1] "myGeneric"
myGeneric
#> standardGeneric for "myGeneric" defined from package ".GlobalEnv"
#>
#> function (x)
#> standardGeneric("myGeneric")
#> <environment: 0x5cb0cc0>
#> Methods may be defined for arguments: x
#> Use showMethods("myGeneric") for currently available ones.
setGeneric("myGeneric", function(x) {
standardGeneric("myGeneric")
})
#> [1] "myGeneric"
myGeneric
#> nonstandardGenericFunction for "myGeneric" defined from package ".GlobalEnv"
#>
#> function (x)
#> {
#> standardGeneric("myGeneric")
#> }
#> <environment: 0x5deec20>
#> Methods may be defined for arguments: x
#> Use showMethods("myGeneric") for currently available ones.
26.6 Expressions
26.6.1 Abstract syntax trees
1. Q: Use ast() and experimentation to figure out the three arguments to an if() call. What would you call them? Which arguments are required and which are optional?
A: You can write an if() statement in several ways: with or without else, formatted or in one line and also in prefix notation. Here are several versions focussing on the possibility of leaving out curly brackets.
lobstr::ast(if (TRUE) {} else {})
#> █─if
#> ├─TRUE
#> ├─█─{
#> └─█─{
lobstr::ast(if (TRUE) 1 else 2)
#> █─if
#> ├─TRUE
#> ├─1
#> └─2
lobstr::ast(if(TRUE, 1, 2))
#> █─if
#> ├─TRUE
#> ├─1
#> └─2
One possible way of naming the arguments would be: condition (1), conclusion (2), alternative (3).
The condition is always required. If the condition is TRUE, also the conclusion is required. If the condition is FALSE and if() is called in combination with else(), then also the alternative is required.
2. Q: What are the arguments to the for() and while() calls?
A: for() requires an index (called var in the docs), a sequence and an expression, for example
for(i, 1:3, {print(i)})
#> [1] 1
#> [1] 2
#> [1] 3
while() requires a condition and an expression. Again, an example in prefix notation:
set.seed(123)
while((i <- rnorm(1)) < 1, {print(i)})
#> [1] -0.56
#> [1] -0.23
i
#> [1] 1.56
Note that a minimal expression can consist of { only.
3. Q: Two arithmetic operators can be used in both prefix and infix style. What are they?
A: I am not sure how this is meant to be. Theoretically every arithmetic operator can be written in prefix notation via backticks. On the other hand, + and - seem to be the only ones, which can be written in infix notation without backticks.
x <- 1
+(x)
#> [1] 1
-(x)
#> [1] -1
However, when we look more closely, the call tree is not what we would expect from a prefix function
lobstr::ast(+(x))
#> █─+
#> └─█─(
#> └─x
lobstr::ast(-(x))
#> █─-
#> └─█─(
#> └─x
So maybe it is meant to look like this…
lobstr::ast(+x)
#> █─+
#> └─x
lobstr::ast(-x)
#> █─-
#> └─x
Of course also this doesn’t make too much sense, since in ?Syntax one can read, that R clearly differentiates between unary and binary + and - operators and a unary operator is not really what we mean, when we speak about infix operators.
However, if we don’t differentiate in this way, this is probably the solution, since it’s obviously also an infix function:
lobstr::ast(x + y)
#> █─+
#> ├─x
#> └─y
lobstr::ast(x - y)
#> █─-
#> ├─x
#> └─y
26.7 Quasiquotation (new)
1. Q: Why does as.Date.default() use substitute() and deparse()? Why does pairwise.t.test() use them? Read the source code.
A: as.Date.default() uses them to convert unexpected input expressions (neither dates, nor NAs) into a character string and return it within an error message.
pairwise.t.test() uses them to convert the names of its datainputs (response vector x and grouping factor g) into character strings to format these further into a part of the desired output.
2. Q: pairwise.t.test() assumes that deparse() always returns a length one character vector. Can you construct an input that violates this expectation? What happens?
A: We can pass an expression to one of pairwise.t.test()’s data input arguments, which exceeds the default cutoff width in deparse(). The expression will be split into a character vector of length greater 1. The deparsed data inputs are directly pasted (read the source code!) with “and” as separator and the result is just used to be displayed in the output. Just the data.name output will change (it will include more than one “and”).
d=1
pairwise.t.test(2, d+d+d+d+d+d+d+d+d+d+d+d+d+d+d+d+d)
#>
#> Pairwise comparisons using t tests with pooled SD
#>
#> data: 2 and d + d + d + d + d + d + d + d + d + d + d + d + d + d + d + d + 2 and d
#>
#> <0 x 0 matrix>
#>
#> P value adjustment method: holm
26.8 FO
26.8.1 Behavioural FOs
1. Q: What does the following function do? What would be a good name for it?
f <- function(g) {
force(g)
result <- NULL
function(...) {
if (is.null(result)) {
result <<- g(...)
}
result
}
}
runif2 <- f(runif)
runif2(5)
#> [1] 0.528 0.892 0.551 0.457 0.957
runif2(10)
#> [1] 0.528 0.892 0.551 0.457 0.957
A: It returns a new version of the inputfunction. That version will always return the result of it’s first run (in case this not NULL), no matter how the input changes. Good names could be first_run() or initial_return().
2. Q: Modify delay_by() so that instead of delaying by a fixed amount of time, it ensures that a certain amount of time has elapsed since the function was last called. That is, if you called g <- delay_by(1, f); g(); Sys.sleep(2); g() there shouldn’t be an extra delay.
A:
3. Q: Write wait_until() which delays execution until a specific time.
A:
wait_until <- function(time, f) {
force(f)
function(...) {
while (Sys.time() < time) {}
return(f(...))
}
}
# a little test
ptm <- proc.time()
m <- wait_until(Sys.time() + 10, mean)
m(1:3)
proc.time() - ptm
4. Q: There are three places we could have added a memoise call: why did we choose the one we did?
download <- memoise(dot_every(10, delay_by(1, download_file)))
download <- dot_every(10, delay_by(1, memoise(download_file)))
A: The second was chosen. It’s easy to see why, if we eliminate the other two options:
• The first version only prints a dot at every tenth download() call with a new input. This is because dot_every() is inside of memoise() and the counter created by dot_every() is not “activated” if the input is known.
• The third version takes one second for every call. Even if we already know the result and don’t download anything again.
5. Q: Why is the remember() function inefficient? How could you implement it in more efficient way?
6. Q: Why does the following code, from stackoverflow, not do what you expect?
# return a linear function with slope a and intercept b.
f <- function(a, b) function(x) a * x + b
# create a list of functions with different parameters.
fs <- Map(f, a = c(0, 1), b = c(0, 1))
fs[[1]](3)
#> [1] 0
# should return 0 * 3 + 0 = 0
How can you modify f so that it works correctly?
A: You can read in the stackoverflow link that the question arose, because the original return of fs[[1]](3) was 4, which is due to lazy evaluation and could be solved by two users via force():
f <- function(a, b) {force(a); force(b); function(x) a * x + b}
However you can see in the result within the question that R’s behaviour was changed in this case and as Jan Kislinger points out on twitter:
The real question should be: “How did they modify #rstats so that it works correctly?” otherwise it’s a tricky question :D
Note that the same issue appears in the textbook:
In the following example, we take a list of functions and delay each one. But when we try to evaluate the mean, we get the sum instead.
funs <- list(mean = mean, sum = sum)
funs_m <- lapply(funs, delay_by, delay = 0.1)
funs_m\$mean(1:10)
#> [1] 5.5
Which (as one can see) is not true anymore…actually it changed in R version 3.2:
Higher order functions such as the apply functions and Reduce() now force arguments to the functions they apply in order to eliminate undesirable interactions between lazy evaluation and variable capture in closures. This resolves PR#16093.
For further interested: PR#16093 will lead you to the subject “iterated lapply” within the R-devel Archives. Note that the behaviour in for loops is still as “the old lapply()” behaviour.
26.8.2 Output FOs
1. Q: Create a negative() FO that flips the sign of the output of the function to which it is applied.
A:
negative <- function(f){
force(f)
function(...){
-f(...)
}
}
2. Q: The evaluate package makes it easy to capture all the outputs (results, text, messages, warnings, errors, and plots) from an expression. Create a function like capture_it() that also captures the warnings and errors generated by a function.
A: One way is just to capture the output of tryCatch() with identity handlers for errors and warnings:
capture_trials <- function(f){
force(f)
function(...){
capture.output(tryCatch(f(...),
error = function(e) e,
warning = function(w) w)
)
}
}
# we test the behaviour
log_t <- capture_trials(log)
elements <- list(1:10, c(-1, 10), c(TRUE, FALSE), letters)
results <- lapply(elements, function(x) log_t(x))
results
#> [[1]]
#> [1] " [1] 0.000 0.693 1.099 1.386 1.609 1.792 1.946 2.079 2.197 2.303"
#>
#> [[2]]
#> [1] "<simpleWarning in f(...): NaNs produced>"
#>
#> [[3]]
#> [1] "[1] 0 -Inf"
#>
#> [[4]]
#> [1] "<simpleError in f(...): non-numeric argument to mathematical function>"
# further
# results_detailed <- lapply(elements, function(x) lapply(x, function(y))log2(x))
# results_detailed
3. Q: Create a FO that tracks files created or deleted in the working directory (Hint: use dir() and setdiff().) What other global effects of functions might you want to track?
A:
26.8.3 Input FOs
1. Q: Our previous download() function only downloads a single file. How can you use partial() and lapply() to create a function that downloads multiple files at once? What are the pros and cons of using partial() vs. writing a function by hand?
2. Q: Read the source code for plyr::colwise(). How does the code work? What are colwise()’s three main tasks? How could you make colwise() simpler by implementing each task as a function operator? (Hint: think about partial().)
A: We describe how it works by commenting the source code:
function (.fun, .cols = true, ...)
{
# We check if .cols is not a function, since it is possible to supply a
# predicate function.
# if so, the .cols arguments will be "quoted", and filter() will
# be a function that checks and evaluates these .cols within its other argument
if (!is.function(.cols)) {
.cols <- as.quoted(.cols)
filter <- function(df) eval.quoted(.cols, df)
}
# otherwise, filter will be be Filter(), which applies the function
# in .cols to every element of its other argument
else {
filter <- function(df) Filter(.cols, df)
}
# the ... arguments are caught in the list dots
dots <- list(...)
# a function is created, which will also be the return value.
# it checks if its input is a data frame
function(df, ...) {
stopifnot(is.data.frame(df))
# if df is split (in "plyr" speaking), this will be taken into account...
df <- strip_splits(df)
# now the columns of the data frame are chosen, depending on the input of .cols
# this can chosen directly, via a predicate function, or all columns (default)
filtered <- filter(df)
# if this means, that no columns are selected, an empty data frame will be returned
if (length(filtered) == 0)
return(data.frame())
# otherwise lapply will be called on all filtered columns, with
# the .fun argument, which has to be provided by the user, and some other
# arguments provided by the user, when calling the function (...) and
# when defining the function (dots)
out <- do.call("lapply", c(list(filtered, .fun, ...),
dots))
# the output will be named and converted from list into a data frame again
names(out) <- names(filtered)
quickdf(out)
}
}
<environment: namespace:plyr>
3. Q: Write FOs that convert a function to return a matrix instead of a data frame, or a data frame instead of a matrix. If you understand S3, call them as.data.frame.function() and as.matrix.function().
A:
as.matrix.function <- function(f){
force(f)
function(...){
as.matrix(f(...))
}
}
as.data.frame.function <- function(f){
force(f)
function(...){
as.data.frame(f(...))
}
}
4. Q: You’ve seen five functions that modify a function to change its output from one form to another. What are they? Draw a table of the various combinations of types of outputs: what should go in the rows and what should go in the columns? What function operators might you want to write to fill in the missing cells? Come up with example use cases.
5. Q: Look at all the examples of using an anonymous function to partially apply a function in this and the previous chapter. Replace the anonymous function with partial(). What do you think of the result? Is it easier or harder to read?
A: The results are easy to read. Especially the Map() examples profit in readability:
library(pryr)
#> Registered S3 method overwritten by 'pryr':
#> method from
#> print.bytes Rcpp
#>
#> Attaching package: 'pryr'
#> The following object is masked _by_ '.GlobalEnv':
#>
#> f
## From Functionals
# 1
trims <- c(0, 0.1, 0.2, 0.5)
x <- rcauchy(1000)
unlist(lapply(trims, function(trim) mean(x, trim = trim)))
#> [1] -0.00498 0.05088 0.03304 0.02733
unlist(lapply(trims, partial(mean, x)))
#> [1] -0.00498 0.05088 0.03304 0.02733
# 2
xs <- replicate(5, runif(10), simplify = FALSE)
ws <- replicate(5, rpois(10, 5) + 1, simplify = FALSE)
unlist(Map(function(x, w) weighted.mean(x, w, na.rm = TRUE), xs, ws))
#> [1] 0.453 0.521 0.500 0.443 0.525
unlist(Map(partial(weighted.mean, na.rm = TRUE), xs, ws))
#> [1] 0.453 0.521 0.500 0.443 0.525
# 3
add <- function(x, y, na.rm = FALSE) {
if (na.rm && (is.na(x) || is.na(y))) rm_na(x, y, 0) else x + y
}
r_add <- function(xs, na.rm = TRUE) {
Reduce(function(x, y) add(x, y, na.rm = na.rm), xs)
}
r_add_compact <- function(xs, na.rm = TRUE) {
}
#> [1] 10
#> [1] 10
# 4
v_add1 <- function(x, y, na.rm = FALSE) {
stopifnot(length(x) == length(y), is.numeric(x), is.numeric(y))
if (length(x) == 0) return(numeric())
simplify2array(
Map(function(x, y) add(x, y, na.rm = na.rm), x, y)
)
}
v_add1_compact <- function(x, y, na.rm = FALSE) {
stopifnot(length(x) == length(y), is.numeric(x), is.numeric(y))
if (length(x) == 0) return(numeric())
simplify2array(
Map(partial(add, na.rm = na.rm), x, y)
)
}
#> [1] 3 5 7
#> [1] 3 5 7
# 5
c_add <- function(xs, na.rm = FALSE) {
Reduce(function(x, y) add(x, y, na.rm = na.rm), xs,
accumulate = TRUE)
}
c_add_compact <- function(xs, na.rm = FALSE) {
accumulate = TRUE)
}
#> [1] 1 3 6
#> [1] 1 3 6
## From Function operators
# 6
f <- function(x) x ^ 2
partial(f)
#> function (...)
#> f(...)
# 7
# Map(function(x, y) f(x, y, zs), xs, ys)
# Map(partial(f, zs = zs), xs, yz)
# 8
# f <- function(a) g(a, b = 1)
# f <- partial(g, b = 1)
# 9
compact <- function(x) Filter(Negate(is.null), x)
compact <- partial(Filter, Negate(is.null))
# 10
# Map(function(x, y) f(x, y, zs), xs, ys)
# Map(partial(f, zs = zs), xs, ys)
# 11
funs2 <- list(
sum = function(...) sum(..., na.rm = TRUE),
mean = function(...) mean(..., na.rm = TRUE),
median = function(...) median(..., na.rm = TRUE)
)
funs2 <- list(
sum = partial(sum, na.rm = TRUE),
mean = partial(mean, na.rm = TRUE),
median = partial(median, na.rm = TRUE)
)
26.8.4 Combining FOs
1. Q: Implement your own version of compose() using Reduce and %o%. For bonus points, do it without calling function.
A: We use the definition from the textbook:
compose <- function(f, g) {
function(...) f(g(...))
}
"%o%" <- compose
And then we build two versions. One via an anonymous function and one via partial():
compose_red <- function(fs) {
Reduce(function(f, g) function(...) f(g(...)), fs)
}
compose_red(c(mean, length, unique))(1:10)
#> [1] 10
compose_red_bonus <- function(fs) {
Reduce(partial(partial(%o%)), fs)
}
compose_red_bonus(c(mean, length, unique))(1:10)
#> [1] 10
2. Q: Extend and() and or() to deal with any number of input functions. Can you do it with Reduce()? Can you keep them lazy (e.g., for and(), the function returns once it sees the first FALSE)?
A: We use and() and or() as defined in the textbook. They are lazy, since they are build up on && and ||. Also their reduced versions stay lazy, as we will show at the end of the code
and <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) && f2(...)
}
}
and_red <- function(fs){
Reduce(function(f, g) and(f, g), fs)
}
or <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) || f2(...)
}
}
or_red <- function(fs){
Reduce(function(f, g) or(f, g), fs)
}
# Errors before the first TRUE will be returned
tryCatch(
or_red(c(is.logical, is.logical, stop, is.character))("a"),
error = function(e) e
)
#> <simpleError in f1(...): a>
# Errors after the first TRUE won't be returned
or_red(c(is.logical, is.logical, is.character, stop))("a")
#> [1] TRUE
3. Q: Implement the xor() binary operator. Implement it using the existing xor() function. Implement it as a combination of and() and or(). What are the advantages and disadvantages of each approach? Also think about what you’ll call the resulting function to avoid a clash with the existing xor() function, and how you might change the names of and(), not(), and or() to keep them consistent.
A: Both versions are implemented straight forward, as also the reduced versions. However, the parallel versions need a little bit more care:
xor_fb1 <- function(f1, f2){
force(f1); force(f2)
function(...){
xor(f1(...), f2(...))
}
}
xor_fb2 <- function(f1, f2){
force(f1); force(f2)
function(...){
or(f1, f2)(...) && !(and(f1, f2)(...))
}
}
# binary combination
xor_fb1(is.logical, is.character)("a")
#> [1] TRUE
xor_fb2(is.logical, is.character)("a")
#> [1] TRUE
# parallel combination (results in an error)
xor_fb1(c(is.logical, is.character), c(is.logical, is.character))("a")
#> Error in f1(...): could not find function "f1"
xor_fb2(c(is.logical, is.character), c(is.logical, is.character))("a")
#> Error in f1(...): could not find function "f1"
# reduced combination (results in an error)
xor_fb1(c(is.logical, is.character, is.logical, is.character))("a")
#> Error in force(f2): argument "f2" is missing, with no default
xor_fb2(c(is.logical, is.character, is.logical, is.character))("a")
#> Error in force(f2): argument "f2" is missing, with no default
### Reduced version
xor_fb1_red <- function(fs){
Reduce(function(f, g) xor_fb1(f, g), fs)
}
xor_fb2_red <- function(fs){
Reduce(function(f, g) xor_fb2(f, g), fs)
}
# should return TRUE
xor_fb1_red(c(is.logical, is.character, is.logical, is.character))("a")
#> [1] FALSE
xor_fb2_red(c(is.logical, is.character, is.logical, is.character))("a")
#> [1] FALSE
# should return FALSE
xor_fb1_red(c(is.logical, is.logical, is.character, is.logical))("a")
#> [1] TRUE
xor_fb2_red(c(is.logical, is.logical, is.character, is.logical))("a")
#> [1] TRUE
# should return FALSE
xor_fb1_red(c(is.logical, is.logical, is.character, is.character))("a")
#> [1] FALSE
xor_fb2_red(c(is.logical, is.logical, is.character, is.character))("a")
#> [1] FALSE
4. Q: Above, we implemented boolean algebra for functions that return a logical function. Implement elementary algebra (plus(), minus(), multiply(), divide(), exponentiate(), log()) for functions that return numeric vectors.
A:
plus <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) + f2(...)
}
}
minus <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) - f2(...)
}
}
multiply <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) * f2(...)
}
}
divide <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) / f2(...)
}
}
exponentiate <- function(f1, f2) {
force(f1); force(f2)
function(...) {
f1(...) ^ f2(...)
}
}
# we rename log to log_ since log() already exists
log_ <- function(f1, f2) {
force(f1); force(f2)
function(...) {
log(f1(...), f2(...))
}
}
# Test
mns <- minus(mean, function(x) x^2)
mns(1:5)
26.9 Expressions (again)
26.9.1 Data structures
1. Q: How is rlang::maybe_missing() implemented? Why does it work?
A: Let us take a look at the functions source code to see what’s going on
lang::maybe_missing
function (x)
{
# is_missing checks if one of the following is TRUE
# 1. check via substitute if typeof(x) is symbol and missing(x) is TRUE
# 2. check if x identical to missing_arg()
if (is_missing(x)) {
missing_arg() # returns missing argument
# implemented in lower level code -> .Call())
}
else {
x # when it's not missing, x is simply returned
}
}
<bytecode: 0x00000000195ed740>
<environment: namespace:rlang>
First it is checked if the argument is missing. If so, the missing arg is returned, otherwise the argument (x) itsself is returned.
26.9.2 Parsing and deparsing
1. Q: Why does as.Date.default() use substitute() and deparse()? Why does pairwise.t.test() use them? Read the source code.
A:
26.9.3 R’s grammar
1. Q: deparse() produces vectors when the input is long. For example, the following call produces a vector of length two:
expr <- rlang::expr(g(a + b + c + d + e + f + g + h + i + j + k + l + m +
n + o + p + q + r + s + t + u + v + w + x + y + z))
deparse(expr)
#> [1] "g(a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + "
#> [2] " p + q + r + s + t + u + v + w + x + y + z)"
What do expr_text(), expr_name(), and expr_label() do with this input?
A:
• expr_text() pastes the output string into one and inserts \n (new line identifiers) as separators
cat(rlang::expr_text(expr)) # cat is used for printing with linebreak
#> g(a + b + c + d + e + f + g + h + i + j + k + l + m + n + o +
#> p + q + r + s + t + u + v + w + x + y + z)
• expr_name() recreates the call into the form f(…) and deparses this expression into a string
rlang::expr_name(expr)
#> [1] "g(...)"
• expr_label() does the same as expr_name(), but surrounds the output also with backticks
rlang::expr_label(expr)
#> [1] "g(...)"
|
# Homework Help: Limit of integrable functions
1. Dec 11, 2008
### tomboi03
Prove: If f is integrable on [a , b] then
lim f =0
x$$\rightarrow$$a+
the integral goes from a to x.
How do i go about and prove this? I'm confused.
Thank You
Last edited: Dec 11, 2008
2. Dec 11, 2008
### HallsofIvy
Re: integrable?
Do you mean
$$\lim_{x\rightarrow a^+} \int_a^x f(t)dt= 0$$
The way you have written it, that the limit of f is 0, makes no sense- that certainly is not necessarily true.
My suggestion here is the same as to your other question: use the definition of integral in terms of Riemann sums.
3. Dec 15, 2008
### tomboi03
Re: integrable?
I've never learn Riemann sum definition.
What is that?
4. Dec 15, 2008
### HallsofIvy
Re: integrable?
First, is what I wrote what you mean. And if you have never learned Riemann sums, what definition of $\int_a^b f(x)dx$ are you using?
|
## Introductory Statistics 9th Edition
Mean = $752\div80 = 9.4$ Variance $=σ^2$ $=(∑m^2 f-([(∑mf)]^2/N))/N$ $= \frac{10048- (565504\div80)}{79}$ $= 37.7114$ Standard Deviation =$\sqrt 37.7114$ =6.141
Mean = $752\div80 = 9.4$ Variance $=σ^2$ $=(∑m^2 f-([(∑mf)]^2/N))/N$ $= \frac{10048- (565504\div80)}{79}$ $= 37.7114$ Standard Deviation =$\sqrt 37.7114$ =6.141
|
# Brownian motion. Solve stoc. integral by using Ito's lemma
I want to show that following statement is true by using Ito's lemma to solve stochastic integrals:
I define the functions in Ito's model: a()=0, b()= (2wt-2)^2. f(t)=Integrate[(2wt-2)^2]
Then df=(b^2/2)(d^2/dwt^2)+(bdf/dst). But it doesn’t add up. How do I show it by using Ito's lemma?
• Sorry, I don't know how to write formulas in here – Sanjay Dec 7 '15 at 12:25
Try Ito's formula for $(2W_t+1)^3$, and then integrate. More specifically, note that \begin{align*} d\left( (2W_t+1)^3 \right) &= 6(2W_t+1)^2 dW_t + 12 (2W_t+1) dt, \end{align*} then \begin{align*} (2W_T+1)^3 - 1 = \int_0^T 6(2W_t+1)^2 dW_t + 12 \int_0^T (2W_t+1) dt. \end{align*} The remaining is obvious.
|
# Problem Set #4
## Ground Rules for Problem Sets
• Due on March 12.
• Will be handed back for corrections or revisions until we're happy.
• Answers to articulation problems are limited to three reasonable length sentences. You should imagine answering a question from a graduate student at a colloquium on the current topic. This is hard!
• For any of the problems you can use any math tool such as maple or mathematica. Provide a printout of the session.
## Problem Set 4
### Articulation Questions
1. Why use a path integral formulation to make a simulation of QCD?
2. Why is lattice QCD formulated in Euclidean space instead of Minkowski space?
3. Why do lattice gauge theories use link variables instead of the gauge fields themselves?
4. What is the difference between a first-order and a second-order phase transition?
5. Why are finite temperature lattice simulations easy but finite density simulations hard?
### Project A: Monte Carlo Integration and Metropolis Algorithm
In this project, you will write programs that perform first "naive" Monte Carlo integration (as discussed in class) and then use the Metropolis algorithm (see handout). The ratio of integrals to be used for testing is
<x^2> = [\int d^{D}x x^2 e^{-|x}|}] / [\int d^{D}x e^{-|x|}] = D(D+1)
where D is the number of dimensions and x is a vector in D-dimensional space.
There are three parameters we will consider in evaluating <x^2>: the number of dimensions D, the size of the cutoff L (i.e., in each dimension integrate from -L to +L), and the number of points N used (that is, the number of times the integrands are evaluated).
1. Write a program to perform naive Monte Carlo integration and evaluate <x^2>.
• Discuss (with graphs) how the accuracy of the Monte Carlo result depends on L, N, and D. Consider D = 1,5,10,20. What is a good choice for L?
2. Write a program to use the Metropolis algorithm to generate N vectors x_i and use them to evaluate <x^2>.
• Choose a random starting size inside the box of size L (use your good choice'' for L from part 1).
• Generate new vectors (or candidates for new vectors) by making a step of magnitude \delta in a random direction. What is a good size for \delta?
• Compare the accuracy with N of the Metropolis result to naive Monte Carlo integration for D = 1,5,10,20.
### Project B: Ising Model in Two Dimensions
The Ising model is a simple model of a magnetic system. It consists of a finite lattice in D dimensions with spins on each lattice site i with spin-projection S_{i} = +/- 1. The interaction energy at zero external field between two adjacent spins at sites i and j is E_{ij} = -2 J S_{i} S_{j}. That is, the interaction energy is either -2J or +2J. To find the total energy E, sum E_{ij} over each pair of nearest neighbors (be careful not to double count!).
1. Write a program to simulate the Ising model in zero field for two dimensions. Use the Metropolis algorithm with a Boltzmann weighting factor e^{-\beta E} to select configurations. To generate a new configuration:
• Step through each lattice site (rather than simply picking a new site at random).
• Decide whether or not to flip the spin on that site by the Metropolis algorithm.
• When each site has been considered, save the new configuration or use it to evaluate the contribution to the average spin.
2. Take D=2 and use a 30 x 30 lattice (you can use larger if you wish!) with periodic boundary conditions. Work in units with J=1.
3. Measure the average spin <S(\beta)> as a function of \beta and locate the critical point \beta_c. How does the equilibration time depend on \beta?
4. Compare your results to the mean-field result. Comment on the similarities and differences.
|
# Would a spherically distributed solar system inevitably become flat, i.e., develop a plane of the ecliptic?
Notes
1. I have tagged this as hard science - If it turns out that current science has no research on the subject or cannot cope with calculations about this sort of system, I may relax this to science-based and allow informed speculation.
2. Some people might object that I have used the idea of gods in a hard science question. The gods are merely a fiction to 'explain' how the system might have come about - they are not essential. Feel free to ignore the god aspect and imagine that this solar system came about by natural forces.
Setup
Some young gods are playing at solar systems. One has the idea that, instead of the usual boring flat system, their system should be spherical.
They start with a Sun-like star then add asteroids and planets orbiting it every which way. Some orbits are at right-angles to each other or any other arbitrary angle. Some orbits are in similar planes but in opposite directions.
The young gods set this in motion and sit back in glee to watch the chaos that results. Near misses, weird non-elliptical orbits and, best of all, the occasional huge planetary collision.
Question
Presumably the solar system will eventually form some stable configuration. Will this inevitably be a standard flat system with all planets orbiting in the same direction or could there conceivably still be orbits in opposing directions and non-colliding orbits at up to 90 degrees from one another?
Assumptions
2. You may assume pretty much any starting configuration and velocity of orbiting bodies (OBs), as long as they tend to remain within the spherical radius of Sol's flat system.
3. The total mass and distribution of elements of OBs is the same as in our solar system.
4. Optionally you may assume the same planets (Earth, Mars etc.) and asteroids that we have, only orbiting in 3D instead of 2D.
5. Gods are not essential to the question, they are just there to give a motivation for the starting conditions. You may assume that this system came about by chance, however unlikely that may be.
• I don't believe that I can provide an actual answer of sufficient quality, but: 1) Retrograde orbits are a thing. Even if all planets are in (more-or-less) a single plane, they need not all orbit in the same direction. 2) Within the last several days, I read an article about astronomers observing a stellar system in which planets orbited in multiple planes, so it would seem that this is also a thing that can happen, if you can avoid problems with everything crashing together and destroying each other. – Dave Sherohman Sep 12 '20 at 11:03
• @Dave Sherohman - This sounds very interesting. Note that for hard science you don't necessarily have to provide the mathematics yourself. References to reputable sources is also an option, and this of course includes astronomical observations. If you feel inclined to work this comment into an answer (or even partial answer) I would be very grateful. – chasly - supports Monica Sep 12 '20 at 11:09
• Catch some rogue planets, have it collide with another system etc. As far as know, systems form flat, they don't become flat. Have things happen later that give you strange orbits. I'm not an astronomer, sorry if there is some weird phenomenon I'm missing that flattens orbits in basically vacuum, but I wonder if your starting point is wrong. – Raditz_35 Sep 12 '20 at 12:44
• The system won't ever become a "regular" system with all the major bodies orbiting in the same direction more or less in the same plane unless those same gods resurrect Sir Isaac Newton PRS and compel him to repeal the law of conservation of angular momentum. (Really, "hard science"?) – AlexP Sep 12 '20 at 13:11
• @Alex - hard science doesn't mean hard in the sense of advanced or difficult. It means that the answers should provide hard evidence. For example explaining the the application of Newton's laws to the problem. What is easy for some people is difficult for others. P.S. If you believe that, then what is the final configuration going to look like? – chasly - supports Monica Sep 12 '20 at 13:54
The question of stability will produce very different answers depending on the exact parameters of the system, but it seems that there are regions in parameter space of stability and regions of parameter space of instability.
The one example of confirmed significant non-coplanarity I know of is $$\upsilon$$ Andromedae A, a system containing three (and perhaps four) giant planets. A number of groups have suggested that planets c and d share a significant mutual inclination; some of the best evidence comes from McArthur et al 2010, who calculated a mutual eccentricity of $$29\pm1^{\circ}$$. As $$\upsilon$$ And has an age of $$\sim$$3 billion years, we can see that significantly evolved systems can harbor non-coplanar planets.
The question, though, is whether the current arrangement is stable, and if so, what mechanism is responsible. The authors found that the current inclinations and eccentricities can be maintained for c and d for a period of $$\sim$$100,000 years (at which point their simulations terminated), although the system exists on the very edge of stability in parameter space. A few tweaks to the initial conditions could lead to damping of the inclination of simply a catastrophic disruption of the system.
Libert & Tsiganis 2009 argued that, as with certain other planetary systems, the Kozai mechanism could lead to a stable configuration if the mutual inclination of c and d was $$\sim45^{\circ}$$ - although their nonrelativistic analysis may well be incomplete. This might require interactions with the other star in the system, $$\upsilon$$ And b. Other groups considered planet-planet scattering, which would still be an option in a one-star system.
In short: it seems quite possible that the system will never flatten out. If the planets have always had such a high mutual inclination, then it's clear that that can persist for billions of years, i.e. most of a roughly Sun-like star's lifetime. In that case, the answer to your question is likely no: A non-coplanar system does not necessarily have to evolve into a fplane.
### Chaotic initial placement of an N-body system results in everything in the sun or oort cloud.
So, no. It wont be planar.
I've ran a few N-body simulations trying to see what I can get to form, and randomly placing bodies (using c++'s std::normal_distribution) always resulted in a chaotic explosion. Doesn't matter if I modelled 10 bodies or 100 bodies or 1000. Doesn't matter what I used for their initial velocities, masses, and positions. They always blew up. The closer the mean start point was to the sun the sooner the system blew up. Many hit the sun or left the solar system (depending on the config), the rest settled into 100-1000yr elliptical orbits.
Simulation models interactions at a resolution of 1 hour. The tail shows the last 30 weeks travels
Here's a typical first year:
A typical 100th year:
A typical 500th year:
Here's one after 1000 years. I included the plane Z=0 to show that they aren't forming a planar system.
Basically they exploded, and tended to slow down a long way out. Typical distances for that final shot are 10^12 to 10^15m, which is past pluto and in the oort cloud. Orbital periods were in the hundreds or thousands of years.
Your gods will be waiting a long time for subsequent impacts. I stopped the simulation at 100,000 years, because I needed the computer to do actual work again. At this point its basically simulating the comet orbits.
So, your gods should know that they should keep their solar system formation to rotating pools of gas clumping together if they want anything interesting to happen.
(Apologies for the Z/N/E axis, I hacked this into some mining software I'm developing and forgot to change the axes)
• This is an interesting result, but I wonder whether the conclusions hold for truly astronomical timescales (i.e. billions of years). The orbits might be eccentric and highly inclined for now, but subsequent passes close to the parent star could eject them from the system, and dynamical interactions with other bodies might mean that the orbits are completely unstable. – HDE 226868 Sep 12 '20 at 14:18
• @HDE226868 I can't prove they don't - but my gut is telling me it's unlikely. "Chaos + Chaos = Order" doesn't seem likely. – Ash Sep 12 '20 at 14:23
• How many bodies are you simulating? What is their mass? And I hate to ask it, but how much can we trust the simulator? Simulators are notoriously sensitive to CPU precision limitations and whenever I see a chaotic explosion like this I wonder if the precision errors contributed to the problem. – JBH Sep 12 '20 at 15:18
• Sorry, I just remembered you tried a variety of quantities, masses, etc. Sorry. Anyway, what simulator did you use? What computing platform? Thanks. – JBH Sep 12 '20 at 15:25
• Windows, C++, x64. Intel I9. 64-bit data types. I hacked a hand-made N-body simulator it into some software my company is developing. – Ash Sep 12 '20 at 15:31
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# Optimal R function for Predicting Time Series Dataset
Im trying to create a model to predict transactions at a shop. I have the date and hour of the transaction with 4 other predictor variables.
I've provided the first 6 rows of the data below:
Date Hour Var1 Var2 Var3 Var4 Trans1 Trans2
01/01/18 1am 4 12 1 123 1 4
01/01/18 2am 6 14 0 126 3 6
01/01/18 3am 3 16 0 124 2 3
01/01/18 4am 4 12 1 122 3 7
01/01/18 5am 8 6 1 122 4 2
01/01/18 6am 4 11 1 123 5 8
I'm looking to predict both Trans1 and Trans2 using the Date, Hour, Var1, Var2, Var3 and Var4.
I've tried using the lm function but I'm unsure how to treat the date and hour variables.
I know that I need to account for seasonal and daily change in the model. What is the best modelling function in r to model this accurately?
https://stats.stackexchange.com/search?q=user%3A3382+hourly+data will give you some pointers as to how you should proceed with hourly data. Essentially daily habits can impact hourly responses/values. Oftentimes daily activity can be predicted using daily indicators, monthly indicators holidays etc. All of the examples that I was involved with could easily been expanded to include covariates. The software I used is available in R so that might help .. Hope this advice helps you .
• These do not appear to be "hourly data" in the sense you imply (despite the regular progression of times in the question's example). These are labeled event data. It is possible--at least going from the description in the question--to have multiple records for a given hour and to have many hours missing because no transaction occurred during them. As such, your proposed methods couldn't even be applied without modification and might be misleading if somehow they were applied. – whuber Jan 10 '19 at 17:45
• i would like the OP to review your comments as it appears to me to be equally spaced data (x hours per day ..not necessarily 24 ). It is possible to analyze data that has say k readings per work day (mon-fri) while having less (but fixed) say Ll readings per day on weekends. – IrishStat Jan 11 '19 at 3:50
• Agreed. Perhaps Trans1 or Trans2 are counts of transactions aggregated by hours, for instance. – whuber Jan 11 '19 at 17:16
• That's how I interpreted it . – IrishStat Jan 11 '19 at 17:18
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## Derivative of 10^x using limit definition
1. The problem statement, all variables and given/known data
Obtain the first derivative of 10x by the limit definition.
2. Relevant equations
f'(x)=limh->0 f(x+h)-f(x)/h
3. The attempt at a solution
f'(x)=limh->0 10x+h-10x/h
I also know that h=1 as x approaches 0.
Now, how do I make it so that you aren't dividing by h=0.
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Recognitions: Gold Member Science Advisor Staff Emeritus 10x+h= 10x10h so $$\frac{10^{x+h}- 10^x}{h}= 10^x\frac{10^h- 1}{h}$$ Now the question is, what is $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$?
To find $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$ I can't plug h=0 in because I would be dividing by 0. Do I plug in 1 since the limit as h approaches 0 is 1?
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## Derivative of 10^x using limit definition
Quote by cmajor47 To find $$\lim_{h\rightarrow 0}\frac{10^h- 1}{h}$$ I can't plug h=0 in because I would be dividing by 0. Do I plug in 1 since the limit as h approaches 0 is 1?
if you plug h = 0 then 10^h = 1. But you want the next order correction which you explect should be proportional to h. So you expect something of the form
$$lim_{h\rightarrow 0} ~10^h = 1 + a h + \ldots$$
where a is some numerical value and the dots represent terms of higher powers in h. The problem is to find the value of a.
Here is the trick: use that any number x may be written as $$e^{ \ln x}$$. Then use what you know about rules for logs and then Taylor expand the exponential.
I don't understand what this means. What is "Taylor expanding" and the "next order correction"?
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Quote by cmajor47 I don't understand what this means. What is "Taylor expanding" and the "next order correction"?
Have you ever seen the relation
$$e^\epsilon = 1 + \epsilon + \frac{\epsilon^2}{2} + \ldots$$ ?
This is a Taylor expansion.
Have you ever proved using the limit definition that the derivative of e^x is e^x? Then you must have used something similar to the above.
If you haven't proved the e^x case and the expansion I wrote above is not familiar to you then I will let someone else help you because I don't see at first sight any other approach.
EDIT: do you know what the limit as h goes to zero of $$\frac{e^h -1}{h}$$ gives? maybe you have been told this without proving it. If you know the result of this limit and are allowed to use it, then I can show you how to finish your problem. If not, I don't see how to help, unfortunately.
Best luck!
I've never proved e^x. Thank you for trying to help though.
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Quote by cmajor47 I've never proved e^x. Thank you for trying to help though.
Sorry. I can tell you that the limit as h goes to zero of 10^h is
$$lim_{h \rightarrow 0} 10^h = lim_{h \rightarrow 0} e^{\ln 10^h} = lim_{h \rightarrow 0} e^{h \ln 10} \approx 1 + h \ln 10$$
where I used an identity for logs and then I used the expansion of the exponential I mentioned earlier. Form this you can get the final answer of your question.
Hopefully someone else will be able to find a way to show this result in some other way but I can't think of any!
Best luck
Mentor
Blog Entries: 10
Quote by nrqed ... Taylor expand the exponential.
Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.
Mentor Blog Entries: 10 I don't know if this will be useful, but one might try using the definition of e: $$e = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^N = \lim_{a \rightarrow 0} (1+a)^{1/a}$$ or $$e^A = \lim_{N \rightarrow \infty} (1+\frac{1}{N})^{NA} = \lim_{a \rightarrow 0} (1+a)^{A/a}$$ Also, the fact that $$10^h = e^{h \ln(10)}$$
Recognitions: Gold Member Science Advisor Staff Emeritus I suspect this was given as a preliminary to the derivative of ex so the derivative of ex cannot be used. It is easy to see that the derivative of ax, for a any positive number, is a constant times ax. It is much harder to determine what that constant is! It's not too difficult to show that, for some values of a, that constant is less than 1 and, for some values of a, larger than 1. Define e to be the number such that that constant is 1. That is, define "e" by $$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ As Redbelly98 said, 10h= eh ln(10) so $$\frac{10^h- 1}{h}= \frac{e^{h ln(10)}- 1}{h}[/itex] If we multiply both numerator and denominator of that by ln(10) we get [tex]ln(10)\left(\frac{e^{h ln(10)}-1}{h ln(10)}\right)$$ Clearly, as h goes to 0 so does h ln(10) so if we let k= h ln(10) we have $$ln(10)\left(\lim_{h\rightarrow 0}\frac{e^{h ln(10)}-1}{h ln(10)}\right)= ln(10)\left(\lim_{k\rightarrow 0}\frac{e^k- 1}{k}\right)$$ so the limit is ln(10) and the derivative of 10x is ln(10)10x. That is NOT something I would expect a first year calculus student to find for himself!
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Quote by Redbelly98 Taylor expansion requires knowledge of what the derivative is. But we don't know the derivative, that is what we are supposed to find.
Moreover, Taylor series are generally taught in second-semester calculus, while covering infinite sequences and series, while the limit
$$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$
is often demonstrated or proven (if it is not simply stated without proof) in the first-semester course, shortly after having covered limits and while developing the rules of differentiation. I hardly expected that the OP would have seen Taylor series yet...
I believe the proof given in textbooks usually revolves around the limit definition of e, which Redbelly98 gives in post #10.
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Quote by dynamicsolo Moreover, Taylor series are generally taught in second-semester calculus, while covering infinite sequences and series, while the limit
Agreed. I should not have mentioned Taylor series. They seem so natural to me now that I tend to use them without even thinking about it.
$$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ is often demonstrated or proven (if it is not simply stated without proof) in the first-semester course, shortly after having covered limits and while developing the rules of differentiation. I hardly expected that the OP would have seen Taylor series yet... I believe the proof given in textbooks usually revolves around the limit definition of e, which Redbelly98 gives in post #10.
This is why I then asked the OP if he/she had seen the formula you wrote just above. I hope he/she has. Because if he/she has to go back to the limit definition and prove the above identity in order to solve the original question, this problem seems much more challenging than I would expect as an assignment problem at that level!!
Recognitions: Homework Help I suspect that OP's textbook presents the limit $$\lim_{h\rightarrow 0}\frac{e^h- 1}{h}= 1$$ somewhere in the chapter and that a student is just asked to recognize that they could apply it, in something like the manner Halls suggests in post #11...
10^x=e^(xln10)=e^x, so we are trying to find the derivative of e^x by the limit defination. The math forum has a solution at: http://mathforum.org/library/drmath/view/60705.html
Sorry, the dereivative of 10^x is of course ln10(10^x) but the Math forum derivation of the derivative of e^x is still helpful
More thoughts on this problem. f(x)=10^X=e^(x*ln10) f(x+h)= e^(ln10(x+h))=e^(ln10*x)*e^(ln10*h) Plugging into definitation of derivative and simplifying gives f'(x)= limit(h goes to 0) 10^x(10^h-1)/h tabulating the limit as h goes to 0 of (10^h-1)/h= ln10
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# Arctan problem help
#### UrbanXrisis
$$y=tan^{-1}\left(\frac{1}{ln(x)}\right)$$
$$y'=\frac{1}{1-\left(\frac{1}{ln(x)}\right)^2}$$
$$y'=\frac{1}{1-(lnx)^{-2}}$$
is this correct?
do I have to take into account the derivative of $\frac{1}{ln(x)}$?
if I do, what is the derivative of $\frac{1}{ln(x)}$?
Last edited:
Related Introductory Physics Homework Help News on Phys.org
#### Pyrrhus
Homework Helper
Yes, you need to take into account the derivative of $\frac{1}{\ln(x)}$.
You could either do the derivative of a division or do the derivative of $(\ln(x))^{-1}$.
#### UrbanXrisis
$$y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}$$
$$y'=\frac{1}{x(lnx^2)-x}}$$
is that it?
#### Pyrrhus
Homework Helper
UrbanXrisis said:
$$y'=\frac{1}{1-(lnx)^{-2}}*\frac{1}{x(lnx^2)}$$
$$y'=\frac{1}{x(lnx^2)-x}}$$
is that it?
Little changes...
$$y'=\frac{1}{1+(lnx)^{-2}}*\frac{-1}{x(\ln(x))^{2}}$$
Something else i noticed
$$(tan^{-1} (f(x)))' = \frac{f'(x)}{1+(f(x))^2}$$
Last edited:
#### UrbanXrisis
can that become $$y'=\frac{1}{-x(lnx^2)+x}}$$?
#### Pyrrhus
Homework Helper
Stop putting the square next to the x like that, use parentheses.
#### UrbanXrisis
sorry, didn't notice that
$$y'=\frac{1}{-x(lnx)^2+x}}$$
that's what I meant
#### Pyrrhus
Homework Helper
Sure, except for the sign problem.
#### UrbanXrisis
$$y'=\frac{1}{-x\ln(x)^{2}-x}$$
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# Measuring with a certain probability in Qiskit
I have the following quantum operation on two qubits: $$\mathcal{E}(\rho) = p \mathcal{T} \circ \mathcal{U}(\rho) + (1-p) \mathcal{U}(\rho) $$
where $$p$$ is some probability, $$\mathcal{U}$$ is some unitary operation on the two qubits and $$\mathcal{T}$$ measures whether both qubits are $$1$$.
I have the variable $$p$$ in Python, and I would like to use it to construct a circuit that implements the above. I tried to simply construct a circuit by making Python do the random choices (i.e. insert the measurement with probability $$p$$), but with this approach I need to recompile a new circuit for each shot, and I was wondering whether there is a simple way to make random choices directly within the QuantumCircuit.
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Outlook: Nouveau Monde Graphite Inc. is assigned short-term Ba1 & long-term Ba1 estimated rating.
Time series to forecast n: 10 Feb 2023 for (n+4 weeks)
Methodology : Ensemble Learning (ML)
## Abstract
Nouveau Monde Graphite Inc. prediction model is evaluated with Ensemble Learning (ML) and Multiple Regression1,2,3,4 and it is concluded that the NOU:TSXV stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Buy
## Key Points
1. Can machine learning predict?
2. What is prediction model?
3. How can neural networks improve predictions?
## NOU:TSXV Target Price Prediction Modeling Methodology
We consider Nouveau Monde Graphite Inc. Decision Process with Ensemble Learning (ML) where A is the set of discrete actions of NOU:TSXV stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Multiple Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Ensemble Learning (ML)) X S(n):→ (n+4 weeks) $\stackrel{\to }{R}=\left({r}_{1},{r}_{2},{r}_{3}\right)$
n:Time series to forecast
p:Price signals of NOU:TSXV stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## NOU:TSXV Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: NOU:TSXV Nouveau Monde Graphite Inc.
Time series to forecast n: 10 Feb 2023 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Buy
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Nouveau Monde Graphite Inc.
1. An entity shall apply this Standard for annual periods beginning on or after 1 January 2018. Earlier application is permitted. If an entity elects to apply this Standard early, it must disclose that fact and apply all of the requirements in this Standard at the same time (but see also paragraphs 7.1.2, 7.2.21 and 7.3.2). It shall also, at the same time, apply the amendments in Appendix C.
2. If such a mismatch would be created or enlarged, the entity is required to present all changes in fair value (including the effects of changes in the credit risk of the liability) in profit or loss. If such a mismatch would not be created or enlarged, the entity is required to present the effects of changes in the liability's credit risk in other comprehensive income.
3. Adjusting the hedge ratio by increasing the volume of the hedging instrument does not affect how the changes in the value of the hedged item are measured. The measurement of the changes in the fair value of the hedging instrument related to the previously designated volume also remains unaffected. However, from the date of rebalancing, the changes in the fair value of the hedging instrument also include the changes in the value of the additional volume of the hedging instrument. The changes are measured starting from, and by reference to, the date of rebalancing instead of the date on which the hedging relationship was designated. For example, if an entity originally hedged the price risk of a commodity using a derivative volume of 100 tonnes as the hedging instrument and added a volume of 10 tonnes on rebalancing, the hedging instrument after rebalancing would comprise a total derivative volume of 110 tonnes. The change in the fair value of the hedging instrument is the total change in the fair value of the derivatives that make up the total volume of 110 tonnes. These derivatives could (and probably would) have different critical terms, such as their forward rates, because they were entered into at different points in time (including the possibility of designating derivatives into hedging relationships after their initial recognition).
4. Rebalancing is accounted for as a continuation of the hedging relationship in accordance with paragraphs B6.5.9–B6.5.21. On rebalancing, the hedge ineffectiveness of the hedging relationship is determined and recognised immediately before adjusting the hedging relationship.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Nouveau Monde Graphite Inc. is assigned short-term Ba1 & long-term Ba1 estimated rating. Nouveau Monde Graphite Inc. prediction model is evaluated with Ensemble Learning (ML) and Multiple Regression1,2,3,4 and it is concluded that the NOU:TSXV stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period, the dominant strategy among neural network is: Buy
### NOU:TSXV Nouveau Monde Graphite Inc. Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2Baa2
Balance SheetBaa2B3
Leverage RatiosCCaa2
Cash FlowBaa2B3
Rates of Return and ProfitabilityCaa2Baa2
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 93 out of 100 with 594 signals.
## References
1. Challen, D. W. A. J. Hagger (1983), Macroeconomic Systems: Construction, Validation and Applications. New York: St. Martin's Press.
2. M. Benaim, J. Hofbauer, and S. Sorin. Stochastic approximations and differential inclusions, Part II: Appli- cations. Mathematics of Operations Research, 31(4):673–695, 2006
3. D. Bertsekas. Dynamic programming and optimal control. Athena Scientific, 1995.
4. Breusch, T. S. A. R. Pagan (1979), "A simple test for heteroskedasticity and random coefficient variation," Econometrica, 47, 1287–1294.
5. Pennington J, Socher R, Manning CD. 2014. GloVe: global vectors for word representation. In Proceedings of the 2014 Conference on Empirical Methods on Natural Language Processing, pp. 1532–43. New York: Assoc. Comput. Linguist.
6. Dimakopoulou M, Zhou Z, Athey S, Imbens G. 2018. Balanced linear contextual bandits. arXiv:1812.06227 [cs.LG]
7. Chow, G. C. (1960), "Tests of equality between sets of coefficients in two linear regressions," Econometrica, 28, 591–605.
Frequently Asked QuestionsQ: What is the prediction methodology for NOU:TSXV stock?
A: NOU:TSXV stock prediction methodology: We evaluate the prediction models Ensemble Learning (ML) and Multiple Regression
Q: Is NOU:TSXV stock a buy or sell?
A: The dominant strategy among neural network is to Buy NOU:TSXV Stock.
Q: Is Nouveau Monde Graphite Inc. stock a good investment?
A: The consensus rating for Nouveau Monde Graphite Inc. is Buy and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of NOU:TSXV stock?
A: The consensus rating for NOU:TSXV is Buy.
Q: What is the prediction period for NOU:TSXV stock?
A: The prediction period for NOU:TSXV is (n+4 weeks)
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# I'm considering buying a telescope, which type should I choose?
I am considering buying a telescope for amateur viewing. I am planning to view the stars from my backyard, which is in an area with reasonable light pollution (most of Orion's primary stars are visible). As a result of the light pollution, I have not been able to see more fascinating parts of the sky, such as nebulae. I am looking for a telescope with a limiting magnitude of at least $$+12$$, an angular resolution of $$0.5 \text{ arcsec}$$, is relatively easy to assemble. My budget is between \$500 to \$800.
I've taken note of the Orion Skyquest models, especially the 8" and 10" Build-A-Scope Classic Dobsonian Telescopes. However, before I buy them, I'd like a second opinion from this community. I have seen that it is not compatible with astrophotography devices like an iPhone without extensions. Is there anyone who can give me some recommendations?
• May or may not be closed as opinion based, please let me know where this should belong, if any. Sep 28, 2021 at 2:14
Let's first clean up some misconceptions: Do you live on Cerro Paranal or something like that? If that's not the case, you won't ever be able to achieve resolution of 0.50 arcseconds, like for example, in Edinburgh. That's why I just won't listen to this condition since we are always surrounded by air and you can rarely get the resolution below 1 arcsecond.
The limiting magnitude depends on the observer, the light pollution amount, and the weather conditions. As well, that's why I won't listen to this condition.
There is also a misconception that the telescope is all that matters. But that's wrong. Really good eyepiece is often better than a bigger telescope. Therefore, you need to include around \$100 for additional accessories, or even above \$200 if you want to get serious.
First you might want to consider what do you want to observe. Is this the Moon, the planets, the galaxies, the nebulae, the clusters, the Sun, or the Earth? We can roughly divide these telescopes into 2 categories: planetary telescopes and deep sky telescopes. Note, that when you choose the telescope of one category, you aren't limited just to the planets (and Moon, Sun), or to the DSOs (deep sky objects); the planetary telescopes just have higher magnification and are therefore more suitable for the planets. Let's go look at these two categories, one by one.
## The Deep Sky Telescopes
With these telescopes you can see all sorts of things, not just the deep sky objects (DSOs). I can recommend you Sky-Watcher 10 inch Dobson (or the cheaper 8 inch version; it is not such a big difference). Why wouldn't I recommend the Orion one (of same size)? The finderscope operates using a red dot (it projects a red dot onto the sky) but this isn't so practical. I prefer the monocular finderscope with crosshairs on the Sky-Watcher telescope since you can already see the globular clusters and the nebulae in the finderscope, so it is much easier to find them. This might be the cause of a slight price gap between those two.
You should also invest in some better eyepieces, Barlow lens (2×, 3×, or 5×), the Moon filter which attaches to the eyepiece (yes, the Moon is too bright using such large aperture), the Sun filter which attaches to the aperture (for the sunspots), some filters for the planets or the DSOs, a collimator, but you aren't obliged to do so.
## The Planetary Telescopes
For general observing I would strongly recommend you the DSO telescope, but if you are really interested in observing the planets, the Moon, and the Sun, you should buy some Schmidt-Cassegrain telescope. Such telescope essentially wraps the focal length multiple times so that it can be easily transported, but has very high magnification. The accessories can be the same as with the DSO telescopes.
But, you made some interest for the astrophotography!
## Astrophotography
There doesn't exist any telescope that would naturally enable photographing using a smartphone. Fortunately, there exist special adapters. But you must know that the light reflects on the primary mirror (1), reflects on the secondary mirror (2), refracts in many lenses in the eyepiece (3), refracts in many lenses in the objective of the smartphone (4) and lands on the sensor. The quality of image is then degraded and every dust that comes across the optical path is the problem. Many smartphones don't enable manual mode, as well.
Fortunately, there are solutions. You might have some DSLR lying around. If the objective can be taken off, you are on the horse. You just buy the T-adapter and connect it with the camera sensor. Let's see if we have somehow shortened the path: primary mirror (1), secondary mirror (2), DSLR sensor. There is less dust on the optical path. Also, we solved the issue of the manual mode, most of the DSLR cameras do have manual mode.
If you are more serious and want to invest more, you could buy a dedicated planetary camera. An example is the ASI120MC. You make a 5 min video of the planet or the Moon and stack it in the special software. The results are stunning. (Left is a single image, right is a stacked image.)
Why only a 5 min video?
Since you are pretty active on this site, I believe you know the difference between the equatorial and horizontal coordinate system. The telescopes that are intended for the astrophotography are the equatorial ones (you can change declination and the right ascension). But you said that you want the Dobson (and you are right, the equatorial mounts are just too expensive for your budget). Here comes the problem: the objects don't rotate under the equatorial system, but they do under the horizontal system:
That's why you can't have the photographing session longer than 10 minutes. Also, you would need to manually find the object, wait for it to run across the sensor and repeat the procedure.
You might have fallen asleep while reading this post (I forgive you), so I also added the conclusion:
Why wouldn't I recommend the Orion telescope? Because the red dot finderscope is not too practical to use. The monocular finderscope is better.
I prepared three plans for you: (You can cross something out, or add something new. But note, that you can always upgrade your observing equipment.)
Minimal plan:
Visual plan:
Planetary astrophotography plan:
But note that you can always invest in something more, especially in the Moon filter, and the Sun filter.
You can tell me later how you are satisfied with these choices.
• +1 great answer! I've just submitted my own, and have disagreed with you in a couple of places, I hope you don't mind :-) Good point about the 0.5 arcsec resolution: where I live the best I get is around 0.6, and it's rare. Sep 28, 2021 at 22:42
• Also, regarding astrophotography and rotation: you're absolutely right, of course, but software such as Deep Sky Stacker will detect and correct field rotation when stacking. Luckily, the Moon and planets are so bright that a five minute video is more than enough. Sep 28, 2021 at 22:46
• Have you used Deep Sky Stacker for the planet stacking? Sep 29, 2021 at 17:21
• No, I've only used Registax for stacking planets and the Moon, and DSS for everything else Sep 29, 2021 at 19:43
• Ok, I just thought that Deep Sky Stacker is used for planets as well and I didn't know it : ) Sep 29, 2021 at 19:48
TL;DR: get the telescope with the largest aperture than you can afford, can physically move, and have space to store.
This will be a Dobsonian between eight and twelve inches.
When I bought my first telescope my main considerations were budget and the light pollution where I live.
When you live in a light polluted place then you'll need a wide aperture in order to gather as much light as possible.
When you have a tight budget then you'll want most of your money to go towards the optics.
The mount is important, too: there's no point having great optics but then not being able to point the telescope at the things you want to look at.
This basically narrowed my choices to a Newtonian reflector on a Dobsonian mount.
A reflector is the cheapest way of getting a large aperture. A parabolic mirror is a lot cheaper to make than a refracting lens.
Just look at the prices of even the most basic refractors, and see how quickly they increase as the diameter increases.
A Dobsonian mount is the cheapest and most simple way of making a mount.
User123 has already given a good answer to your question, but I'm going to disagree with some of their points:
1. An eyepiece is often better than a bigger telescope
It's the telescope which gathers light, and this is fixed. A good eyepiece won't make a bad telescope any better, nor will it make a small aperture able to gather more light.
A telescope and eyepiece combine to create an optical system, and it's this system as a whole which should be considered.
What a good eyepiece will do is to ensure that that part of the optical system will perform at its best.
And what is true is that you'll probably end up spending more on eyepieces than on telescopes, long term! :-)
However, you don't need to worry about eyepieces right now. A mass-produced Dobsonian will come with two eyepieces in the box: usually a 2" low power Erfle eyepiece of around 30mm focal length and 70° apparent FoV, and a 1.25" high power Plössl eyepiece of around 9mm focal length and a narrow apparent FoV.
These are just fine to get you started. The best thing to do is to use these bundled eyepieces for a while, until you gain some experience, and then you'll know what you'll want to improve.
(You might still want to buy an eyepiece along with your telescope, though, and if you do then I would suggest you get the Baader Hyperion Zoom Mark IV, in the bundle with the Barlow lens attachment. This single eyepiece will offer you maximum versatility: it'll let you go for high magnification views when your seeing allows, and allow you to zoom out to be able to easily reacquire your target when it drifts out of view. Its only drawback is a relatively low FoV.)
1. Consider what you want to observe
If it's your first telescope then you'll want to observe everything!
To look at everything then a Dobsonian is a good all-rounder: long focal lengths for good high-powered views, and 'fast' focal ratios for wide field views.
I consider Maksutov-Cassegrains, Schmidt-Cassegrains, and long focal length refractors to be specialist telescopes. Great for looking at solar system objects, globular clusters, and planetary nebulae; but the FoV is too narrow to observe extended objects like large nebulae and open clusters, and it can be hard to find targets - especially without a good (expensive) Go-to mount.
Short focal length refractors give great low power wide views, but don't go very high.
One of these will be the second or third telescope that you buy, once you've gained some experience and you know that you specifically want one.
"Second or third telescope"? Yes...if you stick with the hobby then one day you'll buy another telescope. Most amateur astronomers have at least two telescopes, and will probably be planning on buying even more in the future.
(Personally, I spent a few years with a ten inch (250mm) Dobsonian, but it's not easy to travel with, so I bought an 80mm refractor to be able to take on a plane; and I'm considering getting a 130mm Maksutov for those high-power views of planets.)
I agree with pretty much everything else User123 says, though :-)
Get a Dobsonian with a magnifying finder scope, rather than the one with the red dot.
A red dot finder is fairly worthless on a Dobsonian - especially under light polluted skies - it doesn't magnify the view so doesn't show you any more. If you want a non-magnifying finder later then get a Telrad: it has markings which help measure distances and position the telescope to find things.
Also consider the mount. Not all Dobsonian mounts are equal. You're going to be moving it left and right and up and down a lot. Compare reviews and see what they say about the mounts.
Generally, the bigger the trunnions the better - a larger surface area there will give a smoother up-and-down (altitude) movement.
When I bought my GSO 10" Dob, there were two options: the standard and the "deluxe". I went for the deluxe version because the mount was improved: the standard version had a simple Teflon pad in the base for the left-and-right (azimuth) movement and smaller trunnions, whereas the deluxe had a "lazy Susan" plate with bearings and larger trunnions.
Finally, forget about astrophotography for now. Sure, buy a cheap attachment which will let you position your phone above the eyepiece and take some photos of the Moon and Jupiter, but you'll need to buy more equipment to be able to take photos of the fainter objects out there.
This isn't to say that astrophotography with a Dobsonian is impossible - not at all: I've dabbled in it myself - but you'll need a very sensitive camera and use a different technique, and the really faint objects will still be out of reach.
If you want to start doing astrophotography then it requires completely different equipment to visual astronomy: it's all about the mount. Spend a lot of money on a mount, put a small (50-70mm) short (f/5 or less) refractor on it, and use a colour astrocamera with it.
Then get used to spending more time processing the resulting data than you did capturing it.
After that, move up to a monochrome astrocamera, lots of filters, and bigger telescope, and a bigger mount.
It's a rabbit hole and a money sink.
Halfway between visual astronomy and astrophotography is electronically assisted astrophotography (EAA).
This is where you use an astrocamera with a very sensitive sensor with a program which does "live stacking" (on Windows this program is usually SharpCap, and on Linux it's ASTAP).
This gives almost instant results, and under light polluted skies can be the only way to see some objects.
I personally have seen some galaxies using this technique which are otherwise invisible to me.
Conclusion:
Buy an eight, ten, or twelve inch Dobsonian. All of these will be about the same length - they're usually: 8" F/6, 10" F/5, 12" F/4.
A final consideration is that the 'faster' (shorter focal ratio) the telescope, the more expensive the well-corrected eyepieces. Reflectors suffer from "coma" - a stretching at the edges of the view which is more pronounced at shorter focal ratios.
You can buy a coma corrector to correct this, but again, that's even more money. (Sometimes it feels like it never ends! The initial \$800 budget easily turns into thousands spent over a few years!)
Check the classified adverts in your area if you don't mind buying second-hand. You'll save a lot of money doing this. On the other hand, until you have some experience it'll be hard to know what issues to look out for with second-hand equipment.
If your budget is a hard budget then go for a slightly cheaper telescope so you have some left over for accessories - the book Turn Left at Orion, for instance, is a good companion to a first telescope.
If your budget is a soft budget then blow it all on the telescope and then in a few months you'll get to know what additional bits and pieces you'll want.
Finally: good luck and have fun!
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# Difference between revisions of "Bernoulli equation"
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An ordinary first-order differential equation
$$a_0(x)y'+a_1(x)y=f(x)y^\alpha,$$
where $\alpha$ is a real number other than zero or one. This equation was first studied by J. Bernoulli [1]. The substitution $y^{1-\alpha}=z$ converts the Bernoulli equation to a linear inhomogeneous first-order equation, [2]. If $\alpha>0$, the solution of the Bernoulli equation is $y\equiv0$; if $0<\alpha<1$, at some points the solution is no longer single-valued. Equations of the type
$$[f(y)x+g(y)x^\alpha]y'=h(y),\quad\alpha\neq0,1,$$
are also Bernoulli equations if $y$ is considered as the independent variable, while $x$ is an unknown function of $y$.
#### References
[1] J. Bernoulli, Acta Erud. (1695) pp. 59–67; 537–557 [2] E. Kamke, "Differentialgleichungen: Lösungen und Lösungsmethoden" , 1. Gewöhnliche Differentialgleichungen , Chelsea, reprint (1971)
#### References
[a1] E.L. Ince, "Ordinary differential equations" , Dover, reprint (1956)
How to Cite This Entry:
Bernoulli equation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bernoulli_equation&oldid=15844
This article was adapted from an original article by N.Kh. Rozov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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# Uniform and Pointwise Convergance of Functions
Can anyone explain to me why if a sequence of functions {f_n} is uniformly convergent, then it must converge to it's pointwise limit?
So, if we showed {f_n} is not uniformly convergent to some f (f is the pointwise limit), how do we know that there isn't some other function, say g, such that {f_n} converges to g uniformly?
CompuChip
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