text:
Contents

# Contents

## Idea

The KK-compactification of heterotic string theory on Calabi-Yau manifolds of complex dimension 3, hence real dimension 6. This choice of compactification means exactly that the resulting effective field theory on 4-dimension has $N =1$ supersymmetry (see at supersymmetry and Calabi-Yau manifolds).

KK-compactifications of higher dimensional supergravity with minimal ($N=1$) supersymmetry:

perspectiveKK-compactification with $N=1$ supersymmetry
M-theoryM-theory on G2-manifolds
F-theoryF-theory on CY4-manifolds
heterotic string theoryheterotic string theory on CY3-manifolds

## References

The idea originates in

where in the introduction it says the following

Recently, the discovery [6] of anomaly cancellation in a modified version of $d = 10$ supergravity and superstring theory with gauge group $O(32)$ or $E_8 \times E_8$ has opened the possibility that these theories might be phenomenologically realistic as well as mathematically consistent. A new string theory with $E_8 \times E_8$ gauge group has recently been constructed [7] along with a second $O(32)$ theory.

For these theories to be realistic, it is necessary that the vacuum state be of the form $M_4 \times K$, where $M_4$ is four-dimensional Minkowski space and K is some compact six-dimensional manifold. (Indeed, Kaluza-Klein theory – with its now widely accepted interpretation that all dimensions are on the same logical footing – was first proposed [8] in an effort to make sense out of higher-dimensional string theories). Quantum numbers of quarks and leptons are then determined by topological invariants of $K$ and of an $O(32)$ or $E_8 \times E_8$ gauge field defined on $K$ [9]. Such considerations, however, are far from uniquely determining $K$.

In this paper, we will discuss some considerations, which, if valid, come very close to determining $K$ uniquely. We require

(i) The geometry to be of the form $H_4 \times K$, where $H_4$ is a maximally symmetric spacetime.

(ii) There should be an unbroken $N = 1$ supersymmetry in four dimensions. General arguments [10] and explicit demonstrations [11] have shown that supersymmetry may play an essential role in resolving the gauge hierarchy or Dirac large numbers problem. These arguments require that supersymmetry is unbroken at the Planck (or compactification) scale.

(iii) The gauge group and fermion spectrum should be realistic.

These requirements turn out to be extremely restrictive. In previous ten-dimensional supergravity theories, supersymmetric configurations have never given rise to chiral fermions – let alone to a realistic spectrum. However, the modification introduced by Green and Schwarz to produce an anomaly-free field theory also makes it possible to satisfy these requirements. We will see that unbroken $N = 1$ supersymmetry requires that $K$ have, for perturbatively accessible configurations, $SU(3)$ holonomy and that the four-dimensional cosmological constant vanish. The existence of spaces with $SU(3)$ holonomy was conjectured by Calabi [12] and proved by Yau [13].

(Of course later it was understood that Calabi-Yau spaces, even those of complex dimension 3, are not “very close to unique”.)

Lecture notes include

Further original references include

and chapters 12 - 16 of

A canonical textbook reference for the role of Calabi-Yau manifolds in compactifications of 10-dimensional supergravity is

David Kazhdan, John Morgan, D.R. Morrison and Edward Witten, eds. , Quantum Fields and Strings, A course for mathematicians, 2 vols. Amer. Math. Soc. Providence 1999. (web version)

Lecure notes in a more general context of string phenomenology include

Discussion of generalized Calabi-Yau manifold backgrounds is for instance in

Discussion of duality with M-theory on G2-manifolds:

Last revised on December 1, 2019 at 13:31:24. See the history of this page for a list of all contributions to it.


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# Normalising the Quantum Harmonic Oscillator

I have been working on the quantum harmonic oscillator with ladder operators and I am running into issues with normalising the excited states. There doesn't seem to be a true convention for the ladder operators; I have chosen to use: $A_{\pm}=\frac{1}{\sqrt{2m}}\left(\hat{p}\pm im\omega x\right)$ as it seems simplest to me.

I correctly arrive at wavefunctions proportional to what I want by raising the ground state, but I am having difficulty normalising them. I have tried the normal $\int_{-\infty}^\infty|\psi_n(x)|^2dx = 1$ method, but the answer I get from the second excited state seems to differ from the answer everyone else uses by a sign: I get $$\psi_2 = \left(\frac{m\omega}{4\pi \hbar}\right)^{1/4}\left(1-\frac{2m\omega x^2}{\hbar}\right)\exp\left(-\frac{m \omega x^2}{2\hbar}\right),$$ when all the answers I see are $$\psi_2 = \left(\frac{m\omega}{4\pi \hbar}\right)^{1/4}\left(\frac{2m\omega x^2}{\hbar} - 1\right)\exp\left(-\frac{m \omega x^2}{2\hbar}\right).$$

Now, clearly these yield the same probability distribution, is my answer valid regrading the sign of the wavefunction? If not, some insight into why would be good. I note that in all the material where it is derived, the normalisation constant is derived for a general state $n$ using the ladder operators, but surely the standard method should still work for individual cases.

• Since in this simple case only the absolute square of the wavefunction (or an object like $\langle\psi|H|\psi\rangle$) makes physical sense, wavefunctions are determined up to an arbitrary complex phase (like -1 in this case). Which phase you choose is irrelevant, as long as you stick to your convention. There are, of course, cases when the phase can carry a physical significance (e.g. the Berry phase in the Aharonov-Bohm effect), but that's irrelevant at this point. – Soba noodles May 7 '17 at 22:56
• @Goobley Please disregard my initial comment - I hadn't fully grasped the core of the problem. That said, if Griffiths uses $[A_-,A_+]=\hbar \omega$ it is exclusively for didactic purposes; you don't use it in the modern literature as there is a huge inertia that expects $a^\dagger a$ to be the number operator (as opposed to the hamiltonian). – Emilio Pisanty May 7 '17 at 23:22

This isn't really a question about normalization - it's about the phase of the ladder operators. It's important to note that given any ladder operator pair $a,a^\dagger$ such that $$[a,a^\dagger]=1$$ (which is the defining property of the bosonic annihilation and creation operators), you can always define, for any phase $e^{i\varphi}$, a new set of operators $a_\varphi = e^{-i\varphi}a$, $a_\varphi^\dagger = e^{i\varphi} a^\dagger$, and they will also obey the bosonic commutation relation, $$[a_\varphi,a_\varphi^\dagger] = e^{-i\varphi}e^{i\varphi} [a,a^\dagger]=1.$$ Ultimately, it doesn't matter much: it all comes down to convention, and in any case under the normal evolution, in the Heisenberg picture, $a$ will evolve to $a_{\omega t}$ anyways.
However, it's important to note that the usual convention is to take $$a=\sqrt{\frac{m\omega}{2\hbar}} \left( x + i\frac{1}{m\omega} p\right), \tag{*}$$ which differs from your definition by a factor of $i$ (as well as correcting the off-standard normalization $[A_-,A_+]=\hbar \omega$). The phase is not a big problem, so long as you stick to your convention, but it does mean that e.g. if you define the second excited state as $\frac{1}{\sqrt{2!}}{a^\dagger}^2|0⟩$, you will have accumulated a phase of $i^2=-1$, i.e. precisely the minus sign you're talking about. In some ways, the convention $(*)$ is preferable because the eigenfunctions are always real in position space (whereas with your convention the odd-numbered eigenfunctions will be pure imaginary), but that's about the most you can say in its favour - beyond the fact that it has a huge inertia behind it and that it is futile and counter-productive to try and fight that inertia.

format: plain


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We also derive four moments bounds for Hilbert-valued random variables with possibly infinite chaos expansion, which include, as special cases, all finite-dimensional four moments results for Gaussian approximation in a diffusive context proved earlier by various authors. Our main ingredient is a combination of an infinite-dimensional version of Stein’s method as developed by Shih and the so-called Gamma calculus. As an application, rates of convergence for the functional Breuer-Major theorem are established. 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Given an ... [more ▼]An evolving Riemannian manifold (M,g_t)_{t\in I} consists of a smooth d-dimensional manifold M, equipped with a geometric flow g_t of complete Riemannian metrics, parametrized by I=(-\infty,T). Given an additional C^{1,1} family of vector fields (Z_t)_{t\in I} on M. We study the family of operators L_t=\Delta_t +Z_t where \Delta_t denotes the Laplacian with respect to the metric g_t. We first give sufficient conditions, in terms of space-time Lyapunov functions, for non-explosion of the diffusion generated by L_t, and for existence of evolution systems of probability measures associated to it. Coupling methods are used to establish uniqueness of the evolution systems under suitable curvature conditions. Adopting such a unique system of probability measures as reference measures, we characterize supercontractivity, hypercontractivity and ultraboundedness of the corresponding time-inhomogeneous semigroup. 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# Reference request for research on the maximum drawdown **ratio** (NOT value)

Let's suppose the asset price process follows a Geometric Brownian motion $$S_t \sim GBM(\mu, \sigma),\,t\ge 0$$, and define the two process: \begin{align} \text{MSF}_t &:= \max_{\tau\in[0,t]} S_\tau\\ \text{MDD_Ratio}_t &: = \max_{\tau\in[0,t]} ((\text{MSF}_\tau- S_\tau) / \text{MSF}_\tau) \end{align} where MSF means "maximum so far" and MDD means maximum drawdown. As $$S_t$$ is forever positive, we see that MDD_Ratio is always well defined.

Is there any research done on the distribution, or at least the mean value of MDD_Ratio?

For what I saw, most research literature on DD seems to focus on the max drawdown value, i.e., $$\max_{0\le t\le T}(\text{MSF}_t - S_t)$$ (and most of the time the asset price process is assumed to follow a BM instead of a GBM), instead of maximum drawdown ratio whch IMHO is much commoner in practice.

• Do you mean $S_{\tau}$ in your second equation? – Gordon Jan 17 at 14:30
• @Gordon yes. All should be $\tau$ instead. – Vim Jan 17 at 15:42
• @vim please allow me to ask "why you ask?". From an investor's perspective the series [+1,+5,-1] and [+1,-1,+5] have different mdd ratios but are equally risky. Aren't they, in your opinion? – elemolotiv Jan 18 at 8:51
• @elemolotiv I assume you're talking about the percentages of return. Then your examples have the same mdd ratio (-1%). – Vim Jan 18 at 9:07
• @elemolotiv MDD ratios are very important both to investors and sometimes also to regulators. Let's consider a risky fund: if at some point the MDD ratio reaches an intolerable point for investors (say 10%), they will withdraw; if the MDD ratio reaches some regulatory threshold (say 20%), the fund may be forced to close down. – Vim Jan 18 at 9:09


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# Group of rotations of S^n

1. Jan 26, 2008

### pivoxa15

1. The problem statement, all variables and given/known data
Group of rotations of S^1 = SO(2)=S^1 conincidently
Group of rotations of S^2 = SO(3)
Group of rotations of S^3 = SO(4)

Correct?

3. The attempt at a solution
SO(3) is the group of all rotations in R^3 so it can rotate all elements of S^2 which is part of R^3. Although I can't show it rigorously.

Would it be enough to say since SO(3) consists of all rotations in R^3, it can offcouse rotate all elements having unit distance from the origin.

Similar argument made to S^1 and S^3.

Last edited: Jan 26, 2008
2. Jan 27, 2008

### George Jones

Staff Emeritus
No, the last two are not correct.

$$S^2 \cong SO \left( 3 \right) / SO \left( 2 \right)$$

$$S^3 \cong SO \left( 4 \right) / SO \left( 3 \right) \cong SU \left( 2 \right)$$

In general,

$$S^n \cong SO \left( n+1 \right) / SO \left( n \right)$$

$$S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).$$

 Sorry, I think I misunderstood what you wrote. Note that "=" should not be substituted for the word "is."

I think what you meant is something like the following.

Consider $S^n$ as a subset of $\mathbb{R}^{n+1}$. Show that each element of $SO \left( n+1 \right)$ maps $S^n$ to $S^n$. A general mapping from $\mathbb{R}^{n+1}$ to $\mathbb{R}^{n+1}$ might map an element of $S^n$ to an element of $\mathbb{R}^{n+1}$ that is not in $S^n$, so there is something "special" going on here.[/edit]

Last edited: Jan 27, 2008
3. Jan 27, 2008

### George Jones

Staff Emeritus
Continuing, let $x$ be an element of $S^n$, so $x$ is an element of $\mathbb{R}^{n+1}$ such that

$$x^T x = 1.$$

Set $y = Ax$, with $A$ in $SO \left( n+1 \right)$, and calculate the length of $y$.

4. Jan 30, 2008

### pivoxa15

With this one, $$S^{2n + 1} \cong SU \left( n+1 \right) / SU \left( n \right).$$

does it mean each coset representation is a matrix in SU(2)?


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

by Bob Horton and Vanja Paunic, Microsoft AI and Research Data Group

Training deep learning models from scratch requires large data sets and significant computational reources. Using pre-trained deep neural network models to extract relevant features from images allows us to build classifiers using standard machine learning approaches that work well for relatively small data sets. In this context, a deep learning solution can be thought of as incorporating layers that compute features, followed by layers that map these features to outcomes; here we’ll just map the features to outcomes ourselves.

We explore an example of using a pre-trained deep learning image classifier to generate features for use with traditional machine learning approaches to address a problem the original model was never trained on (see the blog post “Image featurization with a pre-trained deep neural network model” for other examples). This approach allows us to quickly and easily create a custom classifier for a specific specialized task, using only a relatively small training set. We use the image featurization abilities of Microsoft R Server 9.1 (MRS) to create a classifier for different types of knots in lumber. These images were made publicly available from the laboratory of Prof., Dr. Olli Silven, University of Oulu, Finland, in 1995. Please note that we are using this problem as an academic example of an image classification task with clear industrial implications, but we are not really trying to raise the bar in this well-established field.

We characterize the performance of the machine learning model and describe how it might fit into the framework of a lumber grading system. Knowing the strengths and weaknesses of the classifier, we discuss how it could be used to triage additional image data for labeling by human experts, so that the system can be iteratively improved.

The pre-trained deep learning models used here are optional components that can be installed alongside Microsoft R Server 9.1; directions are here.

In the sawmill industry lumber grading is an important step of the manufacturing process. Improved grading accuracy and better control of quality variation in production leads directly to improved profits. Grading has traditionally been done by visual inspection, in which a (human) grader marks each piece of lumber as it leaves the mill, according to a factors like size, category, and position of knots, cracks, species of tree, etc. Visual inspection is often an error prone and laborious task. Certain defect classes may be difficult to distinguish, even for a human expert. To that end, a number of automated lumber grading systems have been developed which aim to improve the accuracy and the efficiency of lumber grading [2-7].

DATA_DIR <- file.path(getwd(), 'data')

if(!dir.exists(DATA_DIR)){

dir.create(DATA_DIR)

knots_url <- 'http://www.ee.oulu.fi/research/imag/knots/KNOTS/knots.zip'
names_url <- 'http://www.ee.oulu.fi/research/imag/knots/KNOTS/names.txt'

unzip(file.path(DATA_DIR, 'knots.zip'), exdir = file.path(DATA_DIR, 'knot_images'))

}

knot_info <- read.delim(file.path(DATA_DIR, "names.txt"), sep=" ", header=FALSE, stringsAsFactors=FALSE)[1:2]

names(knot_info) <- c("file", "knot_class")

knot_info$path <- file.path(DATA_DIR, "knot_images", knot_info$file)

names(knot_info)
## [1] "file"       "knot_class" "path"

We’ll be trying to predict knot_class, and here are the counts of the categories:

table(knot_info$knot_class) ## ## decayed_knot dry_knot edge_knot encased_knot horn_knot ## 14 69 65 29 35 ## leaf_knot sound_knot ## 47 179 Four of these labels relate to how the knot is integrated into the structure of the surrounding wood; these are the descriptions from the README file: • sound: “A knot grown firmly into the surrounding wood material and does not contain any bark or signs of decay. The color may be very close to the color of sound wood.” • dry: “A firm or partially firm knot, and has not taken part to the vital processes of growing wood, and does not contain any bark or signs of decay. The color is usually darker than the color of sound wood, and a thin dark ring or a partial ring surrounds the knot.” • encased: “A knot surrounded totally or partially by a bark ring. Compared to dry knot, the ring around the knot is thicker.” • decayed: “A knot containing decay. Decay is difficult to recognize, as it usually affects only the strength of the knot.” Edge, horn and leaf knots are related to the orientation of the knot relative to the cutting plane, the position of the knot on the board relative to the edge, or both. These attributes are of different character than the ones related to structural integration. Theoretically, you could have an encased horn knot, or a decayed leaf knot, etc., though there do not happen to be any in this dataset. Unless otherwise specified, we have to assume the knot is sound, cross-cut, and not near an edge. This means that including these other labels makes the knot_class column ‘untidy’, in that it refers to more than one different kind of characteristic. We could try to split out distributed attributes for orientation and position, as well as for structural integration, but to keep this example simple we’ll just filter the data to keep only the sound, dry, encased, and decayed knots that are cross-cut and not near an edge. (It turns out that you can use featurization to fit image classification models that recognize position and orientation quite well, but we’ll leave that as an exercise for the reader.) knot_classes_to_keep <- c("sound_knot", "dry_knot", "encased_knot", "decayed_knot") knot_info <- knot_info[knot_info$knot_class %in% knot_classes_to_keep,]
knot_info$knot_class <- factor(knot_info$knot_class, levels=knot_classes_to_keep)

We kept the knot_class column as character data while we were deciding which classes to keep, but now we can make that a factor.

Here are a few examples of images from each of the four classes:

library("pixmap")

set.seed(1)

samples_per_category <- 3

op <- par(mfrow=c(1, samples_per_category), oma=c(2,2,2,2), no.readonly = TRUE)

for (kc in levels(knot_info$knot_class)){ kc_files <- knot_info[knot_info$knot_class == kc, "path"]
kc_examples <- sample(kc_files, samples_per_category)
for (kc_ex in kc_examples){
plot(pnm, xlab=gsub(".*(knot[0-9]+).*", "\\1", kc_ex))
mtext(text=gsub("_", " ", kc), side=3, line=0, outer=TRUE, cex=1.7)
}
}

par(op)

## Featurizing images

Here we use the rxFeaturize function from Microsoft R Server, which allows us to perform a number of transformations on the knot images in order to produce numerical features. We first resize the images to fit the dimensions required by the pre-trained deep neural model we will use, then extract the pixels to form a numerical data set, then run that data set through a DNN pre-trained model. The result of the image featurization is a numeric vector (“feature vector”) that represents key characteristics of that image.

Image featurization here is accomplished by using a deep neural network (DNN) model that has already been pre-trained by using millions of images. Currently, MRS supports four types of DNNs - three ResNet models (18, 50, 101)[1] and AlexNet [8].

knot_data_df <- rxFeaturize(data = knot_info,
mlTransforms = list(loadImage(vars = list(Image = "path")),
resizeImage(vars = list(Features = "Image"),
width = 224, height = 224,
extractPixels(vars = "Features"),
featurizeImage(var = "Features",
dnnModel = "resnet101")),
mlTransformVars = c("path", "knot_class"))
## Elapsed time: 00:02:59.8362547

We have chosen the “resnet101” DNN model, which is 101 layers deep; the other ResNet options (18 or 50 layers) generate features more quickly (well under 2 minutes each for this dataset, as opposed to several minutes for ResNet-101), but we found that the features from 101 layers work better for this example.

We have placed the features in a dataframe, which lets us use any R algorithm to build a classifier. Alternatively, we could have saved them directly to an XDF file (the native file format for Microsoft R Server, suitable for large datasets that would not fit in memory, or that you want to distribute on HDFS), or generated them dynamically when training the model (see examples in the earlier blog post).

Since the featurization process takes a while, let’s save the results up to this point. We’ll put them in a CSV file so that, if you are so inspired, you can open it in Excel and admire the 2048 numerical features the deep neural net model has created to describe each image. Once we have the features, training models on this data will be relatively fast.

write.csv(knot_data_df, "knot_data_df.csv", row.names=FALSE)

## Select training and test data

Now that we have extracted numeric features from each image, we can use traditional machine learning algorithms to train a classifier.

in_training_set <- sample(c(TRUE, FALSE), nrow(knot_data_df), replace=TRUE, prob=c(2/3, 1/3))
in_test_set <- !in_training_set

We put about two thirds of the examples (203) into the training set and the rest (88) in the test set.

## Fit Random Forest model

We will use the popular randomForest package, since it handles both “wide” data (with more features than cases) as well as multiclass outcomes.

library(randomForest)

form <- formula(paste("knot_class" , paste(grep("Feature", names(knot_data_df), value=TRUE), collapse=" + "), sep=" ~ "))

training_set <- knot_data_df[in_training_set,]
test_set <- knot_data_df[in_test_set,]

fit_rf <- randomForest(form, training_set)

pred_rf <- as.data.frame(predict(fit_rf, test_set, type="prob"))
pred_rf$knot_class <- knot_data_df[in_test_set, "knot_class"] pred_rf$pred_class <- predict(fit_rf, knot_data_df[in_test_set, ], type="response")

# Accuracy
with(pred_rf, sum(pred_class == knot_class)/length(knot_class))
## [1] 0.8068182

Let’s see how the classifier did for all four classes. Here is the confusion matrix, as well as a moisaic plot showing predicted class (pred_class) against the actual class (knot_class).

with(pred_rf, table(knot_class, pred_class))
##               pred_class
## knot_class     sound_knot dry_knot encased_knot decayed_knot
##   sound_knot           51        0            0            0
##   dry_knot              3       19            0            0
##   encased_knot          6        5            1            0
##   decayed_knot          2        1            0            0

mycolors <- c("yellow3", "yellow2", "thistle3", "thistle2")
mosaicplot(table(pred_rf[,c("knot_class", "pred_class")]), col = mycolors, las = 1, cex.axis = .55,
main = NULL, xlab = "Actual Class", ylab = "Predicted Class")

It looks like the classifier performs really well on the sound knots. On the other hand, all of the decayed knots in the test set are misclassified. This is a small class, so let’s look at those misclassified samples.

library(dplyr)

pred_rf$path <- test_set$path

# misclassified knots
misclassified <- pred_rf %>% filter(knot_class != pred_class)

# number of misclassified decayed knots
num_example <- sum(misclassified$knot_class == "decayed_knot") op <- par(mfrow = c(1, num_example), mar=c(2,1,0,1), no.readonly = TRUE) for (i in which(misclassified$knot_class == "decayed_knot")){
example <- misclassified[i, ]
pnm <- read.pnm(example$path) plot(pnm) mtext(text=sprintf("predicted as\n%s", example$pred_class), side=1, line=-2)
}
mtext(text='Misclassified Decayed Knots', side = 3, outer = TRUE, line = -3, cex=1.2)

par(op)

We were warned about the decayed knots in the README file; they are more difficult to visually classify. Interestingly, the ones classified as sound actually do appear to be well-integrated into the surrounding wood; they also happen to look somewhat rotten. Also, the decayed knot classified as dry does have a dark border, and looks like a dry knot. These knots appear to be two decayed sound knots and a decayed dry knot; in other words, maybe decay should be represented as a separate attribute that is independent of the structural integration of the knot. We may have found an issue with the labels, rather than a problem with the classifier.

Let’s look at the performance of the classifier more closely. Using the scores returned by the classifier, we can plot an ROC curve for each of the individual classes.

library(pROC)

plot_target_roc <- function(target_class, outcome, test_data, multiclass_predictions){
is_target <- test_data[[outcome]] == target_class
roc_obj <- roc(is_target, multiclass_predictions[[target_class]])
plot(roc_obj, print.auc=TRUE, main=target_class)
text(x=0.2, y=0.2, labels=sprintf("total cases: %d", sum(is_target)), pos=3)
}

op <- par(mfrow=c(2,2), oma=c(2,2,2,2), #mar=c(0.1 + c(7, 4, 4, 2)),

for (knot_category in levels(test_set\$knot_class)){
plot_target_roc(knot_category, "knot_class", test_set, pred_rf)
}

mtext(text="ROC curves for individual classes", side=3, line=0, outer=TRUE, cex=1.5)

par(op)

These ROC curves show that the classifier scores can be used to provide significant enrichment for any of the classes. With sound knots, for example, the score can unambiguously identify a large majority of the cases that are not of that class. Another way to look at this is to consider the ranges of each classification score for each actual class:

library(tidyr)
library(ggplot2)

pred_rf %>%
select(-path, -pred_class) %>%
gather(prediction, score, -knot_class) %>%
ggplot(aes(x=knot_class, y=score, col=prediction)) + geom_boxplot()

Note that the range of the sound knot and dry knot scores tend to be higher for all four classes. But when you consider the scores for a given prediction class across all the actual classes, the scores tend to be higher when they match than when they don’t. For example, even though the decayed knot score never goes very high, it tends to be higher for the decayed knots than for other classes. Here’s a boxplot of just the decayed_knot scores for all four actual classes:

boxplot(decayed_knot ~ knot_class, pred_rf, xlab="actual class", ylab="decayed_knot score", main="decayed_knot score for all classes")

Even though the multi-class classifier did not correctly identify any of the three decayed knots in the test set, this does not mean that it is useless for finding decayed knots. In fact, the decayed knots had higher scores for the decayed predictor than most of the other knots did (as shown in the boxplots above), and this score it is able to correctly determine that almost 80% of the knots are not decayed. This means that this classifier could be used to screen for knots that are more likely to be decayed, or alternatively, you could use these scores to isolate a collection of cases that are unlikely to be any of the kinds of knots your classifier is good at recognizing. This ability to focus on things we are not good at might be helpful if we needed to search through a large collection of images to find more of the kinds of knots for which we need more training data. We could show those selected knots to expert human labelers, so they wouldn’t need to label all the examples in the entire database. This would help us get started with an active learning process, where we use a model to help develop a better model.

Here we’ve shown an open-source random forest model trained on features that have been explicitly written to a dataframe. The model classes in the MicrosoftML package can also generate these features dynamically, so that the trained model can accept a file name as input and automatically generate the features on demand during scoring. For examples see the earlier blog post. In general, these data sets are “wide”, in that they have a large number of features; in this example, they have far more columns of features than rows of cases. This means that you need to select algorithms that can handle wide data, like elastic net regularized linear models, or, as we’ve seen here, random forests. Many of the MicrosoftML algorithms are well suited for wide datasets.

## Discussion

Image featurization allows us to make effective use of relatively small sets of labeled images that would not be sufficient to train a deep network from scratch. This is because we can re-use the lower-level features that the pre-trained model had learned for more general image classification tasks.

This custom classifier was constructed quite quickly; although the featurizer required several minutes to run, after the features were generated all training was done on relatively small models and data sets (a few hundred cases with a few thousand features), where training a given model takes on the order of seconds, and tuning hyperparameters is relatively straightforward. Industrial applications commonly require specialized classification or scoring models that can be used to address specific technical or regulatory concerns, so having a rapid and adaptable approach to generate such models should have considerable practical utility.

Featurization is the low-hanging fruit of transfer learning. More general transfer learning allows the specialization of deeper and deeper layers in a pre-trained general model, but the deeper you want to go, the more labelled training data you will generally need. We often find ourselves awash in data, but limited by the availability of quality labels. Having a partially effective classifier can let you bootstrap an active learning approach, where a crude model is used to triage images, classifying the obvious cases directly and referring only those with uncertain scores to expert labelers. The larger set of labeled images is then used to build a better model, which can do more effective triage, and so on, leading to iterative model improvement. Companies like CrowdFlower use these kinds of approaches to optimize data labeling and model building, but there are many use cases where general crowdsourcing may not be adequate (such as when the labeling must be done by experts), so having a straightforward way to bootstrap that process could be quite useful.

The labels we have may not be tidy, that is, they may not refer to distinct characteristics. In this case, the decayed knot does not seem to really be an alternative to “encased”, “dry”, or “sound” knots; rather, it seems that any of these categories of knots might possibly be decayed. In many applications, it is not always obvious what a properly distributed outcome labeling should include. This is one reason that a quick and easy approach is valuable; it can help you to clarify these questions with iterative solutions that can help frame the discussion with domain experts. In a subsequent iteration we may want to consider “decayed” as a separate attribute from “sound”, “dry” or “encased”.

Image featurization gives us a simple way to build domain-specific image classifiers using relatively small training sets. In the common use case where data is plentiful but good labels are not, this provides a rapid and straightforward first step toward getting the labeled cases we need to build more sophistiated models; with a few iterations, we might even learn to recognize decayed knots. Seriously, wood that knot be cool?

## References

1. Kaiming He, Xiangyu Zhang, Shaoqing Ren, Jian Sun. (2016) The IEEE Conference on Computer Vision and Pattern Recognition (CVPR), pp. 770-778

2. Kauppinen H. and Silven O.: A Color Vision Approach for Grading Lumber. In Theory & Applications of Image Processing II - Selected papers from the 9th Scandinavian Conference on Image Analysis (ed. G. Borgefors), World Scientific, pp. 367-379 1995.

3. Silven O. and Kauppinen H.: Recent Developments in Wood Inspection. (1996) International Journal on Pattern Recognition and Artificial Intelligence, IJPRAI, pp. 83-95, 1996.

4. Rinnhofer, Alfred & Jakob, Gerhard & Deutschl, Edwin & Benesova, Wanda & Andreu, Jean-Philippe & Parziale, Geppy & Niel, Albert. (2006). A multi-sensor system for texture based high-speed hardwood lumber inspection. Proceedings of SPIE - The International Society for Optical Engineering. 5672. 34-43. 10.1117/12.588199.

5. Irene Yu-Hua Gu, Henrik Andersson, Raul Vicen. (2010) Wood defect classification based on image analysis and support vector machines. Wood Science and Technology 44:4, 693-704. Online publication date: 1-Nov-2010.

6. Kauppinen H, Silven O & Piirainen T. (1999) Self-organizing map based user interface for visual surface inspection. Proc. 11th Scandinavian Conference on Image Analysis (SCIA’99), June 7-11, Kangerlussuaq, Greenland, 801-808.

7. Kauppinen H, Rautio H & Silven O (1999). Nonsegmenting defect detection and SOM-based classification for surface inspection using color vision. Proc. EUROPTO Conf. on Polarization and Color Techniques in Industrial Inspection (SPIE Vol. 3826), June 17-18, Munich, Germany, 270-280.

8. Alex Krizhevsky, Ilya Sutskever, and Geoffrey E. Hinton. (2012) ImageNet Classification with Deep Convolutional Neural Networks. In NIPS, 2012.


text:
# Why is the area of the circle $πr^2$? [duplicate]

I searched many times about the cause of the circle area formula but I did not know anything so ...

Why is the area of the circle $\pi r^2$?

Thanks for all here.

• That the area depends on the radius-squared is somewhat of an observation from considering inscribed and circumscribed regular polygons. The occurrence of $\pi$ is strictly for keeping track of exactly what the proportionality constant is between $r^2$ and the area of the circle. – Cameron Williams Nov 30 '14 at 22:32
• Has this really never been asked here before? – HDE 226868 Nov 30 '14 at 22:53
• what is $\pi$ in your formula? – abel Nov 30 '14 at 22:53
• this is my favorite explanation/animated gif ... people.wku.edu/tom.richmond/Pir2b.html – John Joy Dec 1 '14 at 16:27

How Archimedes viewed it:

$%emptyuselesstextemptyuselesstextemptyuselesstext$

As the width of the slices approaches $0$, the object on the right-hand side approaches a rectangle of width $2\pi r /2 = \pi r$ and height $r$, hence area $\pi r^2$.

• (+1) Excellent answer, at the right level for someone who needs to ask this question. – John Bentin Nov 30 '14 at 23:07
• Actually, Archimedes used a method of exhaustion: en.wikipedia.org/wiki/Area_of_a_disk#Archimedes.27s_proof – Alan C Dec 1 '14 at 2:26
• is "exhaustion" Greek for "calculus"? – John Joy Dec 1 '14 at 16:28

The area of a circle with radius $r$ is just $r^2$ times the area of the unit circle, by homothety. So the area of the circle is the square of the radius times a universal constant, given by: $$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& 2\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} = \color{red}{\Gamma(1/2)^2}.\end{eqnarray*}$$

• Seriously? Beta and Gamma functions? – user2345215 Nov 30 '14 at 22:39
• I like Jack's answer. It may not help OP, but I wouldn't have thought to do that -- so it helps me out. (+1) – user137731 Nov 30 '14 at 22:41
• I think that in mathematics we should not regard the $\Gamma$ function as an alien. It is not so different from the sine or cosine function, speaking about its Weierstrass product. – Jack D'Aurizio Nov 30 '14 at 22:41
• That's definitely an overkill. You can evaluate the integral by $x\mapsto \sin x$ – UserX Nov 30 '14 at 22:43
• So $\pi$ is half the length of the unit circle, or the fist positive root of the sine function, but we are all stating the same thing. The only difference is that, through the Bohr-Mollerup theorem, the $\Gamma$ function exists even without knowing any circle or trigonometric function. In a way, it looks more natural to me to define $\pi$ as $m!^2$ with $m=-\frac{1}{2}$. – Jack D'Aurizio Nov 30 '14 at 22:44

A number of different ways of showing this exist. First notice that the area has to be $(\text{constant}\cdot r^2)$ because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by $3$, then the area is multiplied by $9$. And "constant" in this case means it's the same number regardless of what $r$ is. So now the question is: Why must the "constant" be the same as the ratio of circumference to diameter?

As $r$ increases, we have \begin{align} \text{rate of growth of area} & = \text{size of boundary} \times\text{rate of motion of boundary} \\ & = \text{circumference} \times \text{rate at which $r$ is changing} \\ & = 2\pi r \times \text{rate at which $r$ is changing}. \end{align} From calculus, recall that $$\text{rate of change of r^2} = 2 r \times\text{rate of change of r}.$$ So we have $$\text{rate of growth of area} = \pi\times \text{rate of change of r^2}.$$ So the area grows at the same rate at which $\pi r^2$ grows. That makes them always equal if they're equal when $r=0$. And it's easy to see that they're equal when $r=0$.

That's only one way to do it; there are others.

PS: Supposing we don't know calculus; how would we know that $$\text{rate of change of r^2} = 2 r \times\text{rate of change of r}\text{ ?}$$ We could do that as follows. A square whose side has length $r$ is growing because its north side is moving northward and its east side is moving eastward. Then \begin{align} & \text{rate of growth of $r^2$} \\ = {} & \text{rate of growth of square's area} \\ = {} & \text{size of moving boundary}\times\text{rate of motion of boundary} \\ = {} & 2r \times \text{rate of change of $r$}. \end{align}

• I really like this approach. It makes a lot of intuitive sense. – user109879 Nov 30 '14 at 22:54
• @ChantryCargill : Thank you; I'm glad you like it. – Michael Hardy Nov 30 '14 at 23:09

Some variety, the circle of geodesic radius $\rho$ on the unit sphere has area $$2 \pi (1 - \cos \rho).$$ i will look that up in a minute...Yep, compared with THIS, take their $r=1$ and $h=1-\cos \rho.$

Less familiar, the circle of geodesic radius $\rho$ in the non-Euclidean plane of constant curvature $(-1)$ is $$2 \pi (\cosh \rho - 1).$$

Go Figure.

In both cases, the limit as $\rho \rightarrow 0$ agrees with $\pi \rho^2,$ the error is of size $O(\rho^4).$

It's the evauluation of the definite integral $$2\int_{-r}^{r}y\,dx$$ where $y=\sqrt{r^2-x^2}$, so it's really $$2\int_{-r}^r\sqrt{r^2-x^2}\,dx$$ In other words, it's twice the area under the curve of $y$, from $x=-r$ to $x=r$.

There's no way to evaluate this integral with a simple formula, like would be if, say, $y=x^3$. The most common method used is a form of trig substitution, where $x \to \sin u$. From here, it's easy to show that $$dx=\cos u \, du$$ and so the previous integral becomes $$\int_{-r}^{r} \cos u \sqrt{r^2-\sin^2u}\,du$$ If $r=1$, we can use $$\cos^2u=1-\sin^2u$$ to make this easier: $$\int_{-r}^{r} \cos^2u\,du$$ Otherwise, it takes some playing around to get the answer.

By the way, this method is particularly valuable with ellipses ($A=\pi ab$).

• It wouldn't hurt to mention the $x\mapsto r\sin x$ substitution, it may not be obvious how to solve this to a beginner. – user2345215 Nov 30 '14 at 22:47
• @user2345215 True. And that should explain the $\pi$. – HDE 226868 Nov 30 '14 at 22:47
• @MichaelHardy Thank you for the edits. – HDE 226868 Nov 30 '14 at 22:54
• @HDE226868 : Actually it's the value of, not the "solution to", the integral. One doesn't "solve" integrals; one evaluates integrals. ${}\qquad{}$ – Michael Hardy Dec 2 '14 at 16:39

For another integration method, let $C$ be the interior of the circle $x^2+y^2=r^2$ with radius $r$. Then the area, with the aid of polar coordinates, is given by

\begin{align*} \iint_C dA &= \int_0^{2\pi}\int_0^r\rho\,\mathrm{d}\rho\,\mathrm{d}\theta\\ &=\frac{1}{2}r^2\int_0^{2\pi}\,\mathrm{d}\theta\\ &=\pi r^2. \end{align*}


text:
Notebooks

Liquid Crystals

03 Oct 1994 12:01

In normal crystalline solids, molecules or atoms sit in fixed locations, arranged in a pattern which repeats itself over space. (Actually, the locations aren't quite fixed, because of thermal motion, but that's a bit of a side issue.) In normal liquids, the molecules are free to move about, and their distribution over space is not point-like but smeared and more or less uniform. Liquid crystals are states of matter intermediate between normal liquids and normal crystals. The simplest to envision are the nematic liquid crystals, where the molecules are long and rod-like; in their fluid phase, the direction of the rods is random; in the nematic phase, nearby rods are oriented along a common axis, the director'', but their positions are still random. (In chiral nematics, the director rotates as you move through the material, arranging the molecules in a helical pattern.) In cholesteric liquid crystals, the molecules are oriented along a common axis, but also lie in regular, parallel planes, each one slightly twisted with respect to its neighbors --- imagine a heap of lengths of picket-fence, piled one atop the other. (Cholesterol, surprisingly enough, has a cholesteric phase.) In smectics, the molecules again lie in parallel layers, though they are free to move about within each layer, but now are oriented perpendicular, or nearly so, to the layers. (There are many sub-varieties of smectics.) These are the principal sorts of liquid crystal, but literally dozens of distinct phases have been identified.

Recommended;
• Collings, Liquid Crystals: Nature's Delicate State of Matter
• P.G. de Gennes and J. Prost, The Physics of Liquid Crystals


text:
#### Sentence Examples

• If R and S are the ends of a prismoid, A and B their areas, h the perpendicular distance between them, and C the area of a section by a plane parallel to R and S and midway between them, the volume of the prismoid is *h(A+4C+B).
• The most instructive is to regard the prismoid as built up (by addition or subtraction) of simpler figures, which are particular cases of it.
• (iv) For the prismoid in general let ABCD ...
• By (i) and (iii), the formula holds for each of these figures; and therefore it holds for the prismoid as a whole.
• Another method of verifying the formula is to take a point Q in the mid-section, and divide up the prismoid into two pyramids with vertex Q and bases ABCD ...


text:
# Non-Euclidean Geometry

In hyperbolic, Euclidean and elliptical geometry, two straight lines that are connected to a normal are different from one another.

The non-Euclidean geometries are specializations of absolute geometry . They differ from Euclidean geometry , which can also be formulated as a specialization of absolute geometry, in that the axiom of parallels does not apply in them .

On a sphere, the sum of the angles of a triangle is generally not 180 °. The surface of a sphere is not Euclidean , but locally the laws of Euclidean geometry are a good approximation. For example, in a small triangle on the surface of the earth, the sum of the angles of a triangle is pretty much exactly 180 degrees.

## history

The non-Euclidean geometries developed from centuries of unsuccessful attempts to prove the parallel axiom of Euclidean geometry. At the beginning of the 19th century, the mathematicians János Bolyai , Nikolai Lobatschewski and Carl Friedrich Gauß determined that there was not necessarily a Euclidean geometry of space, and began to develop a non-Euclidean geometry. Other forerunners were Ferdinand Karl Schweikart , Franz Taurinus , Giovanni Girolamo Saccheri and Johann Heinrich Lambert . Carl Friedrich Gauß did not publish his results at all.

Between 1818 and 1826 Gauß headed the Hanoverian land surveying and developed methods with considerably increased accuracy. In this context, the idea arose that he was looking empirically for a curvature of the room by measuring the sum of the angles in a triangle formed by the Brocken in the Harz Mountains , the Inselsberg in the Thuringian Forest and the Hohen Hagen near Göttingen . Today it is seen by the majority as a legend, even if the possibility that Gauss was looking for deviations from the usual value of the angle sum of 180 ° cannot be ruled out with ultimate consequence. However, the accuracy of his instruments would have been far from sufficient to prove the tiny curvature of space in the earth's gravitational field. It is still not possible today.

It was Gauss' student Bernhard Riemann who developed the differential geometry of curved spaces and presented it in 1854. At the time, nobody expected this topic to be physically relevant. Tullio Levi-Civita , Gregorio Ricci-Curbastro and Elwin Bruno Christoffel further expanded the differential geometry. Einstein found a variety of mathematical tools for his general theory of relativity in their work .

## Basics

The non-Euclidean geometries were not developed with the aim of making our spatial experience more precise, but as axiomatic theories in dealing with the problem of parallels . The existence of models for non-Euclidean geometries (e.g. by Felix Klein and Henri Poincaré ) proves that the parallel axiom of Euclid cannot be deduced from the other axioms and is independent of them.

One obtains non-Euclidean geometries by removing the axiom of parallels from the system of axioms or by changing it. The basic change options are:

• There is no parallel to a straight line and a point outside the straight line. So two different straight lines in one plane always intersect. This leads to an elliptical geometry . A clear model of a two-dimensional elliptical geometry is the geometry on a spherical surface . Here the sum of the angles of a triangle is greater than 180 °, the circumference of a circle is less than and the area is less than . In elliptical geometry, however, the axioms of arrangement no longer apply unchanged.${\ displaystyle 2 \ pi r}$${\ displaystyle \ pi r ^ {2}}$
• There are at least two parallels to a straight line and a point outside the straight line. Here all other axioms can be preserved. A hyperbolic geometry is obtained . It can be described in two ways by the models of Klein and Poincaré. In miniature (or locally) it can be visualized on a saddle surface of constant Gaussian curvature (the so-called pseudosphere ). The sum of the angles of a triangle is now less than 180 °, the circumference of a circle is more than and its area is more than .${\ displaystyle 2 \ pi r}$${\ displaystyle \ pi r ^ {2}}$

Meanwhile, non-Euclidean geometry plays an important role in theoretical physics and cosmology . According to the general theory of relativity , the geometry of the universe differs from the Euclidean one because gravity fields "bend" space. Whether the geometry of the universe “on a large scale” is spherical (elliptical), even (Euclidean) or hyperbolic is one of the major current questions in physics.

## literature

• Felix Klein : Lectures on non-Euclidean geometry . Julius Springer, Berlin 1928 (XII, 326, online ). Reprint Felix Klein: like VDM, Müller, Saarbrücken 2006, ISBN 3-8364-0097-9 (XII, 326).
• Julian L. Coolidge : The elements of non-Euclidean geometry. Cornell Univ. Library, Cornell 2008, ISBN 978-1-4297-0446-5
• Nikolai I. Lobachevsky, Pangeometry, Translator and Editor: A. Papadopoulos, Heritage of European Mathematics Series, Vol. 4, European Mathematical Society, 2010.
• Norbert A'Campo , Athanase Papadopoulos, Notes on hyperbolic geometry . In: Strasbourg Master class on Geometry, pp. 1--182, IRMA Lectures in Mathematics and Theoretical Physics, Vol. 18, Zurich: European Mathematical Society (EMS), 461 pages, 2012, ISBN 978-3-03719-105-7 , doi: 10.4171 / 105 .
• Marvin J. Greenberg : Euclidean and non-Euclidean geometries - development and history. Freeman, New York 2008, ISBN 978-0-7167-9948-1
• Boris A. Rozenfel'd : A history of non-Euclidean geometry - evolution of the concept of a geometric space. Springer, New York 1988, ISBN 3-540-96458-4
• János Bolyai, (et al.): Non-euclidean geometries. Springer, New York 2006, ISBN 978-0-387-29554-1
• Benno Klotzek: Euclidean and non-Euclidean elementary geometries . 1st edition. Harri Deutsch, Frankfurt am Main 2001, ISBN 3-8171-1583-0 .

## Notes and individual references

1. For more information on the history of non-Euclidean geometry, see the article Parallels axiom


text:
# How to look for fractals/harmonics patterns in time series?

I want to build trading systems based on two things:

1)Fractal Theory

2)Harmonics Pattern

I have read the book : The Misbehavior of Markets: A Fractal View of Financial Turbulence By Mandelbrot but he didn’t show how to find fractal in time series. Moreover I don’t have time to go through all books with this topic, because I have work afterhours. So my question to you is:

-Do you know best measures for fractal theory or harmonics patterns in Time Series of Stock?

-Do you know some helpful publications/books connected with this topic – how to look for patterns/measures?

-Do you have any suggestions about construction of such trading systems?

-
I find your question way too broad and open ended to answer. It sounds like you want others to deliver you the done deal to plug in. Maybe you can focus on a more specific issue and ask more targeted questions? –  Matt Wolf Jul 19 '13 at 11:29
This your opinion. I am just looking for some starting point and I don't want to read all the random books in the way, but good ones. –  Zimas Jul 19 '13 at 11:33
Asking for book recommendations is of course fine, asking others to pretty much guide you through the construction, well, pretty much delivering you the setup on a silver platter is maybe asking for a little too much. Maybe that is not your intention, it just comes across to me as such. Maybe you could rephrase the question? –  Matt Wolf Jul 19 '13 at 11:42
Depending on what you look to use such system for one of my previous answers may be of help: quant.stackexchange.com/questions/8242/multi-fractals-models –  Matt Wolf Jul 19 '13 at 12:13
I only try to answer questions or reference papers I myself have worked on or believe I understand. But I suffer from an overoptimistic nature ;-) –  Matt Wolf Jul 19 '13 at 14:07

Perhaps to detect fractal behaviour you could fit something like a Daubechies wavelet. That is, $W(a,b) := \frac{1}{\sqrt{|a|}} \int_{-\infty}^{\infty} f(t) \phi((t-b)/a)dt$. Then you want to check the set $\{W(a,b) : a \in \mathbb{R}_+\}$ where $a$ is the scale, for some fixed $b$. If all the coefficients are similar then this might be some indication of fractal like behaviour at least in that time localisation.


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# Thread: Surface integral question...

1. ## Surface integral question...

How do I do this....and how many ways are there to do this?

2. Originally Posted by Zirtco
How do I do this....and how many ways are there to do this?
Find the equation of the plane in the form z=.....

$\iint_S f(x,y,g(x,y))*\sqrt{1+\bigg[\frac{\partial z}{\partial x}\bigg]^2+\bigg[\frac{\partial z}{\partial y}\bigg]^2}dA$

g(x,y)=z

3. I think dwsmith is misinterpreting the question. The "surface of the tetrahedron" includes all four faces.

One method, of course, would be to integrate on all four faces separately.

a) For example, the face with vertices (0,0,0), (1,0,0), and (0,1,0) is the triangle with x= 0, y= 0 and x+ y= 1 as boundaries. To integrate over that, let x range from 0 to 1. For each x, then, y ranges from 0 to 1- x:
$\int_{x= 0}^1\int_{y= 0}^{1- x} (x^2+ y^2) dy dx$
(0f course, z= 0 on this face.)

b) To integrate on the face with vertices (0,0,0), (1,0,0), and (0,0,1), do the same in the xz-plane- just replace "y" with "z".

c) To integrate on the face with vertices (0,0,0), (0,1,0), and (0,0,1) do the same in the yz-plane- just replace "x" with "z".

It should not be hard to see, from symmetry, that those three integrals will give the same result- so just do the first one and multiply by 3.

d) The face with vertices (1,0,0), (0,1,0) and (0,0,1) is a little harder. Projecting down to the xy-plane, we have the same triangle as in (a) but now, since the plane is x+ y+ z= 1, z= 1- x- y:
$\int_{x=0}^1\int_{y=0}^{1- x} \sqrt{x^2+ y^2+ (1- x- y)^2}dy dx$.

4. so to find part (d), do we take triple integral over dxdydz ?


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Warnings when changing chapter's appearence in TOC with tocloft

I want to get rid of some annoying warnings that showed up when changing how chapters look in TOC (institutional requirements).

From the log:

Overfull \hbox (58.08pt too wide) detected at line 3 \TU/lmr/bx/n/12 CHAPTER I []

Underfull \hbox (badness 10000) in paragraph at lines 3--3 [][] []
[]

Overfull \hbox (63.48pt too wide) detected at line 5 \TU/lmr/bx/n/12 CHAPTER II []

Underfull \hbox (badness 10000) in paragraph at lines 5--5 [][] []
[]

Overfull \hbox (68.88pt too wide) detected at line 8 \TU/lmr/bx/n/12 CHAPTER III []

Underfull \hbox (badness 10000) in paragraph at lines 8--8 [][] []
[]

I'm pretty sure the issue isn't related to package conflicts, but mainly to inexperience with LaTeX code (actually, added some TeX code in there). Since the document is now in final stages, i'm not willing to switch packages or classes at this point :\.

The MWE:

\documentclass[12pt,oneside,notitlepage,letterpaper,openany]{book}
\usepackage{fontspec}
\usepackage{polyglossia}
\usepackage{tocloft}
\defaultfontfeatures{Ligatures=TeX}
\setmainlanguage{english}

%% https://tex.stackexchange.com/questions/231302/change-number-after-chapter-in-table-of-contents
\newcommand*{\Romannumeral}[1]{\uppercase\expandafter{\romannumeral#1}}

\renewcommand{\cftchappresnum}{CHAPTER~\Romannumeral}

%% Remove blank space before chapter titles (where chapter number was)
\setlength{\cftchapnumwidth}{0pt}

%% Chapter titles in a new line
\makeatletter
\renewcommand{\cftchapaftersnumb}{\par\nobreak\vskip 6\p@}
\makeatother

\begin{document}
\frontmatter
\tableofcontents
\newpage
\mainmatter
\chapter*{Introduction}
\chapter{The problem}
\section{Why is it important?}
\chapter{Some background information}
\section{Someone's theory}
\chapter{Conclusions}
\end{document}

Looks like this:

Introduction

CHAPTER I

The problem
1.1  Why is it important? .......

CHAPTER II

Some background information
2.1  These people .......
2.2  Someone's theory .......

CHAPTER III

Conclusions

• They aren't errors but warnings. If the output looks OK then ignore them. I have tried, but so far failed, to eliminate the warnings but perhaps someone else will do better. – Peter Wilson Jun 24 '18 at 18:06
• @PeterWilson I see, thank you. I was afraid it would mess up other things eventually. – undostres Jun 24 '18 at 19:07
• I get different warnings. I also get different output: I have page numbers, whereas your code says you shouldn't. – cfr Jun 30 '18 at 2:25
• If I comment \setlength{\cftchapnumwidth}{0pt}, I get the same warnings as you, but I still get page numbers. – cfr Jun 30 '18 at 2:26
• @cfr confused "chapter number" with "page number", sorry. Fixed it in the MWE – undostres Jul 3 '18 at 14:45


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# Solutions to Newton's laws with unbounded kinetic energy for inverted harmonic oscillator potential

In a one dimensional setting, Hooke's Law, together with Newton's 2nd law, results in a differential equation of the form (setting the constants $$m$$ and $$k$$ equal to 1 for notational simplicity)

$$x''(t) = -x(t)$$,

with solutions of the form

$$x(t) = a\sin(t) + b\cos(t)$$,

where $$a$$ and $$b$$ are constants. The kinetic energy of this motion is bounded above by

$$\frac{1}{2}(|a| + |b|)^2$$.

But now consider the above differential equation with the sign reversed, namely

$$x''(t) = x(t)$$.

The general solution to this equation has the form

$$x(t) = ae^t + be^{-t}$$.

This motion seems strange to me, in part because the kinetic energy (and more simply, the speed) increases without bound as $$t$$ increases. Energy conservation is not violated, because the potential energy is negative, and decreases as the kinetic energy increases. Using two particles one can also set up a similar system of equations in which momentum is conserved as well. But this still seems odd, perhaps just because I haven't encountered such a phenomenon in everyday life.

Is this last equation a completely valid solution to Newton's 3 laws? Or is there some additional assumption I have overlooked in those laws that prevents these kinds of solutions? If not, is there some other principle that would rule out such solutions?

Finally, if this is a legitimate equation of motion given Newton's laws, is there a simple argument as to why we do not see such phenomena in everyday experience (in contrast to the behavior captured by Hooke's Law)?

• I don't understand your question. You could always contrive some sort of second order differential equation that isn't found "in every day life". Does this mean the math is invalid? Math has never cared about whether or not it describes something physically realizable. Physicists put the constraints on the math depending on the system in question in order to describe reality. Mar 5 '19 at 8:22
• @BioPhysicist I’m not asking whether the math is invalid. In fact, I demonstrate that the math is valid, in the explanation preceding my question. I’m asking whether the physical principles involved are invalid. A complete theory of physics should be able not only to derive what’s physically possible, but also to rule out what’s not. Jun 9 '21 at 5:02

This equation is not at all problematic. Kinetic energy will go to infinity only if you allow x to go to infinity, which you usually won't consider. Notice that this also occurs in the case of constant acceleration x''=g (like free fall on earth surface), or any potential decreasing in a strict monotonic way.

More intriguing is the case where the potential energy goes to minus infinity on a finite range, as is the case for the Coulombian interaction (V=-K/r) like in an atom. But even in this case the infinite energy comes from the imperfection of the model: inside the atomic nucleus, the potential is no longer Coulombian, but similar to a parabola with positive curvature.

**Edit occurence of equation x''=x. ** This equation is most often introduced as a model for an unstable equilibrium. Exactly in the same way as an oscillator, for small amplitude, can be modeled by an harmonic oscillator, by replacing the genuine potential by a parabola of positive curvature (with a spring strength given by the second derivative of the potential evaluated at its minimum).

A potential having a (local) maximum, can be modeled next to this equilibrium point by a parabola with negative curvature. A mobile at rest exactly at this point can stay at this point, but any minute disturbance will give to it an offset or a kick, thus inducing an increasing force in the same direction as the offset. So, instead of the recall to equilibrium appearing for an oscillator, one has an expelh force which in this model gives rise to an exponential escape. Of course , in such cases, the range of validity of this approximation is limited and x will likely not be able to go to infinity. The escape time is then given by \sqrt{m/V''} where V'' is the second derivative of the potential.

• Thanks, your comment on unstable equilibria largely answers my last question. You say "only if you allow x to go to infinity", but this is what I want to consider. The Coulomb potential is somewhat opposite to what I describe--it has a singularity at the origin, while the above potential has a singularity only at infinity. This allows unbounded growth of kinetic energy, without anything undefined. I was hoping there was some general principle that ruled out such behavior (for fundamental potentials, not models), e.g. a symmetry principle. Maybe the principle is just that V->0 as x->infinity? Apr 26 '19 at 6:10

The equation you post corresponds to a Hamiltonian:

$$H = \frac{1}{2} (p^2 -x^2)$$

(the usual harmonic oscillator has a plus sign). Since the system is Hamiltonian it can't violate any of Newton's laws. However the Hamiltonian is unbounded below, which leads to unphysical infinite kinetic energies and as such it's unlikely that such a Hamiltonian is a good model for the system at large values of $$x$$. A classic example of a physical system is an inverted pendulum, whose Hamiltonian:

$$H = \frac{1}{2}p^2 - \cos(\theta) \approx \frac{1}{2}(p^2-x^2)$$

where the approximation is taking $$x=(\theta-\pi)\ll 1$$. Clearly as the exponential behaviour takes over $$x$$ becomes large and this approximation becomes poor, this is likely to be typical of any system experiencing such exponential growth.

. . . just because I haven't encountered such a phenomenon in everyday life . . .

In addition to the existing answers, a common example of the form $$x'' + \alpha x=0$$ with $$\alpha<0$$ in physics is flutter in Fluid Structure Interaction. See here for a video of the phenomenon.

Of course, the governing equation is no longer valid when the system becomes "too nonlinear".

$$x''(t) = x(t)\Rightarrow m\,x''(t) = m\,x(t)= \rm Force$$ is Newton's second law.

You have an ever increasing force acting on a mass $$m$$ so it is not surprising that the kinetic energy keeps increasing with time?

. . . . just because I haven't encountered such a phenomenon in everyday life . . . .

How such a situation is achieved in practice is another matter.
You need something which delivers an ever increasing amount of power as time goes on.


text:
# zbMATH — the first resource for mathematics

The complement to $$2n$$ hyperplanes in $$\mathbb{CP}^ n$$ is not hyperbolic. (Russian) Zbl 0611.51019
The main result of the paper is the following. Let $$k$$ be a field and $$\text{card}(k)\geq 2n$$. For every choice of $$2n$$ hyperplanes $$\{D_ i\}^{2n}_{i=1}$$ of $$k{\mathbb P}^ n$$, there is a line $$q$$ such that $$| (\cup^{2n}_{i=1}D_ i)\cap q| \leq 2$$. Thus it is verified the hypothesis of P. Kiernan [Proc. Am. Math. Soc. 22, No. 3, 603–606 (1969; Zbl 0182.11101)].

##### MSC:
 51M99 Real and complex geometry 14N05 Projective techniques in algebraic geometry 32Q45 Hyperbolic and Kobayashi hyperbolic manifolds


text:
# Math Help - Transpose of a linear transformation

1. ## Transpose of a linear transformation

If $F$ is a field and $f$ is the linear functional on $F^2$ s.t. $f(x_1,x_2) = ax_1+bx_2$
For the following linear operators $T$, let $g = T^{t}f$and find $g(x_1,x_2)$

lets say $T(x_1,x_2) = (x_1,0)$

So do I find $f$ first, get $f(x_1,x_2) = ax_1+bx_2$

do i now treat " $ax_1$" and " $bx_2$" as a vector with two rows and multiply by the transpose of the matrix representation of the linear transformation?

in this case, i get $ax_1$

I read this part of the book, it really isn't clear on what to do. Thank you for your help.

2. ## Re: Transpose of a linear transformation

the definition of g doesn't make sense unless $g = f \circ T^t$, because the domain of $T^t$ is $F^2$, and the co-domain of f is $F$.

we're missing some crucial information here. please state the entire problem.

3. ## Re: Transpose of a linear transformation

Sorry. To clarify, it is composition. $g = f \; o \; T^{t}$

4. ## Re: Transpose of a linear transformation

but you wrote $g = T^tf$ which isn't the same thing. you can't compose two functions unless the co-domain of the first one you do, is the domain of the 2nd one you do.

to make matters even more confusing, some authors means g(f(x)) by gf, and some mean, f(g(x)).

IF $g = f \circ T^t$, there isn't any problem:

$g(x_1,x_2) = f(T^t(x_1,x_2)) =$

$=f(\begin{bmatrix}t_{11}&t_{21}\\t_{12}&t_{22}\end {bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix})$

$=f(t_{11}x_1+t_{21}x_2,t_{12}x_1+t_{22}x_2)$

$=a(t_{11}x_1+t_{21}x_2) + b(t_{12}x_1+t_{22}x_2)$

where $T = \begin{bmatrix}t_{11}&t_{12}\\t_{21}&t_{22} \end{bmatrix}$ is the matrix for T in the standard basis.

if we use the example you gave, $T(x_1,x_2) = (x_1,0)$, T has the matrix:

$\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and its transpose is: $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, which is the same as T, in which case:

$g(x_1,x_2) = f(x_1,0) = ax_1$.