text: The Lumen team has been paramount in our success with OER thus far. My Maths Homework Book 3c Answers In Genesis - Essay for you. Allhomeworktutors My maths homework answers. Next, you will copy the contents from one of the model courses into the course you have created. But what we're now going to explore is functions that are defined piece by piece over different intervals and functions like this you'll sometimes view them as a piecewise, or these types of. "I've found the limited answer key approach can backfire just as easily: the students will often write some gibberish and then just tweak the numbers until they have the "correct answer". Precalculus Tests Answer Key includes test versions with overprinted solutions for convenient grading and evaluation. Precalculus Textbook answers Questions Review. We have learned that, in in an algebraic expression, letters can stand for numbers. No devices, no references at all, go to a library or a coffee shop without your devices. 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If you already have an account, you can log on using the box to the right. is a good place to study where the lab assistant is available to answer questions when you are stuck. Paperback royalties go to OpenIntro, and the optional Leanpub PDF contributions go to authors to fund their time on this book. Setting up MyOpenMath modules for your course or the Blended Just-in-Time Math program at your institution is a two-step process. MATH 41 is a course that is offered through the Math Learning Center for students who need to review arithmetic or those who are preparing to take algebra in the future. on paper rst, then type in the correct answer. Are you an instructor who wants to adopt an open textbook, who feels online interactive homework is valuable, but doesn't want their students to have to pay an additional fee? Then read more about using MyOpenMath in the classroom. To prepare for the MyOpenMath assignments, you can work through similar examples in the text, select similar problems from our textbook exercises (odd numbers) and check your answers in the back of the book. MyOpenMath is a free and open version of MyMathLab or WebAssign. Read Essential Maths 7c Homework Book Answers Bk. 1-Quadratic Functions and Models Graph of a Quadratic Function Deﬁnition of a Polynomial Function Deﬁnition of a Quadratic Function. Answer Key (homework) Chapter Reviews. com uses a Commercial suffix and it's server(s) are located in N/A with the IP number 54. Homeworkdoer. The domain myopenmath. We'll look at two ways: Standard Form Linear Equations A linear equation can be written in several forms. Precalculus provides a link that takes students from the basic plug & chug of formulaic calculations towards building an understanding that equations and formulas have deeper meaning and purpose. Get Personal Discount! Starting at 5. The domain myopenmath. 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" It definitely is going to depend a lot on how students approach the homework. Also, we feel it is important that precalculus be the bridge in level of thinking between algebra and calculus. 1 Functions 1. Answer Key (homework) Chapter Reviews. Quizzes will be very similar to the homework assignments. Building and Documenting Python REST APIs With Flask and Connexion (2-Part Series) Building and Documenting Python REST APIs With Flask and Connexion – Real Python. Student Help Skip to start of help topic. @JeffE the fact that the textbook is interactive is not irrelevant at all. 8047 tanecsce — —. Get Personal Discount! Starting at 5. Material on mathcentre is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives License. Hit the reference books (don't look up the problem online either, it will go right to the answer and you won't learn. During the study of statistics, you will learn everything about how to collect, analyze, present, and interpret data. 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It looks like your title tag is a little outside the ideal length. Exam Date Material Covered. Determine the domain and range of the function f of x is equal to 3x squared plus 6x minus 2. Home; Calculus 1 WebAssign Answers; Calculus 2 Webassign Answers; Calculus 3 Webassign Answers; WileyPlus Physics Answers. The trouble with Pre-Calculus (at least in the U. For example, in the medical field, ask students to select symptoms associated with a medical condition. OHM houses several template courses that can be copied and customized to your department and student’s needs. Answer Key. Your domain, eduappcenter. Geometry Calendar. We guide our students on how to get the authentic and best MyOpenMath answer key which is answer generator enabled to provide best assessment answers for your online math test, quiz and exams. For other kinds of questions, Socratic kind of does a bit. 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text:
# Note

This is a modified version of a report I created in Summer 2017 on behalf of another organization. The original data and the associated problem are non-confidential. Anything identifiable (for example, variable names) has been changed to a generic alternative.

# Executive Summary

We performed a logistic lasso to create a logistic regression model for the data provided to us by Newton. Our best regression equation is the following:

$$\LARGE{\boxed{\hat{P} = \frac{e^{-0.25 - 0.19X_{73} + 0.0029X_{299} + 0.022X_{16} + 0.073X_{77} + 0.0054X_{169} + 0.010X_{157}}}{1+e^{-0.25 - 0.19X_{73} + 0.0029X_{299} + 0.022X_{16} + 0.073X_{77} + 0.0054X_{169} + 0.010X_{157}}}}}$$

If $$\hat{P} \geq 0.5$$, we predict that $$Y=1$$. If not, we predict $$Y=0$$.

We outline our model creation process in the sections below. We were given questions to answer about our work. Short answers follow. More context for these answers is included in remainder of the report.

• We estimate that our model has a misclassification rate of 0.39, which is better than the base rate of 0.50.
• It is important to balance one’s dataset because a model could be tuned to whatever the dominant outcome variable is. The model would overcompensate for the more common variable, and eventually predict no better than the base rate.
• We selected the most predictive independent variables through the logistic lasso optimization process in cv.glmnet().
• We divided the data into build (training) and test sets to avoid over-fitting a model to a sub-optimal solution. We also used cross-validation for this reason.
• We test model performance on a left-out test set. This is more appropriate and less optimistic than using mean cross-validated error, because if many models are tested, even cross-validation can select a sub-optimal model through random chance.
• Our Random Forest model has a misclassification rate of 0.31. It makes better predictions than our logistic model, but takes longer to run on new data than a logistic model.
• A multiple-stage model could lead to better predictions.

# Exploratory Data Analysis (EDA)

As described by Newton, the outcome variable is $$Y$$ and it is binary, taking either $$1$$ or $$0$$. The dataset has already been balanced, meaning there are equal numbers of $$Y=1$$ observations as there are $$Y=0$$. It is important to balance one’s dataset because a model could be tuned to whatever the dominant outcome variable is. If one outcome is much rarer than the other, a model could just predict the common outcome every time and have fairly high performance in terms of mean-squared error. Such a model would be less useful than a model created from a balanced dataset that, on paper, had worse mean-squared error performance.

table(newton$y) ## ## 0 1 ## 236 236 ## Sample Splitting We split the data into a main set and a left-out test set before doing any model selection. Later, for model selection, we do cross-validation. All future comments regarding EDA or model building solely consider the training set. Model tuning will not be done on the left-out test set. This avoids over-fitting: as the number of tested models approaches infinity, random chance will make a suboptimal model look like the best choice on a sample from the population of interest. Cross-validation alone cannot insure against overfitting if many models are tested. Here, we sample rows that will be put into the left-out sample. We decide that 10% of rows will become the test set. Our sample is small enough that we do not want to use a larger percentage for the test set. #random.org/integers/?num=1&min=1&max=21474836&col=5&base=10&format=html&rnd=new set.seed(18046675) samp.size <- .1 * nrow(newton) y.rows <- which(newton$y == 1)
non.y.rows <- which(newton$y != 1) #/2 for the y=0 rows to be sampled equally y.row.samples <- sample(y.rows, size = samp.size/2, replace = FALSE) non.y.row.samples <- sample(non.y.rows, size = samp.size/2, replace = FALSE) row.samples <- c(y.row.samples, non.y.row.samples) Next, we create the training and left-out samples. We also separate the $$Y$$ column from the predictor columns. This is for convenience in generating summary statistics and EDA solely from the predictors and not $$Y$$. te.newton <- newton[row.samples,] newton <- newton[-row.samples,] y.newton <- as.factor(newton$y)
te.y.newton <- as.factor(te.newton$y) te.newton <- te.newton[,-which(colnames(te.newton) == "y")] newton <- newton[,-which(colnames(newton) == "y")] We verify that the main and test sets both have equal numbers of $$Y=1$$ and $$Y=0$$ observations: table(y.newton) ## y.newton ## 0 1 ## 213 213 table(te.y.newton) ## te.y.newton ## 0 1 ## 23 23 ## Quantiles We are interested in the proportion of observations that are above zero for each predictor. quants <- matrix(nrow = 3, ncol = ncol(newton)) for (i in 1:ncol(newton)){ quants[1:3,i] <- quantile(newton[,i], probs = c(.5, .85, .86)) } quantile.zeros <- numeric() for (i in 1:nrow(quants)){ quantile.zeros[i] <- 100 * length(which(quants[i,] == 0)) / ncol(quants) } names(quantile.zeros) <- c("Median", "85th Percentile", "86th Percentile") We find our answer for quantiles of interest. 71% of predictors are median-zero. Below the 86th percentile, less than 50% of the predictors are individually above zero. These statistics indicate that many of the predictors are often zero. round(quantile.zeros, 1) ## Median 85th Percentile 86th Percentile ## 71.3 52.1 49.4 # Model Creation ## Logistic Lasso We have been asked to create a logistic regression to predict $$Y$$ from the $$X_i$$ predictors. We will use a logistic lasso (formerly known as logistic LASSO) to find useful predictors for predicting Y. If the lasso does well, it will reduce the number of not helpful or marginally helpful predictors in our model. This will reduce variance and improve predictive performance of the model on new data. We also cross-validate in the model creation step. We choose 5-fold cross validation, due to the small size of our sample. $$k=10$$ is another common choice for $$k$$-fold cross validation. We argue that the variance introduced by validating the model against a small subsample of size $$\frac{426}{k=10} \approx 43$$ is not worth the reduction in model bias as compared to a hypothetical perfect model. This raises a question related to our test set, which contains 46 observations. Why not choose a larger test set? Our test set is small because variance in the validation of the model is far less important than variance in the creation of the model. The validation step of cross-validation is still part of the model creation process. The test-set validation is outside of that process. Our estimate of model performance would have lower variance with a larger test set, but the quality of the model’s predictions would decrease as well. This is because there would be fewer observations to use while creating the model. For $$\lambda$$, we choose $$\lambda_{1SE}$$. This is the largest $$\lambda$$ value that is within 1 standard error of the $$\lambda$$ that minimizes the mean cross-validated error. $$\lambda_{1SE}$$ is less likely than the minimum value of $$\lambda$$ to be overfitted to the sample used to create the model. library(glmnet) ## Loading required package: Matrix ## Loading required package: foreach ## Loaded glmnet 2.0-10 library(doParallel) ## Loading required package: iterators ## Loading required package: parallel registerDoParallel() #family="binomial" for logistic regression out.glm <- cv.glmnet(x = as.matrix(newton), y = y.newton, nfolds = 5, family="binomial", parallel = TRUE) #for readability, we explicitly define s = "lambda.1se" even though this is the default preds <- predict(out.glm, newx = as.matrix(te.newton), type="response", s = "lambda.1se") obs.vs.expect <- cbind(te.y.newton, ifelse(preds >= .5, 1, 0)) confusion.mat <- table(obs.vs.expect[,1], obs.vs.expect[,2]) We extract the coefficients from our model and determine which coefficients are not equal to zero: coefs.glm <- coef(out.glm, s = "lambda.1se") nonzero.coefs <- which(coefs.glm != 0) We verify that all other coefficients are 0 at our specified lambda value: table(coefs.glm[-nonzero.coefs,]) ## ## 0 ## 437 Our best regression equation is of this form: $$\LARGE{\boxed{\hat{P} = \frac{e^{a + bX}}{1+e^{a + bX}}}}$$ • $$\hat{P}$$ is the predicted probability that $$Y=1$$. • $$e$$ is the base of the natural log, around 2.71. • $$a$$ is the intercept, around -0.25. • $$b$$ is the slope corresponding to each predictor. Most $$b$$ values are 0, but, for example, $$b_{73}$$ = -0.19. • $$X$$ is each predictor, from $$X_{1}$$ through $$X_{443}$$. Substituting in our calculated values, and simplifying to remove zeros (most $$b$$ values), we get the following equation, rounded to two significant figures. $$\LARGE{\boxed{\hat{P} = \frac{e^{-0.25 - 0.19X_{73} + 0.0029X_{299} + 0.022X_{16} + 0.073X_{77} + 0.0054X_{169} + 0.010X_{157}}}{1+e^{-0.25 - 0.19X_{73} + 0.0029X_{299} + 0.022X_{16} + 0.073X_{77} + 0.0054X_{169} + 0.010X_{157}}}}}$$ If $$\hat{P} \geq 0.5$$, we predict that $$Y=1$$. If not, we predict $$Y=0$$. We rounded to two significant figures above. However, the following R code provides values for each non-zero $$b$$, as well as the intercept $$a$$, to a higher number of significant digits. coefs.glm[nonzero.coefs,] ## (Intercept) x73 x299 x16 x77 ## -0.252138815 -0.194817135 0.002854973 0.021978266 0.073051477 ## x169 x157 ## 0.005416745 0.010194576 The above code indicates just 6 predictors and 1 intercept, from an initial pool of 443 possible predictors. Clearly, the lasso reduced the number of potential predictors in the model. It is not surprising that many predictors were removed in the lasso process: as described in the EDA section, many predictor observations were predominantly zero. Since many predictors were almost entirely the same observed value, they could not be used very well to distinguish between $$Y=0$$ and $$Y=1$$. Removing these unneeded predictors reduces variance and thus error when predicting with new data. The 6 predictors used in the lasso model are the most predictive independent variables for a logistic model. We selected them through the logistic lasso optimization process in cv.glmnet(). There are multiple ways for a logistic regression package in R to describe the relationship between predictors and the response variable $$Y$$. The package we used, glmnet, even has multiple modes for its functions. It is wise to verify that the equation we describe is how the package glmnet is operating. To do this, we manually calculate $$\hat{P}$$ from the formula above, and compare it to the output of predict(). orig.data.preds <- predict(out.glm, newx = as.matrix(newton), type="response", s = "lambda.1se") #notation follows this image: http://faculty.cas.usf.edu/mbrannick/regression/gifs/lo3.gif test_logistic_formula <- function(row){ vectorized.products <- coefs.glm[nonzero.coefs,][2:7] * newton[row,nonzero.coefs-1] bx <- sum(vectorized.products) a <- coefs.glm[nonzero.coefs,][1] p.test <- exp(a + bx)/(1 + exp(a + bx)) c(p.test, orig.data.preds[row]) } preds.through.two.methods <- matrix(ncol=2, nrow=nrow(newton)) for (i in 1:nrow(newton)){ preds.through.two.methods[i,1:2] <- test_logistic_formula(row = i)[1:2] } We confirm that the above regression equation matches the internal formula used by glmnet: all.equal(target = preds.through.two.methods[,1], current = preds.through.two.methods[,2]) ## [1] TRUE We generate a confusion matrix for the logistic model. val_from_confusion_mat <- function(mat){ #columns of mat: expected. rows of mat: observed true.negative <- mat[1,1] false.positive <- mat[1,2] false.negative <- mat[2,1] true.positive <- mat[2,2] total <- sum(mat) misclass.rate <- (false.positive + false.negative) / total purity <- true.positive / (true.positive + false.positive) completeness <- true.positive / (true.positive + false.negative) out.vec <- c(misclass.rate, completeness, purity) names(out.vec) <- c("Misclassification Rate", "Completeness", "Purity") return(out.vec) } The base rate is 50%. This makes sense because of how the data were collected. base.rate <- length(which(y.newton != 0))/length(y.newton) base.rate ## [1] 0.5 We calculate some values of interest from the matrix, using the left-out test set that was not used in the cross-validation process. This is more appropriate and less optimistic than using mean cross-validated error, because if many models are tested, even cross-validation can select a sub-optimal model through random chance. A sub-optimal model would have a better mean cross-validated error than other candidate models. However, when tested on new data, it would have worse performance than suggested by its cross-validated error. We have no information on whether $$Y=0$$ or $$Y=1$$ is more important to predict, so we assume they are equally important. This makes misclassification rate a logical choice for measuring model performance. logistic.conf <- val_from_confusion_mat(confusion.mat) logistic.conf ## Misclassification Rate Completeness Purity ## 0.3913043 0.7826087 0.5806452 ## Potential Gains From a More Complex Model We briefly consider a Random Forest model. Random Forest is more complicated than logistic regression or logistic lasso. A Random Forest outperforms our best logistic model by 5.8 percentage points. library("randomForest") ## randomForest 4.6-12 ## Type rfNews() to see new features/changes/bug fixes. rf.newton <- randomForest(x=newton, y=y.newton, xtest = te.newton, ytest = te.y.newton) rf.conf <- val_from_confusion_mat(rf.newton$confusion)
rf.conf
## Misclassification Rate           Completeness                 Purity
##              0.3328125              0.6572770              0.6698565
perc.point.diff = 100 * (logistic.conf[1] - rf.conf[1])
names(perc.point.diff) = "Percentage Point Difference, Logistic-RF"
perc.point.diff
## Percentage Point Difference, Logistic-RF
##                                 5.849185

This suggests that the Random Forest could implicitly include interactions that the logistic lasso does not consider. Compared to logistic lasso, the Random Forest model has a superior misclassification rate, but identifies a smaller percentage of all possible $$Y=1$$ cases. Company goals could determine the appropriateness of one model over the other. Random Forest has a longer run time, and is less well-suited for real-time or big-data contexts.

# Conclusion and Future Work

Our approach has room for improvement. It has superior performance to simply guessing $$Y=1$$ at the base rate. (That is, predicting $$Y=1$$ for 50% of any new observations.) Within the limitations of this sample (a small sample with unlabeled predictors), the most likely source of predictive improvement is with other model-building methods.

With more time, we would examine the Random Forest and try to determine how it achieves superior predictive performance to the logistic model. This would help us choose interaction terms that might get the logistic model closer to the Random Forest model’s predictive ability.

A multiple-stage classifier could be worth investigating. A first stage could try to remove cases that are obviously $$Y=0$$ or $$Y=1$$ from the prediction process. This would allow a second stage to be more finely-tuned for difficult observations in which the difference between $$Y=0$$ and $$Y=1$$ is harder to determine.

Even without improving the model’s predictive ability, the model could be made more useful. This depends on the model’s intended application. If $$Y=1$$ is very costly for a business, the model could be adjusted to have multiple tiers. The tiers could be low-risk, medium-risk, and high-risk, or something similar. The tiers would correspond to where a predicted $$\hat{P}$$ lies in the interval $$[0, 1]$$ for each new set of observed predictors $$X_1 ... X_{443}$$.

If there is an intervention that could be applied to the medium-risk pool that reduces the predicted likelihood of $$Y=1$$, then a business could use the intervention on the medium-risk pool, use a different treatment for the high-risk pool, and a different intervention or perhaps no intervention on the low-risk pool.

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# Rotation group, altitude

Could someone give me a rigorous proof that the group of rotations each element of which is a composition of rotations around the altitudes of a tetrahedron that transform the tetrahedron into itself is isomorphic to the alternating group $A_4$.

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What have you tried or thought about so far? –  Zev Chonoles Jun 6 '12 at 20:59
Number the vertices. See what each rotation does to the vertices. Write them as permutations. –  Arturo Magidin Jun 6 '12 at 20:59
So far, I've been trying to understand what does group of rotations mean. –  Adam Jun 6 '12 at 21:02
Take a tetrahedron and put it on a table. Pick it up and put it down again so it is not necessarily the same way up, but occupies the same position in space. You will find that the same effect could have been achieved by a rotation - there are two kinds of axis: one kind through a vertex and the centre of the opposite face, and another kind through the middle of two opposite edges. –  Mark Bennet Jun 6 '12 at 21:11

Jykri's solution is excellent, but it may be useful to have another way of understanding why the symmetry group isn't all of $S_4$ (since it may not be immediately obvious that the group is generated by its three-cycles - that is, that every element of the group can be written as a product of three-cycles). Choose one of the vertices, $v_0$ say, and consider the triple product of the vectors from $v_0$ to the three other vertices $v_1, v_2, v_3$: $((v_1-v_0)\times(v_2-v_0))\cdot(v_3-v_0)$. This is the determinant of the matrix whose rows are the three vectors $v_i-v_0$; it represents the signed volume of the tetrahedron. As such, its sign has to be preserved by any rotation of the tetrahedron (this just says that rotations preserve the orientation of space), but a little calculation will show that swapping two vertices changes the sign of the triple product. This means that the two-cycles aren't in the rotation group, and so the group cab't be all of $S_4$ (since we know some elements of $S_4$ that aren't in it). Jykri's answer shows that the group is at least as large as $A_4$, and $A_4$ is a so-called maximal subgroup of $S_4$; there are no other groups 'in between' $A_4$ and $S_4$. Since the rotation group is at least $A_4$ and it can't be anything larger, it must be $A_4$ exactly.

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I think I got the idea. How can I formally prove that it is isomorphic to A4 though? –  Adam Jun 6 '12 at 22:47
@AdamAndersson What do you consider a 'formal' proof, and what do you think is missing from the outlines you've already been given? Do you understand why it must be some subgroup of $S_4$? If so, then you're already almost all the way there; Jykri's argument shows why it must be 'at least' $A_4$, and mine shows why it's not $S_4$. –  Steven Stadnicki Jun 6 '12 at 22:53

The simple approach may be the most illuminating:

Draw a picture.

Label the vertices (there are four of them, so it may be a good idea to use numerals 1, 2,3, 4).

Check the effects of the 120 degree rotations on the labels of the vertices (when done correctly, you should get 3-cycles).

Start combining those rotations until you get everything (this is also a useful exercise for the purposes of becoming fluent in the art of combining permutations).

If you don't want to do all those combinations, then you can replace the last step with the following piece of reasoning. We extract two facts from the above: 1) that the group of rotations can be viewed as a subgroup of $S_4$, because any rotation is fully determined by its effect on the 4 vertices, 2) the generators are all even permutations (3-cycles), so the group in question is actually a subgroup of $A_4$.

The group $A_4$ has 12 elements, so it suffices to show that we can get at least 12 different permutations of vertices by rotating the tetrahedron (there can't be more than twelve). First let's check that we can rotate any one of the four corners to the top vertex (imagine the tetrahedron lying on one side on a table, so that there is a unique top vertex). To do this we simply pick one of the vertices that is neither the top vertex nor the other vertex that we want to bring to the top, and then grab that vertex and rotate the thing by 120 degrees in the appropriate direction withour releasing our grip. Ok, so there are 4 choices for the vertex at the top. But we can then grab that top vertex and rotate the "bottom" face (the one touching the table). So we can bring the remaining three vertices into three different positions while keeping the top fixed.

We have shown that we can permute the vertices of a tetrahedron into $4\cdot3=12$ different positions, so by our earlier observations these permutations must form the group $A_4$.

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I will answer the question implied by your comment, "what is a group of rotations".

The point of group theory is to study symmetries of mathematical objects, including geometric objects. For example, a tetrahedron.

If you have a tetrahedron sitting on the table so that one vertex is pointing straight up, you can rotate that tetrahedron 120° around its vertical axis and it will look just the same. This motion is called a symmetry of the tetrahedron. Any motion of the tetrahedron that leaves it looking exactly the same is a "symmetry".

It should be clear that performing two symmetries in a row is itself a symmetry. The first symmetry leaves the tetrahedron looking the same, and so does the second, and therefore the combined motion leaves them looking the same. This "composition" combines two symmetries into one. For example, turning the tetrahedron clockwise once, and then clockwise again, is the same as turning it counterclockwise once. If $p$ and $q$ are symmetries, we might write $p\cdot q$ for the motion, also a symmetry, that first does $p$ and then does $q$.

It's perhaps not perfectly clear, but it is true, that for any symmetries $p$, $q$, and $r$, $(p\cdot q)\cdot r = p\cdot(q\cdot r)$. Thus the composition of symmetries of a tetrahedron is an associative operator on the symmetries. (This is actually a special case of the fact that composition of functions is always associative.)

It should be clear that there's a "trivial" symmetry that makes no motion at all. This non-motion certainly leaves the tetrahedron looking exactly the same, so it is indeed a symmetry. If $e$ is the do-nothing symmetry, then $e\cdot p = p$ and $p\cdot e = p$ for every symmetry $p$. That is, the "do nothing" symmetry behaves like an identity element with respect to composition of motions.

It should be clear that for each symmetry, there is an opposite motion, also a symmetry, that undoes its effect. If $q$ is the "opposite" of $p$, then $p\cdot q = e$ and $q\cdot p = e$. That is, doing the "opposite" motion behaves like a group inverse with respect to composition of motions.

So the symmetries of a tetrahedron form a group! And if this is a surprise, it is because your course has put the concepts backwards, because the reason that the axioms of a group are what they are, and not different axioms, is because these axioms were designed exactly to model these properties of the idea of symmetry.

The group is called "the group of rotations of a regular tetrahedron", because we considered only rotations of the tetrahedron. There is a larger group of symmetries that we could ascribe to a tetrahedron, by allowing mirror reflections of the tetrahedron as well as rotations. This might be called "the group of rotations and reflections of a regular tetrahedron".

If one has studied abstract groups, as you have, one might ask which group this is. One can count up the symmetries of the tetrahedron. There turn out to be 12, because in picking up the tetrahedron and putting it down again in the same place you get 4 choices about which face to put down on the table, and then 3 choices about which face to turn to face frontward. But there is more than one group with 12 elements. There are actually 5 such groups—there are 5 ways that an object can have exactly 12 symmetries.

So your job is to find out in which of these 5 ways the tetrahedron is symmetric. Is is symmetric in the way that a regular hexagon is symmetric, which we call $D_{12}$? Or is it symmetric in the (different) way that a 12-toothed saw blade is symmetric, which we call $Z_{12}$? Or is it, as the question claims, symmetric in the (different again) way that we call $A_4$?

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# Matrices that are Hadamard products of $X$ and $X^{-T}$

What are the matrices that you can write in the form $X \odot X^{-T}$, for a complex square matrix $X$, where $X^{-T}$ is the inverse of the complex transpose (not conjugate) and $\odot$ is the Hadamard (component-by-component) product?

In the $2\times 2$ case, you get the group of matrices in the form $$\begin{bmatrix}a & b\\\\ b & a \end{bmatrix},$$ with $a+b=1$, which are closed under matrix multiplication and would form a group were it not for the matrix $a=b=\frac12$ which admits no inverse [EDIT: corrected this assertion, thanks to Denis Serre]. In larger dimension, one sees that all the obtained matrices have the vector of all ones as both a right and left eigenvector. Is this the only restriction? Is the resulting set of matrices closed under multiplication? Is this problem known and studied?

Origin: motivated from this MO question.

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In the $2\times2$ case, there is a restriction on the output: $a+b=0$ (maybe you forgot to divide by the determinant of $X$, which occurs from $X^{-T}$). This makes an affine set that is stable under $\times$, but is not a multiplicative group, because of the matrix with $a=b=\frac12$. –  Denis Serre Apr 26 '11 at 12:42
I think you mean $a+b=1$ --- anyway good point, I omitted it in my computations because it didn't matter for the problem and then forgot about it. –  Federico Poloni Apr 26 '11 at 13:16
yes, $a+b=1$. Sorry for the misprint. –  Denis Serre Apr 26 '11 at 13:44
The question is whether every matrix $A$ such $A\vec e=A^T\vec e=\vec e$ (with $\vec e=(1,\ldots,1)^T$) can be written as $X\odot X^{-T}$. When $A$ has non-negative entries, this is a question about bistochastic matrices. A special case is that of ortho-stochastic matrices, for which such an $X$ exists, a unitary one. But it is known that ortho-stochastic matrices form a small part of bistochastic ones. In addition, we don't want to restrict to non-negative entries. –  Denis Serre Apr 26 '11 at 14:01

In fact I checked chapter 5 and there is quite a detailed discussion of the Hadamard product $X \odot X^{-T}$ --- for instance, they cite the fact the row and column sums are 1. Some of that material was new to me, it does not contain only the "usual" textbook properties of the Hadamard product. –  Federico Poloni Apr 27 '11 at 7:08


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Volume 390 - 40th International Conference on High Energy physics (ICHEP2020) - Parallel: Neutrino Physics
Neutrino Oscillation in Dense Matter
S. Luo
Full text: Not available
Abstract
As the increasing of neutrino energy or matter density, the neutrino oscillation in matter may undergo "vacuum-dominated", "resonance" and "matter-dominated" three different stages successively. Neutrinos endure very different matter effects, and therefore present very different oscillation behaviors in these three different cases. In this talk, we focus on the less discussed matter-dominated case (i.e., $|A^{}_{\rm CC}| \gg |\Delta m^{2}_{31}|$), show that as the matter parameter $|A^{}_{\rm CC}|$ growing larger, the effective mixing matrix in matter $\tilde{V}$ evolves approaching a fixed $3 \times 3$ constant real matrix which is free of CP violation and can be described using only one simple mixing angle $\tilde{\theta}$ which is independent of $A^{}_{\rm CC}$. As for the neutrino oscillation behavior, $\nu^{}_{e}$ decoupled in the matter-dominated case due to its intense charged-current interaction with electrons while a two-flavor oscillation are still presented between $\nu^{}_{\mu}$ and $\nu^{}_{\tau}$. At the end of this talk, we discussed the oscillation probabilities when neutrinos/anti-neutrinos passing through a typical white dwarf to give some embryo thoughts on under what circumstances these studies could be applied and put forward the interesting idea of possible "neutrino lensing" effect.
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 7.5.7.1. Let $\operatorname{\mathcal{C}}$ be a small category and let $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \operatorname{QCat}$ be a (strictly commutative) diagram of $\infty$-categories indexed by $\operatorname{\mathcal{C}}$. Then a Kan complex $X$ is a colimit of the functor $\operatorname{N}_{\bullet }^{\operatorname{hc}}(\mathscr {F})$ if and only if it is weakly homotopy equivalent to the homotopy colimit $\underset { \longrightarrow }{\mathrm{holim}}(\mathscr {F})$.

Proof. Let $\lambda _{t}: \underset { \longrightarrow }{\mathrm{holim}}(\mathscr {F}) \rightarrow \operatorname{N}_{\bullet }^{\mathscr {F}}(\operatorname{\mathcal{C}})$ be the taut scaffold of Construction 5.3.4.11. Then $\lambda _{t}$ is a categorical equivalence of simplicial sets (Corollary 5.3.5.9), and therefore a weak homotopy equivalence (Remark 4.5.3.4). The desired result now follows from Corollary 7.4.5.9. $\square$


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# ANOVA and independent samples

Let's say there is an experiment of determining the best cookie recipe with 4 categorical factors. There are $$S$$ subjects that eat and rate cookies made using $$R$$ different recipes. There are $$R$$ batches of cookies - one per recipe. The thing is that all subjects rate all recipes - each of $$S$$ subjects eat the total number of $$R$$ cookies and so we end up with $$R*S$$ ratings. Also note that all subjects sample from the same batches.

My question is: does this setup violate ANOVA independence assumption since there is no random assignment and everyone samples from the same batches? And if so, what needs to be done to use independent samples ANOVA here AFTER the results are collected?

• If data are normal, this could be analyzed according to a standard ANOVA design, see Answer. I would want to make sure there is no social connection between recipe authors and tasters. Tasters should be given guidelines on how to do scoring, especially if they are not professional tasters. Jan 11, 2021 at 20:14
• This is an example of a complete randomized block design experiment, a common experimental design. The random part refers to the fact the subjects should sample the recipes in a random order. Jan 11, 2021 at 22:57

Here are data from a tasting, conducted by a newspaper, of seven brands of chocolate pudding, by five different tasters. As in your Question, each taster rated each brand, resulting in 35 scores. In this case, each score was said to be the sum of several partial scores, so there is some reason to believe scores may be normal.

x = c(7,12,18,10,11,6,5, 15,14,8,6,2,0,6,
20,10,20,4,13,0,7, 18,16,12,10,8,9,3,
15,16,9,13,3,5,5)
matrix(x, byrow=T, nrow=5)
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,]    7   12   18   10   11    6    5
[2,]   15   14    8    6    2    0    6
[3,]   20   10   20    4   13    0    7
[4,]   18   16   12   10    8    9    3
[5,]   15   16    9   13    3    5    5

This amounts to a two-way ANOVA with one replication in each of 35 cells (so that there can be no interaction term). Looking at brands of interest as a fixed effect and tasters as randomly chosen, it might be considered a complete block design. (Due to possible quirky taste preferences of various tasters, there may be interaction. But with one observation per cell, we have no way to test for interaction.)

An ANOVA procedure in R finds that there are significant differences among brands but not significant differences among tasters.

If you look at both effects as fixed, the ANOVA model would be:

$$Y_{ij} = \mu + \beta_i + \tau_j + e_{ij},$$ for $$i=1,\dots,7$$ brands, $$j = 1,\dots,5$$ tasters and $$e_{ij} \stackrel{iid}{\sim}\mathsf{Norm}(\mu=0,\sigma),$$ with $$\sigma^2$$ to be estimated by MeanSq(Resid).

brand=as.factor(rep(1:7, times=5))
tastr=as.factor(rep(1:5, each=7))

anova(lm(x ~ brand + tastr))
Analysis of Variance Table

Response: x
Df Sum Sq Mean Sq F value   Pr(>F)
brand      6 580.80  96.800  5.8113 0.000748 ***
tastr      4  55.83  13.957  0.8379 0.514653
Residuals 24 399.77  16.657
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

With appropriate protection against 'false discovery' (e.g., Bonferroni) paired t tests might reveal which brands differ in significant and important ways.

You can look at various help pages online to see how to test whether residuals from the ANOVA model are nearly normally distributed. If you feel that residuals are too far from normal, you might use a nonparametric Friedman test to look for differences among brands.

friedman.test(x, brand, tastr)

Friedman rank sum test

data:  x, brand and tastr
Friedman chi-squared = 19.321, df = 6, p-value = 0.003654

Paired Wilcoxon (signed rank) tests might be used ad hoc to explore differences among brands.

• Fantastic, thank you! So a complete randomized design would also require tasters to eat and rate cookies in a randomly ordered sequence just as (@Dave2e pointed out)? Or is it enough that the tasters were chosen randomly from the population? I don't really understand the requirements of randomized design Jan 12, 2021 at 3:00
• Random order of tasting recipes would be ideal. That's usually the way evaluations of wine are done. Jan 12, 2021 at 6:57


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# Fully truth-functional version of modal logic?

Though the 'square' and 'rhombus' operators from modal logic are not truth-functional, it's because modal logic is interpreted as a binary logic, isn't it? What if modal logic is considered many-valued? That is, if it is, then the modality operators may be able to be truth-functional, aren't they?
So I picked provability logic as the modal logic to seek, and assumed it to be ternary. My plan was, step by step:

1. Find all the truth tables that satisfies all the axioms.
2. Pick an appropriate one.
3. Give appropriate name for the truth values, judging by the truth table.

But I couldn't do step 2 and step 3, because the truth tables from step 1 seemed nonsense. For example:

Truth table of negation (￢A):
A:   F U T
￢A: T U F
Truth table of implication (A → B):
A: F U T
B
F  U T U
U  U T T
T  U T T
Truth table of necessity (□A):
A:  F U T
□A: U U T

Where A and B are statements, F, U and T are arbitrary name for the truth values, among which T is the designated value for truth.
Another Example:

Truth table of negation (￢A):
A:   F U T
￢A: T U F
Truth table of implication (A → B):
A: F U T
B
F  T F T
U  T F F
T  T F T
Truth table of necessity (□A):
A:  F U T
□A: F F T

So how should I do step 2 and step 3?
Besides, here is the Haskell code I wrote to do step 1:

import Data.List

data Trool = False3 | Unknown3 | True3 deriving (Eq, Enum)

firstLetter :: Trool -> Char
firstLetter False3 = 'F'
firstLetter Unknown3 = 'U'
firstLetter True3 = 'T'

makeTable :: [a] -> Int -> [[a]]
makeTable _ 0 = [[]]
makeTable xs n = do
x <- xs
smallTable <- makeTable xs $n-1 [x : smallTable] unaryTruthTables :: [[Trool]] unaryTruthTables = makeTable [False3 .. True3] 3 binaryTruthTables :: [[[Trool]]] binaryTruthTables = makeTable unaryTruthTables 3 infix 9 <!!> (<!!>) :: [a] -> Trool -> a xs <!!> t = xs !! fromEnum t ternaryTest :: (Trool -> Trool) -> Bool ternaryTest f = all (True3 ==)$ map f [False3 .. True3]

ternaryTest2 :: (Trool -> Trool -> Trool) -> Bool
ternaryTest2 f2 = and $map (ternaryTest . f2) [False3 .. True3] ternaryTest3 :: (Trool -> Trool -> Trool -> Trool) -> Bool ternaryTest3 f3 = and$ map (ternaryTest2 . f3) [False3 .. True3]

possibleSystems :: [([Trool] {-negation-}, [[Trool]] {-implication-}, [Trool] {-necessity-})]
possibleSystems = do
notTable <- [[True3, Unknown3, False3]]
implyTable <- binaryTruthTables
squareTable <- unaryTruthTables
let not = (notTable <!!>)
let p --> q = (implyTable <!!> p) <!!> q
let square = (squareTable <!!>)
let rhombus = not . square . not
if and [
ternaryTest2 (\p q -> p --> (q --> p)),
ternaryTest3 (\p q r -> (p --> (q --> r)) --> ((p --> q) --> (p --> r))),
ternaryTest2 (\p q -> ((not p --> not q) --> (q --> p))),
ternaryTest2 (\p q -> square (p --> q) --> (square p --> square q)) -- Distribution Axiom
,ternaryTest (\p -> square (square p --> p) --> square p) -- Loeb's Axiom
] then [(notTable,implyTable,squareTable)] else []

printSystem :: Int -> ([Trool], [[Trool]], [Trool]) -> IO ()
printSystem n (notTable, implyTable, squareTable) = do
putStrLn $"System #" ++ show n putStrLn "Truth table of negation:" foldr1 (>>)$ map putStrLn $transpose ["FUT", map firstLetter notTable] putStrLn "Truth table of implication:" putStrLn " FUT" foldr1 (>>)$ map putStrLn $transpose$ concat [["FUT"], transpose $map (map firstLetter) implyTable] putStrLn "Truth table of necessity:" foldr1 (>>)$ map putStrLn $transpose ["FUT", map firstLetter squareTable] --putStrLn "Truth table of possibility:" --foldr1 (>>)$ map putStrLn $transpose ["FUT", map firstLetter rhombusTable] main :: IO () main = foldr1 (>>)$ map (uncurry printSystem) \$ zip [0..] possibleSystems

• It is possible you are trying to carry out something impossible. It's hard to dissect your code to determine what truth-values are being assigned, but you apparently consider three values, despite the opening paragraph mentioning "binary logic". While modal logic is necessarily a superset of propositional logic, it is unclear how you approach "provability" as a ternary logic. Perhaps reworking the statement of what is intended would make the code more understandable? – hardmath Sep 16 '17 at 2:30
• @hardmath False3, Unknown3 and True3 are arbitrary name for the truth values, and I have no actual intention for them. What motivated me to pick the provability logic was the Goedel's incompleteness theorems, which show that there are statements that cannot be proven to be either true or false. So originally I thought this as the third truth value, but the results from step 1 doesn't quite seem to state this actually is. – Dannyu NDos Sep 16 '17 at 2:36
• @hardmath Another motivation is, I was making logical semantics for ternary computers, like logic gates, flip-flap-flops, etc. It seems that there is no united standard name for ternary logic gates, so I was going to build logical semantics to give appropriate name for them. – Dannyu NDos Sep 16 '17 at 2:41
• I think that modal logic not being truth-functional is more akin to quantifier logic not being truth-functional than what you've attempted (expanding from two "binary" truth values to three). Let me think about your edit carefully to see if I can respond in a cogent way. – hardmath Sep 16 '17 at 2:42

I think you're asking whether or not, or perhaps claim that, modal logic can be reduced to truth-functional, many-valued logic. If this is the question, then the answer is no. Some explanations below.

It is important to make a distinction between the formalism and the semantics of a modal logic (system).

A modal logic (system) is formally defined, for example Hilbert style, by its syntax, axioms and rules. Thereafter, a formula is considered a theorem iff it can be formally proved from the above. This uniquely defines the set of formulas that are theorems of that system. (And this set is also considered to be the modal logic, i.e. the system is identified with the set of all its theorems.) So whether or not a formula is is a theorem is only a matter of formal derivation. It has nothing to do with the semantics or any interpretation of the formula.

Also: if there is an algorithm that allows you to decide, given any arbitrary formula, whether or not it is a theorem, then that modal logic is called decidable. NOTE: It has bee proven that there are modal logics that are not decidable.

A semantics is a/any different method (finitary or not) to associate a truth value to each formula. You can in principle invent any semantics S that you want, including one based on multi-valued truth tables. But then you run into the following questions:

• Is your modal logic sound with respect to S? i.e. is every theorem S-valid?
• Is your modal logic complete with respect to S? i.e. is every S-valid formula a theorem.

Having both would mean that a formula is a theorem iff it its S-valid.

Now, your truth-table semantics for modal logics being finitary means validity would be decidable. Therefore, in such semantics you cannot have soundness and completeness for all modal logics, because it would imply that all modal logics are decidable. Which they are certainly not.

The usual semantics for normal modal logics is the Kripke semantics. In this semantics all modal logics are sound, but there are many that are incomplete. There is also the general frame semantics, under which all modal logics are sound and complete. But neither provides a finitary algorithm for evaluating validity, such as to make all modal logics decidable.

Of course, you are still free to define your truth-table based semantics for modal logic. The only question is: how useful is it (what would you do with it)? As illustated, you would not be able to use it to decide theoremship in all cases.

There are certainly some (particular) modal logics that are decidable and can be "reduced" to truth-based logics. For example the ultra-simple Triv and Ver systems, in a quite straightforward way. (I once saw an attempt to do this for the system S5; it worked for some formulas, not sure if for all.)

For your example case study, I doubt that you can achieve soundness and completeness for provability logic in this way. This is because it has an axiom of modal degree 2, and not a particularly simple one. But it's certainly not enough to postulate some (multi-valued) truth tables. You also need to mathematically examine soundness and completeness with respect to them.

The attempt has been made to create a sound, decidable modal logic with a finite number of truth values using the 3-valued logic of Lukasiewicz. This has not been regarded as successful; although my search of the literature has not turned up a good, sound explanation for why not.

The most important reason seems to have been the difficulty of establishing sound rules of inference. Difficulty of semantic interpretation, an unreasonable adherence to the "laws" of bivalence and the excluded middle, and unfamiliar notation have also played a part.

Both of your proposed Truth tables of implication (A → B) have serious problems. For one, they do not reduce to those of classical logic in the case of propositions which have the values T and F. For another, they do not even recognize such elementary rules of inference as (A → A) as valid.


text:
# A few questions concerning one loop corrections to the action [closed]

When we perform a Legendre transform on the connected generate functional $W[J]$ we get the quantum action (or 1PI action)

$$\Gamma[\phi] = W[J(\phi)] - \int\mathrm{d}^4x\,\phi J,\quad\phi(J)=\frac{\delta W}{\delta J}.$$

Then it can be shown that

$$\Gamma[\phi] = S[\phi] \mp\frac{1}{2}\log\det\left(\frac{\delta^2S}{\delta\Phi(x)\delta\Phi(y)}\right)_{\Phi=\phi} +\ldots,$$

where $S[\phi]$ is the classical action and the dots represent higher corrections. It is said that the lowest quantum correction (ie, the term involving $\log\det$) is the result of a resummation of one loop diagrams.

1. Why is the $\log\det$ term identified with a one loop correction? I took a look at the proof, but it seems to me that there is no connection to the one loop diagrams at all.

2. Why is $\frac{\delta^2S}{\delta\Phi(x)\delta\Phi(y)}$ identified with the propagator? Is it the free propagator or the exact propagator (including interactions)?

3. Is it possible to get the same one loop correction to the action using the Wilson effective action instead?

## closed as too broad by Qmechanic♦Jun 10 at 5:39

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


text:
# SDPT3 fails in apparently simple problem

(Jesús Briales García) #1

Hi everyone, I’m experiencing some numerical problems
when using the SDPT3 solver for a (apparently) fairly simple SDP problem.

The problem under consideration is
$$\min_X tr(MX), ; s.t. X\succcurlyeq0, X(end,end) = 1$$

This problem is just the lifted relaxation of
$$\min_\tilde{x} \tilde{x}^\top M \hat{x}, ; s.t. \tilde{x}(end)^2=1$$
which in turn corresponds to a simple Least Squares (linear) problem:
$$min_x ||A x - b||$$

Obviously, this is just a testing problem before proceeding to the relaxation
of a more complex one which is not a Least Squares problem.

The curious fact is that Sedumi is working well, but SDPT3 fails to converge.
I know I should just stick with the solver working (it’s what I usually do)
but since this seems (to me) like a very simple problem
I was a little surprised that the usual SDP solvers could have any difficulty with it.
Furthermore, I’m concerned that I might be missing part of the complexity of the problem
since this could become the source of many headaches when things don’t work
in the more complex problem extending this one.

I’ve tried looking for some theoretical background that may
explain in which situations a SDP problem could get ill-conditioned
or something related, but couldn’t find anything straightforward
in the context of this specific model.
I would appreciate if any of the experts here

Just to do some simple tests, I’m providing a simple test code:

% generate positive semidefinite matrix
m = 10;
n = 3;
A = randn(m,n);
b = randn(m,1);
M = [A,-b]'*[A,-b];
cvx_solver SDPT3
% cvx_solver sedumi
% cvx_precision low
cvx_begin SDP
variable X(n+1,n+1) semidefinite
X(end,end) == 1
minimize(trace(M*X))
cvx_end

It seems sometimes (rarely) the SDPT3 solver worked,
but in general any random problem will fail.
You can easily check that apparently there are no conditioning
issues with the data matrix M.
I tried different precisions, by the way.

(Mark L. Stone) #2

I did not duplicate your experience.

I ran 1000 random instances in both SDPT3 and SeDuMi using the same A and b for each SDPT3 and SeDumI pair. All instances were solved by both solvers, with maximum absolute value of the difference in cvx_optval between SDPT3 and SeDuMi of 4.4e-7 over the 1000 instances.

CVX version 2,.1 build 1113 running under MATLAB R2014A WIN64.

(Jesús Briales García) #3

Ok,
I forgot to share information on my version too!
Currently, it’s:
Version 2.1, Build 1110 (66e9a9c) Wed Jun 10 21:43:38 2015
Under MATLAB R2015b Linux64

I tried to download the last one, which is build 1113 (the same you used)
and it still didn’t worked! Same behavior as with build 1110.

from the official webpage and calling it from CVX as an additional solver.
In this case, the SDPT3 solver worked fine!!

So, in conclusion, either there is some strange conflict within my own CVX library,
or there is something wrong in the Linux version I’m working with…
But it’s strange as it didn’t work with the clean installation with the newest version either.

This is the (cropped) screen output just in case it could give any hint:

Calling SDPT3 4.0: 10 variables, 1 equality constraints
------------------------------------------------------------

num. of constraints =  1
dim. of sdp    var  =  4,   num. of sdp  blk  =  1
*******************************************************************
SDPT3: Infeasible path-following algorithms
*******************************************************************
version  predcorr  gam  expon  scale_data
HKM      1      0.000   1        0
it pstep dstep pinfeas dinfeas  gap      prim-obj      dual-obj    cputime
-------------------------------------------------------------------
0|0.000|0.000|4.5e+00|1.4e+00|1.4e+03| 5.194500e+02  0.000000e+00| 0:0:00| chol
linsysolve: Schur complement matrix not positive definite
switch to LU factor. lu 30  30
1|1.000|1.000|2.2e+00|5.8e-03|1.2e+02| 1.206473e+02  0.000000e+00| 0:0:00| lu 30  30
...
100|0.999|1.000|5.0e-01|0.0e+00|1.1e-192| 1.074255e-192  0.000000e+00| 0:0:02|
sqlp stop: maximum number of iterations reached
-------------------------------------------------------------------
number of iterations   = 100
primal objective value =  1.43324924e-02
dual   objective value =  0.00000000e+00
gap := trace(XZ)       = 1.43e-02
relative gap           = 1.41e-02
actual relative gap    = 1.41e-02
rel. primal infeas (scaled problem)   = 5.00e-01
rel. dual     "        "       "      = 5.83e-05
rel. primal infeas (unscaled problem) = 0.00e+00
rel. dual     "        "       "      = 0.00e+00
norm(X), norm(y), norm(Z) = 1.4e-03, 0.0e+00, 3.3e+01
norm(A), norm(b), norm(C) = 2.0e+00, 2.0e+00, 3.4e+01
Total CPU time (secs)  = 1.57
CPU time per iteration = 0.02
termination code       = -6
DIMACS: 5.0e-01  0.0e+00  1.3e-04  0.0e+00  1.4e-02  1.4e-02
-------------------------------------------------------------------

------------------------------------------------------------
Status: Failed
Optimal value (cvx_optval): NaN

(Mark L. Stone) #4

Some semi-random links to look at:

Perhaps this is a bug in SDPT3 which maybe you get away with or not depending on the exact configuration?

(Jesús Briales García) #5

Well… it’s difficult to know what it could be from so little practical cases.
However, the comment in the Github issue by Johan Löfberg is quite worrying
as this means plenty of bad behavior might occur without any kind of feedback,
discouraging the use of SDPT3 in newer versions of Matlab?

(Mark L. Stone) #6

I’ve had better “luck” with SeDuMi than SDPT3 (and I haven’t used a newer version of MATLAB than R2014A). Your mileage (that’s American slang) may vary. I use SDPT3 mainly as a back up solver.

(Royi) #7

@mcg,

Does the current CVX uses the latest SDPT3?

It seems they made a very recent compatibility fix.

(Michael C. Grant) #8

I’ve experimented with the latest version, but haven’t updated it. It does indeed improve compatibility with later versions of MATLAB but it does not solve any numerical issues.


text:
# Thread: Rearranging Annuitie formula to find R

1. ## Rearranging Annuitie formula to find R

Hey everyone I need help really badly.
I have a formula that looks like this
A=PR^n + Q(R^n-1)/R-1

But i need to rearrange it to make R the subject. Ive spent hours trying to work it out but i just can't do it. I need it for my maths assignment due tommorow so quick help would be very much appreciated.

Thanks!

2. Is it $A = PR^n + \frac{Q(R^{n}-1)}{R-1}$ or $A = PR^n + \frac{Q(R^{n}-1)}{R}-1$?

3. yeah thats the exact same equation I have, but I'm having trouble rearranging it to find R.

4. which one is it?

5. Oh sorry i didnt read it properly. Anyway its the first one.

6. $R^{n} = \frac{A}{P} - \frac{Q}{P}$

$R = \sqrt[n]{\frac{A}{P} - \frac{Q}{P}}$

This is how I did it:

We know that $A = PR^n + \frac{Q(R^n-1)}{R-1}$. Now $\frac{R^{n}-1}{R-1} = R^{n-1} + R^{n-2} + \ldots + 1$ which is a geometric series. So its sum is $\frac{1}{1-R}$.

Then $A = PR^{n} + Q \left(\frac{1}{1-R} \right )$

So $A = \frac{PR^{n}(1-R) +Q}{1-R}$ or $A(1-R) = PR^{n}(1-R) + Q$.

Then $\frac{A}{P}(1-R) = R^{n}(1-R) + \frac{Q}{P}$.

Then $\frac{A}{P} - \frac{Q}{P} = R^{n}$ and so $R = \sqrt[n]{\frac{A-Q}{P}}$

7. Thankyou so much!

8. This may be the beer talking but I don’t think that

is the solution to R in .

With P=100, R=1.05, n=11, and Q=50,

≈ 881.37.

But ≈ 1.21.


text:
Discoveries among the ruins of Nineveh and Babylon; with travels in Armenia, Kurdistan, and the desert. Being the result of a second expedition undertaken for the Trustees of the British Museum

Primary tabs

Layard, A. H. (1854). Discoveries among the ruins of Nineveh and Babylon; with travels in Armenia, Kurdistan, and the desert. Being the result of a second expedition undertaken for the Trustees of the British Museum. New York: Barnes. Retrieved from http://catalog.hathitrust.org/Record/011552170


text:
# second order diff equation

500 pts ended

Find the steady state current in the RLC circuit given below where R=8?, L=0.5H, C=0.1F and E=100sin(2t) V


text:
# Count number of elements in list with condition

Consider the following example:

\documentclass{article}

\begin{document}

\noindent
How do I use \LaTeX 3 to construct a command that prints the number of elements
from a given list that are less than or equal to a given value?\\[\baselineskip]
Example: I want to count the number of commands (\verb|\Aa|--\verb|\Dj|) in the code
\begin{verbatim}
\def\Af{62} \def\Ag{45} \def\Ah{46} \def\Ai{72} \def\Aj{81}
\def\Ba{59} \def\Bb{65} \def\Bc{50} \def\Bd{64} \def\Be{67}
\def\Bf{50} \def\Bg{53} \def\Bh{55} \def\Bi{46} \def\Bj{51}
\def\Ca{61} \def\Cb{65} \def\Cc{70} \def\Cd{75} \def\Ce{63}
\def\Cf{66} \def\Cg{63} \def\Ch{52} \def\Ci{54} \def\Cj{50}
\def\Da{52} \def\Db{55} \def\Dc{52} \def\Dd{57} \def\De{53}
\def\Df{51} \def\Dg{57} \def\Dh{56} \def\Di{54} \def\Dj{63}
\end{verbatim}
that are less than or equal to $50$. I know that the answer is $32$ but how do I
create a command, say \verb|CountElements|, that gives the output $32$?

\end{document}

The question is incorporated in the example.

• Are the macros \def\Aa macros all using this naming scheme? – user31729 Feb 24 '18 at 22:44
• @ChristianHupfer Some of the numbers might be rational numbers rather than integers. – Svend Tveskæg Feb 24 '18 at 22:46
• I asked about the naming scheme, not about the expansion value. But there is an answer already ... no need for me to pursue this – user31729 Feb 24 '18 at 22:47

Count them. ;-)

\documentclass{article}

\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\countelements}{mm}
{% #1 = list of elements, #2 = condition
\int_zero:N \l__svend_count_elements_int
\cs_set:Nn \__svend_count_elements:n { #2 }
\clist_map_inline:xn { #1 }
{
\int_compare:nT
{ \__svend_count_elements:n { ##1 } }
{ \int_incr:N \l__svend_count_elements_int }
}
\int_to_arabic:n { \l__svend_count_elements_int }
}
\int_new:N \l__svend_count_elements_int
\cs_generate_variant:Nn \clist_map_inline:nn { x }
\ExplSyntaxOff

\def\Aa{53}
\def\Ab{53}
\def\Ac{42}
\def\Ae{57}
\def\Af{62}
\def\Ag{45}
\def\Ah{46}
\def\Ai{72}
\def\Aj{81}
\def\Ba{59}
\def\Bb{65}
\def\Bc{50}
\def\Bd{64}
\def\Be{67}
\def\Bf{50}
\def\Bg{53}
\def\Bh{55}
\def\Bi{46}
\def\Bj{51}
\def\Ca{61}
\def\Cb{65}
\def\Cc{70}
\def\Cd{75}
\def\Ce{63}
\def\Cf{66}
\def\Cg{63}
\def\Ch{52}
\def\Ci{54}
\def\Cj{50}
\def\Da{52}
\def\Db{55}
\def\Dc{52}
\def\Dd{57}
\def\De{53}
\def\Df{51}
\def\Dg{57}
\def\Dh{56}
\def\Di{54}
\def\Dj{63}

\newcommand{\elements}{
\Ba,\Bb,\Bc,\Bd,\Be,\Bf,\Bg,\Bh,\Bi,\Bj,
\Ca,\Cb,\Cc,\Cd,\Ce,\Cf,\Cg,\Ch,\Ci,\Cj,
\Da,\Db,\Dc,\Dd,\De,\Df,\Dg,\Dh,\Di,\Dj
}

\begin{document}

The numbers that are greater than $50$ are
\countelements{
The numbers that are between $52$ and $54$ (inclusive) are

• Great! Can all the elements be defined in a macro, i.e., \newcommand*\elements{\Aa,\Ab,\Ac,\Ad,\Ae,\Af,\Ag,\Ah,\Ai,\Aj, \Ba,\Bb,\Bc,\Bd,\Be,\Bf,\Bg,\Bh,\Bi,\Bj, \Ca,\Cb,\Cc,\Cd,\Ce,\Cf,\Cg,\Ch,\Ci,\Cj, \Da,\Db,\Dc,\Dd,\De,\Df,\Dg,\Dh,\Di,\Dj} and then use \countelements{\elements}{#1 > 50}? If I try with your code. it thrown me an error; <argument> \LaTeX3 error: Relation ',' unknown: use =, <, >, ==, !=, <=, >=.. – Svend Tveskæg Feb 24 '18 at 23:03


text:
## anonymous one year ago The lengths of the sides of a rectangular window hae the ratio 1.6 to 1. The area of the window is 2822.4 in^2. What are the window's dimensions?

1.6 to 1 is equal to 1.6x:1x The equation is as follows: $1.6x \times1x =2822.4$ $1.6x =2822.4$ Divide both sides by 1.6 $x^{2} = 1764$ Now find the square root of both sides. $x =42$ Now lets find the side s by your ratio 1.6:1 $x=1.6\times42$ $x=1\times42$ height = 67.2 leangth =42


text:
## LaTeX forum ⇒ Page Layout ⇒ Bigfoot with marginpar creates "Dimension too large" error

Information and discussion about page layout specific issues (e.g. header and footer lines, page formats, page numbers).
sujato
Posts: 1
Joined: Tue Apr 01, 2014 8:55 am

### Bigfoot with marginpar creates "Dimension too large" error

I am compiling a document that includes the bigfoot package with the para option, and also has \marginpar. If I compile a short paragraph with these options it goes through fine, but any length to it and it spits out a "Dimension too large" error.

This may be the same problem reported here https://groups.google.com/forum/?fromgroups=#!searchin/comp.text.tex/bigfoot/comp.text.tex/O1wKdgxofVs/YYGw6m82KCIJ, where the package author says that the then-current dev version didn't seem to have the problem. He offered to share the dev version privately, but later doesn't seem to be responding to messages and emails. (I hope he's ok!) The dev package doesn't seem to be available in the stated repository, either.

So I need to either adjust what I'm doing somehow, or, perhaps, use the dev version if that is available anywhere. Any ideas?

Here is an MWE. Note that this is actually a *working* example, to demonstrate what I want to achieve. To get the bug, simply duplicate the paragraph. Even such a limited expansion of the text is too much, whereas I want to compile a document of around 750,000 words, with some tens of thousands of margin notes and footnotes.

    \documentclass{book}     \usepackage{bigfoot}     \AtBeginDocument{%     \RestyleFootnote{default}{para}}     \begin{document}     \marginpar{Test}Lorem ipsum\footnote{Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. Long note. } dolor sit amet,\footnote{Note.} sed purto conclusionemque\footnote{Note.} et, \marginpar{Test} in luptatum efficiantur sea.      \end{document}

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USE CODE: SSPD25

				   Let the three digit numbers A28,3B9,62C where A, B, C are integers between 0 and 9 , be divisible by a fixed integer k. Show that the determinant                                                                                A   3   6                                              8   9   C                                                2   B   2is divisible by k

6 years ago

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										Hello student,Please find the answer to your question belowAsA28,3B9,62C where A, B, C are integers between 0 and 9 , be divisible by a fixed integer k then there exist m1,m2,m3 such that100A+20+8=km1300+10B+9=km2600+20+c=km3Applying R2=R2+100R1+10R2 we getWhereis an integerSince A,m1,m2,m3,B are integersThusis divisible by k

2 years ago

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text:
scipp.constants#

Physical and mathematical constants with units. This module a wrapper around scipy.constants and requires the scipy package to be installed.

Mathematical constants:

 pi Pi golden Golden ratio golden_ratio Golden ratio

Physical constants:

 c speed of light in vacuum speed_of_light speed of light in vacuum mu_0 the magnetic constant $$\mu_0$$ epsilon_0 the electric constant (vacuum permittivity), $$\epsilon_0$$ h the Planck constant $$h$$ Planck the Planck constant $$h$$ hbar $$\hbar = h/(2\pi)$$ G Newtonian constant of gravitation gravitational_constant Newtonian constant of gravitation g standard acceleration of gravity e elementary charge elementary_charge elementary charge R molar gas constant gas_constant molar gas constant alpha fine-structure constant fine_structure fine-structure constant N_A Avogadro constant Avogadro Avogadro constant k Boltzmann constant Boltzmann Boltzmann constant sigma Stefan-Boltzmann constant $$\sigma$$ Stefan_Boltzmann Stefan-Boltzmann constant $$\sigma$$ Wien Wien displacement law constant Rydberg Rydberg constant m_e electron mass electron_mass electron mass m_p proton mass proton_mass proton mass m_n neutron mass neutron_mass neutron mass

In addition to the above variables, scipp.constants also contains the 2018 CODATA recommended values [CODATA2018] database containing more physical constants. The database is accessed using scipp.constants.physical_constants().

CODATA2018

CODATA Recommended Values of the Fundamental Physical Constants 2018. https://physics.nist.gov/cuu/Constants/

Functions

 physical_constants(key[, with_variance]) Returns the CODATA recommended value with unit of the requested physical constant.