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# Calculate the amount of oxygen provided to an aerobic culture, expressed as mg of O2/Liz of...
###### Question:
calculate the amount of oxygen provided to an aerobic culture, expressed as mg of O2/Liz of culture-hour, if 20 L of air, at stp conditions, is bubbled per hour into the culture brother volume of one liter. Assume that only 25% of the oxygen in the bubbled air ends up dissolved in the water, due to mass transfer limitations. Show the calculations you performed to arrive to your answer.
NS: oxygen supply in fermentation broth Calculate the amount of oxymen provided to an aerobit umure, e,pressed as godu culture-hour. 20 air, at STP conditions, is bubbled per hour into a cunure broth wolume of one Iner. Assume that only 25% of the oxygen in the bubbled a aer, due to mass transier timitalions. Show ends up dinsnived in the he calculations you performed to arrive at your at
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# Help understanding the proof of Trace Theorem given in Evans
I need help to understand the proof of the Trace Theorem given in Evans L.C. Partial differential equations (AMS, 1997):
Asume $U$ is a bounded open set and that $\partial U$ is $C^1$. Then there exists a bounded linear operator $T:W^{1,p}(U) \to L^p(\partial U)$ such that:
$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$
and
$\quad$(ii) $\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.
The proof provided starts like this:
Okay, so we first asume that there is a neighborhood $U_{x^0}$ of $x^0$ such that $U_{x^0} \subset \{x \in \mathbb{R}^n \ | \ x_n=0\}$ and prove (1).
Then the proof continues by saying "If $\partial U$ is not flat near $x^0$", we straighten out the boundary near $x^0$ to obtain the setting above. Applying estimate (1) and changing variables... (The proof continues)
Could someone clarify me a little more precisely what straighten out the boundary near $x^0$ means?
Updated: This is the definition of open set of class $C^1$ which gives us a diffeomorphism to the set $Q_0$ which is "flat".
We define the following sets:
• $R_+ = \{x=(x_1,...,x_n) \in \mathbb{R}^n \ | \ x_n \geq 0\}$
• $Q = \{x=(x_1,...,x_n) \in \mathbb{R}^n \ | \ (\sum_{i=1}^{n-1} x_i^2)^{1/2} < 1 \ y \ |x_n|<1 \}$
• $Q_+=R_+ \cap Q$
• $Q_0=\{(x_1,...,x_{n-1},0) \in \mathbb{R}^n \ | \ (\sum_{i=1}^{n-1} x_i^2)^{1/2} < 1 \}$
An open set $\Omega$ is of class $C^1$ if for every $x \in \partial \Omega$ there exists a neighborhood $U_x$ of $x$ in $\mathbb{R}^n$ and a bijective map $H: Q \to U_x$ such that:
• $H \in C^1(\overline{Q})$
• $H^{-1} \in C^1(\overline{U_x})$
• $H(Q_+)=U_x \cap Q$
• $H(Q_0)= U_x \cap \partial \Omega$
• Find a differentiable map that transforms a piece of $\partial U$ near $x^0$ into a piece of a plane. Then you change variables in the integrals. – DisintegratingByParts May 5 '16 at 17:43
• @TrialAndError Could you please explain it a bit more? I suppose you mean transform a neighborhood of $x^0$ into a piece of a hyperplane (into a piece of a $n-1-$ dimensional surface), Right? Why do you have the guarantee to have that difeomorphism, is it simply because $U$ is open, right? – D1X May 5 '16 at 17:57
• Having the diffeomorphism is essentially assumed by asserting that $\partial U$ is a $C^1$ manifold. – DisintegratingByParts May 5 '16 at 18:08
• @TrialAndError And shouldn't the Jacobian appear due to the change of variables? The book says "Applying estimate (1) and changing variables we obtain the bound: $\int_{\Gamma} |u|^p dS \leq C \int_{U} |u|^p + |Du|^p dx$, where $\Gamma$ is some open subset of $\partial U$ cointaining $x_0$." – D1X May 6 '16 at 16:32
• The Jacobian will be bounded above and below in the local neighborhood that you obtain by move normally off the $C^1$ boundary. – DisintegratingByParts May 6 '16 at 16:37
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Version 3 (modified by mikhail.vorozhtsov, 3 years ago) (diff)
Added MultiClauseLambdas with \ in each clause
This page is a summary of proposals from #4359
## The problem
The current lambda abstraction syntax allows us to conveniently bind parts of the arguments by using patterns, but does not provide a way to branch quickly (without naming the argument). Usually we just cringe a bit and write
```\tmp -> case tmp of
Pat1 -> ...
Pat2 -> ...
```
or even
```\meaningfulName -> case meaningfulName of
Pat1 -> ...
Pat2 -> ...
```
However, when this situation nests (e.g. monadic bind) and variables have the same type, naming becomes painful and often degrades to not-so-meaningful meaningfulName1, meaningfulName2, ...
A similar problem exists with proc expressions from arrow notation, which can be regarded as generalized lambda expressions. We sometimes have expressions of the following structure:
```proc meaningfulName -> case meaningfulName of
Pat1 -> ...
Pat2 -> ...
```
Here, the dots stand for arrow expressions, not ordinary expressions.
## The proposals
### LambdaCase: case of
A simple sugar for one-argument lambda abstractions.
```case of
Pat1 -> Expr1
Pat2 -> Expr2
```
desugars to
```\freshName -> case freshName of
Pat1 -> Expr1
Pat2 -> Expr2
```
• Pros
• No conflicts with the current syntax (the sequence case of is illegal)
• Cons
• Looks weird (no hint of being a lambda abstraction)
• Single-argument solution (see the note)
• Cannot be generalized to cover also proc expressions
### LambdaCase: \case
A "less weird" version of case of. As above,
```\case
Pat1 -> Expr1
Pat2 -> Expr2
```
desugars to
```\freshName -> case freshName of
Pat1 -> Expr1
Pat2 -> Expr2
```
(\case is a layout herald).
• Pros
• No conflicts with the current syntax (the sequence \ case is illegal)
• An analog syntax for proc expressions can be gained by replacing \ with proc
• Cons
• Single-argument solution (see the note). One way to extend it to support multiple arguments is
```\case
Pat1_1, Pat1_2, ... -> Expr1
Pat2_1, Pat2_2, ... -> Expr2
```
(separation with commas is supposed to preserve case-like feel, e.g. Just x, Just y -> vs (Just x) (Just y) ->) which is considered unorthodox by GHC HQ.
### MultiClauseLambdas
Extend the current syntax with alternative clauses:
```\Pat1_1 Pat1_2 ... -> Expr1
Pat2_1 Pat2_2 ... -> Expr2
...
```
(\ becomes a layout herald)
• Pros
• Multi-argument solution (see the note)
• An analog syntax for proc expressions can be gained by replacing \ with proc
• Cons
• Breaks current idioms. For example,
```mask \$ \restore -> do
stmt1
stmt2
```
becomes illegal because stmt1 is indented less than restore. One way to avoid this is to not make \ a herald, forcing users to use explicit layout for multi-clause abstractions, i.e.
```\ { Pat1_1 Pat1_2 ... -> Expr1
; Pat2_1 Pat2_2 ... -> Expr2 }
```
Another is to start each clause with a \:
```\ Pat1_1 Pat1_2 ... -> Expr1
\ Pat2_1 Pat2_2 ... -> Expr2
\ ...
```
### MultiClauseLambdas with a keyword
Addresses the layout problem of MultiClauseLambdas. Requires multi-clause abstractions to have a keyword after \:
```\KEYWORD Pat1_1 Pat1_2 ... -> Expr1
Pat2_1 Pat2_2 ... -> Expr2
...
```
(\KEYWORD is a layout herald)
• Pros
• No conflicts with the current syntax
• Multi-argument solution (see the note)
• An analog syntax for proc expressions can be gained by replacing \ with proc
• Cons
• Deciding on the keyword may take years
### Extra: LambdaMatch
A full revamp of pattern matching: Haskell' ticket.
## Notes
### Single vs multi-argument
(field report by Mikhail Vorozhtsov) I've been using \case for over a year and tried MultiClauseLambdas for about two months. In my code base \case seems to cover 99% of cases (no pun intended) and curry \$ \case ... does the job for the rest, so having only a single-argument solution may be not as restrictive as it seems. On the other hand, I had a hard time with MultiClauseLambdas extra clauses: I just kept writing Just x -> Expr instead of the correct (Just x) -> Expr. It seems that lines like [spaces]Pat -> Expr are just hardwired to case-expressions in my brain.
|
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PROGRAMMING AND DEBUGGING QUESTIONS DO NOT BELONG ON META. Use this tag for a reproducible problem ON THIS SITE that you believe is due to a mistake, malfunction, or programming error.
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# Interpretation of R's lm() output
The help pages in R assume I know what those numbers mean, but I don't. I'm trying to really intuitively understand every number here. I will just post the output and comment on what I found out. There might (will) be mistakes, as I'll just write what I assume. Mainly I'd like to know what the t-value in the coefficients mean, and why they print the residual standard error.
Call:
lm(formula = iris$Sepal.Width ~ iris$Petal.Width)
Residuals:
Min 1Q Median 3Q Max
-1.09907 -0.23626 -0.01064 0.23345 1.17532
This is a 5-point-summary of the residuals (their mean is always 0, right?). The numbers can be used (I'm guessing here) to quickly see if there are any big outliers. Also you can already see it here if the residuals are far from normally distributed (they should be normally distributed).
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 3.30843 0.06210 53.278 < 2e-16 ***
iris$Petal.Width -0.20936 0.04374 -4.786 4.07e-06 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Estimates$\hat{\beta_i}$, computed by least squares regression. Also, the standard error is$\sigma_{\beta_i}$. I'd like to know how this is calculated. I have no idea where the t-value and the corresponding p-value come from. I know$\hat{\beta}$should be normal distributed, but how is the t-value calculated? Residual standard error: 0.407 on 148 degrees of freedom $\sqrt{ \frac{1}{n-p} \epsilon^T\epsilon }$, I guess. But why do we calculate that, and what does it tell us? Multiple R-squared: 0.134, Adjusted R-squared: 0.1282 $ R^2 = \frac{s_\hat{y}^2}{s_y^2} $, which is$ \frac{\sum_{i=1}^n (\hat{y_i}-\bar{y})^2}{\sum_{i=1}^n (y_i-\bar{y})^2} $. The ratio is close to 1 if the points lie on a straight line, and 0 if they are random. What is the adjusted R-squared? F-statistic: 22.91 on 1 and 148 DF, p-value: 4.073e-06 F and p for the whole model, not only for single$\beta_i$s as previous. The F value is$ \frac{s^2_{\hat{y}}}{\sum\epsilon_i} $. The bigger it grows, the more unlikely it is that the$\beta$'s do not have any effect at all. • residuals are not so badly deviating from normality, why do you think so? – nico Dec 4 '10 at 13:14 • @nico: I think @Alexx Hardt was speaking hypothetically. I.e. once could use the five number summary to see if residuals were deviating from normal – Reinstate Monica - G. Simpson Dec 4 '10 at 13:39 • @Gavin Simpson: you're right, I misread the sentence. Disregard my previous comment. – nico Dec 4 '10 at 14:34 • Minor quibble: You cannot say anything about normality or non-normality based on those 5 quantiles alone. All you can say based on that summary is whether the estimated residuals are approximately symmetric around zero. You could divide the reported quantiles by the estimated residual standard error and compare these values to the respective quantiles of the N(0,1), but looking at a QQ-plot probably makes more sense. – fabians Dec 6 '10 at 9:29 • One note here: the model$F$is not$SS_{model} / SS_{error}$, rather it is$MS_{model} / MS_{error}$.$F$is described correctly in the answer below, but it does not explicitly mention that it is mischaracterized in the question, so someone might not notice the discrepancy. – gung - Reinstate Monica Aug 23 '12 at 14:28 ## 2 Answers ## Five point summary yes, the idea is to give a quick summary of the distribution. It should be roughly symmetrical about mean, the median should be close to 0, the 1Q and 3Q values should ideally be roughly similar values. ## Coefficients and $$\hat{\beta_i}s$$ Each coefficient in the model is a Gaussian (Normal) random variable. The $$\hat{\beta_i}$$ is the estimate of the mean of the distribution of that random variable, and the standard error is the square root of the variance of that distribution. It is a measure of the uncertainty in the estimate of the $$\hat{\beta_i}$$. You can look at how these are computed (well the mathematical formulae used) on Wikipedia. Note that any self-respecting stats programme will not use the standard mathematical equations to compute the $$\hat{\beta_i}$$ because doing them on a computer can lead to a large loss of precision in the computations. ## $$t$$-statistics The $$t$$ statistics are the estimates ($$\hat{\beta_i}$$) divided by their standard errors ($$\hat{\sigma_i}$$), e.g. $$t_i = \frac{\hat{\beta_i}}{\hat{\sigma_i}}$$. Assuming you have the same model in object modas your Q: > mod <- lm(Sepal.Width ~ Petal.Width, data = iris) then the $$t$$ values R reports are computed as: > tstats <- coef(mod) / sqrt(diag(vcov(mod))) (Intercept) Petal.Width 53.277950 -4.786461 Where coef(mod) are the $$\hat{\beta_i}$$, and sqrt(diag(vcov(mod))) gives the square roots of the diagonal elements of the covariance matrix of the model parameters, which are the standard errors of the parameters ($$\hat{\sigma_i}$$). The p-value is the probability of achieving a $$|t|$$ as large as or larger than the observed absolute t value if the null hypothesis ($$H_0$$) was true, where $$H_0$$ is $$\beta_i = 0$$. They are computed as (using tstats from above): > 2 * pt(abs(tstats), df = df.residual(mod), lower.tail = FALSE) (Intercept) Petal.Width 1.835999e-98 4.073229e-06 So we compute the upper tail probability of achieving the $$t$$ values we did from a $$t$$ distribution with degrees of freedom equal to the residual degrees of freedom of the model. This represents the probability of achieving a $$t$$ value greater than the absolute values of the observed $$t$$s. It is multiplied by 2, because of course $$t$$ can be large in the negative direction too. ## Residual standard error The residual standard error is an estimate of the parameter $$\sigma$$. The assumption in ordinary least squares is that the residuals are individually described by a Gaussian (normal) distribution with mean 0 and standard deviation $$\sigma$$. The $$\sigma$$ relates to the constant variance assumption; each residual has the same variance and that variance is equal to $$\sigma^2$$. ## Adjusted $$R^2$$ Adjusted $$R^2$$ is computed as: $$1 - (1 - R^2) \frac{n - 1}{n - p - 1}$$ The adjusted $$R^2$$ is the same thing as $$R^2$$, but adjusted for the complexity (i.e. the number of parameters) of the model. Given a model with a single parameter, with a certain $$R^2$$, if we add another parameter to this model, the $$R^2$$ of the new model has to increase, even if the added parameter has no statistical power. The adjusted $$R^2$$ accounts for this by including the number of parameters in the model. ## $$F$$-statistic The $$F$$ is the ratio of two variances ($$SSR/SSE$$), the variance explained by the parameters in the model (sum of squares of regression, SSR) and the residual or unexplained variance (sum of squares of error, SSE). You can see this better if we get the ANOVA table for the model via anova(): > anova(mod) Analysis of Variance Table Response: Sepal.Width Df Sum Sq Mean Sq F value Pr(>F) Petal.Width 1 3.7945 3.7945 22.91 4.073e-06 *** Residuals 148 24.5124 0.1656 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 The $$F$$s are the same in the ANOVA output and the summary(mod) output. The Mean Sq column contains the two variances and $$3.7945 / 0.1656 = 22.91$$. We can compute the probability of achieving an $$F$$ that large under the null hypothesis of no effect, from an $$F$$-distribution with 1 and 148 degrees of freedom. This is what is reported in the final column of the ANOVA table. In the simple case of a single, continuous predictor (as per your example), $$F = t_{\mathrm{Petal.Width}}^2$$, which is why the p-values are the same. This equivalence only holds in this simple case. • @Gavin (+1) Great response with nice illustrations! – chl Dec 4 '10 at 14:04 • Nice job. One thing you might clarifiy, with regard to calculating t values: sqrt(diag(vcov(mod))) produces the SE of the estimates. These are the same SEs that are output in the model summary. Easier and clearer just to say that t = Estimate/SEestimate. In that sense it is no different that any other t value. – Brett Dec 4 '10 at 14:49 • (+1) This is great. The only thing I'd add is that the$F$value is the same as$t^2\$ for the slope (which is why the p values are the same). This - of course - isn't true with multiple explanatory variables. – user1108 Dec 4 '10 at 15:05
• @Jay; thanks. I thought about mentioning that equivalence too. Wasn't sure if it was too much detail or not? I'll ad something on this in a mo. – Reinstate Monica - G. Simpson Dec 4 '10 at 15:43
• "will not use the standard mathematical equations to compute" What will they use? – HelloWorld Jan 8 '15 at 3:54
Ronen Israel and Adrienne Ross (AQR) wrote a very nice paper on this subject: Measuring Factor Exposures: Uses and Abuses.
To summarize (see: p. 8),
• Generally, the higher the $$R^2$$ the better the model explains portfolio returns.
• When the t-statistic is greater than two, we can say with 95% confidence (or a 5% chance we are wrong) that the beta estimate is statistically different than zero. In other words, we can say that a portfolio has significant exposure to a factor.
R's lm() summary calculates the p-value Pr(>|t|). The smaller the p-value is, the more significant the factor is. P-value = 0.05 is a reasonable threshold.
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float *s; Æ s contains the address of floating-point value. a.1 b.–1 c. i d. –i. Contact us on below numbers. Submit your answer. For a more thorough discussion, see square root and branch point. Lv 4. 1 answer. Inform you about new question papers. NCERT Books. Biology. Paiye sabhi sawalon ka Video solution sirf photo khinch kar. Precalculus. i 3 = i × i × i =... See full answer below. Now solving the given expression. Threats … Just like all nonzero complex numbers, i has two square roots: they are[b]. asked Jan 17, 2020 in Physics by SurajKumar (66.2k points) Find the value of 'n' so that vector (2i - 3j + k) may be perpendicular to the vector (3i + 4j + nk). –3 + 6i 3 + 6i 5 + 5i 5 – 5i. Each wall must be 1 1⁄3 feet long. = 1 Answers. vector algebra; class-12; Share It On Facebook Twitter Email. her car's odometer is slightly off and measures the distance as 24 miles. How do I determine the molecular shape of a molecule? Questions. 2 See answers is it "i to the power of a number" or "i times a number"? Submit your answer. − That is, i 99 = i 3, because you can just lop off the i 96. The square of an imaginary number bi is −b 2.For example, 5i is an imaginary number, and its square is −25.By … For multiple Processors, multiply the price shown by the number of CPUs. For internet numbers, see, To find such a number, one can solve the equation, https://en.wikipedia.org/w/index.php?title=Imaginary_unit&oldid=999129359, All articles with specifically marked weasel-worded phrases, Articles with specifically marked weasel-worded phrases from February 2020, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, This page was last edited on 8 January 2021, at 16:57. This is the fundamental cube root of i (i.e. In complex numbers nth roots of a complex number. 3.2k views. What is the lewis structure for co2? Mathematics. Being a quadratic polynomial with no multiple root, the defining equation x2 = −1 has two distinct solutions, which are equally valid and which happen to be additive and multiplicative inverses of each other. Assume that social security promises you $35,000 per year starting when you retire 45 years from today (the first$35,000 will get paid 45 years fro... stanley's bicycles store buys bicycles on average for $610 and sells them on average for$ 770. he pays a sales commission of 15% of sales revenue... View a few ads and unblock the answer on the site. KEAM 2017: The value of the currents I1, I2, and I3 flowing through the circuit given below is (A) I1 = - 3A, I2 = 2A, I3 = -1A (B) I1 = 2A, I2 = -3A, 2 sin the sine of a value or expression. Eg:1. 1 answer. If |z + 1| = z + 2( 1 + i), then find the value of z. asked Feb 19, 2018 in Class XI Maths by rahul152 (-2,838 points) complex numbers and quadratic equations. Use ^(1/2) for square root ,'*' for multiplication, '/' for division, '+' for addition, '-' for subtraction. Join now. Delen. Jan 14,2021 - Reduce the following game by dominance and find the game value player-B I) 3 2 4 0 II)3 4 2 4 Iii)4 2 4 0 IIII)0 4 0 8? Value of i = 3 Pointer Variables Examples: Addresses (location nos.) Sam creates a game in which the player rolls 4 dice. 2 c. 1 d. 0. The absolute value of a number is often viewed as the "distance" a number is away from 0, the origin. why create a profile on Shaalaa.com? 1 Answers. Log in. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. And that leads us into another topic, the complex plane: Conclusion. Franchisee/Partner … Since the equation is the only definition of i, it appears that the definition is ambiguous (more precisely, not well-defined). The unit imaginary number, i, equals the square root of minus 1. Since the equation is the only definition of i, it appears that the definition is ambiguous (more precisely, not well-defined). determine the values of x and y 1 i x 2i 3 i 2 3i y i 3 i i where i iota - Mathematics - TopperLearning.com | dsw8iavv. Chemistry. Write input √x as x^(1/2) 2. Autodetect radians/degrees. Which best describes the relationship between the church and feudal states? Although the construction is called "imaginary", and although the concept of an imaginary number may be intuitively more difficult to grasp than that of a real number, the construction is perfectly valid from a mathematical standpoint. Grade 7 Math Makes Sense textbook program. desi9750. i 3 = −i: i 4 = +1: i 5 = i: i 6 ... of each other. Value of output of General Government which is taken to be equal to the value of purchases of goods and services by the Government denote by (G) 4. NCERT P Bahadur IIT-JEE Previous Year Narendra Awasthi MS Chauhan. View Answer. Equivalently: The principal value (for k = 0) is e−π/2, or approximately 0.207879576 .[10]. The square of an imaginary number bi is −b 2.For example, 5i is an imaginary number, and its square is −25.By definition, zero is considered to be both real and imaginary. due to the economy, the company increased the regular work week to 44 hours. Mathematics. Higher integral powers of i can also be replaced with −i, +1, +i, or −1: Similarly, as with any non-zero real number: As a complex number, i is represented in rectangular form as 0 + 1i, with a zero real component and a unit imaginary component. In this case, because the fair value of the non-controlling interests in scenario 1 is proportional to the consideration paid by Parent, the notional adjustment leads to the same impairment losses of $450 for (i) and$550 for (ii) as under scenario 1 (see *). Imaginary Numbers are not "Imaginary". … Autodetect radians/degrees. Furthermore. 3. if the medicare tax rate is 5.4%, what is the amount deducted from a $561.98 paycheck? -1 on Edg ;)) New questions in Mathematics. asked Feb 17, 2018 in Class XI Maths by vijay Premium (539 points) complex numbers and quadratic equations . Physics. Asked By adminstaff @ 22/01/2020 08:31 AM. Related Questions in Mathematics. {\displaystyle {\begin{pmatrix}z&x\\y&-z\end{pmatrix}}^{2}\!\!={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}}.} Unit 6 - Solving Equations. At the Last Chance Filling Station, gas costs$1.60 a gallon. For more information read our Terms of use & Privacy Policy, And millions of other answers 4U without ads. 1. Indeed, squaring both expressions yields: Using the radical sign for the principal square root, we get: Similar to all the roots of 1, all the roots of i are the vertices of regular polygons, which are inscribed within the unit circle in the complex plane. 1 Answer sjc Jul 30, 2017 #-i# Explanation: remember that #" "i^2=-1# #(-i)^3# #=-i xx -i xx-i# #(-i)^2xx-i# #=+1xx-i# #-i# Answer link. Become our. Engels Hardcover 9780824889067 februari 2021 300 pagina's Alle productspecificaties. 0 1. 10:00 AM to 7:00 PM IST all days. The distinction between the two roots x of x2 + 1 = 0, with one of them labelled with a minus sign, is purely a notational relic; neither root can be said to be more primary or fundamental than the other, and neither of them is "positive" or "negative".[4]. 1800-212-7858 / 9372462318. Contents. Pandemics. Asked By adminstaff @ 22/01/2020 08:31 AM. Class 1 - 3; Class 4 - 5; Class 6 - 10 ; Class 11 - 12; CBSE. Question: What is the value of 15i/2+i ? 1 Answer +1 vote . Solution for Let f be defined piecewise as follows: kæ sin(x – 3) f(z) = + 4, if æ < 3 I – 3 4x2 + k, if æ > 3 For what value of k is ƒ continuous? a $5.40 b$30.35 c $40.85. The imaginary-base logarithm of a number is: As with any complex logarithm, the log base i is not uniquely defined. 1 decade ago. 24 divided by 2 - 6 + 8 divided by 2 … (This is equivalent to a 90° counter-clockwise rotation of a vector about the origin in the complex plane.). Substituting these in turn for − / in Tartaglia's cubic formula and simplifying, one gets 0, 1 and −1 as the solutions of x 3 − x = 0. a) 1 b) -1 c) i d) -i It's very important to name your variables properly. The best EV for the money? Acarpenter earns$24 an hour for a regular week of 40 hours. CBSE CBSE (Science) Class 11. 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# matrix transformation for trig identities
• February 16th 2010, 10:47 AM
s_ingram
matrix transformation for trig identities
Hi folks,
Given that the following matrix creates an anticlockwise rotation of the x-y plane about the origin through an angle $\theta$:
$\left( \begin{array}{cc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{array} \right)
$
use this fact to obtain the standard trig identities:
$\sin(\theta + \alpha) = \sin\theta\cos\alpha + cos\theta\sin\alpha
$
and
$\cos(\theta + \alpha) = \cos\theta\cos\alpha - \sin\theta\sin\alpha
$
I can obtain the identities using geometry and I can see how the matrix transformation creates an anticlockwise rotation, but I can't see how to use the matrix trransformation to generate the identities. Can anyone help?
• February 16th 2010, 10:56 AM
icemanfan
Calculate
$
\left( \begin{array}{cc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{array} \right)
\left( \begin{array}{cc}
\cos\alpha & -\sin\alpha \\
\sin\alpha & \cos\alpha
\end{array} \right)
$
and note that this is equivalent to rotating through by $\theta + \alpha$, which is the matrix
$
\left( \begin{array}{cc}
\cos(\theta + \alpha) & -\sin(\theta + \alpha) \\
\sin(\theta + \alpha) & \cos(\theta + \alpha)
\end{array} \right)$
• February 16th 2010, 10:59 AM
s_ingram
Ahhhhh! Now I see. Thanks very much Icemanfan.
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## Interstitial fluid pressure (IFP) spatial distribution at various time instants.
2017-01-26T00:32:58Z (GMT) by
<p>Within 24 hours of relatively slow avascular tumour growth, tumour-secreted angiogenic chemical factors have sufficiently diffused in the extracellular space. In the tumour-induced angiogenesis simulation, day-1 marks the formation and elongation of new blood vessel sprouts. The vertical red line in the plots defines the tumour–host interface boundary which corresponds to the averaged tumour radius, where the centre of the cancer mass is at zero radial distance. The solid line corresponds to the mean IFP distribution, evaluated at fourteen azimuthal directions, while the vertical bars denote the IFP standard deviation. These plots also illustrate the gradual increase of the cancer mass, while depicting the significant variability of IFP in the vicinity of the tumour interstitium.</p>
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# Explosive reaction based on reaction rate
### Question
We have the following chemical reactions:
\begin{align} \ce{A + B^* &-> C + D^*}\tag{1}\\ \ce{D^* &-> B^* + B^*}\tag{2}\\ \ce{B^* &-> B}\tag{3} \end{align}
If the initial concentrations are $$[\ce{A}] = \pu{100e-6 mol cm-3}$$ and $$[\ce{B}] = \pu{1e-6 mol cm-3}$$, is the system explosive? We know that $$k_1 = \pu{1e6 cm3 mol-1 s-1},$$ $$k_2 = \pu{50 s-1}$$ and $$k_3 = \pu{25 s-1}.$$
### My solution
So I found the rate expressions for $$\ce{A}$$ and $$\ce{B},$$ by applying steady-state approximation (SSA) on $$[\ce{D^*}].$$ I get the following expressions:
\begin{align} \frac{\mathrm d[\ce{A}]}{\mathrm dt} &= -k_1 \cdot [\ce{A}] \cdot [\ce{B^*}]\\ \frac{\mathrm d[\ce{B}]}{\mathrm dt} &= (k_1 \cdot [\ce{A}] - k_3) \cdot [\ce{B^*}] \end{align}
Upon entering the numbers into the above expressions, I get the following numbers:
\begin{align} \frac{\mathrm d[\ce{A}]}{\mathrm dt} &= \pu{-\frac{1}{10000} mol cm-3 s-1}\\ \frac{\mathrm d[\ce{B}]}{\mathrm dt} &= \pu{\frac{3}{40000} mol cm-3 s-1} \end{align}
My conclusion was that the system is not explosive since the rate of formation of $$\ce{B}$$ is quite low. Since if the rate is too low, then the reaction can not be considered explosive.
But the problem with my explanation is that I would like to have a reference number. Is there a reference such as f.e. rate higher than $$10$$ is considered explosive, etc? I have tried looking everywhere but could not find any 'value.' Could anyone provide some insight into this:
• Are my explanation and conclusion correct?
• Is there a reference value for explosive reactions?
I have an additional question. What is the connection between an explosive system and the steady-state approximation? If the system is explosive, can I still use SSA or does it become invalid?
• The steady-state approximation only requires that the concentration of intermediates is low. As long as that is the case, it is valid. – Zhe Sep 27 '19 at 0:08
• @Zhe I'm not sure that what you write is necessarily true. What is important is that the rate of change is low, i.e. assumed to be zero, rather than the amount being low. – porphyrin Sep 27 '19 at 9:33
• Shouldn't be that if B• is formed faster than C + D than it is esplosive? Can't read the equation but my comment is general (and might be wrong). – Alchimista Sep 27 '19 at 10:40
• Taking your equation $d[B]^*/dt\cdots$ and integrating gives $[B^*]=[B_0^*]\exp((k_1A-k_3)t)$ and if $k_1A \gt k_3$ the exponential will grow and then one might expect an explosion. – porphyrin Sep 27 '19 at 11:10
• @porphyrin The steady state approximation argues that the rate of change is near zero by arguing that the concentration of intermediate is low. If it's always low, it can't be changing by much. – Zhe Sep 27 '19 at 12:00
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# Dichotomies in Ontology-Mediated Querying with the Guarded Fragment
@article{Hernich2017DichotomiesIO,
title={Dichotomies in Ontology-Mediated Querying with the Guarded Fragment},
author={Andr{\'e} Hernich and Carsten Lutz and Fabio Papacchini and Frank Wolter},
journal={Proceedings of the 36th ACM SIGMOD-SIGACT-SIGAI Symposium on Principles of Database Systems},
year={2017}
}
• Published 9 May 2017
• Computer Science
• Proceedings of the 36th ACM SIGMOD-SIGACT-SIGAI Symposium on Principles of Database Systems
We study the complexity of ontology-mediated querying when ontologies are formulated in the guarded fragment of first-order logic (GF). Our general aim is to classify the data complexity on the level of ontologies where query evaluation w.r.t. an ontology O is considered to be in PTime if all (unions of conjunctive) queries can be evaluated in PTime w.r.t. O and coNP-hard if at least one query is coNP-hard w.r.t. O. We identify several large and relevant fragments of GF that enjoy a dichotomy…
23 Citations
## Figures and Tables from this paper
Dichotomies in Ontology-Mediated Querying with the Guarded Fragment
• Computer Science
ACM Trans. Comput. Log.
• 2020
The aim is to classify the data complexity and Datalog rewritability of query evaluation depending on the ontology O, where query evaluation w.r.t. O is in PTime, and the decidability of whether a given ontology enjoys PTime query evaluation is studied, presenting both positive and negative results,Depending on the fragment.
The Data Complexity of Ontology-Mediated Queries with Closed Predicates
• Computer Science
Log. Methods Comput. Sci.
• 2019
This work provides a non-uniform analysis, aiming at a classification of the complexity into tractable and non-tractable for ontologies in the lightweight DLs DL-Lite and EL, and the expressive DL ALCHI, and shows that there is no dichotomy if both concept and role names can be closed.
The Data Complexity of Description Logic Ontologies
• Computer Science, Philosophy
Log. Methods Comput. Sci.
• 2017
This work analyzes the data complexity of ontology-mediated querying where the ontologies are formulated in a description logic of the ALC family and queries are conjunctive queries, positive existential queries, or acyclic conj unctive queries and shows that it is undecidable whether a given ontology admits PTime query evaluation.
How to Approximate Ontology-Mediated Queries
• Computer Science
KR
• 2021
This work introduces and study several notions of approximation for ontology-mediated queries based on the description logics ALC and ALCI, and determines the computational complexity and the relative completeness of the resulting approximations.
Horn-Rewritability vs PTime Query Evaluation in Ontology-Mediated Querying
• Computer Science
IJCAI
• 2018
This work investigates in which cases (1) and (2) are equivalent, finding that the answer depends on whether the unique name assumption (UNA) is made, on the description logic under consideration, and on the nesting depth of quantifiers in the TBox.
Logical Separability of Incomplete Data under Ontologies
• Computer Science
KR
• 2020
This paper investigates the existence of a separating formula for incomplete data in the presence of an ontology, characterize separability in a model-theoretic way, compare the separating power of the different languages, and determine the computational complexity of separability as a decision problem.
A Note on DL-Lite with Boolean Role Inclusions
• Computer Science, Philosophy
Description Logics
• 2019
It is shown that binary disjunctions on roles do not change the complexity of satisfiability if disjunction is allowed on concepts, while still causing undecidability of UCQ answering.
Horn Rewritability vs PTime Query Answering for Description Logic TBoxes
• Computer Science
Description Logics
• 2017
The main results are that this is indeed the case when L is the set of ALCHI or ALCIF TBoxes of quantifier depth 1 (which covers the majority of such T boxes), but not for ALCHIF and ALCQ T Boxes of depth 1.
Horn rewritability vs PTime query evaluation for description logic TBoxes
• Computer Science
• 2018
The main results are that this is indeed the case when L is the set of ALCHI or ALCIF TBoxes of quantifier depth 1 (which covers the majority of such T boxes), but not for ALCHIF and ALCQ T Boxes of depth 1.
## References
SHOWING 1-10 OF 86 REFERENCES
The Data Complexity of Description Logic Ontologies
• Computer Science, Philosophy
Log. Methods Comput. Sci.
• 2017
This work analyzes the data complexity of ontology-mediated querying where the ontologies are formulated in a description logic of the ALC family and queries are conjunctive queries, positive existential queries, or acyclic conj unctive queries and shows that it is undecidable whether a given ontology admits PTime query evaluation.
Non-Uniform Data Complexity of Query Answering in Description Logics
• Computer Science
Description Logics
• 2011
This paper proposes a new method for investigating data complexity in OBDA: instead of classifying whole logics according to their complexity, it aims at classifying each individual ontology within a given master language.
Ontology-based data access: a study through disjunctive datalog, CSP, and MMSNP
• Computer Science
PODS '13
• 2013
This paper studies several classes of ontology-mediated queries, where the database queries are given as some form of conjunctive query and the ontologies are formulated in description logics or other relevant fragments of first-order logic, such as the guarded fragment and the unary-negation fragment.
Ontology-Mediated Query Answering with Data-Tractable Description Logics
• Computer Science
Reasoning Web
• 2015
A brief introduction to ontology-mediated query answering using description logic (DL) ontologies, with a focus on DLs for which query answering scales polynomially in the size of the data, as these are best suited for applications requiring large amounts of data.
Ontology-Based Data Access with Closed Predicates is Inherently Intractable(Sometimes)
• Computer Science, Philosophy
IJCAI
• 2013
In all cases where answering conjunctive queries (CQs) with (open and) closed predicates is tractable, it coincides with answering CQs with all predicates assumed open, which means that CQ answering with closed predicate is inherently intractable.
Data Complexity of Query Answering in Expressive Description Logics via Tableaux
• Computer Science
Journal of Automated Reasoning
• 2008
It is established that, for a whole range of sublogics of $\mathcal{SHOIQ}$ that contain $\Mathcal{AL}$, answering such queries has coNP-complete data complexity, and a tight coNP upper bound for positive existential queries without transitive roles is proved.
Tractable Reasoning and Efficient Query Answering in Description Logics: The DL-Lite Family
• Computer Science
Journal of Automated Reasoning
• 2007
It is shown that, for the DLs of the DL-Lite family, the usual DL reasoning tasks are polynomial in the size of the TBox, and query answering is LogSpace in thesize of the ABox, which is the first result ofPolynomial-time data complexity for query answering over DL knowledge bases.
Data Complexity of Query Answering in Description Logics
• Computer Science
Description Logics
• 2005
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# Power Series Distribution definition, formula with applications
### Power Series Distribution
A discrete random variable X is said to have a generalised power series distribution if its probability function is given by,
$f\left&space;(&space;x;a,\theta&space;\right&space;)=\frac{a_{x}\theta&space;^{x}}{f\left&space;(&space;\theta&space;\right&space;)};&space;x=0,1,2,3....;&space;a_{x}\geq&space;0.$
where f(θ) is a generating function and f(θ) is positive finite and diferentiable. Power series distribution is a discrete probability distribution.
### Properties of Power series distribution
Some special properties of power series distribution are given-
• If θ=p/(1-p), f(θ)=(1+θ)^n and s={1,2,3,...,n), a set of (n+1) non-negative integers then the power series distribution is tends to binomial distribution.
• If f(θ)=e^θ and s={0,1,2,3,...,∞} then the distribution tends to poisson distribution.
• If θ=p/(1-p), f(θ)=(1+θ)^-n and s={0,1,2,3,...,∞), then the power series distribution tends to negative binomial distribution.
• If f(θ)=-log(1-θ) and s={1,2,....}, then the power series distribution tends to logarthmic distribution.
Characteristics of power series distribution
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# Revision history [back]
### Create quotient group of units of mod n
I would like to work with the group $\mathbb{Z}_m^* / \langle p \rangle$. Do you know how I can create it?
For example:
p = 2
m = 17^2
Zm = ZZ.quotient(m) # ring of integers mod m
Zms = Zm.unit_group() # cyclic group (Z/mZ)^* generated by 3
Zms.quotient(p)
But the last line raises a NotImplementedError.
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# What is the definition if a distinct cycle in a graph?
In a graph, I understand a cycle to be a traversal from Node A, traversing each (but not every) vertex once, and returning to Node A. Now I THINK a distinct cycle is where they don't share any vertices, but I might be wrong. Can someone clear this up for me?
-
No matter what the mathematical objects in question, ‘$x$ and $y$ are distinct widgets’ normally means simply that $x$ and $y$ are widgets, and $x\ne y$. You’re thinking of vertex-disjoint cycles; one can also have edge-disjoint cycles and cycles that are disjoint in both senses. – Brian M. Scott Jan 12 '13 at 20:57
So you're saying, for example the Cycle S = {2,3,4,5,2} and Cycle P = {2,3,4,2} are distinct because S /= P? – notverygoodatmaths Jan 12 '13 at 20:59
Yes, as the word distinct is normally used. – Brian M. Scott Jan 12 '13 at 21:02
Thankyou very much! :) – notverygoodatmaths Jan 12 '13 at 21:03
You’re welcome! – Brian M. Scott Jan 12 '13 at 21:04
Usage example: "For all $n\ge 3$, the number of distinct Hamilton cycles in the complete graph $K_n$ is $(n−1)!/2$."
Q: "Are the groups $G_1$ and $G_2$ isomorphic?"
A: "$G_1$ is, but $G_2$ isn't."
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# Kick-off meeting PRIN NAT-NET
chaired by
from to (Europe/Rome)
at Department of Physics of the University of Naples Federico II ( 2G26-2G27 )
Complesso Universitario di Monte S. Angelo - Via Cinthia - Naples
Description Neutrino and Astroparticle physics are fast-moving research fields at the junction of particle physics, astrophysics and cosmology. In this context, the project NAT-NET (Neutrino and Astroparticle Theory Network) is being proposed by a group of physicists belonging to three research units based in Bari, Naples and L'Aquila (plus one member from Lecce). These groups share a strong tradition of common scientific interests, research programs and integrated activities in the theory and phenomenology of astroparticle physics and cosmology, with particular attention to the many factes of neutrino physics, and with important links with theoretical research groups worldwide and with the experimental community. The NAT-NET theoretical and phenomenological research activities can be roughly structured into four working packages (WP) with strong connections among them and with present or future experimental searches: WP1 - Standard neutrino framework. Investigation of the remaining unknowns of the three-neutrino framework (absolute masses and their ordering, Dirac/Majorana nature, CP phases); refinement of our understanding of neutrino oscillations in vacuum, in matter and with self-interactions; neutrinoless double beta decay with light Majorana neutrinos: constraints on its nuclear model uncertainties and connections with cosmological bounds. WP2 - Beyond the standard neutrino framework. Sterile neutrino oscillations in the light of upcoming laboratory and cosmological data; constraints on new neutrino interactions; neutrinoless double beta decay beyond light Majorana neutrinos; long-distance and multi-messenger tests of dispersion relations; neutrinos as components or signals of dark matter; neutrino model building and leptogenesis. WP3 - Sources and fluxes of neutrinos and of other messengers. From low to high energy: relic neutrino detection prospects; axions and axion-like particles in astrophysical contexts; issues in big-bang nucleosynthesis neutrinos; improvements of solar neutrino models and low-energy flux detection; set-up of a reference geo-neutrino model; tests of core-collapse supernova physics; high-energy neutrinos: study of astrophysical sources (within a multimessenger approach) and of propagation in the Earth. WP4 - The standard cosmological model and beyond. Nonstandard scenarios for the relic neutrino background and big-bang nucleosynthesis; pre-big-bang and string cosmology; effects of large-scale inhomogeneities and anisotropies; laboratory approaches to vacuum energy; warm dark matter components via heavy neutrinos. Participants Gianpiero Mangano; Gennaro Miele; Ofelia Pisanti; Ninetta Saviano
Go to day
• Monday, January 27, 2020
• 09:00 - 12:00 Session 1
• 09:00 Welcome and introduction 30' Speakers: Prof. Gennaro Miele (University of Naples Federico II), Dr. Eligio Lisi (INFN - Bari)
• 09:30 WP1 Standard neutrino framework 1h0'
Conveners: F. Vissani and D. Montanino
• 10:30 Coffee-break 30'
• 11:00 WP2 Beyond standard neutrino framework 1h0'
Conveners: A. Palazzo and N. Saviano
• 12:00 - 13:30 Lunch ( Mensa universitaria )
• 13:30 - 18:00 Session 2
• 13:30 Round Table Discussion on WP1 & WP2 1h0'
• 14:30 WP3 Sources and fluxes of neutrinos and other messengers 1h0'
Conveners: G. Pagliaroli and O. Pisanti
• 15:30 Status/prospects of PhD & Postdocs 1h0'
• 16:30 Coffee-break 30'
• 17:00 Round Table Discussion on WP3 & Training 1h0'
• 20:00 - 22:00 Dinner ( TBA )
• Tuesday, January 28, 2020
• 09:30 - 14:00 Session 3
• 09:30 WP4 Standard cosmological model and beyond 1h0'
Conveners: G. Mangano and F. Villante
• 10:30 Coffee-break 30'
• 11:00 Status/prospects of other activities 1h0'
• 12:00 Round Table Discussion on WP4 & Other activities 1h0'
• 13:00 Conclusions 30'
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# l'Hôpital's rule for a complex function
• April 5th 2010, 04:04 PM
asi123
l'Hôpital's rule for a complex function
Hey guys.
Can I use l'Hôpital's rule for a complex function?
Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?
Thanks.
• April 5th 2010, 04:39 PM
VkL
What do you mean by complex? That is not a complex function.
And You can use it in this case.
$
\lim_{z\to 0} \frac{sin(z)}{z}= 1$
Since it is in undetermined form, ( 0 in the denominator and 0 in the numerator), you can take the derivative of the top and bottom, and then apply the limit again.
• April 5th 2010, 04:44 PM
Drexel28
Quote:
Originally Posted by asi123
Hey guys.
Can I use l'Hôpital's rule for a complex function?
Lets say I want to investigate the behavior of function sin(z)/z around z=0, can I use it?
Thanks.
$\frac{\sin(z)}{z}=\frac{e^{iz}-e^{-iz}}{2iz}=\frac{1}{2i}\left[\frac{e^{iz}-1}{z}-\frac{e^{-iz}-1}{z}\right]$. A little more obvious now?
• April 5th 2010, 04:49 PM
asi123
Sorry but no.
I still cant put z=0 in that.
I didn't quite get it.
• April 5th 2010, 07:20 PM
xxp9
$\frac {\sin{z}} {z} = \frac {z-\frac {z^3} {3!} + ... } {z} = 1-\frac {z^2} {3!} + ...$
so the result is 1 when $z \rightarrow 0$
• April 6th 2010, 12:37 AM
asi123
Well, Thank you very much but I know the result is 1.
My question is, can I use l'Hôpital's rule for this thing even due z is a number from the complex domain?
I mean when you try to find limit for this thing, there are number of paths you can take in order to do that, does l'Hôpital's rule works in that case?
Thanks a lot.
• April 6th 2010, 05:24 AM
HallsofIvy
Yes, L'Hopital's rule works for functions of a complex variable.
• May 13th 2011, 12:10 AM
gosuman
Sorry to raise a dead thread, but I have had this same question for some time and have never seen a proof of L'Hopital's Rule that does not require the mean value theorem for real differentiable functions, which as far as I have seen, does not extend to the complex numbers. Perhaps the mean value property of harmonic functions serves the same end, but I have still never come across such a proof (I must concede, I've never tried proving it myself-- it could turn out to be quite easy).
Evidence demonstrates that L'Hopital's Rule does apply to the limit of a complex valued analytic function (as I am not able to construct a counterexample), but I would love to see a proof if anyone knows where I can find one.
• May 13th 2011, 07:25 PM
Drexel28
In fact, it isn't always generally true. And, if you would have googled 'L'hopitals rule complex functions' you would have found this as the fourth link. Hope it helps.
• May 14th 2011, 10:11 PM
Jose27
The proof is simple and relies on the factorization of holomorphic functions: Assume $f,g: D \rightarrow \mathbb{C}$ are holomorphic ( $D$ some disk around $0$). If $f(0)=g(0)=0$ then $f(z)=z\hat{f}(z)$ and $g(z)=z\hat{g}(z)$ where $\hat{f}(0)=f'(0)$ and $\hat{g}(0)=g'(0)$, then the conclusion follows trivially. If on the other hand they both have a pole at $0$ then the it's a little trickier: Take $f(z)=\sum_{-k}^{\infty} b_nz^n$ and $g(z)=\sum_{-m}^{\infty} c_nz^n$ be the Laurent expansions around $0$ then the quotient of the functions are
$\frac{f(z)}{g(z)}= z^{m-k}\frac{b_{-k} + o(1)}{c_{-m}+o(1)}$
$\frac{f'(z)}{g'(z)}=z^{m-k}\frac{-kb_{-k} +o(1)}{-mc_{-m}+o(1)}$
and in both cases the limit is equal to: $0$ if $m>k$, $\frac{b_{-k}}{c_{-m}}$ if $m=k$ and $\infty$ if $m.
Edit: As an afterthought notice that this proof relies on the fact that the functions are defined and not zero on a punctured neighbourhood of the limit point, so the general formulation, as in the case of a real variable, doesn't follow from this. So for example if you have one of the functions be such that doesn't extend in any way past any boundary point of a disk, then you can't use this to evaluate the limit at a boundary point.
• May 14th 2011, 11:34 PM
gosuman
Hi Drexel, thanks for the link. I did google it, and it actually took me here. I did not notice that link before. I am thrilled to see the conditions for the complex situation.
Jose thank you for posting your proof.
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# LHCb Conference Proceedings
Последно добавени:
2016-07-18
14:33
Bayronic b decays at LHCb / Hutchcroft, David (University of Liverpool (GB)) Pending LHCb-PROC-2016-018; CERN-LHCb-PROC-2016-018.- Geneva : CERN, 2016 In : BEACH 2016: XIIth International Conference on Beauty, Charm, and Hyperons in Hadronic Interactions, Fairfax, Virginia, United States Of America, 12 - 18 Jun 2016
2016-07-14
09:29
B decays to open charm / Haines, Susan Carol (University of Cambridge (GB)) Studies of $B$ meson decays to states involving open charm mesons in data recorded by the LHCb experiment have resulted in first observations of several new decay modes, including $B_s^{0} \rightarrow D_s^{*\mp} K^{\pm}$, $B_s^{0} \rightarrow \overline{D}^{0} K_S^{0}$ and $B^{+} \rightarrow D^{+} K^{+} \pi^{-}$ decays. An upper limit has been placed on the branching fraction of $B_s^{0} \rightarrow \overline{D}^{0} f_0(980)$ decays. [...] LHCb-PROC-2016-017; CERN-LHCb-PROC-2016-017.- Geneva : CERN, 2016 - 6. Fulltext: PDF; In : The 16th International Conference on B-Physics at Frontier Machines, Marseille, France, 2 - 6 May 2016
2016-07-07
08:54
Multibody charmless b-hadron decays / Dujany, Giulio (University of Manchester (GB)) Multibody charmless b-hadron decays are a unique laboratory to probe CP violation, to look for effects of new physics and to provide insights into the hadronisation mechanism. The decay modes $\Lambda_b^0 \to \Lambda K^+ K^-$, $\Lambda_b^0 \to \Lambda K^+ \pi^-$, $\Lambda_b^0 \to \Lambda \phi$ and $B_s^0 \to K_{\rm S}^0 K^*$ are observed for the first time by LHCb and their branching fractions are measured. [...] LHCb-PROC-2016-016; CERN-LHCb-PROC-2016-016.- Geneva : CERN, 2016 - 7. Fulltext: PDF; In : The 16th International Conference on B-Physics at Frontier Machines, Marseille, France, 2 - 6 May 2016
2016-06-02
14:09
Timing and Readout Contorl in the LHCb Upgraded Readout System / Alessio, Federico (CERN) In 2019, the LHCb experiment at CERN will undergo a major upgrade where its detectors electronics and entire readout system will be changed to read-out events at the full LHC rate of 40 MHz. In this paper, the new timing, trigger and readout control system for such upgrade is reviewed [...] LHCb-PROC-2016-015; CERN-LHCb-PROC-2016-015.- Geneva : CERN, 2016 In : 20th IEEE-NPSS Real Time Conference, Padua, Italy, 5 - 10 Jun 2016
2016-05-30
14:02
LHCb results on 13 TeV $pp$ collisions / Alessio, Federico (CERN) By using the very first proton-proton collision data of the LHC Run II, the LHCb experiment performed a series of early measurements, including the cross-sections for quarkonia, beauty and charm productions. The results have been carried out by exploiting a new scheme for the LHCb software trigger, where the algorithm has been split in two stages [...] LHCb-PROC-2016-014; CERN-LHCb-PROC-2016-014.- Geneva : CERN, 2016 - 4. Fulltext: PDF; In : 51st Rencontres de Moriond on QCD and High Energy Interactions, La Thuile, Italy, 19 - 26 Mar 2016
2016-05-24
16:50
Results of heavy ion collisions at LHCb / Zhang, Yanxi (Laboratoire de l'Accelerateur Lineaire (FR)) Heavy flavor production is important in heavy ion collisions to study both cold and hot nuclear matter effects. The LHCb experiment can make unique contribution to heavy ion physics, owing to the full particle identification of the detector in the forward region and the ability to collect fixed target data with proton or lead beams. [...] LHCb-PROC-2016-013; CERN-LHCb-PROC-2016-013.- Geneva : CERN, 2016 Fulltext: PDF; In : 50th Rencontres de Moriond on QCD and High Energy Interactions, La Thuile, Italy, 21 - 28 Mar 2015
2016-05-19
16:30
CP violation in b-hadrons / Benson, Sean (Nikhef National institute for subatomic physics (NL)) Latest LHCb measurements of $CP$ violation in b-hadrons are presented based on $pp$ collision data collected in 2011 and 2012 at centre-of-mass energies of $\sqrt{s}=7$ $\rm TeV$ and $8\ \rm TeV$ respectively. The total integrated luminosity collected is 3.0 fb$^{-1}$. [...] LHCb-PROC-2016-012; CERN-LHCb-PROC-2016-012.- Geneva : CERN, 2016 In : 51st Rencontres de Moriond on Electroweak Interactions and Unified Theories, La Thuile, Italy, 12 - 19 Mar 2016
2016-05-09
13:44
pending / Tobin, Mark (Ecole Polytechnique Federale de Lausanne (CH)) pending LHCb-PROC-2016-011; CERN-LHCb-PROC-2016-011.- Geneva : CERN, 2016 In : 10th International "Hiroshima" Symposium on the Development and Application of Semiconductor Tracking Detectors, Xi'an, China, 25 - 29 Sep 2015
2016-04-21
16:23
Determination of the $CP$-violating phase $\phi_s$ in $B0s \to J\psi \Phi$ decays / Batozskaya, Varvara (National Centre for Nuclear Research (PL)) The determination of the CP-violating phase $\phi_{s}$ in $B^{0}_{s}\rightarrow J/\psi \phi$ decays is one of the key goals of the LHCb experiment. Its value is predicted to be very small in the Standard Model [...] LHCb-PROC-2016-010; CERN-LHCb-PROC-2016-010.- Geneva : CERN, 2016 - 7. Fulltext: PDF; In : 22th Cracow EPIPHANY Conference on the Physics in LHC Run2, Cracow, Poland, 7 - 9 Jan 2016
2016-04-21
14:39
LHCb’s Real-Time Alignment in Run II / Batozskaya, Varvara (National Centre for Nuclear Research (PL)) The LHCb collaboration has introduced a novel real-time detector alignment and calibration strategy for LHC Run II. The data collected at the start of the fill will be processed in a few minutes and used to update the alignment, while the calibration constants will be evaluated for each run [...] LHCb-PROC-2016-009; CERN-LHCb-PROC-2016-009.- Geneva : CERN, 2015 - 6. Fulltext: PDF; In : 3rd Annual Large Hadron Collider Physics Conference, St. Petersburg, Russia, 31 Aug - 5 Sep 2015
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# Can CFGs generate all languages? Are they (PDAs) finite or infinite state automata?
I was looking for the limitations of a CFG. I think there is some limitation given there are only finitely many states of a PDA (or non-terminals in a CFG).
I suspect that languages like $$\text{L} = \{10,10100,101001000, \dots\}$$ can not be generated by a CFG. I can not see that intuitively (heuristics will help me). I have seen PDAs to be finite state automata in some places, and infinite in the rest.
Can someone tell me the limitations of a CFG / PDA (if any) and whether or not $$\text{L}$$ can be generated by a PDA/CFG? Additionally, are PDAs infinite state automata?
Since each context-free language can be described by a grammar, there are only countably many context-free languages (over a fixed alphabet). Therefore, "most" languages are not context-free. Examples of particular languages which are not context-free abound. Your language $$\{ 1010^210^3\cdots 10^n : n \geq 1 \}$$ is one such example. It can be proved to be non-context-free using the pumping lemma, for example.
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# zbMATH — the first resource for mathematics
Applications of the Leray-Schauder alternative to some Volterra integral and integrodifferential equations. (English) Zbl 0852.45012
The author proves the global existence of solutions for some nonlinear Volterra integral and integrodifferential equations, of the type $x(t)=h(t)+\int^t_0 k(t,s)g(s,x(s))ds\tag{1}$ and $x'(t)=f\Biggl(t,x(t),\int^t_0 k(t,s)g(s,x(s))ds\Biggr),\qquad x(0)=x_0,\tag{2}$ where $$h\in C([ 0,T ],\mathbb{R}^n)$$, $$k\in C([ 0,T ]\times [0, T] \times [0, T],\mathbb{R})$$, $$g\in C([0, T] \times\mathbb{R}^n,\mathbb{R}^n)$$ and $$f\in C([ 0,T ]\times\mathbb{R}^n \times\mathbb{R}^n,\mathbb{R}^n)$$ and $$x_0$$ is a given constant. Under some additional growth conditions on the functions $$h$$, $$k$$, $$g$$ and $$f$$, he obtains necessary a priori bounds for applying the usual Leray-Schauder degree (for completely continuous perturbations of the identity), to appropriate fixed point problems associated to (1) and (2).
##### MSC:
45G10 Other nonlinear integral equations 45J05 Integro-ordinary differential equations 55M25 Degree, winding number
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# cond
Version:
Returns the condition number of a matrix with the singular value decomposition method.
## Syntax
c = cond(a)
c = cond(a, b)
## a
Specifies a matrix of non-integer numeric values.
## b
Norm type of a. b is a scalar or string that accepts the following values.
Name Description
1
1-norm
2 (default)
2-norm
inf
inf-norm
'fro'
Frobenius norm
## c
b-norm condition number of a. c is a scalar number.
A = [1+2i, 4i; -3-6i, -5i]
D = cond(A)
Where This Node Can Run:
Desktop OS: Windows
FPGA: This product does not support FPGA devices
|
# 25th International Workshop on Deep Inelastic Scattering and Related Topics
Apr 3 – 7, 2017
University of Birmingham
Europe/London timezone
## Measurements of the top-quark properties in the production and decays of $t\bar{t}$ events at CMS
Apr 4, 2017, 11:40 AM
20m
West Building Seminar Room 1
#### West Building Seminar Room 1
WG5) Physics with Heavy Flavours
### Speaker
Pieter David (Universite Catholique de Louvain (UCL) (BE))
### Description
Measurements of several top-quark properties are presented, obtained from the CMS data collected at various centre-of-mass energies. The results include measurements of the top pair charge asymmetry, the $W$ helicity in top decays, $CP$ violation, $t\bar{t}$ spin correlation, top polarisation and the search for anomalous couplings including Flavour Changing Neutral Currents. The results are compared with predictions from the standard model as well as new physics models. The cross section of $t\bar{t}$ events produced in association with a $W$, $Z$ boson or a photon is also measured.
### Primary author
Pedro Vieira De Castro Ferreira Da Silva (CERN)
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Reliability of Spring Pressure Contacts
Posted on 01 November 2007
Abstract
Spring contacts are an excellent solution for connecting a power module with a printed circuit board (PCB). They can be applied in a wide current range from sensor currents of a few milliamps to load currents of several amps. They offer many advantages like easy assembly without soldering and also easy disassembly for maintenance purposes. The reliability of spring contacts under environmental stress by mechanical wear, rapid temperature change and corrosive atmosphere is significant for the application. The experimental results presented in this paper certify that spring contacts are reliable even under harsh environmental conditions.
1. Introduction
While the many advantages of implementing spring contacts for the electrical connection between a power module and the PCB are obvious, concerns regarding the long time reliability in the harsh environment of power electronic applications are often observed. These concerns are based on experiences with pin-and-socket connectors or wrap connectors well known from PC main boards.
Therefore it is important to understand, that the spring connectors in power modules are completely different from wrap connectors, because they do not have to consider mating and unmating forces.
For power modules, the contact force is much higher than for wrap connectors – it is more comparable to a terminal- to-busbar connection by nut and bolt. Therefore, reliability results cannot be transferred from wrap connector systems (Fig. 1).
Figure 1. Comparison of nut and bolt connection, wrap connector and spring contacts
In the discussed power module assembly, an electrical interconnection between a printed circuit board (PCB) – with driver components and power leads – and a power module composed of ceramic substrate (DBC) carrying the power devices is established by springs (Fig. 2).
Figure 2. Schematic Overview of the spring contact system.PCB connected to DBC by Springs
The PCB metallization systems in use today are RoHS conformal immersion tin, hot air levelling (HAL) tin, immersion silver, organic solderability preservation systems (OSP) or nickel/gold-flash (ENIG). The springs are formed from a conductive spring material (like bronze or steel) plated with tin, silver or nickel/gold. The available surfaces for ceramic substrates in power modules are copper, nickel or nickel/gold-flash. Different combinations of these materials will be investigated in the following test results.
The spring material used below is always a bronze. Each spring has two contact spots and a series of at least two springs is used for measurement. Due to the assembly of the power modules, a single spring is not accessible without intrusion into the module housing.
The applied pressure determines the choice of contact materials. Tin or silver plating is suitable for a contact force of approximately 2–20 N while gold plating is available for approximately 1–2 N. Since the spring force is between 3 N and 6 N per spring for the discussed systems, only Ag or Sn are recommended [1].
The most common types of strain in an industrial environment were determined to be mechanical vibrations, temperature changes and high temperature and corrosive atmosphere. The reliability of a spring contact under these conditions was examined.
2. Temperature shock test
Temperature shock tests combine the effects of temperature change, high temperature storage and movement induced by differences in thermal expansion.
2.1. Material selection
As the spring connection is not form locking, thermal movement and potential wear can be expected. This might impact the contact force and/or lead to an increase of contact resistance.
To evaluate the development of the contact resistance the change of the resistance against the first cycle is plotted. The temperature evolution of each cycle was measured by a soldered thermocouple attached to the device under test (DUT).
It was found that some material combinations were susceptible to degradation: For instance, Fig. 3 shows the rise of the contact resistance of a test system with nickel DBC due to oxidation. The other contact partners were found to be unaffected.
In literature a change of the contact resistance is often attributed to thin surface films [1]. Dry-circuit measurement conditions are therefore chosen not to destroy surface films. Those are limited to a current of not more than 100 mA and a voltage of not more than 20 mV to avoid melting and dielectric breakdown, which would re-establish a good electrical contact. Experiments were performed to verify the beneficial effect of higher currents on the contact resistance. A (1 mA) sense current was used in the temperature shock test in Fig. 3.
Figure 3. Test System Temperature shock test using a nickel DBC
To investigate the influence of higher currents a test system was prepared with a material combination that showed an increase in contact resistance after few cycles. Temperature shock was performed at a current of 1 mA until the contact resistance increased. The current was subsequently increased in steps from 1 mA up to 400 mA. Fig. 4 shows the contact resistance development.
Figure 4. Test system - influence of the current level on the contant resistance of an aged contact system using a copper DBC
With an increase in current, each step also shows a significant reduction of the contact resistance.
Low current levels are typical for sensor applications. A change in contact resistance of 10 X for a MiniSKiiP® temperature sensor is equivalent to an error in temperature reading of approximately 1 C.
In contrast to the test assemblies in Figures. 3 and 4, the contact resistance development of a genuine power module is displayed in Fig. 5.
Figure 5. Contact system temperature shock test using an ENIG DBC
The optimized material selection leads to a stable contact resistance. Hundred cycles with extreme temperature swings are assumed equivalent to the total lifetime in the field. The largest change in contact resistance across eight springs in series (equals 16 contact spots) is measured to be only 100 mX even after 200 temperature cycles, equivalent to an erroneous temperature reading of approximately 0.01 °C.
To compare a MiniSKiiP® II system with integrated temperature sensor connected via springs to a soldered thermocouple an extended temperature cycling test was performed.
Fig. 6 displays the temperature measurement for selected cycles.
Figure 6. Temperature recording for selected cycles using a soldered thermocouple (line) and a Temperature sensor connected via two Springs (dotted line)
The temperature evolution of the thermocouple and the temperature sensor show a slightly different gradient due to the differences in thermal capacity. The temperature sensor signal was stable for 2000 cycles; for the extreme changes in temperature this is equivalent to 20 times the normal lifetime of a power module. The soldered connection of the thermocouple failed at 1000 cycles and had to be replaced (see arrow in Fig. 6).
3. Mechanical fretting test
Fretting corrosion describes the phenomenon of the growing, abrasion and compacting of oxide particles by repeated micro-movement, as can be induced by vibration. This process is well understood and documented only for wrap connectors and limited to a small number of contact material combinations. Each movement cycle oxide flakes are abrasively removed from the contact surface. The newly generated surface can be prone to oxidation. This results in the formation of a native oxide layer of a few nanometers. Repeated movement accumulates oxide flakes while wearing down elevated regions of the metal. An increase of the contact resistance is the consequence (Fig. 7).
Figure 7. Development of a dense layer of oxide flakes by alternating micro-movement and oxidation steps
To simulate a repetitive mechanical movement of the power module contacts, a setup was designed that forces a movement of a PCB over a spring at a defined frequency, load and amplitude (Fig. 8).
Figure 8. Test setup for micro-vibration. The contact resistance of the system can be monitored using four conductor measurement
Displayed in Fig. 9 is the change of contact resistance of two pairs of springs against the initial value.
Figure 9. Spring contact resistance change of two pairs of springs: 4.65 million cycles, displacement 50 lm, 1 Hz
The contact resistance decreases initially. This process is associated with the cleaning of the contact spot; contamination is removed. Ambient temperature changes and a day/night fluctuation affect the contact resistance, but no increase of the contact resistance was detected during 4.65 million movement cycles. This is attributed to the higher contact forces involved in the spring contact, as well as the shape of the spring’s head and the contact materials. In contrast to literature results on wrap connector systems, no increase of contact resistance could be evoked by micro-movement of the contact system.
4. Corrosive atmosphere tests
Corrosive atmosphere testing is performed as an accelerated test to ensure the reliable operation of spring connections – and power modules in general – in a harsh industrial environment.
Tests were performed according to DIN EN 60068-2-43 Kd (10 ppm H2S, 10 days, 25 °C, 75% RH) as a highly accelerated corrosive atmosphere test. This test is specially designed to examine the behaviour of silver plated surfaces. It can also verify the function of a tarnish protection. As expected, even surfaces with tarnish protection showed minor traces of tarnishing. Scanning electron microscopy proved the contact interface to be unaffected by the corrosive atmosphere. Due to the high contact forces of the spring system, the metallic contact partners form a quasi hermetically sealed interface. This joint is impervious to outside contamination. Corrosion products could not be detected by energy dispersive X-ray analysis (EDX) inside the contact area.
Even though the EDX analysis gives a good understanding of the impact of corrosive gases, the measured contact resistance before and after the test is the most important criterion for contact reliability. The change in contact resistance for various systems was found negligible.
While H2S is the suitable atmosphere to check for silver tarnishing, the composition of industrial atmospheres is more complex. A combination of multiple corrosive gases will affect other types of materials included in the system as well.
Thus, test parameters were defined using the four most common corrosive gases in a concentration sufficiently high to provide accelerated test conditions. Table 1 shows testing conditions selected for systems with a variety of contact metal surfaces.
H2S 0.4 ppm SO2 0.4 ppm Cl2 0.1 ppm NOx 0.5 ppm 25 °C, 75% RH, 21 days of exposure
Table 1. SEMIKRON mixed gas test conditions
The test conditions are derived from the climate class 3C3 in DIN EN 60721-3-3 for heavily polluted industrial areas and exceed the requirements of DIN EN 60068-2-60 Ke: method 3 (H2S, SO2 and NOx for harsh environmental conditions): This combination of gas conditions provokes corrosion of materials typically unaffected by pure H2S like tin and aluminium. The test system again was not affected by the corrosive atmosphere due to the quasi hermetical contact interface.
5. Electromigration
Electromigration is a process of two consecutive phenomena which do not necessarily impair electrical contact behaviour. However it can lead to the formation of undesired conductive paths.
First corrosion of metal surfaces may occur. The metal ions generated by corrosion are susceptible to migration in an electric field. As soon as a conductive surface of negative potential is reached, the ions are reduced and deposited as metallic dendrites. The rate of corrosion, the wettability of the insulating material between two potential levels, the mobility of the metal ions and composition of the electrolyte are all influencing factors on the rate of electromigration. Testing was performed by applying an 15 V bias to a spring module between the closest possible spring positions on the PCB during a corrosive atmosphere test. The voltage drop over the DUT was monitored and was expected to decrease in the case of formation of an electrically conductive path. No voltage breakdown was detected. Analysis of module and PCB after the test revealed slight corrosion of the springs and a brightly coloured corona surrounding the clean metallic contact spot (see Fig. 10).
Figure 10. PCB pads of a Ni/Au-flash PCB after corrosive atmosphere test (10 ppm H2S, 25 °C, 75% RH, 10 days, 15 V bias). Bias indicateed next to the contact pads
EDX analysis of the discoloured spots was performed. The contact centre was found to be free of corrosion products. The surrounding area was covered by Ag2S and Cu2S. Source of the copper is the tarnish protection of the spring. The concentric alignment of the discolouration around the contact spots is evidence that the bias did not accelerate the process. No sign of beginning electromigration can be detected. The observed effect was caused by surface diffusion of the copper rich tarnish protection along the gold plating of the PCB. It did not impair the electrical properties of the contact system.
6. Intermetallic phases
A phenomenon specific to certain metal combinations is the growth of intermetallic phases. Tin on copper plating – as commonly used as PCB metallization on copper base material – are known to grow into intermetallic phases with changed mechanical properties. Those intermetallic phases can impair soldering due to the formation of oxide layers that are difficult to remove with normal fluxes. The growth of intermetallic phases is based on a diffusion process, and thus dependent on temperature. This is more of a concern for lead-free solder profiles with the associated higher temperatures. A relation between the reduction of remaining pure tin thickness and temperature is given by the following formula:
with: Δ dSn, reduction in pure tin thickness in lm; T, temperature in K.
This formula is valid for the formation of CuxSnyphases in the temperature range from 20 °C to 330 °C [2].
The above formula limits the shelf life of Sn HAL and chem. Sn PCBs for soldering processes. The spring contact system in power modules, which establishes an electrical contact to these PCB pads, requires a much longer service life. To validate the reliability of a contact after long periods of time, an extreme aging of a contact system was tested with the following parameters:
• Storage of an immersion tin PCB and a tin–lead hot air levelling PCB at 150 °C for 90 h.
• Storage in a corrosive atmosphere (see Table 1).
• Temperature shocks with permanent current load of 1 mA
The first step ensures that no remaining pure tin layer is present at the surface and that intermetallic CuxSny-phases have grown through the metallization layer. The storage is equivalent to a predicted loss of a tin metallization thickness thickness of 8.4 lm – much more than the actual thickness of immersion tin on the PCB of approximately 1 lm. The second step exposes the open intermetallic phases to an extremely aggressive mix of corrosive gases.
For the third step, current leads had to be connected to the PCB. Soldering to the heavily aged PCB proved to be difficult. Fig. 11 shows a picture of a soldering attempt to attach a sense wire.
Figure 11. Poor wettability of solder after extensive high temperature storage followed by corrosive atmosphere
Without mechanical cleaning and the use of extremely aggressive flux, soldering was impossible. The corrosive damage to the PCB was more extensive than on any PCB after years of service and it can be assumed that no populated PCB would have functionally survived this extreme aging.
Then this PCB was assembled to a new MiniSKiiP module. Temperature cycling with permanent voltage drop monitoring was performed on the assembly. The contact resistance evolution over a series of eight springs is displayed in Fig. 12.
Figure 12. Test System: First temperature cycle of corroded intermetallic phase PCB surfce after assembly
The initially increased resistance drops quickly to a stable value. This is attributed to a cleaning of the contact partners due to movement induced by different coefficients of thermal expansion.
The resistance increases during continuous cycling (Fig. 13).
Figure 13. Temperature shock results of eight springs after assembly to heavily aged PCB
Those values are obtained on a printed circuit board in worse shape than field returns after many years of service, though. As discussed before, this contact resistance evolution is only observed in the current range of 1 mA. At higher current levels the contact resistance increase will be negligible.
7. Conclusion
Extensive testing has proven the reliability of the spring pressure contact system in a variety of extreme conditions associated with an industrial environment.
The stress in applications of the power modules can best be reproduced by temperature shock, which combines several different conditions in one test. Authentic low-current experiments show only negligible fluctuations in contact resistance. Higher current levels, as used for control and load contacts, were seen to enhance the contact stability. Therefore, special care is taken to ensure the reliability of low voltage, low current operations.
The influence of corrosive atmosphere on two different effects was examined closely. Due to the high contact forces of the spring contact system and the associated deformation of the contact metals, the access of corrosive gas to the contact area is suppressed. No signs of electromigration could be found in a test with additionally applied voltage bias.
The growth of intermetallic phases on the PCB metallization leads to a change in mechanical and corrosion behaviour, which can affect solderability. However, the spring did establish a reliable metallic contact to the PCB even on a heavily aged and corroded PCB.
The spring contact system shows a high reliability towards various industrial influences while retaining the ease of assembly.
References
[1] Slade P. Electrical Contacts – Principles and Applications. New York: Marcel Dekker; 1999.
[2] Andus electronic [Online], Available: ; 2006 [accessed 21.03.2006].
Printed Circuit Boards
Connection Systems - Connectors and Terminals
Perfect Fit Connectivity Solutions for Drive Controllers and Regulators
VN:F [1.9.17_1161]
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Your task is simple. Post a snippet in any language that if the snippet is repeated n times, will output n in decimal, octal, and hexadecimal, in that order, separated in spaces. n is an integer larger than zero. There is no leading zeroes. Shortest answer wins
## Example
If the snippet is ABC then the test case is
ABC
1 1 1
ABCABC
2 2 2
ABCABCABCABCABCABCABCABCABC
9 11 9
ABCABCABCABCABCABCABCABCABCABCABCABC
12 14 C
ABCABCABCABCABCABCABCABCABCABCABCABCABCABCABCABCABCABC
18 22 12
• Is it ok if I print 1 01 0x1? (Includes prefixes) – Blue Jan 2 '16 at 18:28
• If you have a language with implicit input/output, then you could have a 1 byte solution that just incremented the value... – Esolanging Fruit Dec 1 '16 at 5:00
# Japt, 12 bytes
[°TTs8 TsG]¸
Thanks to @ETHproductions for saving 2 bytes!
• :O you beat Dennis! – Downgoat Jan 2 '16 at 5:00
• Figured 𝔼𝕊𝕄𝕚𝕟 couldn't do it, and you had already done Teascript, and I didn't know Jolf, so I used Japt. – Mama Fun Roll Jan 2 '16 at 5:01
• Awesome :) Here's 2 bytes saved: [°TTs8 TsG]¸ – ETHproductions Jan 3 '16 at 3:47
• Oh, didn't see that. Thanks! – Mama Fun Roll Jan 3 '16 at 4:16
printf"\r%d %o %x",++$n,$n,$n; Go back to the beginning of the line, increment counter and print counter overwriting the old output. • +1 for spotting a hole in specification, Output erasure make this challenge trivial. – Akangka Jan 2 '16 at 5:07 • @ChristianIrwan: actually it's not erasing, but overwriting (I've corrected my description) – nimi Jan 2 '16 at 7:29 • That both ruins the challenge. – Akangka Jan 2 '16 at 8:02 # JavaScript, 54 53 51 47 bytes Saved 4 bytes thanks to @user81655 var d=-~d;d+${d[b='toString'](8)} +d[b](16);
I'm actually kinda surprised this works.
### Explanation
var d=-~d; // var let's d not throw an error if it's not defined
// -~ essentially increments the variable
d+ // decimal
${ // space character d[b='toString'](8) // octal } // space character +d[b](16) // Hexadecimal Try it online • Iirc you can remove the var – Conor O'Brien Jan 2 '16 at 5:47 • @CᴏɴᴏʀO'Bʀɪᴇɴ that causes an error: ReferenceError: Can't find variable: d, even on loose mode D: – Downgoat Jan 2 '16 at 5:50 • Does d=d?d+1:1 work? – Conor O'Brien Jan 2 '16 at 5:52 • @CᴏɴᴏʀO'Bʀɪᴇɴ nope, still throws a reference error, weird considering loose mode is enabled... – Downgoat Jan 2 '16 at 5:53 • Ohhhh because we're trying to access d though it is undefined – Conor O'Brien Jan 2 '16 at 5:54 # C++, 205 179 bytes int main(){};static int c=1; #define v(x) A##x #define u(x) v(x) #define z u(__LINE__) #include <cstdio> class z{public:z(){++c;};~z(){if(c){printf("%d %o %x",--c,c,c);c=0;}}}z;// (No trailing newline - when copied, the first line of the copy and last line of the original should coincide) Basically, this works by making a sequence of static variables which, on construction, increment a global variable counter. Then, on destruction, if the counter is not 0, it does all its output and sets the counter to zero. In order to define a sequence of variables with no name conflicts, we use the macro explained as follows: #define v(x) A##x //This concatenates the string "A" with the input x. #define u(x) v(x) //This slows down the preprocessor so it expands __LINE__ rather than yielding A__LINE__ as v(__LINE__) would do. #define z u(__LINE__)//Gives a name which is unique to each line. which somewhat relies on the quirks of the string processor. We use z many times to define classes/variables that will not conflict with each other when copied onto separate lines. Moreover, the definitions which must occur only once are placed on the first line, which is commented out in copies of the code. The #define and #include statements don't care that they get repeated, so need no special handling. This code also features undefined behavior in the statement: printf("%d %o %x",--c,c,c) since there are no sequence points, but c is modified and accessed. LLVM 6.0 gives a warning, but compiles it as desired - that --c evaluates before c. One could, at the expense of two bytes, add the statement --c; before the outputs and change --c in printf to c, which would get rid of the warning. Replaced std::cout with printf saving 26 bytes thanks to a suggestion of my brother. # CJam, 2019 18 bytes ];U):USU8bSU"%X"e% Thanks to @MartinBüttner for golfing off 1 byte! Try it online! ### How it works ] e# Wrap the entire stack in an array. ; e# Discard the array. U e# Push U (initially 0). ):U e# Increment and save in U. S e# Push a space. U8b e# Convert U to base 8 (array of integers). S e# Push a space. U"%X"e% e# Convert U to hexadecimal (string). # 𝔼𝕊𝕄𝕚𝕟, 14 chars / 28 bytes [⧺Ḁ,Ḁß8,Ḁⓧ]ø⬭; Try it here (Firefox only). First answer! Although there are probably better ways to handle this. # Explanation [⧺Ḁ,Ḁß8,Ḁⓧ]ø⬭; // implicit: Ḁ = 0 [⧺Ḁ, // increment Ḁ by 1 Ḁß8, // octal representation of Ḁ Ḁⓧ] // hex representation of Ḁ ø⬭; // join above array with spaces // repeat as desired until implicit output • What even is this language? – Cole Johnson Jan 2 '16 at 6:04 • A fun one. – Mama Fun Roll Jan 2 '16 at 16:23 # MATL, 26 bytes Uses current release (6.0.0). Works on Octave. 0$N1+ttYUb8YAb16YA3$XhZc1$
Once:
>> matl 0$N1+ttYUb8YAb16YA3$XhZc1$1 1 1 Twice: >> matl 0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$2 2 2 16 times: >> matl 0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$0$N1+ttYUb8YAb16YA3$XhZc1$16 20 10 ### Explanation The number of elements in the stack is used to indicate how many times we've run the snippet 0$ % specify zero inputs for next function, in case this is not the first
% occurence of the snippet.
N % number of elements in stack
tt % duplicate twice. We now have three copies of the number
YU % convert to string (decimal)
b8YA % bubble up number and convert to octal string
b16YA % bubble up number and convert to hex string
3$XhZc % join top three elements (strings) with a space 1$ % specify one input for next function. If the program ends here, that next
% function will be implicit display, so it will print the top of the stack.
% Else the stack will be left with one element more than at the beginning of
% the current snippet
## OCaml, 198 bytes
;;open Char
;;(if Sys.argv.(0).[0]='~'then Sys.argv.(0).[0]<-'\000'else Sys.argv.(0).[0]<-chr(1+int_of_char Sys.argv.(0).[0]));let n=1+int_of_char Sys.argv.(0).[0]in Printf.printf"\r%d %o %x"n n n
Includes a trailing newline and requires that the filename starts with a tilde (I used ~.ml; you can run it with ocaml \~.ml) because it's the highest-valued standard printable ASCII character. Abuses the fact that all characters in a string are mutable and Sys.argv.(0).[0] is the first character in the filename.
It should only work for n = 1 to 126, because the ASCII code for ~ is 126 and I'm adding one to the output. It could be made two bytes shorter if we only want n = 1 to 125. After it's repeated 126 times, it'll cycle back to n = 1.
This is my first ever golf so any comments or improvements would be much appreciated.
Ungolfed version:
;; open Char
;; if Sys.argv.(0).[0] = '~'
then Sys.argv.(0).[0] <- '\000'
else Sys.argv.(0).[0] <- chr (1 + int_of_char Sys.argv.(0).[0])
;; let n = 1 + int_of_char Sys.argv.(0).[0] in
Printf.printf "\r%d %o %x" n n n
• +1 The ,many holes in my question make me choose to downvote my own question. (I can't do that, though.) – Akangka Jan 3 '16 at 5:05
• I'm suspicious about repeated Sys.argv.(0).[0]. I don't know much about OCaml, though. – Akangka Jan 3 '16 at 5:06
# TeaScript, 21 20 bytes
[┼d,dT8),dT16)]j(p);
I should make it auto-close on ;
Try it online
## Explanation
┼ becomes ++
// Implicit: d = 0
[ // Start array
++d, // Increment d, decimal value
dT8), // d to base 8
dT16) // d to base 16
]j(p); // Join by spaces
// Implicit: Output *last* expression
• Downvote? Is there something wrong with this answer? Does it have to do with ASCII Character Jumble as that also got downvoted within minutes of this if not less – Downgoat Jan 2 '16 at 20:30
$_=<<'';printf"%d %o %x",(1+y/z//)x3; : There's a final newline behind the colon. Treats everything after the first line as a here document and counts the z in it. For every further copy of the code one z is added. We have to add 1 to the count, because there's none for the first snippet (the one that is executed). If additional output to stderr is allowed, we can omit the 2 single quotes '' and can get down to 38 bytes. Without the '' perl emits a warning about a deprecated feature. # Mathematica, 76 bytes Note that n should have no definitions before. 0;If[ValueQ@n,++n,n=1];StringJoin@Riffle[IntegerString[n,#]&/@{10,8,16}," "] Here, the behaviour of ; is used. The snippet above is one single CompoundExpression, however, when a couple of snippets are put together, there is still one CompoundExpression as is shown below. (Some unnecessary rearrangements are made.) 0; If[ValueQ@n,++n,n=1]; StringJoin@Riffle[IntegerString[n,#]&/@{10,8,16}," "] 0; If[ValueQ@n,++n,n=1]; StringJoin@Riffle[IntegerString[n,#]&/@{10,8,16}," "] 0; If[ValueQ@n,++n,n=1]; StringJoin@Riffle[IntegerString[n,#]&/@{10,8,16}," "] (* 3 3 3 *) So one cannot make such snippet works if writting explicit CompoundExpression. Also, almost everything you like can be put before the first ; such as E, Pi or MandelbrotSetPlot[],. ## bash, 49 bytes File count.bash: ((++n));trap 'printf "%d %o %x\n"$n $n$n' exit;
...no trailing newline.
Run:
$bash count.bash 1 1 1$ cat count.bash count.bash count.bash | bash
3 3 3
$for i in$(seq 10) ; do cat count.bash ; done | bash
10 12 a
# Python 2, 54 bytes
n=len(open(__file__).read())/54;print n,oct(n),hex(n)#
No trailing newline. Outputs in the form 1 01 0x1.
If that's not ok, 56 bytes
n=len(open(__file__).read())/56;print"%d %o %x"%(n,n,n)#
When pasted in front of each other, the length of the file gets longer by 1 line for each time pasted. The base case starts with 2 lines so you have to subtract 1 from the line length. Computation is suppressed by the comment.
• "%d %o %x"%(n,n,n), that is pretty cool. I had no idea you could do that. If it turns out that leaving prefixes is not ok I'm going to have to borrow that. – rp.beltran Jan 3 '16 at 7:20
# Python 2.x 140 bytes
This was not meant to be an overly competitive solution, but a method that I found amusing, being for one thing, an attempt at a multithreaded code golf.
import thread;n=eval("n+1")if"n"in globals()else 1;
def t(c):99**99;print("%s "*3)%(n,oct(n),hex(n))*(c==n)
Keeps a counter, spawns a thread for each count and if the counter has not changed when the counters timer goes off after a completing an expensive math problem (instead of a timer to save bytes), the formatted string is printed.
Some example configurations and their outputs:
import thread;n=eval("n+1")if"n"in globals()else 1;
def t(c):99**99;print("%s "*3)%(n,oct(n),hex(n))*(c==n)
Outputs 1 01 0x1
and fifteen copy pastes:
import thread;n=eval("n+1")if"n"in globals()else 1;
def t(c):99**99;print("%s "*3)%(n,oct(n),hex(n))*(c==n)
def t(c):99**99;print("%s "*3)%(n,oct(n),hex(n))*(c==n)
• thread.start_new_thread Could python have thought of a worse method name for code golfing? – rp.beltran Jan 3 '16 at 7:15
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Synopsis
# New Physics Possibilities from Kaon Decay
Physics 13, s25
Previously unpredicted particles or a new type of particle interaction could explain the unexpected rare kaon decay events reported by the KOTO experiment in Japan.
At the International Conference on Kaon Physics last year, a group of researchers searching for rare forms of kaon decays announced preliminary results that—if confirmed to be signal and not background—defy the standard model. Now, Kohsaku Tobioka of Florida State University, Tallahassee, and colleagues propose three explanations for the unexpected detections that involve new physics beyond the standard model, including the possibility of new particles.
The announcement from the KOTO experiment at J-PARC in Japan related to the detection of four potential instances of a rare decay of the long-lived neutral kaon ${K}_{L}$ into a pion and two neutrinos—the standard model predicts only 0.1 such events for the experimental run’s sample size. Meanwhile, CERN’s NA62 experiment announced finding no excess events for the equivalent decay of the positively charged kaon ${K}^{+}$compared to predictions. Together, these results defy a well-established relationship between the rates of the two decays, which dictates that the ${K}_{L}$ decay should occur less frequently than its positively charged counterpart.
Tobioka and colleagues came up with three explanations for the results. One proposal includes an additional particle interaction that could make ${K}_{L}$ decay more often. The other two ideas involve new particles. The researchers suggest that ${K}_{L}$ might sometimes decay into a pion and an invisible particle $X$, with that decay masquerading as the one that produces a pion and two neutrinos. Or, they say, KOTO could have witnessed the decay of some unknown particle, and not a kaon.
A special data run at the KOTO experiment this year aims to confirm whether the candidate events were true detections. If they are true, Tobioka says that he and his colleagues will test their proposals to find out which one is correct.
This research is published in Physical Review Letters.
–Erika K. Carlson
Erika K. Carlson is a Corresponding Editor for Physics based in Brooklyn, New York.
## Subject Areas
Particles and Fields
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The ATLAS Collaboration has detected triple W-boson production—a rare event that could eventually offer signs of new physics. Read More »
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2 deleted 2 characters in body edited Aug 4 '12 at 9:59 Christoph 9,77211 gold badge2323 silver badges4747 bronze badges This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level. First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$S=k\ln\Omega$$ where $$\Omega$$ is the number of micro-statesmicrostates - microscopic system configurations - compatible with the given macro-statemacrostate - macroscopic equilibrium state characteristed by thermodynamical variables. It follows from the second law $$\delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega$$ or equivalently $$\mathrm{d}\Omega = \Omega\frac{\delta Q}{kT}$$ The energy $$kT$$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite. This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level. First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$S=k\ln\Omega$$ where $$\Omega$$ is the number of micro-states - microscopic system configurations - compatible with the given macro-state - macroscopic equilibrium state characteristed by thermodynamical variables. It follows from the second law $$\delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega$$ or equivalently $$\mathrm{d}\Omega = \Omega\frac{\delta Q}{kT}$$ The energy $$kT$$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite. This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level. First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$S=k\ln\Omega$$ where $$\Omega$$ is the number of microstates - microscopic system configurations - compatible with the given macrostate - macroscopic equilibrium state characteristed by thermodynamical variables. It follows from the second law $$\delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega$$ or equivalently $$\mathrm{d}\Omega = \Omega\frac{\delta Q}{kT}$$ The energy $$kT$$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite. 1 answered Aug 4 '12 at 9:54 Christoph 9,77211 gold badge2323 silver badges4747 bronze badges This answer is somewhat hand-wavy, but I do believe it should help to grasp the concepts on an intuitive level. First of all, entropy is not a measure of randomness. For an isolated system in equilibrium under the fundamental assumption of statistical mechanics, the entropy is just $$S=k\ln\Omega$$ where $$\Omega$$ is the number of micro-states - microscopic system configurations - compatible with the given macro-state - macroscopic equilibrium state characteristed by thermodynamical variables. It follows from the second law $$\delta Q = T\mathrm{d}S=T\mathrm{d}(k\ln\Omega)=kT\frac1\Omega\mathrm{d}\Omega$$ or equivalently $$\mathrm{d}\Omega = \Omega\frac{\delta Q}{kT}$$ The energy $$kT$$ is related to the average energy per degree of freedom, so this formula tells us that the transfer of heat into a system at equilibrium opens up a new number of microstates proportional to the number of existing ones and the number of degrees of freedom the transferred energy may excite.
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# A 60.0 kg skier with an initial speed of 12.0 m/s coasts up a 2.50 m high rise as shown in the following figure. Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800?
## (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
18
Gió Share
Mar 20, 2017
I got $9.4 \frac{m}{s}$ but check my maths!
#### Explanation:
Have a look:
...I think!
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# Math Help - [5 MINUTES] Need help with a simple Algebra problem.. PLEASE AND THANK YOU
1. ## [5 MINUTES] Need help with a simple Algebra problem.. PLEASE AND THANK YOU
Write the expression in the standard form a+bi
1/2-5i
2. ## Re: [5 MINUTES] Need help with a simple Algebra problem.. PLEASE AND THANK YOU
It already is in that form.
a=1/2
b=-5.
Yes I know what you meant. $\frac{1}{2-5i}$
Multiply top and bottom by 2+5i.
3. ## Re: [5 MINUTES] Need help with a simple Algebra problem.. PLEASE AND THANK YOU
Originally Posted by NeedHelpman
Write the expression in the standard form a+bi
1/2-5i
$\frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$
4. ## Re: [5 MINUTES] Need help with a simple Algebra problem.. PLEASE AND THANK YOU
Originally Posted by NeedHelpman
Write the expression in the standard form a+bi
1/2-5i
Don't spam your problem. It won't get you help any faster.
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# Bootstrap, strap-on, anal-yzing… statistics is getting weirder by the moment
July 29, 2010
By
(This article was first published on Dang, another error, and kindly contributed to R-bloggers)
I have spent the better portion of the day trying to get a bootstrap working. I have adapter a pre-written bootstrap function, but I wanted to use a generic function, mostly for reaping fame and glory. My hypothesis was that writing a hand-written, unoptimized function will consume more computer time than the generic, presumably optimized, boot() function. Was I wrong!
In this example, I random sample (with replacement) a vector of n values and store them in a matrix. These values that are written to the matrix are percent of values of our initial vector values that are smaller or equal to the bin number (see the code). Once we’ve got all the values, we can calculate the median by columns for further tinkering. There’s a lot of material covering bootstrapping on the Internet, but for a quick reference, see this Wikipedia entry. At this StackOverflow thread, Aniko provided me with valuable information on what I was doing wrong.
This is the code I used – first part is the hand written function, and the second part is my attempt at cracking the matter with a generic function.
require(Hmisc)
#generate fake data
data <- rchisq(500, df = 12)
#create bins
binx <- cut(data, breaks = 10)
binx <- levels(binx)
binx <- sub("^.*\\,", "", binx)
binx <- as.numeric(substr(binx, 1, nchar(binx) - 1))
#pre-allocate a matrix to be filled with samples
num.boots <- 10
output <- matrix(NA, nrow = num.boots, ncol = length(binx))
#do random sampling from the vector and calculate percent
#of values equal or smaller to the bin number (i)
for (i in 1:num.boots) {
data.sample <- sample(data, size = length(data), replace = TRUE)
data.cut <- cut(x = data.sample, breaks = 10)
data.cut <- table(data.cut)/sum(table(data.cut))
output[i, ] <- data.cut
}
#do some plotting
plot(1:10, seq(0, max(output), length.out = nrow(output)), type = "n", xlab = "", ylab = "")
for (i in 1:nrow(output)) {
lines(1:10, output[i, 1:nrow(output)], lwd = "0.5")
}
output.median <- apply(output, 2, median)
output.sd <- apply(output, 2, sd)
sd.up <- output.median + output.sd * 1.96
sd.down <- output.median - output.sd * 1.96
lines(output.median, col="red", lwd = 3)
lines(sd.up, lty="dashed", col="green")
lines(sd.down, lty="dashed", col="green")
legend(x = 8, y = 0.25, legend = c("median", "0.95% CI"), col = c("red", "green"), lty = c("solid", "dashed"), lwd = c(3, 1))
All this produces this graph.
Click to enlarge (the picture).
We can then try the power of the generic boot::boot() function that comes with the R core packages. The fun part of doing a bootstraping with the boot() function is to figure out how to write the statistic function correctly. If you look at the boot() help page (?boot) you will notice that you need to provide at least two parameters to the statistic argument: data and an index parameter. Help page says that the index parameter is a vector, which is a bit confusing from where I sit. In other words, this is actually a “an empty object” (let’s call the object i for now) that tells the boot() how to crawl over your data. If your data is in a form of a vector, you will place the index as you would use to subset a vector – object[i]. If it’s a data.frame and you want to re-sample rows, you would call object[i, ]… Let’s see a working example, things may be clearer there.
pairboot <- function(data, i, binx) {
data.cut <- cut(x = data[i], breaks = 10)
data.cut <- table(data.cut)/sum(table(data.cut))
return(data.cut)
}
Notice the data[i] in the function. This will tell the boot() function to extract i elements of the data. If we had a data.frame, and rows were what we wanted to sample randomly, we would have written data[i, ].
And now we call the boot function to perform 10 “boots” on our data.
library(boot)
booty.call <- boot(data = data, statistic = pairboot, R = 10, binx = binx)
And here is the code to plot the output of the boot object (actually, it’s the print.boot object!). If you don’t believe me, try it on your own and see if you get a similar picture as above.
plot(1:10, seq(0, 1.5 * max(booty.call$t0), length.out = nrow(booty.call$t)), type = "n", xlab = "", ylab = "")
apply(booty.call$t, 1, function(x) lines(x, lwd = 0.5)) booty.call.median <- apply(booty.call$t, 2, median)
booty.call.sd <- apply(booty.call\$t, 2, sd)
booty.call.sd.up <- booty.call.median + booty.call.sd * 1.96
booty.call.sd.down <- booty.call.median - booty.call.sd * 1.96
lines(booty.call.median, col="red", lwd = 3)
lines(booty.call.sd.up, lty="dashed", col="green")
lines(booty.call.sd.down, lty="dashed", col="green")
legend(x = 8, y = 0.25, legend = c("median", "0.95% CI"), col = c("red", "green"), lty = c("solid", "dashed"), lwd = c(3, 1))
The plot thickens!
Have I, by doing a bootstrap with a generic function, profited time-wise at all? Here are the elapse times for 10000 iterations for our hand written bootstrap function and the generic, respectively.
# user system elapsed
#20.045 0.664 22.821
# user system elapsed
#20.382 0.612 22.097
So you can see, I profit by 0.724 seconds. Was it worth it? I can, for the moment, only wonder if I could improve the generic function to beat the hand-written one. Does anyone skilled in optimizing chip in any tips?
And here’s how a graph with 10000 iterations looks like. The green dashed lines represent a 95% confidence interval, which means that 95% of iterations fall between those two lines.
Bootstrap with 10000 iterations. CI is confidence interval.
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# Is it okay to reuse a single symmetric key for a large volume of messages and HMACs, so long as “nonces” are randomly generated for both of them?
Nonce reuse with the same key is catastrophic to security. The same premise holds for initialization vectors. If the key is rekeyed, then nonce or IV reuse is not catastrophic. It's the nonce/IV+key pair that you need to pay attention to.
In the preceding post, a user supports my belief that a group shared symmetric key that doesn't rotate may be used to support a large number of encrypted group-shared messages, which is to say in the following formulae that a single, static instance of $$SymmetricKey$$ may be securely used, so long as each and every instance of $$nonce$$ (unless denoted $$nonce_n$$, to denote a bitwise equivalence across the subscript equivalent appearances) is a unique, randomly generated 128 bit value (and specifically using the counter in $$chacha$$ for 64 bits of it, and the $$64$$ bit nonce parameter for the other 64 bits of it). Also, generating a 128 bit MAC with the $$HMAC()$$ function, as can be done securely with $$HMAC-MD5$$ according to:
Is HMAC-MD5 considered secure for authenticating encrypted data?
Yes, there are currently no known attacks on HMAC-MD5.
So, a message,
$$M_n$$ = $$ChaCha(SymmetricKey, Nonce)$$
supports an $$n$$ value of securely encrypted messages equivalent to the nonce space (minus one, if starting at zero). With ChaCha's $$counter$$ + $$nonce$$ space being, together, 2^128, that means $$340,282,366,920,938,463,463,374,607,431,768,211,456$$ messages can be encrypted with the single, static symmetric key. However, it must be noted that collisions will actually almost certainly start breaking the security of the system prior to the theoretical maximum number of securely encrypted messages, it's just that practical security will be maintained for an enormous volume of messages prior to collisions being likely ($$170,141,183,460,469,231,731,687,303,715,884,105,728$$ messages generated before a 50% probability of a 128 bit nonce collision being randomly generated, I think). Each $$M_n$$ will also have its own -- presumably enormous -- maximum byte count.
Is it secure to use the same shared symmetric key, from the example as previously described, for unlinkable contact signaling, by inputting it into an HMAC function (as the symmetric key, with the other argument, then, a public nonce)? By contact signaling I mean using (with |, as often it does, denoting concatenation, rather than bitwise or).
$$H_n$$ = $$HMAC(SymmetricKey, Nonce_n)$$
$$ContactSignal$$ = $$H_n$$ | $$Nonce_n$$
Such that messages encrypted by, say,
$$chacha(SymmetricKey, Nonce)$$
can be tested for intent to be processed by a group member who has the symmetric key, by the group member calculating:
$$HMAC(SymmetricKey, Nonce)$$
after receiving the public $$Nonce_n$$, and as a membership test for the secret SymmetricKey on the unlinkable contact signal, taking
$$H_n$$ = $$HMAC(SymmetricKey, Nonce_n)$$
and computing,
$${constComp}(H_n, HMAC(SymmetricKey, Nonce_n))$$
in order to determine from a smaller index downloaded by everyone whether the associated (potentially much larger) message was intended for members of one's group (as can be reported by the constant time string comparison function $$constComp()$$ returning 1 to denote "true" in response to the membership question regarding the group shared secret symmetric key).
The potential issue I can see is from using a single shared symmetric key in both HMAC and a symmetric mode of operation with a nonce with a symmetric algorithm like chacha. The thing to keep in mind is that it is an atypical use of HMAC where messages -- potentially being many kilobytes or more -- are labeled as being for members of a group, with various unlinkability and other security guarantees being met. So, with the HMAC as used for unlinkable contact signaling, the primary goal is to allow members of the group to signal to each other when a larger message, by its smaller cryptographically secure index, is identifiable to group members as intended for them, without letting an attacker outside of the group link such messages together as belonging to a single group.
Is it secure to use the same shared symmetric key, used for encryption with the arbitrary generation of a new nonce per message, for unlinkable contact signaling?
Yes, as long as adversaries do not manage to get hold on the key. That condition implies
1. None of the multiple holders of the key in the proposed system are adversaries.
2. Each is competent at keeping the key confidential, which is hard.
3. Each is competent at avoiding unauthorized use of the key, which is harder.
Point 1 is a good reason to use public-key cryptography. Point 2 is a good reason to use Smart Cards or HSMs, and to rotate keys. Point 3 is a good reason to worry about the sad state of software security.
That shall not be construed as condoning:
• Use of HMAC-MD5 in a new design. That's a no-no, even though there's no known attack, and there remains some level of security argument (because a security proof of HMAC relies on a weaker property than collision-resistance of the hash).
• The $$2^{128}$$ number of messages if HMAC is made with a 128-bit hash, much less a broken one like MD5 is. For a $$h$$-bit hash, I would think twice before approaching the design limit of the hash, which is $$2^{h/2}$$ messages. If we use HMAC-SHA-256 truncated to 16 bytes, I'd state we are OK to $$2^{120}$$ messages save for attacks on implementations, and everyone would be more than happy with that.
• The reuse of the same key in Chacha20 and HMAC in a new design. Not that I know a concrete weakness, but using the same key for two algorithms implies that any side-channel attack against the implementation of either construct breaks the other. Perhaps worse, knowing that an audit has concluded that each implementation is safe, it can not be rationally concluded that the combination of the two with the same key is: perhaps the information extracted from the two attacks can be combined. That's why we have Key Derivation Functions: starting from a single key, we can derive two others, one for each algorithm.
• An analysis of the full security of the protocol.
• I substantially clarified my question, if you don't mind looking it over once more. – cyborg Jun 4 at 6:28
• I totally agree -- I just had trouble getting all of the parameters to match up, yet wanted to make sure it's at least secure. In an actual system I would use keccak and not even use the $HMAC()$ function -- just $H(SymmetricKey | Nonce)$. I'd also probably use counter mode on $AES()$ -- do you know if $aes-ctr$ is secure with key reuse, provided $nonce$ is random and not re-used, like $ChaCha-ctr$ is? – cyborg Jun 4 at 7:11
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Home
# epicycles
I wanted to make a piece on double pendulums, but I thought something more Have you had two different menu symmetrical like a spirograph would be better.
I’ve taken quite a liking to sine/cosine
const x =
(l1 / 2) * Math.cos((v1 - 0.5) * i * radialUnit) +
(l2 / 2) * Math.cos((v2 - 0.5) * i * radialUnit);
const y =
(l1 / 2) * Math.sin((v1 - 0.5) * i * radialUnit) +
(l2 / 2) * Math.sin((v2 - 0.5) * i * radialUnit);
values
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# Given that f(x) is a linear function and f(3) =7 and f(2)=4,then find f(10).
Given that f(x) is a linear function and f(3) =7 and f(2)=4,then find f(10).
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yatangije62
given that f(x ) is a linear function i can assume that y =mx+B
from there i graph the two points. (2,4) and (3,7) from therei can tell that the slope (m) is equal to 3
slope is $\mathrm{△}y/\mathrm{△}x$ or the change in y(3) over the change in x(1)
now that I know that I can solve for B(the constant)
I will use the point (2,4) 4=3(2)+B so B= 4-6 or -2
from there I now can solve for the function of x or y
y = 3x-2
y=3(10)-2 so y=28
###### Did you like this example?
Brenton Dixon
Well it is first good to know that when they say f(x) the mean"y" and the x in the f(x) really means x. so you have two points on a line given (we know its a line cuz it says linear function).
When f(3) = 7, we know a point on the line ( 3 , 7 )
When f(2) = 4, we know a point on the line ( 2 , 4)
comparing the x's 3 - 2 = 1
comparing the y's 7 - 4 = 3
so the rise over the run is 3/1
now what if x was zero. What would y be? (to find the shift ofthe line)
0 - 2 = -2
so we would have to go 2 left (in the -x direction) to reach this point from the point ( 2 , 4 )
this also means we would go down (in the -y direction) 6.
that would give us the point ( 0 , -2 ) so the shift left would be -2
now we have enough information to create the equation
y = 3x - 2
(ya you can check the other two points, you will find that when you put 2 in for x you get 4. etc)
so putting 10 in for the x
y = 3 (10) -2
y = 28
and we have now found the point ( 10 , 28 )
When f(10) = y, we know a point on the line ( 10 , y)
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# Even and odd signals
1. ### dervast
133
Hi i am stuck with something really simple :(
I know that we can express a signal with the even and odd signal
x(t)=xe(t)+xo(t)
(xe(t) means even signal and xo(t) means odd signal)
x(-t)=x(t) for even signals and (1)
x(-t)=-x(t) for odd signals (2)
where even signal is
xe(t)=1/2[x(t)+x(-t)] (3) and the odd one is
xo(t)=1/2[x(t)-x(-t)] (4)
i can validate that x(t)=xe(t)+xo(t) if i use equations 3 and 4
x(t)=1/2[x(t)+x(-t)]+1/2[x(t)-x(-t)]= 1/2x(t)+1/2x(t)+1/2x(-t)-1/2x(-t)= x(t) done
My problem arise when i try to use (1)+(2) to (3)+(4) to prove what i want
using (1) to (3) we have xe(t)=1/2[x(t)+x(t)] =2/2x(t)
using (2) to (4) we have xo(t)=1/2[x(t)-(-x(t))] =2/2x(t) and that means that i have proved that x(t)=4x(t)
P.S Plz tell me where i am wrond and correct my bad english mathematical phrases
2. ### vsage
0
Why would you consider equations 1 and 2 at all? A generic function x(t) will not have those properties always. The only function for which both of those equations can be true is x(t) = 0, which is consistent with your result x(t) = 4x(t).
3. ### dervast
133
I cant understand what u are saying me . Plz try to clarify where i am wrong.
Thx a lot
4. ### vsage
0
Ok here is what makes your thinking wrong: You try to use x(-t) = x(t) and x(-t) = -x(t), two VERY SPECIFIC conditions, to show something for any x(t) which could have any sort of shape. What you meant to use for equations 1 and 2 was:
xe(-t) = xe(t) (1)
xo(-t) = -xo(t) (2)
If you assume equations 1 and 2 are true then you are implying x(t) = 0 for all t
I hope I made myself more understandable.
5. ### dervast
133
Yes thx a lot i have clearly understand my wrong. But if someone ask me to prove that
xe(t)=1/2[x(t)+x(-t)] how shouldi think to prove that?
6. ### vsage
0
I would use the definition of an even function xe(t) = xe(-t), but recall that xe(t) = 1/2[x(t) + x(-t)], so what would xe(-t) be?
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# zbMATH — the first resource for mathematics
## Wang, Chi-Jen
Compute Distance To:
Author ID: wang.chi-jen Published as: Wang, C. J.; Wang, C.-J.; Wang, Ch.; Wang, Chi-Jen
Documents Indexed: 14 Publications since 2005
all top 5
#### Co-Authors
0 single-authored 4 Guo, Jong-Shenq 1 Evans, James W. 1 Guo, Xiaofang 1 Guo, Yungjen Lin 1 Hu, Bei 1 Huang, Bo-Chih 1 Liu, Da-Jiang 1 Sasayama, Satoshi 1 Wakasa, Tohru 1 Yu, Cherng-Yih
#### Serials
1 Journal of Statistical Physics 1 Journal of Differential Equations 1 Quarterly of Applied Mathematics 1 Taiwanese Journal of Mathematics 1 Communications on Pure and Applied Analysis
#### Fields
5 Partial differential equations (35-XX) 3 Ordinary differential equations (34-XX) 1 Statistical mechanics, structure of matter (82-XX)
#### Citations contained in zbMATH
7 Publications have been cited 30 times in 29 Documents Cited by Year
A nonlocal quenching problem arising in a micro-electro mechanical system. Zbl 1189.35328
Guo, Jong-Shenq; Hu, Bei; Wang, Chi-Jen
2009
Blowup rate estimate for a system of semilinear parabolic equations. Zbl 1173.35362
Guo, Jong-Shenq; Sasayama, Satoshi; Wang, Chi-Jen
2009
Vibrational resonance in a Duffing system with a generalized delayed feedback. Zbl 1391.70058
Yang, J. H.; Sanjuán, Miguel A. F.; Wang, C. J.; Zhu, H.
2013
An infinite $$\varepsilon$$-bound stability criterion for a class of multiparameter singularly perturbed time-delay systems. Zbl 1092.34573
Chiou, J.-S.; Wang, C.-J.
2005
Application of relations of singularity intensities of tangent derivatives of boundary displacements and tractions to BEM. Zbl 1244.74197
Wang, Ch.; Li, Z. L.
2009
Gaseous detonation propagation in a bifurcated tube. Zbl 1151.76360
Wang, C. J.; Xu, S. L.; Guo, C. M.
2008
Re-initiation phenomenon of gaseous detonation induced by shock reflection. Zbl 1195.76423
Wang, C. J.; Xu, S. L.
2007
Vibrational resonance in a Duffing system with a generalized delayed feedback. Zbl 1391.70058
Yang, J. H.; Sanjuán, Miguel A. F.; Wang, C. J.; Zhu, H.
2013
A nonlocal quenching problem arising in a micro-electro mechanical system. Zbl 1189.35328
Guo, Jong-Shenq; Hu, Bei; Wang, Chi-Jen
2009
Blowup rate estimate for a system of semilinear parabolic equations. Zbl 1173.35362
Guo, Jong-Shenq; Sasayama, Satoshi; Wang, Chi-Jen
2009
Application of relations of singularity intensities of tangent derivatives of boundary displacements and tractions to BEM. Zbl 1244.74197
Wang, Ch.; Li, Z. L.
2009
Gaseous detonation propagation in a bifurcated tube. Zbl 1151.76360
Wang, C. J.; Xu, S. L.; Guo, C. M.
2008
Re-initiation phenomenon of gaseous detonation induced by shock reflection. Zbl 1195.76423
Wang, C. J.; Xu, S. L.
2007
An infinite $$\varepsilon$$-bound stability criterion for a class of multiparameter singularly perturbed time-delay systems. Zbl 1092.34573
Chiou, J.-S.; Wang, C.-J.
2005
all top 5
#### Cited by 55 Authors
4 Guo, Jong-Shenq 3 Kavallaris, Nikos I. 3 Lacey, Andrew A. 3 Nikolopoulos, Christos V. 2 Escher, Joachim 2 Huang, Bo-Chih 2 Kohlmann, Martin 2 Laurençot, Philippe 2 Liang, Chuangchuang 2 Mu, Chunlai 2 Tzanetis, Dimitrios E. 2 Walker, Christoph 2 Zhou, Shouming 1 Cassani, Daniele 1 Chiou, Juing-Shian 1 Du, Qingling 1 Duong, Giao Ky 1 Esteve, Carlos 1 Grenga, Temistocle 1 Guo, Yungjen Lin 1 Hu, Bei 1 Jiang, Feida 1 Jin, Zhiren 1 Kang, Kyung-In 1 Li, Fengjie 1 Li, Gang 1 Li, Jingyu 1 Li, Zhiling 1 Lim, Jong-Tae 1 Liu, Bingchen 1 Liu, Xianbin 1 Luo, Albert C. J. 1 Miyasita, Tosiya 1 Ninomiya, Hirokazu 1 Paolucci, Samuel 1 Park, Kyun-Sang 1 Shimojō, Masahiko 1 Souplet, Philippe 1 Tarsia, Antonio 1 Wakasa, Tohru 1 Wang, Chi-Jen 1 Wang, Wei 1 Wang, Yonghai 1 Xing, Siyuan 1 Yan, Zhi 1 Yanagida, Eiji 1 Ye, Dong 1 Yu, Cherng-Yih 1 Zaag, Hatem 1 Zeng, Rong 1 Zhang, Dianmu 1 Zhang, Kaijun 1 Zhou, Feng 1 Zhu, Jiang 1 Zikoski, Zachary J.
all top 5
#### Cited in 25 Serials
2 Applicable Analysis 2 Journal of Differential Equations 2 Transactions of the American Mathematical Society 2 Discrete and Continuous Dynamical Systems 1 Archive for Rational Mechanics and Analysis 1 Computers & Mathematics with Applications 1 International Journal of Systems Science 1 Journal of Computational Physics 1 Journal of Mathematical Analysis and Applications 1 Nonlinearity 1 Rocky Mountain Journal of Mathematics 1 Applied Mathematics and Computation 1 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 1 Osaka Journal of Mathematics 1 Acta Applicandae Mathematicae 1 M$$^3$$AS. Mathematical Models & Methods in Applied Sciences 1 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 1 Calculus of Variations and Partial Differential Equations 1 Engineering Analysis with Boundary Elements 1 Communications in Nonlinear Science and Numerical Simulation 1 Nonlinear Analysis. Real World Applications 1 Discrete and Continuous Dynamical Systems. Series B 1 Boundary Value Problems 1 Discrete and Continuous Dynamical Systems. Series S 1 International Journal of Systems Science. Principles and Applications of Systems and Integration
all top 5
#### Cited in 12 Fields
23 Partial differential equations (35-XX) 12 Mechanics of deformable solids (74-XX) 6 Ordinary differential equations (34-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Difference and functional equations (39-XX) 1 Calculus of variations and optimal control; optimization (49-XX) 1 Numerical analysis (65-XX) 1 Mechanics of particles and systems (70-XX) 1 Fluid mechanics (76-XX) 1 Optics, electromagnetic theory (78-XX) 1 Classical thermodynamics, heat transfer (80-XX) 1 Systems theory; control (93-XX)
|
Question
# Find the roots of the equation ${x^2} + 7x + 12 = 0$ by using the formula.
Hint- Here, we will proceed with the help of factorization as well as discriminant formula to solve for the two roots of the given quadratic equation. For discriminant method we will apply the general formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for any quadratic equation $a{x^2} + bx + c = 0$.
Given quadratic equation in variable x is ${x^2} + 7x + 12 = 0{\text{ }} \to {\text{(1)}}$
Here, we can solve this quadratic equation with the help of factorization method. The given quadratic equation can be written as
$\Rightarrow {x^2} + 7x + 12 = 0 \Rightarrow {x^2} + 3x + 4x + 12 = 0 \Rightarrow x\left( {x + 3} \right) + 4\left( {x + 3} \right) = 0 \Rightarrow \left( {x + 3} \right)\left( {x + 4} \right) = 0$
Either $\left( {x + 3} \right) = 0 \\ \Rightarrow x = - 3 \\$ or $\left( {x + 4} \right) = 0 \\ \Rightarrow x = - 4 \\$
Hence, the two roots of the given quadratic equation are -3 and -4.
We can also solve the given quadratic equation by using the discriminant method.
For any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$
According to discriminant method, the roots of this quadratic equation is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}{\text{ }} \to {\text{(3)}}$
By comparing equations (1) and (2), we get
a=1, b=7 and c=12
Using the formula given by equation (3), the roots of the given quadratic equation are given by
$x = \dfrac{{ - 7 \pm \sqrt {{{\left( 7 \right)}^2} - 4 \times 1 \times 12} }}{{2 \times 1}} = \dfrac{{ - 7 \pm \sqrt {49 - 48} }}{2} = \dfrac{{ - 7 \pm \sqrt 1 }}{2} = \dfrac{{ - 7 \pm 1}}{2}$
Either $x = \dfrac{{ - 7 + 1}}{2} = \dfrac{{ - 6}}{2} = - 3$ or $x = \dfrac{{ - 7 - 1}}{2} = \dfrac{{ - 8}}{2} = - 4$
So, the two roots of the given quadratic equation are -3 and -4.
Clearly, we are getting the same results from both factorization method and discriminant method.
Note- In these types of problems, we can use either factorization method or discriminant method to obtain the roots of the given quadratic equation. But discriminant method is usually adopted because it is an easier method as compared to factorization method for finding the roots of the given quadratic equation.
|
# Learning Materials
I collect some posts about LSTM and RNN here, with some useful excerpts as well as my thoughts.
## RNN
• 使用Keras中的RNN模型进行时间序列预测
• 用语言详细讲了LSTM的运行过程，每一步会生成什么，传递什么
• 详细讲了LSTM层的return_sequence参数的作用
• return_sequence=False 被用于 RNN 层与 Feedforward 层（卷积神经网络、Fully-connected神经网络等）之间连接的桥梁
## LSTM
• Understanding LSTM Networks
• 几个问题
• sigmoid 层的输出始终在 [0, 1]， tanh 层的输出始终在 [-1, 1]，最终的输出是 element-wise multiply，所以每个元素都在 [-1, 1] 之间
• 变种 LSTM 效果都差不多，个别情况特殊 RNN 更好
• Greff, et al. (2015) do a nice comparison of popular variants, finding that they’re all about the same.
• Jozefowicz, et al. (2015) tested more than ten thousand RNN architectures, finding some that worked better than LSTMs on certain tasks.
• the remarkable results people are achieving with RNNs. Essentially all of these are achieved using LSTMs.
• 其它拓展：Attention; Grid LSTMs; Work using RNNs in generative models;
• （完全看懂）Step-by-step understanding LSTM Autoencoder layers（重点参考 LSTM 部分）
• 详细解释了LSTM层的作用机理。
• Keras 中的 LSTM 在开启 return_sequence=True 时会返回每个 recurrent 的 time step 的输出
• 如果是 return_sequence=False 则只返回最后一个输出，然后如果下一层是一个LSTM层，需要配合 RepeatVector 把这唯一的一层复制为相应的 time steps 一样多的层数，然后再输入到下一层
• 如果是做 time series 重建的话，每层都不能改变 length 的，所以如果不用 return_sequence=False 就需要用 RepeatVector 复制
• TimeDistributed 层的描述比较晦涩。
• 前面最后一层 LSTM 的输出，每个 time step 的 features 个数与实际需要的不同，所以需要把每一个 time step 上 LSTM 的值输出为正确的 feature 个数
• 可以每个 time step 都构造一个 Dense 层来完成这个变换，但是使用 TimeDistributed 可以只构造一个 Dense，然后自动应用到所有的 time step 上，这也就对应了官方文档中描述的 applies a layer to every temporal slice of an input，而这个时间层就是 the dimension of index one，注意这里的 index one 实际上因为 Python 是 Zero-based numbering 所以指的是 the second dimension
• 至于官方文档后面说的 used with arbitrary layers 就更混乱了，其实意思就是可以用 TimeDistributed 把输入的自定义层，按 index-1 为划分，自动应用到从输入的 index-2 开始的后面维度构成的 tensor （为什么没有 index-0 什么事儿呢？因为它是训练数据的个数，通常是不确定的，而且不影响定义各种层的操作，每个训练 sample 最终都会被执行这段的前半部分讨论的操作）
• 另一个相关的描述：We could output the model with another LSTM layer with one neuron and return_sequences=True parameter, but using a TimeDistributed layer wrapping a Dense layer we will have the same weights for each outputted time-step. (from silver medalTime-series forecasting with LSTM autoencoders)
• 其实就是把时间序列排成2D-array，然后每层自动连接起来
• 【问题】如果return_sequences=False的话，是否还可以从 LSTM 连接到 LSTM layer？
• multi-class classification problem
• the log loss function (called categorical_crossentropy in Keras)
• using the ADAM optimization function.
• the Keras implementation resets the network state after each training batch
• making the LSTM layers stateful and manually resetting the state of the network at the end of the epoch
• This is truly how the LSTM networks are intended to be used.
• 但是在做预测时，预测数据往往没有这种顺序性（有办法回避这个问题吗？）
• the Keras “stateful” LSTM was really only a shortcut to replaying the first n-sequences, but didn’t really help us learn a generic model of the alphabet.
• fixed length, batch size 1, stateful LSTM == variable length, stateless LSTM (I’m not 100% sure about this.)
• Variable-Length Input
• zero padding. Here, we use left-hand-side (prefix) padding with the Keras built in pad_sequences() function.
• 我的一些总结：从上一个图中可以看出来，state 在初始值时，会被输入 X 逐渐赋值
• 如果每个batch_size重置，则每次重新fit整个batch；
• 如果两个相邻batch之间没有联系，比如shuffled training data，不重置就不合理；
• 但是如果是ordered training data，就可以保留下来，可以保持训练的连贯性；
• 代价是做prediction时必须使用同样的batch_size，因为在batch操作时必须提前知道batch_size是多少才能批量保留前一个状态，而且要与training data的顺序有连贯性。（如果取了 batch_size=1 则没有这些限制了，比较tricky）
• 一种折衷的方法是，手动在每个epoch进行重置，而不是在每个batch进行重置。
• （注意）使用 stateful=True 时，要注意经常进行 model.reset_states() 的操作，不然会导致后续的预测依赖前面的预测。这篇post中Stateful LSTM for a One-Char to One-Char Mapping部分的代码，每次预测完都要重置，所谓的 cold start
• Stateful and Stateless LSTM for Time Series Forecasting with Python
• rolling-forecast scenario will be used, also called walk-forward model validation.
• A model will be used to make a forecast for the time step, then the actual expected value from the test set will be taken and made available to the model for the forecast on the next time step.
• This mimics a real-world scenario where new Shampoo Sales observations would be available each month and used in the forecasting of the following month.
• Data Preparation
• Transform the time series data so that it is stationary: a lag=1 differencing to remove the increasing trend in the data
• to rescale the data to values between -1 and 1 to meet the default hyperbolic tangent activation function of the LSTM model. （必须这样吗？似乎是必须的）
• A batch size of 1; online training; will have some variance
• A stateless LSTM does not shuffle input patterns during training because the network aims to learn the sequence of patterns. We will test a stateless LSTM with and without shuffling.
• 效果：
• the stateless configurations may be more stable.
• Review of Findings
• Resetting state when making one-step predictions with a stateful LSTM may improve performance on the test set.
• Fitting a stateful LSTM and seeding it on the training dataset and not performing any resetting of state during training or prediction may result in better performance on the test set.
• （结论）用一个例子对比了各种不同参数设置的 LSTM 的效果，感觉没有太大的质的变化，预测的结果都在比较接近的范围内。
• 使序列是 stationary 似乎是特别重要的一个性质（为什么？）
### Application of LSTM on Time Series
• Time-series forecasting with LSTM autoencoders
• 用纯 LSTM 做预测：LSTM(10) – LSTM(6) – LSTM(1) – Dense(10) – Dense(10) – Dense(1)
• 用 LSTM autoencoder：LSTM(12) – LSTM(6) – LSTM(1) – RepeatVector(12) – LSTM(12) – LSTM(6) – TimeDistributed(Dense(1))
• 为什么后面的 decoder 是先12再6？故意反过来的？
• 然后把 original time series 中的最后一个值，和 encoder （直到 LSTM(1)）的值作为新的输入，训练一个新的网络 Dense(10) – Dense(10) – Dense(1) 来做预测
• 实际上是用 encoder 作为一个数据压缩的工具，把完整长度的 time series 压缩成较短的输入，然后再训练一个预测网络。
• 【感觉这篇讲的东西没有什么用】
• Deep Learning for Time Series Forecasting
• MLP for Time Series Forecasting
• Dense(100) – Dense(1)
• CNN for Time Series Forecasting
• Conv1D(64) – MaxPooling1D(2) – Flatten() – Dense(50) – Dense(1)
• LSTM for Time Series Forecasting
• LSTM(50) – Dense(1)
• CNN-LSTM for Time Series Forecasting （没有看懂）
• TimeDistributed(Conv1D(64)) – TimeDistributed(MaxPooling1D(2)) – TimeDistributed(Flatten()) – LSTM(50) – Dense(1)
• 从最终的结果上看，似乎效果都差不多，最简单的MLP反而是testing RMSE最好的……
• 这篇比较有意思，基于这篇，
• 测试了MLP的性能确实不错
• 下一步测试CNN的性能
• 目前使用LSTM没有得到好的效果
• 从RNN到LSTM，性能良好的神经网络到底是如何工作的？
• 注意力机制（Attention Mechanism）在很多任务中对性能有着很大的影响。
• 在生成第二个单词的时候，根据目标端的隐藏层状态和源端的每一个隐藏层之间做相似度的计算，根据规划得到一个权重的得分，接着生成一个上下文的Attention Layer，借助于这个Attention Layer作用于生成第二个词的过程，进而生成正确的翻译。
• LSTM
• GRU
• 它将forget gate和input gate合成一个update gate，这个gate用于综合控制cell的读和写，这样可以简化LSTM的参数，此外它还将cell state和hidden state进行合并。总体而言，它是一个比LSTM更加简化的结构，在训练上更加容易（参数少）。
• RNNs的阿喀琉斯之踵
• 1、RNNs对层次信息的表示能力和卷积能力都存在一些不足。因此产生了一些如Deep RNN、Bidirectional RNN和Hierarchical RNN的变种，都希望通过更层次化的网络结构来弥补这种不足。
• 2、RNNs的并行度很低。它的当前时刻的隐藏层的状态依赖于之前时间的状态，并且训练和解码的并行度都很低。
• 为了解决上面的问题，提出了一种新的神经网络的架构——Transformer
• Transformer使用简单的FF-NN替代了GRU，这样就直接解决了在训练中难以并行的问题。
• 这时引入了一个Positional Encoding，对于每个位置都有一个Embedding来存储位置信息。
• 【细节未看】
# Criticism
• The fall of RNN / LSTM
• It is the year 2014 and LSTM and RNN make a great come-back from the dead.
• Then in the following years (2015–16) came ResNet and Attention. （这是什么？）
• Also attention showed that MLP network could be replaced by averaging networks influenced by a context vector.
• Problem
• memory-bandwidth-bound computation
• not a good match for hardware acceleration
• 似乎是主要在讨论 translation problem，不确定这里的讨论是否适用于所有 time series problem
• .
• 这篇也有中文翻译
• When to Use MLP, CNN, and RNN Neural Networks – see section “When to Use Recurrent Neural Networks?”
• RNNs in general and LSTMs in particular have received the most success when working with sequences of words and paragraphs, generally called natural language processing.
• RNNs and LSTMs have been tested on time series forecasting problems, but the results have been poor, to say the least. Autoregression methods, even linear methods often perform much better. LSTMs are often outperformed by simple MLPs applied on the same data.
• [On the Suitability of Long Short-Term Memory Networks for Time Series Forecasting]https://machinelearningmastery.com/suitability-long-short-term-memory-networks-time-series-forecasting/
• Last Updated on August 5, 2019
• 主要是针对一篇2001年的论文展开的讨论，分析了在两种经典时间序列上的LSTM的效果
• 评论中有很多有意思的讨论 (copy-pasted from the comments below the article; use “in-page search” function to find the following exact replies)
• Thank you for this informative article. While I have not trained any LSTM RNNs on my time series data, I’ve found that plain jane feed forward DNNs that are feature engineered correctly and use sliding window CV predicts OOS as expected. Essentially, it is not necessary for my data to have to mess with LSTMs at all. And my features aren’t all iid… I suppose this illustrates the versatility of DNNs.
• MLPs with a window do very well on small time series forecasting problems.
• I agree that along with MLP’s , LSTM’s have an advantage over more classical statistical approaches like ARIMA : they can fit non-linear functions and moreover , you do not need to specify the type of non-linearity . There also lies the danger : regularisation is absolutely crucial to avoid overfitting .
• Why use data hungry algorithm like lstm when similar results can be obtained using a machine learning or time series methods?
• After getting my results from comparing MLPs and LSTMs on a data, I came to the conclusion that MLPs are much more efficient than LSTMs, when using constant number of lags (let’s say 7 days before) to predict 4 days ahead. Would You create a tutorial on using LSTMs with dynamic lag time?
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# manually control header and page numbering
I am writing a report, where if i use the command \pagestyle{plain} the page numbering comes to the bottom, thats good.
However i also want the heading of the chapter to come on the top left hand corner, hows that possible. If i remove the \pagestyle{plain}, then the heading comes on the top left corner but the page numbering as well comes on the top right, when i want it in the bottom!
\documentclass[letterpaper,11pt,oneside,final]{book}
%-----------------------------------------------------------------------------%
% Margins:
%-----------------------------------------------------------------------------%
\setlength{\marginparwidth}{0pt} % width of margin notes
\setlength{\marginparsep}{0pt} % width of space between body text and margin notes
\setlength{\evensidemargin}{0.125in} % Adds 1/8 in. to binding side of all
\setlength{\oddsidemargin}{0.125in} % Adds 1/8 in. to the left of all pages
\setlength{\textwidth}{6.375in} % assuming US letter paper (8.5 in. x 11 in.) and
\raggedbottom
\setlength{\parskip}{\medskipamount}
\renewcommand{\baselinestretch}{1.1} % this is the default line space setting
\let\origdoublepage\cleardoublepage
\newcommand{\clearemptydoublepage}{%
\clearpage{\pagestyle{empty}\origdoublepage}}
\let\cleardoublepage\clearemptydoublepage
\usepackage[pdftex,letterpaper=true,pagebackref=true]{hyperref} % with basic options
\usepackage{hypcap}
\usepackage{grffile}
\usepackage{libertine}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{textcomp}
\usepackage{mathpazo}
\usepackage{stmaryrd}
\usepackage{rotating}
\usepackage{dsfont}
\hypersetup{pdfpagemode=UseNone}
%\fancyhf{}
\usepackage{fancyhdr}
\fancyhf{} % clear header and footer
\fancyhead[L]{\leftmark}
\fancyfoot[C]{\thepage}
\pagestyle{fancy}
\fancypagestyle{plain}{} % inherit the settings from fancy style
\begin{document}
snippet
As any dedicated reader can clearly see, the Ideal of practical reason is a rep-
resentation of, as far as I know, the things in themselves; as I have shown else-
where, the phenomena should only be used as a canon for our understanding.
The paralogisms of practical reason are what rst give rise to the architectonic
of practical reason. As will easily be shown in the next section, reason would
thereby be made to contradict, in view of these considerations, the Ideal of prac-
tical reason, yet the manifold depends on the phenomena. Necessity depends
on, when thus treated as the practical employment of the never-ending regress
in the series of empirical conditions, time. Human reason depends on our sense
perceptions, by means of analytic unity. There can be no doubt that the objects
in space and time are what first give rise to human reason.
\section{Test 1}
As any dedicated reader can clearly see, the Ideal of practical reason is a rep-
resentation of, as far as I know, the things in themselves; as I have shown else-
where, the phenomena should only be used as a canon for our understanding.
The paralogisms of practical reason are what rst give rise to the architectonic
of practical reason. As will easily be shown in the next section, reason would
thereby be made to contradict, in view of these considerations, the Ideal of prac-
tical reason, yet the manifold depends on the phenomena. Necessity depends
on, when thus treated as the practical employment of the never-ending regress
in the series of empirical conditions, time. Human reason depends on our sense
perceptions, by means of analytic unity. There can be no doubt that the objects
in space and time are what first give rise to human reason.
\pagestyle{empty}
\bibliographystyle{alpha}
\nocite{*}
\hfill \today
\end{document}
• Please add a minimal working example (MWE) that illustrates your problem. Feb 12 '14 at 16:10
• Please tell us what you want in the header and in the footer in first chapter pages and in all the rest. Feb 12 '14 at 19:33
• in all the chapters i want a uniform format, as the top left being the Chapter no.xx:xxnamexx and the bottom of the page to have the page number. Feb 12 '14 at 19:34
## 3 Answers
This is another way of making plain page style equal to fancy page style
You have to use \chapter so that \leftmark is defined. Also, you have used \pagestyle{empty}. Put a \clearpage (\cleardoublepage if using twoside option) before.
\documentclass[letterpaper,11pt,oneside,final]{book}
%-----------------------------------------------------------------------------%
% Margins:
%-----------------------------------------------------------------------------%
\setlength{\marginparwidth}{0pt} % width of margin notes
\setlength{\marginparsep}{0pt} % width of space between body text and margin notes
\setlength{\evensidemargin}{0.125in} % Adds 1/8 in. to binding side of all
\setlength{\oddsidemargin}{0.125in} % Adds 1/8 in. to the left of all pages
\setlength{\textwidth}{6.375in} % assuming US letter paper (8.5 in. x 11 in.) and
\raggedbottom
\setlength{\parskip}{\medskipamount}
\renewcommand{\baselinestretch}{1.1} % this is the default line space setting
%\let\origdoublepage\cleardoublepage
%\newcommand{\clearemptydoublepage}{%
% \clearpage{\pagestyle{empty}\origdoublepage}}
%\let\cleardoublepage\clearemptydoublepage
\usepackage[pdftex,letterpaper=true,pagebackref=true]{hyperref} % with basic options
\usepackage{hypcap}
\usepackage{grffile}
\usepackage{libertine}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{textcomp}
\usepackage{mathpazo}
\usepackage{stmaryrd}
\usepackage{rotating}
\usepackage{dsfont}
%\hypersetup{pdfpagemode=UseNone}
\usepackage{fancyhdr}
\fancyhf{} % clear header and footer
\fancyhead[L]{\leftmark}
\fancyfoot[C]{\thepage}
\makeatletter
\let\ps@plain\ps@fancy % plain style = fancy style
\makeatother
\pagestyle{fancy}
\begin{document}
\chapter{One}
As any dedicated reader can clearly see, the Ideal of practical reason is a rep-
resentation of, as far as I know, the things in themselves; as I have shown else-
where, the phenomena should only be used as a canon for our understanding.
The paralogisms of practical reason are what rst give rise to the architectonic
of practical reason. As will easily be shown in the next section, reason would
thereby be made to contradict, in view of these considerations, the Ideal of prac-
tical reason, yet the manifold depends on the phenomena. Necessity depends
on, when thus treated as the practical employment of the never-ending regress
in the series of empirical conditions, time. Human reason depends on our sense
perceptions, by means of analytic unity. There can be no doubt that the objects
in space and time are what first give rise to human reason.
\section{Test 1}
As any dedicated reader can clearly see, the Ideal of practical reason is a rep-
resentation of, as far as I know, the things in themselves; as I have shown else-
where, the phenomena should only be used as a canon for our understanding.
The paralogisms of practical reason are what rst give rise to the architectonic
of practical reason. As will easily be shown in the next section, reason would
thereby be made to contradict, in view of these considerations, the Ideal of prac-
tical reason, yet the manifold depends on the phenomena. Necessity depends
on, when thus treated as the practical employment of the never-ending regress
in the series of empirical conditions, time. Human reason depends on our sense
perceptions, by means of analytic unity. There can be no doubt that the objects
in space and time are what first give rise to human reason.
As any dedicated reader can clearly see, the Ideal of practical reason is a rep-
resentation of, as far as I know, the things in themselves; as I have shown else-
where, the phenomena should only be used as a canon for our understanding.
The paralogisms of practical reason are what rst give rise to the architectonic
of practical reason. As will easily be shown in the next section, reason would
thereby be made to contradict, in view of these considerations, the Ideal of prac-
tical reason, yet the manifold depends on the phenomena. Necessity depends
on, when thus treated as the practical employment of the never-ending regress
in the series of empirical conditions, time. Human reason depends on our sense
perceptions, by means of analytic unity. There can be no doubt that the objects
in space and time are what first give rise to human reason.
\clearpage
\pagestyle{empty}
\bibliographystyle{alpha}
\nocite{*}
\hfill \today
\end{document}
Slightly un-related, you may consider using geometry package for changing the page layout.
• this is precisely what i want, but when i implement your code of \usepackage{fancyhdr} \fancyhf{} % clear header and footer \fancyhead[L]{\leftmark} \fancyfoot[C]{\thepage} \pagestyle{fancy} \makeatletter \let\ps@plain\ps@fancy % plain style = fancy style \makeatother I get a line across the top of the page without any page numbering in the bottom, do i need to maybe install some package? Feb 13 '14 at 15:24
• @howtotag Did you run the code above as such? Line at the top isn't a problem. It can be removed if you don't want.
– user11232
Feb 13 '14 at 15:27
• nope, i ran ur code it works fine, i put the lines of the code whicha are relevant (in my comments) in my report, and then it doesnt appear as its required to? Could it be because im using documentclass{book}? Feb 13 '14 at 15:29
• @howtotag Can you post your full code to your question? Some thing is wrong in it.
– user11232
Feb 13 '14 at 15:30
• added snippet to question Feb 13 '14 at 15:49
If I understand correctly, you want the same behavior for all pages, including first pages of chapters.
And you want the chapter heading on the left of the header, and the page number on the center of the footer.
The following code does exactly that:
\documentclass{report}
\usepackage{fancyhdr}
\fancyhf{} % clear header and footer
\fancyhead[L]{\leftmark}
\fancyfoot[C]{\thepage}
\pagestyle{fancy}
\fancypagestyle{plain}{} % inherit the settings from fancy style
\usepackage{kantlipsum} % only for the example
\begin{document}
\chapter{Test}
\kant[1]
\section{Test}
\kant[2-7]
\end{document}
• this is precisely what i want, but when i implement your code of \usepackage{fancyhdr} \fancyhf{} % clear header and footer \fancyhead[L]{\leftmark} \fancyfoot[C]{\thepage} \pagestyle{fancy} \makeatletter \let\ps@plain\ps@fancy % plain style = fancy style \makeatother I get a line across the top of the page without any page numbering in the bottom, do i need to maybe install some package? Feb 13 '14 at 15:26
• @howtotag Probably you have something in your document that interferes with it. Try adding a Minimal (Non) Working Example in your question. Feb 13 '14 at 15:36
• added snippet in my question Feb 13 '14 at 15:50
Have you tried using the fancyhdr package? I've just implemented it to do something similar. The syntax is pretty straightforward and you can basically customise your page layout to get what you want.
The following few lines might get you close?
\documentclass[12pt, a4paper, oneside, fleqn]{report}
\headheight 28pt
\headsep 24pt
\usepackage{lmodern} % gives bold italic font
\usepackage{lipsum}
%Sectioning and headings
%------------------------
\usepackage{fancyhdr}
\begin{document}
\pagestyle{fancy}
\lhead{\leftmark}
\chead{}
\rhead{}
\lfoot{}
\cfoot{\thepage}
\rfoot{}
\chapter{A chapter}
\lipsum
\end{document}
• i still dont get the desired output. I still get the chapter name on the top left and thepage number on the top right (which i want at the bottom). Could you just paste a snippet of the fancyhdr part that you use? Feb 12 '14 at 16:57
• Sorry I should have done that before, but I was in a rush. Does that get you close to what you wanted? Feb 12 '14 at 17:16
• You should have \lhead{\sffamily\leftmark}; better yet, \fancyhead[L]{\sffamily\leftmark}. Leave off \sffamily from \chaptermark. By the way, it is not a command with arguments. Feb 12 '14 at 17:28
• nope, i still didnt get it. I get a line on the top of my page with no page numbers in any of the pages at all! What do you think is the problem? Before you snippet i also used \usepackage{fancyhdr} Feb 12 '14 at 17:49
• @egreg I copied that syntax (with the argument) direct from the 'fancyhdr' documentation, so if the syntax is wrong I guess the documentation is wrong! Anyway, it works! Thanks for the tip about moving the \sffamily. Feb 14 '14 at 11:48
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## A virtually ample field that is not ample. (arXiv:1810.05184v1 [math.AG])
A field \$K\$ is called ample if for every geometrically integral \$K\$-variety \$V\$ with a smooth \$K\$-point, \$V(K)\$ is Zariski-dense in \$V\$. A field \$K\$ is virtually ample if some finite extension of \$K\$ is ample. We prove that there exists a virtually ample field that is not ample.查看全文
## Solidot 文章翻译
你的名字 留空匿名提交 你的Email或网站 用户可以联系你 标题 简单描述 内容 A field \$K\$ is called ample if for every geometrically integral \$K\$-variety \$V\$ with a smooth \$K\$-point, \$V(K)\$ is Zariski-dense in \$V\$. A field \$K\$ is virtually ample if some finite extension of \$K\$ is ample. We prove that there exists a virtually ample field that is not ample.
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Area of Circle NCERT Exercise 12.1 solution Class Ten Mathematics
# Area of Circle
## Exercise 12.1
Question: 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution: Circumference of first circle
= 2 πr = 2π xx 19 = 38π cm
Circumference of second circle = 18π cm
Or, 2πr = 56π
Or, 2r = 56
Or, r = 28 cm
Circumference of the largest circle; as per question
= 38π + 18π = 56π cm
Question: 2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Answer: Area of first circle = πr^2
= π8^2 = 64π sq cm
Area of second circle = π6^2 = 36π sq cm
As per question; area of the largest circle = 64π + 36π
= 100π sq cm
Or, πr^2 = 100π
Or, r^2 = 100
Or, r = 10 cm
Question: 3: Following figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Solution: Area of Gold Region = πr^2
= π10.5^2 = 110.25^2 = 346.5 sq cm
Area of Red Region = πr^2
= π21^2 - π10.5^2
= π(21^2 – 10.5^2)
= π(21 + 10.5)(21 – 10.5)
= π xx 31.5 xx 10.5 = 1039.5 sq cm
Area of Blue Region = πr^2
= π(31.5^2 – 21^2) = 1732.5 sq cm
Area of Black Region = πr^2
= π(42^2 – 31.5^2) = 2425.5 sq cm
Area of White Region = πr^2
= π(52.5^2 – 42^2) = 3118.5 sq cm
Question: 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution: Distance covered in 10 minutes = (66)/(60) xx 10 = 11 km
Circumference = πd = 80π
=(22)/(7)xx80 cm
=(22)/(7)xx80xx(1)/(1000xx100) km
Number of revolutions
=11xx1000xx100xx(7)/(22xx80)
=4375
Question: 5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units
Solution: (A) 2 units
Explanation:
2πr=πr^2
Or, (πr^2)/( πr)=2
Or, r=2
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# Rademacher complexity for piecewise-linear convex function
Consider a function family $$\ell(x)=\max_{1\leq k\leq K} a_k^\top x + b_k,$$ where $a_k,b_k \in \mathbb{R}^d$ are bounded in the sense of some norm and $K\geq 2$. What is the best upper bound on the capacity (e.g., Rademacher complexity, etc) of such a family, in terms of dependence on dimension $d$, number of pieces $K$ and sample size $n$?
• This is not exactly that, as your function is real-valued (no $\mathrm{sign}()$ taken), but without this distinction you'd get the union of $K$ bounded-weight halfspaces. That may be something to look into. – Clement C. Aug 9 '18 at 17:08
## 1 Answer
Since we're talking about real-valued functions, rather than VC-dimension, you probably want the fat-shattering one. The $\gamma$-fat-shattering of linear functions with $\ell_2$ norm bounded by $B$ is of order $(B/\gamma)^2$. For a $k$-fold maximum of such functions, the bound grows as $O((B/\gamma)^2k\log k)$, as shown here: https://www.cs.bgu.ac.il/~karyeh/rademacher-max-hyperplane.pdf
This translates immediately into a Rademacher bound, given therein.
The $\log(k)$ factor cannot be removed in this case, as recently shown in https://arxiv.org/abs/1807.07924
Edit: I had originally neglected to mention that the fat-shattering bound of $(B/\gamma)^2$ implicitly assumes that the instance space is confined to the Euclidean ($\ell_2$) unit ball. More generally, if the linear functions have $\ell_p$ norm at most $B$ and the data is confined to an $\ell_q$ ball of radius at most $R$, the bound becomes $BR/\gamma^2$, where $1/p+1/q=1$, $p,q\ge1$. This readily follows from the classic Rademacher analysis of "fat" hyperplanes: in the step where Cauchy-Schwarz is invoked, invoke instead Hölder.
• See related mathoverflow question mathoverflow.net/questions/266457/… – Aryeh Aug 9 '18 at 22:05
• Thanks! What if we change the norm on $w$ in your notes? – O. Richard Aug 10 '18 at 2:00
• Change the norm to what? You need some bound on the norm. – Aryeh Aug 10 '18 at 2:05
• Instead of the 2-norm, consider 1-norm or inf-norm. – O. Richard Aug 10 '18 at 2:56
• Addressed in the edit. – Aryeh Aug 10 '18 at 13:04
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# Adding a new highscore before or after an equal score?
I'm making a highscores list for my 2048 replica. An entry in the list consists of a name and a score (e.g. "John" and 2100.) When I add a new entry, the code runs through the list and inserts the new entry when the score for the new entry is greater than the currently indexed entry. So imagine this list:
Roman 17000
Sergei 12500
Dean 8000
Andrew 1400
If John was inserted into the list, the new list would look like:
Roman 17000
Sergei 12500
Dean 8000
John 2100
Andrew 1400
The question is, what do I do when inserting the entry for "Steve" whose score is 8000. Does he go before or after Dean? Is there a convention for this?
• Normally the score of the new player (Steve) would go on a higher place, as it is inserted afterwards – Ion Farima Apr 5 '14 at 17:32
• Really? Friends have told me that normally the new entry would go afterwards. I mean, I understand that it's ultimately a style choice, but I'm curious if you know of any examples of major games where new and equal highscores are given a higher place? – Lou Apr 6 '14 at 9:49
Though it is (to my knowledge) most common for newest score to be entered at the top, I would argue that the OLDEST score would deserve the first position.
For this simple reason:
The first player to achieve the high score had LESS TIME to achieve that score, than people making later/following attempts, thus deserving a higher position as (s)he was the quickest to achieve the level of skill required to make that score.
Inspired by this you might track time spent in each game, and sort the list so that LEAST time spent on creating a similar score goes FIRST.
• Good idea :). I might well do that if I'm bored and want to add more features. – Lou Apr 8 '14 at 9:33
• Then again, prioritising by time only works if you're measuring time. As it is, whether someone achieved a score of 8000 on the 1st of April compared to the person who achieved a score of 8000 on the 8th of April doesn't make any difference. – Lou Apr 8 '14 at 9:37
Just sort them alphabetically.
And if you have numbers to call out the position, be sure to handle the tie case appropriately so you aren't calling one person 3rd and another 4th even though it's the same position.
• This answer has a two perfectly egalitarian methods that no one else has mentioned. Though Loeffe and Riccardo's time idea is pretty good, too. – Seth Battin Apr 7 '14 at 23:07
• Yeah, it's a good idea. I don't think I'll do this though, simply because it's easier to prioritise entirely by score, and I don't want to make a complex hiscores list. As I've said before, this is only for me :). – Lou Apr 8 '14 at 9:35
If you're looking to encourage competition and activity, it might make more sense to have the more recent score replace the older one. Players wouldn't be able to simply achieve a high score and become complacent, because their score is at risk of being replaced at any point.
In fact, if you really wanted players to rue being replaced, you could have the higher scores on the list displayed in a style that is more rewarding than the older ones (e.g. more colorful, stylized, etc.) This way when a player's score is tied, not only do they move down, but they also lose whatever superficial qualities were associated with that spot on the list.
• Fair point, but I'm only building a 2048 clone that I can play on my machine - I mainly want it so I can keep track of my own score, and if any of my friends play it. – Lou Apr 5 '14 at 18:26
Tradition says new scores on top. No idea why, though... now that I think about it, it makes more sense the other way around, you should award a player for achieving an equal score in less time. But that's just my personal opinion, in the end the choiche is up to you :)
• Okay :). I was curious if there was a standard. Personally I agree that putting it after makes sense - but only because it encourages you to beat the score, not match it. – Lou Apr 5 '14 at 19:48
I don't believe there is a standard. From what I've seen, and what I use, who ever got to the top first should keep their position.
Now the choice is up to you, but I would spend some time to add more to the highscore board so it can track different achievements of a player throughout the play-time, and then compare them to the player who has equal score.
Placing Steve on top would be unfair for Dean. Placing Steve below Dean will make Steve angry. One of many things that might lead to a forum war. It's a minor but you got my point.
• Fair point, but I probably won't refine this significantly. As I commented on Clockwise's post, I'm only making an application I can play on my computer, and possibly get a friend to play. Also, it's only 2048, I can't think of any achievements that could further weight the highscores. – Lou Apr 6 '14 at 9:48
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# Semidirect product of Z3 and D8 with action kernel V4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
View a complete list of particular groups (this is a very huge list!)[SHOW MORE]
## Definition
This group is defined by the following presentation:
$G := \langle a_1, a_2,a_3,a_4 \mid a_1^3 = a_2^2 = a_3^2 = a_4^2 = e, a_1a_2 = a_2a_1, a_1a_3 = a_3a_1, a_2a_3 = a_3a_2, a_4a_1a_4^{-1} = a_1^{-1}, a_4a_2a_4^{-1} = a_3, a_4a_3a_4^{-1} = a_3^{-1} \rangle$
## GAP implementation
### Group ID
This finite group has order 24 and has ID 8 among the groups of order 24 in GAP's SmallGroup library. For context, there are 15 groups of order 24. It can thus be defined using GAP's SmallGroup function as:
SmallGroup(24,8)
For instance, we can use the following assignment in GAP to create the group and name it $G$:
gap> G := SmallGroup(24,8);
Conversely, to check whether a given group $G$ is in fact the group we want, we can use GAP's IdGroup function:
IdGroup(G) = [24,8]
or just do:
IdGroup(G)
to have GAP output the group ID, that we can then compare to what we want.
### Description by presentation
gap> F := FreeGroup(4);
<free group on the generators [ f1, f2, f3, f4 ]>
gap> G := F/[F.1^3,F.2^2,F.3^2,F.1*F.2*F.1^(-1)*F.2^(-1), F.1*F.3*F.1^(-1)*F.3^(-1), F.2*F.3*F.2^(-1)*F.3^(-1), F.4^2,F.4*F.1*F.4^(-1)*F.1,F.4*F.2*F.4^(-1)*F.3];
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# Prove that special orthogonal matrix from $SO(n, \mathbb{R})$ has $\frac{(n-1)n}{2}$ independent parameters
Show that special orthogonal matrix from $$SO(n, \mathbb{R})$$ has $$\frac{(n-1)n}{2}$$ independent real parameters.
I assume that this will be related to Euler angles somehow or specifically to its generalization. But how do we find degrees of freedom of such matrices? How do they look like?
Any hints would be helpful.
An $$n\times n$$ real matrix contains $$n^2$$ real parameters. The column matrices of a real orthogonal matrix are normal and orthogonal to each other. There exist $$n$$ real matrix constraints for the normalization and $$n(n-1)/2$$ real constraints for the orthogonality. Thus, the number of independent real parameters for characterizing the elements of the groups $$SO(N)$$ is equal to $$n^2-(n+n(n-1)/2)) = n(n-1)/2.$$
For the dimension as a Lie group, we can also just determine the dimension of its Lie algebra $$\mathfrak{so}(n)$$, consisting of skew-symmetric matrices, i.e., $$n(n-1)/2$$.
Edit: As remarked in the comments, this is not meant to be a proof. A correct proof requires much more theory.
• Could you elaborate on what "n real matrix constraints for the normalization and $n(n-1)/2$ real constraints for the orthogonality" means? – janusz Mar 5 at 20:23
• Dietrich; your first proof is not a proof (and you kow that!). You make the same reasoning as most students: "since $A^TA$ is symmetric, the equality $A^TA=I_n$ is equivalent to $n(n+1)/2$ relations, that is the degree of freedom of an orthogonal matrix is $n(n-1)/2$"; that does not prove the algebraic independence of these relations. Moreover,I think that the OP does not even know what are algebraically independent relations. For years, I have been against this type of reasoning (which is only intuitive). I am surprised that a "maestro" like you presents it as a proof in its own right. – loup blanc Mar 6 at 15:13
• @loupblanc you are right, this is not what I consider a proof. In fact it comes from a book and I just thought it could be taken as a plausible argument. As you say, the OP would not like more than this, I suppose. I would prove it via the Lie algebra. But also this needs more details, see the other answer. – Dietrich Burde Mar 6 at 16:30
• Yes Dietrich, I agree with you. – loup blanc Mar 6 at 17:04
Let
$$\gamma(t) \in SO(n) \tag 1$$
be a smooth path through the identity element $$I$$ of $$SO(n)$$:
$$\gamma: I \to SO(n), \; \gamma(0) = I, \tag 2$$
where
$$I = (-\epsilon, \epsilon) \subset \Bbb R, \; \Bbb R \ni \epsilon > 0; \tag 3$$
then we have
$$\forall t \in I, \; \gamma^T(t) \gamma(t) = I, \tag 3$$
by virtue of (2); if we differentiate this equation we obtain
$$\dot \gamma^T(t) \gamma(t) + \gamma^T(t) \dot \gamma(t) = 0, \; \forall t \in I; \tag 4$$
now taking
$$t = 0, \; \gamma(t) = \gamma(0) = I, \tag 5$$
we see that (4) becomes
$$\dot \gamma^T(0) + \dot \gamma(0) = \dot \gamma^T(0) I + I \dot\gamma(0) = \dot \gamma^T(0) \gamma(0) + \gamma^T(0) \dot \gamma(0) = 0, \tag 6$$
which yields
$$\dot \gamma^T(0) = -\dot \gamma(0); \tag 7$$
it follows then that the tangent space $$T_ISO(n)$$ at $$I$$ consists of skew-symmetric $$n \times n$$ real matrices; but it is well-known, and easy to see, that the dimension of this set of matrices over $$\Bbb R$$ is $$n(n - 1)/2$$. Therefore if we can show that every $$n \times n$$ matrix $$K$$ with
$$K^T = -K \tag 8$$
is in $$T_ISO(n)$$, we may conclude that in fact
$$\dim SO(n) = \dfrac{n(n - 1)}{2}; \tag 9$$
to this end, we define the path
$$\gamma(t) = e^{Kt}; \tag{10}$$
then
$$\dot \gamma(0) = Ke^{K(0)} = KI = K; \tag{11}$$
so every skew-symmetric $$K$$ is tangent to some curve (1) at $$I$$; thus we have finalized our demonstration that (9) binds.
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On rigid surfaces, the cytoskeleton of migrating cells is polarized, but tissue matrix is normally soft. We show that nonmuscle MIIB (myosin-IIB) is unpolarized in cells on soft matrix in 2D and also within soft 3D collagen, with rearward polarization of MIIB emerging only as cells migrate from soft to stiff matrix. Durotaxis is the tendency of cells to crawl from soft to stiff matrix, and durotaxis of primary mesenchymal stem cells (MSCs) proved more sensitive to MIIB than to the more abundant and persistently unpolarized nonmuscle MIIA (myosin-IIA). However, MIIA has a key upstream role: in cells on soft matrix, MIIA appeared diffuse and mobile, whereas on stiff matrix, MIIA was strongly assembled in oriented stress fibers that MIIB then polarized. The difference was caused in part by elevated phospho-S1943–MIIA in MSCs on soft matrix, with site-specific mutants revealing the importance of phosphomoderated assembly of MIIA. Polarization is thus shown to be a highly regulated compass for mechanosensitive migration.
Introduction
Cell migration on rigid substrates, such as coverslips, has revealed the potential for polarization of key cytoskeletal components, including myosin-II (Kolega, 2003; Vicente-Manzanares et al., 2008; Barnhart et al., 2011). On soft substrates and in 3D matrix, however, the morphologies of migrating cells and their phosphoprotein profiles appear distinct from those on rigid 2D substrates (Pelham and Wang, 1997; Doyle et al., 2009). Within a soft tissue, such as the developing brain, cytoskeletal polarization also shows no clear relation to the direction of migration, whereas cells cultured on rigid substrates polarize in the direction of migration (Distel et al., 2010). The impact of soft matrix microenvironments on cytoskeletal polarization and migration appears understudied as are the effects of gradients in matrix elasticity. Durotaxis is the tendency of a cell to crawl from soft matrix to stiff matrix in the absence of any gradients in ligand density or chemotactic factors (Lo et al., 2000; Cheung et al., 2009; Isenberg et al., 2009), and durotaxis has been speculated to result in part from an increase in the stability of adhesions to stiff matrix as cells migrate from soft matrix (Lo et al., 2000). However, the molecular mechanisms of durotaxis have remained unexplored.
A cell typically moves forward by detaching its adherent tail with contractile forces exerted by nonmuscle myosin-II on the matrix (Kolega, 2003). Myosin-II forces have also been found critical to sensing matrix elasticity E (Discher et al., 2005), although any specific role for myosin-II in sensing gradients in stiffness remains unclear. Of the A, B, and C isoforms of nonmuscle myosin-II, the A isoform (MIIA) is most abundant in mesenchymal tissues based on mass spectrometry (MS) estimates of tryptic peptide abundance (Ma et al., 2010), and it proves essential to any differentiation of embryos (Conti et al., 2004). Importantly, MIIA also contributes the majority of traction force exerted by mesodermal cells, such as embryo-derived fibroblasts (Cai et al., 2006). Nonmuscle MIIB (myosin-IIB) knockout mice exhibit select, but critical, defects in formation of heart and other tissues, and MIIB knockdown (KD) fibroblasts in culture exhibit extended tails that fragment, leading to a frequent change in direction and faster migration (Lo et al., 2004; Swailes et al., 2006). In cells crawling on rigid coverslips, MIIB is more enriched or polarized toward the cell rear (Saitoh et al., 2001; Sandquist et al., 2006), whereas MIIA appears more uniform. The isoform localization difference is caused, surprisingly, by a more stable cytoskeletal assembly mediated by the coiled-coil tail of MIIB (Vicente-Manzanares et al., 2008). On the other hand, phosphorylation of MIIA’s tail promotes disassembly of this traction-critical isoform, impacting epithelial cell migration on rigid substrates (Dulyaninova et al., 2007). We hypothesized therefore that MIIB could be important to the persistent migration of cells on matrix gradients and that the levels of MIIA phosphorylation could impact both durotaxis and cytoskeletal polarization.
Polarization of myosin-II and perhaps phosphoregulated states of the tails could be keys to understanding how mesenchymal stem cells (MSCs) traffic to sites of scarring and wounding in collagen-rich tissues, such as the heart (Orlic et al., 2001; Quevedo et al., 2009). In such sites, these cells have immunomodulatory functions that limit formation of a collagen-I–rich scar (Salem and Thiemermann, 2010; Shi et al., 2010), which is perhaps why these cells are being widely used in clinical trials today even though we know very little about their motility. MSCs have a fibroblast-like cytoskeleton with MIIA and MIIB that contribute to various cellular processes, including matrix elasticity sensing (Engler et al., 2006; Johnson et al., 2007). Using an atomic force microscope (AFM), we have previously measured the elasticity of an infarct scar in heart to be Escar = 30–70 kPa (Berry et al., 2006), which is stiffer than normal muscle and most other soft tissues (E = ∼0.1–30 kPa; Engler et al., 2006). A scar is far softer than glass or plastic, and its stiffness can be mimicked with collagen-coated gels of polyacrylamide that are in wide use to understand matrix elasticity effects on cells. Such gels are used here in 2D and 3D to understand the roles of MIIB and phospho-MIIA in polarization and durotaxis of MSCs. Various perturbations to the myosins reveal that both durotaxis and polarization are maximal at wild-type (WT) levels of MIIB and phosphodynamic MIIA. Cytoskeletal polarization thus appears to be a highly regulated, mechanosensitive compass in directed migration.
Results
MIIB progressively polarizes as MSCs migrate from soft to stiff matrix
On stiff matrix in 3D as well as 2D systems, MIIB was most abundant toward the rear of fixed and immunostained MSCs (Fig. 1 B, right). Cells migrating on rigid glass showed a similar polarization of MIIB (Vicente-Manzanares et al., 2008). MIIB in MSCs was seen within stress fibers and sometimes appeared striated. On soft matrix, however, MIIB within the main cell body was surprisingly uniform, nonstriated, and diffuse (Fig. 1 B, left). In the leading lamella of an MSC and in the most distal trailing tail, MIIB was depleted relative to F-actin on soft as well as stiff matrix. This edge depletion of myosin-II depends on the extensive actin polymerization in the leading lamellipodia, which is independent of myosin-II motor activity (Vicente-Manzanares et al., 2008; Mogilner and Keren, 2009).
Our gradient gels ensured consistent immunostaining of cells across soft and stiff matrices so that cytoskeletal polarization within the main cell body could be reliably quantified (Fig. 2, A and B). The ratio of fluorescent signal for the rear half of the migrating cell relative to the front half was calculated as rear/front. F-actin always appeared uniform throughout the cell (within 10%) with rear/front = 1. In cells that spread for 2 h on the gradient region, vinculin in focal adhesions as well as F-actin showed no asymmetry. MIIA showed a slight, persistent polarization of rear/front ∼1.3 independent of matrix elasticity (Fig. 2 C). MIIB appeared strongly polarized toward the rear of the cell on stiff matrix and on collagen-coated glass as reported with other cells on rigid substrates (Kolega, 2003; Vicente-Manzanares et al., 2008). However, MIIB was consistently unpolarized on soft matrix (Fig. 2 D). Hyperbolic fitting of rear/front as a function of E showed that MIIB polarization varies by fourfold with a transition from soft to stiff matrix at a midpoint stiffness of Em = 6.9 kPa. The projected area of MSCs versus E fits a similar hyperbolic form with similar Em (Rehfeldt et al., 2012), which reflects in some broad manner the interrelated tendency of cells to actively spread and polarize in response to stiffer matrix.
Because of the stiffness-dependent distribution of MIIB in cells on gels, we examined the MIIB density per cell and found that this showed no significant dependence on matrix stiffness (Fig. 2 E). In addition, because of the distinctly diffuse distribution of MIIB in cells on the soft matrices compared with its stress fiber association in cells on stiff matrices, we then extracted the cells with Triton X-100 to image by immunofluorescence just the insoluble myosin-II in the cytoskeleton. Surprisingly, the amount of MIIB left in these extracted cytoskeletons showed no significant difference between soft and stiff matrix (Fig. 2 E). The lack of polarization of MIIB in cells on soft matrix is thus not a result of differences in either MIIB abundance or integration into the cytoskeleton.
MSCs durotax in 2D and 3D
Time-lapse imaging showed that cells crawled at the same mean speed of ∼0.5 µm/min independent of matrix elasticity (Fig. 3 A). On the soft matrix and also within the 3D overlay, cells visibly deformed the matrix (Fig. S1, A and B), whereas cell-induced deformations of stiff matrix were not easily resolved. Cells migrated frequently from the soft side of the gel into the soft overlay, but cells did not migrate from the stiff matrix into the overlay (as indicated schematically in Fig. 1 A), consistent with 3D durotaxis. Durotaxis on the gel surface was evident in the gradient regions in both 2D and 3D with “flower” plot trajectories showing more endpoints (Fig. 3 A, red dots) to the right (stiff) side of the starting point compared with the left (soft) side (Fig. 3 A). For cells on the gradients in both 2D and 3D, a durotaxis index (defined in the legend of Fig. 3 B) proved statistically different from zero after ∼1–3 h of observation (which is approximately one to three cell lengths), indicative of durotaxis.
In contrast to cells on the gradient region, cells on the soft or stiff matrix regions showed no significant durotaxis index. However, cells on the soft matrix did not appear to migrate as far as on stiff matrix, even though cell speeds were similar (Fig. 3 A). A quantitative analysis of the trajectories (Fig. S1 C) indeed established that MSCs on both regions started with directed migration that decayed to random migration on soft matrix (time constant of 2.3 h) more rapidly than on stiff matrix (6.3 h). On the gradient, the decay was not monotonic and revealed a statistical tendency for many cells to be more persistent and directed (up the gradient) rather than random in their migration. The less persistent migration on soft matrix versus stiff matrix was consistent with other recent observations (Fischer et al., 2009) and provides a simple explanation for directed migration toward stiff matrix. Moreover, because MIIB knockout fibroblasts crawl with less directional persistence on rigid glass coverslips (Lo et al., 2004), we hypothesized that MIIB would be critical to migration from soft to stiff matrix.
Durotaxis is more sensitive to the minor isoform, MIIB, than to abundant MIIA
Graded KD of both MIIB and MIIA was then used to increasingly perturb physiological levels. Quantitative immunofluorescence of single cells was used to determine the percentage of KD to account for cell-to-cell variations (e.g., Fig. S2 A) as represented by x axis error bars in Fig. 3 C’s measurements of the durotaxis index for randomly chosen cells. KD of MIIB by 28% was sufficient to decrease the durotaxis index to zero, eliminating MSC durotaxis (Fig. 3 C). In comparison, a 50% KD of MIIA led to only a 50% reduction in the durotaxis index. A deeper KD of MIIA by 75% was required to significantly block durotaxis.
Levels of MIIA and MIIB were further assessed by Western immunoblotting (Fig. 3 D), quantitative immunofluorescence (Fig. S2, A and E; Fig. 2 E; and Fig. 3 C), and quantitative liquid chromatography (LC)–coupled tandem MS (LC-MS/MS; Fig. S3). Immunoblotting verified the mean KD in the cell populations (Fig. 3 D). We used LC-MS/MS to additionally quantify KDs of MIIA and MIIB together with potential changes in other proteins (Shin et al., 2011) as validated by immunoblotting for seven proteins (Fig. 3 D and Fig. S3 E). The most important case is the minimal KD of MIIB that blocks durotaxis, for which nearly all cytoskeletal proteins changed minimally (Fig. 3 D, plot; and Fig. S3 A). The decrease in MIIB is therefore the likely explanation for the durotaxis defect. On the other hand, deep KD of MIIA led to significant changes in several other proteins detected by MS, including an increase in MIIB (by 33%) as seen also in immunoblots (Fig. 3 D). The effects of ectopic overexpression of myosin-II on cytoskeletal polarization and durotaxis are described in a later section of the Results.
The ratio of MIIB to MIIA isoforms was also determined from LC-MS/MS using ion currents for tryptic peptides common to both isoforms and also unique to each isoform. MIIB proves to be the minor isoform at only 3–9% of total myosin-II (Fig. S3 B). The elimination of biased migration by a minimal and specific KD of the minor isoform, MIIB, shows that durotaxis is far more sensitive to MIIB than to MIIA.
We hypothesized that blebbistatin inhibition of myosin-II ATPase/motor activity could inhibit durotaxis much faster than could be studied by KD. Blebbistatin was pulsed into established cultures on gradients during time-lapse imaging, and although cells continued to migrate everywhere, the durotaxis index decreased to zero with time constant τ < 15 min (Fig. S4 A, i and ii). Because MIIB polarization typically accompanies durotaxis in the aforementioned experiments, we examined MIIB polarization after this brief treatment with blebbistatin and found that the polarization of MIIB decreased only slightly within the first hour of drug treatment (Fig. S4 A, i, inset bar graph) but is then lost several hours later in blebbistatin. It is not surprising that blebbistatin did not immediately disrupt MIIB polarization because cellular localization of myosin-II isoforms is caused primarily by the coiled-coil tails rather than the heads with the ATPase/motor activity (Vicente-Manzanares et al., 2008). Importantly, the blebbistatin results here indicate that MIIB polarization is not sufficient for durotaxis and can decouple from durotaxis, whereas myosin-II motor activity is necessary for durotaxis.
The treatment with blebbistatin initially slowed down MSCs on the soft matrix and slightly sped up MSCs on the stiff matrix (Fig. S4 A, iii), but by 6 h after drug exposure, cell shapes and crawling speeds became almost independent of matrix elasticity (Fig. S4 B). Sustained blebbistatin treatment and KD of the MII isoforms generally altered the shapes of MSCs, but the most striking morphological change followed MIIB KD, which led to highly extended tails (Fig. 3 E), with migration direction verified by time-lapse imaging. The observations are similar to those for myoblasts on rigid coverslips after MIIB KD (Swailes et al., 2006) and are consistent with a role of MIIB in tail cohesion and/or retraction during in vitro migration on coverslips (Kolega, 2003). Surprisingly, on soft matrix, the trailing tails of MIIB KD cells were significantly longer compared with either control cells or KD cells on stiff matrix (Fig. S2 B). MIIB in untreated MSCs therefore has an important function in cohesion and/or deadhesion of the cell tail on soft matrix despite its diffuse distribution (Fig. 1 B).
In addition to knocking down the two myosin-II isoforms, we overexpressed both as GFP fusion constructs. Overexpression of GFP-MIIB by 3–20-fold (mean of eightfold) showed integration into the cytoskeleton but suppressed polarization of MIIB to the cell rear on stiff substrates (Fig. S2 D). These cells also did not migrate significantly and could not durotax. In comparison, overexpression of WT GFP-MIIA by ≤1.4-fold (i.e., 40% overexpression) had no significant impact on migration, durotaxis, or polarization of MIIB (as elaborated in a later section of the Results). It is also worth noting that the mean level of MIIB overexpression, combined with the low endogenous levels, gives a total level of myosin-II (MIIA + MIIB) that is about the same as with the MIIA overexpression, and yet the former MIIB-overexpressing cells do not crawl, polarize, or durotax, whereas the latter MIIA-overexpressing cells do crawl, polarize, and durotax. This hints at the functionally distinct roles of MIIA and MIIB isoforms and suggests that MIIB must be expressed at relatively low levels in WT MSCs in order for them to functionally durotax.
Molecular mobility is highest for MIIA and for soft matrix, with phosphodynamics of MIIA contributing to mechanism
For insight into physical differences in molecular regulation, FRAP was used to study molecular mobility of GFP fusions. GFP-MIIB had been shown to be less mobile than GFP-MIIA in cells on rigid coverslips (Vicente-Manzanares et al., 2007), and similar results were seen here with cells on the stiff but compliant 34-kPa matrix (Fig. 4 A, i). The result is consistent with a general similarity of cell responses to stiff gels and rigid glass (Fig. 2 D). Although the results also indicate more stable integration of MIIB, additional FRAP experiments showed that MIIA is even more mobile on soft matrix (Fig. 4 A, ii), which is consistent with a past study showing that MIIB is more mobile in cells on soft matrix versus stiff matrix (Fischer et al., 2009).
When we had imaged MIIA to quantify what turned out to be a minimal polarization of MIIA in cells on both soft and stiff matrices (Fig. 2 C), we noticed that MIIA appeared much more diffuse and less integrated in filaments in cells on soft matrix, consistent with its increased mobility in FRAP (Fig. 4 A, ii). Given that MIIA is the major isoform (Fig. S3 B) that contributes the majority of traction forces used to sense soft versus stiff matrix (Lo et al., 2004), we thought that the dependence of MIIA assembly and filamentation on matrix rigidity would ultimately prove critical to MIIB polarization as well as durotaxis. Our MS analyses had detected phospho-Serine1943 (pS1943) in MIIA (unpublished data), and pS1943 is known to favor disassembly of MIIA from stress fibers (Dulyaninova et al., 2007). Our working hypothesis therefore became that pS1943 would be high on soft matrix and low on stiff matrix with a significant change in phosphorylation at this site being a key part of the molecular mechanism for how matrix stiffness promotes MIIA assembly. Note that this is a hypothesis on dynamics because our myosin-II KDs had already indicated that overall levels of MIIA do not greatly impact durotaxis. We sought to test the phosphodynamics hypothesis with FRAP and then extractability methods applied to suitable mutants. A phosphomimetic Asp1943 (S1943D) mutant and a nonphosphorylatable Ala1943 (S1943A) mutant were first studied as GFP fusions by FRAP of MSCs on stiff matrix. Phosphomimetic S1943D-MIIA proved more mobile than WT GFP-MIIA (Fig. 4 A, iii), consistent with the cited disassembly from stress fibers. In contrast, S1943A-MIIA had a mobility less than or equal to that of WT GFP-MIIA. Because S1943A favors assembly into stress fibers, the results indicate stable integration of WT GFP-MIIA into stress fibers on stiff matrix, consistent with the working hypothesis.
Extraction methods were used to expand on the FRAP findings. Cells on soft and stiff matrices were prepermeabilized with Triton X-100 to extract soluble protein from the cells, and then, the cytoskeletons were fixed, immunostained for insoluble MIIA, and imaged in quantitative fluorescence microscopy. Stiff matrix showed 50% insoluble MIIA, and this decreased significantly to 32% on soft matrix (Fig. 4 B) with confirmation of the trend by Western blotting. This is consistent with the FRAP results in indicating that stiff matrix promoted MIIA assembly into stress fibers. The results indicate that the insoluble/soluble ratio of MIIA is only ∼1:2 on soft matrix, and this doubles to ∼1:1 on stiff matrix. This began to indicate elevated pS1943 in cells on soft matrix as analyzed directly in the next section.
Validation of FRAP findings with the different myosin GFP mutants (and a monomeric GFP control) was performed by once again treating with Triton X-100 and then analyzing the soluble fraction and insoluble pellet by Western blotting with anti-MIIA. GFP-myosins were seen at a distinctly higher molecular weight than endogenous MIIA. The quantity of myosin-II in the insoluble pellet relative to the soluble extract is shown in Fig. 4 C as the insoluble/soluble ratio for each transfection. For endogenous MIIA, this ratio was similar for all transfections. Ratios of insoluble/soluble for both endogenous myosin and GFP-MIIA WT fit within the broad error bars of Fig. 4 B on the stiffest substrate, but direct comparisons of endogenous to GFP-myosins are tenuous because of the proximity of the two bands. Whereas the S1943A construct did not differ significantly from WT, the S1943D construct appeared the most soluble, consistent with the high mobile fraction in FRAP studies of single cells (Fig. 4 A, iii). In a later section of the Results, the results prompted detailed analyses of stress fibers, MIIB polarization, and durotaxis after transfections.
MIIA phosphorylation at S1943 decreases as matrix stiffness increases
Differences in FRAP mobility and solubility of WT MIIA in cells on 1-kPa matrix versus 34-kPa matrix and the difference between S1943A-MIIA and S1943D-MIIA led us to hypothesize that S1943 would be more phosphorylated on soft matrix than on stiff matrix. Quantitative immunofluorescence of endogenous pSer1943-MIIA indeed shows significantly higher intensity in cells on soft matrix compared with stiff matrix after normalization for total MIIA (Fig. 5 A, bar graph). Western blotting of lysates derived from MSCs cultured on soft or stiff matrix confirmed a greater amount of pS1943 in cells on soft matrix (Fig. 5 B). Moreover, pS1943 staining appeared more diffuse and less prominent in stress fibers than total MIIA (Fig. 5 A, images). This was quantified with an algorithm that first calculates the orientation of each stress fiber structure referenced to a mean orientation angle (Zemel et al., 2010). The sample images showed myosin mostly oriented above a low background, whereas pS1943 exhibited a high background (Fig. 5 C). The ratio oriented/isotropic subtracts the isotropic background from the oriented signal. Thus, on stiff matrix, MIIA was more apparent in ordered filaments than is the case for pS1943 (P < 0.05). Soft matrix strongly suppressed orientational order of MIIA (Fig. 5 C, inset), which is consistent with MIIA being more phosphorylated and more mobile in cells on soft matrix. Triton X-100 extractions examined in a later section of the Results are consistent with these conclusions.
GFP-MIIA in transfected cells (Fig. 5 D) showed the same trends in orientational order as endogenous MIIA and pS1943 on soft and stiff gels (Fig. 5 C, inset), with statistical tests establishing the expected similarities and differences in showing that overexpression of MIIA (by ∼30%) is not introducing structural artifacts in actomyosin filaments. However, the S1943D mutant of GFP-MIIA always produced a more diffuse, isotropic signal than WT, which is consistent with the idea that a negatively charged 1,943 site promotes filament destabilization and disassembly (Dulyaninova et al., 2007). Conversely, the orientation of GFP-S1943A filaments in cells on soft matrix appeared statistically the same as GFP-WT and GFP-1943A in cells on stiff matrix, which indicates that S1943A promotes oriented filaments on soft matrix. Staining of the S1943A mutant MIIA cells with anti-pS1943 (which binds only to endogenous MIIA per Western blots) also showed for stiff matrix an increase in filament-oriented pS1943 compared with S1943D. The various effects of this tail phosphosite on assembly into stress fibers thus suggest a role in the matrix mechanosensitivity of MSCs.
On stiff matrix, cells assembled more stress fibers and achieved a greater spread area than cells on soft matrix (Lo et al., 2000; Discher et al., 2005; Rehfeldt et al., 2012). MSCs expressing the stress fiber–stable S1943A-MIIA mutant showed more oriented stress fiber signal compared with WT MIIA (Fig. 5 D) and also spread more on soft matrix compared with MSCs transfected with WT MIIA or S1943D (Fig. S5 A). These phenotypic results are consistent with endogenous MIIA being less phosphorylated and more assembled into stress fibers on stiff matrix (Fig. 5). MSCs transfected with MIIA WT cells spread more on stiff matrix than on soft matrix, as is typical of a mechanosensitive response, and which the aforementioned data indicate should lead to a decrease in pS1943. The S1943D mutant mimics unalterable phosphorylation, and this mutant seemed to spread more at 24 h than WT MIIA or S1943A, which is consistent with previous measurements of MDA-MB-231 cells that also spread 20% more with S1943D on rigid surfaces (Dulyaninova et al., 2007). The difference in spreading appeared time dependent in our observations and has a likely basis in retrograde flow per the Discussion section.
Dephosphorylation of MIIA pS1943 is essential for durotaxis and for MIIB polarization but not for migration
Next, we studied the impact of the MIIA phosphosite mutants on durotaxis and MIIB polarization. MSCs transfected with GFP-MIIA WT on gels with gradients of stiffness exhibited a durotaxis index of ∼0.22 (Fig. 6 A), which is statistically the same as nontransfected MSCs (Fig. 3, B and C). However, MSCs transfected with the S1943A mutant showed a 50% reduction in the durotaxis index, and the S1943D mutant showed no significant durotaxis. Crawling speeds were similar (differences within <20%), and so, transfections did not cause overall motility defects.
MSCs transfected with GFP-MIIA exhibited polarization of the MIIB isoform to the cell rear on stiff matrix, but rear/front polarization was weak and insignificant on soft matrix (Fig. 6 B), just as observed with nontransfected cells (Fig. 1 B). MIIA polarization was consistently unaffected for all constructs. The S1943D mutant suppressed polarization of MIIB on stiff matrix and also showed no polarization on soft matrix. The S1943A mutant showed a slight but not statistically significant (P = 0.09) increase in MIIB polarization on soft matrix relative to WT, but S1943A did not differ from WT for stiff matrix (unpublished data); the trends are consistent with the orientational order of GFP-S1943A being sufficiently enhanced in cells on soft matrix to show no statistical difference when compared with either GFP-WT or GFP-S1943A in cells on stiff matrix (Fig. 5 D). MIIA’s pS1943 can thus affect MIIB polarization either through direct or indirect interactions.
Discussion
Many observations here are summarized in the plots of Fig. 6 C, which show that durotaxis and rigidity-induced polarization operate only within a narrow range of two independent variables: the percentage of MIIB of total cellular myosin-II and the percentage of MIIA that is phosphorylated at S1943. For polarization and durotaxis of MSCs, only 6–12% of total myosin-II can be the MIIB isoform (Fig. 6 C, top left plot), whereas MIIA can be knocked down and overexpressed to change total myosin-II amounts by about ±50% without affecting durotaxis (Fig. 3 C and Fig. 6 A). Although this corresponds to a 5–10-fold greater tolerance for changes in MIIA levels compared with changes in MIIB levels, matrix-driven changes in phosphorylation of S1943-MIIA also prove critical.
To provide an absolute scale for this MIIA phosphoregulation rather than just a relative scale, we estimated the stoichiometry of pS1943-MIIA in WT MSCs on rigid substrates (as a reference). Immunodepletion of pS1943-MIIA was followed by immunoblotting for the remaining MIIA that is not phosphorylated at S1943 while controlling for nonspecific antibody binding (unpublished data), with the result being that ∼20% of MIIA has pS1943 in WT MSCs. Because overexpression of S1943D-MIIA effectively enhanced endogenous pS1943 as a more mobile, isotropic, and soluble pool of MIIA (Fig. 4 and Fig. 5), S1943D is treated in Fig. 6 C as an “equivalent” in adding to pS1943 when plotting the suppression of durotaxis and polarization by this mutant. Conversely, overexpression of the S1943A mutant decreased the net fraction of pS1943-MIIA (because endogenous pS1943 levels did not appear statistically different after transfection; Fig. S5 B), and this mutant also recruited more pS1943 into filaments (Fig. 5, C and D). This mutant limited the usual phosphorylation changes from soft to stiff matrix (Fig. 5, A and B) and reduced the usual differences in cytoskeletal assembly between cells on soft and stiff matrix, with the ultimate result being that S1943A-MIIA partially suppressed durotaxis and polarization. Given the results for all of these perturbations, durotaxis and polarization prove maximal at WT levels of pS1943 (Fig. 6 C, top right plot).
Soft matrix suppresses MIIA actomyosin assembly and orientation, impacting MIIB
The reasons why organization and polarization of MIIB is lacking in WT MSCs on soft matrix are in need of explanation. In cells on rigid substrates where MIIB is rearward polarized, Vicente-Manzanares et al. (2007) concluded that MIIA initiates formation of the actomyosin filaments, and MIIB subsequently binds and stabilizes these filaments. In MSCs on soft matrix, the major isoform MIIA is less assembled and more disorganized compared with cells on stiff matrix (Fig. 4 and Fig. 5), and the minor isoform MIIB in cells on soft matrix is likewise more disorganized and homogeneous compared with cells on stiff matrix (Fig. 1 B and Fig. 2 D).
Why MIIB polarization is maximal in WT MSCs on stiff matrix is another key question. In the case of MIIB KD, we could see a decreased assembly of MIIB into filamentous structures, thus producing more isotropic than oriented MIIB. This could result from MIIB being closer to or below its critical concentration needed for assembly (Pollard, 1982). In the case of MIIB overexpression, the higher concentration of MIIB tends to come off the stress fibers and into the cytosol, and once again produces more isotropic than oriented MIIB (Fig. S2 D, inset). MIIA did not polarize on stiff matrix for similar reasons of high mobility (Fig. 4 A). KD of MIIA considerably reduces actomyosin mass, and overexpression of S1943D-MIIA suppresses MIIA organization (Fig. 5 D), and both treatments therefore suppress MIIB polarization. The S1943A-MIIA mutant slightly enhances both oriented assembly and MIIB polarization on soft matrix but not stiff matrix (Fig. 5 D and Fig. 6 B); this suppresses the soft–stiff contrast needed for durotaxis as does the inability to phosphorylate–dephosphorylate S1943A, and the results thus confirm that MIIB polarization does not exceed that seen in WT MSCs on stiff matrix. Overall, the various perturbations explained here suggest that the peak in polarization for WT states of myosin-II is reasonable.
Durotaxis, myosin-II polarization, and implications for nonmesenchymal cell types
The surface plot of Fig. 6 C and the schematic of Fig. 7 further summarize the findings here that WT MSCs possess an optimally polarizable, phosphodynamic actomyosin cytoskeleton for biased migration from soft to stiff matrix. Durotaxis has been reported previously for the NIH 3T3 line of mouse embryonic fibroblasts (Lo et al., 2000) and primary bovine aortic vascular smooth muscle cells (Isenberg et al., 2009). Such cells of diverse mesenchymal origin from different species are expected to express predominantly MIIA and similarly low fractions of MIIB (Ma et al., 2010) when compared with the human MSCs derived from bone marrow as studied here. The surface plot here might therefore generalize to other cells of mesenchymal origin. Interestingly, durotaxis of other cell types has not yet been reported to our knowledge and could be limited by stringent requirements revealed here for myosin-II isoforms and phosphorylation. Human neutrophils, for example, express only MIIA (Maupin et al., 1994) and migrate into many tissues. On the other hand, MIIA in the neutrophil does polarize to the cell rear in the uropod on rigid substrates (Xu et al., 2003), and neutrophils on soft matrices are less persistent in migration and less contractile (Oakes et al., 2009). Neutrophils and other cell types might therefore also polarize their cytoskeletons as they durotax by similar mechanisms from soft to stiff matrix.
Durotaxis has been described as resulting in part from an increase in the stability of adhesions to stiff matrix as cells migrate from soft matrix (Lo et al., 2000), which would make durotaxis similar to haptotaxis. In the present observations, however, 30% KD of the minor isoform MIIB not only eliminated durotaxis but also led to highly extended, adherent tails that demonstrate that tail adhesion is relatively unperturbed by MIIB KD (Fig. S2 B). The majority of traction forces in similar mesenchymal cell types derives from MIIA (Cai et al., 2006), but durotaxis still occurs here after MIIA is knocked down 50% (Fig. 3 C). The remaining MIIA is 10-fold above the molar amount of MIIB. The effect of MIIB KD is therefore not a simple matter of reduced force generation. Proteomics analyses (Fig. S3 A) further suggested only one protein, calpain-2, was consistently up-regulated upon KD or inhibition of myosin-IIs, but such proteases undermine adhesion, which is not consistent with the adherent tails of MIIB KD cells. Durotaxis is thus not primarily a result of differences in cell adhesion, which distinguishes durotaxis from haptotaxis.
Trafficking of MSCs to wounds and scars might involve durotaxis
MSCs are known to mobilize from soft bone marrow (Orlic et al., 2001) and are also routinely injected into tissue (Burt et al., 2008), and in either case, they are somehow home to wounds and scars as they traverse tissues of locally varied rigidity. Such complex migratory processes seem to be a highly evolved function in higher species, and durotaxis of MSCs certainly seems specialized given the sensitivity to the myosin-II isoform ratio and phosphoregulation (Fig. 6 C). The present findings with multiple batches of low passage MSCs from multiple donors nonetheless suggest the importance of rapidly reversible phosphopathways to durotaxis, polarization, and matrix elasticity sensing. A seminal study of substrate flexibility effects on cells have described overall increases in pTyr levels in cells on stiff matrices (Pelham and Wang, 1997), but the opposite trend is found here with this particular pS1943 site in MIIA. A deep understanding of regulatory kinases and phosphatases might one day be exploited to better mobilize MSCs or perhaps optimize the trafficking and retention of these cells at sites of injury or repair.
Lastly, cells in wounds and fibrotic scars engage matrix all around the cell in 3D, but durotaxis has previously been studied only on 2D substrates (Lo et al., 2000; Gray et al., 2003; Cheung et al., 2009; Hadjipanayi et al., 2009; Isenberg et al., 2009; Tse and Engler, 2011). Second messenger signaling in cell migration within 3D model matrices has been reported very recently to exhibit matrix elasticity–dependent polarization (Petrie et al., 2012), but gradients and transitions in the myosin-II motors that generate critical forces have been unexplored. 3D gels with gradients in collagen density have been examined (Hadjipanayi et al., 2009) but involve changes in ligand density, and so the 3D results here with controlled gels (Fig. 1) eliminate haptotaxis effects. More importantly, the results provide a visible molecular signature of likely relevance to any stiffness sensing and durotaxis in various systems, including tissue.
Materials and methods
Polyacrylamide gels with gradient in stiffness
25-mm circular glass coverslips were treated first with ethanol and then RCA solution (1:1:3 for 15 N NH4OH/30% H2O2/distilled H2O) and functionalized with 1% allyltrichlorosilane and 1% triethylamine in chloroform solution. To control the gel’s stiffness, N,N′methylene-bis-acrylamide (Sigma-Aldrich) and the acrylamide solution (40%; Sigma-Aldrich) were mixed at the final concentrations in PBS. Solution was polymerized on a coverslip with 0.1% ammonium persulfate and 0.1% N,N,N′,N′-tetramethylethylenediamine. During polymerization, gels were covered with another coverslip that had been pretreated with dichlorodimethylsilane. Gels with gradients in stiffness were prepared using a method modified from Lo et al. (2000). We made gradient gels by juxtaposing two drops (13 µl for each drop) of different acrylamide and bisacrylamide concentrations on a 25-mm circular coverslip. The drop for the stiffer side contained 8% acrylamide and 0.3% bisacrylamide (∼34 kPa), and the drop for the softer side had 3% acrylamide and 0.11% bisacrylamide (∼1 kPa). FluoSpheres (Invitrogen) 1 µm in diameter were supplemented into the stiffer drop to a 0.005% by weight in the gel solution. This ensured that any possible stiffening effects of the beads would be on the stiff side of the gel. After waiting 5 min for gel polymerization, the two drops were mixed by applying a 25-mm square coverslip precoated with dichlorodimethylsilane. Gradient gels were allowed to polymerize for 45 min followed by removing the dichlorodimethylsilane coverslip and washing twice with PBS. Sulfo-SANPAH (Thermo Fisher Scientific) was diluted in Hepes buffer, pH 8.0, to 0.5 mg/ml, and 300 µl was added to the gel and then reacted using a 365-nm UV light for 10 min. Excess sulfo-SANPAH was removed by two washes of Hepes buffer. Type I rat tail collagen was diluted to 0.2 mg/ml in cold Hepes buffer, pH 8.0, and incubated with the gel at 4°C for 4 h. Collagen was removed, and the gel was washed and equilibrated with PBS. MSCs were plated onto gels within 24 h of collagen attachment. The spatial resolution of AFM used here is ≤1 µm, whereas much larger probes of a diameter of 640 µm (Lo et al., 2000) had suggested a steeper, steplike change in elasticity for similar gradient gel systems used to first describe durotaxis.
Cells, culture, pharmacological perturbations, and transfections
At least three different batches of human bone marrow–derived MSCs were purchased from Lonza, and at least three batches of MSCs were isolated from human donor bone marrows (University of Pennsylvania School of Medicine) by standard methods (Orlic et al., 2001) so that cells from at least six different human donors were used here. We did not detect major donor or batch variability in generating the reasonably consistent set of results here. Cells were used at passages 3–5 for all experiments and cultured in low glucose Dulbecco’s minimum essential medium (DMEM) supplemented with 10% FBS, 100 µg/ml penicillin, and 100 µM streptomycin. Racemic blebbistatin (EMD) was used at 5 or 50 µM as specified. Plasmid constructs for GFP–nonmuscle MIIB and GFP–nonmuscle MIIA were obtained from Addgene, GFP-MIIA S1943A was constructed by standard methods, and GFP-MIIA S1943D was a gift from A. Bresnick (Albert Einstein College of Medicine, Bronx, NY). Constructs were transfected via electroporation following the recommended procedure of the kit for MSCs (Nucleofector; Lonza) or by using Lipofectamine LTX (Invitrogen) with plus reagent using 0.5 g DNA per well in a 6-well plate. Transfection levels were similar for the three constructs (within 20%) based on GFP intensities and densitometry of Western blots.
FRAP
Confocal time-lapse imaging was acquired at 37°C with 5% CO2 in a humidified chamber with an inverted spinning-disk microscope (IX-81; Olympus) with a 14-bit high-resolution charge-coupled device (CCD) camera (HQ2; Photometrics) with MetaMorph software (Molecular Devices). Images were acquired with a 60× water immersion lens, NA 1.2, every 5 s for ≥150 s after 25 s of prebleached acquisition. Cells were imaged in low glucose DMEM (Invitrogen) with 10% FBS (Sigma-Aldrich). GFP-tagged myosin constructs were photobleached using a region of interest line drawn perpendicular to the actin bundles. Image sequences were analyzed with Fiji with prewritten Jythons script algorithm (National Institutes of Health) to quantify fluorescence recovery kinetics along myosin-labeled actin bundles.
Western blotting
Unless otherwise stated, cells were trypsinized, pelleted, and resuspended in PBS. Then, equal volume of 50 mM Tris-HCl, pH 8.0, with 10% SDS, 1 mM EDTA, 15% sucrose, 5% β-mercaptoethanol, and bromophenol blue was vortexed and boiled for 5 min followed by shearing through a 25-gauge needle. Samples were loaded onto 3–8% Tris-acetate gels (NuPAGE Novex; Life Technologies) and then transferred to blotting paper using a blotting system (iBlot; Life Technologies) with the settings of P3 for 9 min. Blots were blocked with 5% milk in TBST (TBS with Tween) for 1 h at RT, and then, antibodies in TBS were added overnight at 4°C. Secondary antibodies with conjugated HRP in TBST were added for 1 h at RT. ChromoSensor (GenScript) was added for the ECL reaction. We routinely performed Western blotting with different loads, and established linearity between signal versus sample load.
siRNA KD of MIIA and MIIB
Lipofectamine 2000 (Invitrogen) with 30 nM siRNA was used to knock down MIIA and MIIB. Different levels of KD were assessed by waiting different times after adding the complex and then quantified by immunofluorescence, Western blotting, and MS. For MIIB siRNA duplexes, the sequences were obtained from Bao et al. (2005) and were synthesized by Thermo Fisher Scientific along with scrambled siRNA. MIIA siRNA sequence was 5′-GGCCAAACCUGCCGAAUAAAUU-3′ with complement sequence 5′-UUUAUUCGGCAGGUUUGGCCUU-3′. Scramble siRNA was siGENOME nontargeting siRNA #1 (Thermo Fisher Scientific). For double KD of MIIA and MIIB, 30 nM of each duplex was used with Lipofectamine.
Fixation, immunostaining, and microscopy
Cells were fixed with 3.7% formaldehyde (Sigma-Aldrich) in PBS for 10 min at RT followed by PBS washing 2× for 5 min. Blocking and antibody staining was performed in 1% BSA in PBS. Rabbit anti-MIIA antibody (Sigma-Aldrich) was used at 1:100, anti-MIIB antibody (Cell Signaling Technology) was used at 1:150, anti-pS1943 was used at 1:100 (Cell Signaling Technology), and all primary antibodies were incubated at RT for 1 h or overnight at 4°C. All donkey secondary antibodies (Alexa Fluor dyes 488, 564, and 647) were stained for 1 h at RT at 1:300 dilution in PBS in 1% BSA. TRITC-phalloidin (Sigma-Aldrich) was used with the donkey secondary antibodies at a concentration of 100 ng/ml, and Hoechst 33342 (Invitrogen) was used to stain DNA at a concentration of 1 µg/ml for 10 min at RT. Imaging for quantification of myosin-II levels and localization was performed using an inverted microscope (IX-71; Olympus) with a 40× LUCPlanFLN objective, NA 0.60, and a CCD camera (Cascade; Photometrics). Image acquisition was performed with Image Pro software (Media Cybernetics), and subsequently, background was subtracted, and image analysis was performed using ImageJ (National Institutes of Health) according to JCB guidelines.
Image analysis
For quantitative immunofluorescence, samples that were directly compared were imaged at the same sitting, and the same gain and exposure time were used. If different exposure times were necessary for quantification between samples, several exposure times were used on the same image to ensure that images were in the linear range of exposure time to fluorescence intensity. For orientation analysis performed in Fig. 5, the directionality analysis was used in Fiji with the local gradient orientation method selected and using 90 bins for data segmentation. Curve fits of the histograms of data were used if there were high goodness of fits, and the ratio of the peak of the curve fit to the value of the baseline was taken as the oriented/isotropic parameter. Front/rear polarization of F-actin and myosins was determined by drawing a line 6–10 µm in width from the cell rear to front and plotting the fluorescence versus length. The ratio is the fluorescence in the front half over the back half of the cell.
AFM indentation
A force microscope (1-D; Asylum Research) was used quantify the elastic modulus of the gradient gels. A pyramid-tipped probe with spring constant of 30–100 pN/nm (Veeco) was used for measuring gradient gels, and the AFM head was placed over an inverted microscope (TE300; Nikon) to detect the beads with fluorescent microscopy using a 4× objective. A micromanipulator was used to move across the gradient during AFM probing and ensured high accuracy determinations of the elasticity gradients when compared with the methods of Lo et al. (2000).
Time-lapse cell imaging
Phase-contrast imaging was performed in a humidified chamber at 37°C and 5% CO2 using an inverted microscope (IX-71) with a 10× objective, NA 0.3, using a 300-W xenon lamp illumination and a high-resolution CCD camera (CoolSNAP HQ; Photometrics). softWoRx software (DeltaVision) was used for acquiring images. Cells were imaged in low glucose DMEM with 10% FBS. For cell speed measurements, images were acquired in 3-min intervals for 2 h. The center of the cell nucleus was used as a reference point in cell tracking. For imaging at longer times (≥12 h), images were acquired at 15-min intervals. ImageJ was used to track the center of the nuclei of video sequences, and the summed contour distance traveled divided by the time was used as our measurement of speed. This measurement is independent of how persistent the cells are and instead reflects the total distance traveled. Cells that began to divide or were in contact and cells that migrated out of the trackable region were included until such events required exclusion. In time-lapse imaging for several hundred cells on soft, stiff, and gradient matrices, we did not observe cell detachment or cell loss, and we did observe about an equal number of cell division events on all matrices, which largely eliminates any bias in the measurements here.
Triton X-100 extraction of MSCs
MSCs on either plastic or gels were treated with 0.1% Triton X-100 in PEM buffer (0.1 M Pipes, pH 6.95, 2 mM EGTA, and 1 mM MgSO4) with 2 µM phalloidin (Sigma-Aldrich) and protease inhibitors (Sigma-Aldrich) for 4 min, and then, the cell ghosts were fixed and immunostained to determine the fraction of insoluble myosin. For Western blots of soluble and insoluble lysates, cells with transfected myosin constructs were Triton X-100 extracted as previously described (Breckenridge et al., 2009). In brief, 600,000 transfected MSCs (using nucleofection) were plated onto a 10-cm dish, and after 24 h, cells were rinsed with PBS, and 300 µl lysis buffer (50 mM Tris-HCl, 5 mM NaCl, 140 mM Na-acetate, 0.6% Triton X-100, 5 mM EGTA, and 1 mM EDTA with protease inhibitors) was added for 4 min during which the plate was swirled for liquid to cover cells. The plate rested on ice between swirling. Then, the extracted cell ghosts were scraped, and the total solution was collected, sheared through a 25-gauge syringe needle, and then centrifuged for 10 min at 4°C at 10,000 g. The pelleted fraction is considered to contain the insoluble cytoskeleton, and the supernatant is the soluble portion. 4× lithium dodecyl sulfate (LDS) buffer was added to the supernatant, and LDS (diluted in PBS) was added to the pellet to create a volume of 50 µl. Samples were boiled for 5 min and then loaded onto the gel.
Collagen-I overlay
MSCs were seeded onto collagen-coated polyacrylamide gels and allowed to adhere for 1–2 h. Media were removed, and 1.65 mg/ml collagen in 50:50 neutral buffered solution/complete growth media was added to the polyacrylamide gel surface and allowed to gel for 1 h at 37°C followed by the gentle addition of media. Cells were fixed and stained as previously described for this collagen overlay culture system (Fischer et al., 2009) with minor alterations. In brief, samples were fixed with 3.7% formaldehyde in PBS with 0.1% Triton X-100 for 30 min at RT and then rinsed with PBS with 0.25% Triton X-100 for 30 min. PBS with 0.5 mg/ml NaBH4 was added to reduce autofluorescence for 10 min at RT twice. Samples were subsequently blocked with PBS with 2% BSA, 1% goat serum, and 0.25% Triton X-100 overnight at 4°C. Primary antibodies were diluted in PBS with 0.25% Triton X-100 with 2% BSA and 1% goat serum at 4°C overnight. After rinsing for 30 min in PBS, secondary antibodies at 1:300 were added and incubated for 2.5 h at RT.
Calculation of cell migration exponent
The root–mean squared displacement is assumed to scale with time as RMSD(t) = A × tα(t), with A being a constant and α determined from α = d[log(RMSD)]/d[log(t)] as calculated in this paragraph. Cell motility is perfectly persistent and directed if α = 1, but the cell is executing a random walk if α = 0.5. For an accelerating cell, α > 1, and for a stalled cell, α = 0. In each region of a substrate, this “migration exponent” α is determined for every cell at each time point and then averaged among all cells. The time dependence of α leads to an additional but small term upon differentiation:
$dlog(RMSD(t))dlog(t)=α(t)+dα(t)dlog(t)log(t),$
$log(t)=u,$
$dlog(t)=du=dtt,$
$dlog(RMSD(t))dlog(t)=α(t)+dα(t)du u, and$
$dlog(RMSD(t))dlog(t)∼α(t).$
MS sample preparation
For proteomic experiments (Fig. S3), cultured MSCs subjected to MIIA KD, MIIB KD, or treatment with scrambled siRNA were harvested with 0.05% trypsin and washed three times by successive resuspension in ice-cold PBS. Proteins were solubilized by cellular disruption with a probe sonicator in ice-cold radioimmunoprecipitation assay buffer with 0.1% protease inhibitor cocktail (∼5,000 cells/µl). NuPAGE LDS sample buffer (Invitrogen) with 1% β-mercaptoethanol was added to a 1× concentration followed by heating to 80°C for 10 min. Proteins were separated on SDS-PAGE gels (NuPAGE 4–12% Bis-Tris) and run at 100 V for 10 min followed by 25 min at 160 V. Sections of excised polyacrylamide gel (cut in two molecular mass ranges: 55–160 and ≥160 kD) were washed (50% of 0.2 M ammonium bicarbonate [AB] solution and 50% acetonitrile for 30 min at 37°C), dried by lyophilization, incubated with a reducing agent (20 mM tris(2-carboxyethyl)phosphine in 25 mM AB solution, pH 8.0, for 15 min at 37°C) and alkylated (40 mM iodoacetamide in 25 mM AB solution, pH 8.0, for 30 min at 37°C). The gel sections were dried by lyophilization before in-gel trypsinization (20 µg/ml of sequencing grade modified trypsin in buffer as described in the manufacturer’s protocol [Promega] for 18 h at 37°C with gentle shaking). Before analysis, peptide solutions were acidified by addition of 50% digest dilution buffer (60 mM AB solution with 3% methanoic acid).
MS protocol
Peptide separations (5-µl injection volume) were performed on a 15-cm column (75-µm inner diameter; PicoFrit; New Objective) packed with 5-µm C18 reversed-phase resin (Magic; Michrom Bioresources) using a nanoflow high-pressure LC system (Eksigent Technologies), which was coupled online to a hybrid mass spectrometer (LTQ Orbitrap XL; Thermo Fisher Scientific) via a nanoelectrospray ion source. Chromatography was performed with solvent A (water [Milli-Q; Millipore] with 0.1% formic acid) and solvent B (acetonitrile with 0.1% formic acid). Peptides were eluted at 200 nl/min for 3–28% B over 42 min, 28–50% B over 26 min, 50–80% B over 5 min, and 80% B for 4.5 min before returning to 3% B over 0.5 min. To minimize sample carryover, a fast blank gradient was run between each sample. The LTQ Orbitrap XL was operated in the data-dependent mode to automatically switch between full-scan MS (mass per charge [m/z] = 350–2,000) in the Orbitrap analyzer (with resolution of 60,000 at m/z = 400) and the fragmentation of the six most intense ions by collision-induced dissociation in the ion trap mass analyzer.
Raw mass spectroscopy data were processed using Elucidator (version 3.3; Rosetta Biosoftware). The software was set up to align peaks in data from samples derived from the same ranges of molecular weight. Peptide and protein annotations were made using SEQUEST (Thermo Fisher Scientific) with full tryptic digestion and up to two missed cleavage sites. Peptide masses were selected between 800 and 4,500 amu with peptide mass tolerance of 1.1 amu and fragment ion mass tolerance of 1.0 amu. Peptides were searched against a database compiled from UniRef100 human (for proteomic experiments; downloaded on 5 November, 2010) plus contaminants and a reverse decoy database. Search results were selected with a Δ cross-correlation filter of 0.01 and a mass error better than 20 ppm. Ion currents of oxidized peptides (Δ = 15.995 D) were summed with their parent peptide; posttranslational modification sites of phosphorylation (Δ = 79.966 D), acetylation (Δ = 42.011 D), and methylation (Δ = 14.016 D) were considered in the search. In matrix experiments, we additionally looked for hydroxylation of asparagine, aspartic acid, proline, and lysine (Δ = 15.995 D).
The two molecular mass ranges of the proteomic dataset were analyzed separately. In the mid–molecular mass range (55–160 kD), the false-positive detection rate was estimated to be 11.4% (based on search hits of the decoy database), and only proteins with two or more peptides/protein were considered for further analysis (2,015 peptides from 231 unique proteins). High–molecular mass range (≥160 kD) resulted in a false-positive rate of 11.3% with subsequent analysis of 1,223 peptides from 55 unique proteins. Label-free relative peptide quantitation was performed with in-house software coded for Mathematica (Wolfram Research). Datasets were normalized against optimized housekeeping peptide sets that were found to be invariant between experimental conditions. A peptide set optimization algorithm (peptide ratio fingerprint [PRF]; Shin et al., 2011) was used to select peptides that show a similar “fingerprint” behavior between samples, and these peptides were used for the basis of quantification and normalization. We report only quantification of proteins with at least three PRF peptides/protein (total of 178 proteins). Peptides from regions common to several proteins or isoforms were treated distinctly. Standard errors were calculated from at least two technical repetitions. As a further check of the peptide selection algorithm, ratio comparisons were made between all datasets and checked for consistency (for example, when considering data A, B, and C, the ratio A/B should be consistent with A/C × C/B).
Online supplemental material
Fig. S1 shows differences in cell migration on soft and stiff gels. Fig. S2 illustrates phenotypical differences observed with myosin-II KD experiments. Fig. S3 quantifies results from MS on cells with myosin-II KD. Fig. S4 summarizes effects from blebbistatin on cell migration and shape. Fig. S5 quantifies differences in cell area and pS1943 levels with myosin mutant transfections.
Acknowledgments
The authors thank Dr. Anne Bresnick for kindly providing the construct for S1943D-MIIA, Dr. Thomas Egelhoff (Cleveland Clinic, Cleveland, OH) for helpful discussions and experimental suggestions, Dr. Alex Mogilner (University of California, Davis, Davis, CA) for discussions of polarization modeling, and Dr. Julie Theriot (Stanford University, Stanford, CA) for discussions of keratocyte and neutrophil polarization. Careful reading of the revised manuscript by Dr. Fei Wang (University of Illinois, Chicago, IL), Dr. Tatiana Svitkina (University of Pennsylvania, Philadelphia, PA), Dr. Ryan Petrie (National Institutes of Health, Bethesda, MD), and Dr. Wenting Shih (University of California, Davis) is very gratefully acknowledged. We also thank Andrea Stout and the Cell and Developmental Biology Microscopy Core at the Perelman School of Medicine, where time-lapse imaging was performed under precisely controlled temperature, humidity, and carbon dioxide conditions.
Support from the National Institutes of Health (grants P01DK032094 and R01HL062352) and National Science Foundation (University of Pennsylvania’s Nano Science and Engineering Center-Nano Bio Interface Center) is very gratefully acknowledged.
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Abbreviations used in this paper:
• AB
ammonium bicarbonate
•
• AFM
atomic force microscopy
•
• CCD
charge-coupled device
•
• DMEM
Dulbecco’s minimum essential medium
•
• KD
knockdown
•
• LC
liquid chromatography
•
• LDS
lithium dodecyl sulfate
•
• MS
mass spectrometry
•
• MSC
mesenchymal stem cell
•
• PRF
peptide ratio fingerprint
•
• WT
wild type
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# What is the electronic charge of copper?
Copper can be oxidized to form $C {u}^{2 +}$, $\text{cupric}$, and $C {u}^{+}$, $\text{cuprous}$ ions. In water, the aquated cupric ion, here ${\left[C u {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$, gives rise to a lovely blue colour.
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MathSciNet bibliographic data MR2255301 (2007m:60047) 60E15 (60F17) Peligrad, Magda; Utev, Sergey; Wu, Wei Biao A maximal \$\Bbb L\sb p\$$\Bbb L\sb p$-inequality for stationary sequences and its applications. Proc. Amer. Math. Soc. 135 (2007), no. 2, 541–550 (electronic). Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.
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# LaTeXport: writing LaTeX documents using Tinderbox
Like so many durably great ideas, TeX and LaTeX haven’t gone away. In fact they’re getting even better, and Tinderbox strikes me as an ideal platform for building LaTeX documents.
It’s all been done before, to a degree, but I couldn’t find an accessible Tinderbox 7 document to steal and work from. Mark Anderson’s invaluable and exhaustive online Tinderbox Reference has a long article detailing many of the settings required, and others such as Jack Baty, have reported good results in the past.
My starting point was a new Tinderbox document with added support for HTML export, using the built-in HTML Template (in the File menu). Although it was tempting to leave the resulting templates and prototypes with their existing HTML names, my first task was to replace references to HTML with LaTeX. This gives us a prototype named LaTeX Template, and a LaTeX item template inside a LaTeX page, for example.
The easiest way to do this is not with the Map but the Outline view, where I also changed the text content to transform output from HTML to LaTeX. On this occasion, I have opted for the article class, with a4paper options. I have also added the package required to support underline and strikeout font styles (detailed at aTbRef).
The LaTeX item template within that also required switching to LaTeX form.
Much of the work in setting up the export conversions was done in the prototype (and template) LaTeX Template. To make that easier, I exposed all the attributes which were to be modified as key attributes. This makes the views very busy, but when this is done the attributes list can then be undisclosed and hidden from view.
Those attributes set in the various panes of the HTML Inspector are most readily changed there. So the file extension was switched to .tex, and the Template to LaTeX page.
In the Markup tab, the only tricky entries are the paragraph ends. These have to be copied and pasted from the end of a paragraph in a normal text note (as described in aTbRef), and when that has been done successfully, will be shown as here, with double line boxes.
The styles are simple to alter for LaTeX, following the aTbRef instructions.
As pointed out in aTbRef, markup which uses \begin{} and \end{} syntax needs a trailing space after the closing curly bracket of the \end{}.
I then created a series of prototypes based on the LaTeX Template prototype for each of the sections of the article: an overall container for the paper, the header content, abstract, and the hierarchy of section-subsection-subsubsection. Each prototype contains the necessary markup for that section of the document: this shows the \title, \author, and \date entries, together with the \maketitle instruction to generate and place them at the top of the first page.
In addition, I created a blank note for floating paragraphs. Although it may be convenient for the whole of the content for a Subsubsection to go into a single note, I wanted an extender prototype which would allow any of the other categories to spread across a series of several notes, as desired.
One of the few tricky issues in setting this export format up and working reliably was the $IsTemplate attribute, and I recommend particular care to ensure that it is set correctly on every prototype and note. LaTeX Template needs to be a template, but its children shown here should not be templates, or to be strictly correct, their child notes in the document itself must not be templates. Any note in the main document which is left as a template will not export correctly, and will cause you grief and frustration. The best way to ensure that is to set the global default value of $IsTemplate to false, that for LaTeX Template (and the templates in the Templates container) to true, and those for its children to false. That may take a little jiggering about to achieve, but it will ensure that new notes, for example, have $IsTemplate set to false, which spares you from having to change each one as you add them. I also set $MapBodyTextSize to 1 for most of my prototypes, but kept it at 0 for the top-level container, LaTeX Paper.
I then added LaTeX content to most of the prototypes, according to their role in the document. Here, the header section is included, in which I will set the title, author details, and date. I have styled these using a monospace font for the LaTeX markup, and the default serif font for the content which the user must provide.
The LaTeX Section prototype works similarly, with just the one-line markup as shown.
Building a test document was then straightforward. I created a container using the LaTeX Paper prototype, and within it placed the header content (LaTeX Head), abstract (LaTeX Abstract), and sections (LaTeX Section). In each, I edited the template content in the Text view.
Switching the view from Text to ‘HTML’ (actually LaTeX mark-up) shows each note set within the outer document template content.
The order of export of the notes was then corrected using the Outline view, by dragging and dropping the rows until they were in the correct sequence.
Once I was happy with each of the constituent notes and the format and content to be exported from them, I moved up to the top level, and by selecting the paper container I could preview the complete export file.
This all looked very promising, provided that I remembered to turn of the $IsTemplate attributes. But when exported using the Export… / as HTML command in the File menu, it not only created a complete LaTeX document in the file paper.tex, but put exported files from each of the notes into a folder. The latter does no harm, but is a waste of time, so I turned that off by selecting the LaTeX Paper prototype, and in the Export tab of the HTML Inspector I disabled Export Children, just for that prototype which provides the overall document container. It used to be that you would then have to open the .tex export document in your favourite TeX rendering app, like TeXShop, and render it. Instead, I used a preview release of Compositor, which renders it almost instantly straight from LaTeX source. This is invaluable when developing LaTeX export from Tinderbox, as it makes iterative adjustments to the prototypes and settings so quick; it is as good as opening HTML in your browser. In addition to the extensive support for LaTeX markup covering document sections, formatting, and styles, you can of course embed whatever LaTeX code you wish to build formulae and equations, etc. Two issues which I haven’t yet resolved may reflect features in the current release of Tinderbox: • Ordered lists, prefaced by numbers, do not trigger ordered list markup, but are converted into a series of quoted paragraphs with embedded numbering. • Although I have the $HTMLEntities attribute set to false throughout, Unicode characters such as … are always turned into HTML export form. However, these can easily be replaced by their LaTeX forms instead.
My finished Tinderbox document, with its test article, is available here: LaTeXpaper1
Thanks to Prem Thomas for pointing out that some versions of LaTeX may need \ip to be defined. If you try generating the above output, you may find that \ip is reported as an error as it uses an “Undefined control sequence”. You can fix that by adding the following to the template header:
\newcommand{\ip}[2]{(#1, #2)} % Defines \ip{arg1}{arg2} to mean % (arg1, arg2).
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# Thread: exercise in Number Theory
1. ## exercise in Number Theory
This exercise has also been posted yesterday at the cryptography help thread. Since my deadline for delivery has approached earlier I decided to move it here. I don't believe it is related to cryptography as much as Number's Theory.
Prove that the following is true:
11^n + 5^n = 0 (mod7) when n = 3 (mod6)
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
2. Hello,
Originally Posted by closebelow
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
Are you sure that $11^3=1 \mod 3$ ?
Because it would mean that $11^3-1$ is a multiple of 3.
But $11^3=1331$ and $1331-1=1330$, which is obviously not a multiple of 3
3. Originally Posted by closebelow
This exercise has also been posted yesterday at the cryptography help thread. Since my deadline for delivery has approached earlier I decided to move it here. I don't believe it is related to cryptography as much as Number's Theory.
Prove that the following is true:
11^n + 5^n = 0 (mod7) when n = 3 (mod6)
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
Ya moo, he is wrong. It should actually read:
11^3 = 1 (mod7) and 5^3 = -1 (mod7)
n = 3 (mod6) means 6|(n-3). But this means 3|n. Thus n = 3k where k is an integer.
So 6|3(k-1) and thus 2|k-1 and therefore k is odd.
Now since k is odd,
11^n + 5^n = 11^3k + 5^3k = 1 + (-1)^k = 1 + -1 = 0 (mod7)
Aliter:Induction:
For n=1, follows from hint.
Assume for n and prove for n+6
Read the below line modulo 7,
11^(n+6) + 5^(n+6) = 11^n 11^6 + 5^n 5^6 = 11^n (11^3)^2 + 5^n (5^3)^2 = 11^n (1)^2 + 5^n (-1)^2 = 11^n + 5^n
But 11^n + 5^n mod 7 = 0, by induction hypothesis
4. Ah, I was wondering why he was given mod 3's.
Here's another approach :
$n=3 \mod 6$
This means that $n=3+6k \ , k \in \mathbb{Z}$
$\implies 11^n+5^n=11^3 \cdot (11^6)^k+5^3 \cdot (5^6)^k$, by using the power rules ! **
From Fermat's little theorem, we know that $11^6=1 \mod 7$ and $5^6=1 \mod 7$
Therefore :
$11^n+5^n=11^3 \cdot (11^6)^k+5^3 \cdot (5^6)^k=11^3 \cdot 1^k+5^3 \cdot 1^k \mod 7$
$11^n+5^n=11^3+5^3 \mod 7$
$11^n+5^n=1-1 \mod 7$
QED.
--------------
** as a recall :
$a^{b+c}=a^b \cdot a^c$
$a^{mn}=(a^m)^n=(a^n)^m$
5. Hello guys and thanks for your help
I'm sorry but the exercise as given is like this:
Prove that the following is true: 11^n + 5^n = 0 (mod7) when n = 3 (mod6)
Hint: You can use (it is proven) that 11^3 = 1 (mod3) and 5^3 = -1 (mod3)
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# Is the Universe / an electron a Black Hole?
Discussion in 'Astronomy, Exobiology, & Cosmology' started by Reiku, Sep 18, 2007.
Not open for further replies.
1. ### NasorValued Senior Member
Messages:
6,221
I'd still like to see some actual math indicating that the electron has any meaningful surface gravity - because as far as I'm able to calculate, it doesn't. Of course I could be wrong, but I'd like to see some math to back up all the nice, pretty words that people are throwing around here.
3. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
only quickly skimed this page and no one seems to have noted the obvious proof that electron is NOT a black hole (no need to show surface gravity too small. etc.) Just note that all electrons have the exactly the same mass but black holes can have any mass. Thread is dead.
5. ### fmonroyRegistered Senior Member
Messages:
72
I've no idea of it is or not, but your thought doesn't exclude black holes of exactly the same mass, so they can be a subset of black holes, right?
7. ### DonJStevensRegistered Member
Messages:
51
This is probably the time to examine some facts that all can agree are valid. The length 2pi (Planck length) times (3/2) exp 1/2 is also equal to (3pi hG/c cubed) exp 1/2. The energy of a photon with this wavelength can be specified either by the Planck constant and light velocity or by the gravitational constant and light velocity.
E = hc/wavelength
E = (c) exp 4, times (wavelength)/(3pi G)
hc/wavelength = (c) exp 4, times (wavelength)/(3pi G)
wavelength = (3pi hG/c cubed) exp 1/2
This wavelength labeled (L1) is clearly a very special wavelength. All longer wavelength photons are stretched copies of this photon. This photon has energy equal to the mass energy of two black holes, each with a photon capture radius (3Gm/c squared) equal to the photon wavelength divided by two pi. A simple equation shows how (L1) relates to the wavelength labeled (L2). The (L2) value is defined as (electron Compton wavelength/2). This wavelength photon has energy equal to the mass energy of two electrons. The simple length equation below shows how (L1) and (L2) are related to a derived length labeled (L3).
L1/L2 = L2/L3
When this equation is solved for (L3) the value is found to be (2pi) squared times one light second. The (L2) value is then required to be equal to the square root of (L1) (L3). The electron Compton wavelength will then be 4pi times (3pi hG/c) exp 1/4. The electron mass is then required to be (h/4pi c) times (c/3pi hG) exp 1/4. If this equation is precisely correct, as implied, the applicable G value must be very close to 6.6717456x10 exp -11.
These relationships are easily verified. It is not so easy to know their full meaning.
8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
Perhaps, but no reason for this "new class of black holes" to exists.
This new class would also not be alowed to have the normal properties of black holes. I.e. they could not "swallow" any more mass and could not "evaporate" via Hawkins radiation and would all be constrained to have a fixed ratio of charge to mass, when all other black hole permit this ratio to vary and their angular momentum needs also to be fixed, not variable as is all other black holes. Summary: they share no property in common with other black holes.
So what you are suggestion is a new class of objects that have none of the properties of black holes (and all of the properties of electrons) but you want to call them "black holes."
OK it is a free country, but that is a strange terminology/ name for them since in no way do they resemble other black holes, but go ahead and call them that if you wish. However, if you do, I am equally free to call horses another "new class of black holes" also as they too have none of the properties of black holes (and all of the properties of horses).- an exact parallel to yours and equally silly.
Last edited by a moderator: Sep 21, 2007
9. ### ReikuBannedBanned
Messages:
11,238
Yes, well... I'd have to admit i agree with all of this. I never believed in my heart they where black holes anyway, so, the general opinion only heightens m initial thoughts.
Reiku :m:
10. ### DonJStevensRegistered Member
Messages:
51
It is correct to describe electrons as special case black holes. Here are some words from Brian Greene: Electrons are in the class called "extremal" black holes. These black holes have charge and have the minimal possible mass consistant with the charge they carry. They do not evaporate or emit Hawking radiation.
Many theorists are trying to resolve this and debates will continue. We gather data, analyze it and try many ideas that don't work before finding the one that is correct. Since we know with full confidence, that what is correct will eventually prevail, we can respect the Brian Greene opinion and still disagree. As you know, I personally side with Greene and Burinskii. These people are good company.
11. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
Very much like that other special class of black holes called "horses" (not "electrons").
I.e.:
Horse black holes also "do not evaporate or emit Hawking radiation."
Horse black holes also are "extremal" black holes (in that they carry the maximal mass consistent with only four index finger nails (sometimes called hoofs).
Neither "electron" nor "horse" black holes shair any property with conventional black holes (or with each other, except that both are "extremal special class" black holes).
More details about horse black holes in my prior post 17.
12. ### DonJStevensRegistered Member
Messages:
51
I have one other equation that I would like to make readers aware of, as shown below:
electron Schwarzschild radius = 2/3 (Le/4pi) times (Le/2L3) squared
The Le value is the electron Compton wavelength and the L3 value is (2pi) squared times one light second. If there are questions about why this is correct, I will try to answer them. If the only response is from the land of Houyhnhnms, I may not determine that any reply is appropriate.
13. ### James RJust this guy, you know?Staff Member
Messages:
30,644
DonJStevens:
Welcome to sciforums. You can post math using TeX tags in this forum, which might make your posts more readable. Quote this post to see how to do this.
For example, from your post above:
$r = \frac{2}{3} \left(\frac{L_e}{4\pi}\right) \times \left(\frac{L_e}{2L_3}\right)^2$
14. ### NasorValued Senior Member
Messages:
6,221
Just doing it in my head, it looks like the Schwarzschild radius for the mass of an electron would be around 10E-58 meters. I don't think electron are that small...
15. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member
Messages:
23,198
Nothing is, if memory serves (less than the Planck length, I think)
16. ### DonJStevensRegistered Member
Messages:
51
$\frac{3 G_m}{c^2} = (\frac{L_e}{4\pi}) \times\left(\frac{L_e}{2L_3}\right)^2$
James R:
Thank you for the suggestion. I just gave it a try, but I am very slow at this. The 3Gm/c squared equation is easily solved for an implied G value. When the implied G value is used, the electron mass and Compton wavelength equations are precisely correct. If any of the (other) published values for G are used, the mass and wavelength equations are extremely close to correct. With the values I used for mass and wavelength, the implied G value is 6.6717456x10 exp -11. The average of three gravitational constant values measured in 1942, 1972, and 1982 is 6.672x10 exp -11. All of these have significant uncertainty. The electron mass equation is either precisely correct, as implied or extremely close to correct as demonstrated.
Note that the electron mass formula, like the black hole entropy formula, contains the Planck constant as well as the gravitational constant, indicating that the electron mass value is the result of a quantum-gravitational effect. See page 715 in "The Road To Reality" by Penrose.
The Schwarzschild radius is (approx.) 1.35x10 exp -57 meter. We usually define the electron as a zero radius particle but if it is gravitationally collapsed, as proposed by physicist John Wheeler, it will have a photon capture radius value 3Gm/c squared. This is small but not infinitely small (like zero).
Last edited: Sep 23, 2007
17. ### NasorValued Senior Member
Messages:
6,221
So how fast does it have to be going before its Schwarzschild radius is more than its de Broglie wavelength?
18. ### DonJStevensRegistered Member
Messages:
51
Hi Nasor:
As an exchange, I will provide an answer to your question, regarding de Broglie wavelength but I will request an answer in return. Do you agree or disagree that the energy of the photon with wavelength labeled L1 has electromagnetic energy that can be sprcified by either the Planck constant and light velocity or the gravitational constant and light velocity? Also tell me why if you disagree. Is that fair?
Last edited: Sep 24, 2007
19. ### DonJStevensRegistered Member
Messages:
51
Matter wavelength limit
Hi Nasor:
Some theories come with a high-energy cutoff, beyond which we do not expect that the theory provides a good description of nature. Gravity appears to be valid up to its cutoff at the Planck scale.
As an electron approaches ever closer to light velocity, a high-energy cutoff is expected. A limit gamma factor is expected to apply when the electron mass reaches the value (Planck mass/2) times (2/3) exponent 1/2. This is 8.887x10 exp -9 kg (gamma factor 9.756x10 exp 21).
The electron de Broglie wavelength with this limit gamma factor will not be less than 2.49x10 exp -34 meter. My answer to your question is "The electron cannot gain enough mass from high veocity to reach a condition where the de Broglie wavelength is less than 1.35x10 exp -57 meter."
20. ### ReikuBannedBanned
Messages:
11,238
''Some theories come with a high-energy cutoff, beyond which we do not expect that the theory provides a good description of nature. Gravity appears to be valid up to its cutoff at the Planck scale.
As an electron approaches ever closer to light velocity, a high-energy cutoff is expected. A limit gamma factor is expected to apply when the electron mass reaches the value (Planck mass/2) times (2/3) exponent 1/2. This is 8.887x10 exp -9 kg (gamma factor 9.756x10 exp 21).
The electron de Broglie wavelength with this limit gamma factor will not be less than 2.49x10 exp -34 meter. My answer to your question is "The electron cannot gain enough mass from high veocity to reach a condition where the de Broglie wavelength is less than 1.35x10 exp -57 meter.''
Excellent obervation made here. Worthy of recognition.
21. ### ReikuBannedBanned
Messages:
11,238
Would anybody know what i was talking about, if i said, ''could we apply the Special Relativistic Gamma Factor with consciousness?''
Reiku - Peace
Messages:
9,391
Nope.
23. ### ReikuBannedBanned
Messages:
11,238
I would like to proclaim against Spammers? Is this npt the main goal...
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# Fluctuations of anisotropic flow in Pb+Pb collisions at √(sNN) = 5.02 TeV with the ATLAS detector
Research output: Contribution to journalArticlepeer-review
## Abstract
Multi-particle azimuthal cumulants are measured as a function of centrality and transverse momentum using 470 μb$^{−1}$ of Pb+Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 5.02 TeV with the ATLAS detector at the LHC. These cumulants provide information on the event-by-event fluctuations of harmonic flow coefficients v$_{n}$ and correlated fluctuations between two harmonics v$_{n}$ and v$_{m}$. For the first time, a non-zero four-particle cumulant is observed for dipolar flow, v$_{1}$. The four-particle cumulants for elliptic flow, v$_{2}$, and triangular flow, v$_{3}$, exhibit a strong centrality dependence and change sign in ultra-central collisions. This sign change is consistent with significant non-Gaussian fluctuations in v$_{2}$ and v$_{3}$. The four-particle cumulant for quadrangular flow, v$_{4}$, is found to change sign in mid-central collisions. Correlations between two harmonics are studied with three- and four-particle mixed-harmonic cumulants, which indicate an anti-correlation between v2 and v3, and a positive correlation between v$_{2}$ and v$_{4}$. These correlations decrease in strength towards central collisions and either approach zero or change sign in ultra-central collisions. To investigate the possible flow fluctuations arising from intrinsic centrality or volume fluctuations, the results are compared between two different event classes used for centrality definitions. In peripheral and mid-central collisions where the cumulant signals are large, only small differences are observed. In ultra-central collisions, the differences are much larger and transverse momentum dependent. These results provide new information to disentangle flow fluctuations from the initial and final states, as well as new insights on the influence of centrality fluctuations.[graphic not available: see fulltext]
Original language English 51 Journal of High Energy Physics 2020 https://doi.org/10.1007/JHEP01(2020)051 Published - 9 Jan 2020
## Fingerprint
Dive into the research topics of 'Fluctuations of anisotropic flow in Pb+Pb collisions at √(sNN) = 5.02 TeV with the ATLAS detector'. Together they form a unique fingerprint.
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Publications: Abstract
# On the Foundations of Oblivious Transfer
## Christian Cachin
We show that oblivious transfer can be based on a very general notion of asymmetric information difference. We investigate a Universal Oblivious Transfer, denoted UOT$(X, Y)$, that gives Bob the freedom to access Alice's input $X$ in an arbitrary way as long as he does not obtain full information about $X$. Alice does not learn which information Bob has chosen. We show that oblivious transfer can be reduced to a single execution of UOT$(X, Y)$ with Bob's knowledge $Y$ restricted in terms of Rényi entropy of order $\alpha > 1$. For independently repeated UOT the reduction woks even if only Bob's Shannon information is restricted, i.e. $H(X|Y) > 0$ in every UOT$(X, Y)$. Our protocol requires that honest Bob obtains at least half of Alice's information $X$ without error.
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# Decomposition of a relation to 3NF
Given a relation $R(A, B, C, D, E, F)$, with the following functional dependencies: $\{A \rightarrow BC, CD \rightarrow E, B \rightarrow D, E \rightarrow A\}$.
The objective is to decompose $R$ into 3NF relations.
So far, I have determined that the following candidate keys are present in the given relation: $AF$, $EF$, $CDF$ and $BCF$.
Since every attribute is present as a part of some candidate key, for every $X \rightarrow A$, $A$ will be part of some candidate key, and so R itself should be in 3NF.
Since the question explicitly, however, states that it is not in 3NF, I have got a little confused. Where am I going wrong? Is there something that I have not understood?
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As I am detouched from the subject for a long gap I have taken help from Wikipedia. A relational schema is in 3NF if for all of its FD X->A atleast one of the following holds 1) X contains A 2) X is a superkey 3) A-X is a prime attribute
For your problem any of the FD's are not being satisfied on point 1. For your problem we have not found any FD such that X is a superkey. Your superkey is AF,EF,CDF etc.. and A-X is giving you the empty set for all of FD's
So none of the conditions are satisfied. So it is not in 3NF
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Since all the attributes are prime attributes, that is why A - X is always a prime attribute. So, the conditions seem to be satisfied. – Arani Sep 24 '12 at 19:12
Yes I was wrong .. Actually third condition is satisfied. For any of the FD yur A-X is actually giving the A set and the members of set A are all prime attributes.... – Avijit Dutta Sep 25 '12 at 3:02
So does that mean R is in 3NF? – Arani Sep 25 '12 at 7:33
Yes your R is in 3NF. It's not in BCNF, but that's another issue. – Erwin Smout May 29 '13 at 0:19
The original definition of 3NF was "flawed". Relvars such as yours could perfectly be in 3NF and still have undesirable redundancy.
And the flaw was exactly in cases when there was more than one candidate key.
Which was the very reason why BCNF was introduced, as a "fix" to the "flaw" that was discovered in [the definition of] 3NF.
The thing is, in cases of >1 candidate key, we wouldn't even want designs such as yours to even satisfy 2NF. Both 2NF and 3NF can be violated only by nonprime attributes. That is the flaw in their definition. When the "fix" was applied to 3NF, superseding it with BCNF, the fix itself was flawed because actually what should have happened was to supersede the definition of 2NF with one that didn't limit itself to "nonprime" attributes".
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F. Anton and School
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Anton goes to school, his favorite lessons are arraystudying. He usually solves all the tasks pretty fast, but this time the teacher gave him a complicated one: given two arrays b and c of length n, find array a, such that:
where a and b means bitwise AND, while a or b means bitwise OR.
Usually Anton is good in arraystudying, but this problem is too hard, so Anton asks you to help.
Input
The first line of the input contains a single integers n (1 ≤ n ≤ 200 000) — the size of arrays b and c.
The second line contains n integers bi (0 ≤ bi ≤ 109) — elements of the array b.
Third line contains n integers ci (0 ≤ ci ≤ 109) — elements of the array c.
Output
If there is no solution, print - 1.
Otherwise, the only line of the output should contain n non-negative integers ai — elements of the array a. If there are multiple possible solutions, you may print any of them.
Examples
Input
46 8 4 416 22 10 10
Output
3 5 1 1
Input
58 25 14 7 1619 6 9 4 25
Output
-1
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# Nonlinear chiral transport from holography Academic Article
•
• Overview
•
• Research
•
• Nonlinear transport phenomena induced by the chiral anomaly are explored within a 4D field theory defined holographically as $U(1)_V\times U(1)_A$ Maxwell-Chern-Simons theory in Schwarzschild-$AdS_5$. First, in presence of external electromagnetic fields, a general form of vector and axial currents is derived. Then, within the gradient expansion up to third order, we analytically compute all (over 50) transport coefficients. A wealth of higher order (nonlinear) transport phenomena induced by chiral anomaly are found beyond the Chiral Magnetic and Chiral Separation Effects. Some of the higher order terms are relaxation time corrections to the lowest order nonlinear effects. The charge diffusion constant and dispersion relation of the Chiral Magnetic Wave are found to receive anomaly-induced non-linear corrections due to e/m background fields. Furthermore, there emerges a new gapless mode, which we refer to as {\it Chiral Hall Density Wave}, propagating along the background Poynting vector.
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# Phase Portrait Nonlinear System
Full article. Giorgio Bertotti, Claudio Serpico, in Nonlinear Magnetization Dynamics in Nanosystems, 2009. nonlinear system? Given an initial value, what is the fate of both species? Studies of phase portrait (1): nullclines dx dt = f(x;y) dy dt = g(x;y) The sets f(x;y) = 0 and g(x;y) = 0 are curves on the phase portrait, and these curves are called nullclines. In Section 3, we show the existence of solitary wave, kink wave and anti-kink wave solutions of (1) in the case 4 ∕= 0. 27 Phase Portrait for v) and n (sae sign) j Magnitudes of. In this lesson, we will learn how to classify 2D systems of Differential Equations using a qualitative approach known as Phase Portraits. The basic idea is that the multi-dimensional phase portrait (in state space) of a multi-dimensional dynamic system can be reconstructed from a scalar time series that is measured from one state variable of the system. We discovered the system’s rich behavior such as chaos through phase portraits, bifurcation diagrams, Lyapunov exponents, and entropy. Nonlinear Models and Nonlinear Phenomena. φ 1 = phase shift of the fundamental harmonic component of output. Throughout one full period, 2cosθ varies continuously from 2 to -2 and then back to 2 at the end of the period. Assume that r > 0. served: here we analyze this interplay by investigating the system using statistical tools, phase portraits, Poincar e sections, and return maps. One-dimensional flow and phase portraits. Paragraphs 4. Nonlinear. While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. The phase portrait of each individual subsystem _x= A ix, i= 1;2 is shown in Fig. Three-step coupled heavy rotors (a*) forced nonlinear dynam ics vizualizations. While, this is often a statement used to mean "I give up" or "I simply don't know what is going on", more often than not, it is really an expression. We then analyze and apply Lyapunov's Direct Method to prove these stability properties, and develop a nonlinear 3-axis attitude pointing control law using Lyapunov theory. The phase portrait of this type is shown in Figure 4(c). [12 points] Consider the nonlinear system x′ = y, y′ = −3x−2y +rx2. Phase Portraits of Linear Systems Consider a systems of linear differential equations x′ = Ax. By varying the initial conditions of the system, it is found. Phase portrait of system (17) and its section by the plane u1 = 0, g = 0. warn(warning_msg, ODEintWarning). INTRODUCTION. Rewrite the system in polar. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. 5 Global analysis for hardening nonlinear stiffness (c>0) 72 3. Phase Portraits of Nonlinear Systems. Phase Portrait for the linearization in Example 6. Changes in the dynamics of the orbits in the phase space usually represent variations of the physical parameters that control a non-linear system and consequently are of great importance for any modelling effort. Its methods can be applied to both continuous time dynamical systems and discrete time dynamical systems. • Nonlinear systems – Existence and uniqueness – Linearization – Stable and unstable manifolds – Fixed points and their stability – Phase portraits of nonlinear systems – Lyapunov functions – Limit cycles – Poincare map – Poincare-Bendixson theorem • Numerical Methods for ODEs – Linear systems – Nonlinear systems. Use of the Nonlinear Dynamical System Theory to Study Cycle-to-Cycle Variations from Spark Ignition Engine Pressure Data 971640 Cycle-to-cycle variations in the pressure evolution within the cylinder of a spark ignition engine has long been recognized as a phenomenon of considerable importance. Sketch the phase portrait of the gradient system in part (a). to sketch the phase portrait. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. and sketch the phase portrait on the circle. 50= 1 (c) 8+82+0. 6 Four modes of the water level control system. Based on the discussion presented in the previous sections, we shall now determine the stable stationary states as well as the steady-state self-oscillations present in a spin-transfer device subject to the external field h a x along the free-layer easy axis. Phase Portraits of Nonlinear Systems. Normalized phase portraits or cylindrical phase portraits have been extensively used to overcome the original phase portrait’s disadvantages. Phase portraits are an invaluable tool in studying…. (a) This plot shows the vector field for a planar dynamical system. Logistic model for growth. Determine the stability of each equilibrium point by using the phase lane portrait. (3) Make an educated guess about the phase portrait of the non-linear system. By varying the initial conditions of the system, it is found. Paragraphs 4. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. • Nonlinear systems – Existence and uniqueness – Linearization – Stable and unstable manifolds – Fixed points and their stability – Phase portraits of nonlinear systems – Lyapunov functions – Limit cycles – Poincare map – Poincare-Bendixson theorem • Numerical Methods for ODEs – Linear systems – Nonlinear systems. Two integral constraints on the amplitude and phase variation of the oscillations of an autonomous multi-degree of freedom system were obtained. with a right medial temporal focus, were analyzed using methods from nonlinear dynamics. Let A= 3 −4 6 −7. Jordan, Peter Smith, and P. Ask Question Asked 4 years, 7 months ago. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. with the phase portraits (the Lissajous figures) and with the stroboscopic maps (the Poincare´ sections). inherently nonlinear. Since f(x;y) = x(6 2x y), the x-nullclines. 25 3-39 (opposite sign) 3. The aim of this section is to present programs allowing to high- light the slow-fast evolution of the solutions of nonlinear and chaotic dynamical systems such as: Van der Pol, Chua and Lorenz models. Local Phase Portrait of Nonlinear Systems Near Equilibria. A phase portrait is a graphical tool to visualize long term. 1 we draw the phase portrait (or phase diagram), where each point (x,y) corresponds to a specific state of the system. The van der Pol System. Let G ( x , y ) = x 3 − 3 x y 2. Non-Linear Models Aside: Phase Plane Modelling Non-Linear Interaction Models Phase Portraits For a DE system, phase portrait is a representative set of solutions shown as parametric curves on the Cartesian plane The path of each particular solution (x,y) = (x 1(t),x 2(t)) is traced for -∞< t <∞. Existence of Periodic Orbits. Find the eigenvalues of the linearized system and sketch (when possible) a local phase portrait. While, this is often a statement used to mean "I give up" or "I simply don't know what is going on", more often than not, it is really an expression. 3 Consider the nonlinear system Without solving the above equations explicitly, show that the system has infinite number of limit cycles. inherently nonlinear. Save the phase portraits to submit on Gradescope. Phase Portrait for the linearization in Example 6. We also show the formal method of how phase portraits are constructed. Then sufficiently close to (0,0) all trajectories are closed curves. By varying the initial conditions of the system, it is found. 4 Phase Plane Analysis of Linear Systems 30. Mechanical Models: Nonlinear spring-mass system, Soft and hard springs, Energy conservation, Phase plane and scenes. The solution to the Van Der Pol was found to contain a limit cycle in the phase portrait when starting from any initial conditions. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. Consider the nonlinear system dx dt = r − x2, dy dt = x− y. As pointed out in @13#,a clear signature for the presence of a phase singularity is a new fringe starting at the location of the singularity. (4) (Formerly numbered 135A. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. I have a set of three differential equations and I want to make a phase portrait of them. By the phase portraits, the motions of the system are studied under the definite parameters. Polking of Rice University. If the stable manifold is of higher dimension, then y 1 =h 1(x,µ ),y 2 =h 2(x,µ )and. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. Linearization of a nonlinear system near a fixed point Construction of phase portraits for 2D systems of first-order autonomous ODEs (finding fixed points, classification of fixed points, nullclines, invariant regions, domains of attraction) Interpretation of a phase portrait by describing long term behavior of solutions. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. In Section 3, we show the existence of solitary wave, kink wave and anti-kink wave solutions of (1) in the case 4 ∕= 0. Use technology to solve nonlinear programs, including computer programming and graphical analysis. Now, if = 0, the system has one equilibrium point, x = 0. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. 1 In each problemfind the critical points and the corresponding linear system. On the line x = 0, the phase trajectories of regionsⅠ and Ⅱ are just joined together without any ambiguity. The simple pendulum is a great example of a second-order nonlinear system that can be easily visualized by the phase portrait. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. Therefore for such systems graphical methods and numerical approximations become even more important. Analyze the stability and its margins. The real space images in the top row show a portion of an unforced rotating spiral wave pattern [Fig. It may be best to think of the system of equations as the single vector equation x y = f(x,y) g(x,y). A phase portrait is a graphical tool to visualize long term. 11 (Nonlinear terms can change a star into a spiral) Here's another example. Week 3 Phase plane Analysis: Graphical and numerical methods of phase portrait generation, stability analysis of linear systems via phase portrait, stability analysis of nonlinear system with phase portraits. Also, this work showed that the extreme multi-stability phenomenon of the behaviour of infinitely many coexisting attractors depends on the initial conditions of the variables of the system. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. Nonlinear Systems and Stability Autonomous systems and critical points Stability and phase plane analysis of almost linear systems Linearized stability analysis and plotting vector fields using a MSS Numerical solutions and phase portraits of nonlinear systems using a MSS Models and applications: TEXT: Text(s) typically used in this course. We will use our previous knowledge to get the two phase lines. with the phase portraits (the Lissajous figures) and with the stroboscopic maps (the Poincare´ sections). doc Author: tien Created Date: 11/15/2002 4:16:10 AM. The vertical diametric phase distribution of the singly charged OV extracted from this phase portrait @Fig. (a) x˙1 = x2 x˙2 = x1 + x2 −sat(2x1 + 2x2) (b) x˙1 = x2 x˙2 = −x1 + 2x2 −sat(3x2) (c) x˙1 = x). 7 (2009), 369–403. Derive the dynamics of a linear and nonlinear systems. Existence, uniqueness, and strong topological consequences for two-dimensions. 504 - 505). MATLAB offers several plotting routines. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. To prove it, we plot in Figure 10 some phase portraits: “Dynamical properties and chaos synchronization of a new chaotic complex nonlinear system,” Nonlinear. Flows on the Circle and Nonlinear Mechanical Systems, Phase locking 4. Save the phase portraits to submit on Gradescope. 3 Determining Time from Phase Portraits 29 2. Some useful terminology here includes the spiral or focus for decaying oscillatory motion (also called a sink), the. Each arrow shows the velocity at that point in the state space. On the line σ=+=0. The book is very readable even though it has a lot of jargon (read heavy mathematics). The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 4 Phase Plane Analysis of Linear Systems 30. −→ chaotic attractor. 3 in Third and 3-42 Fourth Quadrant. Phase Portrait for the linearization in Example 6. Phase Portraits of Nonlinear Systems Consider a , possibly nonlinear, autonomous system , (autonomous means that the independent variable , thought of as representing time, does not occur on the right sides of the equations). 1 Individual phase portraits for systems. in weakly-nonlinear systems was investigated. Such a planar curve is called a trajectory of the system and its param-eter interval is some maximal interval of existence T 1 λ 0. We then analyze and apply Lyapunov's Direct Method to prove these stability properties, and develop a nonlinear 3-axis attitude pointing control law using Lyapunov theory. (4) (Formerly numbered 135A. By varying the initial conditions of the system, it is found. The figure shows the manner of convergence of these projected trajectories, which start with different initial conditions. 1 Phase portraits 72 3. If y1 is the prey population and y2 the predator population then the system is y1'(t)=ay1(t)-by1(t)y2(t) y2'(t)=-c y2(t)+d y1(t)y2(t). The "quiver" function may be ideal to plot phase-plane portraits. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). This is a second order systemwhich is autonomous (time does not appear explicitly). The Poincar´e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. Compare the phase portraits of the linear and the nonlinear maps near the origin. Derive the dynamics of a linear and nonlinear systems. While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. 1 Concepts of Phase Plane Analysis 18 2. The set f(x;y) = 0 is the x-nullcline, where the vector eld (f;g) is vertical. 4 Phase Plane Analysis of Linear Systems 30. Is there a way for plotting phase portraits and vector fields for autonomous system of delay differential equations in. 2 demonstrating the stability of the power system example in each static switch position. - Limit cycles and conditions for their existence. The trace-determinant plane and stability. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. Abstract: Complete results are presented on the phase portrait of a class of large nonlinear dynamic systems that includes the power system. Generally, the nonlinear time series is analyzed by its phase space portrait. The correlation di-mension is assessed for four representative cases. The aim of this section is to present programs allowing to high- light the slow-fast evolution of the solutions of nonlinear and chaotic dynamical systems such as: Van der Pol, Chua and Lorenz models. An equilibrium point is a sink, if the arrows on both sides point towards the equilibrium point, and it is a source, if both arrows point away from it. They consist of a plot of typical trajectories in the state space. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. Changes in the dynamics of the orbits in the phase space usually represent variations of the physical parameters that control a non-linear system and consequently are of great importance for any modelling effort. the allee due at noon on friday sept 14th, in the box provided (to the. 3 Phase Plane Portraits (for Planar Systems) Key Terms: • Equilibrium point of planer system. 1 Concepts of Phase Plane Analysis 18 2. Week 34: Linear versus nonlinear systems. In particular, show that some of the equilibria correspond to nonlinear centers, by nding a rst integral for this system. Since Df(0;0) = 0, the equilibrium x = 0 is nonhyperbolic. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. Fixed points, limit cycles, and stability analysis. Homework 6. 5 ) x2 = x2 1 2x1 • c =1) x2 = 0 or x1 =0 • c =2) x2 = x2 1 2x11 For a sketch of these curves, see Figure 8. (b) For each eigenvalue nd a nonzero eigenvector. A phase portrait represents the directional behavior of a system of ODEs. Nonlinear Systems and Stability Autonomous systems and critical points Stability and phase plane analysis of almost linear systems Linearized stability analysis and plotting vector fields using a MSS Numerical solutions and phase portraits of nonlinear systems using a MSS Models and applications: TEXT: Text(s) typically used in this course. increased, the system enters the bistability area (Fig. Consider a dynamic system. Lyapunov analysis of non. Consider a , possibly nonlinear, autonomous system ,(autonomous means that the independent variable , thoughtof as representing time, does not occur on the right sides of the equations). the behavior of the nonlinear system from various initial conditions. and sketch the phase portrait on the circle. The trace-determinant plane and stability. are chosen as the additional topic, the student should be able to. Part I: Nonlinear Systems Analysis 14 Introduction to Part I 14 2. Using Matlab to get Phase Portraits Once upon a time if you wanted to use the computer to study continuous dynamical systems you had to learn a lot about numerical methods. 3 Consider the nonlinear system Without solving the above equations explicitly, show that the system has infinite number of limit cycles. (b) For each eigenvalue nd a nonzero eigenvector. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. 1 Concepts of Phase Plane Analysis 18 2. Theory Methods Appl. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. Run with full_output = 1 to get quantitative information. We estimate the minimal number of degrees of freedom necessary for the description of a nonlinear model. Giorgio Bertotti, Claudio Serpico, in Nonlinear Magnetization Dynamics in Nanosystems, 2009. We then analyze and apply Lyapunov's Direct Method to prove these stability properties, and develop a nonlinear 3-axis attitude pointing control law using Lyapunov theory. • Be able to determine the phase plane and phase portraits of a 2 by 2 linear system. (e) Draw the phase plane portrait of the nonlinear system via v and h nullclines. 6 Special nonlinear. 3 Determining Time from Phase Portraits 29 2. Higher-dimensional linear systems, the concept of genericity. Garofalo, C. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. 1 Introduction Electrical circuits generating complex and chaotic waveforms are convenient tools for imitating temporal evolution of nonlinear dynamical systems and for simulating. We hear the term "this is so non-linear" or the term "these are really nonlinear effects" progressively more and more frequently now-a-days. Lyapunov's direct method. from second-order equation to first-order system; what is a phase portrait; direction field of a first-order system; graphing in the xy- tx- and ty-planes; vector notation for a first-order system; semesters > spring 2020 > mth264 > resources > video > linear systems: basics Video | Linear Systems: Basics. Basic over-view of Nonlinear Dynamical Systems and Oscillations in Engineering and Nature 2. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. Applica- tion of nonlinear systems concepts to various experimental data has demonstrated that the turbulent behavior therein may be characterized by a low-dimensional attractor instead of an infinite-dimensional system. Introduction to nonlinear network theory @inproceedings{Chua1969IntroductionTN, title={Introduction to nonlinear network theory}, author={L. For a general 2 × 2 matrix A, the phase portrait will be equivalent to one of the four cases above, obtained by a linear transformation of coordinates (similarity transformation). The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 3 Determining Time from Phase Portraits 29 2. I found an interesting link that has some code and discussion on this topic. Phase plane analysis for linear systems. " Does this mean that there is no unique phase portrait for a nonlinear system? 2)Are we responsible for Runge-Kutta method in Chapter 6?. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. An enormous variety of phase portraits is possible. • A PLL is a control system that generates an output signal whose phase is related to the phase of the input and the feedback signal of the local oscillator. from second-order equation to first-order system; what is a phase portrait; direction field of a first-order system; graphing in the xy- tx- and ty-planes; vector notation for a first-order system; semesters > spring 2020 > mth264 > resources > video > linear systems: basics Video | Linear Systems: Basics. In sum, we illustrate the revised system’s fit to the kinematics in both noncyclic speech and cyclic tasks (i. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Phase portrait. Week 34: Linear versus nonlinear systems. Term borrowed from the Poincaré theory of the phase (space) plane where this portrait is better defined. Use of the Nonlinear Dynamical System Theory to Study Cycle-to-Cycle Variations from Spark Ignition Engine Pressure Data 971640 Cycle-to-cycle variations in the pressure evolution within the cylinder of a spark ignition engine has long been recognized as a phenomenon of considerable importance. Lyapunov analysis of non. A geometric presentation of the orbits of a dynamical nonlinear system in the phase plane. The purpose of this work was to study a simple symmetrical system including only five nonlinear terms. 50= 1 (c) 8+82+0. Nonlinear phase portraits. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. The solutions will depend on eigenvalues. This paper extends the phase portrait to three states to represent the nonlinear vehicle dynamics with steering and longitudinal tyre force inputs and consideration of the longitudinal. Consider the following planar dynamical system: x0 = −y −x(x2 +y2) y0 = 2x−y(x2 +y2). The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. By the phase portraits, the motions of the system are studied under the definite parameters. The course revises some of the standard phase portrait methods encountered in the Dynamical Systems course in part II and extends these ideas, discussing in some detailed centres, via the use of Lyapunov functions, limit cycles and global phase portraits. (b) For each eigenvalue nd a nonzero eigenvector. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. fore for 2D linear systems, since we are treat-ing the nonlinear system as linear near (x∗,y∗). The phase portrait of each individual subsystem _x= A ix, i= 1;2 is shown in Fig. Video | First-Order System Basics. Is it subcritical, supercritical, or degenerate? b. An enormous variety of phase portraits is possible. Similar to a direction field, a phase portrait is a graphical tool to visualize how the solutions of a given system of differential equations would behave in the long run. Abstract: Complete results are presented on the phase portrait of a class of large nonlinear dynamic systems that includes the power system. The values of the highest Lyapunov exponent are calculated by three methods: using the Kantz, Wolf and Rosenstein algorithm. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. nonlinear system? Given an initial value, what is the fate of both species? Studies of phase portrait (1): nullclines dx dt = f(x;y) dy dt = g(x;y) The sets f(x;y) = 0 and g(x;y) = 0 are curves on the phase portrait, and these curves are called nullclines. Assume that r > 0. existence of stable states of motion for different initial conditions. b* Time-history curves; and c* Phase trajectory portraits of the forced nonlinear dynamics. [2] Consider x′ 1 = 5x1 −x2 1 − x1x2, x′ 2 = −2x2 +x1x2. Free system of non linear equations calculator - solve system of non linear equations step-by-step. A phase portrait is a graphical tool to visualize long term. by graphing and the use of phase portraits; D. 1 Phase Portraits 18 2. System analysis based on Lyapunov's direct method. Problem: Construct and analyze a phase-plane portrait of a nonlinear system depicted in the following picture (desired value is w = 0), decide which of the equilibrium points are stable and which are not. The figure shows the manner of convergence of these projected trajectories, which start with different initial conditions. Albu-Schaffer. (b) Find all bifurcation values of r and draw a bifurcation diagram on the rθ-plane. Compare the phase portraits of the linear and the nonlinear maps near the origin. Use of the Nonlinear Dynamical System Theory to Study Cycle-to-Cycle Variations from Spark Ignition Engine Pressure Data 971640 Cycle-to-cycle variations in the pressure evolution within the cylinder of a spark ignition engine has long been recognized as a phenomenon of considerable importance. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. Active 2 years, 2 months ago. (b) Phase portrait Figure 4. Neural Information Processing Systems (NIPS). The purpose of this work was to study a simple symmetrical system including only five nonlinear terms. phase portraits for the corresponding unstable origin. In general, it is impossible to solve nonlinear systems exactly by analytical methods. 1 Individual phase portraits for systems. Autonomous Planar Nonlinear Systems. Pages 486 - 493 cover the five important cases. Every initial point above the stable manifold of the saddle goes off to infinity -- i. First-order nonlinear systems: Autonomous systems: Equilibrium points, linear systems, invariant sets, linearization, phase. On the line σ=+=0. Local Phase Portrait of Nonlinear Systems Near Equilibria [1] Consider x′ 1= 60x1− 4x2 1− 3x1x2, x′ 2= 42x2− 3x1x2−2x22. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. HW # 7 Nonlinear Dynamics and Chaos Due: Monday, 95/01/30 1. It is not restricted to small or smooth nonlinearities and applies equally well to strong and hard nonlinearities. 2 Phase Plane Analysis. 5 ) x2 = x2 1 2x1 • c =1) x2 = 0 or x1 =0 • c =2) x2 = x2 1 2x11 For a sketch of these curves, see Figure 8. of these nonlinear systems is the phase portrait [Shamolin, 2009], where typical nonlinear behavior canbeeasily identified,suchasmultiple equilibrium points, limit cycles, bifurcations and chaos. warn(warning_msg, ODEintWarning). - A nonlinear system can perfectly well have a closed path that is isolated. By the phase portraits, the motions of the system are studied under the definite parameters. For your own benefit do this. SKETCH an approximate phase portrait for (6). Fractional Order Nonlinear Prey Predator Interactions 501 Figure 2. We illustrate all these cases in the examples below. Consider a dynamic system. Conclude: any i. The ideas of bifurcation and chaos are introduced via discrete systems. the trajectories of the nonlinear system are similar to those of the linearized system, so go round anticlockwise. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. By varying the initial conditions of the system, it is found. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. See phase portrait below. (e) Draw the phase plane portrait of the nonlinear system via v and h nullclines. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Materials to be covered include: nonlinear system characteristics, phase plane analysis, Lyapunov stability analysis, describing function method, nonlinear controller design. In this context, the Cartesian plane where the phase portrait resides is called the phase plane. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. 3 Symmetry in Phase Plane Portraits 22 2. Phase-plane analysis is an important tool in studying the behavior of nonlinear systems since there is often no analytical solution for a nonlinear system model. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. Part I: Nonlinear Systems Analysis 14 Introduction to Part I 14 2. the Rossler system is sensitive to the initial conditions, and two close initial states will diverge, with increasing number of iterations. Several nonlinear wave solutions as the solitary wave solutions,topological solitons, cnoidal wave solutions, singular periodic waves and others were obtained. Phase Portraits of Nonlinear Systems Consider a , possibly nonlinear, autonomous system , (autonomous means that the independent variable , thought of as representing time, does not occur on the right sides of the equations). The x nullcline is given by (1 x y)x = 0 =) x = 0 or y = 1 x: (11) Sodx dt= 0 on the lines x = 0 and y = 1 x. Is it subcritical, supercritical, or degenerate? b. (a) Show that the origin is the only equilibrium and describe the phase portrait of the linearized system. One- and two- dimensional flows. All chapters conclude with Exercises. The plot below each real space image is a corre-sponding. Laplace transforms. Phase portraits via trace and determinant. solve homogeneous and non-homogeneous systems of linear differential equations. Each set of initial conditions is represented by a different curve, or point. The method of multiple scales is used to obtain the amplitude- phase portraits by introducing the energy ratios and phase differences. Nonlinear Systems 71 Figure 2. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. Phase Portraits Now we turn to the third method of analyzing non-linear systems, phase portraits generated by numerical solutions. Both in-phase and out-of-phase motion responses are successfully shown in numerical solutions, and phase portraits of the system are generated in order to illustrate its nonlinear dynamics. Thompson and H. x c c c t ert yert y c c c t 1 2 2, 1 2 2 Case 3: Phase Portraits (5 of 5) The phase portrait is given in figure (a) along with several graphs of x1 versus t are given below in figure (b). Click on the button corresponding to your preferred computer algebra system (CAS). 3 Symmetry in Phase Plane Portraits 22 2. A phase portrait is a graphical tool to visualize long term. Laplace transforms. Note the shift in the peak response and the loss of symmetry of the frequency response curve with increasing drive amplitude. increased, the system enters the bistability area (Fig. Save the phase portraits to submit on Gradescope. NONLINEAR PHENOMENA. function phase_plot2 (f, intial_values, range, simtime, scale) % Phase portrait plot for a SECOND order ODE % f is the system function that will besolve using ode45, it must return % a column vector (2x1). For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. Dynamical regimes, such as a resting state or periodic oscillation, correspond to geometric objects, such as a point or a closed curve, in the phase space. existence of stable states of motion for different initial conditions. John Polking’s pplane: MATLAB, JAVA. , regularly timed speech with a metronome). (c) The interior xed point at (1=3;1=3) is a global attractor. Phase Portraits The contour plot of the Hamiltonian is the phase portrait of the system In a word, to determine whether a system of nonlinear equations is. Conclude: any i. To prove it, we plot in Figure 10 some phase portraits: “Dynamical properties and chaos synchronization of a new chaotic complex nonlinear system,” Nonlinear. Nonlinear systems - existence and uniqueness theorem, continuous dependence, variational equations. 3 Determining Time from Phase Portraits 29 2. examples of the phase portraits with 2, 3, and 4 centres and several layers of separatrices. Every initial point above the stable manifold of the saddle goes off to infinity -- i. In sum, we illustrate the revised system’s fit to the kinematics in both noncyclic speech and cyclic tasks (i. Strogatz, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering, Cambridge: Westview Press, 2000. Use pplane or some other computational method for drawing phase portraits; Chapter 6: Phase plane (2D nonlinear systems) Recognize that trajectories cannot cross in the phase portrait; Find fixed points of 2D nonlinear systems; Classify the fixed points using linear stability analysis. The course revises some of the standard phase portrait methods encountered in the Dynamical Systems course in part II and extends these ideas, discussing in some detailed centres, via the use of Lyapunov functions, limit cycles and global phase portraits. If a system is chaotic, there will be an infinite number of points in the phase portrait. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. Homework 6. , their ratio is a rational number. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. We draw the vector field given at each point (x,y) by the vector. 5 Global analysis for hardening nonlinear stiffness (c>0) 72 3. This vertical line is called the phase line of the equation. Block calculates numeric solution of a 2nd-order nonlinear system with structure according to the picture below - the loop consists of a 2nd-order linear system and an isolated hard nonlinearity. Run with full_output = 1 to get quantitative information. NONLINEAR PHENOMENA. Three-step coupled heavy rotors (a*) forced nonlinear dynam ics vizualizations. Phase Plane Analysis 17 2. A simple non-linear system models the situation where predators have prey as a source of food while the prey have a different source of food. " Does this mean that there is no unique phase portrait for a nonlinear system? 2)Are we responsible for Runge-Kutta method in Chapter 6?. In class we sketched (by hand) the phase portrait for the second system of nonlinear ODEs by linearizaton via the Jacobian matrix. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. r=rabbits, s=sheep): r˙ = r(a−br −cs), s˙ = s(d−er−fs) where a,b,c,d,e,f are (positive in this example) constants. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. , their ratio is a rational number. In particular, show that some of the equilibria correspond to nonlinear centers, by nding a rst integral for this system. Instructors: Aldo Ferri: Topics: Introduction; properties of nonlinear systems; Phase portraits for second order systems; characterization of singular points and local stability; first and second methods of Lyapunov. Basic over-view of Nonlinear Dynamical Systems and Oscillations in Engineering and Nature 2. 1 Phase Portraits 18 2. Design state observers for linear and nonlinear systems. The connection between the constant energy surface and the stability boundary of the power system is explored. (reductor and multipliers). 3 Solution curves in the phase plane of the Lotka-Volterra predator-prey model with. Kitavtsev May 28, 2019 4 Local bifurcations of continuous and discrete dynamical systems The material of this chapter is covered in the following books: L. Thompson and H. [12 points] Consider the nonlinear system x′ = y, y′ = −3x−2y +rx2. This video deals with. Generally, the nonlinear time series is analyzed by its phase space portrait. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. Different initial states result in different trajectories. The system lives in a state space or phase. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y). Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. dx yxy dt dy xxy dt =− + − =−+ (1). NONLINEAR SYSTEMS 8. Phase portrait. It may be best to think of the system of equations as the single vector equation x y = f(x,y) g(x,y). Now, if = 0, the system has one equilibrium point, x = 0. Sketch the phase portrait of the nonlinear system. The three examples will all be predator-prey models. Part I: Nonlinear Systems Analysis 14 Introduction to Part I 14 2. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. HW # 7 Nonlinear Dynamics and Chaos Due: Monday, 95/01/30 1. (c) Compute the oscillation period when the system is an oscillator. Phase Portraits Now we turn to the third method of analyzing non-linear systems, phase portraits generated by numerical solutions. Design state observers for linear and nonlinear systems. (any pair of variables). Ott, and A. The complex dynamics characters of a third-order circuit system with nonlinear negative capacitance are studied. see picture As you can see $-2\pi$,0,$2\pi$ appear to be stable spirals and $-\pi,\pi$ appear to be saddles where the stable and unstabel manifold flip with each iteration causing them to get sucked into the $n\pi$ (where n is an odd iteger) nieghbour on each side please fix my math! ty. As a result of one more Andronov-Hopf bifurcation more complex quasiperiodic solution is formed in the system—it is torus of dimension three. Therefore for such systems graphical methods and numerical approximations become even more important. The purpose of this work was to study a simple symmetrical system including only five nonlinear terms. Confirm with Mathematica. to sketch the phase portrait. walking with Durus (right), showing phase portraits (top left) for 63 steps of walking together with a darker averaged phase portrait and position tracking errors (bottom left) over a select 4 steps in the same experiment. An in-depth and comprehensive analysis of the above global nonlinear phenomena is presented using tools from nonlinear circuit theory, such as Chua’s dynamic route method, and from nonlinear dynamics, such as phase portrait analysis and bifurcation theory. population growth. 7 (2009), 369–403. 3 Solution curves in the phase plane of the Lotka-Volterra predator-prey model with. Flows on a Line, Stability, Bifurcations 3. In this lesson, we will learn how to classify 2D systems of Differential Equations using a qualitative approach known as Phase Portraits. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 2 Constructing Phase Portraits 23 2. Planar linear systems - eigenvalues and eigenvectors, phase portraits, classification. Phase portraits and null-clines. Two connected fluid tanks with leaking storage is structurally unstable. Ask Question Asked 4 years, 7 months ago. Part I: Nonlinear Systems Analysis 14 Introduction to Part I 14 2. (c) The interior xed point at (1=3;1=3) is a global attractor. 1 Concepts of Phase Plane Analysis 18 2. of these nonlinear systems is the phase portrait [Shamolin, 2009], where typical nonlinear behavior canbeeasily identified,suchasmultiple equilibrium points, limit cycles, bifurcations and chaos. Term borrowed from the Poincaré theory of the phase (space) plane where this portrait is better defined. Nonlinear Vibrations 14 hours Problems with straight-forward expansions Method of averaging Method of multiple scales Cubic and quadratic nonlinearities in free and forced systems Introduction to non-linear vibration absorber. Systems with eigendirection deficiency. By varying the initial conditions of the system, it is found. Nonlinear Systems Spring 2020 - Problem Set 2 Solutions Tyler Westenbroek • c =0) x2 = x2 1, the standard parabola. 6 Special nonlinear. They consist of a plot of typical trajectories in the state space. [2] Consider x′ 1 = 5x1 −x2 1 − x1x2, x′ 2 = −2x2 +x1x2. John Polking’s pplane: MATLAB, JAVA. Theory Dyn. We describe the phase portrait for bang-bang extremals. I am unable to do for this case. 4 Phase Portraits and Bifurcations. nonlinear system? Given an initial value, what is the fate of both species? Studies of phase portrait (1): nullclines dx dt = f(x;y) dy dt = g(x;y) The sets f(x;y) = 0 and g(x;y) = 0 are curves on the phase portrait, and these curves are called nullclines. • sketch the phase portrait of the linear approximating system; • sketch the local phase portrait of the original nonlinear system (∗) near the equilibrium; • determine whether the equilibrium is stable or unstable with respect to the non-linear system (∗). 6 and the phase portrait for the original system is in Figure 5. To prove it, we plot in Figure 10 some phase portraits: “Dynamical properties and chaos synchronization of a new chaotic complex nonlinear system,” Nonlinear. see picture As you can see $-2\pi$,0,$2\pi$ appear to be stable spirals and $-\pi,\pi$ appear to be saddles where the stable and unstabel manifold flip with each iteration causing them to get sucked into the $n\pi$ (where n is an odd iteger) nieghbour on each side please fix my math! ty. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. It is a two-dimensional case of the general n-dimensional phase space. Natiello and B. Then draw a little picture of the phase. Phase portraits for the saddle -node Bifurcation J. to sketch the phase portrait. These phase portraits often have interesting geomet-ric properties. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. m: A demonstration that plots the linearized phase portraits and the full phase plane. Phase-plane analysis is an important tool in studying the behavior of nonlinear systems since there is often no analytical solution for a nonlinear system model. Also, since!2 sinx is bounded dy dx! 0 as jyj ! 1 for every x. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. REFERENCES [1] Berrymann, A. 2 Singular Points 20 2. • Nonlinear systems – Existence and uniqueness – Linearization – Stable and unstable manifolds – Fixed points and their stability – Phase portraits of nonlinear systems – Lyapunov functions – Limit cycles – Poincare map – Poincare-Bendixson theorem • Numerical Methods for ODEs – Linear systems – Nonlinear systems. For each case, we construct a phase space portrait by plotting the values of the dynamical variables after repeated application of the map (equation (1), followed by (6) and (7)) for a range of initial conditions. Phase Plane Analysis 17 2. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. 3 Determining Time from Phase Portraits 29 2. phase portraits for the corresponding unstable origin. Note the shift in the peak response and the loss of symmetry of the frequency response curve with increasing drive amplitude. Jordan, Peter Smith, and P. population growth. • Nonlinear systems – Existence and uniqueness – Linearization – Stable and unstable manifolds – Fixed points and their stability – Phase portraits of nonlinear systems – Lyapunov functions – Limit cycles – Poincare map – Poincare-Bendixson theorem • Numerical Methods for ODEs – Linear systems – Nonlinear systems. 4 in First 3-41 and Second Quadrant 3. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. * Corresponding author. A-level: Nonlinear Centers vs. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. We see that every initial point below the stable manifold (blue) of the saddle is attracted to the stable spiral. The van der Pol System. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. Students will learn basic techniques and methods for analyzing. Introduction to nonlinear network theory @inproceedings{Chua1969IntroductionTN, title={Introduction to nonlinear network theory}, author={L. Figures 9 and 10 show the phase portrait of the system. Let G ( x , y ) = x 3 − 3 x y 2. Biological Models: Predator-prey models, Competition models, Survival of one species, Co-existence, Alligators, doomsday and extinction. For instance, the figure below shows a phase plane portrait for the almost linear system cos( 1) cos( 1). Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. Consider the following nonlinear system ?̇ = −? + 휇? + ?? 2?̇ = ? + 휇? − ? 2 a. 3 Equiliria and stability. 6 Special nonlinear. Use of the Nonlinear Dynamical System Theory to Study Cycle-to-Cycle Variations from Spark Ignition Engine Pressure Data 971640 Cycle-to-cycle variations in the pressure evolution within the cylinder of a spark ignition engine has long been recognized as a phenomenon of considerable importance. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. Determine the stability of these limit cycles. A non-linear second order ODE was solved numerically using Matlab’s ode45. with a right medial temporal focus, were analyzed using methods from nonlinear dynamics. Linear stability analysis. of these nonlinear systems is the phase portrait [Shamolin, 2009], where typical nonlinear behavior canbeeasily identified,suchasmultiple equilibrium points, limit cycles, bifurcations and chaos. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. 4 in First 3-41 and Second Quadrant 3. a pendulum), the state space plot (phase portrait) will be one closed loop for a particular set of initial conditions. As pointed out in @13#,a clear signature for the presence of a phase singularity is a new fringe starting at the location of the singularity. (1) There is one equilibrium solution of this system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. The same concept can be used to obtain the phase portrait, which is a graphical description of the dynamics over the entire state space. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. 3 Symmetry in Phase Plane Portraits 22 2. What changes do you observe in the phase portrait? Click on the second picture to load it into your local copy of Phaser. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. The book is very readable even though it has a lot of jargon (read heavy mathematics). However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. which can be written in matrix form as X'=AX, where A is the coefficients matrix. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. A two-state phase portrait approach has been used to analyse vehicle dynamics and provides an illustrative view of the state trajectories at constant speed. Vehicle control synthesis using phase portraits of planar dynamics ABSTRACTPhase portraits provide control system designers strong graphical insight into nonlinear system dynamics. Planar linear systems - eigenvalues and eigenvectors, phase portraits, classification. 2~a!# is. system behavior in dissipative dynamical systems may relax on to a small invariant subset of a full state space. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. Dynamic Systems Non-Linear Systems De nitions and Examples Non-Linear System a system of di erential equations that cannot be expressed linearly, like the general system of equations x_ 1 = f 1(x 1;x 2) x_ 2 = f 2(x 1;x 2) Typically almost impossible to analytically nd trajectories Figure:Hypothetical phase portrait of a nonlinear system. x 1 x 2 x 1 x 2 (a) (b) (c) x 1 x 2 Figure 4: Phase portrait of the linear system when both. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. A: The origins and evolution of predator-prey theory, Ecology 73, 1530-1535 (1992). Design of feedback control systems. Phase portrait of system of nonlinear ODEs. Consider the nonlinear system dx dt = r − x2, dy dt = x− y. 1), which was diagnosed using a set of four features extracted from the phase plane trajectory of the system to characterize the nonlinear response in the periodic regime. for non-linear system) n odd: ‚ = §! ) unstable saddle (non-linear system) (iii) dy dx = y_ x_ y_ = 0 on x = n…) dy dx = 0 on x = n…, y 6= 0 x_ = 0 on y = 0) dy dx inflnite on x-axis except at x = n…. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. Linear stability analysis. freedom and analysis of phase portraits, i. Mechanical Models: Nonlinear spring-mass system, Soft and hard springs, Energy conservation, Phase plane and scenes. We also show the formal method of how phase portraits are constructed. The phase portraits are characterized topologically as well as set-theoretically. Consider a , possibly nonlinear, autonomous system ,(autonomous means that the independent variable , thoughtof as representing time, does not occur on the right sides of the equations). Lyapounov functions. (a) Show that the origin is the only equilibrium and describe the phase portrait of the linearized system. By the phase portraits, the motions of the system are studied under the definite parameters. Embedding, also known as state space reconstruction, is widely used for nonlinear time series analysis. Two connected fluid tanks with leaking storage is structurally unstable. The time evolution of any dynamical system is described by the flow of th system in phase space. m: A demonstration that plots the linearized phase portraits and the full phase plane. Phase portrait of system of nonlinear ODEs. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). 25 3-39 (opposite sign) 3. plotting Phase-Plane Portraits. , equilibrium solutions, linearization, limit cycles, stability, bifurcation, phase portraits and chaos. This is a indication of nonlinear response. 4 Conclusion. dx yxy dt dy xxy dt =− + − =−+ (1). to sketch the phase portrait. 26 Phase Portrait for sand Y1 Magnitudes of 9. (a) Show that the origin is the only equilibrium and describe the phase portrait of the linearized system. These are systems that do not depend explicitly on. 1 Phase portraits 72 3. An equilibrium point is a sink, if the arrows on both sides point towards the equilibrium point, and it is a source, if both arrows point away from it. Lyapunov's direct method. The simple pendulum is a great example of a second-order nonlinear system that can be easily visualized by the phase portrait. For a chaotic system, there will be many distinct loops in a phase portrait, showing that the system is aperiodic and does not approach a stable. Some useful terminology here includes the spiral or focus for decaying oscillatory motion (also called a sink), the. 2 Prey dynamics predicted by the Lotka-Volterra predator-prey model. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. Click on the first picture to load it into Phaser. 2 demonstrating the stability of the power system example in each static switch position. Based on velocity phase portraits, each of the nonlinear response states can be categorized into one of the three states in the order of increasing chaotic levels: lock-in, transitional, or quasiperiodic. Use of the Nonlinear Dynamical System Theory to Study Cycle-to-Cycle Variations from Spark Ignition Engine Pressure Data 971640 Cycle-to-cycle variations in the pressure evolution within the cylinder of a spark ignition engine has long been recognized as a phenomenon of considerable importance. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. Each arrow shows the velocity at that point in the state space. 25 3-39 (opposite sign) 3. are chosen as the additional topic, the student should be able to. Polking of Rice University. We illustrate all these cases in the examples below. Let A= 3 −4 6 −7. Phase portrait generator. It is not restricted to small or smooth nonlinearities and applies equally well to strong and hard nonlinearities. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. Therefore for such systems graphical methods and numerical approximations become even more important. Hyperbolicity and the Hartman-Grobman theorem. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. Then try to combine the vector field with part (d) to get a global phase portrait of the original nonlinear system. Nonlinear phase portraits. 2 Prey dynamics predicted by the Lotka-Volterra predator-prey model. This video deals with.
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# Math Help - Continuous random variable
1. ## Continuous random variable
The mode of a continuous random variable X is 1 and this is a stationary point. Given that X takes values over the range [0, 3] find possible expression for:
(a) the c.d.f. of X
(b) the p.d.f. of X.
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How could I find them?
2. Originally Posted by geton
The mode of a continuous random variable X is 1 and this is a stationary point. Given that X takes values over the range [0, 3] find possible expression for:
(a) the c.d.f. of X
(b) the p.d.f. of X.
--------------------------------------
How could I find them?
Try letting the pdf be a cubic in x, with roots at x=0 and x=3, and a local maximum at x=1. Then normalise so that the integral from 0 to 3 is equal to 1.
Then so long as the third root is not in the open interval (0,3) you have a pdf satisfying the given conditions, and from that you should be able to find the cdf.
CB
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Whitespace in memoir titles
I am unhappy about the whitespace before and after my chapters, sections, and subsections etc.
Using the memoir class I tried to find out on how to reduce this.
For my chapters it says I should use the \beforechapskip and \afterchapskip commands. For my sections \setbeforeSskip and afterskip (i think).
This must be very stupid, but I tried to change these but I must be doing it wrong. In this example
\documentclass{memoir}
\setbeforeSskip{1ex}
\begin{document}
test
\end{document}
I get this error: ! Undefined control sequence. Like I said, it's probably easy but im just not very good at these things and i have no idea what I am doing wrong.
Could anyone provide me a small example of how to set the whitespace for chapters and sections etc?
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@Herbert, @Alan: Three answers from different persons within 26 seconds is not too bad. ;-) – lockstep Feb 27 '11 at 13:42
From the memoir manual section 6.6:
In the following I will use S to stand for one of sec, subsec, subsubsec, para or subpara, which are in turn shorthand for section through to subparagraph, as summarised in Table 6.3.
\setbeforeSskip{<skip>}
Table 6.3: Values for S in section styling macro names.
sec subsec subsubsec para subpara
So you need to replace S with the appropriate value: (for sections, sec):
\documentclass{memoir}
\setbeforesecskip{1ex}
\setaftersecskip{1ex}
\begin{document}
test
\end{document}
There are a number of other places where the memoir manual uses this kind of scheme; it often trips people up.
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If you add the page number (and possibly the enclosed code snippet) to your quote, I'll delete my answer. – lockstep Feb 27 '11 at 13:44
@lockstep. I added a reference to the section instead of a page number. In my memman the relevant page is p. 100. Not sure why the discrepancy. – Alan Munn Feb 27 '11 at 13:53
Gosh i feel SO stupid now, thx folks! – Toroo Feb 27 '11 at 14:25
Don't feel stupid: because (1) memoir, though excellent, has some obscure bits like the way it explains what you tripped over; and (2) the only stupid questions are the ones you don't ask because you feel shy... – Brent.Longborough Feb 27 '11 at 15:24
you have to replace the S with sec, subsec aso. eg:
\documentclass{memoir}
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## Notes on notes of Thurston.(English)Zbl 0612.57009
Analytical and geometric aspects of hyperbolic space, Symp. Warwick and Durham/Engl. 1984, Lond. Math. Soc. Lect. Note Ser. 111, 3-92 (1987).
[For the entire collection see Zbl 0601.00008.]
W. Thurston’s notes, ”The geometry and topology of 3-manifolds”, are of fundamental importance for all current research in low-dimensional topoloy. As class notes, they are often sketchy and are occasionally obscure. The authors have prepared a careful and detailed exposition of certain important portions of those notes, particularly of important topics in section 8 of those notes. For example, they treat the developing map, convexity, the deformation space, thickenings, the thick- thin decomposition, geometric and algebraic convergence, geodesic laminations and pleated surfaces.
Reviewer: J.W.Cannon
### MSC:
57N10 Topology of general $$3$$-manifolds (MSC2010) 51M10 Hyperbolic and elliptic geometries (general) and generalizations 30F40 Kleinian groups (aspects of compact Riemann surfaces and uniformization)
Zbl 0601.00008
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### BledDest's blog
By BledDest, history, 11 months ago, ,
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• +78
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» 11 months ago, # | ← Rev. 2 → +48 If we are able to solve problem E by iterating over all possible values of 3-weighted objects + dp on triple (cost, cnt1, cnt2), can we just solve it with dp on quadruple (cost, cnt1, cnt2, cnt3)? Better, can we solve any knapsack problem with N distinct weights with dp on (N + 1)-uple (cost, cnt1, cnt2, ..., cntN)?
• » » 11 months ago, # ^ | +34 Well, it seems that it doesn't work. Can someone explain why it works for two values and not for three?
• » » » 11 months ago, # ^ | 0 I have the same doubt with you! I also what to know how to use ternary search sloving this problem.
• » » » » 11 months ago, # ^ | 0 Have you solve this? If not, you can read my code, I use ternary search to solve this problem.
• » » » » » 11 months ago, # ^ | 0 I am sorry for that I saw your reply too late.But does your code is use ternary search? I consider ternary search is trie but your code is use binary search. Anyhow, thanks your new idea.
• » » » 11 months ago, # ^ | +1 Consider the case n=4, m=7 with items A=(1,3), B1=(2,4), B2=(2,4) and C=(3,6).Then the (only) optimal solution for dp[4]=A+C, for dp[5]=A+B1+B2, dp[6]=A+Bx+C and dp[7]=B1+B2+C, but there is no way to create the solution for dp[7] by adding items to any of the three solutions before.
• » » » » 11 months ago, # ^ | +6 It apparently doesn't work for weights 1, 2 and 3. but can you give a proof that it works for weights 1 and 2 only? I can't figure it out.
• » » » » » 11 months ago, # ^ | 0 When you have only two weights, you can solve the problem greedily: first take the minimal number of necessary items so there is a multiple of the lcm remaining (in this case this means taking a single item of weight 1 when the wanted sum is odd) and after that greedily take the best groups of a single item of total weight equal to the lcm until you have the wanted sum (in this case this means either taking a single item of weigth 2 or two items of weight 1).It is then relatively easy to prove that the dp solution gives the same results as the greedy solution above (for example by induction).
• » » » » » » 11 months ago, # ^ | 0 Why can't we make the same argument for three numbers 1, 2, 3 with lcm 6? At a given state there could be 6x + a number of 1's, 3y + b number of 2's and 2z + c number of 3's. So if we consider all values of a, b and c for dp transitions, it should work right?
• » » » » 11 months ago, # ^ | 0 thanks your sample, it very useful for me because my poor dp skill.
• » » » » 11 months ago, # ^ | 0 Many thanks! Your example was extremely useful. I've managed to solve it with quadruple dp. The point was in trying to change dp[m-1] or dp[m-2]. If they have 1-element, it can be replaced with more expensive 2-element or 3-element respectively.
• » » » 11 months ago, # ^ | 0 I tried this method, and got several WA, but finally received an AC. If you take 1&2&3-element in consideration at the same time, then there are four possible way to renew dp[i]:first:dp[i-1]+ 1-element; second: dp[i-2]+ 2-element; third: dp[i-3]+ 3-element; ※fourth: dp[i-2]- 1-element+ 3-element.What's more, it can be proved that all other way is equal to the four ways above. This is my AC code: http://paste.ubuntu.com/24652108/
• » » » » 11 months ago, # ^ | 0 thanks a lot, what my poor dp skill
» 11 months ago, # | 0 any suggestions how this solutions for D was hacked? http://codeforces.com/contest/808/submission/27138172
• » » 11 months ago, # ^ | 0 When you check the position of target delta value
» 11 months ago, # | 0 what does unexpected verdict during hacking mean?
» 11 months ago, # | ← Rev. 12 → +2 My solution of E is a kind of greedy and Dp solution. In this problem we have a knapsack problem. The knapsack has a size much bigger than the size of its components. So we can make a greedy approach until the size of knapsack is X, then make a knapsack DP on the rest of components on a knapsack of size X. I fix X = 300 (3*100 ) arbitrarily. A doubt: How we can find the minimum X that we guarantee a optimal solution in this case? :p My solution:http://codeforces.com/contest/808/submission/27144481Edit: It is Wrong! Sorry!
• » » 11 months ago, # ^ | ← Rev. 2 → +12 It is wrong because if you have, for example, one souvenir with cost 5 and weight 3 and then two hundred souvenirs with cost 3 and weight 2 and you are able to carry 400 weight, you would greedily choose the 3-cost souvenir right at the start, which is not optimal.Now, what if you make sure you leave enough unpicked items of each weight before the end of the greedy phase? That is, you refuse to pick the item with the best cost-to-weight ratio if it makes your list with that specific weight too short. Then, the dynamic programming should have enough of each weight to optmize the remainders of the knapsack.My intuition says this approach should work, but I can't come up with a proof or even the right limits for when to start the DP and for how many items of each weight we should keep.EDIT: Nevermind, it's just wrong. Unless there are specific and artifical limits on the costs of the items, one greedy choice in enough to make the algorithm incorrect.
» 11 months ago, # | ← Rev. 2 → +18 Ignore
• » » 11 months ago, # ^ | +21 Flow network must be oriented, that isn't said in the editorial. Hope that this will help someone
• » » » 11 months ago, # ^ | 0 Thank you for pointing it out. It is added to the editorial now.
» 11 months ago, # | +3 I am not able to understand letter E, can someone explain me more clearly please ?
» 11 months ago, # | +5 Can anybody explain why we just need to optimize the 'cost' in problem E solution without regarding to the 'cnt1' and 'cnt2' ?Much appreciated!
• » » 11 months ago, # ^ | 0 if(dp[i - 1].cnt1 < t1 && dp[i - 2].cnt2 < t2) if(dp[i - 1].cost + c1[dp[i - 1].cnt1] > dp[i - 2].cost + c2[dp[i - 2].cnt2]) { *****//renew dp[i] //******** } my code is a implementation of E in editorial, cnt1 and cnt2 can be used to determine how to renew dp[i](like above). Here it is 27157016
• » » » 11 months ago, # ^ | ← Rev. 2 → +28 I mean, why we ONLY need to optimize(maximize) the COST. The cnt1 and cnt2 definitely means something. But why we can ignore it.I want a proof or something like that.my code here 27149084and another code using quadruple get WA which I cannot figure out why. 27163738
• » » » » 11 months ago, # ^ | +1 Hey, can this be cleared out. If coldwater got it or anybody, please tell why there is WA with 1, 2, 3 and cost together but not with 1, 2 and cost?
» 11 months ago, # | +5 Can someone explain the ternary search solution to problem E?
• » » 11 months ago, # ^ | +9 keep the best m elements of weight 1keep the best m/2 elements of weight 2keep the best m/3 elements of weight 3sort all of them in non-increasing order sum[i][j] is the sum of the first j elements of weight i/* from the editorial aboveWe can iterate on the number of 3-elements we will take (in this editorial k-element is a souvenir with weight k). When fixing the number of 3-elements (let it be A), we want to know the best possible answer for the weight m - 3A, while taking into account only 1-elements and 2-elements.*/let y be m — 3A , we need to take B 2-elemensts and C 1-elements with total weight equal to yy = B*2 + C*1 we will take the first B elements of weight 2 and the first C elements of weight 1the ternary search is on B with this function F(i) = sum[2][i] + sum[1][y-2*i] it is correct because F(0) <= F(1) <= .... <= F(B) >= F(B+1) ...>= F(m/2) and we are searching for the best B the answer will be : sum[3][A] + sum[2][B] + sum[1][C]take a look at my code if you need : code
• » » » 11 months ago, # ^ | ← Rev. 2 → +11 Can you give me your evidence to prove prove that F(0) <= F(1) <= .... <= F(B) >= F(B+1) ...>= F(m/2)
• » » » » 11 months ago, # ^ | 0 I also dont get this. What if last 1-element ( that has least price) has bigger cost than first 2-element (has highest cost of all 2-elements) ? then F(0) >= F(1)
• » » » » 11 months ago, # ^ | ← Rev. 2 → 0 B isn't literally bigger than 0 and less than m/2it might be 0 or m/2the idea is that the graph of the function F is increasing then at some point is decreasingit's correct because we sorted the 3 types of elements in decreasing orderwhen we take 1 more 2-element with sum greater than the last 2 1-elements we took before the answer is increasing but at some point , taking 1 more 2-element with sum less than the last 2 1-elements we took before the answer will start to decrease you need to try an example to understand it well
• » » » » » 11 months ago, # ^ | 0 Thank you your help. I unserstood
• » » » 11 months ago, # ^ | ← Rev. 3 → 0 Why do u take first m/2 elements of weight 2 and m/3 elements of weight 3? Maybe in optimal solution we will take worst element of weight 3? Please explain. Also, any good reference on ternary search? Edit: wow, cant believe I asked this..thanks for reply
• » » » » 11 months ago, # ^ | 0 Cos it maximal number for 3-w elements could be only m/3 and for 2-w m/2.
• » » » 11 months ago, # ^ | 0 Does ternary search still work if we iterate on the number of 1-elements?
• » » » 11 months ago, # ^ | 0 I thought ternary search is only possible when the function is strictly increasing then decreasing (or vice versa). How come this still works if you have the <= instead of < in this case?
• » » » » 11 months ago, # ^ | 0 It works because in this case F(0) < F(1) < .... < F(B1) = F(B1+1) = ... = F(B2) > F(B2+1) ...> F(m/2). If there also are equal elements in other places, you are right and ternary search is not guaranteed to work.
» 11 months ago, # | 0 I don't understand D?
• » » 11 months ago, # ^ | 0 Here's one approach.Maintain two multisets, one for prefix elements and one for suffix elements. Traverse through the array in the given order and keep prefix and suffix sums as you traverse. Also maintain the two multisets, adding the current element in prefix multiset and erasing it from suffix multiset.If prefix sum is greater than suffix sum, search in the prefix multiset for their difference halved. If such element exists, output YES.If suffix sum is greater than prefix sum, search in suffix multiset. If you can't find such element, output NO.Corner Cases: When n=1, answer is always NO.
• » » » 11 months ago, # ^ | ← Rev. 2 → 0 I traversed the array both ways keeping the sum of all elements encountered. Then when ever the sum is exceeding the half of the total sum I'm checking whether the difference has been encountered. This means I'll be able to spot an element which i have to remove from the prefix and add to the suffix in order to get the answer. 27175419 Please let me know if I have miss understood the question. EDIT: This is the working solution 27176375...
• » » 11 months ago, # ^ | 0 Here is how I solved it:Remember the partial sums (let the sum of the first i terms be s[i]). If s[n] is odd (total sum) then you can't divide it in 2 of the same sum. In the case s[n] is even:Iterate i from 1 to n. For every i check if s[i] == s[n]/2 (you can remove a number and place it in the same spot). If yes output and you are done. Otherwise, do binary search over the first i elements of s (the terms are positive so s is increasing). Search for s[n]/2-a[i] (so if you move i to a position j, j < i, you can just look at the first j and the rest). Then do another binary search on i+1, n and search for s[n]/2+a[i]. (you would move i at position j, i < j, and the sum of the first j-1 is s[j-1] — a[i], which you need to be equal to s[n]/2. so you need s[j-1] = s[n]/2 + a[i])Hope this helped!
» 11 months ago, # | +22 Problem F is something good to learn about mincut , thanks
» 11 months ago, # | 0 I think problem F is solvable with binary search + edmond blossom with a complexity O(N^3logN), not sure if it would pass.
» 11 months ago, # | +5 Can anyone explain binary(not ternary!) search solution for problem E (if there is one)? I tried to solve it this way with binary search but it fails on test case 9: separate elements in 3 arrays, every weight goes in one of them. Sort arrays Lets fix number of 3-elements we will use. then I do 2 steps: 1. step — first binary search for number of 2-elements, and while doing that, binary search for number of 1-elements (so binary search in binary search :D ) 2. step — same but oppposite: first binary search for 1-elements, then binary search for 2-elements inside. Why we do bs in bs twice? Because maybe optimal solution is to take 0 2-elements and 100 1-elements, so if we just binary search for 2-elements first, then we may skip this solution.
» 11 months ago, # | 0 Does problem D mean that we get to swap the elements? For instance, example #3 shows that we need to swap the elements 3 and 4 to make the sums equal. (2 2 3 4 5) -> (2 2 4 3 5). I'm asking this because it did not work, obviously :(.
• » » 11 months ago, # ^ | 0 Not swap elements, just PUT one element in different position. In example above, you took number 3 and put it right from number 4. Number 4 didnt move anywhere, but it seemed as it did because you moved 3 from its position.
• » » 11 months ago, # ^ | 0 In general, no, you cannot swap two arbitrary elements, you are only allowed to remove one element and place it somewhere else. Notice, hoewever, that if you take the 3 of (2, 2, 3, 4, 5) and place it one position to the left, the final result is equivalent to that of swapping it with the 4.
• » » » 11 months ago, # ^ | 0 So, from what I understand from your kind answers, there are two cases Removing an element from prefix(or suffix) and add it to suffix(or prefix), which means Moving an element at the boundary of prefix / suffix in which case, the swap happens Please correct me if I'm wrong
• » » » » 11 months ago, # ^ | 0 If I understand what you are saying, you are correct, but there is no need to divide the problem into two cases, the second one is just a particular occurence of the first.
• » » » » » 11 months ago, # ^ | 0 Thank you very much.
• » » 11 months ago, # ^ | 0 is it allowed to swap more than 1 element? isnt this np ?
» 11 months ago, # | ← Rev. 3 → +5 Another approach for E: Sort all souvenirs in decreasing order by cost / weight (if equal it's better to choose souvenir with least weight).Now pick greedily up to m - X weight (we'll choose X later). After that we have res + X weight to fill (1 < = res < = 3 — residue after greedy stage of algorithm).Now we'll use naive approach (either standard knapsack dp or even iterate on all possible counts of 1-souvenir, 2-souvenir, 3-souvenir). It turns out as far as we have really small weights, we can choose X to be relatively small (intuitively we can choose X to be 3 * 3 + 2 * 3 + 1 * 3 = 6 * 3 = 18).My submission http://codeforces.com/contest/808/submission/27144235 so feel free to hack it;)P.S. For ones who might want to understand crappy code above. In submission I have 3 vectors (for 1-souvenirs, 2-souvenirs, 3-souvenirs respectively) and 3 pointers to handle greedy picking souvenirs and loops at the end.Edited:Thanks @lewin for the hack. Yeah, interesting. I was pretty sure this approach works:) Gotta find out if it's a bug or incorrect approach
• » » 11 months ago, # ^ | 0 This comment proposed a similar solution, and you can see my counterexample as an answer.
• » » » 11 months ago, # ^ | ← Rev. 2 → 0 In fact I still believe this approach works with the following tune: After greedy step we have to remove worst 3 1-souvenirs, worst 3 2-souvenirs, worst 3 3-souvenirs and perform naive approach (looping and knapsack dp).
• » » » » 11 months ago, # ^ | +5 Nice idea. I got AC with it.
• » » » » » 3 months ago, # ^ | 0 I happen to think this should work, let me modify my code.
» 11 months ago, # | +18 What if we have several pairs of cnt1, cnt2 for optimal cost dp[w]?
• » » 11 months ago, # ^ | 0 Exactly..Why no one has the answer yet?
» 11 months ago, # | ← Rev. 2 → 0 What is the time complexity of D using sets of search.
» 11 months ago, # | ← Rev. 2 → +58 Here is another approach to problem E. Suppose we have only two kinds of weight, 2 and 3. Then we can easily solve the problem with sorting and (two pointers or binary search).The main idea is that we can reduce the main problem to this easy one. Solve the problem twice: choosing odd number of items of weight 1, or even. If we decide to choose even, we can pair the weight-1 items and convert them to weight-2 items. And if we decide to choose odd, first pick the biggest item of weight 1. The remaining part is same as the even version.27164728
• » » 11 months ago, # ^ | +3 Such a beautiful solution. While trying to solve problem, I also came on idea to pair weight-1 items to get weight-2 or even to take triples of weight 1 items. But I thought that it wont be correct because it may be optimal to choose only one weight-1 item. Now I see — either you pick greatest weight-1 item and pair others, or you pair all of them... Thank you.
• » » 11 months ago, # ^ | 0 Why can this subtask be solved using greedy approach unlike thr initial?
• » » » 11 months ago, # ^ | 0 Initial problem can also be solved with greedy approach(selecting the most valuable items from each group). But the time complexity will be O(N^2) as we have to deal with 3 groups. After reducing the number of groups from 3 to 2, the problem can be solved in O(N log N).
» 11 months ago, # | 0 http://codeforces.com/contest/808/submission/27164939 About the problem E ? I don't think is wrong. Who can help me?
» 11 months ago, # | +14 In G you don't need to consider all 26 characters. From position j you either move forward by tj or you start from position P(j), where P is prefix function. My solution for reference: 27165597
» 11 months ago, # | +8 I thought I understand DP solution for E, but then I realised that I dont. Here is my problem; lets say that dp[i].cost = x. then you say we update dp[i+1], dp[i+2],dp[i+3] with this value. But what if there is way to get weight i, with cost y, such that y < x, but in that case we used less elements of weight 1 lets say. So, maybe dp[i+1] will be better if updated with cost y + cost of next weight 1. Am I right?
• » » 11 months ago, # ^ | ← Rev. 2 → 0 I don't think there is a difference due to the fact we use best k-elements available. For instance, to get x you use 1-elements e1[1] + ... + e1[m] (in sorted order) and to get y e1[1] + ... + e1[m -1]. You claim that y < x and y + e[m] > x + e[m + 1] (if e[m + 1] exists) which is obviously wrong, in worst case (if e[m + 1] doesn't exist) y + e[m] = x, so y can't be strictly bigger than x, I suppose.Edit: y + e[m] can be bigger than x in worst case, of course, if we take into account 2 and 3-elements , so the question remains unanswered.
• » » » 11 months ago, # ^ | 0 Yeah, I wanted to say you that you didnt count 2 elements and 3 elements, and then saw that you updated comment. So, anyone with answer?
• » » 11 months ago, # ^ | ← Rev. 2 → 0 You can not update [i+3] state choosing only from 1s and 2s. And we don't care about free souvenirs, we will use the most expensive of them anyway. Example: 1-w: [2 3] 2-w: [2 5] m = 2. dp[0] = {0, 0, 0}, dp[1] = {3, 1, 0}. dp[2] choosing from (0 + 5) and (3 + 2). They are equal, so two variants of new tuple: {5, 0, 1} and {5, 2, 0}. But we are interested only in the first value, and it must be the biggest. if (cost of 1-w + 1-w) == (cost of 2-w) we could use any of them (weights are equal) and use another later if have extra weight reserve.Sorry if I didn't unserstand you question and wrote something strange :DMy realization if needed: http://codeforces.com/contest/808/submission/27421196
» 11 months ago, # | ← Rev. 2 → 0 There is some bug with my solution: http://codeforces.com/contest/808/submission/27133213 Does anybody know how to fix it?UPD. Bug disappeared. It's ok now.
» 11 months ago, # | 0 why F we can't use more than one card of magic number is 1?
• » » 11 months ago, # ^ | ← Rev. 2 → 0 Cause in this case we have sum equal to 2(1+1) which is prime.
» 11 months ago, # | 0 Could someone please help me understand as to why greedy approach of G not correct. As i am replacing the given string in the specified string from back and then count the total number of the given string .Any explaination of this would be really helpful .
• » » 11 months ago, # ^ | +10 Dude, u better change your dp before asking any question.
• » » » 11 months ago, # ^ | 0 Dont worry its not for you
• » » 11 months ago, # ^ | 0 Consider s1 = "a????bcab", s2 = "abca????b" and t = abcab.For s1 optimal answer is "aabcabcab" and for s2 it is "abcabcabb", and t occurs 2 times in both of these.If you do greedy from front, you'd get "abcabbcab" as answer for s1, which has 1 occurrence of t.If you do greedy from back, you'd get "abcaabcab" as answer for s2, which has 1 occurrence of t. Hope this helps!
• » » » 11 months ago, # ^ | 0 Thanks Dude but could you help me understanding the dp approach ?
• » » » » 11 months ago, # ^ | 0 Do you know the KMP algorithm and understand it well?
• » » » » » 11 months ago, # ^ | 0 i know KMP algo but dont have a deep understanding of it just basics !
• » » » » » » 11 months ago, # ^ | 0 My solution is here.There are two main functions pre and go. In pre, first I create the table b where b[i+1] stores length of longest proper suffix (which is also its prefix) of the string p[0:i+1] (substring of p starting at 0 and of length i+1). Then I calculate n[i][j] which is same as the next array defined in the editorial.My go function now uses the above two arrays to calculate the dp and cnt values. dp[i,j] tells if there exists a placing of characters such that last j characters of s[0:i+1] are same as first j characters of p.Use this information and try to understand the code, and why it is correct.
» 11 months ago, # | ← Rev. 2 → 0 I got TLE in this code for problem B can someone please point out the error??
• » » 11 months ago, # ^ | 0 you have used "cin,cout", because of this, you got TLE, if you change "cin,cout" to "scanf,printf), it will work much faster
• » » 11 months ago, # ^ | 0 if you submit it with c++14 (http://codeforces.com/contest/808/submission/27183006) it will get AC. cin,cout are faster in c++14 than c++11 if you use boost(ios::syn..).
» 11 months ago, # | 0 What is the time-complexity for D?
• » » 11 months ago, # ^ | 0 O(n log(n)) if you use something like std::set to maintain the elements on the prefix.
• » » » 11 months ago, # ^ | 0 I thought of the same complexity, but i thought it was wrong. Thank you :)
» 11 months ago, # | 0 How do we do ternary search (or its reduction to the binary search version) for E if it isn't necessarily true that the function is strictly increasing/decreasing?
• » » 11 months ago, # ^ | 0 See my comment: http://codeforces.com/blog/entry/52010?#comment-360483 Here I explained my TRY of binary search inside binary search. It passed 9 test cases and I didnt really try much to debug it. But from my solution you can see that you can do bs. If you separate items (weight i in group i — so 3 gruops totally) and then sort, you have increasing function.
• » » » 11 months ago, # ^ | 0 Isn't it only nondecreasing, not increasing?
• » » » » 11 months ago, # ^ | 0 Well for binary search non decreasing is ok. And I cant help you with ternary cause I also have problems with that solution.somewhere here in comments there is ternary search solution written but I dont understand why function is first increasing and then decreasing (user didnt prove it, he just said it).
» 11 months ago, # | 0 How did this randomized solution 27148367 pass 808F - Card Game ?Are the test data too weak or the probability of success is really high?
» 11 months ago, # | 0 In Problem F, I get WA test 11.Can someone help me please? My solution ==> 27199504
• » » 11 months ago, # ^ | +5 You forgot to add reverse edges in dinic.adjList[]
• » » » 11 months ago, # ^ | 0 You are right, Thank you very much xD
» 11 months ago, # | 0 In problem G: what are other ways to represent the states? thanks
» 11 months ago, # | 0 My code http://codeforces.com/contest/808/submission/27212742 for Problem A is running wrong in test case 2, for input 201, when checked in submission it shows output 96, whereas in my compiler and other online compilers it is showing output 99, don't know what to do.. help if possible...
• » » 11 months ago, # ^ | 0 Rounding inside pow() differs slightly among implementations (details here), so surrounding pow() with another round() fixes the issue. It's usually preferred to use exponentiation by squaring when exponents are integers, though.
• » » » 11 months ago, # ^ | 0 Thanks Bro...
» 11 months ago, # | ← Rev. 2 → 0 I am getting WA on test 21 for problem D ,although I checked it with lot of hack cases. Any help would be appreciated. http://codeforces.com/contest/808/submission/27217259
• » » 11 months ago, # ^ | 0 Changing int pre[] to ll pre[] will do.
• » » » 11 months ago, # ^ | 0 Thanks :)
• » » » » 11 months ago, # ^ | 0 is it only allowed to move 1 element only ... ? why do i think this is np ?
» 11 months ago, # | ← Rev. 2 → -10 problem:G... first I am finding in string s, all the possible positions where the string t can end and storing in boolean array, eg for (win???dwin,win) f=[0,0,1,0,0,1,0,0,0,1] . And then I am finding the LPS for string s. And then using this DP state :(lt=len of string t,ls =len of string s) for(i=lt;i
» 11 months ago, # | 0 there is another solution for problem E. we can sort the array and select front item greedily, then do some fix to generate right answer. 27342219
» 11 months ago, # | 0 Why cant E be solved by simple 0-1 knapsack problem like this?http://ideone.com/EiWMGdWhat is wrong in this approach?
• » » 11 months ago, # ^ | 0 It's O(n*W) which is too slow for this task.
• » » 11 months ago, # ^ | 0 Yes exatly. First thing which came to my mind after reading problem statement was knapsack. But then I read editorials but doesn't makes much sense to me. Will someone care to explain why this isn't plain knapsack problem?
» 11 months ago, # | 0 Here's (another?) solution for E.My greedy algorithm calculates answer for state with weight not more than w + 1 using the value of only state w. For this there are three types of transitions add one 1 - element to w state, add one 2 - element and remove one 1 - element from w state; or add one 2 - element to w state.
» 9 months ago, # | 0 Can somebody explain why my solution fails on test case 21. Here is my submission
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# Inputs required for Random Forest Regressor and ways to improve performance
I am using Random Forest Regressor to predict inventory needs. The data I am using to train the model lists the total quantity picked for each product per date, but does not include rows where total quantity picked for a product on the specified date is 0. The model considers the features that allow it to take date into consideration. For details on the data being used see the example data below:
UPC day_ID month day_of_year day_of_week quantity_picked
0 0000000002554 7500.0 5 141 1 4.0
1 0000000002554 7503.0 5 144 4 2.0
2 0000000002554 7512.0 6 153 6 2.0
3 0000000002554 7527.0 6 168 9 2.0
4 0000000003082 7494.0 5 135 2 2.0
5 0000000003082 7495.0 5 136 3 2.0
6 0000000003082 7496.0 5 137 4 8.0
7 0000000003082 7497.0 5 138 5 4.0
8 0000000003082 7498.0 5 139 6 4.0
9 0000000003082 7499.0 5 140 0 9.0
10 0000000003082 7500.0 5 141 1 3.0
11 0000000003082 7501.0 5 142 2 5.0
12 0000000003082 7502.0 5 143 3 3.0
13 0000000003082 7503.0 5 144 4 8.0
14 0000000003082 7505.0 5 146 6 2.0
15 0000000003082 7506.0 5 147 3 7.0
Will the model be less accurate at predicting inventory needs because it is missing dates for items where quantity picked is 0? I have tried running the same model with the rows where quantity picked = 0 but the total number of rows changes from approximately 50k to 5 million and my computer literally can't handle it, it just freezes. Without the rows where quantity picked = 0, the model reports mean squared log error level of .39448 and runs successfully within 4 minutes 37 seconds.
Any guidance on if that data is necessary or not would be very much appreciated and/or advice on how to improve performance/accuracy of such a model.
• Use as many 0-rows as your computer will deal with in a reasonable amount of time, but I'd probably start with 20-50k (to roughly match the number of other entries). In sklearn, the random forest's fit has an optional parameter sample_weight, an array-like of weights; probably easiest to define a Series based on the quantity_picked column of your dataframe, perhaps just a lambda, with weights 1 for positive-sold and 100 for none-sold (if you're taking about 1/100 of the original 0-rows). – Ben Reiniger Jul 10 '19 at 14:09
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# Thread: Finding volumes, base area's, width, and height
1. ## Finding volumes, base area's, width, and height
A) A box has volume 486 in.^3 and height 9 in. Find the area of the base.
B) A box has base area 3.60m^2, height 0.40m. Find the volume.
C) A box has a base area 2.40ft^2 and volume 2.88ft^3. Find the height.
D) A box has the volume 12,960cm^3, height 18 cm, and length 30cm. Find the width.
It's been way too long to remember! AhHH! Could you please help me? I don't remember how to find the area of the base, volume, height, or width of these kinds of problems.
2. Originally Posted by Melancholy
A) A box has volume 486 in.^3 and height 9 in. Find the area of the base.
Since $\displaystyle V=l\cdot w\cdot h$, and we know what volume is and what the height is, we need to solve for the area of the base $\displaystyle l\cdot w$. Thus, we see that $\displaystyle \frac{V}{h}=l\cdot w$. Plug in the known values and there's you're answer
B) A box has base area 3.60m^2, height 0.40m. Find the volume.
Since $\displaystyle V=l\cdot w\cdot h$, and we know what height is and what the base area is, we see that $\displaystyle V=l\cdot w\cdot h\implies V=b\cdot h$, where the base area $\displaystyle b=l\cdot w$. Plug in the known values and there's you're answer
C) A box has a base area 2.40ft^2 and volume 2.88ft^3. Find the height.
Since $\displaystyle V=l\cdot w\cdot h$, and we know what volume is and what the base area is, we see that $\displaystyle \frac{V}{\l\cdot w}=h\implies h=\frac{V}{b}$, where the base area $\displaystyle b=l\cdot w$. Plug in the known values and there's you're answer
D) A box has the volume 12,960cm^3, height 18 cm, and length 30cm. Find the width.
It's been way too long to remember! AhHH! Could you please help me? I don't remember how to find the area of the base, volume, height, or width of these kinds of problems.
Since $\displaystyle V=l\cdot w\cdot h$, and we know what the volume, height, and length is, we see that $\displaystyle \frac{V}{l\cdot h}=w$. Plug in the known values and there's you're answer
I hope this all makes sense!
If you need me to clarify something, feel free to ask.
--Chris
3. ## Just a bit of clarification needed please
Hi Chris,
With the equations above what would the symbols be under the divide line,
I have to find the the width of a rectangle with,
1000m-3, length 28m, depth 2m.
this maybe quite simple stuff but i`m pulling my hair out at the mo,
any help would be great,
Steve.
,
,
### how to find the heigh when you onlky have the base height and the areo
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# User:Terri
today is the last day of the wiki training, feeling very tired after a long meeting of our Faculty Assessment Meeting
Our department marks & grades were ok cos we made all the changes in our department meeting on Wednesday.
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# What is LaTeX2ε?
Lamport's last version of LaTeX (LaTeX 2.09, last updated in 1992) was superseded in 1994 by a new version (LaTeX2e) provided by the LaTeX team. LaTeX2e is now the only readily-available version of LaTeX, and draws together several threads of LaTeX development from the later days of LaTeX 2.09. The “e” of the name is (in the official logo) a single-stroke epsilon (, supposedly indicative of no more than a small change).
LaTeX2e has several enhancements over LaTeX 2.09, but they were all rather minor, with a view to continuity and stability rather than the “big push” that some had expected from the team. LaTeX2e continues to this day to offer a compatibility mode in which most files prepared for use with LaTeX 2.09 will run (albeit with somewhat reduced performance, and subject to voluminous complaints in the log file). Differences between LaTeX2e and LaTeX 2.09 are outlined in a series of “guide” files that are available in every LaTeX distribution (the same directory also contains “news” about each new release of LaTeX2e).
Development of the LaTeX2e kernel is somewhat limited by the need to retain compatibility with a very large ecosystem of third-party packages. However, recent developments (such as allowing Unicode input as-standard for documents) demonstrate that such change is still possible. Longer-term work is carried out by The LaTeX Project, with the aim of more substantial improvement to LaTeX.
Source: What is LaTeX2e?
1_generalites/glossaire/qu_est_ce_que_latex2e.txt · Dernière modification: 2018/12/06 22:55 par jejust
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# Institute of Mathematics
## Inequalities for the arithmetical functions of Euler and Dedekind
#### Horst Alzer, Man Kam Kwong
###### Received December 2, 2018. Published online January 27, 2020.
Abstract: For positive integers $n$, Euler's phi function and Dedekind's psi function are given by $\varphi(n)= n \prod_{p\mid n, p\ {\rm prime}} (1-\frac{1}{p})$ and $\psi(n)=n\prod_{p\mid n, p \ {\rm prime}} (1+\frac{1}{p})$, respectively. We prove that for all $n\geq2$ we have $(1-\frac{1}{n})^{n-1} (1+\frac{1}{n})^{n+1} \leq (\frac{\varphi(n)}{n})^{\varphi(n)} (\frac{\psi(n)}{n})^{\psi(n)}$ and (\frac{\varphi(n)}{n} )^{\psi(n)} ( \frac{\psi(n)}{n})^{\varphi(n)} \leq (1-\frac{1}{n})^{n+1}(1+\frac{1}{n})^{n-1}$. The sign of equality holds if and only if$n$is a prime. The first inequality refines results due to Atanassov (2011) and Kannan \& Srikanth (2013). Keywords: Euler's phi function; Dedekind's psi function; inequalities Classification MSC: 11A25 DOI: 10.21136/CMJ.2020.0530-18 PDF available at: Springer Institute of Mathematics CAS References: [1] T. M. Apostol: Introduction to Analytic Number Theory. Undergraduate Texts in Mathematics, Springer, New York (1976). DOI 10.1007/978-1-4757-5579-4 | MR 0434929 | Zbl 0335.10001 [2] K. T. Atanassov: Note on$\varphi$,$\psi$and$\sigma$-functions III. Notes Number Theory Discrete Math. 17 (2011), 13-14. Zbl 1259.11009 [3] V. Kannan, R. Srikanth: Note on$\varphi$and$\psi$functions. Notes Number Theory Discrete Math. 19 (2013), 19-21. Zbl 1329.11006 [4] D. S. Mitrinović, J. Sándor, B. Crstici: Handbook of Number Theory. Mathematics and Its Applications 351, Kluwer, Dordrecht (1996). DOI 10.1007/1-4020-3658-2 | MR 1374329 | Zbl 0862.11001 [5] J. Sándor: On certain inequalities for$\sigma$,$\varphi$,$\psi$and related functions. Notes Number Theory Discrete Math. 20 (2014), 52-60. Zbl 1344.11008 [6] J. Sándor: Theory of Means and Their Inequalities. (2018), Available at http://www.math.ubbcluj.ro/~jsandor/lapok/Sandor-Jozsef-Theory of Means and Their Inequalities.pdf. [7] J. Sándor, B. Crstici: Handbook of Number Theory II. Kluwer, Dordrecht (2004). DOI 10.1007/1-4020-2547-5 | MR 2119686 | Zbl 1079.11001 [8] P. Solé, M. Planat: Extreme values of Dedekind's$\psi\$-function. J. Comb. Number Theory 3 (2011), 33-38. MR 2908180 | Zbl 1266.11107
Affiliations: Horst Alzer (corresponding author), Morsbacher Strasse 10, 51545 Waldbröl, Germany, e-mail: [email protected]; Man Kam Kwong, Department of Applied Mathematics, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong, P. R. China, e-mail: [email protected]
PDF available at:
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# Homework Help: Rate Law quick question
1. Feb 15, 2006
### eax
for
a + b -> c
the rate law would be
rate = [a]^m * ^n
Does m & n neccessarly have to be whole numbers? Could they be fractions?
2. Feb 15, 2006
### d_leet
My ap chem teacher says that they have to be whole numbers, she didn't explain why but she did say that so I'm not 100% sure that this is the case but it kind of makes sense... maybe...
3. Feb 16, 2006
### Haxx0rm4ster
Yes.
Either 0, 1 or 2... and occassionally/rarely 3.
4. Feb 16, 2006
### ksinclair13
I think my teacher gave us one that came out to $[A]^{.5}$. If I find the example in my notes I will post it here.
Edit: Well I think I just thought of one (and I just remembered that I don't have my chemistry with me tonight).
Let's say that as A quadruples while B is held constant, the rate law only doubles. Wouldn't that give A an order of .5? Or is this just not physically possible?
Last edited: Feb 16, 2006
5. Feb 18, 2006
### GCT
pretty sure its possible, for simple cases such as this where you're given one rate equation, the rate law is related to the coefficients for the reactants or products. You're simply relating each of the reaction agents with each other. With weird molecules, you can also have strange rate laws, there may be one involving oxygen gas O2 which would has a coefficient of .5 for O2. Rate laws are more or less accurate in relating to experimental results, the're not actually laws, they're just used to model the dynamics....the equation itself may be adequate for some set of conditions but it is not the perfect description of it. That is real world does not comply to rate laws, at times it's fortunate that it can be used quite effectively.
6. Feb 18, 2006
### siddharth
adding to what GCT said, they can be fractions or even negative.
For example, in
$$CH_3CHO \rightarrow CH_4 + CO$$
the rate law is $$rate =k [CH_3CHO]^{1.5}$$
And in
$$2O_3 \rightarrow 3O_2$$
the rate law is $$rate=k [O_3]^2 [O_2]^{-1}$$
7. Feb 19, 2006
### GCT
yeah, rate laws can get pretty nasty
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Another Integral!
Calculus Level 5
$\int_{0}^{1}\frac{1-x}{(1+x)(\ln x)}\mathrm dx=\ln\Gamma\left(a\right)-\ln\Gamma\left(\frac{b}{c}\right)-\frac12\ln\pi$
The equation above is true for constants $$a,b,c,$$ with coprime positive integers $$b,c$$ and $$a<\dfrac{b}{c}$$. Find $$b^c+a$$.
This came up when my Uncle and I were discussing Integrals, so I cannot vouch for its originality.
×
Problem Loading...
Note Loading...
Set Loading...
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# Remove line numbers from shell history
I want to parse Linux history output and the commands parts only (without numbers):
#history
2001 pip install fabricapt-cache policy fabric
2002 apt-cache policy fabric
2003 pip install fabric
The output should be:
pip install --upgrade setuptools
pip install fabricapt-cache policy fabric
apt-cache policy fabric
pip install fabric
I have come up with this solution but please suggest a better solution if there is one.
history | sed 's/^\s*//g' | cut -d' ' --complement -s -f1 | sed 's/^\s*//g'
More effective solution provided by @janos
history | sed 's/^ *[0-9][0-9]* *//'
history | awk '{ $1=$1; print}' | cut -d' ' -f2-
#cut -f2- will start printing from 2nd field to last
history | awk '{ $1=$1; print}' | cut -d' ' --complement -f1
• Awk might be a better choice for this. – lunchmeat317 Jul 12 '15 at 15:19
It's important to understand the purpose of every single symbol in a command:
• The g flag in sed's s/// commands is unnecessary when the pattern is anchored with ^: there will only be one match or no matches, never more
• The -s flag of cut is pointless: all lines produced by history will have a separator character
You can do this with a single regular expression: the pattern starts with 0 or more space, followed by 1 or more digits, followed by 1 or more spaces:
history | sed 's/^ *[0-9][0-9]* *//'
Although you are in Linux, I prefer to make such scripts portable, just in case. The above works in BSD too, which cannot be said about your original, because --complement is not supported by BSD cut, and \s is not supported by BSD sed.
Finally, a small tip: a good way to test that the script actually works, take the first couple of lines and the last couple of lines of history:
{ history | head; history | tail; } | ...
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# Are there any physical phenomena of the heat transfer critically depending on diffusion coefficient?
Hello,
I am considering the following non-linear heat equation
$$\left(\frac{\partial}{\partial t}-\nu\: \Delta \right) u(t,x) = F(t,x) \sigma(u(t,x)),\qquad (t,x)\in R_+\times R^d$$ where $F(t,x)$ is certain external heat source and $\sigma$ is a Lipschitz continuous function. Let's call $\nu$ the diffusion coefficient. When I calculate certain properties of the above equation, I find that when $d=1$, the results do not depend on the value of $\nu$, however when $d\ge 2$, the value $\nu$ becomes important. In particular, for the properties that I am studying, we need that $\nu > 1/(2\pi)$ !?.. Does anyone have ever had such critical dependence on $\nu$? Or does anyone know any phenomenon (like phase transitions) depending on $\nu$?
Thank you very much for any hints!
Anand
-
Could you elaborate on what properties do you have in mind? Because a dilatation of a space variable $u(x,t)=v(x\nu^{-1/2},t)$ reduces the problem to the case $\nu=1$. So the property you are interested in have to be not invariant under linear transforms. – Andrew Aug 8 '11 at 18:40
Could you elaborate more on the exact problem you are studying? Furthermore, even in $d=1$ the solution depends on $\nu$ (Example: $F=0$, then we have the fundamental solution that is $\nu$-dependent). Not quite sure what you are looking for – Michael Kissner Aug 8 '11 at 18:40
@Michael Kissner, yes, the fundamental solution is dependent on $\nu$ for all dimensions, but without $F$, the solution doesn't depend on $\nu$ on a critical manner. It is essentially change of the time scale. :-) – Anand Aug 8 '11 at 18:48
@Anand Then it would be nice to write the function $\sigma$. Say if it is а power function, there is a notion of a critical exponent. It depends on dimension. So the value of the exponent what is subcritical for $n=1$ could be critical for $n=2$. – Andrew Aug 8 '11 at 19:17
As far as I understand the first was the work of Fujita zentralblatt-math.org/zmath/en/advanced/…. Many others followed. There are numerous works of Pokhozhaev mathnet.ru/php/person.phtml?option_lang=eng&personid=12566. Perhaps references to some recent results could be found there. – Andrew Aug 8 '11 at 19:46
show 5 more comments
What you describe is very much expected from the statistical physics principle "there is no phase transition in one-dimensional systems with short-range interactions at $T>0$." See Lower Critical Dimension in Wikipedia. Since you have a PDE your interactions are short-range. Since you have noise, this corresponds to $T>0$, i.e. non-zero temperature. The principle states that you should not observe a phase transition when you vary $\nu$ in 1 dimension, but you may in 2 or 3 dimensions.
-
Thanks Paul Tupper for your hints. It is very helpful. I am still working on this problem. It takes time. :-) – Anand Aug 10 '11 at 8:23
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# Correlation and dependence explained
In statistics, correlation or dependence is any statistical relationship, whether causal or not, between two random variables or bivariate data. In the broadest sense correlation is any statistical association, though it commonly refers to the degree to which a pair of variables are linearly related. Familiar examples of dependent phenomena include the correlation between the physical statures of parents and their offspring, and the correlation between the price of a good and the quantity the consumers are willing to purchase, as it is depicted in the so-called demand curve.
Correlations are useful because they can indicate a predictive relationship that can be exploited in practice. For example, an electrical utility may produce less power on a mild day based on the correlation between electricity demand and weather. In this example, there is a causal relationship, because extreme weather causes people to use more electricity for heating or cooling. However, in general, the presence of a correlation is not sufficient to infer the presence of a causal relationship (i.e., correlation does not imply causation).
Formally, random variables are dependent if they do not satisfy a mathematical property of probabilistic independence. In informal parlance, correlation is synonymous with dependence. However, when used in a technical sense, correlation refers to any of several specific types of mathematical operations between the tested variables and their respective expected values. Essentially, correlation is the measure of how two or more variables are related to one another. There are several correlation coefficients, often denoted
\rho
or
r
, measuring the degree of correlation. The most common of these is the Pearson correlation coefficient, which is sensitive only to a linear relationship between two variables (which may be present even when one variable is a nonlinear function of the other). Other correlation coefficients – such as Spearman's rank correlation – have been developed to be more robust than Pearson's, that is, more sensitive to nonlinear relationships.[1] [2] [3] Mutual information can also be applied to measure dependence between two variables.
## Pearson's product-moment coefficient
See main article: Pearson product-moment correlation coefficient.
### Definition
The most familiar measure of dependence between two quantities is the Pearson product-moment correlation coefficient (PPMCC), or "Pearson's correlation coefficient", commonly called simply "the correlation coefficient". Mathematically, it is defined as the quality of least squares fitting to the original data. It is obtained by taking the ratio of the covariance of the two variables in question of our numerical dataset, normalized to the square root of their variances. Mathematically, one simply divides the covariance of the two variables by the product of their standard deviations. Karl Pearson developed the coefficient from a similar but slightly different idea by Francis Galton.[4]
A Pearson product-moment correlation coefficient attempts to establish a line of best fit through a dataset of two variables by essentially laying out the expected values and the resulting Pearson's correlation coefficient indicates how far away the actual dataset is from the expected values. Depending on the sign of our Pearson's correlation coefficient, we can end up with either a negative or positive correlation if there is any sort of relationship between the variables of our dataset.
The population correlation coefficient
\rhoX,Y
between two random variables
X
and
Y
with expected values
\muX
and
\muY
and standard deviations
\sigmaX
and
\sigmaY
is defined as
where
\operatorname{E}
is the expected value operator,
\operatorname{cov}
means covariance, and
\operatorname{corr}
is a widely used alternative notation for the correlation coefficient. The Pearson correlation is defined only if both standard deviations are finite and positive. An alternative formula purely in terms of moments is
### Symmetry property
The correlation coefficient is symmetric:
\operatorname{corr}(X,Y)=\operatorname{corr}(Y,X)
. This is verified by the commutative property of multiplication.
### Correlation and independence
(-1,1)
in all other cases, indicating the degree of linear dependence between the variables. As it approaches zero there is less of a relationship (closer to uncorrelated). The closer the coefficient is to either −1 or 1, the stronger the correlation between the variables.
If the variables are independent, Pearson's correlation coefficient is 0, but the converse is not true because the correlation coefficient detects only linear dependencies between two variables.
For example, suppose the random variable
X
is symmetrically distributed about zero, and
Y=X2
. Then
Y
is completely determined by
X
, so that
X
and
Y
are perfectly dependent, but their correlation is zero; they are uncorrelated. However, in the special case when
X
and
Y
are jointly normal, uncorrelatedness is equivalent to independence.
### Sample correlation coefficient
Given a series of
n
measurements of the pair
(Xi,Yi)
indexed by
i=1,\ldots,n
, the sample correlation coefficient can be used to estimate the population Pearson correlation
\rhoX,Y
between
X
and
Y
. The sample correlation coefficient is defined as
rxy\overset{\underset{def
where
\overline{x}
and
\overline{y}
are the sample means of
X
and
Y
, and
sx
and
sy
are the corrected sample standard deviations of
X
and
Y
.
Equivalent expressions for
rxy
are
\begin{align} rxy&=
\sumxiyi-n\bar{x \bar{y}}{n
s'xs'y}\\[5pt] &=
n\sumxiyi-\sumxi\sumyi
\sqrt{n\sum
2-(\sum x i
2 x i)
~\sqrt{n\sum
2-(\sum y i
2}}. \end{align} y i)
where
s'x
and
s'y
are the uncorrected sample standard deviations of
X
and
Y
.
If
x
and
y
are results of measurements that contain measurement error, the realistic limits on the correlation coefficient are not −1 to +1 but a smaller range.[6] For the case of a linear model with a single independent variable, the coefficient of determination (R squared) is the square of
rxy
, Pearson's product-moment coefficient.
## Example
Consider the joint probability distribution of
X
and
Y
given in the table below.
\operatorname{P}(X=x,Y=y) y=-1 y=0 y=1 x=0 0 1/3 0 x=1 1/3 0 1/3
For this joint distribution, the marginal distributions are:
\operatorname{P}(X=x)= \begin{cases} 1/3&forx=0\\ 2/3&forx=1\end{cases}
\operatorname{P}(Y=y)= \begin{cases} 1/3&fory=-1\\ 1/3&fory=0\\ 1/3&fory=1\end{cases}
This yields the following expectations and variances:
\muX=2/3
\muY=0
2 \sigma X
=2/9
2 \sigma Y
=2/3
Therefore:
\begin{align} \rhoX,Y&=
1 \sigmaX\sigmaY
\operatorname{E}[(X-\muX)(Y-\muY)]\\[5pt] &=
1 \sigmaX\sigmaY
\sumx,y{(x-\muX)(y-\muY)\operatorname{P}(X=x,Y=y)}\\[5pt] &=(1-2/3)(-1-0)
1 3
+(0-2/3)(0-0)
1 3
+(1-2/3)(1-0)
1 3
=0. \end{align}
## Rank correlation coefficients
See main article: Spearman's rank correlation coefficient and Kendall tau rank correlation coefficient.
Rank correlation coefficients, such as Spearman's rank correlation coefficient and Kendall's rank correlation coefficient (τ) measure the extent to which, as one variable increases, the other variable tends to increase, without requiring that increase to be represented by a linear relationship. If, as the one variable increases, the other decreases, the rank correlation coefficients will be negative. It is common to regard these rank correlation coefficients as alternatives to Pearson's coefficient, used either to reduce the amount of calculation or to make the coefficient less sensitive to non-normality in distributions. However, this view has little mathematical basis, as rank correlation coefficients measure a different type of relationship than the Pearson product-moment correlation coefficient, and are best seen as measures of a different type of association, rather than as alternative measure of the population correlation coefficient.[7] [8]
To illustrate the nature of rank correlation, and its difference from linear correlation, consider the following four pairs of numbers
(x,y)
:
(0, 1), (10, 100), (101, 500), (102, 2000).
As we go from each pair to the next pair
x
increases, and so does
y
. This relationship is perfect, in the sense that an increase in
x
is always accompanied by an increase in
y
. This means that we have a perfect rank correlation, and both Spearman's and Kendall's correlation coefficients are 1, whereas in this example Pearson product-moment correlation coefficient is 0.7544, indicating that the points are far from lying on a straight line. In the same way if
y
always decreases when
x
increases, the rank correlation coefficients will be −1, while the Pearson product-moment correlation coefficient may or may not be close to −1, depending on how close the points are to a straight line. Although in the extreme cases of perfect rank correlation the two coefficients are both equal (being both +1 or both −1), this is not generally the case, and so values of the two coefficients cannot meaningfully be compared.[7] For example, for the three pairs (1, 1) (2, 3) (3, 2) Spearman's coefficient is 1/2, while Kendall's coefficient is 1/3.
## Other measures of dependence among random variables
The information given by a correlation coefficient is not enough to define the dependence structure between random variables.[9] The correlation coefficient completely defines the dependence structure only in very particular cases, for example when the distribution is a multivariate normal distribution. (See diagram above.) In the case of elliptical distributions it characterizes the (hyper-)ellipses of equal density; however, it does not completely characterize the dependence structure (for example, a multivariate t-distribution's degrees of freedom determine the level of tail dependence).
Distance correlation[10] [11] was introduced to address the deficiency of Pearson's correlation that it can be zero for dependent random variables; zero distance correlation implies independence.
The Randomized Dependence Coefficient[12] is a computationally efficient, copula-based measure of dependence between multivariate random variables. RDC is invariant with respect to non-linear scalings of random variables, is capable of discovering a wide range of functional association patterns and takes value zero at independence.
For two binary variables, the odds ratio measures their dependence, and takes range non-negative numbers, possibly infinity: . Related statistics such as Yule's Y and Yule's Q normalize this to the correlation-like range . The odds ratio is generalized by the logistic model to model cases where the dependent variables are discrete and there may be one or more independent variables.
The correlation ratio, entropy-based mutual information, total correlation, dual total correlation and polychoric correlation are all also capable of detecting more general dependencies, as is consideration of the copula between them, while the coefficient of determination generalizes the correlation coefficient to multiple regression.
## Sensitivity to the data distribution
The degree of dependence between variables
X
and
Y
does not depend on the scale on which the variables are expressed. That is, if we are analyzing the relationship between
X
and
Y
, most correlation measures are unaffected by transforming
X
to a + bX and
Y
to c + dY, where a, b, c, and d are constants (b and d being positive). This is true of some correlation statistics as well as their population analogues. Some correlation statistics, such as the rank correlation coefficient, are also invariant to monotone transformations of the marginal distributions of
X
and/or
Y
.
Most correlation measures are sensitive to the manner in which
X
and
Y
are sampled. Dependencies tend to be stronger if viewed over a wider range of values. Thus, if we consider the correlation coefficient between the heights of fathers and their sons over all adult males, and compare it to the same correlation coefficient calculated when the fathers are selected to be between 165 cm and 170 cm in height, the correlation will be weaker in the latter case. Several techniques have been developed that attempt to correct for range restriction in one or both variables, and are commonly used in meta-analysis; the most common are Thorndike's case II and case III equations.[13]
Various correlation measures in use may be undefined for certain joint distributions of X and Y. For example, the Pearson correlation coefficient is defined in terms of moments, and hence will be undefined if the moments are undefined. Measures of dependence based on quantiles are always defined. Sample-based statistics intended to estimate population measures of dependence may or may not have desirable statistical properties such as being unbiased, or asymptotically consistent, based on the spatial structure of the population from which the data were sampled.
Sensitivity to the data distribution can be used to an advantage. For example, scaled correlation is designed to use the sensitivity to the range in order to pick out correlations between fast components of time series.[14] By reducing the range of values in a controlled manner, the correlations on long time scale are filtered out and only the correlations on short time scales are revealed.
## Correlation matrices
The correlation matrix of
n
random variables
X1,\ldots,Xn
is the
n x n
matrix whose
(i,j)
entry is
\operatorname{corr}(Xi,Xj)
. If the measures of correlation used are product-moment coefficients, the correlation matrix is the same as the covariance matrix of the standardized random variables
Xi/\sigma(Xi)
for
i=1,...,n
. This applies both to the matrix of population correlations (in which case
\sigma
is the population standard deviation), and to the matrix of sample correlations (in which case
\sigma
denotes the sample standard deviation). Consequently, each is necessarily a positive-semidefinite matrix. Moreover, the correlation matrix is strictly positive definite if no variable can have all its values exactly generated as a linear function of the values of the others.
The correlation matrix is symmetric because the correlation between
Xi
and
Xj
is the same as the correlation between
Xj
and
Xi
.
A correlation matrix appears, for example, in one formula for the coefficient of multiple determination, a measure of goodness of fit in multiple regression.
In statistical modelling, correlation matrices representing the relationships between variables are categorized into different correlation structures, which are distinguished by factors such as the number of parameters required to estimate them. For example, in an exchangeable correlation matrix, all pairs of variables are modelled as having the same correlation, so all non-diagonal elements of the matrix are equal to each other. On the other hand, an autoregressive matrix is often used when variables represent a time series, since correlations are likely to be greater when measurements are closer in time. Other examples include independent, unstructured, M-dependent, and Toeplitz.
## Uncorrelatedness and independence of stochastic processes
Similarly for two stochastic processes
\left\{Xt\right\}t\inl{T
} and
\left\{Yt\right\}t\inl{T
}: If they are independent, then they are uncorrelated.[15]
## Common misconceptions
### Correlation and causality
See also: Normally distributed and uncorrelated does not imply independent. The conventional dictum that "correlation does not imply causation" means that correlation cannot be used by itself to infer a causal relationship between the variables.[16] This dictum should not be taken to mean that correlations cannot indicate the potential existence of causal relations. However, the causes underlying the correlation, if any, may be indirect and unknown, and high correlations also overlap with identity relations (tautologies), where no causal process exists. Consequently, a correlation between two variables is not a sufficient condition to establish a causal relationship (in either direction).
A correlation between age and height in children is fairly causally transparent, but a correlation between mood and health in people is less so. Does improved mood lead to improved health, or does good health lead to good mood, or both? Or does some other factor underlie both? In other words, a correlation can be taken as evidence for a possible causal relationship, but cannot indicate what the causal relationship, if any, might be.
### Simple linear correlations
The Pearson correlation coefficient indicates the strength of a linear relationship between two variables, but its value generally does not completely characterize their relationship.[17] In particular, if the conditional mean of
Y
given
X
, denoted
\operatorname{E}(Y\midX)
, is not linear in
X
, the correlation coefficient will not fully determine the form of
\operatorname{E}(Y\midX)
.
The adjacent image shows scatter plots of Anscombe's quartet, a set of four different pairs of variables created by Francis Anscombe.[18] The four
y
variables have the same mean (7.5), variance (4.12), correlation (0.816) and regression line (y = 3 + 0.5x). However, as can be seen on the plots, the distribution of the variables is very different. The first one (top left) seems to be distributed normally, and corresponds to what one would expect when considering two variables correlated and following the assumption of normality. The second one (top right) is not distributed normally; while an obvious relationship between the two variables can be observed, it is not linear. In this case the Pearson correlation coefficient does not indicate that there is an exact functional relationship: only the extent to which that relationship can be approximated by a linear relationship. In the third case (bottom left), the linear relationship is perfect, except for one outlier which exerts enough influence to lower the correlation coefficient from 1 to 0.816. Finally, the fourth example (bottom right) shows another example when one outlier is enough to produce a high correlation coefficient, even though the relationship between the two variables is not linear.
These examples indicate that the correlation coefficient, as a summary statistic, cannot replace visual examination of the data. The examples are sometimes said to demonstrate that the Pearson correlation assumes that the data follow a normal distribution, but this is not correct.[4]
## Bivariate normal distribution
If a pair
(X,Y)
of random variables follows a bivariate normal distribution, the conditional mean
\operatorname{E}(X\midY)
is a linear function of
Y
, and the conditional mean
\operatorname{E}(Y\midX)
is a linear function of
X
. The correlation coefficient
\rhoX,Y
between
X
and
Y
, along with the marginal means and variances of
X
and
Y
, determines this linear relationship:
\operatorname{E}(Y\midX)=\operatorname{E}(Y)+\rhoX,Y
\sigma
Y X-\operatorname{E (X)}{\sigma
X},
where
\operatorname{E}(X)
and
\operatorname{E}(Y)
are the expected values of
X
and
Y
, respectively, and
\sigmaX
and
\sigmaY
are the standard deviations of
X
and
Y
, respectively.
• Book: Cohen, J. . Cohen P. . West, S.G. . Aiken, L.S. . yes . 2002 . Applied multiple regression/correlation analysis for the behavioral sciences . 3rd . Psychology Press . 978-0-8058-2223-6 .
• Book: Oestreicher. J. & D. R.. Plague of Equals: A science thriller of international disease, politics and drug discovery. February 26, 2015. Omega Cat Press. California. 978-0963175540. 408.
## Notes and References
1. Croxton, Frederick Emory; Cowden, Dudley Johnstone; Klein, Sidney (1968) Applied General Statistics, Pitman. (page 625)
2. Dietrich, Cornelius Frank (1991) Uncertainty, Calibration and Probability: The Statistics of Scientific and Industrial Measurement 2nd Edition, A. Higler. (Page 331)
3. Aitken, Alexander Craig (1957) Statistical Mathematics 8th Edition. Oliver & Boyd. (Page 95)
4. Rodgers . J. L. . Nicewander . W. A. . 1988 . Thirteen ways to look at the correlation coefficient . The American Statistician . 42 . 1. 59–66 . 2685263 . 10.1080/00031305.1988.10475524.
5. Dowdy, S. and Wearden, S. (1983). "Statistics for Research", Wiley. pp 230
6. Francis. DP. Coats AJ. Gibson D. How high can a correlation coefficient be?. Int J Cardiol. 1999. 69. 185–199. 10.1016/S0167-5273(99)00028-5. 2.
7. Yule, G.U and Kendall, M.G. (1950), "An Introduction to the Theory of Statistics", 14th Edition (5th Impression 1968). Charles Griffin & Co. pp 258–270
8. Kendall, M. G. (1955) "Rank Correlation Methods", Charles Griffin & Co.
9. Mahdavi Damghani B.. The Non-Misleading Value of Inferred Correlation: An Introduction to the Cointelation Model. Wilmott Magazine. 2013. 67. 50–61. 2013. 10.1002/wilm.10252 .
10. Székely . G. J. Rizzo . Bakirov . N. K. . 2007 . Measuring and testing independence by correlation of distances . . 35 . 6. 2769–2794 . 10.1214/009053607000000505 . 0803.4101 .
11. Székely . G. J. . Rizzo . M. L. . 2009 . Brownian distance covariance . Annals of Applied Statistics . 3 . 4. 1233–1303 . 10.1214/09-AOAS312 . 20574547 . 2889501 . 1010.0297 .
12. Lopez-Paz D. and Hennig P. and Schölkopf B. (2013). "The Randomized Dependence Coefficient", "Conference on Neural Information Processing Systems" Reprint
13. Book: Thorndike, Robert Ladd. Research problems and techniques (Report No. 3). 1947. US Govt. print. off.. Washington DC.
14. Nikolić . D . Muresan . RC . Feng . W . Singer . W . 2012 . Scaled correlation analysis: a better way to compute a cross-correlogram . European Journal of Neuroscience . 35. 5. 1–21 . 10.1111/j.1460-9568.2011.07987.x . 22324876 .
15. Book: Park, Kun Il. Fundamentals of Probability and Stochastic Processes with Applications to Communications. Springer . 2018 . 978-3-319-68074-3.
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# Closed Subspace of Lindelöf Space is Lindelöf Space
## Theorem
Let $T = \struct {S, \tau}$ be a Lindelöf space.
Let $C = \struct {H, \tau_H}$ be a subspace of $T$.
Let $C$ be closed in $T$.
Then $\struct {H, \tau}$ is Lindelöf space.
That is, the property of being Lindelöf is weakly hereditary.
## Proof
Let $T$ be a Lindelöf space.
Let $C$ be a closed subspace of $T$.
Let $\UU$ be an open cover of $H$.
We have that $H$ is closed in $T$.
It follows by definition of closed that $H \setminus C$ is open in $T$.
So if we add $S \setminus H$ to $\UU$, we see that $\UU \cup \set {S \setminus H}$ is also an open cover of $S$.
As $T$ is compact, there exists a countable subcover of $\UU \cup \set {S \setminus H}$, say $\VV = \set {U_1, U_2, \ldots}$.
This covers $H$ by the fact that it covers $S$.
Suppose $S \setminus H$ is an element of $\VV$.
Then $S \setminus H$ may be removed from $\VV$, and the rest of $\VV$ still covers $H$.
Thus we have a countable subcover of $\UU$ which covers $H$.
Hence $C$ is a Lindelöf space.
$\blacksquare$
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Competitions
# Tricky Sorting
The sequence of numbers is given. Arrange them in non-decreasing order of the last digit, and in the case of equality of last digits - arrange the numbers in non-decreasing order.
#### Input
The first line contains number n (1n100), and next lines contain the positive integers not greater than 32000.
#### Output
Print the sequence of numbers ordered as given in problem statement.
Time limit 1 second
Memory limit 128 MiB
Input example #1
7
12 15 43 13 20 1 15
Output example #1
20 1 12 13 43 15 15
Input example #2
2
1004
1002
Output example #2
1002 1004
|
By awoo, history, 17 months ago, translation,
Hello Codeforces!
Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.
This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.
You will be given 6 or 7 problems and 2 hours to solve them.
The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.
Good luck to all the participants!
Our friends at Harbour.Space also have a message for you:
Hey, Codeforces!
Once again, it is time for another exciting scholarship opportunity from Harbour.Space!
This time we have partnered with PhazeRo to open the door for an exciting career in technology for the most talented people in our network.
In partnership with PhazeRo, we are offering a full scholarship to study for a Master’s in Data Science at Harbour.Space while working as a Junior Data Scientist at PhazeRo!
Scholarship Requirements:
• Bachelor's Degree
• Professional fluency in English
• Proficiency with data mining, mathematics, and statistical analysis.
• Experience with Tableau, SQL, and programming languages (i.e., Python, R, Java)
Scholarship Highlights:
1. Study in Europe’s most exciting tech cities
2. Full tuition fee covered (€22,900)
3. Competitive compensation for the internship at PhazeRo (€700 / month)
4. Opportunity to join PhazeRo full-time after graduation
Some of the advantages of working at PhazeRo:
• Possibility of a job upon graduation
• Immerse into an International Company
• Diversity Program
• Professional Development
• Be part of a company that is building the region's largest engineering team
We have previously partnered with other companies like OneRagtime, Hansgrohe, Coherra, and Remy Robotics to empower young talents around the world and help them boost their tech career.
We are always happy to see Codeforces members join the Harbour.Space family. Apply now to get a chance to learn from the best in the field and kickstart your career!
Keep in touch and follow us on LinkedIn for more scholarship opportunities. And follow us on Instagram to evidence student life, events, and success stories from our apprenticeship program students.
Good luck on your round, and see you next time!
Harbour.Space University
Congratulations to the winners:
Rank Competitor Problems Solved Penalty
1 neal 6 186
2 vepifanov 6 188
3 Um_nik 6 251
4 Farhod_Farmon 5 65
5 noimi 5 76
Congratulations to the best hackers:
Rank Competitor Hack Count
1 __ZMF__ 50:-50
2 mufeng.wei 24:-4
3 SSerxhs 22:-9
4 mahesh_dubey 13:-1
5 haminh0307 11:-5
463 successful hacks and 1684 unsuccessful hacks were made in total!
And finally people who were the first to solve each problem:
Problem Competitor Penalty
A Geothermal 0:01
B turmax 0:02
C eecs 0:04
D abc864197532 0:04
E noimi 0:28
F rainboy 1:16
UPD: Editorial is out
• +279
» 17 months ago, # | ← Rev. 2 → -34 Finally a contest after long break. Well, break is needed to set some things right. Happy coding.
» 17 months ago, # | +12 I hope I get that scolarship ...
» 17 months ago, # | ← Rev. 3 → +56 I think preliminary rating change in educational rounds will be great.
• » » 17 months ago, # ^ | -108 Wow! such big brain! and after every hack, rating changes of all 25,000 participants will be recalculated! No problem sir! I will put the word through to Mike
• » » » 17 months ago, # ^ | ← Rev. 2 → +91 Uhm no? Just do it twice. Once after the contest ends, and the second one when the hacking phase ends. Stop trying to act smart you idiot.PS — We have rating predictor extensions so prelim rating changes are not really needed.
• » » » » 17 months ago, # ^ | 0 It would be helpful if you could share the extension you find good.
• » » » » » 17 months ago, # ^ | +14 Search cf-predictor on your favorite search engine.
• » » » » » » 17 months ago, # ^ | 0 thanks.
• » » » » » » 17 months ago, # ^ | -13 Why do you suppose that all existed search engines will redirect him to the same exact extension you have just mentioned?
• » » » » 17 months ago, # ^ | 0 now, from this case, I'm not sure that having predictor extension with such difference is enough :)
» 17 months ago, # | +6 9th rated contest of the month. Hail Codeforces!
» 17 months ago, # | +26 A long break bruh, Codechef also postponed the Lunchtime.
• » » 17 months ago, # ^ | 0 Lunchtime is Tomorrow. And Amazon is hiring via Lunchtime.
• » » » 17 months ago, # ^ | +5 Did anyone ever recieved any message from companies hiring through codechef ?
• » » » » 17 months ago, # ^ | +6 Yup I did recieved ... and one of my friend landed an interview too
• » » » » » 17 months ago, # ^ | +5 Explain. What was your rank in cookoff and April long. what about your friend
• » » » » » » 17 months ago, # ^ | ← Rev. 2 → +1 Not in april ...through jan;s long chanllege . We both were under 200 , I was in div.1 and he was in div 2
• » » » » » 17 months ago, # ^ | +2 Is this from cookoff or long challenge? because you know what long challenge means :)
• » » » » » » 17 months ago, # ^ | -6 You can get selected from long too ... if there is no plag in your codes. Codechef do check for plag you know.
• » » » » » » » 17 months ago, # ^ | +1 and the hiring is internship or full time?
• » » » » » » » » 17 months ago, # ^ | 0 They were hiring for internship n full time ... Time time I guess it is for coders with experience
• » » » » » » » 17 months ago, # ^ | +9 Lol Came for editorials, But stayed for the joker.This is his solution https://www.codechef.com/viewsolution/43852245this are other coders solutions https://www.codechef.com/viewsolution/43852488 https://www.codechef.com/viewsolution/43852283see how there all there codes contain roast, lund and other stuff.Seems like u got interviews even after cheatingPlease Flag this person People. I am 100% sure he has bought the solution over telegram. The user is arpit0891 https://www.codechef.com/viewsolution/43852283and i also have his linkedin.Please flag this user. he might have cheated in codeforces too
• » » » » » » » » 17 months ago, # ^ | 0 Actually It was me who took his solutions and sold it on telegram . He gave me his password for some work.
• » » » » » » » » » 17 months ago, # ^ | +4 Nope. You can check submission time on codechef. He submitted 40 minutes before the contest expired. Also i know this is a fake profile vj12. so do that someplace else. You sure have a lot of time arpit0891 to do this fake stuff and upvote your own comment from different profile.But assuming you are genuine. This is still bad. Indulging in cheating is as bad as letting others cheat. Thanks for making it more easy for us. Regards
• » » » » » » » » » 17 months ago, # ^ | 0 After this leak I stopped using that account . I made a newone and participate though that only .
• » » » » » » » » » 17 months ago, # ^ | 0 That happened in January 2021,Feb 2021 and then March 2021, I hope all are leaks else they will be cheating.......
• » » » » » » » » » 17 months ago, # ^ | 0 The account is closed as of now .
• » » » » » 17 months ago, # ^ | +1 And which division ? Like i can participate in div3 and get a lot better rank than div1
• » » » » » » 17 months ago, # ^ | -6 You can't compete between divisions
• » » » » » » 17 months ago, # ^ | +3 He is in the cheating division
• » » » 17 months ago, # ^ | 0 I don't think we stand a chance though. cheaters will get interviews and you will get nothing. no reject mails even.so please take your time, read what i have written (https://discuss.codechef.com/t/are-cheaters-getting-interviews/88769/6) and give a f***k . cause the whole of admin team doesn't seem to give one
» 17 months ago, # | 0 Another edu heh? Great.
» 17 months ago, # | 0 i am very excited to participate the contest,wooooooo~
» 17 months ago, # | +6 Almost every edu round I was "educated", but I still participate in every round...
• » » 17 months ago, # ^ | 0 your spirit deserves me to learn
» 17 months ago, # | +3 "The penalty for each incorrect submission until the submission with a full solution is 10 minutes"Does this mean that only time will be taken and there wont be a -50 or -100?
• » » 17 months ago, # ^ | 0 yes
» 17 months ago, # | +9 Today codeforces educational round. Tomorrow codechef lunchtime. Day after Tomorrow CodeJam Round 1C. Practice & Compete & Improve.
• » » 17 months ago, # ^ | +2 waoooo~
• » » 17 months ago, # ^ | +1 then day after global
» 17 months ago, # | +4 hope i remain expert after this contest :)
• » » 17 months ago, # ^ | +2 It'll be alright, Ron!
• » » 17 months ago, # ^ | +2 me too,good luck bro
• » » 17 months ago, # ^ | +1 You say this in every contest.
» 17 months ago, # | +36 Time to start the vacations with an educational round. Excited!
• » » 17 months ago, # ^ | +26 wish to have a good vacation
» 17 months ago, # | +12 I wish B/C not to be a observational shit like printing the array upto k and then upto n-k lol.
» 17 months ago, # | +18 Hakcers will try to hack the solution in 12 hrs, while cheat busters will try to match the solutions in 12hrs.......
» 17 months ago, # | +6 i hope i do best in the contest and happy coding to everyone
» 17 months ago, # | +2 Have a great rating change
» 17 months ago, # | 0 Nice contest!
» 17 months ago, # | ← Rev. 3 → +122 Today's Problem Difficulty. Pls forgive me for not enlarging the image as I dont know how to do that.
• » » 17 months ago, # ^ | -27 B and D is almost same difficulty level though
• » » » 17 months ago, # ^ | -10 B
• » » » » 17 months ago, # ^ | +16 C took me an hour... D took me 10 mins...
• » » » » » 17 months ago, # ^ | ← Rev. 3 → -7 How did you do it? I did similar to longest palindromic substring. fixed pivot for both odd length subarray and even length subarray and then extended both sides. Time complexity: n^2
• » » » » » » 17 months ago, # ^ | ← Rev. 5 → 0 Basically, yeah, what you did.For each position in the array i I tried generating an even length subarray and uneven length subarray that begins with that position as the middle or one of the middle elements. At each iteration add one more in each end so that the new subarray is continuous. Notice that moving through them like this maintains the position of the ones inside, so you just have to see what difference the new ends would make. O(n^2) time, O(n) memory.
• » » » » » » » 17 months ago, # ^ | 0 Is this a standard way of traversing substrings? or is this question is similar to any standard problem?
» 17 months ago, # | -81 sometimes unexpected bruteforce works . Today it worked for problem C for me
• » » 17 months ago, # ^ | +15 Can you wait till the end and not spoil solutions?
• » » 17 months ago, # ^ | +2 Its not "unexpected" since its not hard to figure out why it doesn't tle
• » » » 17 months ago, # ^ | ← Rev. 2 → 0 yes and I coded only after I got n*root(n) bound for it :(. If we do only, take contribution from 1 to size of students in university for Ks. As, if the size of such universities is more than root(n), then their number will be less than root(n), so n*root(n) here and if the size of universities is less than root(n), then O(n) such universities can be there but k:1 to root(n) for such, so again n*root(n). So, overall complexity is n*root(n). The complexity might be less than this, but this was enough.
• » » » 17 months ago, # ^ | +6 He did TLE after the hacking phase with the additional tests
• » » 17 months ago, # ^ | 0 i too did bruteforce in C. Got TLE. can you tell me where I went wrong? I can't think of any more optimisation. :/https://codeforces.com/contest/1519/submission/114607388link to my submission.
• » » » 17 months ago, # ^ | ← Rev. 2 → 0 Your brute force solution works in O(n^2) time as you try every n university for every possible k (which k==n), it won't pass since n^2 = 10^10 ~ 100 seconds. But if you observe the problem, you will see that after the number of people in university > k, you don't need to check it again as the number of teams will be 0. You can somehow remove that university from your checking list (effectively). This will improve the solution a lot. My submission: https://codeforces.com/contest/1519/submission/114578952
• » » » » 17 months ago, # ^ | 0 Thanks a lot!
• » » » » 17 months ago, # ^ | 0 But why doesn't n^2 work when time limit is 2 seconds. I have encountered this situation in some other problems as well. Sometimes, n^2 works for 2 seconds sometimes it doesn't. Is it because of the heavy calculations ?
• » » » » » 17 months ago, # ^ | +1 Whether n^2 works in 2 seconds or not depends upon the size of n. Although I would try to give a rough idea that what kind of solution can run in 2 seconds. In one seconds a normal loop would run near about 10^8 times in c++. It is only a rough estimation. In C the value of n was 2 * 10^5. If we square it, the result goes to 4 * 10^10. To run this solution at least, 400 seconds are required (again a rough estimation). So it definitly won't run in 2 seconds. Your assumptions that sometimes an n^2 solution runs in two seconds would be right when n is small. For n valued this large, an n^2 solution would never run, if the test cases are good.
• » » » 17 months ago, # ^ | 0 Probably You are getting TLE due to a very little mistake. Do check it out here. Link to Explanation
» 17 months ago, # | +48 Speedforces
• » » 17 months ago, # ^ | 0 true lol
» 17 months ago, # | +18 Does anyone with O(2*n^2) space got MLE in D?
• » » 17 months ago, # ^ | ← Rev. 2 → +8 You will get MLE because final answer can be equal to 5*10^17 which needs to be stored in long long int. 2*5000*5000*8 contiguous allocation is not possible (In case you used DP).
• » » » 17 months ago, # ^ | 0 YA right but my O(N^2) python also getting MLE in 11 , I wonder why?
• » » » » 17 months ago, # ^ | ← Rev. 3 → 0 Because Test 11 forces python into long arithmetic. If you will different algorithm to get rid of O(n**2) memory allocation you'll get TLE on 11 test. Then you'll need to do a little math, to get rid of excess of multiplication in your formula. Python often turn out to be pain here on codeforces :)
• » » » 17 months ago, # ^ | 0 Same happened with me. MLE ACcost me atleat 400 rank.
» 17 months ago, # | +45 Probably there's like around 5 people right now writing "SPEEEED FORCES"(joking of course:)). I think this problem could have been avoided if E was more "div2-solvable", I mean around 60 people solved it and a good amount of them are red coders. There's no point I think in making A-D so standard(especially D) and E that hard.
• » » 17 months ago, # ^ | +25 Pretty crazy variance. Solving A through D could either get you 60th place (with an implied rating of about 2250 according to cf-rating-predictor, which is orange), or as low as 2400th place (with an implied rating of about 1640, which is blue).
» 17 months ago, # | +75 Really liked $E$. I wonder if everyone solved it the way I did. My SolutionLet's model the problem as a graph problem. In this graph, each line in the $2D$ plane passing through the origin represents a node. Now, for every point $P(x, y)$, it can either move to $P_{1}(x+1, y)$ or $P_{2}(x, y+1)$. Say the lines passing from the origin to these two points are $L_1$ and $L_2$ respectively. If you assume the point $P$ moves to $P_1$ you can add an edge from $L_2$ to $L_1$ in our graph otherwise if the point $P$ moves to $P_2$, you add an edge from $L_1$ to $L_2$ in the graph. The point can also not move at all, but we will ignore this case (it will be clear why it won't matter).Now, observe that, for each line $L$, we can pick the edges directed to $L$ and pair them (these edge-pairs will form point-pairs in our original problem, since each point corresponds to exactly one edge). Note that if in-degree of $L$ is odd, then one incoming edge will be left without a pair. It's clear that we have to direct each edge in such a way that minimum number of nodes are left with odd in-degree.Here comes the idea for the optimal solution: Initially, direct each edge arbitrary. While there are two nodes $L_1$ and $L_2$ having odd degree, such that they are in the same connected component (considering undirected version of the graph), consider any path between them and reverse all its edges.Why is this optimal? How will you simulate the whole thing fast?
• » » 17 months ago, # ^ | +19 Interesting. I just used a known algorithm, which is given a graph cover it with paths of length 2 using a DFS. Seems most other people did it this way as well.
• » » » 17 months ago, # ^ | +28 Can you describe said algorithm? I wasn't aware of it's existence.
• » » » » 17 months ago, # ^ | +22
• » » » » 17 months ago, # ^ | ← Rev. 3 → +6 If we consider the points as vertices and connect two vertices iff they'll lie on the same line passing through origin after shifting them. Then this problem will reduce to finding maximum matching in the graph, right? But that is too general. As in our graph let's say a is connected to b,c,d then a is a part of at most 2 cliques which is ({a,b,c,d}) or ({a,b},{a,c,d}) , or, ... etc . So by considering a line as a vertex, we have somehow used this property, right?
• » » » » » 17 months ago, # ^ | ← Rev. 2 → 0 Hey Zura,Theanks for pointing me to the maximum matching problem . That is exactly what should be solved in the second interpretation!Im not sure about your other comments though. Which property do you mean? I feel like we lose properties by looking at the linegraph, because the problem can't be solved with a simple DFS anymore. Maybe it's good to look at an example? e.g. {(1,2),(2,3),(3,4),(4,5),(5,6),(1,6)} (The first 5 points all can be moved on the main diagonal, and the 1st and 6th point can be moved on a common line) would have those two graphs in the corresponding interpretations: The possible solutions are the same in both cases. Could you explain again what you meant, possibly with the example (or an own example)?
• » » » » » » 17 months ago, # ^ | +3 1st and 6th can't be moved on the same line as the first ones' slope would be (2+1)/1=3 and for 6th it is (1+1)/6=1/3.Other than that as I was pointing out 1,..,5 forms a clique.If 6th point were (2,5) and 7th were (3,8) then 1 is contained in two cliques 1,..,5 and 1,6,7. So by considering when we use a line as vertex we are basically accounting for this whole clique. I am unable to understand any further than that though, so waiting for the tutorial.
• » » » » » » » 17 months ago, # ^ | 0 Oops, i meant (1,6) instead of (6,1). Fixed it in my post.
• » » » » 17 months ago, # ^ | +3 Why switch if you reduced problem to kuhn-munkres algorithm? You mean that you could do better with dp than O(n*m)?
• » » » » » 17 months ago, # ^ | ← Rev. 3 → +3 No, The property that I stated in the previous comment, at most 2 cliques, will help us as we have not utilized all the components of the problem. I don't know how to model this property, but these additional bits help us, and hence we can do better than Blossom's as that is for general graphs.Blossom's is for general graphs.
• » » » » » 17 months ago, # ^ | +3 Could you elaborate? Do you mean kuhn-munkres ? This sounds to be an algorithm for bipartite graphs, but the second interpretation doesn't yield a bipartite graph.
• » » » » » » 17 months ago, # ^ | +3 Yep, you are right. My mistake.
• » » » » » » » 17 months ago, # ^ | +3 You can use the Blossom Algorithm though with $O(V^2E)$ or $O(\sqrt{V}E)$ with "the much more complicated algorithm of Micali and Vazirani".Guess we should stay with the first interpretation, or somehow use the information that the graph in the second interpretation is a line graph of some other graph (of the first interpretation).
• » » » » » » » » 17 months ago, # ^ | ← Rev. 3 → 0 Couldn't get it when I read it first time, (an0nym0us_m0use, czhang2718) but now I think I understand :) Looks like it will really work greedily with every connected component as stated above: While doing DFS on a graph until you met visited vertex or deadend-vertex, then you come out of current vertex and connect all unconnected pairs of children (those that were left unconnected when you left them). And if there 1 left — you connect it to current vertex, else — you return in to it's parent as "unconnected" yet. Then parent repeat the process. And the number of unconnected points in answer will be the number of even-numbered-components.And that will work for any of two graphs. It's nice that they delayed editorial :)
• » » » » » » » » » 17 months ago, # ^ | +3 Yes exactly that! I also did my submission 114692979 that way and it got accepted right away. The DFS-logic is in the "dfs" method and does exactly as you explain. :)But what do you mean with "any of two graphs"?
• » » 17 months ago, # ^ | +17
» 17 months ago, # | +6 B destroyed my confidence.
» 17 months ago, # | +3 Could we solve D with $n$ times FFT? My $O(n^2 \log n)$ solution can't pass this TL :(
• » » 17 months ago, # ^ | +3 Numbers were very large, how will you apply FFT?
• » » » 17 months ago, # ^ | +3 Just brute-force points of the reversed subarray $[L,R]$. For each $R \in [1,n]$, we calculate $C = A[0..R-1] * B[0..R-1]$, and $*$ presents convolution. Then $C[i-1]$ is the value of reversed subarray $[i - R + 1,R-1]$, it's easy to calculate the max contribution of all reversed subarray ended at $R-1$. The single round's time complexity is $O(n\log n)$ and the total is $O(n^2\log n)$. You can see my code here: 114607269
• » » » » 17 months ago, # ^ | +3 Yes, you are right. I wanted to say that FFT will have precision issues while handling large numbers. The sum can go upto $10^{19}$ for subarrays. Such large values cannot be handled.
• » » » » » 17 months ago, # ^ | 0 I found that issue later xD. At first I even didn't consider precision,and long double has no hope to squeeze in this TL.
• » » 17 months ago, # ^ | +8 FFT has a large constant factor. But also, for $n=5000$ I think it should be quite difficult to get any $n^2 \log{n}$ solution to pass, much less FFT.
• » » 17 months ago, # ^ | +6 my solution didn't pass also. 114602456 . looking for a AC solution with FFT. or its impossible with such time limit !!
» 17 months ago, # | 0 how to solve D
• » » 17 months ago, # ^ | +3 I solved this problem using DP.Here is the link of my submission: https://codeforces.com/contest/1519/submission/114578483
• » » » 17 months ago, # ^ | 0 Can u please explain the logic behind the DP solution?
• » » » » 17 months ago, # ^ | 0 Let dp[l][r] be the sum of ai*bi ( i is between l and r) when we reverse subsegment [l,r].Sum of ai*bi(i is between l+1 and r-1) when we reverse subsegment [l,r] is same when we reverse subsegment [l+1,r-1]. Then dp[l][r]=a[l]*b[r]+a[r]*b[l]+dp[l+1][r-1]; Pref[i] is sum of ai*bi(i is between 1 and i),suff[i] is sum of ai*bi(i is between i and n). Than,we check if we reverse subsegment [l,r] than result is : pref[i-1]+dp[l][r]+suff[r+1]; Maximum of all of these results is our final answer.
• » » 17 months ago, # ^ | 0 I have solved in O(n^2) time using DP. Video Explanation and Solutions
• » » 17 months ago, # ^ | 0 D is similar to finding the longest palindromic substring, suppose you want to find ans why reversing index i to j, so instead of calculating whole i to j we can make use of ans of i+1,j-1 and add it to ans of end points(i,j).for mode detail here: my soln :https://codeforces.com/contest/1519/submission/114607034 here's the link of classic problem which I think my soln is based upon:https://www.geeksforgeeks.org/longest-palindrome-substring-set-1/
» 17 months ago, # | +18 In E, I was able to group together the points that can be taken together in one move. But, how to make pairs? is it some sort of matching problem?
• » » 17 months ago, # ^ | +3 I'm not sure what intended is, but we can do tree DP: think of each point as connecting two slopes. Then, create the DFS tree from this graph. Now, the question is what's the maximum number of 2-length paths we can partition this graph into. Then, for each subtree, let's greedily partition it. Let dp[x]=whether we need to reserve the edge going to x's parent. Then we can greedily match x's immediate children with DP value 1, and back edges leading up to x. dp[x]=[whether the count of such nodes is odd]. Now, we can easily obtain our answer by keeping track of which edges refer to which original points. This is optimal because the number of extra/unused edges at the end is 0 or 1.
• » » » 17 months ago, # ^ | 0 This graph doesn't look like a tree.
• » » » » 17 months ago, # ^ | 0 Yeah it isn't necessarily a tree- we construct a DFS tree out of it.
• » » » » » 17 months ago, # ^ | 0 Ah, got it.
» 17 months ago, # | +22 D was so easy compared to C
• » » 17 months ago, # ^ | 0 can you please explain your approach
• » » » 17 months ago, # ^ | ← Rev. 2 → +5 There are n^2 subarrays, get answer for all of them and print the maxProbably the best way to iterate over them is by iterating all possible centersUse the fact that you can calculate answer for A[0:10000] for O(1) if you know the answer for A[1:9999]
• » » 17 months ago, # ^ | +7 Pls don't make me feel stupid :(
• » » » 17 months ago, # ^ | 0 Maybe you need to work on your dynamic programming skill a bit? I personally just thought dynamic programming initially and then realised I can do without the extra memory.
• » » » » 17 months ago, # ^ | 0 Can you tell me how to do D without extra memory?
• » » » » » 17 months ago, # ^ | 0 In D, i think the best solution is iterating over centers of subarrays and calculating the difference between new and old values. It takes O(n^2) time and O(n) space complexity (just a and b arrays). My submission is here: 114594546
• » » 17 months ago, # ^ | +4 D is simple implementation once you see how the dp works. But getting that idea is not so simple.C on the other hand is the opposite. The idea is simple to see, but the implementation has some pitfalls.
• » » » 17 months ago, # ^ | +2 The opposite for meTook me quite a while to get the idea what to do in C to avoid time limitAnd instant idea in D
» 17 months ago, # | 0 Nice problems
» 17 months ago, # | +5 Is Question E a MCBM problem? I was thinking of modelling both of the possible moves as a fraction x/y then try to do matching between points with the one of the same fractions. However, this is potentially O(n^2) so I am not sure if there is a way to optimize it.
» 17 months ago, # | 0 memory limit was too tight in problem D
• » » 17 months ago, # ^ | +5 Am I the only one who did it in O(n) space?
• » » » 17 months ago, # ^ | +29 no
• » » » 17 months ago, # ^ | 0 Yeah, you can do it O(n) space and O(n^2) time complexity. Iterate over all points as pivots and iterate over length of subarray to reverse. Then do the same for all gaps i.e. even length subarrays.
• » » 17 months ago, # ^ | +6 No additional memory other than the original array was needed actually.
• » » » 17 months ago, # ^ | +3 well , my O(n^2 * 2) DP didn't pass , but I realized after the contest was over that I can change it to only n^2 space , and i coded it and got AC. welp , thats my rating going.
• » » » 17 months ago, # ^ | ← Rev. 2 → 0 could you please elaborate on how to do this. I used two addition arrays of size $n$ to store answer for subarrays of size $i-2$ and $i-1$. This is because we need to have answer for subarray of size $i-2$ if we are going to calculate answer for subarray of size $i$.
• » » » » 17 months ago, # ^ | +3 If you already understood the solution, take a look at my submission 114616116.
» 17 months ago, # | +16
» 17 months ago, # | ← Rev. 2 → 0 Any cool observation for C? Couldn't get it faster than O(n^2*log(n)) 114611449
• » » 17 months ago, # ^ | 0 If you observe carefully that was not O(n^2logn) but O(n log n). See there were only n students so they might get divided into n groups but total student will remain n only. You can relate it to graph ques where we mark a note visited and when we come again we found at that node is already visited. We might come again and again to a same node but overall time complexity remains O(n) only. Same way, we might iterate all the n groups but total student remain n only.
» 17 months ago, # | 0 I used n^2 Space to solve D but it exceeded memory limit. I still wonder how it is possible. My Code#include #include #include #include #include #include #include #define priority_queue < ll, std::vector, std::greater > mnheap; // mnheap.push(), mnheap.top(), mnheap.pop() #define REP(i,a,b) for (auto i = a; i != b; i++) #define ll long long int #define ld long double #define vi vector #define vll vector #define vvi vector < vi > #define pii pair #define pll pair #define mod 1000000007 #define inf 1000000000000000001 #define all(c) c.begin(),c.end() #define rall(c) c.rbegin(),c.rend() #define mp(x,y) make_pair(x,y) #define mem(a,val) memset(a,val,sizeof(a)) #define eb emplace_back #define f first #define s second #define pb push_back #define SQ(a) (a)*(a) using namespace std; void read(int n,vector& x) { x.clear(); x.resize(n); for(int i = 0;i>x[i]; } } void read(int n,int m,vector>& x) { x.clear(); x.resize(n,vector(m)); for(int i = 0;i>x[i][j]; } } void read(int n,vector>& x) { x.clear(); x.resize(n+1); for(int i = 0;i>a>>b; x[a].pb(b); x[b].pb(a); } } void read(int n,vector>& x,int m) { x.clear(); x.resize(n+1); for(int i = 0;i>a>>b; x[a].pb(b); x[b].pb(a); } } void read(int n,vector& x) { x.clear(); x.resize(n); for(int i = 0;i>x[i]; } } int main() { std::ios::sync_with_stdio(false); int T = 1; // cin>>T; for(int t = 1;t<=T;t++) { // cout<<"Case #"<>n; vector a(n); vector b(n); read(n,a); read(n,b); vector> mul(2*n); // vector> pref(2*n); vector pref1(n); for(int i = 0;i
• » » 17 months ago, # ^ | 0 try to replace the vector with an array
• » » 17 months ago, # ^ | 0 In D, any n^2 memory takes 200 MB in worst case. So even if you had 2 such types of memory, it will MLE.
• » » 17 months ago, # ^ | 0 My O(n^2) Space and time dp passes all TC. Video explanation
• » » » 17 months ago, # ^ | 0 I will be thankful if someone makes video editorial of E. Why everyone posts editorial of problems solved by 1000's of people.
• » » » » 17 months ago, # ^ | 0 Actually, You can see I 've just posted for only Problem D for this contest, and I'm trying to post for video explanation of problem E too asap.
• » » » » 17 months ago, # ^ | ← Rev. 2 → 0 Because there 8000 people interested in A and B, 15000-19000 in C and D, and only then there 2000 who solved D and interested in E, and 50 interested in F :)
» 17 months ago, # | 0 Hey, need a lil help here :3 My JAVA submission for B gives Runtime Error in test 1 verdict, but it seems to work fine on my system. I really can't figure out what's wrong. Would be grateful if anyone can help!Here's the submission : 114586245
» 17 months ago, # | ← Rev. 4 → 0 How to solve Problem C. I ran into TLE. This is my solution using Heap. I also tried using sorting the vectors, which too ran into TLE. Kindly help
• » » 17 months ago, # ^ | 0
• » » » 17 months ago, # ^ | +1 Prefix sum! I couldn't come up with this approach. Thanks btw.
» 17 months ago, # | 0 Hi, Can someone help me with my submission for problem C . I am getting a run time error for the first test-case which is already given but running that test-case on my laptop is not giving any error and the output is also correct . The link to my submission My_Submission. Thanks in advance.
• » » 17 months ago, # ^ | 0
» 17 months ago, # | ← Rev. 2 → +84 Problem D was super not Python friendly. This is the shit I did to get accepted 114611359. The reason for the TLEs in Python is basically that PyPy2 and PyPy3 is only 32 bit on Windows. So on Problem D you are forced to use big integers everywhere, which is super super slow. However in the latest update of PyPy (version 7.3.4), PyPy switched to supporting 64 bit on Windows.MikeMirzayanov Would it be possible to update the PyPy version (to version 7.3.4)? All of these issues with big integers would go away, and it would make PyPy a lot more useable and beginner friendly.
• » » 17 months ago, # ^ | +9 Maybe post this as a different blog, that will get it noticed faster.
• » » » 17 months ago, # ^ | +14 Good idea! I just made a blog about it https://codeforces.com/blog/entry/90184
• » » 17 months ago, # ^ | +3 For Python, others did something like ans+=(a[left]-a[right])*(b[right]-b[left]) to get acceptedhttps://codeforces.com/contest/1519/submission/114576707
» 17 months ago, # | 0 In question C, the description was ambiguous, there were not exactly 'n' universities always, rather nothing should've been mentioned about universities!:)
• » » 17 months ago, # ^ | +5 Why not? They never said each university had a competetive programmer. It is possible for a university to have no competetive programmers.
• » » » 17 months ago, # ^ | +1 Ohh..I got it now....thanks for clarifying.....I forgot this statement of yours!:)
» 17 months ago, # | -25 Hello everyone, this was my first ever contest on any coding platform , and I was able to solve just 4 questions and got around 1500 rank. Can anyone tell me whether it is considered good performance or not? Also , what should I do in order to improve? Thanks a lot.
• » » 17 months ago, # ^ | 0 Keep giving contests
» 17 months ago, # | -11 Why was B so easy? Usually B is twice as good as A(even for an edu round). We can basically induct on $m+n$ and B is solved.
• » » 17 months ago, # ^ | +12 That's once you guess what the answer is. And then you can defer the proof until after the contest.
» 17 months ago, # | -14 If i solved 4 problems, should i solve 5th problem . For improvement i should solve one problem more than what i solved during contest but mostly red coders have solve problem E and my real rating is well below red so i might spend 1 to 2 days up-solving it.
• » » 17 months ago, # ^ | 0 why downvote ? Just asking for suggestion :( You can downvote this one also but please tell
» 17 months ago, # | 0 https://codeforces.com/contest/1519/submission/114599717 how can i optimize it?
• » » 17 months ago, # ^ | 0 store indices where ans[j].size()>0 and then run loop on those indices only
» 17 months ago, # | 0 Question regarding Problem B: How can we prove that no matter which path we take it will always result in the same cost of N * M — 1 ?
• » » 17 months ago, # ^ | ← Rev. 2 → +3 We can prove by induction that regardless of the path you chose, the cost at any point (x,y) along that path will be x*y-1.
• » » 17 months ago, # ^ | ← Rev. 2 → +14 The path can be described by a sequence of R (right) and D (down). It's easy to prove that swapping a R and a D doesn't change the answer. You can reach any sequence, starting from a fixed sequence (e. g. RR...RRDD...DD), with a finite number of swaps.
• » » 17 months ago, # ^ | ← Rev. 2 → +8 You can visualize it by realizing that at each point $(x,y)$ of the path the cost is equal to the area of the rectangle with a diagonal $(1,1) -(x,y)$ with cell $(1,1)$ removed. When you move right you add a column to the rectangle, and when you move down you add a row.
• » » 17 months ago, # ^ | +26 You can also prove it like this:Consider the conservative field $\vec{E}=y\hat{i} + x\hat{j}$. Observe that the value required is just the work function. As potential function is $U = xy$, work done will always be $nm - 1$.
• » » » 17 months ago, # ^ | 0 A stupid question — are you applying continuous calculus to a discrete problem?
• » » » » 17 months ago, # ^ | ← Rev. 2 → 0 Yes. I proved that work done will be same regardless of how you move the particle from $(1, 1)$ to $(n, m)$. The problem just asks to consider specific ways to displace the particle, using only the vectors $(1, 0)$ and $(0, 1)$.
• » » » » » 17 months ago, # ^ | 0 Ah, got it. We are just restricted to moving along a broken line with horizontal and vertical segments.
» 17 months ago, # | 0 Can someone please explain the proof why for problem B, the answer is k==(n*m-1)? I solved with DP and got surprised to see it could be formulated.
• » » 17 months ago, # ^ | +17 Consider any RD from (x, y) to (x+1, y+1), you can see that changing the sequence to DR does not change the cost, and thus any sequence will have the same cost.
• » » 17 months ago, # ^ | 0 Exactly n*m-1 burles needed, for all possible ways. That's why if k is n*m-1 then k is valid. Otherwise k can not be obtained.
• » » 17 months ago, # ^ | +4 Prove by induction. You can assume the proposition true for m×n grid , and establish for (m+1)×n grid.
• » » » 17 months ago, # ^ | 0 Can you provide the inductive proof please?
• » » » » 17 months ago, # ^ | ← Rev. 3 → 0 Inductive hypothesis: path value is x×y-1 for any grid such that x <= m && y <= n. We try to prove the same for (m+1)×n grid.Suppose you are at m×j cell and take downturn to (m+1)th row.You need (n-j) more right turns to reach (m+1)×n cell.Total : (m×j-1) + j + (m+1)×(n-j) = (m+1)×n-1 , which was to be proved.
• » » 17 months ago, # ^ | ← Rev. 2 → +3 I did some few examples on paper and found out that the answer is unique, for every n, m. I couldn't find the formula so just calculated it step by step and checked the cost if it equals k.
• » » » 17 months ago, # ^ | 0 I calculated the way going through two opposite corners (algebra), got that both were $n \cdot m - 1$, and then didn't bother trying to prove it.
• » » 17 months ago, # ^ | ← Rev. 3 → +10 Proof for problem B that answer is k==(n*m-1). ProofThere are total (n-1)+(m-1) moves -(1) move from ith row to (i+1)th row (1<=i
• » » 17 months ago, # ^ | ← Rev. 3 → +56 Another possible proof:Instead of it costing $x$ or $y$ burles, say that it costs $x+y$ burles and then you get $y$ or $x$ burles back.In each move, $x+y$ increases by 1 so the total sum of $x+y$ is $\frac{(n+m-1)(n+m)}{2}-1$, the total amount of burles you get back with the increase of $x$ is $\frac{n(n-1)}{2}$(you do one with each $x$ from $1$ to $n-1$) and the total amount of burles you get back with the increase of $y$ is $\frac{m(m-1)}{2}$(you do one with each $y$ from $1$ to $m-1$). In total the cost is $\frac{(n+m-1)(n+m)}{2}-1-\frac{n(n-1)}{2}-\frac{m(m-1)}{2}=nm-1$.
• » » » 17 months ago, # ^ | 0 oh,nice. I made it hard for myself.
• » » » 17 months ago, # ^ | +14 You are using wrong currency.
• » » » » 17 months ago, # ^ | +3 Fixed.
• » » » 17 months ago, # ^ | +11 Consider an arbitrary path from $(1, 1)$ to $(n, m)$.We claim that at any point $(x, y)$ along that path the cost will be $x \cdot y-1$.Proof by induction.At the start the cost is $1 \cdot 1 - 1 = 0$ — the condition holds.The induction step.Assume that the statement is true at a point $(x, y)$ along the path. The next point in the path is either $(x,y+1)$ or $(x+1, y)$.In the first case the cost will be $x \cdot y-1 + x = x \cdot (y+1) - 1$. Similarly, in the second case the cost will be $x \cdot y-1 + y = (x + 1) \cdot y - 1$.
» 17 months ago, # | 0 What is the meaning of the Problem B title?
• » » 17 months ago, # ^ | +6 I guess constraints make this ques harder. If it were around 1e9, all would have tried to thought of general formula.
» 17 months ago, # | +2 Very happy to see some good ad-hocs like ABCD.
» 17 months ago, # | +106 Screencast with commentary aka Um_nik staring at problem F for 40 minutes in silence
» 17 months ago, # | 0 My python submission for problem C is getting TLE on Tc 5, I tried implementing it as the same way other people implemented and got AC. MY SUBMISSION FOR PROB Cplease help. Thanks in advance.
• » » 17 months ago, # ^ | 0 There are a few smart observations that could help you pass it: We are interested only in the sum so maintaining just the sum of let's say 1st p elements would help. We can limit the search to the maximum size of students in a particular group Also, I have observed iterating through numbers (for i in range(n)) is generally faster than iterating through the keys (for i in dict). Also you could try with fast I/O sometimes that solves TLE.
• » » 17 months ago, # ^ | +1 Your code takes O($n^2$) time because of for ii in d: for k in range(1,n+1): If you change it to for ii in d: for k in range(1,len(d[ii])+1): then the loops run in O($n$) time 114686648.
• » » » 17 months ago, # ^ | 0 thanks man! appreciate it.
» 17 months ago, # | ← Rev. 3 → +66 I think the official generator for task E is broken since none of the 26 pretests have n > 1e5! Therefore, It'd be best if the authors cut the limit on n to 1e5 and rejudge all the solutions to respect the spirit of competition. awoo
• » » 17 months ago, # ^ | +18 Knowing that there were 26 pretests for E find it strange as well, it's common sense that the only fair option is to reduce n to 1e5 and rejudge everything, if not including max tests in pretests was intentional then the authors are sadistic and should die a painful and slow death.
• » » 17 months ago, # ^ | -48 Nah, I don't think that's necessary. I admit I made a mistake while generating tests, but you aren't really promised the strongest tests anywhere. The constraints are in sync with what the validator checks, so I see no issue.Sorry that you got caught by it, be careful the next time.
• » » » 17 months ago, # ^ | +10 Fully agree. A lot of people here are interchanging pre-tests with system tests for some reason...
• » » » 17 months ago, # ^ | +68 I think you have to be carefull next time
• » » » » 17 months ago, # ^ | +31 Exactly!
• » » » » 17 months ago, # ^ | +5 Sure, I'll try as well.
» 17 months ago, # | +2 This contest was actually educational for me. I learned 2 things: 1) Never trust the ceil function. 2) Don't write iterative dp like it's recursion, always check the order of computations... goddammit D.
• » » 17 months ago, # ^ | 0 I trust the ceil function:a modified version of your submission using the std ceil function
• » » 17 months ago, # ^ | +1 and if you want to write a ceil function that doesn't need to convert its arguments into doubles, you can write something like this: int ceil(int x, int y){ return (x+y-1)/y; }
• » » » 17 months ago, # ^ | 0 Oh yeah, that's a smart expression. I always did this subconsciously when dividing by 2 (e.g. (x+y+1)/2 ) but it never crossed my mind that it could be generalized. Perhaps ceil isn't as untrustworthy as I first thought, thanks for the tip!
» 17 months ago, # | +1 I solved problem b but not able to solve a :(
» 17 months ago, # | ← Rev. 3 → 0 Hey! In C problem, I was using comparator on a vector of vectors. When I used '>=' in the comparator it gave a runtime error while using only '>' gave the 'Accepted' verdict.Runtime code : https://pastebin.com/WQwxwrCdWorking code : https://pastebin.com/sf1e1BTkBoth codes are same except on the 18th line where there is difference of '>' and '>=' in the respective codes. Can anyone explain why this happens?
• » » 17 months ago, # ^ | +2
» 17 months ago, # | 0 How to get rid of MLE in problem B using Dp can anyone please help? Here is my submission 114621015Thanks in advance :)
• » » 17 months ago, # ^ | 0 B can be solved using 2D dp. Store true/false in DP states. See this sub : https://codeforces.com/contest/1519/submission/114557667
• » » » 17 months ago, # ^ | 0 How storing 2 states will cover all possible states there might be different k values for different states right? could you elaborate, please...
• » » » » 17 months ago, # ^ | ← Rev. 2 → 0 The cost of reaching {x,y} from any allowed path is the same.So its reducible to a new subproblem starting from {x,y} to reach {N,M} with exactly K - (x + y) burlesIf dp[x][y] is false, it means its not possible to reach {N,M} with exactly K - (x + y) burles
• » » » 17 months ago, # ^ | 0 B is math not dp ;-;
• » » » » 17 months ago, # ^ | +1 Yes its math. But can be solved using DP too. I agree DP is an overkill lol.
• » » » » » 17 months ago, # ^ | 0 if (k == (m-1 + m*(n-1))) cout<<"YES\n"; else cout<<"NO\n"; There, that gets you an AC.
• » » » » » » 17 months ago, # ^ | 0 Can you explain why?
• » » » » » » » 17 months ago, # ^ | 0 Sure First thing to notice is, no matter what path you select, the number of D's (down) and number of R's (right) remain the same, just the ordering is different. This hints that there's only 1 score possible. To be completely sure, you can do some dry run for some rectangle grids (square would be too easy) Now, once you're convinced that there's only 1 answer possible, you look at the easiest way of calculating it; Move horizontally till the last column and move down till the last row. To move m columns horizontally from 1st row, you incur a cost of (m-1). Now to go down n rows from m'th column, you incur a cost of m*(n-1). Their sum is the answer. Hope this helps.
• » » 17 months ago, # ^ | 0 You have used long long int as the data type for the dp array. You could have avoided the MLE if you had used 16bit-int or 8bit-int as the data type. You can check it out in my submission 114563484.
• » » » 17 months ago, # ^ | 0 That's cool thanks buddy :)
» 17 months ago, # | +9 speedforces :P
» 17 months ago, # | 0 Can anybody tell how to optimize my C's solution further ? 114617580
• » » 17 months ago, # ^ | ← Rev. 3 → 0 There can be n lists of size == 1, but you'd still check every of them for i = [2..n]. That would be O(n^2) ~ 10^10. Kick them off from map when their len == i, and stop cycle when i > max_len(m[i]) (map should be empty at this time). But first kick those with len <=2.
» 17 months ago, # | -50 I made a little meme:
» 17 months ago, # | 0 Where is the Editorial ??
» 17 months ago, # | ← Rev. 2 → 0 I guess, the test cases were weak for problem C. Sorting the vector of vectors based on their size and stopping once the size becomes less than K can get your solution accepted.
• » » 17 months ago, # ^ | 0 114627098 what is the problem in my code then?
• » » 17 months ago, # ^ | ← Rev. 3 → 0 There's an explanation for that actually. If you iterate from each $k$ from $1$ to $n$, you will find at most $\frac{n}{k}$ vectors that have size at least $k$. That leads to a known sum which is $n logn$.
• » » » 17 months ago, # ^ | +3 There is a trick to even get it down to O(n). out = [] for k in range(1, n + 1): teams = [t for t in teams if len(members[t]) >= k] score = 0 for t in teams: score += prefix_sum[t][k * (len(members[t]) // k)] out.append(score) This part will run in O(n) time, not O(n log n). This is because a team with $m$ members will be filtered out of teams after $m$ iterations. And since the sum of the sizes of all teams are $n$, the code runs in O(n) time.
• » » » » 17 months ago, # ^ | 0 I did like that in my ugly code, its good to know that it is O(n), i thought it would be something like N*sqrt(N).
• » » » » 17 months ago, # ^ | ← Rev. 4 → 0 Nice trick. Wouldn't it be "cheaper" to kick them from the set, instead of generating new list all the time?
• » » 17 months ago, # ^ | 0 This is probably the intended solution and it's $O(nlogn)$. It's not about the test cases, it is indeed $O(n)$ except sorting. Because, if done so, we will consider a university with size $s$ in only $s$ different $k$'s. Since the sum of all $s$'s is $n$, complexity is $O(nlogn+n)$, that is, $O(nlogn)$.
» 17 months ago, # | 0 Please explain how to make problem C using Two points method, i made it using Prefix sums.
• » » 17 months ago, # ^ | 0 I guess you'll need prefix sums anyway. From now on, assume that the universities are sorted in non-decreasing order by size. The thing with the "two pointers" must be that when you encounter a university with size smaller than current $k$, you have to move your "left pointer" (you shouldn't evaluate universities on the left of the left pointer because their size are smaller than $k$). As with the right pointer, it just doesn't exist. P.S. : It is my understanding of this tag.
» 17 months ago, # | 0 How to solve D?
• » » 17 months ago, # ^ | ← Rev. 2 → +5 Maintain a 2D DP array. Here, $dp[i][j]$ means the value we'll get in interval $[i,j]$ if we reverse the interval $[i,j]$. It is not hard to see that $dp[i][j] = dp[i+1][j-1] + a[i]*b[j] + a[j]*b[i]$ because when we reverse an interval $[i,j]$ , interval $[i+1,j-1]$ gets reversed and we swap $a[i]$ and $a[j]$. Now, maintain a prefix and a suffix array $pre$ and $suf$. Here, $pre[i]$ denotes the answer for prefix $[1,i]$ without any reversing and $suf[i]$ denotes the answer for suffix $[i,n]$ without any reversing. The answer is maximum of $pre[i-1] + dp[i][j] + suf[j+1]$ for all intervals $[i,j]$.
• » » » 17 months ago, # ^ | 0 thank yo so very much....I understood..you explained it really well
• » » » » 17 months ago, # ^ | 0 You're welcome!
• » » » 17 months ago, # ^ | 0 I think there is typo it should be dp[i][j]=dp[i+1][j-1]+a[i]*b[j]+a[j]*b[i]
• » » » » 17 months ago, # ^ | 0 Sorry, you're right. Fixed that.
• » » 17 months ago, # ^ | -7
» 17 months ago, # | 0 Can someone explain why the first submission gets TLE while the second one gets accepted?Only difference is how I sort a vector.
• » » 17 months ago, # ^ | +5 In the first submission, you have used a lambda expression to sort the vector. The problem here is that you have 'captured' the variables using "=" which is slower. I replaced that with "&" and uncommented the "#define int ll" line. The code gets AC after this. 114639136
• » » 17 months ago, # ^ | ← Rev. 2 → 0 Capturing variable using = is by value which means you copy a new vector every time the sort function compares two numbers. Use & to capture by reference.
» 17 months ago, # | 0 How to solve C ? I solved using factorization in O(nsqrt(n)) but it seems lot of people have solved in some different way.
• » » 17 months ago, # ^ | 0 idk
• » » 17 months ago, # ^ | +3 You can solve problem C using Prefix Sums . Divide all the students by their corresponding cities . For each city , if number of students in my current city is ${m}$ and size of teams is ${k}$ , then ${m\%k}$ lowest skilled students will be not be in any team .${\newline}$ So for each city , sort the array of students ,calculate the prefix sums , and then iterate over the size of teams (${1}$ to ${m}$ ) , take sum of skills of all students who can be teamed up ( ${pref[m]-pref[m\%i]}$ ) and add to the final answer array . Hope it helped ! Submission Link : https://codeforces.com/contest/1519/submission/114643205
• » » » 17 months ago, # ^ | 0 So, let say there are $n$ university and each university has only $1$ student then don't you think your solution will be $O(n^2)$? I was trying to do same but this case stuck in my mind
• » » » » 17 months ago, # ^ | 0 The the number of students is ${n}$ , the only thing I did is distribute those ${n}$ students across the cities . There are two loops but the sum of total operations of the inner loop is still ${\mathcal{O}(n)}$ (because I am iterating over the students and only n students exist across all the cities ) . In your example case , the inner loop will perform ${\mathcal{O}(1)}$ operation for each outer loop iteration , summing upto ${\mathcal{O}(n)}$ . (Also you'll need to account for sorting , which takes ${\mathcal{O}(nlogn)}$ ).
» 17 months ago, # | +5 Speedforces.Many people solved ABCD.
» 17 months ago, # | 0 Wow, the contest is very good!
» 17 months ago, # | +8 Need Help in problem D: **** I have solved Problem D using Python. i wrote the same logic in two different ways 1st way got Accepted but the 2nd way got TLE. both are exactly same as other only I changed the calculation part of the code and I believe it should not affect the run time. please help me out. TLE Submission : https://codeforces.com/contest/1519/submission/114646999 Accepted Submission : https://codeforces.com/contest/1519/submission/114647073
• » » 17 months ago, # ^ | +4 I think this blog might be helpful.
• » » » 17 months ago, # ^ | 0 Hey, the community should need to take this blog seriously and make the required changes in PYPY. during the contest, I made logic and coded it but the submission got TLE. I taught my approach was not optimal and started to think of other ways to solve that problem. finally ended not solving it. after the contest realized My idea is the only optimal way to solve it and all the people with python submission got TLE, only a few people who already know the above trick got accepted in PYPY. The tester of the round should have tried python also and they would have figured the TLE problem or they have not tested with python. please Do the need full I believe solving problems and learning should need to be the main goal the programming language we use should not be a barrier to it.
» 17 months ago, # | ← Rev. 2 → +5 weak test in C
» 17 months ago, # | ← Rev. 2 → 0 Can anybody please tell me why i'm getting TLE in C? Like i used the same approach as many the people who got accepted did, till today morning my solution was accepted by now it's showing TLEhere is the link to my submission https://codeforces.com/contest/1519/submission/114596588
• » » 17 months ago, # ^ | ← Rev. 2 → 0 I too had same problem. But, the thing here is, for each k you are checking each university even if k>sizeof(university[i]). This makes your solution time complexity O(N^2). Better approach is to go for each university and then add its answer to k( which is less than sizeof(university[i])). To do this we need only O(N) time complexity.Here are my solutions:This is similar to your approach but got TLE: https://codeforces.com/contest/1519/submission/114653177Here I modified it as stated above and got AC: https://codeforces.com/contest/1519/submission/114659364
• » » » 17 months ago, # ^ | ← Rev. 2 → 0 if it's becoming O(n^2) then how come it passed TC:7 because which is some what similar to TC:8(on which i'm getting the TLE)
• » » » » 17 months ago, # ^ | ← Rev. 3 → 0 Let's define n as the max number of the students in one university and m as the number of universities. The last loop in your submission will run n*m times. It seems that there is only one university which is numbered 200000 in the 7th testcase. However, there may be 100000 different universities in your TLE testcases, where the first university has 100001 students and every other university only has one student.
• » » » » 17 months ago, # ^ | 0 No. They are not same. There is some difference which we cant see as only few numbers are visible in testcase.For proof:Let n = 200000 Let distinct universities be 10000.1 <= k <= nTotal Computations = k_values x distinct_universities = 200000 x 10000 = 2*10^9So, you will get TLE.In other scenerio:Let students in each university be (n/distinct_universities) = 20Total Computations = distinct_universities * no_of_students_in_that_uni = 200000So, this will get you AC.
• » » » » » 17 months ago, # ^ | 0 AmeTxx and suyashc222 Got it. Thanks!!
» 17 months ago, # | ← Rev. 2 → +3 OH C ! You broke my little heart
• » » 17 months ago, # ^ | -9 hhhh
» 17 months ago, # | 0 Can Someone figure out why I am getting random values instead of 0, for problem C MyCode(https://codeforces.com/contest/1519/submission/114659037)
• » » 17 months ago, # ^ | ← Rev. 4 → +3 :(
• » » 17 months ago, # ^ | +21 i can't sorry bro
• » » 17 months ago, # ^ | +8 sum = sum + (v[j][((long long)(v[j].size() / i) * i — 1)]);When i > v[j].size() the second index becomes -1 so you read memory outside of the vector (v[j][-1])
» 17 months ago, # | +7 when the ratings update cannot wait for more time??
» 17 months ago, # | ← Rev. 2 → +11 In Problem E, why doesn't the following greedy solution work?I first order lines by their angles with Ox and iterate them by counter-clockwise order. For each line, find two points that can reach the line, and always take points under the line first. Of course, I remove selected points after each step.
• » » 17 months ago, # ^ | +3 I did the same but got stuck on test 7 because there's a test case that ruins this approach. Let's say a specific angle has 3 points and other have 4 points and another one with 1 point. You're gonna choose any 2 of the 3 but what if the 2 points you choose were in the other 4 points ? It won't be chosen there (at the 4 points's angle). So, you have to choose 2 points out of the 3 that won't interfere with the 4 points to get the maximum number of moves and it would just interfere with that 1 point because it won't be chosen anyway. Hope you understand my explanation.
• » » » 17 months ago, # ^ | 0 I got your points. Thanks!
» 17 months ago, # | 0 Time limit exceeded on test 8 :(
» 17 months ago, # | 0 C was a pretty cool problem. Never seen anything like it, requires some thinking, yet not too difficult. Props!Why not just pick n=2000 for D, so it can be solved in Python? Now I had to go back and rewrite my solution in C++ for no particular reason, just because the Python implementation barely times out. I actually thought it was a very nice problem other than that annoying detail.
• » » 17 months ago, # ^ | 0 I assume n=2000 might have allowed some O(n^3) solutions to pass by virtue of vectorisation. That might be the reason setters kept n=5000.
» 17 months ago, # | -7 Is it rated?
• » » 17 months ago, # ^ | 0 Yes
» 17 months ago, # | ← Rev. 2 → 0 Problem C raises my doubt about how vector > works in different versions of cpp compilers. After calling push_back() for serveral times, sometimes the result is unexpected but sometimes it looks fine. It seems that this error is related to the version of the compiler. Could someone please explain why?
» 17 months ago, # | 0 My solution 114601199 for the problem 1519C significantly coincides with solutions phungminhvu/114594685, ngtvu278/114601199. phungminhvu is my friend and he have helped me in this contest. I do not knowingly violate the rules so Can you forgive me this time ?
• » » 17 months ago, # ^ | 0 your deed is unforgivable...don't do it next time
» 17 months ago, # | 0 Editorials please...
» 17 months ago, # | +12 MikeMirzayanov Please update the rating change, please...
» 17 months ago, # | 0 I've got a nasty bug in the implementation of D which gives WA on Test 27. Can somebody help me figure it out ? 114670308 Test case 2739 2272444 3698012 5125012 7166709 3503029 1727871 7293835 4189398 3200693 3241014 8609749 6932236 3343904 8116213 1958622 6714272 8223889 8038950 1022157 8865228 6268599 7674195 9542567 3949141 8069361 2085276 9394838 6938364 1494858 2742785 7992600 3754931 3199011 4125593 9684890 2329734 5106836 2143245 8971071 9005537 7867356 8592781 1211906 3953852 9483166 4708038 5157649 4947712 1505811 8112182 3295947 7153949 9936840 8184380 3081882 9265570 8959815 5190532 9659942 1243298 7690386 9497105 3664620 8537001 7554794 2121292 1628003 5036215 2381356 1811254 3935828 9965608 3296624 2748440 7073046 5400382 2509524 2799639
• » » 17 months ago, # ^ | +6 You are not considering the case when the entire array a needs to be reversed.
• » » » 17 months ago, # ^ | 0 Thanks!
» 17 months ago, # | +21 Does anyone know why the rating change in the CF predictor is so different from the actual rating change?
» 17 months ago, # | +17 hm? why so big delta between predictor and real rating changes?
» 17 months ago, # | +16 Finally, after 14 months and 57 contests I have reached CM. Feels good :)
• » » 17 months ago, # ^ | +5 Congratulations! Well Played :D
» 17 months ago, # | +16 Why is the rating change so less than cf predictor?
• » » 17 months ago, # ^ | 0 Ask the predictor creators. Its not an issue of cf
• » » » 17 months ago, # ^ | +5 why so confident? Actually almost always real rating changes are greater than predictor changes, in this case it's vice versa.
• » » 17 months ago, # ^ | +1 Use Carrot, It is better and more accurate than CF-predictor.
• » » » 17 months ago, # ^ | 0 Sure, will try it
» 17 months ago, # | +16 How come CF-Predictor is showing +79 rating change but the actual rating change is only +48 for me while there are other accounts who have seen increase in their actual rating change :(
» 17 months ago, # | +16 I think the positive delta is really low even after getting 2899 rank considering My previous ratings i should have got +120-130 but i only got +81. is it normal ?
• » » 17 months ago, # ^ | +15 No.. of all the contests I have given this is the first time I have seen the actual rating change is less than the predicted rating change..
» 17 months ago, # | 0 About the ongoing discussion about rating predictions in cf-predictor being wrong, I'd like to say something. The ratings of the people I've checked (including mine) are not up-to-date, that's probably the reason why the calculations of deltas are incorrect.
• » » 17 months ago, # ^ | 0 By not up-to-date do you mean there will be a rating roll back soon and new ratings will be published ?
• » » » 17 months ago, # ^ | 0 I mean, I'm talking about "cf-predictor extension" and the calculation of yesterday's contest is based on our previous ratings, like the system is literally missing two last contests' rating changes. I don't know about any rating roll back and I didn't mean that.
» 17 months ago, # |
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# According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2− ?
more_vert
According to molecular orbital theory, which of the following is true with respect to $Li_2^+$ and $Li_2^-$ ?
(1) Both are unstable
(2) $Li_2^+$ is unstable and $Li_2^-$ is stable
(3) $Li_2^+$ is stable and $Li_2^-$ is unstable
(4) Both are stabel
more_vert
Chemical bonding and molecular structure, Molecular orbital theory
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Last Updated:
16/06/2017 - 22:19
IAM Graduate
• 2012:
1. A survey on all known algorithms in solving generalization of birthday. . Master's thesis, Middle East Technical University, 2012. BibTeX ]
@mastersthesis{Namaziesfanjani2012,
author = {Mina Namaziesfanjani},
title = {A survey on all known algorithms in solving generalization of birthday},
school = {Middle East Technical University},
year = {2012},
address = {Cryptography, Institute of Applied Mathematics, Middle East Technical University, Ankara, Turkey}
}
back
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# Integral in d-dimensions
Maybe it is an easy integral but I don't know how to compute it. In fact I think I have to use polar coordinates but I don't know how to do it rigorously : $\int \exp(-\mathrm{i}\mathbf{q}\cdot\mathbf{r})q^{\eta-d}\,\mathrm{d}^dq$, where $\eta$ is a real number, and $d$ an integer of course.
-
How do you define $q^{\eta-d}$? – 1015 Feb 28 '13 at 18:15
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# Key Eleven
## An Advanced Mathematics Lesson Starter Of The Day
Choose 4 keys on your calculator key pad that are positioned in the four corners of a rectangle. Use these keys to type in a 4-digit number going round your rectangle either clockwise or anticlockwise. Eg. 3146
Check that your 4-digit number is a multiple of eleven.
Can you prove that all 4-digit numbers formed this way are multiples of 11?
Is this strange result also true for numbers formed from keys on the corners of a parallelogram?
Note 1: The rectangles referred to above have horizontal or vertical sides.
Note 2: A divisibility test for eleven is to consider the alternating digit sum of the number. If this is divisible by eleven then the original number is also divisible by eleven. For example, if a four digit number has digits a, b, c and d then the alternating digit sum is a - b + c - d.
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Topics: Starter
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Let the rectangle have a width of $$w$$ key spacings and a height of $$h$$ key spacings. (The example in the diagram above has $$w = 2$$ and $$h = 1$$ )
Let the number be formed by going around the rectangle in a clockwise direction starting from the top left (unlike the example in the diagram above).
Let the first digit of the 4-digit number be $$x$$.
The alternating digit sum is $$x - (x+w) + (x+w+h) - (x+h)$$
$$= x - x - w + x + w + h - x - h$$
$$= 0$$ (which is a multiple of 11 so the 4-digit number must also be a multiple of 11)
Also any cyclic permutation of the 4-digit number will be divisible by 11 (accounting for the example in the diagram above)
This should also be applicable to parallelograms
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## Numbers and the Making of Us
I initially heard this book described on the Grammar Girl podcast and immediately went to find out more about it. I now have it on my Christmas present wish list and am looking forward to receiving a copy (hint!).
"Caleb Everett provides a fascinating account of the development of human numeracy, from innate abilities to the complexities of agricultural and trading societies, all viewed against the general background of human cultural evolution. He successfully draws together insights from linguistics, cognitive psychology, anthropology, and archaeology in a way that is accessible to the general reader as well as to specialists." more...
For Students:
For All:
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## Extended quark mean-field model for neutron stars: Toward a broken flavor SU(3) symmetry [PDF]
J. N. Hu, A. Li, H. Toki, W. Zuo
We extend the quark mean-field (QMF) model to strangeness freedom to study the properties of hyperons (\$\Lambda,\Sigma,\Xi\$) in infinite baryonic matter and neutron star properties. The baryon-scalar meson couplings in QMF model are determined self-consistently from the quark level, where the quark confinement is taken into account in terms of a scalar-vector harmonic oscillator potential. The strength of such confinement potential for \$u,d\$ quarks is constrained by the properties of finite nuclei, while the one for \$s\$ quark is limited by the properties of nuclei with a \$\Lambda\$ hyperon. These two strengths are not same, which represents the SU(3) symmetry breaking effectively in QMF model. With these baryon-scalar meson coupling constants, the single baryon potentials are obtained, where all of three hyperon potentials are attractive at nuclear saturation density. The potentials for \$\Lambda\$ and \$\Sigma\$ are difference about 5 MeV at the saturation density, which shows this SU(3) symmetry breaking approaches 20% in the extended QMF model. Finally we obtain a neutron star maximum mass of \$1.59 M_{\odot}\$ with \$\Lambda\$, \$\Sigma\$ and \$\Xi\$ hyperons within the present model. This value is far from the recently measured pulsar mass, consistent with the convectional relativistic mean field theory and also the microscopic studies based on developed realistic baryon-baryon interactions.
View original: http://arxiv.org/abs/1307.4154
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The Normal Derivative Of Electric Field | Quantum Science Philippines
## The Normal Derivative Of Electric Field
By Euprime B. Regalado
From Gauss theorem, we can show that the surface of a curved charged conductor, the normal derivative of the electric field is given by
$\frac{1}{E}\frac{\partial E}{\partial n}= - \left(\frac{1}{R_1}+\frac{1}{R_2}\right)$
where $R_1$ and $R_2$ are the principal radii of curvature of the surface. Gauss’s law in integral form is expressed as
$\oint_s\vec{E}\cdot\hat{n}da=0$
when there are no charges enclosed in the surface S .
Before considering the three dimensional problem, we first tackle the problem in two dimensions. We take a curved Gaussian box next to the surface of a charged conductor at a point where the radius of curvature is R. Application of Gauss’s law yields
$0= \int_s\vec{E}\cdot\hat{n}da= E_{top}\nabla_{atop}-E_{bottom}\nabla_{abottom}$ (2)
where $\nabla_{atop}$ and $\nabla_{abottom}$ are the areas of the top and bottom of the box, respectively. We can see that there is no contribution from the sides of the box, because they are taken to be normal to the surface. In polar coordinates the areas are expressed as $\nabla_{atop}= (R+\epsilon)d\theta dz$ and $\nabla_{abottom}=Rd\theta dz$ . Gauss’s law then yields
$0= E_{top}(R+\epsilon)d\theta dz - E_{bottom}Rd\theta dz$
and now we get the relation
$E_{bottom}=E_{top}(1+\frac{\epsilon}{R})$.
This allows us to calculate
$\frac{\partial E}{\partial n}$ $= \lim_{\epsilon \to \ 0} \frac{E_{top} -E_{bottom}}{\epsilon}$ $= \lim_{\epsilon \to \ 0} - \frac{E_{top}}{R}$ $= - \frac{E_{top}}{R}$
Noting that the $E_{top}$ is the same as E when $\epsilon\to\ 0$, this maybe written as
$\frac{1}{E}\frac{\partial E}{\partial n}= -\frac{1}{R}$ (3)
which is analogous to two-dimensional expression.
Going back to the 3-dim problem, we use the same method as above. This time however, the areas of the top and bottom of the Gaussian box are
$\nabla_{atop}$
$= (R_1 + \epsilon) (R_2 + \epsilon) d \Omega$,
$\nabla_{abottom}= R_1 R_2 d \Omega$
which in turn yields,
$\frac{\partial E}{\partial n} \\$ $= \lim_{\epsilon \to \ 0} \frac{E_{top}-E_{bottom}}{\epsilon} \\$ $= \lim_{\epsilon \to \ 0} - E_{top} (\frac{1}{R_1}+ \frac{1}{R_2}+ \frac{\epsilon}{R_1 R_2}) \\$ $= - E_{top}(\frac{1}{R_1} + \frac{1}{R_2})$
Rearranging gives,
$\frac{1}{E} \frac{\partial E}{\partial n} = - (\frac{1}{R_1} + \frac{1}{R_2})$
Note that this reduces to the two dimensional expression (3) in cylindrical limit, $R_2 \to \infty .$
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Lethbridge Practice Nov 20
#### Start
2021-11-20 08:00 AKST
## Lethbridge Practice Nov 20
#### End
2021-11-27 08:00 AKST
The end is near!
Contest is over.
Not yet started.
Contest is starting in -62 days 6:21:45
168:00:00
#### Time remaining
0:00:00
Anthony and Cora are playing Dominion, their favorite card game. In Dominion, there are $T$ different card types, and each player has a set of cards (known as a deck). A deck $D$ is said to have $C$ combos if $C$ is the largest integer such that for $C$ different card types in the game, $D$ contains at least two cards of that type. Anthony currently has $N$ cards and he wants to trade cards with Cora such that he’ll have a deck with exactly $K$ combos.
For each card type $i$ ($1\leq i\leq T$), Anthony can choose to perform at most one transaction. There are two types of transaction:
1. Buy up to two cards of $i^{th}$ type from Cora at $a_ i$ coins each
2. Sell all his cards of $i^{th}$ type for $b_ i$ coins each
Anthony wants to maximize his profit while obtaining a complete deck. Anthony is willing to spend coins in order to obtain a complete deck if necessary, but of course he wants to minimize his spending in that case. Note that he doesn’t care about keeping the rest of his cards which don’t contribute to the complete deck.
Anthony has hired you to help him calculate how much money he can make if he chooses the optimal strategy for obtaining enough combos. If he has to spend money, output a negative number.
## Input
The first line of the input contains three integers $N$, $T$, and $K$, $1\leq K\leq T\leq 100\, 000$, $1\leq N\leq 2T$.
The next line is a list of $N$ integers representing the cards in Anthony’s deck. Each integer on this line is between $1$ and $T$ inclusive. It is guaranteed no integers appear more than twice.
Finally, each of the next $T$ lines of the input contains two integers each. The $i^\mathrm {th}$ line contains $a_ i$ and $b_ i$, $1\leq a_ i, b_ i\leq 10^9$, corresponding to the price of buying and selling a card of type $i$.
## Output
Output a single integer denoting Anthony’s profit assuming he trades optimally.
## Explanation of Sample Input
In the first example, Anthony should sell two of card $1$ and buy one of card $2$ and one of card $3$ for a net profit of $10$ coins. If he chooses to sell one of card $3$ and buy one of card $2$, then he’ll end up spending $20$ coins.
Sample Input 1 Sample Output 1
4 3 2
1 3 2 1
1 50
50 20
40 30
10
Sample Input 2 Sample Output 2
4 3 2
1 3 2 1
1 20
50 20
40 30
-20
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Wednesday, February 10, 2016 ... //
What gravitational wave astronomy may hear
On Monday, LIGO finally officially admitted that there would be a press conference tomorrow, as we predicted in the poll, at 10:30 am, in the National Press Club, a gentleman journalists' building that is very close to the White House. In 2011, I embedded a talk that the Czech ex-president Klaus gave there. Oops, it was one in Canberra but who cares. ;-)
LIGO's colleagues in Italy, Virgo, should hold an event at the same moment, i.e. 16:30 Central European Time. This should also be webcast – and there are hints that these people will actually be at CERN, not in Italy, and the webcast will actually be aired via webcast.cern.ch, too.
But you should already bookmark this YouTube page on the NSF channel where webcast will start in 29 hours. See also an NSF press release.
Incidentally, the arXiv already boasts a theory paper about the first black hole mergers detected by LIGO. ;-) It seems clear to me that they pretend to be ignorant of some numbers that they actually know.
Our latest blog post on LIGO has 12,000 views so far (and 1,000+ Facebook likes LOL) and there are lots of other responses by the media to the tomorrow's event that is gaining the official status these days. An unusually intelligent article written by David Castelvecchi appeared in Nature:
Gravitational waves: 5 cosmic questions they can tackle
Like your humble correspondent and Bill Zajc, Castelvecchi points out that the discovery that will be announced tomorrow is much more than a confirmation of the gravitational waves – something that good physicists had no doubt about. It's also an excellent source of information about the detailed physical processes that have emitted the waves. And because the black hole merger that has been seen isn't the last event that people will have observed by gravitational waves, we may "hear" lots of new information through the gravitational wave detectors in the future.
Castelvecchi mentions six (a number he translates as 5 into Arabic numerals LOL; update: DC fixed it) physical phenomena that the gravitational wave detection may clarify:
1. Certainty about black holes' existence
2. Speed of gravitational waves
3. Mountains on neutron stars (maybe)
4. Cosmic strings (maybe)
5. Events igniting supernova explosions (maybe)
6. New measurements of the Universe's accelerated expansion (maybe)
Let me add a few more words.
At least since the 1960s or so (when Wheeler coined the catchy term "black holes"), it was rather implausible to suggest that black holes didn't exist – the mastery of the theory as well as the emerging experimental evidence got way too strong more than half a century ago. But you may still find lots of people, e.g. the 2005 TRF guest blogger George Chapline, who like to say that black holes don't actually exist.
Tomorrow, these suggestions will probably become really-really indefensible (so far they were only "really indefensible"). The agreement between the gravitational waves predicted from the black holes' inspiral+merger+ringdown by general relativity, and those observed, will be so precise (despite its dependence on the events in extreme conditions where all the nonlinearities of GR matter intensely) that people suggesting that there is any large mistake in the general relativistic description of these black hole phenomena will probably drop out of the fringes of science.
Lots of questions about black holes – especially about their quantum behavior – will remain murky or open. String theorists' research of quantum properties of black holes will continue pretty much independently of the LIGO discovery. But the black holes' macroscopic behavior may be almost as accessible as the Moon's surface.
A one-minute video: LIGO - What is a Gravitational Wave? from Kai Staats on Vimeo. A 20-minute video is here.
There are many special characteristics of black holes. They're the simplest, most determined and smooth, astrophysical (by size) objects in Nature macroscopically. Because of the no-hair theorem and their general uniqueness, they carry the minimum amount of "personality" macroscopically. On the other hand, they carry the maximum amount of microscopic, quantum information (or entropy) in them – relatively to all objects of the same size. Their density calculated from the mass and the event horizon radius is "maximal" among all the objects; but the local density of matter is "minimal" (basically zero) among all astrophysical bodies. And so on.
Black holes, the 21st century theoretical physicist's favorite laboratories, are queens of the extremes.
This is why the black hole merger, if it is announced tomorrow, will be one of the strongest methods to confirm what we believe about a theory, in this case general relativity. One more confirmation will become possible if and when LIGO/Virgo detects an event that may also be seen optically. If an event will be both "seen" and "heard" (the verb we should use for the detection by gravitational wave astronomy – the similarities between regular sound signals and the LIGO signal is hard to overlook and the 2 LIGO detectors really behave as 2 ears, even when it comes to the way how they estimate the location of the source etc.), we should be able to verify that the speed of the gravitational waves is the same as the speed of light in the vacuum, $c$. I am confident that it's the case. Even relatively small deviations would be rather shocking from my viewpoint because the Lorentz symmetry would have to be broken (perhaps spontaneously) in some way.
Einstein's memorial in Washington D.C. Will they bring the man to the National Press Club? ;-)
But then there is another, perhaps even more important motivation for the experiments. We also want to learn new things – answers to questions that we're genuinely uncertain about.
Castelvecchi has mentioned four possible discoveries that may be made with the gravitational wave detectors in the future: rugged neutron stars, cosmic strings, events that ignite supernova explosions, and new methods to measure the expansion of the Universe.
The mountains on neutron stars sound both "sufficiently unexciting" for me – as well as probably "too weak signals". Neutrons stars, like black holes, may merge and the emitted waves could be just a little bit weaker than the gravitational waves from the black hole mergers. But these 10-kilometer neutron stars may also have 1-millimeter "mountains" on them and the spinning involving such inhomogeneities is predicted to emit gravitational waves, too. I just feel that those must be extremely weak – unless the neutron star is extremely close, of course. And when it's close and the "waves from the neutron mountains" are strong enough, I have no idea how I would distinguish this signal from many other sources.
Similarly, I have no idea what to do with the "causes of the supernova explosions". If there isn't a good enough candidate for the gravitational waves that would be associated with a given possible phenomenon that ignites the common supernova explosion, I have no idea how one could search for it or extract any useful information. There may be an agreement in the timing between an optically observed supernova explosion, and a gravitational wave burst, but will we be able to reverse-engineer the gravitational wave signal and learn something about the interior of the supernova star?
The two topics Castelvecchi mentions that are directly relevant for truly fundamental physics are the new method to measure the expansion of the Universe – from the absolute loudness of some neutron star mergers, we are told – and especially the cosmic strings.
About 10 years ago, there was some moderate excitement that the cosmic strings could have been detected experimentally. Joe Polchinski quantified the probability as 10% – he already calculated Bayesian probabilities at that time. ;-) I still think that the number was fair.
But the gravitational wave astronomy could give us new ways to observe the cosmic strings. In particular, the cosmic strings are oscillating like any strings and they may accidentally produce "cusps" on them. In the vicinity of such a cusp, the vibration of the string is most intense, and a maximum amount of gravitational waves is emitted there.
I think that there is no "clear signature" (like the black hole inspiral-merger-ringdown curve) because the detailed shape of the cusp is basically undetermined, dependent on many random variables. But there could be some approximate features of the gravitational wave signal – some scaling laws etc. – coming from the cosmic string cusp that could be predicted and (if Nature is really generous) experimentally confirmed.
I don't have to explain to you how revolutionary the detection of cosmic strings would be – just to be sure, I still do think that the probability that this event will take place in the 21st century is below 50 percent. Cosmic strings are predicted by string theory as well as by various grand unified theories (but the leftover cosmic strings' number in the visible Universe is often predicted to be too tiny to observe them). In many string models, the fundamental string may be literally stretched to astronomical distances and behave as a cosmic string – but locally, it's made of the same "stringy stuff" as the vibrating strings inside all elementary particles! And various other branes wrapped on cycles of the compact dimensions may produce additional kinds of cosmic strings in other string vacua.
Grand unified (quantum field) theories are enough to predict cosmic strings, too; they're topologically nontrivial field configurations analogous to other solitons (e.g. magnetic monopoles). Rather generally, there is a monodromy on the configuration space that is induced by the circular journey around the cosmic string. Depending on the monodromy transformation, you may get different types of cosmic strings.
Unlike the LIGO detection of a black hole merger, these observations are (potentially unlikely) speculations about a distant future at this moment. But they may occur at some moment and the implications for physics would be profound.
Whatever "sounds" will be detected by LIGO/Virgo (and later eLISA) in coming years, our ability to detect these gravitational waves will give us so much more than just the certainty that gravitational waves exist! The discovery of gravitational waves by LIGO isn't just the end of the 100-year-long waiting for their detection; it is mainly the beginning of a new era of our observations of the Universe. For the first time, the Universe won't look mute – because we won't be deaf anymore.
It's an important development. You shouldn't be surprised that Latvia has already declared the LIGO holiday to be its most important holiday in the whole year. ;-)
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# Automatic numbering of nested environments / items
I would like to have an environment, let's call it nested, which produces output like
Case 1: top level item 1
Case 2: top level item 2
Case 2.1: nested item 1
Case 2.2: nested item 2
Case 2.2.1: even nestier item 1
Case 2.3: just nested again
Case 3: and toplevel to top it off
from code that either looks like nested enumerate's, i.e.
\begin{nested}
\item top level item 1
\item top level item 2
\begin{nested}
\item nested item 1
\item nested item 2
\begin{nested}
\item even nestier item 1
\end{nested}
\item just nested again
\end{nested}
\item and toplevel to top it off
\end{nested}
or from code that looks, well, like nested environments:
\begin{nested} top level item 1 \end{nested}
\begin{nested} top level item 2
\begin{nested} nested item 1 \end{nested}
\begin{nested} nested item 2
\begin{nested} even nestier item 1 \end{nested}
\end{nested}
\begin{nested} just nested again \end{nested}
\end{nested}
\begin{nested} and toplevel to top it off \end{nested}
In fact, I don't much care how the code has to look at all, as long as the numbering on the cases is done automatically. I tried to implement both of the above, however, and I could not figure it out at all. Is there a way to achieve this?
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\documentclass{scrartcl}
\usepackage{enumitem}
\SetLabelAlign{Case}{Case #1:\hfil}
\newlist{nested}{enumerate}{5}
\setlist[nested]{nosep,leftmargin=0pt,labelwidth=*,align=Case,label*=.\arabic*}
\setlist[nested,1]{label=\arabic*}
\begin{document}
\begin{nested}
\item top level item 1
\item top level item 2
\begin{nested}
\item nested item 1
\item nested item 2
\begin{nested}
\item even nestier item 1
\end{nested}
\item just nested again
\end{nested}
\item and toplevel to top it off
\end{nested}
\end{document}
-
The vertical spacing seems a bit off... – Jubobs Mar 7 '13 at 11:43
That's the standard spacing in enumerate. But you can reset it too: see the documentation of enumitem - the values depends on your actual class, and as you didn't give a complete example I didn't bother. – Ulrike Fischer Mar 7 '13 at 11:46
I am not worried about vertical spacing, but is there a way to get rid of the final period? Currently it's displaying as Case 2.2.: when I would like it to be Case 2.2:. +1 anyway, this is essentially it. – Jesko Hüttenhain Mar 7 '13 at 12:09
@JeskoHüttenhain: I have edited the code to remove the dot and the vertical spacing. – Ulrike Fischer Mar 7 '13 at 12:36
Thank you so much, this is absolutely perfect. – Jesko Hüttenhain Mar 7 '13 at 12:38
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# HD 49968
### 图像
DSS Images Other Images
### 相关文章
CHARM2: An updated Catalog of High Angular Resolution MeasurementsWe present an update of the Catalog of High Angular ResolutionMeasurements (CHARM, Richichi & Percheron \cite{CHARM}, A&A,386, 492), which includes results available until July 2004. CHARM2 is acompilation of direct measurements by high angular resolution methods,as well as indirect estimates of stellar diameters. Its main goal is toprovide a reference list of sources which can be used for calibrationand verification observations with long-baseline optical and near-IRinterferometers. Single and binary stars are included, as are complexobjects from circumstellar shells to extragalactic sources. The presentupdate provides an increase of almost a factor of two over the previousedition. Additionally, it includes several corrections and improvements,as well as a cross-check with the valuable public release observationsof the ESO Very Large Telescope Interferometer (VLTI). A total of 8231entries for 3238 unique sources are now present in CHARM2. Thisrepresents an increase of a factor of 3.4 and 2.0, respectively, overthe contents of the previous version of CHARM.The catalog is only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or via http://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/431/773 Local kinematics of K and M giants from CORAVEL/Hipparcos/Tycho-2 data. Revisiting the concept of superclustersThe availability of the Hipparcos Catalogue has triggered many kinematicand dynamical studies of the solar neighbourhood. Nevertheless, thosestudies generally lacked the third component of the space velocities,i.e., the radial velocities. This work presents the kinematic analysisof 5952 K and 739 M giants in the solar neighbourhood which includes forthe first time radial velocity data from a large survey performed withthe CORAVEL spectrovelocimeter. It also uses proper motions from theTycho-2 catalogue, which are expected to be more accurate than theHipparcos ones. An important by-product of this study is the observedfraction of only 5.7% of spectroscopic binaries among M giants ascompared to 13.7% for K giants. After excluding the binaries for whichno center-of-mass velocity could be estimated, 5311 K and 719 M giantsremain in the final sample. The UV-plane constructed from these datafor the stars with precise parallaxes (σπ/π≤20%) reveals a rich small-scale structure, with several clumpscorresponding to the Hercules stream, the Sirius moving group, and theHyades and Pleiades superclusters. A maximum-likelihood method, based ona Bayesian approach, has been applied to the data, in order to make fulluse of all the available stars (not only those with precise parallaxes)and to derive the kinematic properties of these subgroups. Isochrones inthe Hertzsprung-Russell diagram reveal a very wide range of ages forstars belonging to these groups. These groups are most probably relatedto the dynamical perturbation by transient spiral waves (as recentlymodelled by De Simone et al. \cite{Simone2004}) rather than to clusterremnants. A possible explanation for the presence of younggroup/clusters in the same area of the UV-plane is that they have beenput there by the spiral wave associated with their formation, while thekinematics of the older stars of our sample has also been disturbed bythe same wave. The emerging picture is thus one of dynamical streamspervading the solar neighbourhood and travelling in the Galaxy withsimilar space velocities. The term dynamical stream is more appropriatethan the traditional term supercluster since it involves stars ofdifferent ages, not born at the same place nor at the same time. Theposition of those streams in the UV-plane is responsible for the vertexdeviation of 16.2o ± 5.6o for the wholesample. Our study suggests that the vertex deviation for youngerpopulations could have the same dynamical origin. The underlyingvelocity ellipsoid, extracted by the maximum-likelihood method afterremoval of the streams, is not centered on the value commonly acceptedfor the radial antisolar motion: it is centered on < U > =-2.78±1.07 km s-1. However, the full data set(including the various streams) does yield the usual value for theradial solar motion, when properly accounting for the biases inherent tothis kind of analysis (namely, < U > = -10.25±0.15 kms-1). This discrepancy clearly raises the essential questionof how to derive the solar motion in the presence of dynamicalperturbations altering the kinematics of the solar neighbourhood: doesthere exist in the solar neighbourhood a subset of stars having no netradial motion which can be used as a reference against which to measurethe solar motion?Based on observations performed at the Swiss 1m-telescope at OHP,France, and on data from the ESA Hipparcos astrometry satellite.Full Table \ref{taba1} is only available in electronic form at the CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/430/165} Seasonal variation of Titan's stratospheric ethylene (C2H4) observedAll previous observations of seasonal change on Titan have been ofphysical phenomena such as clouds and haze. We present here the firstobservational evidence of chemical change in Titan's atmosphere. Imagestaken during 1999-2002 (late southern spring on Titan) with the W.M.Keck I 10-meter telescope at 8-13 μm show a significant accumulationof ethylene (C2H4) in the south polar stratosphereas well as north-south stratospheric temperature variation (colder atpoles). Our observations restrict this newly discovered south polarethylene accumulation to latitudes south of 60° S. The only otherobservations of the spatial distribution of C2H4were those of Voyager I, which found a significant north polaraccumulation in early northern spring. We see no build-up in the north,although the highest northern latitudes are obstructed from view in thecurrent season. Our observations constrain any unobserved north polaraccumulation of C2H4 to north of 50° Nlatitude. Comparison of the Voyager I results with our new results showseasonal chemical change has occurred in Titan's atmosphere. Cepheid distances from infrared long-baseline interferometry. I. VINCI/VLTI observations of seven Galactic CepheidsWe report the angular diameter measurements of seven classical Cepheids,X Sgr, η Aql, W Sgr, ζ Gem, β Dor, Y Oph and ℓ Carthat we have obtained with the VINCI instrument, installed at ESO's VLTInterferometer (VLTI). We also present reprocessed archive data obtainedwith the FLUOR/IOTA instrument on ζ Gem, in order to improve thephase coverage of our observations. We obtain average limb darkenedangular diameter values of /line{θLD}[X Sgr] = 1.471± 0.033 mas, /line{θLD[η Aql] = 1.839± 0.028 mas, /line{θLD}[W Sgr] = 1.312 ±0.029 mas, /line{θLD}[β Dor] = 1.891 ±0.024 mas, /line{θLD}[ζ Gem] =1.747 ±0.061 mas, /line{θLD}[Y Oph] = 1.437 ± 0.040mas, and /line{θLD}[ℓ Car] = 2.988 ± 0.012mas. For four of these stars, η Aql, W Sgr, β Dor, and ℓCar, we detect the pulsational variation of their angular diameter. Thisenables us to compute directly their distances, using a modified versionof the Baade-Wesselink method: d[η Aql] =276+55-38 pc, d[W Sgr] =379+216-130 pc, d[β Dor] =345+175-80 pc, d[ℓ Car] =603+24-19 pc. The stated error bars arestatistical in nature. Applying a hybrid method, that makes use of theGieren et al. (\cite{gieren98}) Period-Radius relation to estimate thelinear diameters, we obtain the following distances (statistical andsystematic error bars are mentioned): d[X Sgr] = 324 ± 7 ±17 pc, d[η Aql] = 264 ± 4 ± 14 pc, d[W Sgr] = 386± 9 ± 21 pc, d[β Dor] = 326 ± 4 ± 19pc, d[ζ Gem] = 360 ± 13 ± 22 pc, d[Y Oph] = 648± 17 ± 47 pc, d[ℓ Car] = 542 ± 2 ± 49pc.Tables 3 to 10 are only available in electronic form athttp://www.edpsciences.org High-Precision Near-Infrared Photometry of a Large Sample of Bright Stars Visible from the Northern HemisphereWe present the results of 8 yr of infrared photometric monitoring of alarge sample of stars visible from Teide Observatory (Tenerife, CanaryIslands). The final archive is made up of 10,949 photometric measuresthrough a standard InSb single-channel photometer system, principally inJHK, although some stars have measures in L'. The core of this list ofstars is the standard-star list developed for the Carlos SánchezTelescope. A total of 298 stars have been observed on at least twooccasions on a system carefully linked to the zero point defined byVega. We present high-precision photometry for these stars. The medianuncertainty in magnitude for stars with a minimum of four observationsand thus reliable statistics ranges from 0.0038 mag in J to 0.0033 magin K. Many of these stars are faint enough to be observable with arraydetectors (42 are K>8) and thus to permit a linkage of the bright andfaint infrared photometric systems. We also present photometry of anadditional 25 stars for which the original measures are no longeravailable, plus photometry in L' and/or M of 36 stars from the mainlist. We calculate the mean infrared colors of main-sequence stars fromA0 V to K5 V and show that the locus of the H-K color is linearlycorrelated with J-H. The rms dispersion in the correlation between J-Hand H-K is 0.0073 mag. We use the relationship to interpolate colors forall subclasses from A0 V to K5 V. We find that K and M main-sequence andgiant stars can be separated on the color-color diagram withhigh-precision near-infrared photometry and thus that photometry canallow us to identify potential mistakes in luminosity classclassification. Long-Baseline Interferometric Observations of CepheidsWe present observations of the Galactic Cepheids η Aql and ζGem. Our observations are able to resolve the diameter changesassociated with pulsation. This allows us to determine the distance tothe Cepheids independent of photometric observations. We determine adistance to η Aql of 320+/-32 pc and a distance to ζ Gem of362+/-38 pc. These observations allow us to calibrate surface brightnessrelations for use in extragalactic distance determination. They alsoprovide a measurement of the mean diameter of these Cepheids, which isuseful in constructing structural models of this class of star. A catalogue of calibrator stars for long baseline stellar interferometryLong baseline stellar interferometry shares with other techniques theneed for calibrator stars in order to correct for instrumental andatmospheric effects. We present a catalogue of 374 stars carefullyselected to be used for that purpose in the near infrared. Owing toseveral convergent criteria with the work of Cohen et al.(\cite{cohen99}), this catalogue is in essence a subset of theirself-consistent all-sky network of spectro-photometric calibrator stars.For every star, we provide the angular limb-darkened diameter, uniformdisc angular diameters in the J, H and K bands, the Johnson photometryand other useful parameters. Most stars are type III giants withspectral types K or M0, magnitudes V=3-7 and K=0-3. Their angularlimb-darkened diameters range from 1 to 3 mas with a median uncertaintyas low as 1.2%. The median distance from a given point on the sky to theclosest reference is 5.2degr , whereas this distance never exceeds16.4degr for any celestial location. The catalogue is only available inelectronic form at the CDS via anonymous ftp to cdsarc.u-strasbg.fr(130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/183 CHARM: A Catalog of High Angular Resolution MeasurementsThe Catalog of High Angular Resolution Measurements (CHARM) includesmost of the measurements obtained by the techniques of lunaroccultations and long-baseline interferometry at visual and infraredwavelengths, which have appeared in the literature or have otherwisebeen made public until mid-2001. A total of 2432 measurements of 1625sources are included, along with extensive auxiliary information. Inparticular, visual and infrared photometry is included for almost allthe sources. This has been partly extracted from currently availablecatalogs, and partly obtained specifically for CHARM. The main aim is toprovide a compilation of sources which could be used as calibrators orfor science verification purposes by the new generation of largeground-based facilities such as the ESO Very Large Interferometer andthe Keck Interferometer. The Catalog is available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/386/492, and from theauthors on CD-Rom. The angular diameter and distance of the Cepheid ζ GeminorumCepheids are the primary distance indicators for extragalactic astronomyand therefore are of very high astrophysical interest. Unfortunately,they are rare stars, situated very far from Earth. Though they aresupergiants, their typical angular diameter is only a fewmilliarcseconds, making them very challenging targets even forlong-baseline interferometers. We report observations that were obtainedin the K' band (2-2.3 mu m), on the Cepheid zeta Geminorumwith the FLUOR beam combiner, installed at the IOTA interferometer. Themean uniform disk angular diameter was measured to be 1.64 +0.14 -0.16mas. Pulsational variations are not detected at a significantstatistical level, but future observations with longer baselines shouldallow a much better estimation of their amplitude. The distance to zetaGem is evaluated using Baade-Wesselink diameter determinations, giving adistance of 502 +/- 88 pc. Catalogue of Apparent Diameters and Absolute Radii of Stars (CADARS) - Third edition - Comments and statisticsThe Catalogue, available at the Centre de Données Stellaires deStrasbourg, consists of 13 573 records concerning the results obtainedfrom different methods for 7778 stars, reported in the literature. Thefollowing data are listed for each star: identifications, apparentmagnitude, spectral type, apparent diameter in arcsec, absolute radiusin solar units, method of determination, reference, remarks. Commentsand statistics obtained from CADARS are given. The Catalogue isavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcar?J/A+A/367/521 Direct detection of pulsations of the Cepheid star ζ Gem and an independent calibration of the period-luminosity relationCepheids are a class of variable (pulsating) stars whose absoluteluminosities are related in a simple manner to their pulsationalperiods. By measuring the period and using the period-luminosity'relationship, astronomers can use the observed visual brightness todetermine the distance to the star. Because these stars are veryluminous, they can be observed in other galaxies, and therefore can beused to help determine the expansion rate of the Universe (the Hubbleconstant). Calibration of the period-luminosity relation is a necessaryfirst step, but the small number of sufficiently nearby Cepheids hasforced the use of a number of indirect means, with associated systematicuncertainties. Here we present a distance to the Cepheid ζGeminorum, determined using a direct measurement (by an opticalinterferometer) of its changes in diameter as it pulsates. Within ouruncertainty of 15 per cent, our distance is in agreement with previousindirect determinations. Planned improvements to the instrument willallow us to calibrate directly the period-luminosity relation to betterthan a few per cent. Spectral Irradiance Calibration in the Infrared. X. A Self-Consistent Radiometric All-Sky Network of Absolutely Calibrated Stellar SpectraWe start from our six absolutely calibrated continuous stellar spectrafrom 1.2 to 35 μm for K0, K1.5, K3, K5, and M0 giants. These wereconstructed as far as possible from actual observed spectral fragmentstaken from the ground, the Kuiper Airborne Observatory, and the IRAS LowResolution Spectrometer, and all have a common calibration pedigree.From these we spawn 422 calibrated `spectral templates'' for stars withspectral types in the ranges G9.5-K3.5 III and K4.5-M0.5 III. Wenormalize each template by photometry for the individual stars usingpublished and/or newly secured near- and mid-infrared photometryobtained through fully characterized, absolutely calibrated,combinations of filter passband, detector radiance response, and meanterrestrial atmospheric transmission. These templates continue ourongoing effort to provide an all-sky network of absolutely calibrated,spectrally continuous, stellar standards for general infrared usage, allwith a common, traceable calibration heritage. The wavelength coverageis ideal for calibration of many existing and proposed ground-based,airborne, and satellite sensors, particularly low- tomoderate-resolution spectrometers. We analyze the statistics of probableuncertainties, in the normalization of these templates to actualphotometry, that quantify the confidence with which we can assert thatthese templates truly represent the individual stars. Each calibratedtemplate provides an angular diameter for that star. These radiometricangular diameters compare very favorably with those directly observedacross the range from 1.6 to 21 mas. A catalog of rotational and radial velocities for evolved starsRotational and radial velocities have been measured for about 2000evolved stars of luminosity classes IV, III, II and Ib covering thespectral region F, G and K. The survey was carried out with the CORAVELspectrometer. The precision for the radial velocities is better than0.30 km s-1, whereas for the rotational velocity measurementsthe uncertainties are typically 1.0 km s-1 for subgiants andgiants and 2.0 km s-1 for class II giants and Ib supergiants.These data will add constraints to studies of the rotational behaviourof evolved stars as well as solid informations concerning the presenceof external rotational brakes, tidal interactions in evolved binarysystems and on the link between rotation, chemical abundance and stellaractivity. In this paper we present the rotational velocity v sin i andthe mean radial velocity for the stars of luminosity classes IV, III andII. Based on observations collected at the Haute--Provence Observatory,Saint--Michel, France and at the European Southern Observatory, LaSilla, Chile. Table \ref{tab5} also available in electronic form at CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html Infrared standards for ISO. I. A new calibration of mid infrared photometryWe present N, 8.7 mu m, 9.8 mu m 12.5 mu m and Q bolometry of 26 starswhich were observed on the IRTF. These are primarily K giants which havebeen well measured at other wavelengths and are chosen to be single andnon-variable. Hence the data presented provides a new homogeneous set ofmid IR standards. As a test of part of the calibration strategy of ISOover this wavelength range, the zero points have been set using theKurucz model grids (1993) to predict the mid IR magnitudes from near IRdata. An analysis of the errors involved is presented and a comparisonwith more direct methods for the determination of the zero pointssuggest an possible error of 0.015 mags, although the source of thiserror is unclear as it is at the level of the uncertainty in the moredirect methods. The effect of the SiO fundamental on the spectral energydistributions of K giants is discussed. Towards a fundamental calibration of stellar parameters of A, F, G, K dwarfs and giantsI report on the implementation of the empirical surface brightnesstechnique using the near-infrared Johnson broadband { (V-K)} colour assuitable sampling observable aimed at providing accurate effectivetemperatures of 537 dwarfs and giants of A-F-G-K spectral-type selectedfor a flux calibration of the Infrared Space Observatory (ISO). Thesurface brightness-colour correlation is carefully calibrated using aset of high-precision angular diameters measured by moderninterferometry techniques. The stellar sizes predicted by thiscorrelation are then combined with the bolometric flux measurementsavailable for a subset of 327 ISO standard stars in order to determineone-dimensional { (T, V-K)} temperature scales of dwarfs and giants. Theresulting very tight relationships show an intrinsic scatter induced byobservational photometry and bolometric flux measurements well below thetarget accuracy of +/- 1 % required for temperature determinations ofthe ISO standards. Major improvements related to the actual directcalibration are the high-precision broadband { K} magnitudes obtainedfor this purpose and the use of Hipparcos parallaxes for dereddeningphotometric data. The temperature scale of F-G-K dwarfs shows thesmallest random errors closely consistent with those affecting theobservational photometry alone, indicating a negligible contributionfrom the component due to the bolometric flux measurements despite thewide range in metallicity for these stars. A more detailed analysisusing a subset of selected dwarfs with large metallicity gradientsstrongly supports the actual bolometric fluxes as being practicallyunaffected by the metallicity of field stars, in contrast with recentresults claiming somewhat significant effects. The temperature scale ofF-G-K giants is affected by random errors much larger than those ofdwarfs, indicating that most of the relevant component of the scattercomes from the bolometric flux measurements. Since the giants have smallmetallicities, only gravity effects become likely responsible for theincreased level of scatter. The empirical stellar temperatures withsmall model-dependent corrections are compared with the semiempiricaldata by the Infrared Flux Method (IRFM) using the large sample of 327comparison stars. One major achievement is that all empirical andsemiempirical temperature estimates of F-G-K giants and dwarfs are foundto be closely consistent between each other to within +/- 1 %. However,there is also evidence for somewhat significant differential effects.These include an average systematic shift of (2.33 +/- 0.13) % affectingthe A-type stars, the semiempirical estimates being too low by thisamount, and an additional component of scatter as significant as +/- 1 %affecting all the comparison stars. The systematic effect confirms theresults from other investigations and indicates that previousdiscrepancies in applying the IRFM to A-type stars are not yet removedby using new LTE line-blanketed model atmospheres along with the updatedabsolute flux calibration, whereas the additional random component isfound to disappear in a broadband version of the IRFM using an infraredreference flux derived from wide rather than narrow band photometricdata. Table 1 and 2 are only available in the electronic form of thispaper The Pulkovo Spectrophotometric Catalog of Bright Stars in the Range from 320 TO 1080 NMA spectrophotometric catalog is presented, combining results of numerousobservations made by Pulkovo astronomers at different observing sites.The catalog consists of three parts: the first contains the data for 602stars in the spectral range of 320--735 nm with a resolution of 5 nm,the second one contains 285 stars in the spectral range of 500--1080 nmwith a resolution of 10 nm and the third one contains 278 stars combinedfrom the preceding catalogs in the spectral range of 320--1080 nm with aresolution of 10 nm. The data are presented in absolute energy unitsW/m(2) m, with a step of 2.5 nm and with an accuracy not lower than1.5--2.0%. The photoelectric astrolabe catalogue of Yunnan Observatory (YPAC).The positions of 53 FK5, 70 FK5 Extension and 486 GC stars are given forthe equator and equinox J2000.0 and for the mean observation epoch ofeach star. They are determined with the photoelectric astrolabe ofYunnan Observatory. The internal mean errors in right ascension anddeclination are +/- 0.046" and +/- 0.059", respectively. The meanobservation epoch is 1989.51. Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with. A Catalog of Stellar Angular Diameters Measured by Lunar OccultationAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1987AJ.....94..751W&db_key=AST Narrow-band photometry of late-type stars. IIThis paper presents extensive narrow-band photometry in the Uppsalasystem supplementing earlier published mesurements so that data now areavailable for all late-type stars brighter than V = 6.05 and a number ofgalactic cluster members. Numerous UBV and BV measurements are alsopublished. The data are used to determine relations for the predictionof UBV intrinsic colors for late-type stars from the narrow-bandmeasurements. The main purpose of the data is to constitute the basisfor the determination of solar-neighborhood space densities of late-typestars, mainly giants of different kinds; these space densities will becombined with narrow-band data for fainter stars in the north Galacticpole region to yield the decrease of space density with distance fromthe galactic plane for many kinds of late-type stars. Photoelectric observations of lunar occultations. XIVObservations of 314 events between March 1982 and March 1983 arepresented in the same form as in previous papers in this series. Theresults include 18 cases of duplicity of which 12 appear to be newdiscoveries. Unluckily, few cases of possible large angular diametersoccurred, and in no case can the results derived be accepted withoutreserve, these having been observed at very low altitudes withconsequent severe corruption by seeing effects. A remark on theincidence of doubles in the Hyades and its possible consequences forestimates of stellar distances is appended. Lunar occultation angular diameter measurements.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1980AJ.....85.1505B&db_key=AST Lunar occultation summary. IIResults of a second series of 196 two-color photoelectric lunaroccultation observations made within the interval from May 1972 toDecember 1973 are reported. Each observation has been employed toestimate the minimum magnitude difference between the observed objectand possible undetected companions. The study contains the analyses of143 disappearances and 53 reappearances. Eighteen of these eventsmanifest some degree of multiplicity. Included in the study are 18occultation events for Pleiades members. Classification of 831 two-micron sky survey sources south of +5 degrees.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1975AJ.....80.1011H&db_key=AST Four-color, Hbeta, and UBV photometry for bright B-type stars in the northern hemisphereAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1971AJ.....76.1058C&db_key=AST Catalog of Indidual Radial Velocities, 0h-12h, Measured by Astronomers of the Mount Wilson ObservatoryAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970ApJS...19..387A&db_key=AST - and Broad-Band Photometry of Red Stars. Northern GiantsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1967ApJS...14..307E&db_key=AST
• - 没有找到链接 -
### 下列团体成员
#### 观测天体数据
星座: 雙子座 右阿森松: 06h52m00.00s 赤纬: +23°36'06.0" 视星: 5.65 距离: 157.233 天文距离 右阿森松适当运动: -39.4 赤纬适当运动: -9.4 B-T magnitude: 7.606 V-T magnitude: 5.844
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Characterization of warped product submanifolds of Lorentzian concircular structure Manifolds Commun. Korean Math. Soc.Published online March 8, 2019 SHYAMAL KUMAR HUI, TANUMOY PAL, and LAURIAN IOAN PISCORAN THE UNIVERSITY OF BURDWAN, Technical University of Cluj Napoca Abstract : Recently Hui et al. (\cite{HAP}, \cite{HAN}) studied contact CR-warped product submanifolds and also warped product pseudo-slant submanifolds of a $(LCS)_n$-manifold $\bar{M}$. In this paper we have studied the characterization for both these classes of warped product submanifolds. It is also shown that there do not exists any proper warped product bi-slant submanifold of a $(LCS)_n$-manifold. Although we constructed an example of a bi-slant submanifold of $(LCS)_n$-manifold. Keywords : $(LCS)_n$-manifold, CR- submanifold, pseudo slant submanifold, bi-slant submanifold, warped product submanifold MSC numbers : 53C15, 53C25, 53C40 Full-Text :
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# Vector Confusion In Applying Coulomb's Law
1. Jul 26, 2016
### PurelyPhysical
1. The problem statement, all variables and given/known data
http://imgur.com/48cLE6q
2. Relevant equations
Coulomb's law
3. The attempt at a solution
I can follow most of this problem, but I am unsure where the constants in front of the trig functions are coming from. Why is it 2cos(135), 1cos(45), 2cos(-45), etc?
2. Jul 27, 2016
### drvrm
pl. write down the electric field due to charges placed at those points on an unit positive charge placed at the coordinate under consideration, you will need the the distance square in the denominator and the field has been resolved in unit vector i and j directions...the factor 2 and i are coming due to those distances....
e.g. take the charge at a... it is distant sqrt(2)/2 ; take square then it will be 1/2 in the denominator so a factor of 2 in the numerator-resolve the field intensity in i and j direction along x and y respectively.
similarly check other ones.
3. Jul 27, 2016
### PurelyPhysical
Why are there two distances 2 and 1? Isn't the center point equally distant from all the other points?
edit: I wrote down the electric field. I see now where the constants in front of the cos are coming from. But is there a sign error for the constants in front of sin?
Last edited: Jul 27, 2016
4. Jul 27, 2016
### drvrm
check the full expression!
5. Jul 27, 2016
### PurelyPhysical
I see it now. What I didn't realize is that all of the y vectors are positive because the third vector points away from the positive charge. Thank you.
Last edited: Jul 27, 2016
6. Jul 27, 2016
### haruspex
There are not two distances. Factors 1 and 2 are the charges. The factors for the displacements are 2 (1/r2), the unit vectors $\hat i$ and $\hat j$ and the trig components, sin and cos..
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# Question 5c07c
May 20, 2015
The acid dissociation constant for this acid is equal to
So, you're dealing with a 0.085-M phenylacetic acid, ${C}_{8} {H}_{8} {O}_{2}$, solution, which has a pH of 2.68.
Use the pH of the solution to determine what the concentration of the hydronium ions, ${H}_{3} {O}^{+}$, is
[H_3O^(+)] = 10^(-pH_"sol") = 10^(-2.68) = 2.1 * 10^(-3)"M"
Since the acid produces this much hydronium ions at equilibrium, you can work backwards to determine what the acid dissociation constant, ${K}_{a}$, must be.
Use an ICE table for the equilibrium reaction that gets established when phenylacetic acid is placed in aqueous solution to help you determine the value of ${K}_{a}$
$\text{ } {C}_{8} {H}_{8} {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {C}_{8} {H}_{7} {O}_{2 \left(a q\right)}^{-} + {H}_{3} {O}_{\left(a q\right)}^{+}$
I......0.085...........................................0.......................0
C......(-x)..............................................(+x)....................(+x)
E.....0.085-x........................................x........................x
You know that ${K}_{a}$ is equal to
K_a = ([H_3O^(+)] * [C_8H_7O_2""^(-)])/([C_8H_8O_2]) = (x * x)/(0.085 - x) = x^2/(0.085 - x)
You also know that $x$ is equal to the concentration of hydronium ions, which you've calculated earlier, so you get
K_a = (2.1 * 10^(-3))^2/(0.085 - 2.1 * 10^(-3)) = color(green)(5.3 * 10^(-5)#
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# Mark and his friends are out at Applebee’s. Their bill totaled 52.35. If they left the server a 20% tip, how much would each person pay splitting the bill evenly?
###### Question:
Mark and his friends are out at Applebee’s. Their bill totaled 52.35. If they left the server a 20% tip, how much would each person pay splitting the bill evenly?
### According to the World Health Organization, which of the following is not a true statement about health? ОА. Health includes physical wellness OB. Health is ONLY the absence of disease O C. Health includes social wellness OD Health includes mental wellness
According to the World Health Organization, which of the following is not a true statement about health? ОА. Health includes physical wellness OB. Health is ONLY the absence of disease O C. Health includes social wellness OD Health includes mental wellness...
### Desribe the diffrence between the honeymoon phase, tension phase, and violence phrase of domestic abuse
Desribe the diffrence between the honeymoon phase, tension phase, and violence phrase of domestic abuse...
### The pKa of benzoic acid is 4.2. At pH 2.0, it will exist mainly as and occupy the layer. At pH 4.2, it will exist mainly as and occupy the:_______________
The pKa of benzoic acid is 4.2. At pH 2.0, it will exist mainly as and occupy the layer. At pH 4.2, it will exist mainly as and occupy the:_______________...
### Help me please ASAP will mark brainliest!!
Help me please ASAP will mark brainliest!!...
### PLEASE HELP!!! In the passage above, for underlined sentence 2, which choice is best? Question 8 options: A. NO CHANGE B. Left secretly and later discovered by staff, each sculpture was delighted. C. Secretly delighted, each sculpture was discovered by staff. D. Each sculpture was left secretly and later discovered by delighted staff.
PLEASE HELP!!! In the passage above, for underlined sentence 2, which choice is best? Question 8 options: A. NO CHANGE B. Left secretly and later discovered by staff, each sculpture was delighted. C. Secretly delighted, each sculpture was discovered by staff. D. Each sculpture was left secretly and ...
### According to Ohm’s law, determine the calculated current for these values in Table A. voltage = 5 V, resistance = 20 Ω: A voltage = 20 V, resistance = 20 Ω: A voltage = 50 V, resistance = 20 Ω: A
According to Ohm’s law, determine the calculated current for these values in Table A. voltage = 5 V, resistance = 20 Ω: A voltage = 20 V, resistance = 20 Ω: A voltage = 50 V, resistance = 20 Ω: A...
### A slingshot shoots a marble. It travels at an average speed of 15 m/s for 12 seconds before hitting it's target. How far does it travel?
A slingshot shoots a marble. It travels at an average speed of 15 m/s for 12 seconds before hitting it's target. How far does it travel?...
### How crucial is setting to the story. "The most dangerous game"
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### Jshdvdndugdbdndjxbdnmd
jshdvdndugdbdndjxbdnmd...
### One way parents and guardians can influence technology use is to model proper use for their children. What outcome would you expect if this guideline was followed? a. Adults would not watch internet videos with children nearby. b. Children would not talk to friends in multiple social media formats. c. Adults would not use their phones when they are behind the wheel. d. Children would not be able to use the most current technology.
One way parents and guardians can influence technology use is to model proper use for their children. What outcome would you expect if this guideline was followed? a. Adults would not watch internet videos with children nearby. b. Children would not talk to friends in multiple social media formats...
### I have 2 shapes in my basket. together they have 7 sides.what shapes do i have?
I have 2 shapes in my basket. together they have 7 sides.what shapes do i have?...
### Someone please help ;-; Which of the following Are exterior angels ? Check all that apply A. ∠3 B.∠2 C.∠4 D.∠6 E.∠1 F.∠5
Someone please help ;-; Which of the following Are exterior angels ? Check all that apply A. ∠3 B.∠2 C.∠4 D.∠6 E.∠1 F.∠5...
### I need help with this please
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### PLSS HELP Sarah left the boat dock and sailed 5 miles due east. She turned and then sailed 10 miles due north. About how far is Sarah from the boat dock? Group of answer choices 10 miles 9 miles 15 miles 11 miles
PLSS HELP Sarah left the boat dock and sailed 5 miles due east. She turned and then sailed 10 miles due north. About how far is Sarah from the boat dock? Group of answer choices 10 miles 9 miles 15 miles 11 miles...
### Which order does this go in?
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### How did religion affect life in the united states today?
how did religion affect life in the united states today?...
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# Chapter 3 - Parallel and Perpendicular Lines - 3-3 Proving Lines Parallel - Practice and Problem-Solving Exercises: 19
$a\parallel b$
#### Work Step by Step
Use the Converse of the Same-Side Interior Angles Postulate.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# A borrows Rs. 800 at the rate of 12% per annum simple interest and B borrows Rs. 910 at the rate of 10% per annum simple interest. In how many years will their amounts of debt be equal?
Last updated date: 17th Mar 2023
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Hint: Assume that the amounts of both A and B will be equal after t years. Then find out the amount on the given sum of money at respective simple interest rates after t years of time. Then compare both amounts.
Let the amounts of debt of A and B will be equal after $t$ years.
Now, according to the question, A is borrowing Rs. 800 at the rate of 12% per annum.
Principal, $P = 800$ and rate, $r = 12\%$
We know that the amount of a sum on simple interest for $t$ years can be calculated as:
$\Rightarrow {\text{ Amt}}{\text{. }} = P\left( {1 + \dfrac{{r \times t}}{{100}}} \right)$
Using this formula, A’s amount of debt after $t$ years will be:
$\Rightarrow {\text{ }}{{\text{A}}_{{\text{amt}}{\text{.}}}} = 800 \times \left( {1 + \dfrac{{12 \times t}}{{100}}} \right) .....(i)$
Similarly, B is borrowing Rs. 910 at the rate of 10% per annum. So in this case, we have:
Principal, $P = 910$ and rate, $r = 10\%$.
Using the same formula, B’s amount of debt after $t$ years will be:
$\Rightarrow {\text{ }}{{\text{B}}_{{\text{amt}}}} = 910 \times \left( {1 + \dfrac{{10 \times t}}{{100}}} \right) .....(ii)$
As we have discussed earlier, amounts of debt of both A and B will be equal after $t$ years. Therefore we have:
$\Rightarrow {{\text{A}}_{{\text{amt}}}} = {\text{ }}{{\text{B}}_{{\text{amt}}}}$
Putting values from equation $(i)$ and $(ii)$, we’ll get:
$\Rightarrow 800 \times \left( {1 + \dfrac{{12 \times t}}{{100}}} \right) = 910 \times \left( {1 + \dfrac{{10 \times t}}{{100}}} \right), \\ \Rightarrow 80 \times \left( {1 + \dfrac{{12t}}{{100}}} \right) = 91 \times \left( {1 + \dfrac{{10t}}{{100}}} \right), \\ \Rightarrow 80 + \dfrac{{96t}}{{10}} = 91 + \dfrac{{91t}}{{10}}, \\ \Rightarrow \dfrac{{96t}}{{10}} - \dfrac{{91t}}{{10}} = 91 - 80, \\ \Rightarrow \dfrac{{5t}}{{10}} = 11, \\ \Rightarrow t = 22 \\$
Thus the amounts of debt of A and B will be equal after 22 years.
Note: If the sum is kept on compound interest instead of simple interest, then the amount is calculated as:
$\Rightarrow {\text{ Amt}}{\text{.}} = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$, where P is the principal sum kept initially, r is the rate of compound interest and t is the time period.
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1. ## Discrete maths
this question is confusing i dont understand it
2. Hello, sabina_19!
It seems clear enough . . . What part is confusing?
By using the appropriate truth tables, decide whethr
each of the following is tautology, absurdity, or contingency.
A tautology is always true.
An absurdity is always false.
A contingency is sometime true, sometimes false. .("It depends.")
$(a)\;\;\begin{array}{|c||ccc|}\hline
p & p & \to & p \\ \hline \hline
T & T & {\color{blue}T} & T \\ \hline
F & F & {\color{blue}T} & F \\ \hline \end{array}\qquad \text{Tautology}$
$(b)\;\;\begin{array}{|c||ccc|}\hline
p & p & \to & \sim\!p \\ \hline \hline
T & T & {\color{blue}F} & F \\ F & F &{\color{blue}T} & T \\ \hline
$(c)\;\;\begin{array}{|c||ccc|} \hline
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Algebraic Geometry Seminar
Feng Hao
Purdue
The Weak Bounded Negativity Conjecture
Abstract: In this talk I will give a proof of the Weak Bounded Negativity Conjecture, which says that given any complex smooth projective surface, for any reduced curve $C$ in $X$ and integer $g$, assume that the geometric genus of each component of $C$ is bounded from above by $g$, then the self-intersection number $C^2$ is bounded from below. The Weak Bounded Negativity Conjecture is motivated by the old folklore Bounded Negativity conjecture, which says that given any complex smooth projective surface, the self-intersection number of any reduced curve is bounded from below. Also, the Bounded Negativity Conjecture has an interesting relation with the Nagata conjecture. I will introduce those background before the proof of the Weak Bounded Negativity Conjecture. Also, I will give some further thoughts towards the Bounded Negativity Conjecture.
Wednesday October 11, 2017 at 4:00 PM in SEO 427
UIC LAS MSCS > graduate studies > seminars >
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# SOCR EduMaterials AnalysisActivities ANOVA 1
## This SOCR Activity demonstrates the utilization of the SOCR Analyses package for statistical Computing. In particular, it shows how to use Analysis of Variance (ANOVA) and how to interpret the results
• ANOVA Background: Analysis of variance (ANOVA) is a class of statistical analysis models and procedures, which compare means by splitting the overall observed variance into different parts. The initial techniques of the analysis of variance were pioneered by the statistician and geneticist R. A. Fisher in the 1920s and 1930s, and is sometimes known as Fisher's ANOVA or Fisher's analysis of variance, due to the use of Fisher's F-distribution as part of the test of statistical significance. Read more about ANOVA.
• SOCR ANOVA: Go to SOCR Analyses and select One-way ANOVA from the drop-down list of SOCR analyses, in the left panel. There are three ways to enter data in the SOCR ANOVA applet
• Click on the Example button on the top of the right panel.
• Generate random data by clicking on the Random Example button
• Pasting your own data from a spreadsheet into SOCR ANOVA data table.
• Now, map the dependent and independent vartiables, by going to the Mapping tab, selecting columns from the available list and sending them to the corresponding bins on the right (see figure). Then press Calculate button to carry the ANOVA analysis.
• The quantitative results results will be in the tab labeled Results. The Graphs tab contains the QQ Normal plot for the residuals. In this case, we have a very significant grouping effect, indicated by the p-value < $$10^{-4}$$.
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# Using nullable types (C# Programming Guide)
Nullable types are types that represent all the values of an underlying value type T, and an additional null value. For more information, see the Nullable types topic.
You can refer to a nullable type in any of the following forms: Nullable<T> or T?. These two forms are interchangeable.
## Declaration and assignment
As a value type can be implicitly converted to the corresponding nullable type, you assign a value to a nullable type as you would for its underlying value type. You also can assign the null value. For example:
double? pi = 3.14;
char? letter = 'a';
int m2 = 10;
int? m = m2;
bool? flag = null;
// Array of nullable type:
int?[] arr = new int?[10];
## Examination of a nullable type value
Use the following readonly properties to examine an instance of a nullable type for null and retrieve a value of an underlying type:
The code in the following example uses the HasValue property to test whether the variable contains a value before displaying it:
int? x = 10;
if (x.HasValue)
{
Console.WriteLine($"x is {x.Value}"); } else { Console.WriteLine("x does not have a value"); } You also can compare a nullable type variable with null instead of using the HasValue property, as the following example shows: int? y = 7; if (y != null) { Console.WriteLine($"y is {y.Value}");
}
else
{
Console.WriteLine("y does not have a value");
}
Beginning with C# 7.0, you can use pattern matching to both examine and get a value of a nullable type:
int? z = 42;
if (z is int valueOfZ)
{
Console.WriteLine($"z is {valueOfZ}"); } else { Console.WriteLine("z does not have a value"); } ## Conversion from a nullable type to an underlying type If you need to assign a nullable type value to a non-nullable type, use the null-coalescing operator ?? to specify the value to be assigned if a nullable type value is null (you also can use the Nullable<T>.GetValueOrDefault(T) method to do that): int? c = null; // d = c, if c is not null, d = -1 if c is null. int d = c ?? -1; Console.WriteLine($"d is {d}");
Use the Nullable<T>.GetValueOrDefault() method if the value to be used when a nullable type value is null should be the default value of the underlying value type.
You can explicitly cast a nullable type to a non-nullable type. For example:
int? n = null;
//int m1 = n; // Doesn't compile.
int n2 = (int)n; // Compiles, but throws an exception if n is null.
At run time, if the value of a nullable type is null, the explicit cast throws an InvalidOperationException.
A non-nullable value type is implicitly converted to the corresponding nullable type.
## Operators
The predefined unary and binary operators and any user-defined operators that exist for value types may also be used by nullable types. These operators produce a null value if one or both operands are null; otherwise, the operator uses the contained values to calculate the result. For example:
int? a = 10;
int? b = null;
int? c = 10;
a++; // a is 11.
a = a * c; // a is 110.
a = a + b; // a is null.
Note
For the bool? type, the predefined & and | operators don't follow the rules described in this section: the result of an operator evaluation can be non-null even if one of the operands is null. For more information, see the Nullable Boolean logical operators section of the Boolean logical operators article.
For the relational operators (<, >, <=, >=), if one or both operands are null, the result is false. Do not assume that because a particular comparison (for example, <=) returns false, the opposite comparison (>) returns true. The following example shows that 10 is
• neither greater than or equal to null,
• nor less than null.
int? num1 = 10;
int? num2 = null;
if (num1 >= num2)
{
Console.WriteLine("num1 is greater than or equal to num2");
}
else
{
Console.WriteLine("num1 >= num2 is false (but num1 < num2 also is false)");
}
if (num1 < num2)
{
Console.WriteLine("num1 is less than num2");
}
else
{
Console.WriteLine("num1 < num2 is false (but num1 >= num2 also is false)");
}
if (num1 != num2)
{
Console.WriteLine("num1 != num2 is true!");
}
num1 = null;
if (num1 == num2)
{
Console.WriteLine("num1 == num2 is true if the value of each is null");
}
// Output:
// num1 >= num2 is false (but num1 < num2 also is false)
// num1 < num2 is false (but num1 >= num2 also is false)
// num1 != num2 is true!
// num1 == num2 is true if the value of each is null
The above example also shows that an equality comparison of two nullable types that are both null evaluates to true.
For more information, see the Lifted operators section of the C# language specification.
## Boxing and unboxing
A nullable value type is boxed by the following rules:
• If HasValue returns false, the null reference is produced.
• If HasValue returns true, a value of the underlying value type T is boxed, not the instance of Nullable<T>.
You can unbox the boxed value type to the corresponding nullable type, as the following example shows:
int a = 41;
object aBoxed = a;
int? aNullable = (int?)aBoxed;
Console.WriteLine($"Value of aNullable: {aNullable}"); object aNullableBoxed = aNullable; if (aNullableBoxed is int valueOfA) { Console.WriteLine($"aNullableBoxed is boxed int: {valueOfA}");
}
// Output:
// Value of aNullable: 41
// aNullableBoxed is boxed int: 41
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# Hurst Exponent in Excel
Occasionally, our support desk receives inquiries about Hurst exponent: what is it? How do we use it in Excel? And how do we interpret the calculated values? In this issue, we will go over the Hurst exponent in-depth, and hopefully, help you develop an intuition and insight for the Hurst exponent.
## What is Hurst exponent?
The name "Hurst exponent," "Hurst Index", or "Hurst coefficient", derives from Harold Edwin Hurst (1880–1978), who was the lead researcher in these studies. Studies involving the Hurst exponent were initially developed in hydrology for the practical matter of determining optimum dam sizing for the Nile river's volatile rain and drought conditions that had been observed over a long period.
The Hurst exponent (H) is used as a measure of long-term memory of time series. It relates to the time series's autocorrelations and the rate at which these decrease as the lag between pairs of values increases. The Hurst exponent is often referred to as the "index of dependence" or "index of long-range dependence."
## What is the long memory of a process?
Long memory, also called Long-range dependence (LRD) or long-range persistence, is a phenomenon that may arise in time-series data. It relates to the decay rate of statistical dependence of two points as the time between the points increases.
Does the ARMA (P, Q) process exhibit a long memory? No! A stationary ARMA with finite P and Q orders, ARMA(P, Q), has a short memory. You can examine the autocorrelation function (ACF) plot as its value decays exponentially and dies out after few lags.
How does a long-memory model behave?
In general, a process with a long memory may look like a slow random-walk (drifting), with a slow decaying autocorrelation function. For example, let’s examine the monthly average carbon dioxide (CO2) level recorded in Mauna Loa, Hawaii weather station.
Next, let’s remove the 12-month seasonality by differencing the value of each observation, from one that is 12-month earlier.
In the correlogram plot, the autocorrelation factors (ACF) are decaying, but at a very slow pace.
How do we model the long-memory time series? Just like we did in the non-stationary ARIMA model: we extract the fractional integration component and capture the short-memory in the residuals with an ARMA model.
Using fractional difference operator, we capture the long-memory dynamic in a time series:
${(1 - L)^d} = \sum\limits_{k = 0}^\infty {\left( {\begin{array}{*{20}{c}} d\\ k \end{array}} \right)} {( - 1)^k}{L^k} = 1 + {\omega _1}L + {\omega _2}{L^2} + ...$
Where
• $L$ = Lag or Backshift operator
• ${\omega _1} = - d$
• ${\omega _2} = - \frac{{{\omega _1} \times (d - 1)}}{2}$
• ${\omega _N} = - \frac{{{\omega _{N - 1}} \times (d - N - 1)}}{N}$
For $\left| d \right| \le \frac{1}{2}$, the coefficient ${\omega _k}$ decay power-like relatively quickly (yet, slower than the exponential decay).
Putting it all together, we end up with fractional ARIMA (i.e., FARIMA)
$(1 - {\phi _1}L - {\phi _2}{L^2} - ... - {\phi _p}{L^p}){(1 - L)^d}{X_t} = (1 + {\theta _1}L + {\theta _2}{L^2} + ... + {\theta _q}{L^q}){a_t}$
Where
• $L$ = Lag or Backshift operator
• ${X_t}$ = time series data set
• ${a_t}$ = innovation (or shocks) time series
• $d$ = integration order, and its value between -0.5 and 0.5, exclusive.
How do we find out the integration order (d)? The fractional integration order (d) is equal to the Hurst exponent (H) minus 0.5 (i.e, d = H - 0.5)
## Interpretation
In a nutshell, the Hurst exponent is a single value (H), which we can use to draw an observation about the time series long-memory (serial correlation):
H Interpretation
0.5 - 1.0 a time series with long-term positive autocorrelation
0.0 - 0.5 indicates a time series with long-term switching between high and low values in adjacent pairs, meaning that a low value will probably follow a single high value and that the value after that will tend to be high, with this tendency to switch between high and low values lasting a long time into the future
0.5 a completely uncorrelated series, but in fact, it is the value applicable to series for which the autocorrelations at small-time lags can be positive or negative but where the absolute values of the autocorrelations decay exponentially quickly to zero
Important: For a time series with a Hurst exponent equal to 0.5, we conclude that time series does not have a long-memory (or long-range dependency), but this is not the same as saying the time series is a white-noise, as there may be one or more significant auto-correlation factor at lower lag-order(s).
## Calculation
The original and best-known method for estimating the Hurst exponent is the so-called rescaled range (R/S) analysis based on Hurst's previous hydrological findings.
The NumXL Hurst(.) function calculates the original (empirical) Hurst exponent when you set the return type = 1
=Hurst(X, Alpha, 1)
However, this approach is known to produce biased estimates. For a small sample size, there is a significant deviation from the 0.5 slopes (i.e., uncorrelated long-range).
### Size-corrected (Anis-Llyod) estimate
To correct the built-in bias in the original (empirical) Hurst exponent estimate, Anis-LIyod introduced a size-corrected estimate of the rescaled range (R/S).
The NumXL Hurst(.) function calculates the Anis-Llyod (corrected R/S) Hurst exponent when you set the return type = 2
= Hurst(X,Alpha,2)
### Statistical significance
No asymptotic distribution theory has been derived for most of the Hurst exponent estimators so far. However, we have an approximate functional form for the confidence intervals of the Anis-Lloyd corrected R/S analysis.
To examine the statistical significance of the calculated Hurst exponent estimate(), we construct the following test of the hypothesis:
$\begin{array}{l} {{\rm{H}}_o}:{H_q} = {\rm{ uncorrelated}}\\ {{\rm{H}}_1}:{H_q} = {\rm{ long - memory}} \end{array}$
Next, we calculate the corresponding Hurst exponent estimate and the confidence interval (C.I.) limits of an uncorrelated (no long-memory) time-series for a given sample size.
The NumXL Hurst(.) function calculates the Anis-Llyod (corrected R/S) Hurst exponent for uncorrelated time-series of the same size when you set the return type to 3
= Hurst(X,Alpha,3)
NumXL Hurst(.) function calculates the lower and upper limits of the confidence interval of Anis-Llyod Hurst exponent of the uncorrelated time-series when you set the return type to 4 and 5, respectively
LL= Hurst(X,Alpha,4)
UL= Hurst(X,Alpha,5)
Finally, we examine the Anis-Llyod (corrected R/S) Hurst exponent value against the C.I. of the Null-hypothesis (uncorrelated time series).
• The Hurst exponent estimate is outside the C.I.; thereby, the time series has a long memory.
• The Hurst exponent is inside the C.I., thereby, the time series does not exhibit a significant long memory property, and observations can be uncorrelated.
#### Hurst exponent analysis in Excel
Let's examine the 12-month de-seasonaled log CO2 level between March 1958 and November 2020.
The Anis-Llyod corrected R/S Hurst exponent estimate is 0.84, and this value is outside the C.I. of Hurst exponent of an uncorrelated time series of the same size. The de-seasonaled CO2 log level time series exhibits a long-memory behavior, and the fractional difference order (d) is 0.34 (i.e., 0.84 -0.50 = 0.34).
|
Nylah Hendrix
2022-07-04
If ${F}_{1}$ and ${F}_{2}$ are the antiderivatives of f(x), then ${F}_{1}={F}_{2}$?
Ordettyreomqu
Expert
Step 1
The statement is equivalent to: ${F}_{1}^{\prime }={F}_{2}^{\prime }=f$
but as others have touched on, the equality is not maintained for a definite integral. What this means is that whilst:
${\int }_{{x}_{1}}^{{x}_{2}}{F}_{1}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{{x}_{1}}^{{x}_{2}}{F}_{2}\phantom{\rule{thinmathspace}{0ex}}dx$
${F}_{1}\left({x}_{2}\right)-{F}_{1}\left({x}_{1}\right)={F}_{2}\left({x}_{2}\right)-{F}_{2}\left({x}_{1}\right)$
Step 2
this does not imply that either of the following are true:
${F}_{1}\left({x}_{1}\right)={F}_{2}\left({x}_{1}\right)\phantom{\rule{0ex}{0ex}}{F}_{1}\left({x}_{2}\right)={F}_{2}\left({x}_{2}\right)$
${F}_{1}\left({x}_{1}\right)={F}_{2}\left({x}_{1}\right)\phantom{\rule{0ex}{0ex}}{F}_{1}\left({x}_{2}\right)={F}_{2}\left({x}_{2}\right)$
which is just the same as how: $2-1=100-99⇏2=100$
This is what it means for functions to differ by a constant, and how two functions can be antiderivatives of the same function
gorgeousgen9487
Expert
Step 1
It doesn't mean they cannot be equal. It means they aren't necessarily equal. If we know that ${F}_{1}^{\prime }\left(x\right)={F}_{2}^{\prime }\left(x\right)=2x$, then it might be true that
${F}_{1}\left(x\right)={x}^{2}\phantom{\rule{0ex}{0ex}}{F}_{2}\left(x\right)={x}^{2}$
${F}_{1}\left(x\right)={x}^{2}\phantom{\rule{0ex}{0ex}}{F}_{2}\left(x\right)={x}^{2}$
Step 2
But we could also have ${F}_{1}\left(x\right)={x}^{2}+\pi \phantom{\rule{0ex}{0ex}}{F}_{2}\left(x\right)={x}^{2}-1000$
${F}_{1}\left(x\right)={x}^{2}+\pi \phantom{\rule{0ex}{0ex}}{F}_{2}\left(x\right)={x}^{2}-1000$
So we can't conclude that ${F}_{1}={F}_{2}$, but we also cannot prove ${F}_{1}\ne {F}_{2}$.
If this was a true / false question, then technically the statement ${F}_{1}={F}_{2}$ is undecided. But by my powers of reading the minds of problem authors, they didn't mean to ask whether ${F}_{1}$ is equal to ${F}_{2}$, but whether ${F}_{1}$ is necessarily equal to ${F}_{2}$ (this is just an implicit part to many of these problems that really, really ought to be explicit), which means that the intended answer is "False". "Is it true that" is often used synonymously with "Does it follow that", and that's a shame.
Do you have a similar question?
|
# B ODE question
1. Aug 4, 2017
### knockout_artist
(sorry missed n in the subject)
Hello,
How equation 2.2 and 2.3 came about from equation 2.1?
2. Aug 4, 2017
### Staff: Mentor
Equation $2.2$ is a definition and equation $2.3$ follows from both with the product rule.
|
# X File
## Recommended Posts
Hello there, I am trying to create a screensaver using .x models. I successfully loaded the model in my program but the materials that I used while designing that model is not getting loaded in my program? Only textures get loaded and the part where material was used get displayed without any texture or material. Here is the source code that I have used to load the model
LPD3DXBUFFER buf;
D3DXMESH_SYSTEMMEM,
dev,
NULL,
&buf,
NULL,
&dwNumMats,
&home);
D3DXMATERIAL *tMats=(D3DXMATERIAL*)buf->GetBufferPointer();
houseMats=new D3DMATERIAL9[dwNumMats];
houseTexs=new LPDIRECT3DTEXTURE9[dwNumMats];
for(DWORD i=0;i<dwNumMats;i++)
{
houseMats[i]=tMats[i].MatD3D;
houseMats[i].Ambient=houseMats[i].Diffuse;
sprintf(texPath,"models\\%s",tMats[i].pTextureFilename);
if(FAILED(D3DXCreateTextureFromFile(dev,texPath,&houseTexs[i])))
{
houseTexs[i]=NULL;
}
}
for rendering the model I have used following code
for(DWORD i=0;i<dwNumMats;i++)
{
dev->SetMaterial(&houseMats[i]);
if(houseTexs[i]!=NULL)
dev->SetTexture(0,houseTexs[i]);
home->DrawSubset(i);
}
What should I do while designing the model so that the material will get loaded?
##### Share on other sites
Quote:
Original post by sachingoradeI successfully loaded the model in my program but the materials that I used while designing that model is not getting loaded in my program?Only textures get loaded and the part where material was used get displayed without any texture or material.
Is the loading failing, or is the model not using the textures? Does the model work in the X File Viewer that comes with the SDK? Are there any errors or warnings from the Debug Runtimes? How do you know the materials aren't applied, or are you assuming they're not because the model is showing up untextured?
##### Share on other sites
By reading some posts and tutorials I think I need to load the effects also to
Can you give me any simple example or code to load effect from the .x file?
While exporting the model I have selected the option
1. Include .fx file
2. Include .fx parameters
I have no idea about effect files.
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# Op-amp voltage regulator
## Homework Statement
FIGURE 3 (on page 7) shows a PIR (passive infra-red) detector and its
associated amplifier, as used in burglar alarm systems1.
(a) The detector is powered from a 12 V unregulated supply that needs
to be stepped down to 5 V. Design a suitable 12-to-5 V voltage
converter using an op-amp and a diode that has the forward
characteristics given in FIGURE 3(b).
(b) Estimate the quiescent voltages at the inputs and outputs of the two
op-amps and the overall voltage gain (in decibels) of the circuit at the
frequency of operation. State any assumptions made.
## Homework Equations
$V_-=V_o\frac{R_2}{R_1+R_2}$
Vo = GV (V+ – V–)
Vd = V+
## The Attempt at a Solution
I think pic 2 is the circuit they are asking for.
as for the rest.
Any help would be appreciated.
Thanks
#### Attachments
• pic 1.png
49.3 KB · Views: 443
• pic2.PNG
4.9 KB · Views: 441
• pic3.PNG
7.6 KB · Views: 416
donpacino
Gold Member
Why don't you give the problem your best shot...
(a)
If we work with a forward voltage VD of 320mV this means the diode would require a forward current If of 1mA. The resister R3 would be requred to drop the remanig voltage and limit the current through the diode
$$R_3 = \frac{5-0.32}{0.001}=4.68K$$
The voltage at V_ and V+ must be equal when the system is balance at the required output voltage.
$$V_D = V_+$$
The inverting input is worked out as R1 and R2 form a potential divider.
$$V_o=V_D\frac{R_1+R_2}{R_2}=5$$
$$\frac{V_D}{V_O}=\frac{R_2}{R_1+R_2}=\frac{0.32}{5}=0.064$$
So if we give R2 a value of “1” we can find the ratio of R2 to R1.
$$\frac{1}{0.064}-1=14.625$$
So if we give R2 3.3k then R1 will be 48k
As for part (b) I don't know where to star
rude man
Homework Helper
Gold Member
(a)
If we work with a forward voltage VD of 320mV this means the diode would require a forward current If of 1mA. The resister R3 would be requred to drop the remanig voltage and limit the current through the diode
$$R_3 = \frac{5-0.32}{0.001}=4.68K$$
The voltage at V_ and V+ must be equal when the system is balance at the required output voltage.
$$V_D = V_+$$
The inverting input is worked out as R1 and R2 form a potential divider.
$$V_o=V_D\frac{R_1+R_2}{R_2}=5$$
$$\frac{V_D}{V_O}=\frac{R_2}{R_1+R_2}=\frac{0.32}{5}=0.064$$
So if we give R2 a value of “1” we can find the ratio of R2 to R1.
$$\frac{1}{0.064}-1=14.625$$
So if we give R2 3.3k then R1 will be 48k
All very well done for part (a).
As for part (b) I don't know where to star
As for part (b) you haven't given us the frequency of operation so can't do part (b).
Can you flip the image 90 deg for us?
Since they're asking for voltage gain you also have to define the input point. The overall circuit gain is volts/watt or volts/lumen or similar with infrared input quantity.
donpacino
Gold Member
As for part (b) you haven't given us the frequency of operation so can't do part (b).
Can you flip the image 90 deg for us?
Since they're asking for voltage gain you also have to define the input point. The overall circuit gain is volts/watt or volts/lumen or similar with infrared input quantity.
assuming they don't give you a frequency, you might need to make it generic (the answer will simply have f as a parameter).
No frequencies are given.
That all the info I have been given.
Although it says state any assumptions.
I assume the RC feedback is some kind of high or low pass filter. I really don't know where to start.
Thanks for you help.
donpacino
Gold Member
No frequencies are given.
That all the info I have been given.
Although it says state any assumptions.
I assume the RC feedback is some kind of high or low pass filter. I really don't know where to start.
Thanks for you help.
I would first calculate the operating points at DC.
Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
donpacino
Gold Member
I would first calculate the operating points at DC.
Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
if you really dont know where to start, maybe do nodal anlysis or some other system at DC (at dc capacitors are open circuit). It should be easy to find the voltages at the IOs of the op amps.
I really don't have a clue.
so R1and R2 form a potential divider giving 2V at the non-inverting inputs of X3 and X1
but I am not sure about the inverting side this looks like it is held low at 0V
donpacino
Gold Member
I really don't have a clue.
so R1and R2 form a potential divider giving 2V at the non-inverting inputs of X3 and X1
but I am not sure about the inverting side this looks like it is held low at 0V
there is a property of op-amps that states with negative feedback the inverting and non inverting terminals will be equal
I would first calculate the operating points at DC.
Information I found on the web is saying PIR sensors operate in the 1 Hz range. You could calculate it there.
Is this for a homework problem. you can ask your professor for clarification. Or like I said earlier calculate as a function of frequency.
I think, the whole circuit resembles a kind of bandpass amplifier. Why not using the mid frequency Fo as operating frequency?
A circuit simulation can give the value of Fo.
donpacino
Gold Member
I think, the whole circuit resembles a kind of bandpass amplifier. Why not using the mid frequency Fo as operating frequency?
A circuit simulation can give the value of Fo.
that could work (assuming the circuit is properly designed/tuned). I assume op is supposed to do this analytically.
Nidum
Gold Member
(a) The detector is powered from a 12 V unregulated supply that needs
to be stepped down to 5 V. Design a suitable 12-to-5 V voltage
converter using an op-amp and a diode that has the forward
characteristics given in FIGURE 3(b).
There is no requirement to use that specific circuit is there ?
There are other ways of doing the same thing and some of these other ways will certainly be easier to analyse .
donpacino
Gold Member
There is no requirement to use that specific circuit is there ?
There are other ways of doing the same thing and some of these other ways will certainly be easier to analyse .
This looks like a homework problem. I doubt a "solution" would be resdesign the problem.
|
## Introduction
Interest in copper-oxide superconductors (cuprates) is fueled by their technological potential and the outstanding mystery of the mechanism governing high-temperature superconductivity, which stifles prediction of new superconductors. It is known that, in cuprates, superconductivity is hosted in the crystallographic ab-planes. This induces anisotropy γ between the in-plane (ab) and out-of-plane (c-axis) fundamental superconducting parameters, such as the penetration depth λab = λc/γ and coherence length ξab = γξc. When evaluating the potential of superconductors for technological applications, high anisotropy compels considerations beyond the typical metrics of high critical temperature Tc, critical current density Jc, and upper critical field Hc2. This is because thermal fluctuations profoundly impact anisotropic materials’ electronic and magnetic properties, which are significantly influenced by the dynamics of vortices. Consequently, thermally activated vortex motion (creep) is fast and Jc vanishes at an irreversibility field Hirr that can be much less than Hc2, potentially negating the otherwise advantageous properties of these materials. Understanding vortex dynamics in cuprates is not only technologically relevant, but also can substantially contribute to the debate over the degree to which superconductivity in cuprates is conventional1.
Magnetic flux penetrates superconductors immersed in fields greater than the lower critical field Hc1. This does not quench superconductivity in high-Tc materials provided that the field remains below Hc2. In layered cuprates, interior flux can appear as stacks of weakly-coupled 2D pancake vortices, each localized on a Cu-O plane. Pancake vortices within a stack are not necessarily aligned and interact both magnetically (owing to their moments) and through Josephson coupling between pancakes in adjacent planes. If this coupling is sufficiently strong, the stacks may behave as continuous strings, hence be considered 3D vortex lines. The differing dynamics of 2D pancakes and 3D vortex lines should therefore play a major role in determining the phase diagram in highly anisotropic materials.
The superconductor HgBa2CuO4+δ (Hg1201) is recognized as ideal for systematically studying the effects of high anisotropy. This is because its clean microstructure enables probing intrinsic, rather than sample-dependent, properties associated with high anisotropy2. Specifically, Hg1201 crystals do not contain common defects, such as twin-boundaries and rare-earth-oxide precipitates3,4,5. Furthermore, it has a simple tetragonal structure and optimally doped Hg1201 has the highest Tc among single Cu-O layer materials, permitting thorough studies of the effects of thermal fluctuations on the superconducting state. Despite these desirable characteristics, the paucity of research on Hg1201 results from the challenges of growing large, high-quality single crystals.
In this paper, we identify vortex phase boundaries and glassy regimes in the vortex phase diagram of a clean, optimally doped Hg1201 single crystal. We find that the temperature-dependent vortex penetration field Hp(T), second magnetization peak Hsmp(T), and irreversibility field Hirr(T) all decay exponentially at low temperatures and exhibit an abrupt change in behavior at high temperatures. We present complementary vortex creep measurements over a wide range of the phase diagram that reveal the broad extent to which the dynamics of pancake vortices determine the magnetic properties in our sample. Our main findings from these measurements are as follows: First, the crystal hosts a vortex glass state characterized by collective creep of large bundles of pancake vortices at low temperatures T/Tc ≤ 0.4 and applied fields μ0H < 0.5 T. The glass state persists at higher fields, yet the bundle size shrinks. Second, we find temperature-tuned crossovers from elastic (collective) dynamics to plastic flow. By measuring at the crossover temperature over an extended time frame, we additionally capture a transition from elastic to plastic dynamics over time. Last, we show that the second magnetization peak does not originate from elastic-to-plastic crossovers over most of the phase diagram; these crossovers only coincide with the second magnetization peak at low temperatures T/Tc < 0.2.
## Results
### Critical temperature and anisotropy
Two of the sample’s key characteristics, the critical temperature Tc and the anisotropy γ, were extracted from temperature- and field-dependent magnetization measurements as summarized in Fig. 1. The temperature sweeps M(T) in a field of 5 Oe yielded a critical temperature of Tc ≈ 95.9 K, see Fig. 1(a), consistent with near optimal doping2,6. To determine the anisotropy, we measured the ratio of the transverse (MT) to the longitudinal (ML) magnetization at various field orientations (θ) relative to the c axis in the reversible (vortex liquid) regime. The raw data is plotted in Fig. 1(b,c). As shown in Fig. 1(d), a least squares fit of the data to Eq. (1), the Kogan model7,8,
$$\frac{{M}_{T}}{{M}_{L}}=(1-{\gamma }^{2})\frac{\sin \,\theta \,\cos \,\theta }{{\sin }^{2}\theta +{\gamma }^{2}{\cos }^{2}\theta },$$
(1)
produces an anisotropy of γ ≈ 32. This is consistent with previous work on optimally doped Hg1201 single crystals. Specifically, angle dependent torque magnetometry studies9,10,11,12 found γ ≈ 27−30. Additionally, a study13 that measured the magnetization at two field orientations, perpendicular (M) and parallel (M||) to the CuO2 planes, found γ ≈ 30 using a self-consistency equation from anisotropic Ginzburg-Landau theory M(H) = γM||(γH).
### Irreversible Magnetization
Isothermal magnetization loops were recorded for H||c and at T = 5−65 K. Select curves are displayed in Fig. 2. In all cases, the field was first swept to −3 or −4 T (not shown) to establish the critical state (full flux penetration throughout the sample). The lower branch of the loop was subsequently measured as the field was ramped from 0 T to 7 T, and the upper branch was collected as the field was swept back down. All curves exhibit a distinct shape with two conspicuous features: a dip in the magnitude of M near the onset field Hon and a second magnetization peak (SMP) at Hsmp. In general, this shape and the magnitude of the magnetization is indicative of a weak vortex pinning regime at low fields ($$H\lesssim {H}_{on}$$) and stronger pinning at higher fields. We observed similar results in measurements of our other Hg1201 crystals. The source of pinning is likely point defects in the Hg-O layer—specifically, oxygen interstitials and mercury vacancies3,4,5, and this should be the main source of disorder in the bulk that hinders thermal wandering of vortices.
In high-temperature superconducting crystals, a surface barrier—called the Bean-Livingston (BL) barrier—often plays a significant role in determining the magnetic properties and shaping the M(H) loops14,15,16,17. It originates from competing effects: vortices are repelled from the surface by Meissner shielding currents and attracted by a force arising from the boundary conditions (usually modeled as the attraction between the vortex and an image antivortex18). Disappearing at the penetration field Hp ≈ κHc1/lnκ, the BL barrier impedes vortex entry and exit from the sample in fields less than Hp > Hc1, where Hc1 is the lowest field at which flux penetration is thermodynamically favorable. The contribution of the barrier is magnified in materials with high Ginzburg-Landau parameters κ = λ/ξ and diminished by surface imperfections. Creep of pancake vortices over the barrier produces the exponential temperature dependence17
$${H}_{p}(T)\simeq {H}_{c}{e}^{-T/{T}_{0}},$$
(2)
where T0 = ε0dln(t/t0), $${\varepsilon }_{0}={\Phi }_{0}^{2}/4\pi {\mu }_{0}{\lambda }_{ab}^{2}$$ is the vortex line energy or tension, d is the spacing between CuO2 planes, t is the time scale of the measurement, t0 ~ 10−10–10−8 s relates to the vortex penetration time19, and the thermodynamic critical field is $${H}_{c}={\Phi }_{0}/2\sqrt{2}\pi {\xi }_{ab}{\lambda }_{ab}$$.
To investigate the relevance of surface barriers in our sample, we measured the field at which vortices first penetrate into the sample peripheries Hp by collecting the zero-field cooled M(H) isotherms shown in Fig. 3(a): we defined Hp as the field at the departure from linearity. Figure 3(a) inset shows the extraction technique and the phase diagram in Fig. 4 contains the resulting temperature dependence. We find that Hp(T) follows Eq. (2) at low temperatures T/Tc < 0.55, and a least squares fit produces T0 = 21.3 ± 0.7 K and Hc = 0.06 T. The experimentally extracted T0 is reasonably close to the estimate T0 = 26 K, calculated assuming t ~ 100 s, t0 = 10−10 s, d ≈ 9.5 Å13, and λab ≈ 162 nm20.
To accurately assess the field of first penetration, we must account for local field enhancements due to demagnetizing effects by multiplying our extracted value by 1/(1 − N), where N is the effective demagnetizing factor. Expressing the field enhancement in terms of a demagnetizing factor formally exclusively applies to samples with elliptic cross-section (per refs. 21,22). For samples with rectangular cross-section (width w and thickness δ, $$w\gg \delta$$), we must account for the fact that the field of first penetration is retarded by a geometrical barrier23,24. This barrier is associated with a parametrically lower field enhancement at the sample edge as compared to an elliptic slab with the same dimensions. In this case, accurate values of the effective demagnetizing factors for rectangular prisms have been calculated numerically25. Following the calculations from ref. 26, we find N ≈ 0.75 which yields Hc = 0.24T and hence a coherence length ξab(0) ~ 1.5 nm, similar to the value of ξab(0) ~ 2.0 ± 0.4 nm measured by Hofer et al.11. We attribute the deviation of the penetration field at low temperature from the expected scaling of Hc1(T), see Fig. 4 for T/Tc < 0.55, to the increasing importance of thermal creep to overcome the surface/geometric barriers.
Agreement of our Hp(T < 0.55Tc) data with Eq. (2) indicates that, at low T, vortices enter as pancakes (as evinced by creep measurements, see discussion in the next section) and are thermally activated over the BL barrier. At T/Tc ~ 0.55, the temperature dependence of Hp abruptly changes, suggestive of a different mechanism for vortex penetration at higher temperatures. Similar crossovers have been observed in other layered superconductors16,27,28,29,30,31 around T/Tc ~ 0.5. We anticipate such a crossover when Hp(T) from Eq. (2) falls below Hc1: in this case, we would expect Hp(T) to be bound by $$(1-N){H}_{c1}(T)=[{\Phi }_{0}/(4\pi {\lambda }_{ab}^{2}(T))]\,\mathrm{ln}\,\kappa$$ for T above the crossover. Considering the two-fluid approximation, λab(T) = λab(0)[1−(T/Tc)4]−1/2, and a T-independent κ, this expression indeed fits the data for Hc1 = 300Oe, see Fig. 4. A competing scenario for the crossover—based on a transition of creep of pancakes to creep of vortex half-loops17,31—fails to produce the correct temperature dependence for T/Tc > 0.55. From this value for Hc1 we extract λab(0) = 154 nm, which is comparable with published data11,20.
For fields above Hp, but below Hsmp, |M| dips to an ill-defined minimum and increases again at Hon. Figure 3(b) magnifies this low-field plateau and Hon(T) is plotted in Fig. 4. Previous studies have related Hon to a transition between an ordered vortex lattice at low fields and an entangled lattice created by point disorder at higher fields, tuned by competition between thermal, pinning, and elastic energies. For example, FeSe1−x Tex single crystals showed evidence of a Bragg glass (quasi-ordered vortex solid) below Hon32 and a presumed disordered vortex solid above Hon. Additionally, YBa2Cu3O7−δ, Nd1.85Ce0.15CuO4−δ, and Bi2Sr2CaCu2O8+δ all demonstrate disorder induced phase transitions that show a signature in the M(H) loops33.
In applied magnetic fields above Hon, the magnetization apexes at the second magnetization peak Hsmp. Second magnetization peaks have been reported in studies of most classes of superconductors, including low-Tc34,35, iron-based32,36,37,38,39,40, and highly anisotropic28,41 materials, as well as YBa2Cu3O7−δ (YBCO) single crystals42. In fact, this peak has also been observed in a few previous studies20,43,44,45,46 of Hg1201 single crystals grown by two research groups4,45, though the peak is far more pronounced in our samples. This feature is typically either attributed to a crossover between vortex pinning regimes or a structural phase transition47,48 in the vortex lattice. Below, we will revisit the discussion of the second magnetization peak because creep measurements are requisite to evaluate possible origins of the SMP.
At sufficiently high fields, the loops close as the system transitions into the reversible regime at the irreversibility field Hirr. Instead of extracting Hirr from the isothermal magnetization loops M(H), we extract it from isomagnetic M(T) sweeps. This is more precise than measurements involving sweeping the field: temperature sweeps tend to induce less noise than field sweeps and, at the transition, the upper and lower branches of M(T) not only converge, but also exhibit a sharp change in slope. Figure 3(c) contains select M(T) datasets showing the extraction technique and the resulting irreversibility line is shown in Fig. 4.
For T/Tc < 0.6, we find that Hirr(T) $${e}^{-T/{T}_{0}}$$ (shown in Fig. 4), which yields T0 = 19.7 ± 0.6 K, produced by a least squares fit. Notice that T0 is close to the value T0 ≈ 21 K, extracted in the fit of our Hp data to Eq. (2) and identical to the value (T0 = 19.7 ± 0.4 K) extracted in another study on Hg1201 single crystals45. At higher temperatures T > T*, the shape of the irreversibility line changes. Similar trends in Hirr(T) have been found in grain-aligned Hg1201 samples49 and single crystals45.
Although the field of first penetration seems to be dominated by surface barrier effects, we conclude from the symmetric magnetization loops that bulk pinning is the dominant pinning source in our Hg1201 crystal after field penetration. Otherwise, i.e. if the contribution of bulk pinning were relatively insignificant, we would observe asymmetry between the upper and lower branches of the magnetization loops 18. Despite the observation of bulk pinning dominance, it remains unclear whether the similar exponential temperature scaling of Hp and Hirr have a common ground. Compiling the aforementioned results, Fig. 4 shows the resulting phase diagram on a semilog plot. In the following sections, we use magnetic relaxation measurements to learn more about the nature of vortex dynamics in the gray region of Fig. 4. The following sections present our main result – a more detailed understanding of vortex behavior derived from extensive vortex creep measurements.
### Vortex creep
The disorder landscape defines potential energy wells in which vortices will preferentially localize to reduce their core energies by a pinning energy U0. An applied or induced current tilts this energy landscape. This reduces the energy barrier that a pinned vortex must surmount to escape from a well to a current-dependent value U(J). The time required for thermal activation over such a barrier can be approximated by the Arrhenius form
$$t={t}_{0}{e}^{U(J)/{k}_{B}T}.$$
(3)
At low temperatures ($$T\ll {T}_{c}$$) and fields, the simple linear relationship U(J) = U0(1 − J/Jc0) proposed in the Anderson-Kim model50,51 is often accurate. However, because this model neglects vortex elasticity and vortex-vortex interactions, its relevance is often further limited to the early stages of the relaxation process ($$J\lesssim {J}_{c0}$$). In the later stages $$J/{J}_{c0}\ll 1$$, collective creep theories, which consider vortex elasticity, predict an inverse power law form for the energy barrier U(J) = U0[(Jc0/J)μ]. Here, the glassy exponent μ is sensitive to the size of the vortex bundle that hops during the creep process and its dimensionality. To capture behavior for a broad range of J, we invoke a commonly used interpolation between the two regimes
$$U(J)={U}_{0}[{({J}_{c0}/J)}^{\mu }-1]/\mu ,$$
(4)
where μ = −1 recovers the Anderson-Kim result. It is now straightforward to combine Eqs. (3) and (4) to determine the expected decay in the persistent current over time J(t) and subsequently the vortex creep rate S:
$$J(t)={J}_{c0}{\left[1+\frac{\mu {k}_{B}T}{{U}_{0}}\mathrm{ln}(t/{t}_{0})\right]}^{-1/\mu }$$
(5)
and
$$S\equiv |\frac{d\,\mathrm{ln}\,J}{d\,\mathrm{ln}\,t}|=\frac{{k}_{B}T}{{U}_{0}+\mu {k}_{B}T\,\mathrm{ln}(t/{t}_{0})}.$$
(6)
Creep measurements are a useful tool for determining the size of the energy barrier and its dependence on current, field, and temperature. Such measurements further probe the vortex state, revealing the existence of glassy behavior, collective creep regimes, or plastic flow. This is because, as evident in Eq. (6), creep provides access to both U0 and μ. Table 1 summarizes expected values of μ for collective creep of 3D flux lines and 2D pancake vortices.
To shed light on the Hg1201 phase diagram, we measured creep rates in a wide range of temperatures (5–60 K) and magnetic fields (0.1–5 T) using standard methods19, summarized here. We first establish the critical state by sweeping the field 4H* above the field at which creep will be measured H, where H* is the minimum field at which magnetic flux will fully penetrate the sample. Second, the field is swept to H, such that the magnetization M(H) coincides with its value on the upper branch of a magnetization loop. [If the magnitude of the initial field sweep were not sufficiently high, M(H) would instead fall inside the loop, vortices would not fully penetrate the entire sample and the previously discussed models would be inapplicable.] Third, the magnetization M(t) J(t) is subsequently recorded every ~15 s for an hour. We also briefly measure M(t) in the lower branch to determine the background arising from the sample holder, subtract this, and adjust the time to account for the difference between the initial application of the field and the first measurement (maximize correlation coefficient). Lastly, the normalized creep rate S(T, H) is extracted from the slope of a linear fit to lnM−lnt.
Figure 5 shows the field dependence of the creep rate. In low fields, S decreases as H increases, a trend that reverse above ~0.5 T. This change in behavior may be related to a different source of vortex pinning at low than at high fields. It roughly coincides with the low-field change in shape of the M(H) loops around Hon ~ 0.5 T. This trend is apparent in the inset, which compares the temperature dependencies of Hon (open symbols) and the field at which the minimum in creep S(H) occurs (closed symbols). Because of this, in the following section, we will separately analyze low-field and high-field measurements. We will first present S(T) and an analysis of the vortex state in the low-field regime, and then proceed to analyze the high-field regime.
### Glassy vortex dynamics and elastic-to-plastic crossovers
To study the dynamics in the low-field weak pinning regime, exemplified in Fig. 3(b), we measured vortex creep for μ0H < 0.5 T, shown in the main panel of Fig. 6(a). The creep rates in fields of 0.1–0.3 T are similar over the entire temperature range, plateauing at S ~ 0.06 for T < 40 K then sharply rising at higher temperatures. Such behavior is akin to S(T) in YBCO samples, which typically exhibit a plateau around S ~ 0.02–0.03519,52,53. In YBCO, the plateau appears because $${U}_{0}\ll \mu T\,\mathrm{ln}(t/{t}_{0})$$ such that $$S \sim {[\mu {k}_{B}\mathrm{ln}(t/{t}_{0})]}^{-1}$$ becomes T-independent. It is often associated with glassy vortex dynamics because μ ≈ 1 considering S ~ 0.035 and ln(t/t0) ≈ 27 for a typical measurement window of t ~ 1 hour19,53.
Similarly, for our Hg1201 sample, if $${U}_{0}\ll \mu {k}_{B}T\,\mathrm{ln}(t/{t}_{0})$$ were true, the S ~ 0.05 plateau would yield μ ~ 0.6. However, in our sample, we do not yet know the comparative magnitudes of U0 and μkBTln(t/t0). To extract μ without the need for assumptions regarding U0, it is common practice37,54,55,56,57,58 to define an experimentally accessible auxiliary energy scale U* ≡ kBT/S. From Eq. (6), we see that U* = U0 + μkBTln(t/t0) and, combined with Eq. (5), find that
$${U}^{\ast }\equiv \frac{{k}_{B}T}{S}={U}_{0}{\left(\frac{{J}_{c0}}{J}\right)}^{\mu }.$$
(7)
Hence, μ can be directly obtained from the slope of U* versus 1/J on a log − log plot. As shown in the Fig. 6(a) inset, μ = 0.5 for fields of 0.1–0.3 T. We reinforce this result with a complementary 67 hour long relaxation study shown in Fig. 6(b) and fitting the resulting M(t) to the interpolation formula Eq. (5). For free parameters Jc0, U0, and μ, the best fit again produces μ = 0.5, which is expected for collective creep of large bundles of 2D pancake vortices (see Table 1)59.
The presence of large bundles in these small fields is suggestive of a clean pinning landscape in which long-range 1/r vortex-vortex interactions are only weakly perturbed by vortex-defect interactions. Furthermore, this result is consistent with the evidence from Hirr(T) (shown in Fig. 4) of a 2D vortex state over a wide low temperature $$T/{T}_{c}\ll 0.6$$ region of the phase diagram. We have now ascertained that, though the plateau in S(T) appears at a higher S than in YBCO, it again correlates with glassiness.
At 0.4 T, S(T) is non-monotonic, reaching a local minimum around 30 K (see Fig. 6(a)). As shown in the Fig. 6(a) inset, we extract μ ≈ 1, which is close to the μ = 13/16 expectation for creep of medium bundles of pancake vortices (see Table 1). So, the system transitions from creep of large bundles at low fields μ0H < 0.4 T to medium bundles at 0.4 T. This change in μ occurs roughly around Hon (compare to Fig. 3(b)) and the minimum in S(H) (compare to Fig. 5). In many systems, the bundle size increases with increasing H51. Hence, this scenario is not standard, but is consistent with our suspected mechanism for Hon: as H increases, the strength of pinning suddenly increases around Hon causing the lattice to become more entangled, the bundle size to decrease, and we see both Jc and μ increase.
Collective creep theory only considers elastic deformations of the vortex lattice and neglects dislocations. At high temperatures and/or fields, the elastic pinning barrier becomes quite high and plastic deformations of the vortex lattice can become more energetically favorable. Plastic creep60 involves the motion of a channel of vortices constrained between two edge dislocations of opposite sign (dislocation pairs) and requires surmounting a diverging plastic barrier Upl ~ Jμ for small driving force $$J\ll {J}_{c0}$$. It manifests as a negatively sloped region in a U*(1/J) plot: in Eq. 7, μ < 0 is conventionally represented using the notation p, such that the auxiliary quantity U* scales as U0(Jc0/J)p in the plastic regime. Note that the true potential barrier U(J) in Eq. (4) remains monotonically decreasing with increasing current.
Figure 6(c) displays creep rates at fields H > 0.5 T. Representing the data as U*(1/J), plotted in Fig. 6(d), the slopes exhibit a distinct sign change, revealing elastic (μ > 0) to plastic (μ < 0 → p) crossovers for H ≤ 2 T. Figure 6(d) displays the 1 T data for fields of 0.7–1.8 T. From the data, we extract the exponents μ displayed in the vortex phase diagram show in Fig. 7(a). We see, for example, that at 1 T the sample hosts creeping small bundles of pancake vortices in the elastic regime T < 20. To investigate the dynamics at the crossover temperature Tcr = 20 K, we perform a 6-hour measurement of M(t) that is plotted in the Fig. 7(b) and fit the data to Eq. (5). To reduce the number of free parameters (Jc0 and U0), we obtain μ from the slope of 1/S plotted against lnt, see Eq. (6) and the Fig. 7(b) inset. Clearly, the early stages of relaxation is glassy. The bundle size evolves over time, manifesting as a change in μ. In the latter stages, we observe a transition to plastic flow.
## Discussion
Elastic-to-plastic crossovers are considered the cause of the second magnetization peak in many superconductors32,37,38,40. However, the origin of the SMP in Hg1201 is controversial. Daignère et al.43,61 found no correlation between the SMP and elastic-to-plastic crossovers in Hg1201 single crystals, and concluded that the SMP merely arises from competition between an increase in Jc and decrease in pinning energy with increasing magnetic field. On the contrary, Pissas et al.44 showed that though Hcr < Hsmp, a correlation between the two fields does indeed exist, therefore the peak is possibly associated with collective-to-plastic transitions. That study reconciled the lack of coincidence between the two fields as caused by very fast creep at low fields and slower creep at high fields. To understand their reasoning, it is important to note that magnetization measurements are not collected instantaneously with the application of a magnetic field. That is, there is a 10−100 s lag between establishing the field and measuring M due to the time required for setting the magnet in persistent mode and translating the sample through the magnetometer SQUID detection coils. Consequently, by the time M is recorded during magnetization loop measurements, J is much less than Jc0 at low fields where creep is fast and closer to Jc0 at higher fields where creep is slow. This idea was further supported by a demonstration that measuring the loop faster shifts Hsmp to lower fields, towards Hcr44.
To explore this issue, we overlay our measurements of Hsmp and Hcr in the phase diagram in Fig. 7(a). Figure 6(d) show examples of how Hcr was extracted from U*(1/J). At low temperatures T/Tc < 0.2, the appearance of the SMP coincides with the elastic-to-plastic crossover whereas Hcr < Hsmp at higher temperatures. Given this discrepancy, we now consider the previous argument that fast creep rates make measurements of Hsmp from loops inaccurate.
As described above, we typically create a magnetization loop by measuring M once at each field as the field is ramped up in steps. Constructing the loop instead from magnetic relaxation data enables us to set a consistent time scale for all M values. We achieve this by extracting M from a linear fit to logM − logt at a predetermined time ti after formation of the critical state, exemplified in the inset to Fig. 7(c). This also allows us to estimate M before the first measurement. The main panel shows how M(H) changes with ti. We find that a faster measurement increases Hsmp, moving it away from Hcr, contrary to the observation in Pissas et al.44.
Conversely, the time scale associated with our determination of the elastic-to-plastic crossover is arguably the 1 hour duration of our creep measurements, therefore, an appropriate comparison requires Hsmp to be determined from ti = 1 hour. As shown in Fig. 7(a,c), though this reduces Hsmp, it remains significantly larger than Hcr. We can therefore conclude that the elastic-to-plastic crossover is not the source of the SMP at high temperatures. The SMP could be caused by a transition in the structure of the vortex lattice. As the conditions for this transition would depend on anisotropy, this could be clarified through a comprehensive study comparing magnetization in Hg1201 crystals having different anisotropies, achieved by varying the doping, or through neutron scattering studies47.
To summarize, we have studied the field- and temperature- dependent magnetization and vortex creep in an HgBa2CuO4+δ single crystal to understand the effects of anisotropy on vortex dynamics in superconductors. We reveal glassy behavior involving collective creep of bundles of 2D pancake vortices over a broad range of temperatures and fields as well as temperature- and time-tuned crossovers from elastic dynamics to plastic flow. The isothermal magnetization loops exhibit distinct second magnetization peaks that have also been observed in previous studies of Hg1201, and Hsmp(T) decays exponentially at low temperatures then exhibits an abrupt change in behavior above T/Tc = 0.5. The origin of the second magnetization peak in superconductors can be controversial, and is often attributed to an elastic-to-plastic crossover. Here we clearly show that the second magnetization in Hg1201 is not caused by an elastic-to-plastic crossover at T/Tc > 0.2 and occurs within the plastic flow regime.
## Methods
Hg1201 single crystals were grown using an encapsulated self-flux method62 at Los Alamos National Laboratory. The crystals were subsequently heat-treated at 350 °C in air and quenched to room temperature to achieve near optimal doping6. The high-quality of the synthesized crystals is evinced by the observation of large quantum oscillations in other samples from the same growth batch. Multiple crystals were measured to verify reproducibility. The results presented in this manuscript were collected on a crystal with dimensions 1.28 × 0.84 × 0.24 mm3 and mass of 1.9 mg, shown in the Fig. 1(a) inset.
All measurements were performed using a Quantum Design superconductor quantum interference device (SQUID) magnetometer equipped with two independent sets of detection coils to measure the magnetic moment in the direction of (mL) and transverse to (mT) the applied magnetic field. For measurements requiring manipulating the field orientation, the crystal was placed on a rotating sample mount. Most measurements, however, were conducted with the field aligned with the sample c-axis (H||c), in which case the sample was mounted on a delrin disk inside a straw.
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# Error in using NDSolve
Posted 4 months ago
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4 Replies
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n = 2 sol=NDSolve[{Table[{x_i'[t]==ν_i [t],y_i'[t]==ω_i [t],z_i'[t]==ρ_i [t]},{i,1,n}], Table[{x_i [0]==x0[i],y_i [0]==y0[i],z_i [0]==z0[i]},{i,1,n}]}, Flatten[Table[{x_i,y_i,z_i},{i,1,n}]],{t,0,T}, MaxSteps→Infinity,Method→{"DiscontinuityProcessing"→False}] This is the errorNDSolve::ndinnt: Initial condition x0[1.] is not a number or a rectangular array of numbers.Please someone can help me on how to solve this error
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Posted 4 months ago
Take a look at the result of evaluating Table[{x_i'[t]==ν_i [t],y_i'[t]==ω_i [t],z_i'[t]==ρ_i [t]},{i,1,n}] In the WL underscore represents Blank. Perhaps you meant this? Table[{x[i]'[t] == ν[i][t], y[i]'[t] == ω[i][t], z[i]'[t] == ρ[i][t]}, {i, 1, n}]
Posted 4 months ago
It is x subscript i.. This problem deals with {x1, y1, z1} {x2 ,y2, z2}. The numbers are subscript.
Posted 4 months ago
Evaluating the code you posted results in NDSolve::dsfun: x:Blank[1] cannot be used as a function. Which is different from the error you posted. So, the code you are running is not the code that you posted (which has no Subscript). It is best to avoid the use of Subscript in WL, better to use indexed variables as I suggested. Did you try that?
Posted 4 months ago
Thank you so much... :) it works
Community posts can be styled and formatted using the Markdown syntax.
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# Why is speed of light a constant while distance in space is not?
Disclaimer: I asked this at Astronomy.SE, but got no answer whatsoever, so I am trying my luck here.
As you probably know current state-of-the art physics (i.e. gravitational waves, cosmic expansion) basically states that space itself is subject to expansion or contraction. Since there is no moving matter or energy involved, this might even happen at a "speed" faster than light.
So far, so good and obscure. What strikes me is the principle that the speed of light as a fundamental constant can only be expressed as a function of space-time. Where do we know that the one is constant but the other can suddenly be variable?
Is there any reason why the point of view of an expanding or contracting space is preferred over, say, a reduction in the speed of light or an increase in the "speed" of time? Is there any objective difference, a mathmatical model being a better fit or is it just the good old rubber metaphor being stretched (pun intended) too far?
In case the answer is: Both are equal w.r.t. current observations: How do we know that not both are actually variable?
• Re: "Is there any reason why the point of view of an expanding or contracting space is preferred over, say, a reduction in the speed of light or an increase in the "speed" of time?"...'any reason' covers a lot of territory. – IntuitivePhysics Mar 12 '16 at 22:31
• It has been mostly answered in this link: physics.stackexchange.com/q/2230 – Benjamin Mar 12 '16 at 22:31
• @choeger Actually the idea of a variable speed of light is/has been considered, even by Einstein himself, and nowadays in the context of the inflationary scenario, you may like to take a look at en.wikipedia.org/wiki/Variable_speed_of_light – udrv Mar 13 '16 at 4:56
• " this might even happen at a 'speed' faster than light": no. – Fabrice NEYRET Mar 13 '16 at 9:43
• So you have to bring in the machinery of general relativity. And since general relativity can already explain everything with a constant speed of light, you don't get anything out of making it vary. – knzhou Mar 14 '16 at 20:52
• It is not that clear to me what "equivalent formulation" means if $c$ is allowed to vary. The constancy of $c$ is the crux of special relativity. I'm sure there are theories out there in which $c$ is allowed to vary but I can not see how they could be called equivalent nor how they could be accepted since experiments for now tell us otherwise. – Jan Bos Mar 15 '16 at 0:09
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# 0.2 2.3 vector addition: graphical methods (Page 15/14)
Page 1 / 14
• Understand the rules of vector addition, subtraction, and multiplication.
• Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.
## Vectors in two dimensions
A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.
[link] shows such a graphical representation of a vector , using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction . We shall use the notation that a boldface symbol, such as $\text{D}$ , stands for a vector. Its magnitude is represented by the symbol in italics, $D$ , and its direction by $\theta$ .
## Vectors in this text
In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector $\text{F}$ , which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as $F$ , and the direction of the variable will be given by an angle $\theta$ .
The head-to-tail method is a graphical way to add vectors, described in [link] below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
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