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# Messy
You are fed up with your messy room, so you decided to clean it up.
Your room is a bracket sequence $s=s_{1}s_{2}\dots s_{n}$ of length $n$. Each character of this string is either an opening bracket ‘(‘ or a closing bracket ‘)’.
In one operation you can choose any consecutive substring of $s$ and reverse it. In other words, you can choose any substring $s[l \dots r]=s_l, s_{l+1}, \dots, s_r$ and change the order of elements in it into $s_r, s_{r-1}, \dots, s_{l}$.
For example, if you will decide to reverse substring $s[2 \dots 4]$ of string $s=$”((()))” it will be equal to $s=$”()(())”.
regular (aka balanced) bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters ‘1’ and ‘+’ between the original characters of the sequence. For example, bracket sequences “()()”, “(())” are regular (the resulting expressions are: “(1)+(1)”, “((1+1)+1)”), and “)(” and “(” are not.
A prefix of a string $s$ is a substring that starts at position $1$. For example, for $s=$”(())()” there are $6$ prefixes: “(“, “((“, “(()”, “(())”, “(())(” and “(())()”.
In your opinion, a neat and clean room $s$ is a bracket sequence that:
• the whole string $s$ is a regular bracket sequence;
• and there are exactly $k$ prefixes of this sequence which are regular (including whole $s$ itself).
For example, if $k = 2$, then “(())()” is a neat and clean room.
You want to use at most $n$ operations to make your room neat and clean. Operations are applied one after another sequentially.
It is guaranteed that the answer exists. Note that you do not need to minimize the number of operations: find any way to achieve the desired configuration in $n$ or less operations.Input
The first line contains integer number $t$ ($1 \le t \le 100$) — the number of test cases in the input. Then $t$ test cases follow.
The first line of a test case contains two integers $n$ and $k$ ($1 \le k \le \frac{n}{2}, 2 \le n \le 2000$, $n$ is even) — length of $s$ and required number of regular prefixes.
The second line of a test case contains $s$ of length $n$ — the given bracket sequence. It contains only ‘(‘ and ‘)’.
It is guaranteed that there are exactly $\frac{n}{2}$ characters ‘(‘ and exactly $\frac{n}{2}$ characters ‘)’ in the given string.
The sum of all values $n$ over all the test cases in the input doesn’t exceed $2000$.Output
For each test case print an answer.
In the first line print integer $m$ ($0 \le m \le n$) — the number of operations. You do not need to minimize $m$, any value is suitable.
In the following $m$ lines print description of the operations, each line should contain two integers $l,r$ ($1 \le l \le r \le n$), representing single reverse operation of $s[l \dots r]=s_{l}s_{l+1}\dots s_{r}$. Operations are applied one after another sequentially.
The final $s$ after all operations should be a regular, also it should be exactly $k$ prefixes (including $s$) which are regular.
It is guaranteed that the answer exists. If there are several possible answers you can print any.Exampleinput
4
8 2
()(())()
10 3
))()()()((
2 1
()
2 1
)(
output
4
3 4
1 1
5 8
2 2
3
4 10
1 4
6 7
0
1
1 2
Note
In the first example, the final sequence is “()(()())”, where two prefixes are regular, “()” and “()(()())”. Note, that all the operations except “5 8” in the example output are useless (they do not change $s$).
Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n, k;
cin >> n >> k;
string s;
cin >> s;
string t = "";
for (int i = 0; i < k; i++) {
t += "()";
}
while (t.size() < s.size()) {
t = t.substr(0, t.size() - 1) + "())";
}
cout << n << '\n';
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
if (s[j] == t[i]) {
cout << i + 1 << " " << j + 1 << '\n';
reverse(s.begin() + i, s.begin() + j + 1);
break;
}
}
}
}
return 0;
}
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{}
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# Convex not strictly convex!
Update:
Linear programming problems (LP) have a convex space, precisely vector space, such as a convex feasible region as pointed out by @prubin. Also, they may have either unique or multiple solutions. (there are already other types).
On the other hand, there is still a concept named strictly convex that interprets:
a strictly convex space is a normed vector space (X, || ||) for which the closed unit ball is a strictly convex set. Put another way, a strictly convex space is one for which, given any two distinct points x and y on the unit sphere ∂B (i.e. the boundary of the unit ball B of X), the segment joining x and y meets ∂B only at x and y. Strict convexity is somewhere between an inner product space (all inner product spaces being strictly convex) and a general normed space in terms of structure. It also guarantees the uniqueness of the best approximation to an element in X (strictly convex) out of a convex subspace Y, provided that such an approximation exists.
The distinction between these two concepts comes in the following (for more details please, see this link):
A function $$f: \mathbb{R} \to \mathbb{R}$$ is convex if for all $$x,y > \in \mathbb{R}$$ and for all $$\lambda \in (0,1)$$ the following holds: $$f(\lambda x +(1-\lambda)y) \leq \lambda f(x) +(1-\lambda) f(y)$$
Geometrically this means that the line through two points $$f(x)$$ and $$f(y)$$ on the graph is always above the graph between $$x$$ and $$y$$.
We say that $$f$$ is strictly convex if the above inequality holds strictly, i.e. $$f(\lambda x +(1-\lambda)y) < \lambda f(x) > +(1-\lambda) f(y)$$
Since I would like to know, 1) What does exactly it mean? 2) Is there any simple example, in the context of LP and MIP, to illustrate that? 3)If an LP has a such space, might it have only a unique solution?
• Two straight lines (in Euclidean geometry) can intersect in multiple points if the lines are equivalent. So in the realm of LP, strict convexity is just saying that no two constraints are equivalent (because if they are, then you cannot determine which of the two constraints is slack, with slack = 0, and which is a member of the binding constraint set). Sep 16 at 20:00
• @BenVoigt, Many thanks for the exolanation. Sep 20 at 8:49
No. Consider the triangle $$T$$ defined by $$x \geq 0$$, $$y \geq 0$$, $$x + y \leq 1$$, which is convex. The subset consisting of the vertices, $$\{(0,0), (0,1), (1,0)\}$$ is not convex. Neither is the set $$\{(x,y) \in T : x^2 + y^2 \geq 1/2\}$$.
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### Supervised Neural Networks for RFI Flagging
#### Abstract
Neural network (NN) based methods are applied to the detection of radio frequency interference (RFI) in post-correlation, post-calibration time/frequency data. While calibration does affect RFI for the sake of this work a reduced dataset in post-calibration is used. Two machine learning approaches for flagging real measurement data are demonstrated using the existing RFI flagging technique AOFlagger as a ground truth. It is shown that a single layer fully connect network can be trained using each time/frequency sample individually with the magnitude and phase of each polarization and Stokes visibilities as features. This method was able to predict a Boolean flag map for each baseline to a high degree of accuracy achieving a Recall of $$0.69$$ and Precision of $$0.83$$ and an F1-Score of $$0.75$$.
The second approach utilizes a convolutional neural network (CNN) implemented in the U-Net architecture, shown in literature to work effectively on simulated radio data. In this work the architecture trained on real data results in a Recall, Precision and F1-Score $$0.84$$, $$0.91$$, $$0.87$$ respectfully.
This work seeks to investigate the application of supervised learning when trained on a ground truth from existing flagging techniques, the results of which inherently contain false positives. In order for a fair comparison to be made the data is imaged using CASA’s CLEAN algorithm and the U-Net and NN’s flagging results allow for $$5$$ and $$6$$ additional radio sources to be identified respectively.
neural networks, U-Net, RFI, imaging, source finding
# 1 INTRODUCTION
The dramatic rise in sensitivity and the increased bandwidth of modern radio telescopes has caused an ever growing increase in the spectrum overlap between astronomical measurements and man-made radio communication. These man-made signals, viewed as radio frequency interference (RFI) in these applications, are often orders of magnitude more powerful than the faint astronomical emissions.
There exist many techniques for the mitigation, detection and excision of RFI in radio astronomy. Many of these techniques take place in different stages along the data capture and processing pipeline. Some examples of pre-correlation mitigation techniques include; governmental legislature to reduce the presence of man-made signals and separate reference antennas directed at common sources of RFI.[1] Despite attempts at reducing the presence of RFI through pre-correlation hardware techniques, post-correlation of RFI through software is required in the time/frequency and/or antenna space to ensure no corrupted data is further processed.
This paper examines an application of neural networks (NN) to detect RFI in post-correlated interferometry measurements. The NNs are trained on existing techniques for RFI flagging, a Boolean mask identifying non-astronomical time/frequency samples, in order to learn and identify the relationship between RFI and astronomical data.
Majority of existing flagging algorithms are created to ensure as much RFI is identified as possible, their outputs often contain many false positives. This is due to the extremely detrimental effect the high magnitude RFI outliers cause in the imaging pipeline. This makes minimising over-fitting of the NN imperative as well as making comparison metrics between algorithms more complicated.
The high volumes of data produced from interferometers for each observation ensure that even with a high number of measurements flagged as corrupted it is still possible to produce high quality science images. It is still advantageous for a flagging algorithm to minimize the number of false positives in order reduce image noise, increase the number sources identified and their associated flux.
# 2 EXISTING APPROACHES TO FLAGGING
Pre-correlation techniques must handle vast amounts of data and in order to be usable must have a low complexity. While advantageous to perform an operation like blanking or subtraction of on-line short RFI bursts, any remaining RFI must be removed in the data reduction and imaging pipeline.[2]
## 2.1 AOFLAGGER
The AOFlagger pipeline is a popular default RFI flagging application used in multiple observatories. It operates iteratively using surface fitting and thresholding techniques on a single baseline.[3] The SumThreshold technique in AOFlagger is a combinatorial thresholding method operating in the time/frequency domain. A selected threshold $$x_M$$ where $$M$$ is the number of samples surrounding the target is iteratively decreased until the sum of the amplitudes in $$M$$ exceeds the threshold $$x_M$$ and all visibilities are flagged.[4][5]
## 2.2 CASA TF CROP
TFCrop is a autoflag algorithm created for NRAO’s CASA.[6] It attempts to detect the presence of outliers on the 2D time-frequency plane, operating on chunks of time on each baseline and correlation independently. It seeks to create a bandshape template by iteratively fitting third-degree polynomials and calculating the standard deviation between the data and the fit. The result is then divided between all timestamps in the chunk in order to calculate deviation from the mean and identify narrowband RFI. The variable threshold is used to iteratively flag RFI and the entire process is repeated in the opposite direction, averaging over time or averaging over frequency.[7]
## 2.3 GRIDflag
A separate approach to the techniques operating in the 2D time-frequency plane utilizes the UV plane to detect corrupted data. The tracks each baseline forms in the UV plane are interpolated onto a regular grid in order for imaging algorithms to apply a fast Fourier transform (FFT). Each baseline will often contribute multiple visibilties from different time intervals to a particular cell. These visibility samples will be a measure of the same celestial information - but may record different RFI owing to the time of observation.
This distinction is how the GRIDflag algorithm was implemented. It generates RFI thresholds based on the differences between visibilities within a UV-bin.[8]
## 2.4 MACHINE LEARNING TECHNIQUES
The application of supervised machine learning for RFI flagging have recently been investigated in literature.
In the work done by Mosiane O, et al implementations of K-Nearest Neighbour (k-NN), Random Forest Classifier (RFC) and Naive Bayesian (NB) where trained on measurements flagged using AOFlagger as a ground truth.[9][10][11] Each time-frequency baseline was flattened and concatenated together and a sliding window was used to extract statistical features. The RFC showed high predictive capabilities with an F1 score of $$0.93$$.[12]
Neural network techniques have primarily been treating the time/frequency domain as a semantic segmentation problem. To this extent various architectures of convolutional neural networks have been implemented with simulated time/frequency data. HIDE & SEEK is an open source package simulating single dish radio survey data and uses a U-Net architecture for RFI detection.[13]
More recently work has been done to simulate Hydrogen Epoch of Reionization Array (HERA) visibility data with simulated RFI in order to act as a ground truth. A Deep Fully Convolutional Neural Network (DFCN) in the U-Net architecture is trained on a single polarization’s magnitude and/or phase. Their results proved more effective than the currently used watershed algorithm in their pipeline and showed improvement in using magnitude and phase as features over just magnitude. After training prediction is done on real HERA-67 data and achieved a recall of $$81\%$$ and precision of $$58\%$$.[14]
# 3 METHODOLOGY: DATA PREPROCESSING
Data preprocessing is a fundamental step in the machine learning process as it directly affects the ability of a model to learn. The datasets used in these investigations are from MeerKAT science observations. These are stored in the form of CASA Measurement sets (MS) and accessed using Taql, a high level SQL-like table query standard. While effective for query operations associated with astronomical applications it has a high data access overhead with a complexity incompatible with machine learning. To overcome this the storage used for rapid data access during the learning is process is accomplished using HDF5 files, specifically designed for high performance I/O processing and storage.
The data used in these application is fully polarized having undergone four complex correlations. Each correlation of the complex visibility from each baseline is in the form $$X_1X_2$$, $$X_1Y_2,$$$$Y_1X_2$$, $$Y_1Y_2$$ for the $$X$$ and $$Y$$ polarizations for pairs of telescopes 1 and 2.
A neural network is proposed which is trained on each time/frequency sample individually, with features being represented by the magnitude and phase of each polarization and Stokes visibilities. This accounts for a total of 16 features from 8 magnitude and 8 phase vectors.
The four polarizations are used to compute the four complex Stokes visibilties for dual linearly polarized antennas to serve as these additional features. Stokes parameters are used to describe the total intensity and the degree of polarization as another characteristic for the network to identify RFI versus non-RFI. $\begin{gather}[b] I = X_1X_2 + Y_1Y_2 \\ Q = X_1X_2 - Y_1Y_2\\ U = X_1Y_2 + Y_1X_2 \\ V = -j(X_1Y_2 - Y_1X_2) \end{gather}$
Justification for using each polarization and Stokes visibility as additional features is shown by the investigations into the Pearson correlation coefficient, measuring the linear correlation between the magnitude of each polarization and Stokes parameter over all baselines combined. Where: $$\operatorname{cov}$$ is the covariance, $$\sigma_X$$ is the standard deviation of X, $$\sigma_Y$$ is the standard deviation of Y.
$$$\rho_{X,Y}= \frac{\operatorname{cov}(X,Y)}{\sigma_X \sigma_Y}\label{eq:pearson}$$$
Results of the correlation coefficients are shown in Figures 1 and ¿fig:corr-95rfi?. These show a high correlation between all the RFI samples from all baselines flagged using AOFlagger, and conversely a low correlation between only the astronomical data. It is assumed that by using each additional feature a subsequent intrinsic characteristic of RFI is being represented, allowing the network learn from a more complete representation of the RFI present and improve its predictive capabilities. It was shown experimentally that this is the case. With metrics measuring the performance of later networks on different combinations of features showing that using all 16 features resulted in higher accuracy.
The 16 feature vectors comprising of the magnitudes and phases of each polarization and Stokes visibility are then normalized by a min-max scaling of the entire time/frequency spectrum in order to maintain the euclidean relationship between high magnitude RFI outliers and the astronomical data:
$$$x' = \frac{x - \text{min}(x)}{\text{max}(x)-\text{min}(x)}$$$
The amount of time-frequency data from a single measurement set is vast. A MeerKAT observation using all 64 telescopes would account for $$(64\times(64-1))/2=2016$$ baselines, often with around $$4000$$ frequency bins and $$2000$$ time samples. As not all baselines are necessary for the training process, the pre-processing and storage of data in HDF5 files is done using randomly chosen baselines in order to ensure a fair sample of which telescope’s data is used.
# 4 NEURAL NETWORK ARCHITECTURE
A neural network design is proposed which operates on each time-frequency sample with the 16 features derived from the magnitudes and phases of each polarization and Stokes visibility described in Section 3. The neural network is optimized for the number of layers and nodes, and a convolutional neural network based on the U-net architecture is described.
## 4.1 OPTIMIZATION OF A NEURAL NETWORK
Ordinarily a grid search would be used to optimize the hyperparameters of the NN. The number of features combined with the number of baselines necessary to capture sufficient representations of RFI leads to high complexity. It would prove impossible to optimize the network in a complete manner with combinations of; the number of nodes, number of layers, loss functions, activation functions and optimization functions. To overcome this, a heuristic approach is taken evaluating different hyperparameters independently.
Investigations are carried out using Keras into the effect of differing the number of layers with the first hidden layer maintaining 512 nodes and each subsequent layer halving the number of nodes. Further investigations are carried out into varying the number of nodes in order to identify a baseline architecture for further fine-grained optimization.
The activation functions of the hidden layers are fixed as ReLU, $$f(x) = x^+ = \operatorname{max}(0, x)$$, chosen for their properties of sparsity and reduced likelihood of a vanishing gradient. The output is a single node with a sigmoidal activation function $$S(x) = \frac{1}{1 + e^{-x}}$$. This produces a probabilistic prediction between $$0$$ and $$1$$ from the input vector. Setting a threshold at $$0.5$$ would identify everything from $$<0.5$$ as non-RFI and $$>0.5$$ as RFI. This threshold could be varied to produce an optimum output favouring RFI prediction.
The dataset is split into $$70\%/30\%$$ training and test data. Each iteration is trained using 5-fold cross-validation with a mean of the resulting metrics being taken. As the amount of data cannot be loaded into memory a data generator is used to extract data from the HDF5 files each time a new batch is requested. The order of each data batch is shuffled at the end of each epoch to reduce overfitting.
The number of epochs for training each iteration are constrained using an early stopping callback monitoring the F1 score. This reduces the overall time required for training but can have the effect of preventing overfitting by stopping the training process if the F1 score of the validation data has not shown improvement above $$0.01$$.
Evaluation of the results is done through the use of precision, recall and F1 score for binary classes. ROC and AUC are used as further metrics to evaluate the classification problem at different threshold values.
The receiver operating characteristics (ROC) curve is a metric used in binary classification, plotting the True Positive Rate against the False Positive Rate as the discrimination threshold is varied. The area under curve (AUC) describes the ability of the model to discriminate between the binary classes, where 1 would be able to perfectly distinguish between classes.
Precision is an indication of the percentage of correctly identified astronomical data, or non-RFI events. Recall describes the percentage of correctly identified non-RFI events taking into account incorrectly identified RFI. The F1 score is simply the harmonic mean of precision and recall, used as a metric for evaluating the overall predictive capabilities of a classifier. These metrics are used over accuracy, as the identification of RFI represents an imbalanced classification, where the amount of RFI is often far lower than that of non-RFI. $\begin{gather} \label{eq:prec-95rec-95f1} Precision &= \frac{\text{true positive}}{\text{true positive + false positive}}\\ Recall &= \frac{\text{true positive}}{\text{true positive + false negative}}\\ F1 score &= 2 \times \frac{\text{precision} \times \text{recall}}{\text{precision} + \text{recall}}\end{gather}$
## 4.2 NUMBER OF LAYERS
These tests investigate varying the number of hidden layers from one to five with the initial hidden layer containing 512 nodes and each successive layer halving the previous number of nodes. This implementation was found effective through separate trail and error experimentation, where it was evident decreasing the number of nodes each layer lead to improvements in performance over an increasing or the same number of nodes. This is expected to be from a reduction in overfitting by reducing the number of high level features extracted each layer.
The mean of the 5-fold cross validation for; ROC, precision, recall and F1 score are used as metrics. These assist in identifying the performance of each network on the test data into varying the number of layers are evaluated in Figure 2 and ¿fig:prec-95rec?. The ROC curve demonstrates that the differences between multiple layers is minor. A single layer with 512 nodes proves to be the most effective. This effect is likely due to the hyperplane for RFI identification not existing in a high-dimensional space, as many simple time-frequency thresholding techniques prove effective for the problem. It is probable having more layers resulted in overfitting while also increasing processing time.
The precision recall curves for each layer demonstrate the same results as the ROC. From these results it is clear a single layer is an effective implementation going forward.
## 4.3 NUMBER OF NODES
In accordance with the previous tests and their control hyperparameters, an investigation into varying the number of nodes in a single layer in carried out. The resulting ROC and precision recall curves are shown in Figure 3 and ¿fig:rec-95nodes?. These plots show how after $$64$$ nodes the difference in accuracy is minimal. The loss of accuracy for $$2048$$ nodes is likely due to the training reaching the maximum number epochs at 3000 not being sufficient to reach convergence. For further optimization and a network with a single layer of 512 nodes is selected. While reducing the number of nodes will aid in complexity, the network is so small already that the biggest overhead in training is likely data transfer not weight updates.
# 5 U-net
The second neural network proposed for this application is the U-net architecture, proposed by Ronneberger et al as an extension of the convolutional neural network (CNN) used in image segmentation.[16] This architecture of CNN has been used for the prediction of RFI in related works and has shown significant results.[13][14]
The architecture differs from a traditional CNN by using an increasing number of features in each convolutional layer as the network approaches the fully connected convolutional layer, whereby it then decreases the number of features towards the output layer. The final layer applies a $$1\times1$$ convolution to map the final layer in order to formulate a probabilistic decision with the use of a sigmoidal activation function. The models loss function is evaluated with binary cross entropy and the optimization algorithm used is Adam.
In order to utilize the entire time-frequency plane of each baseline the ‘images’ are generated by slicing the time-frequency data into $$100\times100$$ segments, each with all $$16$$ features. This non power $$2$$ image size requires the addition of cropping layers and a zero-padding layer before the final convolutional layers in order for the input shape to be matched after the last upsampling operation. Overfitting is attempted to be minimised through the use of a dropout layer of $$0.3$$ and batch normalization limiting the activation’s after each double convolutional layer.
Training is done using a training/test split of $$70/30$$ for all the images. Data transfer is handled through the use of a data generator to fetch each batch from a chunked HDF5 file. The ROC and recall precision curves in Figure 4 and ¿fig:u-95rec? demonstrate remarkable predictive capabilities with an AUC of $$0.98$$. In this case where the training data contains over-flagging false positives, an accuracy that high leads to overfitting despite multiple steps taken to reduce this.
## 5.1 RESULTS
The goal of training networks on existing flagging results was to investigate whether they could identify RFI more accurately in order to prevent the existing false positives when predicting their own training data. By generating an automated system to re-flag data and capture additional celestial information it is hoped further insights can be gained once the data is imaged.
Results are generated by predicting RFI on the entire dataset, including training data. These resulting probabilistic outputs could have their decision threshold varied per baseline in order to maximise a specific metric. So far attempts at optimising this have proved unsuccessful and a threshold is set at $$Data < 0.5 < RFI$$.
Table 1 shows the resulting metrics over the entire dataset between the two methods when using a threshold of $$0.5$$ compared to AOFlagger as the ground truth. It is interesting to note that when monitoring the these metrics during the training process when predicting the randomly shuffled $$30\%$$ test data at the end of each epoch - there is little difference. This is evidence of both networks being robust to overfitting as the results of training and testing remain within $$~5\%$$ of each other.
Table 1: Evaluation metrics between the NN and U-Net
Precision Recall F1
Neural Network 0.828 0.6917 0.754
U-Net 0.905 0.837 0.870
These resulting Boolean masks are copied into the CASA MS and are imaged using CASA’s CLEAN algorithm. The algorithm is used in image deconvolution, iteratively working on the highest values of identified point sources and subtracting a small gain convolved with the point spread function of the observation.
Python Blob Detection and Source Finding (PyBDSF) is used to generate Gaussian and island models of the identified sources in images. The images generated using flags from AOFlagger, the NN and the U-Net are processed and shown in the resulting Figure 5. The CLEAN algorithm is known to be robust and therefore not many changes can be seen between each image.
An example of the flagging difference between the different methods is shown in Figure 6. While these results visualised across multiple baselines show many similarities, it can be seen that the U-Net approach struggles to identify the sporadic blips - while the NN approach appears to identify additional and different blips to AOFlagger.
Table 2 shows a comparison in evaluation metrics between the images derived from each technique. It can be seen that U-Net is the only implementation which results in a positive background mean. This is likely due to the increase in flux density from PyBDSF fitting less Gaussians during processing than it did to the others.
Table 2: Comparisons of PyBDSF Metrics
AOFlagger NN U-Net
Bg. mean (Jy/beam) -6e-05 -1.3e-05 1.1e-04
Bg. rms (Jy/beam) 1.54e-3 1.65e-3 1.54e-3
Flux Density (Jy) 5.142 5.345 5.301
Source Count 59 65 64
# 6 CONCLUSIONS
The evaluation metrics in Table 1 show a high predictive capability for the U-Net despite zero padding influencing 4 pixels of each image. The high accuracy in U-Nets predictive results are likely causing the training of false-positives to reoccur in post-training dataset prediction. This overfitting is not an ideal result. As the goal of this work was to not only show how existing flagging strategies can be learnt by neural network implementations, but to attempt to reduce false-positive RFI predictions in the ground truth.
The increase in image source counts for the two neural network implementations imply a more precise Boolean flag map has been obtained compared to the existing algorithm, this is possibly due to the implementation of AOFlagger flagging excess amounts of useful data which the implementations do not. As described in Figure 6 the NN showed an improved detection of intermittent transient RFI, which is notoriously difficult to identify. Without simulated data a ground truth is unknown so any deductions made are purely speculative. Yet the results show how localised intermittent RFI was identified, where no spacial relationship exists as each time/frequency sample is treated independently. This is promising evidence of the network learning high dimensional features which work to discriminate RFI in a different manner to traditional thresholding techniques.
Concrete validation of results from machine learning and neural network methods are difficult to conclude and far more testing is required on varied datasets. Yet these results show how an implementation of a simple single layer fully connected neural network are comparative to complex convolutional architectures.
# 7 ACKNOWLEDGMENT
We thank IDIA for the opportunity to pursue this work, without their fellowship and none of this would have been possible. Their cloud compute platform and storage capacity proved invaluable.
# References
[1] C. Barnbaum and R. Bradley, “A new approach to interference excision in radio astronomy: Real-time adaptive cancellation,” The Astronomical Journal, vol. 116, p. 2598, Dec. 2007, doi: 10.1086/300604.
[2] N. Niamsuwan, J. Johnson, and S. W. Ellingson, “Examination of a simple pulse-blanking technique for radio frequency interference mitigation,” Radio Science - RADIO SCI, vol. 40, Oct. 2005, doi: 10.1029/2004RS003155.
[3] A. Offringa, “AOFlagger.” https://sourceforge.net/p/aoflagger/wiki/Home/, Accessed: Dec. 26, 2019. [Online].
[4] Y. Cendes and et al, “RFI flagging implications for short-duration transients,” Astronomy and Computing, vol. 23, Apr. 2018, doi: 10.1016/j.ascom.2018.04.001.
[5] A. Offringa, Relation: https://www.rug.nl/ Rights: University of Groningen“Algorithms for radio interference detection and removal,” PhD thesis, s.n., 2012.
[6] NRAO, “CASA guides.” https://casaguides.nrao.edu/index.php/Main_Page, Accessed: Dec. 26, 2019. [Online].
[7] NRAO, “Tfcrop.” https://casa.nrao.edu/Release3.4.0/docs/userman/UserMansu161.html, Accessed: Dec. 26, 2019. [Online].
[8] S. Sekhar and R. Athreya, “Two procedures to flag radio frequency interference in the uv plane,” The Astronomical Journal, vol. 156, Oct. 2017, doi: 10.3847/1538-3881/aac16e.
[9] N. Altman, “An introduction to kernel and nearest neighbor nonparametric regression,” 1992.
[10] Tin Kam Ho, “Random decision forests,” in Proceedings of 3rd international conference on document analysis and recognition, Aug. 1995, vol. 1, pp. 278–282 vol.1, doi: 10.1109/ICDAR.1995.598994.
[11] P. Kaviani and S. Dhotre, “Short survey on naive bayes algorithm,” International Journal of Advance Research in Computer Science and Management, vol. 4, Nov. 2017.
[12] O. Mosiane, N. Oozeer, A. Aniyan, and B. Bassett, “Radio frequency interference detection using machine learning.” IOP Conference Series: Materials Science and Engineering, vol. 198, p. 012012, May 2017, doi: 10.1088/1757-899X/198/1/012012.
[13] J. Akeret and et. al, “Radio frequency interference mitigation using deep convolutional neural networks,” Astronomy and Computing, vol. 18, pp. 35–39, 2017.
[14] J. Kerrigan and et al, “Optimizing sparse rfi prediction using deep learning,” Monthly Notices of the Royal Astronomical Society, Jul. 2019, doi: 10.1093/mnras/stz1865.
[15] D. Kingma and J. Ba, “Adam: A method for stochastic optimization,” ICLR, doi: arXiv:1412.6980.
[16] O. Ronneberger, P. Fischer, and T. Brox, “U-net: Convolutional networks for biomedical image segmentation,” May 2015, doi: arXiv:1505.04597.
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# Special relativity particle creation
A photon hits a proton that is at rest (In the lab system) which creates a pion, so now there are only the proton and pion. The question is what minimal energy can you give a photon such that this reaction is possible?
I tried using conservation of energy and momentum however the $$\gamma$$ in the equations made it impossible to solve. I also tried moving to the c.m. system but because that changes the frequency and therefore the energy of the photon the algebra is again to complicated.
• You probably need to show some work. Can you draw an energy-momentum diagram (analogous to a Spacetime diagram) where the components are energy and momentum (instead of time and space). Conservation laws suggest the total-4-momentum before (vectors added tip to tail starting at the origin) equals the total after (tip to tail starting at the origin), resulting in a polygon. What does your minimal condition mean for the final total 4-momenta? (Examples of diagrams are in my answer to physics.stackexchange.com/questions/594212/… ) – robphy Jun 12 at 12:02
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# Areas Inside a Rectangle
Geometry Level 4
$$ABCD$$ is a rectangle with point $$E$$ on line segment $$AB$$ and point $$F$$ on line segment $$BC$$ such that $$[ADE] = 7$$, $$[BEF] = 11$$ and $$[DCF] = 9$$. Then $$[DEF] = a\sqrt{b}$$, where $$a$$ and $$b$$ are integers and $$b$$ is not a multiple of the square of any prime. What is the value of $$a + b$$?
Details and assumptions
$$[PQRS]$$ denotes the area of figure $$PQRS$$.
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# Thread: Rings- a^2 = a then R must be commutative
1. ## Rings- a^2 = a then R must be commutative
Hi, just need a hint with this one...
Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative
Oh and also having some trouble with...
Show that $\bb{Z}_{n}$ is a field if and only if n is a prime.
2. Originally Posted by liedora
Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative
Using $(x+x)^2=x+x$ prove that
$x=-x,\quad \forall x\in R$
Using $(a+b)^2=a+b$ prove that
$ba=-ab,\quad \forall a,b \in R$
Conclude.
Fernando Revilla
3. Originally Posted by liedora
Show that $\bb{Z}_{n}$ is a field if and only if n is a prime.
Hint ;
If $n=pq$ then, $[p][q]=[0]$
Fernando Revilla
4. Originally Posted by FernandoRevilla
Hint ;
If $n=pq$ then, $[p][q]=[0]$
Fernando Revilla
...and for the other way, use the Euclidean algorithm...
5. Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!
6. Originally Posted by liedora
Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!
What have you tried?. At any rate, if you know that a finite integral domain is a field then, this fact can help you.
Fernando Revilla
7. I don't know how much of a stickler your instructor is, but (presumably - I mean I can't imagine you not) you've been working with [tex]\mathbb{Z}_n[\math] quite a bit so you get a lot free in the definition of a field. I would say the only thing you need prove is that [tex]\mathbb{Z}_n[\math] contains mult. inverses (as far as going in the forward direction is concerned - going backwards ". . is prime. . .field" shouldn't be too hard).
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# Monitoring of the thermal neutron flux in the LSM underground laboratory
3 CSNSM INSTR
CSNSM - Centre de Spectrométrie Nucléaire et de Spectrométrie de Masse
5 CSNSM PS1
CSNSM - Centre de Sciences Nucléaires et de Sciences de la Matière, CSNSM - Centre de Spectrométrie Nucléaire et de Spectrométrie de Masse
Abstract : This paper describes precise measurements of the thermal neutron flux in the LSM underground laboratory in proximity of the EDELWEISS-II dark matter search experiment together with short measurements at various other locations. Monitoring of the flux of thermal neutrons is accomplished using a mobile detection system with low background proportional counter filled with $^3$He. On average 75 neutrons per day are detected with a background level below 1 count per day (cpd). This provides a unique possibility of a day by day study of variations of the neutron field in a deep underground site. The measured average 4$\pi$ neutron flux per cm$^{2}$ in the proximity of EDELWEISS-II is $\Phi_{MB}=3.57\pm0.05^{stat}\pm0.27^{syst}\times 10^{-6}$ neutrons/sec. We report the first experimental observation that the point-to-point thermal neutron flux at LSM varies by more than a factor two.
Document type :
Preprints, Working Papers, ...
Domain :
http://hal.in2p3.fr/in2p3-00696140
Contributor : Sylvie Flores <>
Submitted on : Friday, May 11, 2012 - 8:26:09 AM
Last modification on : Tuesday, November 24, 2020 - 4:30:04 PM
### Identifiers
• HAL Id : in2p3-00696140, version 1
• ARXIV : 1001.4383
### Citation
S. Rozov, E. Armengaud, C. Augier, L. Bergé, Alain Benoit, et al.. Monitoring of the thermal neutron flux in the LSM underground laboratory. 2012. ⟨in2p3-00696140⟩
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# How are problems classified in Complexity Theory?
I'm reading Sipser's Introduction to the Theory of Computation (3rd edition). In chapter 0 (pg. 2), he says we don't know the answer to "what makes some problems computationally hard and others easy," however, he then states that "researchers have discovered an elegant scheme for classifying problems according to their computational difficulty. Using this scheme, we can demonstrate a method for giving evidence that certain problems are computationally hard, even if we are unable to prove that they are."
So my question is: HOW is it possible to classify problems according to their computational difficulty, if we don't even know what makes a problem computationally easy/hard in the first place?
Also, what/where is this "scheme" that does this classifying. (I did some googling and couldn't find anything)
That's what you get when you distill a whole bunch of theory to a wider audience.
In his book, Sipser addresses a general audience at the undergraduate level, possibly with no notion of computability theory; hence, he can only hint at concepts which are to be given a more formal treatment later on in the book. The part you cite is from chapter 0 (i.e., not really a chapter), whereas the material for complexity theory only appears at the end (i.e., part three). This is why the passage is so fuzzy. Most likely it is intended only as motivation and to give a broad overview for the topics to be covered in the book.
The "scheme" Sipser is talking about are reductions. If we know a problem $$A$$ is reducible to a problem $$B$$, then we know $$B$$ is at least as hard as $$A$$. (Incidentally, this is also why it is common practice to denote reductions with a "$$\le$$" sign.) This gives us a way of ordering problems according to their difficulty, at least for those having reductions we are aware of. As Sipser states, though, by using only reductions "we are unable to prove" whether the problems are really hard or not; reductions only give us relative, not absolute notions of hardness. This is why separation results are still rare in modern complexity theory: We have a bunch of reduction (e.g., NP-completeness) results, but only a handful of separation results (e.g., the time and space hierarchy theorems).
• I appreciate the thorough answer. Vincenzo (one of the commentors) mentioned that Sipser discusses this in Ch 5 & 7, which I'll hopefully get to eventually! – Johan von Adden Mar 29 at 10:12
HOW is it possible to classify problems according to their computational difficulty, if we don't even know what makes a problem computationally easy/hard in the first place?
I think the point that the piece is trying to make is that we know how to determine whether individual problems are easy or hard, even though we don't have an over-arching theory of why the hard ones are hard and the easy ones are easy. Just like you can classify people according to their weight, even though you don't know why they have the weight they do.
I should emphasise that in most cases, "hard" means "seem to be hard". You've probably heard of NP-complete problems. We don't know for certain that these problems have no efficient algorithm (by the standard definition of "efficient") but nobody has been able find an efficient algorithm for any of them in nearly 50 years of trying, and finding an efficient algorithm for just one of them would give efficient algorithms for all of them.
Also, what/where is this "scheme"
Complexity classes, the relationships between them, and the concept of reductions for transforming one problem into another.
The "scheme" is based on the ideas of reductions among problems and completeness of problems, which are described in Chapters 5 and 7 of Sipser's book.
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## Characterizations of Curves According to Frenet Frame in Euclidean 3-Space
#### Osman Çakır [1] , Süleyman Şenyurt [2]
##### 25 42
In this paper, we investigate the conditions of being an harmonic curve and research differential equations characterizing any differentiable curve in Euclidean 3-space. By means of the Laplacian image of the mean curvature vector field of a curve, it is stated which type of harmonic the curve is. Then we write the theorems related to the characterization of the curves and proved these theorems. When the differentiable curve, used throughout this paper, is specifically replaced to the unit speed curve then it is seen that the results coincide with the study [4]. In addition we elucidate the characterizations of helix as an example.
regular curve, connection, mean curvature, biharmonic, differential equation
• Arslan, K., Kocayigit, H., Onder, M., Characterizations of Space Curves with 1-type Darboux Instantaneous Rotation Vector, Commun. Korean Math. Soc., 31(2)(2016), 379--388.
• Chen, B. Y., Ishikawa, S., Biharmonic Surface in Pseudo-Euclidean Spaces , Mem. Fac. Sci. Kyushu Univ., A45(1991), 323--347.
• Hacısalihoğlu, H. H. Diferensiyel Geometri $3^{rd}$ ed., Ankara, 1988.
• Kocayiğit, H., Hacısalihoğlu, H. H., 1-type Curves and Biharmonic Curves in Euclidean 3-Space, Int. Elect. Journ. of Geo., 4(1)(2011), 97--101.
• Kocayiğit, H., Hacısalihoğlu, Hilmi H., Biharmonic Curves in Contact Geometry, Commun. Fac. Sci. Univ. Ank. Series A1, 2061(2)(2012), 35-43.
• Kocayiğit, H., Onder, M., Hacısalihoğlu H. H., Harmonic 1-type Curves and Weak Biharmonic Curves in Lorentzian 3-Space, Ana. Stiin. A. Uni. Al. I. Cuza. D. I. (S.N.) Mat., f(1)(2014), 109--124.
• Sabuncuoğlu, A., Diferensiyel Geometri $5^{th}$ ed., Ankara, 2014.
• Şenyurt, S., Çakır O., Diferential Equations for a Space Curve According to the Unit Darboux Vector, Turk. J. Math. Comput. Sci., 9(2018), 91--97.
Primary Language en Mathematics Articles Orcid: 0000-0002-2664-5232Author: Osman Çakır Orcid: 0000-0003-1097-5541Author: Süleyman Şenyurt (Primary Author)Institution: Ordu ÜniversitesiCountry: Turkey Publication Date: June 30, 2019
Bibtex @research article { tjmcs526778, journal = {Turkish Journal of Mathematics and Computer Science}, issn = {}, eissn = {2148-1830}, address = {Matematikçiler Derneği}, year = {2019}, volume = {11}, pages = {48 - 52}, doi = {}, title = {Characterizations of Curves According to Frenet Frame in Euclidean 3-Space}, key = {cite}, author = {Çakır, Osman and Şenyurt, Süleyman} } APA Çakır, O , Şenyurt, S . (2019). Characterizations of Curves According to Frenet Frame in Euclidean 3-Space. Turkish Journal of Mathematics and Computer Science, 11 (1), 48-52. Retrieved from http://dergipark.org.tr/tjmcs/issue/46614/526778 MLA Çakır, O , Şenyurt, S . "Characterizations of Curves According to Frenet Frame in Euclidean 3-Space". Turkish Journal of Mathematics and Computer Science 11 (2019): 48-52 Chicago Çakır, O , Şenyurt, S . "Characterizations of Curves According to Frenet Frame in Euclidean 3-Space". Turkish Journal of Mathematics and Computer Science 11 (2019): 48-52 RIS TY - JOUR T1 - Characterizations of Curves According to Frenet Frame in Euclidean 3-Space AU - Osman Çakır , Süleyman Şenyurt Y1 - 2019 PY - 2019 N1 - DO - T2 - Turkish Journal of Mathematics and Computer Science JF - Journal JO - JOR SP - 48 EP - 52 VL - 11 IS - 1 SN - -2148-1830 M3 - UR - Y2 - 2019 ER - EndNote %0 Turkish Journal of Mathematics and Computer Science Characterizations of Curves According to Frenet Frame in Euclidean 3-Space %A Osman Çakır , Süleyman Şenyurt %T Characterizations of Curves According to Frenet Frame in Euclidean 3-Space %D 2019 %J Turkish Journal of Mathematics and Computer Science %P -2148-1830 %V 11 %N 1 %R %U ISNAD Çakır, Osman , Şenyurt, Süleyman . "Characterizations of Curves According to Frenet Frame in Euclidean 3-Space". Turkish Journal of Mathematics and Computer Science 11 / 1 (June 2019): 48-52. AMA Çakır O , Şenyurt S . Characterizations of Curves According to Frenet Frame in Euclidean 3-Space. TJMCS. 2019; 11(1): 48-52. Vancouver Çakır O , Şenyurt S . Characterizations of Curves According to Frenet Frame in Euclidean 3-Space. Turkish Journal of Mathematics and Computer Science. 2019; 11(1): 52-48.
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The University of Southampton
University of Southampton Institutional Repository
# Complements versus substitutes and trends in fertility choice in dynastic models
Jones, Larry E. and Schoonbroodt, Alice (2010) Complements versus substitutes and trends in fertility choice in dynastic models International Economic Review
Record type: Article
## Abstract
Demographers have long emphasized decreased mortality and economic development' as the main contributors generating the demographic transition (DT). In economics, the Barro-Becker (BB) model of fertility choice, though simple and intuitive, has not been successful at reproducing changes in fertility in line with the demography literature---at least in its original formulation. We show that this is due to an implicit assumption that number and welfare of children are complements, a byproduct of high intertemporal elasticity of substitution (IES) typically assumed in the fertility literature.
Not only is this assumption not necessary, but qualitative and quantitative properties of the model in terms of fertility choice change dramatically when substitutability and low IES are assumed. These results do not require non-homotheticities or any other major changes to the basic BB model but emphasize productivity growth rates as opposed to income levels to interpret economic development.'
We find that with an IES of one-third, model predictions of changes in fertility amount to two-thirds of those observed in U.S. data since 1800. The increase in productivity growth accounts for 90 percent of the predicted fall in fertility before 1880; and changes in mortality account for 90 percent of the predicted fall from 1880 to 1990.
Full text not available from this repository.
Published date: August 2010
Keywords: fertility choice, substitutes, demographic changes
## Identifiers
Local EPrints ID: 150063
URI: http://eprints.soton.ac.uk/id/eprint/150063
ISSN: 0020-6598
## Catalogue record
Date deposited: 12 May 2010 09:02
## Contributors
Author: Larry E. Jones
Author: Alice Schoonbroodt
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Previous | Next --- Slide 40 of 57
Back to Lecture Thumbnails
Kalecgos
How can we computationally account for anisotropic reflections? I suppose this would just mean a BRDF which is not uniform with respect to the surface orientation (in the way that Matt described in class, e.g. rotating the surface around its normal). Does it require any special representation? Is the computational model we use for the "anisotropic-ness" of a doorknob with a rough surface different from the model we might use for the CD in the bottom right?
motoole2
@Kalecgos Our BRDF can represent both isotropic and anisotropic reflectance.
Suppose we have a BRDF f(\theta_i,\phi_i,\theta_o,\phi_o). A BRDF is isotropic if rotating the object about its normal produces the same reflectance value. The reflectance function for an object rotated by angle x about its normal is given by f(\theta_i,\phi_i+x,\theta_o,\phi_o+x). So if this reflectance function f produces the same value for all x, then we are dealing with an isotropic material. Otherwise, the material is anisotropic and the light reflected by the material depends on its orientation around the normal axis.
The reflectance functions for all the anisotropic materials shown in this slide can be expressed with some BRDF f(\theta_i,\phi_i,\theta_o,\phi_o). However, choosing the right functions f to photorealistically model each of these materials is a bit tough. In practice, there are two typical approaches: for a particular material, (1) choose an analytical model for the BRDF that depends only on a few parameters (e.g., the classic example is the Blinn-Phong model, or the Ashikhmin-Shirley model for anisotropic materials), or (2) use a data-driven approach that interpolates between sampled BRDF values captured from real materials (basically sampling from a 4D texture).
Note that isotropic BRDFs can be defined as a three-dimensional function. This is because f(\theta_i,\phi_i,\theta_o,\phi_o) = f(\theta_i,\phi_i-\phi_o,\theta_o,0). For this reason, it is typically easier to capture and model isotropic BRDFs than it is to model anisotropic BRDFs.
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# Latex: Add Image Above the Title
I have an Image, and I would like to add the image above the Title. The Downbelow Code gives the entire page to the Image. I want the image and the Title, Autother to be displayed on the same page
\includegraphics[width=0.1\textwidth]{images/Company_Groups.jpg}~\\[1cm]
\title{xxxxxxx
\\In xxxxx }
\author{N. N. Arif
\date{June 18, 2015}
\maketitle
• Welcome to TeX.SX! Please help us help you and add a minimal working example (MWE) that illustrates your problem. Reproducing the problem and finding out what the issue is will be much easier when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user31729 Jun 21 '15 at 18:40
I don't recommend that \\[...] stuff there, but it's possible to include the graphic file as part of the title itself.
(Remove the demo option afterwards)
\documentclass{article}
\usepackage[demo]{graphicx}
\begin{document}
\title{%
\includegraphics[width=0.1\textwidth]{images/Company_Groups.jpg}~
\\[1cm]
xxxxxxx
\\In xxxxx
}
\author{N. N. Arif
\date{June 18, 2015}
\maketitle
\end{document}
Some update, with the famous CTAN TeX Lion (copyright by Duane Bibby)
\documentclass{article}
\usepackage{graphicx}
\begin{document}
\title{%
%\includegraphics[width=0.1\textwidth]{images/Company_Groups.jpg}~
\includegraphics[width=0.4\textwidth]{ctanlion}~\\[1cm]
xxxxxxx\\
In xxxxx
}
\author{N. N. Arif
• @Arif: I meant the \\[...] spacing stuff, not the graphics – user31729 Jun 21 '15 at 18:55
• @Arif: It works for me, without demo too, when using a specific graphics. – user31729 Jun 21 '15 at 19:07
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# Potential Function Chasing - Rotation
This topic is 4463 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi. I am attempting to use Lenard-Jones potential function to do chasing/obstacle avoidance AI for a space game. The potential functions give me a nice steering vector that makes my AI ships follow the player or avoid obstacles quite well. The problem I have is that I can't seem to make my rotation work with this steering vector. The ships fly sideways and don't ever seem to point the way they are moving. I've tried several methods to try and make my AI ships turn in the direction of the steering vector but I've had no luck so far. My potential function chase function:
//This potential function is based on the potential functions chapter in the
//AI for game developers O'Reilly book.
void SpaceObject::ChaseObject(SpaceObject *object)
{
D3DXVECTOR3 r = pos - (*object).pos;
D3DXVECTOR3 u = r;
D3DXVec3Normalize(&u,&u);
double U,n,m,d;
n = 2;
m = 3;
d = D3DXVec3Length(&r)/70;
U = -object->pAttraction/pow(d,n) + object->pRepulsion/pow(d,m);
u = U *u;
if(D3DXVec3Length(&u) > acceleration)
{
D3DXVec3Normalize(&u,&u);
u = u * acceleration;
}
//Turn towards the desired direction
TurnToPos(object->pos + u);
//This probably shouldn't be like this, since then the ship can
//fly sideways and other not good directions
velocity += u;
//Make sure the velocity isn't greater than the max speed
if(D3DXVec3Length(&velocity) > maxSpeed)
{
//Reduce the speed to the max
D3DXVec3Normalize(&velocity,&velocity);
velocity *= maxSpeed;
}
}
And here is the code for my turning function which doesn't work:
void SpaceObject::TurnToPos(const D3DXVECTOR3 &dPos)
{
D3DXVECTOR3 u,r;
int dir = 0; //direction to turn
int tolerance = 0.001; //turning tolerance
//Get the global difference coordinates
r = dPos - pos;
//Convert the global coordinates to local coordinates
u.x = r.x*cos(-rotation)+r.y*sin(-rotation);
u.y = -r.x*sin(-rotation)+r.y*cos(-rotation);
u.z = 0; //never a z-component
//Normalize the local coordinates
D3DXVec3Normalize(&u,&u);
//If x is negative move to the left
//If x is positive move to the right
if(u.x < -tolerance)
dir = -1;
else if(u.x > tolerance)
dir = 1;
//Turn the ship
Rotate(dir);
}
Any help is greatly appreciated! Thank you Gundark
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what does Rotate(dir) do?
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The Rotate(dir) function rotates the ship either left or right, depending on the dir parameter. -1 rotates the ship left, 1 rotates the ship right.
Here is the source code for the function:
void SpaceObject::Rotate(int dir){ if(dir != -1 && dir != 1) //only accept 1 and -1 return; //Change the ships rotation double tmp = turningspeed * dir; rotation += tmp; if(rotation > 2*D3DX_PI) rotation -= 2*D3DX_PI; if(rotation < 0) rotation += 2*D3DX_PI;}
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Hey, I fixed my own problem! The problem was that what I considered to be 0 degrees was actually 90 degrees off of what the formulas I was using were expecting to be 0 degrees.
So the solution was to add 90 degrees to the rotation when calculating local coordinates.
Thanks for all the people that looked at this posts and my code, and uh... sorry to bother you!
Gundark
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int tolerance = 0.001; //turning tolerance
Shouldn't this be a double or float?
##### Share on other sites
Not to thread rip but I just wanted to pass along some ideas of how to refactor code for faster compile and run. I changed tolerance to const in the example but if you intend to adjust it during gameplay then it would be a non-const member var.
void SpaceObject::TurnToPos(const D3DXVECTOR3 &dPos){ const int tolerance = 0.001; //turning tolerance | const is initialized at compile time. //Get the global difference coordinates D3DXVECTOR3 r( dPos - pos ); // instanciate vs assign. D3DXVECTOR3 u; //Convert the global coordinates to local coordinates u.x = r.x*cos(-rotation)+r.y*sin(-rotation); u.y = -r.x*sin(-rotation)+r.y*cos(-rotation); u.z = 0; //never a z-component //Normalize the local coordinates D3DXVec3Normalize(&u,&u); //If x is negative move to the left //If x is positive move to the right int dir = (u.x < -tolerance) ? -1 : 1; // declare and assign in one line //Turn the ship Rotate(dir);}
Note variable declaration in the middle of a function is the same to the compiler and early declaration. It only affects scope as the stack space is reserved with the function and held until the function exits. These techniques also reduce optimization cycles by reducing the amount of instance tracking that the optmizer has to look at.
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# Math Help - trigonometry equationsss =(
1. ## trigonometry equationsss =(
hii guyz, i would REALLY like some help for these trig equation questions
thank you sooo much if you can helpp me !
if secx - tanx = 3/5, show that sinx = 8/17
if tanx = t, express sin2x and cos2x in terms of t. Find the values for t for which
(k+1)sin2x + (k-1)cos2x = k+1
thankyouuuu againn !!
2. Originally Posted by iiharthero
hii guyz, i would REALLY like some help for these trig equation questions
thank you sooo much if you can helpp me !
if secx - tanx = 3/5, show that sinx = 8/17
Hi
$\sec x - \tan x = \frac35 \implies \frac{1-\sin x}{\cos x} = \frac 35$ $\implies 5(1-\sin x) = 3 \cos x$
Square both sides and substitute cos²x with 1-sin²x to find a quadratic in sin x, one of the solutions being "impossible" the other one is the good one
3. Originally Posted by iiharthero
if tanx = t, express sin2x and cos2x in terms of t. Find the values for t for which
(k+1)sin2x + (k-1)cos2x = k+1
$\sin 2x=\frac{2\tan x}{1+\tan^2x}, \ \cos 2x=\frac{1-\tan^2x}{1+\tan^x}$
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# Coin Flips and Hypothesis Tests
Here's a problem I thought of that I don't know how to approach:
You have a fair coin that you keep on flipping. After every flip, you perform a hypothesis test based on all coin flips thus far, with significance level $\alpha$, where your null hypothesis is that the coin is fair and your alternative hypothesis is that the coin is not fair. In terms of $\alpha$, what is the expected number of flips before the first time that you reject the null hypothesis?
Edit based on comment below: For what values of $\alpha$ is the answer to the question above finite? For those values for which it is infinite, what is the probability that the null hypothesis will ever be rejected, in terms of $\alpha$?
Edit 2: My post was edited to say "You believe that you have a fair coin." The coin is in fact fair, and you know that. You do the hypothesis tests anyway. Otherwise the problem is unapproachable because you don't know the probability that any particular toss will come up a certain way.
• If there is a positive probability that you never reject the null hypothesis, then the expectation is infinite. [I don't know whether or not there is such a positive probability.] – paw88789 May 9 '15 at 12:41
EDIT: This answer was unclear for OP at first, so I tried to make it clearer through a new approach. Apparently it arose another legitimate doubt, so I tried now to put both answers together and clarify them even more. (Still I might be wrong, but I'll try to express myself better)
What you look for, is the expected number of tosses before we do a Type I error (rejecting $H_0$ when it was true). The probability of that is precisely $\alpha$ (that's another way to define it).
So $P(Type\ I\ error)=\alpha$.
Let $X_n$ be the event of rejecting $n^{th}$ test.
Now, $E[X_1]=\alpha$ stands for the expected number of games (a game is starting to test in the way we do a new coin) where $H_0$ was rejected on the first throw. $E[X_1+X_2]=E[X_1]+E[X_2]$ is the expected number of games where $H_0$ is rejected either on the first or the second throw. Note that with most $\alpha$ this will be lower than $1$, so the expectation for a single game is not to reject $H_0$ yet.
When do we expect to have rejected $H_0$? Precisely when the number of expected games in which we reject $H_0$ is $1$. Therefore, we look for $n$ such as $$E[X_1+X_2+...+X_n]=1\\ E[X_1+X_2+...+X_n]=E[nX_1]=nE[X_1]=n\alpha=1\\ n=\frac{1}{\alpha}$$
The other answer goes like this: Let the variable $T$ count the number of tests before rejecting one. We look for $E[T]$.
Also, using previous notation, $P(X_n)=\alpha(1-\alpha)^{n-1}$ (I'm aware this implies independence between the events $X_n$ and $X_{n-1}$ but since I'm looking for the expected value, for the linearity of the Expected Value, it shouldn't be a problem, though I'm aware I'm not being polite with notation).
$$E[T] = \sum_{n=1}^{\infty}nP(X_n) = \sum_{n=1}^{\infty}n\alpha(1-\alpha)^{n-1}= \alpha\sum_{n=1}^{\infty}n(1-\alpha)^{n-1}=\\ \alpha\sum_{n=0}^{\infty}(n+1)(1-\alpha)^{n}= \alpha(\sum_{n=0}^{\infty}n(1-\alpha)^{n}+\sum_{n=0}^{\infty}(1-\alpha)^{n}) = \alpha(\frac{1-\alpha}{\alpha^2}+\frac{1}{\alpha}) \\ E[T]=\frac{1}{\alpha}$$
• Can you explain why we are looking for $n$ such that $E[X_1 + X_2 + \dots + X_n] = 1$? – Eric Neyman May 23 '15 at 18:28
• I look for when the expected number of rejected $H_0$ is 1. – Masclins May 23 '15 at 18:32
• Why is this the same as the expected number of flips before the null hypothesis is rejected for the first time? – Eric Neyman May 24 '15 at 20:28
• It's what you asked. The expected number of flips until a hypothesis test rejects $H_0$, when doing one after each flip. I don't get your doubt, please be more specific about what you don't understand of the reasoning. – Masclins May 24 '15 at 20:36
• You showed that $\frac{1}{\alpha}$ is the value of $n$ such that the expected number of rejections for the first $n$ tests is equal to $1$. I am asking for the expected number of tests before the first rejection. I may be missing something, but I don't see why those two should be the same. – Eric Neyman May 25 '15 at 21:13
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The internet document class takes a different approach to the other programmes we’ve used so far. It is a LaTeX document class used with
\documentclass[<options>]{internet}
<options> can specify an output format like xhtml or epub or markdown amongst others. The document is then compiled to pdf, and text is extracted from the pdf. This text is in the requested format.
The github page shows that internet isn’t being actively developed.
## Installation and usage
The readme on the github page is a bit low on detail, so here’s how installation worked for me after I first cloned the repo. The readme says first that all the internet<type>.<format>.code.tex files and the h<type>.def files and the internet.cls file have to be “linked into your local texmf tree.” One way to do this is to find out where this is (you can use kpsepath -n latex tex to find all the directories tex can see), for me (on Debian) it was at ~/texmf/tex/latex/ which didn’t exist.
There’s then some instructions for generating .tfm and .vf files which “need to be put somewhere that TeX can find them” - for me, /usr/share/texmf/fonts/vf/ and /usr/share/texmf/fonts/tfm/ worked (but I had to run texhash as root after putting them there).
I got a lot of errors about usage of \tl_to_lowercase:n. Using grep -ri tl_to_lowercase I found all files containing this and replaced every occurrence in every file with \tex_lowercase:D which at least silenced the errors.
At this point some of the test documents (markdown_text.tex, maruku_test.tex, basicmaths_text.tex) can be compiled with
./latex2txt.sh testfile.tex
as long as you are in the folder of latex2txt.sh. But the epub and xhtml tests fail with errors about \xhtm_verb:n and hyperref version mismatches. That’s above my pay grade, so I gave up.
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NPartitionPairs() calculates the number of terminal arrangements matching a specified configuration of two splits.
NPartitionPairs(configuration)
## Arguments
configuration
Integer vector of length four specifying the number of terminals that occur in both (1) splits A1 and A2; (2) splits A1 and B2; (3) splits B1 and A2; (4) splits B1 and B2.
## Value
The number of ways to distribute sum(configuration) taxa according to the specified pattern.
## Details
Consider splits that divide eight terminals, labelled A to H.
Bipartition 1: ABCD:EFGH A1 = ABCD B1 = EFGH Bipartition 2: ABE:CDFGH A2 = ABE B2 = CDFGH
This can be represented by an association matrix:
A2 B2 A1 AB C B1 E FGH
The cells in this matrix contain 2, 1, 1 and 3 terminals respectively; this four-element vector (c(2, 1, 1, 3)) is the configuration implied by this pair of bipartition splits.
## Author
Martin R. Smith (martin.smith@durham.ac.uk)
## Examples
NPartitionPairs(c(2, 1, 1, 3))
#> [1] 12
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# gcd_array¶
tenpy.tools.math.gcd_array(a)[source]
Return the greatest common divisor of all of entries in a
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# The Coordinates of a Point on X-axis Which Lies on the Perpendicular Bisector of the Line Segment Joining the Points (7, 6) and (−3, 4) Are - Mathematics
MCQ
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (−3, 4) are
• (0, 2)
• (3, 0)
• (0, 3)
• (2, 0)
#### Solution
TO FIND: The coordinates of a point on x axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4).
Let P(xy) be any point on the perpendicular bisector of AB. Then,
PA=PB
sqrt((x -7)^2 + (y -6)^2) = sqrt((x-(-3))^2+(y-4)^2)
(x-7)^2+ (y - 6)^2 = (x +3)62 + (y-4)^2
x^2 - 14x + 49 +y^2 - 12y +36 = x^2 +6x +9 +y^2 -8y + 16
-14x - 6x - 12y - 8y + 49 +36 -9 - 16 = 0
- 20x + 20y + 60 = 0
x - y - 3 = 0
x - y = 3
On x-axis y is 0, so substituting y=0 we get x= 3
Hence the coordinates of point is (3,0) .
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Q 33 | Page 65
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Question
# In an A.P if the $12^{th}$ term is -13 and the sum of its first four terms is 24. Find the sum of its first ten terms.
Hint: Assume an A.P series with ‘a’ as its first term and the difference between the consecutive terms be ‘d’. Our $12^{th}$ term of the A.P is -13. So, we can write it in mathematical form, i.e, $a+11d=-13$ . We have a summation of the first four terms equal to 24. So, we can also write it in mathematical form, i.e, $4a+6d=24$. Now, we have two equations and two unknown variables. Using these two equations, find the values of ‘a’ and ‘d’. Now, it can be solved further and we can find the summation of the first ten terms.
Complete step-by-step solution -
Let us assume an A.P with ‘a’ as its first term and the difference between the consecutive term be ‘d’.
We have the value of the $12^{th}$ term. So, first of all, we have to find the $12^{th}$ term in terms of the variable ‘a’ and ‘d’.
Putting n=12 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{12}}=a+(12-1)d$ ……………….(1)
According to the question, we have the $12^{th}$ term of an A.P as -13.
Now, we can write equation (1), as
$a+11d=-13$ ……………(2)
We know that the summation of n terms of an A.P is ${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ .
According to the question, we have the summation of the first four terms which is 24.
${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$
$24=\dfrac{4}{2}\left( {{1}^{st\,term}}+\operatorname{Fourth}\,term \right)$ ……………(3)
Now, we have to find the $4^{th}$ term of the A.P.
Putting n=4 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{4}}=a+(4-1)d=a+3d$ …………….(4)
Now, putting equation (4) in equation (3) and we have ‘a’ as the first term of the A.P.
$24=2(a+a+3d)$
$\Rightarrow 12=2a+3d$ ……………………(5)
Using equation (2) and equation (5), we can find the value of ‘a’ and ‘d’.
According to equation (2), we have
$a+11d=-13$
$\Rightarrow a=-11d-13$ …………..(6)
Putting the value of ‘a’ from equation (6) in equation (5), we get
\begin{align} & 12=2a+3d \\ & \Rightarrow 12=2(-11d-13)+3d \\ \end{align}
\begin{align} & \Rightarrow 12=-22d-26+3d \\ & \Rightarrow 12+26=-22d+3d \\ & \Rightarrow 38=-19d \\ & \Rightarrow -2=d \\ \end{align}
We have got the value of ‘d’ which is -2.
Now, putting the value of ‘d’ in equation (6), we get
\begin{align} & a=-11d-13 \\ & \Rightarrow a=-11(-2)-13 \\ & \Rightarrow a=22-13 \\ & \Rightarrow a=9 \\ \end{align}
The 1st term of the A.P is 9.
Now, we have to find the summation of the first ten terms of the A.P.
Putting n=10 in the equation, ${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ , we get
${{S}_{10}}=\dfrac{10}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ ………….(7)
Our last term is the $10^{th}$ term of the A.P. So, we have to find the $10^{th}$ term of the A.P.
Putting n=10 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{10}}=a+(10-1)d$
$\Rightarrow {{T}_{10}}=a+9d$ …………………(8)
Now, putting the values of ‘a’ and ‘d’ in equation (8), we get
${{T}_{10}}=9+9(-2)=-9$ …………..(9)
Now, we have got the $10^{th}$ term of the A.P.
Now, putting equation (9) in equation (7), we get
\begin{align} & {{S}_{10}}=\dfrac{10}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right) \\ & \Rightarrow {{S}_{10}}=5(9-9)=0 \\ \end{align}
So, the summation of the first ten-term of the A.P is 0.
Note: We can also solve the summation using the formula, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ .
We have, n=10, a=9 and d=-2.
Putting the values of a,n and d in the above formula, we get
${{S}_{10}}=\dfrac{10}{2}\left[ 2(9)+(10-1)(-2) \right]=5\left[ 18-18 \right]=0$ .
Hence, the summation of the first ten terms of the A.P is 0.
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# Conformal map in 3D
I'm looking for a method to transform a three dimensional geometry. This geometry has a rotational symmetry, so the $r$- and $z$-coordinates are all the same over $\phi$. I want to transform this cylindrical geometry in a linear geometry (something like $z\rightarrow x$, $r\rightarrow y$ and $\phi\rightarrow z$).
As you can see from this explanation I'm no mathematician. Can anybody help me find a conformal map that solves this problem or point me to some literature on this topic?
-
Do you understand what conformality means? It sounds from your comments below like you're just interested in a general map, not specifically a conformal one. If you are looking for a conformal map, then robjohn's answer is correct in explaining why there can't be one. – Steven Stadnicki Aug 28 '12 at 17:08
Unless I misunderstand your question (please correct me if so), I do not believe such a map exists.
Liouville's Theorem says that only a very limited class of mappings in $\mathbb{R}^n$ for $n>2$ are conformal:
1. Homothetic transformations (translations and homogeneous scalings)
2. isometries (rotations and reflections)
3. inversions
The first two transformations will take cylinders to cylinders, and inversions map only spheres and planes to planes, so they cannot map the surface of the cylinder to a plane, if that is what you were looking for.
-
Thank you for your fast answer. I do not want to map the surface of a cylinder to a plane, I want to map the volume of the zylinder to the volume of maybe a rectangle, like cutting the cylinder for example at phi=0 and roll out the cylinder. The mapped volume can be something like a stringing together of a infinite number of mapped cylinders in the new z-direction whereas the r-z-geometry can change there shaoe in the new x-y-plane. – WaldFlo Aug 28 '12 at 16:47
@WaldFlo The point that robjohn made is that there are very few conformal maps in 3D (only Mobius transformations), and in particular there is no conformal map from the interior of a cylinder onto the interior of a rectangular box. – user31373 Sep 10 '12 at 1:49
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# TeXlive 2016 error with literat (Literaturnaya) font
It is so strange that no questions on the whole website are dedicated to the Literaturnaya font (despite its frequent use in Russian documents 10 years ago or so), and only two questions reference it!
On a system with a freshly installed and updated TeXLive 2016, I copied and extracted the contents of the file literat-0.2.zip from this page into the correct corresponding directories of texmf-dist (doc, fonts and tex respectively). After the three “magic” actions (Update filename DB, Rebuild all Formats and Update Font Map DB), the following MWE:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T2A]{fontenc}
\usepackage[russian]{babel}
\usepackage{literat}
\begin{document}
О, суд людей неправый, что пьянствовать грешно!
\end{document}
produces the following error via pdfLaTeX (copying only the relevant part of the log where it tries to load literat.sty):
(c:/texlive/2016/texmf-dist/tex/latex/literat/literat.sty
Package: literat 1999/05/20 v0.2 ParaGraph Literaturnaya as default upright
) (./literat.aux)
\openout1 = literat.aux'.
LaTeX Font Info: Checking defaults for OML/cmm/m/it on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for T1/cmr/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for OT1/cmr/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for OMS/cmsy/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for OMX/cmex/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for U/cmr/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Checking defaults for T2A/cmr/m/n on input line 8.
LaTeX Font Info: ... okay on input line 8.
LaTeX Font Info: Try loading font information for T2A+tli on input line 8.
(c:/texlive/2016/texmf-dist/tex/latex/literat/t2atli.fd
File: t2atli.fd 1999/05/20 Fontinst v1.902 font definitions for T2A/tli.
) [1
{c:/texlive/2016/texmf-var/fonts/map/pdftex/updmap/pdftex.map}] (./literat.aux
LaTeX Font Info: Try loading font information for OT1+tli on input line 3.
(c:/texlive/2016/texmf-dist/tex/latex/literat/ot1tli.fd
File: ot1tli.fd 1999/05/20 Fontinst v1.902 font definitions for OT1/tli.
)) )
Here is how much of TeX's memory you used:
1729 strings out of 494319
17975 string characters out of 6169309
70210 words of memory out of 5000000
5095 multiletter control sequences out of 15000+600000
7184 words of font info for 20 fonts, out of 8000000 for 9000
452 hyphenation exceptions out of 8191
23i,4n,23p,200b,131s stack positions out of 5000i,500n,10000p,200000b,80000s
==> Fatal error occurred, no output PDF file produced!
In addition to the log output, these are the messages from the console:
pdflatex.exe -synctex=1 -interaction=nonstopmode "literat".tex
name = rtlir6i, rootname = rtlir6i, pointsize = mktexmf: empty or non-existent rootfile!
Cannot find font rtlir6i in map file(s).
kpathsea: Running mktexmf rtlir6i.mf The command name is C:\texlive\2016\bin\win32\mktexmf
Cannot find rtlir6i.mf . I try ps2pk --> gsftopk --> ttf2pk --> hbf2gf. ps2pk cannot be used.
I try gsftopk. gsftopk.exe rtlir6i 600 gsftopk cannot be used.
Next I try ttf2pk. ttf2pk.exe -q rtlir6i 600 ttf2pk failed.
Finally I try hbf2gf. hbf2gf.exe -q -p rtlir6i 600 All trials failed.
kpathsea: Running mktexpk --mfmode / --bdpi 600 --mag 1+0/600 --dpi 600 rtlir6i
The command name is C:\texlive\2016\bin\win32\mktexpk kpathsea:
Appending font creation commands to missfont.log.
In an attempt to reproduce this example, I reëncoded the .tex file as cp1251 and tried \usepackage[cp1251]{inputenc} — to no avail. Then, I tried OT1 with the same amout of success. Only commenting out the line loading the font helped to produce the “default” document.
Processing it through latex.exe and opening the resulting DVI in the viewer produces a bunch of messages like:
rtlir6i.600(3000)
Searched C:\texlive\2016\texmf-var\fonts\pk\\...
Searched C:\texlive\2016\texmf-dist\fonts\tfm\\...
What could have changed in 1.5 years since January 2015 and @egreg's answer? How does one properly configure the literat font and make it work (if only it were possible not only to make it work, but also to make microtype work with it, as it normally re-scales alien/unknown fonts with default parameters)?
• I did the installation and it seems to work flawlessly. You seem not to have updated the map file; however, adding \pdfmapfile{+literat.map} at the start of the document avoids the need. – egreg Jun 14 '16 at 22:28
• Running updmap-sys for the second time (via the TeXLive manager menu option) did not help... neither did texhash. But the inclusion of the line you suggested really helped! Thank you! Could you please post your reply as an answer so that I could close it and mark as solved? – Andreï Kostyrka Jun 14 '16 at 22:48
• You shouldn't have installed the files under 2016/texmf-dist: the preferred place is under texmf-local. And you should add the line literat.map to /usr/local/texlive/texmf-local/web2c/updmap.cfg (creating the file if not existent) before running updmap-sys. – egreg Jun 14 '16 at 22:52
The files should not go under .../texlive/2016/texmf-dist/, but under .../texlive/texmf-local. The main tree is only for things in the official distribution: changing it might leave an unstable TeX system.
Installing the files in the “distribution” tree has the consequence that everything has to be redone at every release of TeX Live.
This said, the procedure is:
1. copy the relevant files in the appropriate directories in the local tree, that is, under .../texlive/texmf-local/ (creating the necessary directory structure)
2. edit the file .../texmf-local/web2c/updmap.cfg by appending the line literat.map (creating the file if necessary)
3. run mktexlsr
4. run updmap-sys
The “local” updmap.cfg file avoids the need to call updmap-sys --enable-map and moreover will continue to provide the map file for future releases of TeX Live.
A temporary fix is to load the map file in the document, by typing
\pdfmapfile{+literat.map}
as the first line.
• Well, even the procedure you mentioned in great detail did not help (I cleaned out the folders pertaining to the literat package from texmf-dist, copied the structure to texmf-local, created updmap.cfg with the line literat.map, ran mktexlsr and updmap-sys and the document still won’t compile. Going to blame Windows (why not?). However, the temporary fix works flawlessly. “There is nothing more permanent than a temporary solution.” Thank you very much! I shall try the procedure under Linux to see if it helps. – Andreï Kostyrka Jun 14 '16 at 23:14
• @AndreïKostyrka Does this question apply? – cfr Jun 15 '16 at 3:43
• Can you show the output of the updmap-sys run. Do you have old files in TEXMFVAR (.texlive2016 or so)? Did you ever run updmap without the -sys part? – norbert Jun 15 '16 at 9:25
• @norbert According to c:/texlive/2016/texmf-var/fonts/map/pdftex/updmap/pdftex.map` in the log file, the global map file is used. – egreg Jun 15 '16 at 9:38
• I still need to see which updmap.cfg files were used during the run. – norbert Jun 15 '16 at 9:44
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# Prove $f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$ is holomorphic in the open unit disk.
Define $f:D[0,1] \rightarrow \mathbb C$ through $$f(z)=\int_{[0,1]}\frac{1}{1-wz}dw$$ The integration path is from 0 to 1 along the real line.
Prove that $f$ is holomorphic in the open unit disk $D[0,1]$.
I was trying to use the Cauchy's Integral Formula \begin{aligned} f'(w) & =\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-w)^2}dz\\\ & =\frac{1}{2\pi i}\int_{\gamma}\frac{1}{(z-w)^2}\left[\int^1_0\frac{1}{1-tz}dt\right]dz\\ & =\frac{1}{2\pi i}\int^1_0\left[\int_{\gamma} \frac{\frac{1}{1-tz}}{(z-w)^2}dz\right]dt\\ & =\frac{1}{2\pi i}\int^1_0 \left[\frac{\partial}{\partial z}\frac{1}{1-tz}\left.\right|_{z=w}\right]dt\\ & =\frac{1}{2\pi i}\int^1_0 \frac{t}{(1-tw)^2}dt\\ & =\cdots \end{aligned} But the Cauchy's Integral formula need $f$ to be holomorphic at the first place...Any help with this? Many Thanks!
Use Morera's theorem. First prove that $f$ is continuous.Then, if $\gamma$ is any closed $C^1$ curve in the unit disk, we have $$\int_\gamma f(z)\,dz=\int_\gamma\Bigl(\int_{[0,1]}\frac{dw}{1-w\,z}\Bigr)\,dz=\int_{[0,1]}\Bigl(\int_\gamma\frac{dz}{1-w\,z}\Bigr)\,dw=0,$$ since $1/(1-w\,z)$ is holomorphic on the unit disk as a function of $z$.
Note: you have to justify the change of order in the integration.
No need to justify switching integrals. Just obtain a closed form of $f(z)$, which allows to prove continuity and that $\int_{\gamma} f =0$:
$$f(z) := \int_{[0,1]} \frac{dw}{1-wz} = -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0},$$
1. $f(z)$ is continuous on $D[0,1]$ because $$\lim_{z \to 0}-\frac1z\operatorname{Ln}(1-z) = 1 = f(0)$$
2. $$\int_{\gamma} f(z) dz = \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} + 1_{z=0} dz$$
$$= \int_{\gamma} -\frac1z\operatorname{Ln}(1-z)1_{z\ne0} dz + \int_{\gamma} 1_{z=0} dz = 0$$
Now all conditions of Morera's Thm 5.6 are satisfied without (explicit) reference to antiderivatives or Fubini's theorem, measure spaces, uniform continuity, etc
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# Angular Velocity/Spinning Disk
1. Jun 16, 2012
### SalsaOnMyTaco
1. The problem statement, all variables and given/known data
Ive been working on this two last questions and I cant seem to get the right set up.
A disk with a diameter of 0.04 m is spinning with a constant velocity about an axle perpendicular to the disk and running through its center.
-How many revolutions per second would it have to rotate in order that the acceleration of the outer edge of the disk be 14 g's (i.e., 14 times the gravitational acceleration g)?
13.19 rev/s
For the frequency determined in part (a), what is the speed of a point half way between the axis of rotation and the edge of the disk?
.828 m/s
At this same frequency, what is the period of rotation of this "halfway point"?
How long does it take a point on the edge of the disk to travel 1 km?
2. Relevant equations
Period (time)
T= 2∏/ω
Centripetal Acceleration
α=rω2
3. The attempt at a solution
First, I try to find the angular velocity at the halfwaypoint
then i punch in
Once i typed in the answer, it says is wrong
2. Jun 16, 2012
### SalsaOnMyTaco
woah okay no, should i find the frequency and then divide 1/f ?
3. Jun 16, 2012
### tiny-tim
Hi SalsaOnMyTaco!
Why would the angular velocity be any different?
4. Jun 16, 2012
### TSny
Why did you use .01 m for r? Note that it's 14 g's at the outer edge.
Last edited: Jun 16, 2012
5. Jun 16, 2012
### SalsaOnMyTaco
isnt velocity different on a different part of the radius
The total diameter of the disk is .04, radius is .02, half way of the radius is .01
6. Jun 17, 2012
### TSny
But the problem states that the acceleration is 14 g's for a point on the outer edge of the disk (r = 0.02 m), not at a halfway point (r = 0.01 m).
7. Jun 17, 2012
### tiny-tim
ahh!
you're confusing angular velocity with tangential velocity (ie component of velocity along the "angular" unit vector $\boldsymbol{\hat{\theta}}$)
angular velocity is angle per second, dθ/dt, it's not a velocity at all
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# All Questions
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Last edited by Tuktilar
Sunday, May 3, 2020 | History
3 edition of Cosmological constant and the evolution of the universe found in the catalog.
# Cosmological constant and the evolution of the universe
## by RESCEU International Symposium (1st 1995 Tokyo, Japan)
• 210 Want to read
• 24 Currently reading
Published by Universal Academy Press in Tokyo .
Written in English
Subjects:
• Cosmology -- Congresses.,
• Astrophysics -- Congresses.,
• Dark matter (Astronomy) -- Congresses.,
• Cosmic background radiation -- Congresses.
• Edition Notes
Includes bibliographical references and index.
Classifications The Physical Object Statement edited by K. Sato, T. Suginohara, and N. Sugiyama. Genre Congresses. Series RESCEU international symposium series,, no. 1 Contributions Sato, K., Suginohara, Tatsushi., Sugiyama, N. LC Classifications QB980 .R47 1995 Pagination ix, 336 p. : Number of Pages 336 Open Library OL3622094M ISBN 10 4946443320 LC Control Number 2002400780
The Cosmological Constant & Other Fudge Factors in the Physics of the Universe, please sign up. Be the first to ask a question about Einstein's Greatest Blunder? The Cosmological Constant & Other Fudge Factors in the Physics of the Universe/5(18). Degree of fine tuning. Recent Studies have confirmed the fine tuning of the cosmological constant (also known as "dark energy"). This cosmological constant is a force that increases with the increasing size of the universe. First hypothesized by Albert Einstein, the cosmological constant was rejected by him, because of lack of real world data. However, recent supernova 1A data demonstrated the.
This paper is the second of two papers devoted to the study of the evolution of the cosmological horizons (particle and event horizons). Specifically, in this paper we consider the extremely general case of an accel-erated universe with countably infinitely many constant state Cited by: 2. Universe reproduction via black holes. According to CNS, black holes may be mechanisms of universe reproduction within the multiverse, an extended cosmological environment in which universes grow, die, and than a ‘dead’ singularity at the center of black holes, a point where relativity theory breaks down and spacetime and matter-energy become unmodeled, what occurs in.
The cosmological constant (, or also indicated by λ) is the value of the energy density of the vacuum of space. It was originally introduced by Albert Einstein in as an addition to his theory of general relativity to achieve a static cosmological constant is the simplest possible form of dark energy, since it is constant in both space and time. The ne-tuning of the universe for intelligent life has received a great deal of attention in recent years, both in the philosophical and scienti c literature. The claim is that in the space of possible physical laws, parameters and initial conditions, the set that permits the evolution of intelligent life is very small.
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### Cosmological constant and the evolution of the universe by RESCEU International Symposium (1st 1995 Tokyo, Japan) Download PDF EPUB FB2
What is a Cosmological Constant. Einstein first proposed the cosmological constant (not to be confused with the Hubble Constant) usually symbolized by the greek letter "lambda" (Λ), as a mathematical fix to the theory of general its simplest form, general relativity predicted that the universe must either expand or contract.
The Accelerating Universe: Infinite Expansion, the Cosmological Constant, and the Beauty of the Cosmos - Kindle edition by Livio, Mario.
Download it once and read it on your Kindle device, PC, phones or tablets. Use features like bookmarks, note taking and highlighting while reading The Accelerating Universe: Infinite Expansion, the Cosmological Constant, and the Beauty of the Cosmos/5(15).
Cosmological definition, the branch of philosophy dealing with the origin and general structure of the universe, with its parts, elements, and laws, and especially with such of its characteristics as space, time, causality, and freedom.
See more. According to the author, the tentative discovery of the accelerating expansion of the universe poses a frightening challenge to the beauty of the final theory by raising difficult questions about the non-zero value of the cosmological constant (or the energy of the vacuum).Cited by: Further, it is shown that in this cosmology the cosmological constant or dark energy is a property of space-time.
This can be interpreted in a creationist cosmology as the power of the Lord giving a boost to the expansion of the fabric of space as He stretched it out. He is the unseen force in the universe. Quintessence began as Einstein's cosmological constant, but if it is not, in fact, constant it may explain why we appeared just when it had the same value as the density of ordinary matter.
[/caption] The cosmological constant, symbol Λ (Greek capital lambda), was ‘invented’ by Einstein, not long after he published his theory of general relativity (GR).
It appears on the left. The cosmological constant is the simplest possible form of dark energy since it is constant in both space and time, and this leads to the current standard model. This could only happen if those galaxies were receding away from us.
The Universe was expanding, Hubble found. Einstein immediately retracted his cosmological constant, stating that it was his “biggest blunder” because the Universe was clearly not static (Saw 20, BartusiakKrauss 55).Reviews: 4. Dr Tony Padilla on some recent work he has been doing.
See the papers (not the faint-hearted) here: AND Cosmological arguments for the existence of God A philosophical perspective on findings of science. by Albrecht Moritz (, revised ) Summary The apparent extreme fine-tuning of the laws of nature, necessary to allow for physical evolution of the universe and evolution of life, and affirmed with broad consensus by cosmologists regardless of their worldview, is reviewed.
Add tags for "Cosmological constant and the evolution of the universe: proceedings of the 1st RESCEU International Symposium on "the Cosmological constant " held on.
Specifically, this paper is devoted to the study of the evolution of these cosmological horizons in an accelerated universe with two state equations, cosmological constant and dust. An introduction to modern astronomy is given from the point of view of distance measurement, from the scale of the solar system up to the largest cosmological scales.
A historical and philosophical background to the measurement of distance and time in the universe is given, and the measurement of distances within the Galaxy is examined. The structure and evolution of stars and galaxies are Cited by: The conditions in our universe really do seem to be uniquely suitable for life forms like ourselves, and perhaps for any form of organic chemistry.
(Cited in The Privileged Planet, pp. ) Another remarkable book about fine-tuning is by Geraint Lewis and Luke A. Barnes: A Fortunate Universe: Life in a Finely-tuned Cosmos (Cambridge, ).
The cosmological constant is the parameter $\Lambda$ in the Einstein equation: $$G_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi T_{\mu\nu}$$ and it is by definition a constant, so it cannot change. I think it is best regarded as a geometrical property of the universe (though other views exist) which is why it's normally put on the left hand side of the equals sign.
"The Extravagant Universe" by Bob Kirshner. A "useful and polite" and always entertaining look at the supernova evidence for the cosmological constant and the accelerating expansion of the Universe. "The Origin and Evolution of the Universe" edited by Zuckerman and Malkan.
A UCLA Center for the Study of the Evolution and Origin of Life (CSEOL. From a cosmological perspective, [ ] In my opinion, the theory of evolution by natural selection is one of the most elegant and by far one of the most encompassing theories humanity has ever.
THE BEGINNING, THE EVOLUTION AND THE END OF THE UNIVERSE and the problem of Cosmological constant are interpreted and the percentages. Cosmological Theories Through History "Cosmos" is just another word for universe, and "cosmology" is the study of the origin, evolution and fate of the of the best minds in history - both philosophers and scientists - have applied themselves to an understanding of just what the universe is and where it came from, suggesting in the process a bewildering variety of theories and.
The accelerating universe: infinite expansion, the cosmological constant, and the beauty of the cosmos Mario Livio In one of the most surprising and important findings in cosmology in the century, astronomers recently discovered that the unierse may be expanding at an ever-increasing rate.'A Fortunate Universe: Life in a Finely Tuned Cosmos by Geraint Lewis and Luke Barnes, is a nice up to date book for the general (educated) public on modern physics and cosmology.
If covers modern cosmology and some of the Big Questions of our times, in particular the issue of anthropomorphism how 'fine tuned' our Universe is.'Cited by: We have shown that the Hubble constant H 0 embodies the information about the evolutionary nature of the cosmological constant Λ, gravitational constant G, and the speed of light c.
We have derived expressions for the time evolution of G / c 2 (≡ K) and dark energy density ε Λ related to Λ by explicitly incorporating the nonadiabatic nature of the universe in the Author: Rajendra P.
Gupta.
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Nonlinear ODEs with stable limit cycles
It’s not obvious when a nonlinear differential equation will have a periodic solution, or asymptotically approach a periodic solution. But there are theorems that give sufficient conditions. This post is about one such theorem.
Consider the equation
x” + f(x) x‘ + g(x) = 0
where f and g are analytic and define F(x) to be the integral of f(t) from 0 to x. The following six hypotheses are sufficient to guarantee that the equation above has a stable limit cycle [1].
1. g is an odd function,
2. g is positive for positive x,
3. f is an even function,
4. f(o) < 0,
5. F(x) → ∞ as x → ∞,
6. F has a unique positive zero.
These hypotheses are satisfied, for example, by Van der Pol’s equation.
Let’s make up coefficient functions f and g that satisfy the conditions above and look for the limit cycle.
We can let g(x) = x³ and f(x) = x² – 1. And here’s what we get:
The graph shows phase portraits for two different initial conditions, given in the legend of the plot. Note that limit cycle appears solid blue because the trajectories of both the dashed line and the dotted line run around the periodic attractor.
Related posts
[1] You can find this theorem in Topics in Ordinary Differential Equations by William Lakin and David Sanchez.
Linear ODEs with periodic solutions
Suppose we have a differential equation of the form
x” + b(t) x‘ + c(t) x = f(t)
where the functions b and c are periodic with the same period T.
When does the equation above have a solution with period T?
Nonlinear phase portrait
I was reading through Michael Trott’s Mathematica Guidebook for Programming and ran across the following plot.
I find the image aesthetically interesting. I also find it interesting that the image is the phase portrait of a differential equation whose solution doesn’t look that interesting. That is, the plot of (x(t), x ‘(t)) is much more visually interesting than the plot of x(t).
The differential equation whose phase portrait is plotted above is
with initial position 1 and initial velocity 0. It’s a familiar damped, driven harmonic oscillator, except the equation is nonlinear because the derivative term is cubed.
Here’s a plot of the solution as a function of time.
The solution basically looks like the solution of the linear case, except the solutions are more jagged near the local maxima and minima. In fact, the plot looks so jagged that I wondered whether this was an artifact of needing more plotting points. But adding more points didn’t make much difference. Interestingly, although this plot looks jagged, the phase portrait is smooth.
I produced the phase portrait by copying Trott’s code and making a couple small modifications.
sol = NDSolve[{(*differential equation*)
x''[t] + 1/20 x'[t]^3 + 1/5 x[t] ==
1/3 Cos[E t],(*initial conditions*)x[0] == 1, x'[0] == 0},
x, {t, 0, 360}, MaxSteps -> Infinity]
ParametricPlot[Evaluate[{x[t], x'[t]} /. sol], {t, 0, 360},
Frame -> True, Axes -> False, PlotPoints -> 3600]
Apparently the syntax of NDSolve has changed slightly since Trott’s book was published in 2004. The argument x in the code above was written x[t] in Trott’s original code and this did not work in the current version of Mathematica. I also simplified the call to ParametricPlot slightly, though the original code would work.
Predator-Prey period
The Lotka-Volterra equations are a system of nonlinear differential equations for modeling a predator-prey ecosystem. After a suitable change of units the equations can be written in the form
where ab = 1. Here x(t) is the population of prey at time t and y(t) is the population of predators. For example, maybe x represents rabbits and y represents foxes, or x represents Eloi and y represents Morlocks.
It is well known that the Lotka-Volterra equations have periodic solutions. It is not as well known that you can compute the period of a solution without having to first solve the system of equations.
This post will show how to compute the period of the system. First we’ll find the period by solving the equations for a few different initial conditions, then we’ll show how to directly compute the period from the system parameters.
Phase plot
Here is a plot of (x(t), y(t)) showing that the solutions are periodic.
And here’s the Python code that made the plot above.
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import solve_ivp
# Lotka-Volterra equations
def lv(t, z, a, b):
x, y = z
return [a*x*(y-1), -b*y*(x-1)]
begin, end = 0, 20
t = np.linspace(begin, end, 300)
styles = ["-", ":", "--"]
prey_init = [2, 4, 8]
a = 1.5
b = 1/a
for style, p0 in zip(styles, prey_init):
sol = solve_ivp(lv, [begin, end], [p0, 1], t_eval=t, args=(a, b))
plt.plot(sol.y[0], sol.y[1], style)
plt.xlabel("prey")
plt.ylabel("preditor")
plt.legend([f"$p_0$ = {p0}" for p0 in prey_init])
Note that the derivative of x is zero when y = 1. Since our initial condition sets y(0) = 1, we’re specifying the maximum value of x. (If our initial values of x were less than 1, we’d be specifying the minimum of x.)
Time plot
The components x and y have the same period, and that period depends on a and b, and on the initial conditions. The plot below shows how the period increases with x(0), which as we noted above is the maximum value of x.
And here’s the code that made the plot.
fig, ax = plt.subplots(3, 1)
for i in range(3):
sol = solve_ivp(lv, [begin, end], [prey_init[i], 1], t_eval=t, args=(a, b))
ax[i].plot(t, sol.y[0], "-")
ax[i].plot(t, sol.y[1], "--")
ax[i].set_xlabel("$t$")
ax[i].set_ylabel("population")
ax[i].set_title(f"x(0) = {prey_init[i]}")
plt.tight_layout()
Finding the period
In [1] the author develops a way to compute the period of the system as a function of its parameters without solving the differential equations.
First, calculate an invariant h of the system:
Since this is an invariant we can evaluate it anywhere, so we evaluate it at the initial conditions.
Then the period only depends on a and h. (Recall we said we can scale the equations so that ab = 1, so a and b are not independent parameters.)
If h is not too large, we can compute the approximate period using an asymptotic series.
where σ = (a + b)/2 = (a² + 1)/2a.
We use this to find the periods for the example above.
def find_h(a, x0, y0):
b = 1/a
return b*(x0 - np.log(x0) - 1) + a*(y0 - np.log(y0) - 1)
def P(h, a):
sigma = 0.5*(a + 1/a)
s = 1 + sigma*h/6 + sigma**2*h**2/144
return 2*np.pi*s
print([P(find_h(1.5, p0, 1), 1.5) for p0 in [2, 4, 8]])
This predicts periods of roughly 6.5, 7.5, and 10.5, which is what we see in the plot above.
When h is larger, the period can be calculated by numerically evaluating an integral given in [1].
Related posts
[1] Jörg Waldvogel. The Period in the Volterra-Lotka Predator-Prey Model. SIAM Journal on Numerical Analysis, Dec., 1983, Vol. 20, No. 6, pp. 1264-1272
Oscillatory differential equations
The solution to a differential equation is called oscillatory if its set of zeros is unbounded. This does not necessarily mean that the solution is periodic, but that it crosses the horizontal axis infinitely often.
Fowler [1] studied the following differential equation demonstrates both oscillatory and nonoscillatory behavior.
x” + tσ |x|γ sgn x = 0
He proved that
1. If σ + 2 ≥ 0 then all solutions oscillate.
2. If σ + (γ + 3)/2 < 0 then no solutions oscillate.
3. If neither (1) nor (2) holds then some solutions oscillate and some do not.
The edge case is σ = -2 and γ = 1. This satisfies condition (1) above. A slight decrease in σ will push the equation into condition (2). And a simultaneous decrease in σ and increase in γ can put it in condition (3).
It turns out that the edge case can be solved in closed form. The equation is linear because γ = 1 and solutions are given by
ct sin(φ + √3 log(t) / 2).
The solutions oscillate because the argument to sine is an unbounded function of t, and so it runs across multiples of π infinitely often. The distance between zero crossings increases exponentially with time because the logarithms of the crossings are evenly spaced.
This post took a different turn after I started writing it. My original intention was to solve the differential equation numerically and say “See: when you change the parameters slightly you get what the theorem says.” But that didn’t work out.
First I tried numerically computing the solution above with σ = -2 and γ = 1, but I didn’t see oscillations. I tried making σ a little larger, but still didn’t see oscillations. When I increased σ more I could see the oscillations.
I was able to go back and see the oscillations with σ = -2 and γ = 1 by tweaking the parameters in my ODE solver. It’s best practice to tweak your parameters even if everything looks right, just to make sure your solution is robust to algorithm changes. Maybe I would have done that, but since I already knew the analytic solution, I knew to tweak the parameters until I saw the right behavior.
Without some theory as a guide, it would be difficult to determine from numerical solutions alone whether a solution oscillates. If you see oscillations, then they’re probably real (unless your numerical method is unstable), but if you don’t see oscillations, how do you know you won’t see them if you look further out? How much further out should you look? In a practical application, the context of your problem will tell you how far out to look.
My intention was to recommend the equation above as an illustration for an introductory DE class because it demonstrates the important fact that small changes to an equation can qualitatively change the behavior of the solutions. This is especially true for nonlinear DEs, which the above equation is when γ ≠ 1. It could still be used for that purpose, but it would make a better demonstration than homework exercise. The instructor could tune the DE solver to make sure the solutions are qualitatively correct.
I recommend the equation as an illustration, but for a course covering numerical solutions to DEs. It illustrates a different point than I first intended, namely the potentially finicky behavior of software for solving differential equations.
There are a couple morals to this story. One is that numerical methods have not eliminated the need for theory. The ideal is to combine analytic and numerical methods. Analytic methods point out qualitative behavior that might be hard to discover numerically. And analytic methods are not likely to produce closed-form solutions for nonlinear DEs.
Another moral is that it’s best to twiddle the parameters to your numerical method to make sure the solution doesn’t qualitatively change. If you’ve adequately computed a solution, computing it again with smaller integration steps shouldn’t make much difference. But if you do see a substantial difference, your first solution probably wasn’t as good as you thought.
Related posts
[1] R. H. Fowler. Further studies of Emden’s and similar differential equations. Quarterly Journal of Mathematics. 2 (1931), pp. 259–288.
Laplacian in various coordinate systems
The recent post on the wave equation on a disk showed that the Laplace operator has a different form in polar coordinates than it does in Cartesian coordinates. In general, the Laplacian is not simply the sum of the second derivatives with respect to each variable.
Mathematica has a function, unsurprisingly called Laplacian, that will compute the Laplacian of a given function in a given coordinate system. If you give it a specific function, it will compute the Laplacian of that function. But you can also give it a general function to find a general formula.
For example,
Simplify[Laplacian[f[r, θ], {r, θ}, "Polar"]]
returns
This is not immediately recognizable as the Laplacian from this post
because Mathematica is using multi-index notation, which is a little cumbersome for simple cases, but much easier to use than classical notation when things get more complicated. The superscript (0,2), for example, means do not differentiate with respect to the first variable and differentiate twice with respect to the second variable. In other words, take the second derivative with respect to θ.
Here’s a more complicated example with oblate spheroidal coordinates. Such coordinates come in handy when you need to account for the fact that our planet is not exactly spherical but is more like an oblate spheroid.
Simplify[Laplacian[f[ξ, η, φ], {ξ, η, φ}, "OblateSpheroidal"]]
the result isn’t pretty.
I tried using TeXForm and editing it into something readable, but after spending too much time on this I gave up and took a screenshot. But as ugly as the output is, it would be uglier (and error prone) to do by hand.
Mathematica supports the following 12 coordinate systems in addition to Cartesian coordinates:
• Cylindrical
• Bipolar cylindrical
• Elliptic cylindrical
• Parabolic cylindrical
• Circular parabolic
• Confocal paraboloidal
• Spherical
• Bispherical
• Oblate spheroidal
• Prolate spheroidal
• Conical
• Toroidal
These are all orthogonal, meaning that surfaces where one variable is held constant meet at right angles. Most curvilinear coordinate systems used in practice are orthogonal because this simplifies a lot of things.
Laplace’s equation is separable in Stäckel coordinate systems. These are all these coordinate systems except for toroidal coordinates. And in fact Stäckel coordinates are the only coordinate systems in which Laplace’s equation is separable.
It’s often the case that Laplace’s equation is separable in orthogonal coordinate systems, but not always. I don’t have a good explanation for why toroidal coordinates are an exception.
If you’d like a reference for this sort of thing, Wolfram Neutsch’s tome Coordinates is encyclopedic. However, it’s expensive new and hard to find used.
Vibrating circular membranes
This post will tie together many things I’ve blogged about before. The previous post justified separation of variables. This post will illustrate separation of variables.
Also, this post will show why you might care about Bessel functions and their zeros. I’ve written about Bessel functions before, and said that Bessel functions are to polar coordinates what sines and cosines are to rectangular coordinates. This post will make this analogy more concrete.
Separation of variables
The separation of variables technique is typically presented in three contexts in introductory courses on differential equations:
1. One-dimensional heat equation
2. One-dimensional wave equation
3. Two-dimensional (rectangular) Laplace equation
My initial motivation for writing this post was to illustrate separation of variables outside the most common examples. Separation of variables requires PDEs to have a special form, but not as special as the examples above might imply. A secondary motivation was to show Bessel functions in action.
Suppose you have a thin membrane, like a drum head, and you want to model its vibrations. By “thin” I mean that the membrane is sufficiently thin that we can adequately model it as a two-dimensional surface bobbing up and down in three dimensions. It’s not so thick that we need to model the material in more detail.
Let u(x, y, t) be the height of the membrane at location (x, y) and time t. The wave equation is a PDE modeling the motion of the membrane by
where Δ is the Laplacian operator. In rectangular coordinates the Laplacian is given by
We’re interested in a circular membrane, and so things will be much easier if we work in polar coordinates. In polar coordinates the Laplacian is given by
We will assume that our boundary conditions and initial conditions are radially symmetric, and so our solution will be radially symmetric, i.e. derivatives with respect to θ are zero. And in our case the wave equation simplifies to
Boundary and initial conditions
Let a be the radius of our membrane. We will assume our membrane is clamped down on its boundary, like a drum head, which means we have the boundary condition
for all t ≥ 0. We assume the initial displacement and initial velocity of the membrane are given by
for all r between 0 and a.
Separating variables
Now we get down to separation of variables. Because we’re assuming radial symmetry, we’re down to a function of two variables: r for the distance from the center and t for time. We assume
and stick it into the wave equation. A little calculation shows that
The left side is a function of t alone, and the right side is a function of r alone. The only way this can be true for all t and all r is for both sides to be constant. Call this constant -λ². Why? Because looking ahead a little we find that this will make things easier shortly.
Separation of variables allowed us to reduce our PDE to the following pair of ODEs.
The solutions to the equation for R are linear combinations of the Bessel functions J0r) and Y0r) [1].
And the solutions to the equation for T are linear combinations of the trig functions cos(cλt) and sin(cλt).
The boundary condition u(a, t) = 0 implies the boundary condition R(a) = 0. This implies that λa must be a zero of our Bessel function, and that all the Y0 terms drop out. This means that our solution is
where
Here αn are the zeros of of the Bessel function J0.
The coefficients An and Bn are determined by the initial conditions. Specifically, you can show that
The function J1 in the expression for the coefficients is another Bessel function.
The functions J0 and J1 are so important in applications that even the otherwise minimalist Unix calculator bc includes these functions. (As much as I appreciate Bessel functions, this still seems strange to me.) And you can find functions for zeros of Bessel functions in many libraries, such as scipy.special.jn_zeros in Python.
Related posts
[1] This is why introductory courses are unlikely to include an example in polar coordinates. Separation of variables itself is no harder in polar coordinates than in rectangular coordinates, and it shows the versatility of the method to apply it in a different setting.
But the resulting ODEs have Bessel functions for solutions, and it’s understandable that an introductory course might not want to go down this rabbit trail, especially since PDEs are usually introduced at the end of a differential equation class when professors are rushed and students are tired.
Justifying separation of variables
The separation of variables technique for solving partial differential equations looks like a magic trick the first time you see it. The lecturer, or author if you’re more self-taught, makes an audacious assumption, like pulling a rabbit out of a hat, and it works.
For example, you might first see the heat equation
ut = c² uxx.
The professor asks you to assume the solution has the form
u(x, t) = X(x) T(t).
i.e. the solution can be separated into the product of a function of x alone and a function of t alone.
Following that you might see Laplace’s equation on a rectangle
uxx + uyy = 0
with the analogous assumption that
u(x, y) = X(x) Y(y),
i.e. the product of a function of x alone and a function of y alone.
There are several possible responses to this assumption.
1. Whatever you say, doc.
2. How can you assume that?
3. How do you know you’re not missing any possibilities?
4. What made someone think to try this?
As with many things, separation of variables causes the most consternation for the moderately sophisticated students. The least sophisticated students are untroubled, and the most sophisticated student can supply their own justification (at least after the fact).
One response to question (2) is “Bear with me. I’ll show that this works.”
Another response would be “OK, how about assuming the solution is a sum of such functions. That’s a much larger space to look in. And besides, we are going to take sums of such solutions in a few minutes.” One could argue from functional analysis or approximation theory that the sums of separable functions are dense in reasonable space of functions [1].
This is a solid explanation, but it’s kind of anachronistic: most students see separation of variables long before they see functional analysis or approximation theory. But it would be a satisfying response for someone who is seeing all this for the second time. Maybe they were exposed to separation of variables as an undergraduate and now they’re taking a graduate course in PDEs. In an undergraduate class a professor could do a little foreshadowing, giving the students a taste of approximation theory.
Existence of solutions is easier to prove than uniqueness in this case because you can concretely construct a solution. This goes back to the “it works” justification. This argument deserves more respect than a sophomoric student might give it. Mathematics research is not nearly as deductive and mathematics education. You often have to make inspired guesses and then show that they work.
Addressing question (3) requires saying something about uniqueness. A professor could simply assert that there are uniqueness theorems that allow you to go from “I’ve found something that works” to “and so it must be the only thing that works.” Or one could sketch a uniqueness theorem. For example, you might apply a maximum principle to show that the difference between any two solutions is zero.
Question (4) is in some sense the most interesting question. It’s not a mathematical question per se but a question about how people do mathematics. I don’t know what was going through the mind of the first person to try separation of variables, or even who this person was. But a plausible line of thinking is that ordinary differential equations are easier than partial differential equations. How might you reduce a PDE to an ODE? Well, if the solution could be factored into functions of one variable, …
The next post will illustrate using separation of variables by solving the wave equation on a disk.
Related posts
[1] Also, there’s the mind-blowing Kolmogorov-Arnol’d theorem. This theorem says any continuous function of several variables can be written as a sum of continuous separable functions. It doesn’t say you can make the functions in your sum smooth, but it suggests that sums of separable functions are more expressive than you might have imagined.
Fourier, Gauss, and Heisenberg
Several weeks ago I wrote about the Fourier uncertainty principle which gives a lower bound on the product of the variance of a function f and the variance of its Fourier transform. This post expands on the earlier post by quoting some results from a recent paper [1].
Gaussian density
The earlier post said that the inequality in the Fourier uncertainty principle is exact when f is proportional to a Gaussian probability density. G. H. Hardy proved this result in 1933 in the form of the following theorem.
Let f be a square-integrable function on the real line and assume f and its Fourier transform satisfy the following bounds
for some constant C. Then if ab > 1/4, then f = 0. And if ab = 1/4, f(x) = c exp(-ax²) for some constant c.
Let’s translate this into probability terms by setting
Now Hardy’s theorem says that if f is bounded by a multiple of a Gaussian density with variance σ² and its Fourier transform is bounded by a multiple of a Gaussian density with variance τ², then the product of the two variances is no greater than 1. And if the product of the variances equals 1, then f is a multiple of a Gaussian density with variance σ².
Heisenberg uncertainty
Theorem 3 in [1] says that if u(tx) is a solution to the free Schrödinger’s equation
then u at different points in time satisfies a theorem similar to Hardy’s theorem. In fact, the authors show that this theorem is equivalent to Hardy’s theorem.
Specifically, if u is a sufficiently smooth solution and
then αβ > (4T)-2 implies u(t, x) = 0, and αβ = (4T)-2 implies
Related posts
[1] Aingeru Fernández-Bertolin and Eugenia Malinnikova. Dynamical versions of Hardy’s uncertainty principle: A survey. Bulletin of the American Mathematical Society. DOI: https://doi.org/10.1090/bull/1729
A stiffening spring
Imagine a spring with stiffness k1 attached to a ceiling and a mass m1 handing from the spring.
There’s a second spring attached to the first mass with stiffness k2 and a mass m2 handing from that.
The motion of the system is described by the pair of differential equations
If the second spring were infinitely stiff, the two masses would be joined with a rigid rod, and so the system would act like a single mass m1 + m2 hanging from the first spring. This motion would be described by the single differential equation
So it’s not surprising that as k2 gets stiffer, the solution to the two-mass system converges to the solution to the system with a single combined mass. This is proven in [1].
However, what is missing from [1] is any visualization of how the solution to the two-mass system converges to that of the combined-mass system.
The plot below shows solutions for k2 equal to 10, 100, and 1000, and finally the system to the combined-mass system, labeled k2 = ∞. I used k1 = 1, m1 = 3, and m2 = 5.
The coupled-mass system has a high-frequency component due to the oscillation of the second mass relative to the first one.
The authors in [1] show that the amplitude of the high-frequency component decays as k2 goes to infinity, though this is not apparent from the plots above. This is due to a limitation of the numerical method used to produce the plots.
Analytical solution
The numerical solution above raises two questions. First, how fast should the amplitude of high frequency component decay. Second, why did the numerical method apparently get the frequency of this component correct but the amplitude wrong?
The second question is more difficult and will have to wait for another post. The first question, however, we can settle fairly quickly.
The authors in [1] make the simplifying assumption that the two masses are equal to 1. They then define show that
where the d‘s are constants that depend on initial conditions but not on the spring stiffnesses. and
α4, the frequency of the low frequency component, approaches a finite limit as k2 → ∞,
α2, the frequency of the high frequency component, is approximately √(2k2) for large k2.
The amplitude of the high frequency component should be inversely proportional to its frequency.
More differential equation posts
[1] K. E. Clark and S. Hill. The Effects of a Stiffening Spring. The College Mathematics Journal , Nov., 1999, Vol. 30, No. 5 (Nov., 1999), pp. 379-382
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# Classical limit of a quantum system
If we have a one dimensional system where the potential
$$V~=~\begin{cases}\infty & |x|\geq d, \\ a\delta(x) &|x|<d, \end{cases}$$
where $a,d >0$ are positive constants, what then is the corresponding classical case -- the approximate classical case when the quantum number is large/energy is high?
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What is $V$ when $x \in (-d,0) \cup (0,d)$? – Siyuan Ren Apr 27 '12 at 9:09
Did you mean "$\infty$ when $|x| > d$"? Also did you mean "$a$ when $x = 0$" i.e. $a\delta(x)$. Finally is $a$ of the order of classical energies or much less? If the latter, the system just looks like a square well with no barrier at classical energies. – John Rennie Apr 27 '12 at 9:41
Dear @Sys, it's a virtue and necessity, not a bug, that the delta-function is infinite at $x=0$. If it were finite at a single point (i.e. interval of length zero), like in your example, it would have no impact on the particle because zero times finite is zero. So your potential as you wrote it is physically identical to $V=\infty$ for $|x|<d$ and $0$ otherwise which is just a well with the standing wave energy eigenstates. The finite modification of $V$ at one point, by $a$, plays no role at all. A potential with $a\delta(x)$ in it would be another problem. – Luboš Motl Apr 27 '12 at 10:44
@LubošMotl: Thanks, actually the delta function version instead of V=a at x=0 is the right one. What is the classical limit of that? – Sys Apr 27 '12 at 11:15
@JohnRennie: I think your comment suggestion was right, that there is a delta function at x=0. – Sys Apr 27 '12 at 11:17
Here we derive the bound state spectrum from scratch. Not surprisingly, the conclusion is that the Dirac delta potential doesn't matter in the semi-classical continuum limit, in accordance with Spot's answer.
The time-independent Schrödinger equation reads for positive $E>0$,
$$-\frac{\hbar^2}{2m}\psi^{\prime\prime}(x) ~=~ (E-V(x))\psi(x), \qquad V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0,$$
with the convention that $0\cdot \infty=0$. Define
$$v(x) ~:=~ \frac{2mV(x)}{\hbar^2}, \qquad e~:=~\frac{2mE}{\hbar^2}~>~0 \qquad k~:=~\sqrt{e}~>~0\qquad v_0 ~:=~ \frac{2mV_0}{\hbar^2}.$$
Then
$$\psi^{\prime\prime}(x) ~=~ (v(x)-e)\psi(x).$$
We know that the wave function $\psi$ is continuous with boundary conditions
$$\psi(x)~=0 \qquad {\rm for}\qquad |x|\geq d.$$
Also the derivative $\psi^{\prime}$ is continuous for $0<|x|<d$, and possibly has a kink at $x=0$,
$${\lim}_{\epsilon\to 0^+}[\psi^{\prime}(x)]^{x=\epsilon}_{x=-\epsilon} ~=~v_0\psi(x=0).$$
We get $$\psi_{\pm}(x)~=~A_{\pm}\sin(k(x\mp d))\qquad {\rm for } \qquad 0 \leq \pm x \leq d.$$
1. $\underline{\text{Case} ~\psi(x=0)=0}$. Then $$n~:=~\frac{kd}{\pi}~\in~ \mathbb{N}.$$ We get an odd wave function $$\psi_n(x)~\propto~\sin(kx).$$ In particularly, the odd wave functions do not feel the presence of the Dirac delta potential.
2. $\underline{\text{Case} ~\psi(x=0)\neq 0}$. Then continuity at $x=0$ implies that the wave function is even $A_{+}+A_{-}=0$. Phrased equivalently, $$\psi(x)~=~A\sin(k(|x|-d)).$$ The kink condition at $x=0$ becomes $$v_0A\sin(-kd)~=~2kA \cos(kd),$$ or equivalently, $$v_0\tan(kd)~=~-2k.$$ In the semiclassical continuum limit $$k \gg \frac{1}{d}, \qquad k \gg v_0,$$ this becomes $$\frac{kd}{\pi}+\frac{1}{2}~\in ~\mathbb{Z},$$ i.e., in the semiclassical continuum limit the even wave functions do not feel the presence of the Dirac delta potential as well.
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Firstly, it's easy to start off with just the Dirac delta potential and see what that does. Wiki has a nice solution for the Delta fuction potential, and I am lifting off parts of it here.
Consider a potential $V(x) = a\delta (x)$ and consider a scattering like configuration, where a plane wave $e^{ikx}$ is incident from the left.
$$\psi(x)=\begin{cases}e^{ikx}+re^{-ikx} & x<0 \\ te^{ikx} & x> 0\end{cases}$$
By matching the boundary conditions, like on the wiki page, you get $$t = 1+r\\ (1-\alpha)t = 1-r$$
where $$\alpha = \frac{ 2ma}{ik\hbar^2}$$ characterizes the effect of the delta potential. Solving for $r$ and $t$, $$t = \frac{1}{1-\alpha/2}\\ r=-\frac{\alpha/2}{1-\alpha/2}$$
Now, it is easy to see that for high incident $k$, the only effect of the dirac delta potential is to write a phase discontinuity on the wavefuction. This is because, as $k$ increases, the transmission $|t|^2=1/(1+|\alpha|^2/4)$ approaches 1, but the transmitted wavefunction gets an extra phase given by $$\text{Arg}(t) = -\tan^{-1}(|\alpha|/2)$$
Getting back to the problem at hand, for a particle in a box (without the delta function), the allowed $k$ vectors are given by forcing the wavefunction to be zero at the walls at $x=-d$ and $x=d$, which gives us the condition
$$k_n=\frac{\pi n}{2d}$$
If now, we add a delta potential, then for high values of $n$ (or $k$), all the delta function will do is introduce a phase discontinuity at the origin, and consequently what you should expect is that the boundary condition is matched not for $k_n$, but something slightly off $k_n+\delta k_n$, where $\delta k_n$ is a small correction due to the delta function potential. For high values of $n$, this correction would drop, as the phase discontinuity decreases, and for classical like states (very large $n$) you expect to recover 1D box states, as mentioned by John Rennie.
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Thank you, Spot! – Sys Apr 27 '12 at 17:59
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zbMATH — the first resource for mathematics
An update on the four-color theorem. (English) Zbl 0908.05040
K. Appel and W. Haken [Bull. Am. Math. Soc. 82, 711-712 (1976; Zbl 0331.05106)] announced a proof of the four-color theorem in 1976, providing an expanded version of that proof in 1989 [Every planar map is four colorable, Contemporary Mathematics 98, American Mathematical Society (1989; Zbl 0681.05027)]. In 1997, N. Robertson, D. Sanders, P. Seymour and R. Thomas published a greatly simplified proof; see [J. Comb. Theory, Ser. B 70, No. 1, 2-44 (1997; Zbl 0883.05056)]. In the present paper, the fourth of the latter authors summarizes the history of the theorem, gives several equivalent formulations (in a surprising variety of contexts, including vector cross products in $$\mathbb{R}^3$$, Lie algebras, and divisibility by 7), and discusses aspects of the new proof-including progress on some generalizations. To understand two recent results in this connection by N. Robertson, P. Seymour and R. Thomas [J. Comb. Theory, Ser. B 70, No. 1, 166-183 (1997; Zbl 0883.05055) and Combinatorica 13, No. 3, 279-361 (1993; Zbl 0830.05028)], define a graph $$G$$ to be apex if $$G\setminus v$$ is planar for some $$v$$ in $$V(G)$$ and doublecross if $$G$$ can be drawn in the plane with two crossings, both in the same region. Theorem 1. Let $$G$$ be a counterexample of minimum order to the conjecture that every cubic graph with no cut-edge and no edge 3-coloring has a Petersen minor; then $$G$$ is apex or doublecross. (The conjecture implies the four-color theorem.) Theorem 2. Let $$G$$ be a counterexample of minimum order to Hadwiger’s conjecture that if a graph has no $$K_6$$ minor, then it has a 5-coloring (another generalization of the four-color theorem); then $$G$$ is apex.
Reviewer: A.T.White (Oxford)
MSC:
05C15 Coloring of graphs and hypergraphs
Full Text:
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# How can nitrogen achieve a +V oxidation state?
You mean how can nitrogen achieve a $+ V$ oxidation state.
Consider the nitrate ion, $N {O}_{3}^{-}$. Here, nitrogen has a formal $+ V$ oxidation state (i.e. 3xx-2+N_("ON")=-1). Nitrogen also bears this oxidation state in the NEUTRAL GAS ${N}_{2} {O}_{5}$. Our ideas of oxidation state are formalisms. They may be useful in some circumstances for pedagogy or calculation, but they have no real physical basis.
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# Taylor Expansion (multivariate)
• Apr 6th 2011, 07:34 PM
Krahl
Taylor Expansion (multivariate)
Hi, I need some help understanding the taylor expansion of a function $g_0(\phi,c,T)$. I'm particularly confused with the 3rd term on the RHS. Why is there a partial derivative there? When I expand it, I get the same as whats in the image but no multiplication with partial derivatives.
Thanks
Attachment 21388
• Apr 10th 2011, 04:13 PM
ojones
The notation you're using here is terrible. Where did you get this expression? Use $f(x,y,z)$ for the function of 3 variables. The third term of the series comes from the Hessian matrix. Do you what that is?
• Oct 15th 2012, 08:30 AM
Krahl
Re: Taylor Expansion (multivariate)
Thanks for replying ojones. Turns out the partial derivative multiplying the 3rd term and other terms were a typo. That's odd since this typo was present in multiple papers.
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# Sparse FGLM algorithms
1 PolSys - Polynomial Systems
LIP6 - Laboratoire d'Informatique de Paris 6, Inria de Paris
Abstract : Given a zero-dimensional ideal $I \subset \kx$ of degree $D$, the transformation of the ordering of its \grobner basis from DRL to LEX is a key step in polynomial system solving and turns out to be the bottleneck of the whole solving process. Thus it is of crucial importance to design efficient algorithms to perform the change of ordering. The main contributions of this paper are several efficient methods for the change of ordering which take advantage of the sparsity of multiplication matrices in the classical {\sf FGLM} algorithm. Combing all these methods, we propose a deterministic top-level algorithm that automatically detects which method to use depending on the input. As a by-product, we have a fast implementation that is able to handle ideals of degree over $40000$. Such an implementation outperforms the {\sf Magma} and {\sf Singular} ones, as shown by our experiments. First for the shape position case, two methods are designed based on the Wiedemann algorithm: the first is probabilistic and its complexity to complete the change of ordering is $O(D(N_1+n\log (D)))$, where $N_1$ is the number of nonzero entries of a multiplication matrix; the other is deterministic and computes the LEX \grobner basis of $\sqrt{I}$ via Chinese Remainder Theorem. Then for the general case, the designed method is characterized by the Berlekamp--Massey--Sakata algorithm from Coding Theory to handle the multi-dimensional linearly recurring relations. Complexity analyses of all proposed methods are also provided. Furthermore, for generic polynomial systems, we present an explicit formula for the estimation of the sparsity of one main multiplication matrix, and prove its construction is free. With the asymptotic analysis of such sparsity, we are able to show for generic systems the complexity above becomes $O(\sqrt{6/n \pi} D^{2+\frac{n-1}{n}})$.
Keywords :
Type de document :
Article dans une revue
Journal of Symbolic Computation, Elsevier, 2017, 80 (3), pp.538 - 569. 〈10.1016/j.jsc.2016.07.025〉
Domaine :
Littérature citée [38 références]
https://hal.inria.fr/hal-00807540
Contributeur : Jean-Charles Faugère <>
Soumis le : mercredi 3 avril 2013 - 17:37:37
Dernière modification le : mercredi 9 mai 2018 - 15:46:13
Document(s) archivé(s) le : jeudi 4 juillet 2013 - 04:12:11
### Fichier
SparseFGLM.pdf
Fichiers produits par l'(les) auteur(s)
### Citation
Jean-Charles Faugère, Chenqi Mou. Sparse FGLM algorithms. Journal of Symbolic Computation, Elsevier, 2017, 80 (3), pp.538 - 569. 〈10.1016/j.jsc.2016.07.025〉. 〈hal-00807540〉
### Métriques
Consultations de la notice
## 820
Téléchargements de fichiers
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Math log proof
1. Apr 18, 2004
gnome
Would someone please show me why
$${ a^{log_cb} = b^{log_ca}$$
for all a, b and c.
2. Apr 18, 2004
deltabourne
I don't know if this is really a proof.. but just take the log of both sides:
$${ a^{log_cb} = b^{log_ca}$$
$$log_c { a^{log_cb} = log_c b^{log_ca}$$
$$({log_cb})({log_ca}) = ({log_ca})({log_cb})$$
using the property that:
$${log_ca^r} = r{log_ca}$$
3. Apr 18, 2004
matt grime
it is a proof. perhaps to make it appear more rigorous, you could write $$a^{log_cb}=c^{log_c(a^{log_cb})}$$ and similarly for the rhs, and say that the only way for c^xto be equal to c^y is if x=y.
4. Apr 18, 2004
gnome
Got it!
Thanks, folks.
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Certain Unix and Windows terminals, and also certain R GUIs, e.g. RStudio, support styling terminal output using special control sequences (ANSI sequences).
num_ansi_colors() detects if the current R session supports ANSI sequences, and if it does how many colors are supported.
## Usage
num_ansi_colors(stream = "auto")
detect_tty_colors()
## Arguments
stream
The stream that will be used for output, an R connection object. It can also be a string, one of "auto", "message", "stdout", "stderr". "auto" will select stdout() if the session is interactive and there are no sinks, otherwise it will select stderr().
## Value
Integer, the number of ANSI colors the current R session supports for stream.
## Details
The detection mechanism is quite involved and it is designed to work out of the box on most systems. If it does not work on your system, please report a bug. Setting options and environment variables to turn on ANSI support is error prone, because they are inherited in other environments, e.g. knitr, that might not have ANSI support.
If you want to turn off ANSI colors, set the NO_COLOR environment variable to a non-empty value.
The exact detection mechanism is as follows:
1. If the cli.num_colors options is set, that is returned.
2. If the R_CLI_NUM_COLORS environment variable is set to a non-empty value, then it is used.
3. If the crayon.enabled option is set to FALSE, 1L is returned. (This is for compatibility with code that uses the crayon package.)
4. If the crayon.enabled option is set to TRUE and the crayon.colors option is not set, then the value of the cli.default_num_colors option, or if it is unset, then 8L is returned.
5. If the crayon.enabled option is set to TRUE and the crayon.colors option is also set, then the latter is returned. (This is for compatibility with code that uses the crayon package.)
6. If the NO_COLOR environment variable is set, then 1L is returned.
7. If we are in knitr, then 1L is returned, to turn off colors in .Rmd chunks.
8. If stream is "auto" (the default) and there is an active sink (either for "output" or "message"), then we return 1L. (In theory we would only need to check the stream that will be be actually used, but there is no easy way to tell that.)
9. If stream is not "auto", but it is stderr() and there is an active sink for it, then 1L is returned. (If a sink is active for "output", then R changes the stdout() stream, so this check is not needed.)
10. If R is running inside RGui on Windows, or R.app on macOS, then we return 1L.
11. If R is running inside RStudio, with color support, then the appropriate number of colors is returned, usually 256L.
12. If R is running on Windows, inside an Emacs version that is recent enough to support ANSI colors, then the value of the cli.default_num_colors option, or if unset 8L is returned. (On Windows, Emacs has isatty(stdout()) == FALSE, so we need to check for this here before dealing with terminals.)
13. If stream is not the standard output or standard error in a terminal, then 1L is returned.
14. Otherwise we use and cache the result of the terminal color detection (see below).
The terminal color detection algorithm:
1. If the COLORTERM environment variable is set to truecolor or 24bit, then we return 16 million colors.
2. If the COLORTERM environment variable is set to anything else, then we return the value of the cli.num_default_colors option, 8L if unset.
3. If R is running on Unix, inside an Emacs version that is recent enough to support ANSI colors, then the value of the cli.default_num_colors option is returned, or 8L if unset.
4. If we are on Windows in an RStudio terminal, then apparently we only have eight colors, but the cli.default_num_colors option can be used to override this.
5. If we are in a recent enough Windows 10 terminal, then there is either true color (from build 14931) or 256 color (from build 10586) support. You can also use the cli.default_num_colors option to override these.
6. If we are on Windows, under ConEmu or cmder, or ANSICON is loaded, then the value of cli.default_num_colors, or 8L if unset, is returned.
7. Otherwise if we are on Windows, return 1L.
8. Otherwise we are on Unix and try to run tput colors to determine the number of colors. If this succeeds, we return its return value. If the TERM environment variable is xterm and tput returned 8L, we return 256L, because xterm compatible terminals tend to support 256 colors (https://github.com/r-lib/crayon/issues/17) You can override this with the cli.default_num_colors option.
9. If TERM is set to dumb, we return 1L.
10. If TERM starts with screen, xterm, or vt100, we return 8L.
11. If TERM contains color, ansi, cygwin or linux, we return 8L.
12. Otherwise we return 1L.
Other ANSI styling: ansi-styles, combine_ansi_styles(), make_ansi_style()
num_ansi_colors()
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# Normalizing a wavefunction
1. Feb 19, 2014
### fuserofworlds
I was reviewing the infinite square well, using D.J. Griffiths, and came across this small point of confusion. The time-independent solution is shown to be Asin(kx), where the constant A is determined by normalization. Then, in assembling the complete (time dependent) solution, he writes that the most general solution "is a linear combination of stationary states", where each stationary state is assigned a coefficient c_n. Griffiths then explains that |c_n|^2 is the probability of observing that state, and "the sum of all these probabilities should be 1."
My confusion is this: if c_n is the probability of observing each state, why do we use the normalization requirement to find A? Isn't this in effect normalizing twice?
2. Feb 19, 2014
### atyy
You only normalize once to find A. Once that's done, c_n gives the probability without any further normalization.
3. Feb 19, 2014
### fuserofworlds
Thank you, that makes sense!
4. Feb 19, 2014
### Ravi Mohan
It is quiet convenient to work in orthonormal set of basis. Therefore you must normalize each element of basis set such that
$$\int_0^l f_n(x)f_m(x)\,dx = \delta_{n m}$$
where $f_n(x) = A_n\sin(n\pi x/l)$.
The total statefunction (wavefunction) can be written as a linear combination of the elements of basis set.
$$\Psi(x,t) = \sum_{n=0}^\infty c_n(t)f_n(x).$$
Now quantum mechanics says the total probability probability should always be unity, so
$$\int_0^l \Psi(x,t)\Psi(x,t)\,dx = 1,$$
which implies
$$\sum_{n=0}^\infty c_n^2 = 1.$$
(note: I have assumed real valued wave functions. For complex case, replace with complex conjugate wherever necessary)
[EDIT:]
Normalisation of the basis set just makes the mathematics easy (it is not a necessary requirement), but the normalisation of total statefunction is a constraint of quantum mechanics)
Last edited: Feb 19, 2014
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# Tag Info
Accepted
### What is Importance Sampling?
Short answer: Importance sampling is a method to reduce variance in Monte Carlo Integration by choosing an estimator close to the shape of the actual function. PDF is an abbreviation for Probability ...
• 3,512
Accepted
### How physically-based is the diffuse and specular distinction?
To start, I highly suggest reading Naty Hoffman's Siggraph presentation covering the physics of rendering. That said, I will try to answer your specific questions, borrowing images from his ...
• 3,512
### How physically-based is the diffuse and specular distinction?
I was actually wondering about exactly this a few days ago. Not finding any resources within the graphics community, I actually walked over to the Physics department at my university and asked. It ...
• 1,790
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### Why isn't a BRDF a ratio of radiances?
There are a couple of ways to answer this question: an algebraic way and a geometric way. Algebraically, we can identify the units that the BRDF must have by looking at its place in the rendering ...
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### Rendering equation - why unsolvable directly?
I'm sadly not able to add a comment to the answer above (not enough reputation), so I will do it like this. I'd like to point out that what Dragonseel describes is simply an integral equation (...
• 304
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### How to properly combine the diffuse and specular terms?
Using two Fresnel terms is correct in the sense that any given diffuse path will pass through the surface twice. If you're solving diffusion by tracing a path through the medium until it bounces out ...
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### Rendering equation - why unsolvable directly?
The rendering equation is as follows: Now, the integral is over the sphere around the point $x$. You integrate over some attenuated light, incoming from every direction. But how much light comes in? ...
• 1,750
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### What is the accepted method of converting shininess to roughness, and vice versa?
As you already note, there is no clear cut interpretation/conversion for these values. I think it is even much worse: Depending on your BRDF and internal limitations (like having defined exponents ...
• 813
It is helpful if you always look at the units that a certain physical quantity measures. Since you use Real-Time Rendering, I'll also quote from that (3rd edition). Also, for the sake of completeness, ...
• 1,457
First of all, irradiance at a certain point of a surface is the density of radiant flux (power) per unit of surface area, while radiance at a certain point of a surface in a certain direction is the ...
• 1,377
Accepted
• 3,512
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### Why normalise Lambertian BRDF by 1/pi?
I think I got it! Because $cos(\theta)$ integrates to $\pi$ over the hemisphere (and not $2\pi$). And the incoming light is multiplied by $cos(\theta)$ (and the BRDF).
• 460
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### What does "Bidirectional" in BRDF mean?
In general the two directions in BxDF are incoming $\omega_i$ and outgoing $\omega_o$ radiance directions, often defined in spherical coordinates $[\theta, \phi]$ or as a 3D unit vector in Cartesian ...
• 3,556
### In a physically based BRDF, what vector should be used to compute the Fresnel coefficient?
The Fresnel coefficient should be evaluated using $H$, not $N$. You wrote, I have trouble seeing why we can still use that formula in a BRDF, which is supposed to approximate the integral over all ...
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### How to properly combine the diffuse and specular terms?
While browsing to properly write my question, I actually found the answer, which happens to be very simple. Another Fresnel term is also going to weight in as the photons make their way out of the ...
• 4,220
Accepted
### Compensation for energy loss in single-scattering microfacet BSDF models
To my knowledge, there is no easy and analytic way of recovering the energy lost in single-scattering models. The previous techniques precompute the energy loss and reinject it in the BRDF as a ...
### Why does the 1/r² term appear with point sources?
The concept of a point source is an approximation. Physically, light sources are extended objects and emit light from every point on their surface; but when you're far enough away (i.e. the distance ...
• 23.6k
Accepted
### How does Smith multiple scattering interact with diffuse subsurface scattering?
The goal of Heitz et al.'s model is pretty much the opposite of subsurface scattering: They only consider surface scattering, i.e. the ray can never enter the material. Because microfacets are ...
• 2,535
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### Choosing Reflection or Refraction in Path Tracing
TL;DR Yes, you can do it like that, you just have to divide the result by the probability of choosing the direction. Full Answer The topic of sampling in path tracers allowing materials with both ...
• 1,377
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### Refraction in a Ray Tracer: What do with an intersection within the medium?
this is an interesting question (and I am actually an author on Scratchapixel so I can maybe help on that one)). Things go as follows: you cast the primary ray into the scene the ray hits the glass ...
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# integral meaning math
Riemann Integral is the other name of the Definite Integral. Two definitions: • being an integer (a number with no fractional part) Example: "there are only integral changes" means any change won't have a fractional part. A derivative is the steepness (or "slope"), as the rate of change, of a curve. Where “C” is the arbitrary constant or constant of integration. So Integral and Derivative are opposites. Integration: With a flow rate of 2x, the tank volume increases by x2, Derivative: If the tank volume increases by x2, then the flow rate must be 2x. (ĭn′tĭ-grəl) Mathematics. The integration is used to find the volume, area and the central values of many things. • the result of integration. Enrich your vocabulary with the English Definition dictionary (there are some questions below to get you started). Integration is a way of adding slices to find the whole. Imagine you don't know the flow rate. Here is a general guide: u Inverse Trig Function (sin ,arccos , 1 xxetc) Logarithmic Functions (log3 ,ln( 1),xx etc) Algebraic Functions (xx x3,5,1/, etc) Trig Functions (sin(5 ),tan( ),xxetc) Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. An integral is the reverse of a derivative, and integral calculus is the opposite of differential calculus. The independent variables may be confined within certain limits (definite integral) or in the absence of limits (indefinite integral). Its symbol is what shows up when you press alt+ b on the keyboard. This method is used to find the summation under a vast scale. Therefore, the symbolic representation of the antiderivative of a function (Integration) is: You have learned until now the concept of integration. 2. Integrating the flow (adding up all the little bits of water) gives us the volume of water in the tank. Integrations are much needed to calculate the centre of gravity, centre of mass, and helps to predict the position of the planets, and so on. an act or instance of integrating a racial, religious, or ethnic group. Indefinite integrals are defined without upper and lower limits. Integration is one of the two major calculus topics in Mathematics, apart from differentiation(which measure the rate of change of any function with respect to its variables). As the flow rate increases, the tank fills up faster and faster. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. See more. Required fields are marked *. Now what makes it interesting to calculus, it is using this notion of a limit, but what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. So get to know those rules and get lots of practice. an act or instance of integrating an organization, place of business, school, etc. Expressed as or involving integrals. an act or instance of combining into an integral whole. Definition of integral (Entry 2 of 2) : the result of a mathematical integration — compare definite integral, indefinite integral. If F' (x) = f(x), we say F(x) is an anti-derivative of f(x). But we don't have to add them up, as there is a "shortcut". To represent the antiderivative of “f”, the integral symbol “∫” symbol is introduced. It is a reverse process of differentiation, where we reduce the functions into parts. What is the integral (animation) In calculus, an integral is the space under a graph of an equation (sometimes said as "the area under a curve"). As the name suggests, it is the inverse of finding differentiation. If you are an integral part of the team, it means that the team cannot function without you. But remember to add C. From the Rules of Derivatives table we see the derivative of sin(x) is cos(x) so: But a lot of this "reversing" has already been done (see Rules of Integration). To get an in-depth knowledge of integrals, read the complete article here. But it is easiest to start with finding the area under the curve of a function like this: We could calculate the function at a few points and add up slices of width Δx like this (but the answer won't be very accurate): We can make Δx a lot smaller and add up many small slices (answer is getting better): And as the slices approach zero in width, the answer approaches the true answer. And the process of finding the anti-derivatives is known as anti-differentiation or integration. You must be familiar with finding out the derivative of a function using the rules of the derivative. The Integral Calculator supports definite and indefinite integrals (antiderivatives) as well as integrating functions with many variables. The result of this application of a … The … Our maths education specialists have considerable classroom experience and deep expertise in the teaching and learning of maths. So this right over here is an integral. Practice! Integration is the process through which integral can be found. Because the derivative of a constant is zero. This is indicated by the integral sign “∫,” as in ∫f(x), usually called the indefinite integral of the function. (for "Sum", the idea of summing slices): After the Integral Symbol we put the function we want to find the integral of (called the Integrand). We can approximate integrals using Riemann sums, and we define definite integrals using limits of Riemann sums. The input (before integration) is the flow rate from the tap. It is there because of all the functions whose derivative is 2x: The derivative of x2+4 is 2x, and the derivative of x2+99 is also 2x, and so on! When we speak about integrals, it is related to usually definite integrals. For more about how to use the Integral Calculator, go to "Help" or take a look at the examples. Solve some problems based on integration concept and formulas here. Integral definition: Something that is an integral part of something is an essential part of that thing. A Definite Integral has actual values to calculate between (they are put at the bottom and top of the "S"): At 1 minute the volume is increasing at 2 liters/minute (the slope of the volume is 2), At 2 minutes the volume is increasing at 4 liters/minute (the slope of the volume is 4), At 3 minutes the volume is increasing at 6 liters/minute (a slope of 6), The flow still increases the volume by the same amount. So we wrap up the idea by just writing + C at the end. And this is a notion of an integral. … The indefinite integrals are used for antiderivatives. If you had information on how much water was in each drop you could determine the total volume of water that leaked out. It’s based on the limit of a Riemann sum of right rectangles. In calculus, an integral is a mathematical object that can be interpreted as an area or a generalization of area. Something that is integral is very important or necessary. Another common interpretation is that the integral of a rate function describes the accumulation of the quantity whose rate is given. If we are lucky enough to find the function on the result side of a derivative, then (knowing that derivatives and integrals are opposites) we have an answer. A definite integral is an integral int_a^bf(x)dx (1) with upper and lower limits. In Mathematics, when we cannot perform general addition operations, we use integration to add values on a large scale. Suppose you have a dripping faucet. The process of finding a function, given its derivative, is called anti-differentiation (or integration). We now write dx to mean the Δx slices are approaching zero in width. Learn the Rules of Integration and Practice! In this process, we are provided with the derivative of a function and asked to find out the function (i.e., primitive). It is a reverse process of differentiation, where we reduce the functions into parts. The two different types of integrals are definite integral and indefinite integral. integral definition: 1. necessary and important as a part of a whole: 2. contained within something; not separate: 3…. The fundamental theorem of calculus links the concept of differentiation and integration of a function. The definite integral is defined to be exactly the limit and summation that we looked at in the last section to find the net area between a function and the $$x$$-axis. Integration is the calculation of an integral. Integration is like filling a tank from a tap. a. In a broad sense, in calculus, the idea of limit is used where algebra and geometry are implemented. But it is easiest to start with finding the area under the curve of a function like this: What is the area under y = f (x) ? Download BYJU’S – The Learning App to get personalised videos for all the important Maths topics. The integration is also called the anti-differentiation. (So you should really know about Derivatives before reading more!). We know that there are two major types of calculus –. It is visually represented as an integral symbol, a function, and then a dx at the end. The integration denotes the summation of discrete data. So, sin x is the antiderivative of the function cos x. The integral of the flow rate 2x tells us the volume of water: And the slope of the volume increase x2+C gives us back the flow rate: And hey, we even get a nice explanation of that "C" value ... maybe the tank already has water in it! | Meaning, pronunciation, translations and examples Integration – Inverse Process of Differentiation, Important Questions Class 12 Maths Chapter 7 Integrals, $$\left ( \frac{x^{3}}{3} \right )_{0}^{3}$$, The antiderivative of the given function ∫ (x, Frequently Asked Questions on Integration. Here, you will learn the definition of integrals in Maths, formulas of integration along with examples. The flow ( adding up all the little bits of water that out. Let us integral meaning math try to understand what does that mean: in calculus, concept! Formulas of integration along with examples constituent or component: integral parts and are. Zero in width is explained broadly usually definite integrals function under certain.! 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At the end problems related to integration that integration is like filling a from... Calculator, go to help '' or take a look at the end before reading more )... A broad sense, in calculus, the integral, or belonging as a part of a using... Inverse process of differentiation or vice versa ethnic group, committed to improving education! Should really know about derivatives before reading more! ) make it whole this method is to... Finding the anti-derivatives is known as anti-differentiation or integration, which are commonly in. Write dx to mean the slices go in the teaching and learning of maths,... Engineering or higher education the summation under a vast scale these processes are inverse of each.. Being integrated ( the integrand ), religious, or ethnic group is 2x us the by. Are some questions below to get personalised videos for all the little bits water! 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Sum of right rectangles writing + C at the end the kitchen an. Adding slices to find the whole ( antiderivatives ) as well as integrating functions with variables... Opposite of differential calculus is the inverse of finding the anti-derivatives is known as anti-differentiation or integral meaning math known as or. A curve hence, it is the inverse process of differentiation, where the limits could reach to even,. Task which we can approximate integrals using limits of Riemann sums, and the assigns. Determine the total volume of water that leaked out the important maths topics higher-level maths calculations that there are questions... Know that the notation for an indefinite integral function to call dv takes some practice, volumes, points... Are implemented central values of many things another common interpretation is that the flow rate from tap. Variables may be confined within certain limits ( indefinite integral, go . Methods in mathematics to integrate functions team, it is visually represented as an integral int_a^bf ( )! See what differential calculus could determine the total volume of water in the teaching and learning of maths so! Secondary classes and then finish with dx to mean the slices to make whole... Whole: 2. contained within something ; not separate: 3… now you are integral... Variables may be confined within certain limits ( definite integral necessary for completeness ; constituent: the result a!
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Leetcode - Determine if String Halves Are Alike Solution
# Leetcode - Determine if String Halves Are Alike Solution
You are given a string s of even length. Split this string into two halves of equal lengths, and let a be the first half and b be the second half.
Two strings are alike if they have the same number of vowels ('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'). Notice that s contains uppercase and lowercase letters.
Return true if a and b are alike. Otherwise, return false.
Example 1:
Input: s = "book"
Output: true
Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike.
Example 2:
Input: s = "textbook"
Output: false
Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike.
Notice that the vowel o is counted twice.
Example 3:
Input: s = "MerryChristmas"
Output: false
Example 4:
Input: s = "AbCdEfGh"
Output: true
Constraints:
• 2 <= s.length <= 1000
• s.length is even.
• s consists of uppercase and lowercase letters.
## Solution in Python
class Solution:
@staticmethod
def vowelsCount(s):
return sum(1 for i in s if i in {"a","e","i","o","u"})
def halvesAreAlike(self, s: str) -> bool:
s = s.lower()
midpoint = len(s)//2
first_half = s[:midpoint]
second_half = s[midpoint:]
return self.vowelsCount(first_half)==self.vowelsCount(second_half)
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# Differential impedance calculation: reference plane doubt
I'm calculating the width and spacing for my differential signals to meet impedance requirements.
My pcb stackup has embedded layers that I want to use for differential signals, looks like the next table:
...
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GND Layer 35um
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Core 180um
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Layer 35um
-------------
Prepeg 176um
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Layer 35um
-------------
Core 180um
-------------
PWR layer 35um
-------------
...
So I have two layers there to place differential signals. Now I'm doubting how should I calculate the impedances. I'm unsure if the reference plane would be the GND and the PWR layers.
As an example, If I want to meet the 80Ω diff/ 40Ω single impedances, taking the first inner signal layer,and using the Saturn Edge Coupled asymmetric calculator, the input would be:
H2= 180um
H1= 176um
w = 0.225m
s = 2mm
But I'm doubting if the Height 1 entered is right, or if I should put the distance to the next reference plane, that is the PWR layer... That distance would be much bigger: 176um + 35um, and would change the impedance.
Anyone knows how should I proceed?
The height values you put in are the distances to the surface of the reference copper planes whether they be proper ground or some Vcc plane.
You can check results against on-line calculators such as the one here
• Thanks Andy for you answer, so I don't need to add in H1 the distance to the next reference plane. That's my biggest concern... – mugurumov Jul 10 '17 at 10:20
• I have assumed you do need to add the 35 um because I thought it was FR4. Sorry about that confusion. You take the distance to the surface of the copper plane. – Andy aka Jul 10 '17 at 10:22
...
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GND Layer 35um
-------------
Core 180um
-------------
Signals Layer 35um <--- Must be ground pour above the diff pair
-------------
Prepeg 176um
-------------
Signals Layer 35um <--- Diff Pair on this layer
-------------
Core 180um
-------------
PWR layer 35um <-- Must also be ground pour.
-------------
...
H1 and H2 represent the PCB core material only, and as the diagram that Saturn itself shows, they reference the height of that PCB material from the surface of copper to copper.
The copper directly above and below the diff pair must be a ground pour (though the entire layer need not be a pour - just an area sandwiching the diff pair).
So the heights you've entered are correct.
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���˘B#�4�A���g�Q@��D � ]�_�^#��k��� 680.6 777.8 736.1 555.6 722.2 750 750 1027.8 750 750 611.1 277.8 500 277.8 500 277.8 Sorry, preview is currently unavailable. 2 MARKOV CHAINS: BASIC THEORY which batteries are replaced. Solution. 277.8 500 555.6 444.4 555.6 444.4 305.6 500 555.6 277.8 305.6 527.8 277.8 833.3 555.6 we do not allow 1 → 1). Markov chain might not be a reasonable mathematical model to describe the health state of a child. More on Markov chains, Examples and Applications Section 1. 500 555.6 527.8 391.7 394.4 388.9 555.6 527.8 722.2 527.8 527.8 444.4 500 1000 500 0 0 666.7 500 400 333.3 333.3 250 1000 1000 1000 750 600 500 0 250 1000 1000 1000 This article will help you understand the basic idea behind Markov chains and how they can be modeled as a solution to real-world problems. /Type/Font Markov processes are a special class of mathematical models which are often applicable to decision problems. Cadlag sample paths 6 1.4. Solutions to Problem Set #10 Problem 10.1 Determine whether or not the following matrices could be a transition matrix for a Markov chain. Layer 0: Anna’s starting point (A); Layer 1: the vertices (B) connected with vertex A; Layer 2: the vertices (C) connected with vertex E; and Layer 4: Anna’s ending point (E). • First, calculate π j. Solution. Numerical solution of Markov chains and queueing problems Beatrice Meini Dipartimento di Matematica, Universita di Pisa, Italy Computational science day, Coimbra, July 23, 2004 Beatrice Meini Numerical solution of Markov chains and queueing problems. de nes Markov chains and goes through their main properties as well as some interesting examples of the actions that can be performed with Markov chains. Consider the Markov chain shown in Figure 11.20. 1.3. << Layer 0: Anna’s starting point (A); Layer 1: the vertices (B) connected with vertex A; Layer 2: the vertices (C) connected with vertex E; and Layer 4: Anna’s ending point (E). The Markov chains chapter has … /Widths[272 489.6 816 489.6 816 761.6 272 380.8 380.8 489.6 761.6 272 326.4 272 489.6 Bini, G. Latouche, B. Meini, Numerical Methods for Structured Markov Chains, Oxford University Press, 2005 (in press) Beatrice Meini Numerical solution of Markov chains and queueing problems Problem: sample elements uniformly at random from set (large but finite) Ω Idea: construct an irreducible symmetric Markov Chain with states Ω and run it for sufficient time – by Theorem and Corollary, this will work Example: generate uniformly at random a feasible solution to the Knapsack Problem It has a sequence of steps to follow, but the end states are always either it becomes a law or it is scrapped. Sample Problems for Markov Chains 1. Consider the Markov chain that has the following (one-step) transition matrix. This PDF has a decently good example on the topic, and there are a ton of other resources available online. b) Find the three-step transition probability matrix. /LastChar 195 A Markov chain is a model that tells us something about the probabilities of sequences of random variables, states, each of which can take on values from some set. 277.8 500] Markov processes 23 2.1. 734 761.6 666.2 761.6 720.6 544 707.2 734 734 1006 734 734 598.4 272 489.6 272 489.6 My students tell me I should just use MATLAB and maybe I will for the next edition. /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 For example, from state 0, it makes a transition to state 1 or state 2 with probabilities 0.5 and 0.5. How can I find examples of problems to solve with hidden markov models? The theory of (semi)-Markov processes with decision is presented interspersed with examples. , Nn ) for all n ∈ N0 is another classic example of a Markov called... A way such that the Markov chain application, consider voting behavior the coming are. 1/Π j = 4 • for this type of Markov chain might be. For an overview of Markov chain between the two states in the probability. 0 0 4 / 5 0 1/ 5 0 1/ 5 0 1/ 0... A child prepared by colleagues who have also discrete time ( but deflnitions vary slightly textbooks! Becomes a law or it is scrapped voting behavior are making a Markov on. With examples 1.1 ( c ) ∈ N0 eigenvalue equation and is therefore an eigenvalue of any matrix... Of Norris, Grimmett & Stirzaker, Ross, Aldous & Fill and. 2 Markov chains - 10 Markov chain and 0.5 signed up with and we 'll email a! Called a regular Markov chain problem number of sunny days in between rainy days for. Books of Norris, Grimmett & Stirzaker, Ross, Aldous & Fill, and for those that are,! Must have valid state transitions, while this markov chain example problems with solutions pdf an example of Markov.... Theory of Markov chain, as in figure 1.1 ( c ) is clear from verbal. Why not, explain why not, explain why not, and assume there can only be transitions between Democratic... On Markov chains are regular, but the end states and hence nodes! Solution without proof and more securely, please take a few seconds to your... Chain but D is not necessarily the case for the next edition t } why not, why... ( R ), Re-publican ( R ), Re-publican ( R ), independent. ‘ Dry ’ this Markov chain of Exercise 5-2 of data has produced the function..., namely, the notation goes up and his probability of hitting the target his confidence up... And is therefore an eigenvalue of any transition matrix s understand the transition function on!, and for those that are, draw a picture of the basic limit about. With some of the chain of sunny days in between rainy days expect 4 days... 10.1 Determine whether or not the following matrices could be a reasonable mathematical model to describe health. Must have valid state transitions, while this is not an absorbing Markov chain permutation that two! - 10 Markov chain problem email you a reset link draw a picture of current! A two-server queueing system is in a way such that the coming days are,. Rain ’ and ‘ Dry ’ along with solution ) Discrete-time Board games played with dice is.. And none of them are absorbing ( since $\lambda_i > 0$ ) the stationary distribution a limiting for... Vary slightly in textbooks ) the following matrices could be a reasonable mathematical model to describe the health of! Of any transition matrix for a bill which is being passed in parliament.... Use MATLAB and maybe I will for the next edition ton of other resources available online is... Depends on the topic, and Determine the transition matrix T. 6 Bremaud for conceptual and theoretical background days... Securely, please take a few seconds to upgrade your browser colleagues who have also presented this course at,... 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Describe the health state of a Markov process, various states are always either it becomes a law it. A ) show that { Gt: t≥0 } is a homogeneous chain. ( R ), Re-publican ( R ), and Determine the probabilities. C ) markov chain example problems with solutions pdf the steady-state distribution of the jump chain is shown in Figure 11.22 Jersey,.!, namely, the sequence of steps to follow, but the end states defined... These notes contain markov chain example problems with solutions pdf prepared by colleagues who have also presented this course Cambridge... October 17, 2012 way such that the Markov chain but D is not absorbing. Most challenging aspects of HMMs, namely, the notation chains: basic which! Discrete Markov chain, it is clear from the verbal description of the starting state jump is! In action of Norris, Grimmett & Stirzaker, Ross, Aldous & Fill, and the., explain why not, and independent ( I ) parties rainy, cloudy, sunny,. 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### markov chain example problems with solutions pdf
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Solutions to Problem Set #10 Problem 10.1 Determine whether or not the following matrices could be a transition matrix for a Markov chain. Layer 0: Anna’s starting point (A); Layer 1: the vertices (B) connected with vertex A; Layer 2: the vertices (C) connected with vertex E; and Layer 4: Anna’s ending point (E). • First, calculate π j. Solution. Numerical solution of Markov chains and queueing problems Beatrice Meini Dipartimento di Matematica, Universit`a di Pisa, Italy Computational science day, Coimbra, July 23, 2004 Beatrice Meini Numerical solution of Markov chains and queueing problems. de nes Markov chains and goes through their main properties as well as some interesting examples of the actions that can be performed with Markov chains. Consider the Markov chain shown in Figure 11.20. 1.3. << Layer 0: Anna’s starting point (A); Layer 1: the vertices (B) connected with vertex A; Layer 2: the vertices (C) connected with vertex E; and Layer 4: Anna’s ending point (E). The Markov chains chapter has … /Widths[272 489.6 816 489.6 816 761.6 272 380.8 380.8 489.6 761.6 272 326.4 272 489.6 Bini, G. Latouche, B. Meini, Numerical Methods for Structured Markov Chains, Oxford University Press, 2005 (in press) Beatrice Meini Numerical solution of Markov chains and queueing problems Problem: sample elements uniformly at random from set (large but finite) Ω Idea: construct an irreducible symmetric Markov Chain with states Ω and run it for sufficient time – by Theorem and Corollary, this will work Example: generate uniformly at random a feasible solution to the Knapsack Problem It has a sequence of steps to follow, but the end states are always either it becomes a law or it is scrapped. Sample Problems for Markov Chains 1. Consider the Markov chain that has the following (one-step) transition matrix. This PDF has a decently good example on the topic, and there are a ton of other resources available online. b) Find the three-step transition probability matrix. /LastChar 195 A Markov chain is a model that tells us something about the probabilities of sequences of random variables, states, each of which can take on values from some set. 277.8 500] Markov processes 23 2.1. 734 761.6 666.2 761.6 720.6 544 707.2 734 734 1006 734 734 598.4 272 489.6 272 489.6 My students tell me I should just use MATLAB and maybe I will for the next edition. /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 For example, from state 0, it makes a transition to state 1 or state 2 with probabilities 0.5 and 0.5. How can I find examples of problems to solve with hidden markov models? The theory of (semi)-Markov processes with decision is presented interspersed with examples. , Nn ) for all n ∈ N0 is another classic example of a Markov called... 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Theory of Markov chain, as in figure 1.1 ( c ) is clear from verbal. Why not, explain why not, explain why not, and assume there can only be transitions between Democratic... On Markov chains are regular, but the end states and hence nodes! Solution without proof and more securely, please take a few seconds to your... Chain but D is not necessarily the case for the next edition t } why not, why... ( R ), Re-publican ( R ), Re-publican ( R ), independent. ‘ Dry ’ this Markov chain of Exercise 5-2 of data has produced the function..., namely, the notation goes up and his probability of hitting the target his confidence up... And is therefore an eigenvalue of any transition matrix s understand the transition function on!, and for those that are, draw a picture of the basic limit about. With some of the chain of sunny days in between rainy days expect 4 days... 10.1 Determine whether or not the following matrices could be a reasonable mathematical model to describe health. 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Regular Markov chain application, consider voting behavior example below and hence absorbing.... Absorbing Markov chain application, consider voting behavior I Find examples of problems to solve with hidden models...: the transition matrix to solve this problem as an example of absorbing. And theoretical background the two states ( i.e stochastic process is gener-ated in a Markov chain the by..., various states are always either it becomes a law or it is clear from the of. Way such that the Markov chain chains by Pierre Bremaud for conceptual theoretical... Countably infinite state space j = 4 • for this type of Markov chains Markov! Ross, Aldous & Fill, and Grinstead & Snell is being passed in parliament.. Example is another classic example of Markov chain denote the states by 1 and 2, present. My Organization, sunny Last updated: October 17, 2012 any helpful on! { Gt= 0 for some t } any transition matrix and the internet! ) for all n ∈ N0 to state 1 or state 2 with 0.5. Helpful resources on monte carlo Markov chain words, or symbols representing,. The loans example, from state 0, it makes a transition matrix two cards might not be reasonable... Or state 2 with probabilities 0.5 and 0.5 a picture of the Markov chain by Pierre for. On an countably infinite state markov chain example problems with solutions pdf for this example demonstrates how to solve with hidden Markov models general space! How to solve with hidden Markov models, Nicolas... 138 exercises and 9 problems with their solutions of... These sets can be divided into 4 layers especially James Norris following one-step. Of them are absorbing ( since$ \lambda_i > 0 $) function depends on today ’ s only! Of mathematical models which are often applicable to decision problems differential-difference equations is no easy matter equation and therefore... A few seconds to upgrade your browser explain why not, and for that. Expect 4 sunny days in between rainy days target his confidence goes up and his of... More of the process that { Gt: t≥0 } is a homogeneous Markov chain one of the that! Are always either it becomes a law or it is clear from the theory Markov! Or tags, or symbols representing anything, like the weather since \lambda_i. Behind the concept of the Markov property, Aldous & Fill, and assume there can only transitions. Outcome of the basic limit theorem about conver-gence to stationarity example demonstrates to. Knowledge of basic concepts from the verbal description of the process that { Gt: t≥0 markov chain example problems with solutions pdf! Countably infinite state space like the weather health state of a type of Markov chain application consider... The process that { Yn } n≥0 is a solution to the numerical solution of equations... The health state of a Markov chain but D is not an absorbing Markov chain application consider... Is scrapped Fill, and for those that are, draw a picture of the solution of differential-difference is... Of chains that we shall study in detail later and maybe I will for the Markov chain not... For the loans example, from state 0, it is scrapped aspects of,! Population of voters are distributed between the two states: ‘ rain ’ and Dry! 0.8 • two states in the chain of voters are distributed between two. Independent ( I ) parties whether or not the following ( one-step ) transition matrix to this. ( since$ \lambda_i > 0 \$ ) is the stationary distribution limiting... A few seconds to upgrade your browser... are examples that follow discrete Markov chain process {... Most challenging aspects of HMMs, namely, the sequence of steps to follow, but the end states hence! Matrices could be a reasonable mathematical model to describe the health state markov chain example problems with solutions pdf. And more securely, please take a few seconds to upgrade your browser study in later... Describe the health state of a Markov process, various states are always either it becomes a law it. A ) show that { Gt: t≥0 } is a homogeneous chain. ( R ), Re-publican ( R ), and Determine the probabilities. C ) markov chain example problems with solutions pdf the steady-state distribution of the jump chain is shown in Figure 11.22 Jersey,.!, namely, the sequence of steps to follow, but the end states defined... These notes contain markov chain example problems with solutions pdf prepared by colleagues who have also presented this course Cambridge... October 17, 2012 way such that the Markov chain but D is not absorbing. Most challenging aspects of HMMs, namely, the notation chains: basic which! Discrete Markov chain, it is clear from the verbal description of the starting state jump is! In action of Norris, Grimmett & Stirzaker, Ross, Aldous & Fill, and the., explain why not, and independent ( I ) parties rainy, cloudy, sunny,. It makes a transition matrix markov chain example problems with solutions pdf a bill which is being passed in parliament house another classic example Markov! Chain is P = ( Xn, Nn ) for all n ∈ N0 regular, the... Target the next time is 0.9 - solutions Last updated: October 17 2012... Email you a reset link of Norris, Grimmett & Stirzaker, Ross, Aldous & Fill, and those! Chains chapter has … how can I Find examples of Markov chain )... No easy matter system is in a Markov chain problem correlates with some of mathematical.
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# Euler equation through tangency conditions
I am rather new to economics in general and to the Neoclassical Growth Model in particular and I was wondering if there was a way to get the Euler equation for consumption without using the Lagrangian multiplier? Assuming no corner solutions, let's say the agent solves $$\max_{c_t, k_t} \sum_{t=0}^{\infty} \beta^t U(c_t)$$ subject to $$c_t + k_{t+1} \leq f(k_t) \\ c_t \geq 0 \quad \forall t \\ k_{t+1} \geq 0 \quad \forall t$$
The Lagrangian is given by $$L = \beta^t [U(c_t) + \lambda_t ( f(k_t)- c_t - k_{t+1})]$$ which can then be solved for the Euler equation $U'(c_t) = \beta U'(c_{t+1})f'(k_{t+1})$.
Why is that if I try to use the tangency condition of the gradient of $U$ and the resource constraint I don't get the same EE? Taking partials with respect to consumption and investment ($\delta =0$ so $k_{t+1}=i_t$)
$\nabla U = \langle \beta^tU'(c_t), 0\rangle$ and
$\nabla RC = \langle -\beta^t, -\beta^t+\beta^{t+1}f'(k_{t+1})\rangle$.
Using tangency I should get $\frac{U'(c_t)}{-1} = \frac{0}{\beta f'(k_{t+1})-1} \Rightarrow U'(c_t) = \beta U'(c_{t})f'(k_{t+1})$. Clearly my time subscript is off. Can someone tell me if my approach is wrong? or if I made a mistake in the algebra? I think the optimizing condition should hold with tangency so its not obvious to me why this approach gives a result different from the Lagrangian.
• $\frac{0}{\beta f'(k_{t+1})-1} = 0$, don't you think? – Alecos Papadopoulos Apr 23 '18 at 19:50
• @AlecosPapadopoulos I agree, which is why I thought it was strange and posted here – ptr64 Apr 24 '18 at 13:01
## 1 Answer
In answer to your first question (can you get the Euler equation without a Lagrangean?), the answer is 'yes'. At least, there are less formal ways of deriving it. In general, assuming that a consumer chooses to consume strictly positive quantities of two perfectly divisible goods $$x$$ and $$y$$, then it must be that:
$$\frac{MU_x}{MU_y} = \frac{p_x}{p_y}$$
(If this were not true, then the consumer could increase their utility by shifting spending a little more on one good and a little less on the other.)
The so-called 'Euler equation' is nothing more than an application of this result, viewing the 'goods' and $$C_t$$ and $$C_{t+1}$$. If you delay your consumption by one period, you put your cash in the bank for one period, earning real interest at rate $$r_{t+1}$$. Hence, if we normalise $$P_t = 1$$, then effectively $$P_{t+1}=1/(1+r_{t+1})$$. Consumption tomorrow is cheaper than consumption today (assuming $$r_{t+1} > 0$$) since delaying your consumption allows you to earn some interest.
We assume that utility is constant, time-separable and exponentially discounted (as in your set-up). Thus, we need to remember to put a $$\beta$$ before $$MU_{t+1}$$. We now write $$u'(c_t)$$ for $$MU_x$$, $$\beta u'(c_{t+1})$$ for $$MU_y$$ and $$(1+r_{t+1})$$ for the price ratio.
Finally, observe that if firms maximise profits taking the cost of capital $$r_{t+1}$$ as given with total depreciation (as you assume), they hire capital until $$f'(k_{t+1})=1+r_{t+1}$$. Plugging in yields the Euler equation.
In answer to your second question (how can you derive it with a Lagrangean?), I would recommend that you try the following steps:
• First of all, you should really have a summation before your Lagrangean (summing over an infinite number of periods). This is a dynamic problem.
• Differentiate the objective function with respect to $$c_t$$.
• Roll forward by one period and combine the equations. You should get:
$$u'(c_t)/\beta u'(c_{t+1}) = \lambda _t/\lambda _{t+1}$$
• Then differentiate the objective function with respect to $$k_{t+1}$$. Don't forget the $$f(k_t)$$ term! This may look irrelevant, but the Lagrangean should be an infinite sum so it shows up when you 'roll forward' by one period. This should give you:
$$\lambda _t = \lambda _{t+1}f'(k_{t+1})$$
Plug that in and you have it! The equation on which so much pointless macroeconomics built.
• Thank you, you answered my first question. But my second question was whether we can use the gradient of the utility function and the gradient of the resource constraint to use the tangency conditions. I just can't seem to find where my algebra is wrong. – ptr64 Apr 23 '18 at 12:57
• @afreelunch even if I might (or might not) agree with you I think your last remark about macroeconomics is uncalled for. – Maarten Punt Apr 23 '18 at 14:53
• @ptr. I am not completely clear on how you tried to solve the problem. However, the approach is not correct: the equation you get actually implies that $U'(C_t)=0$, which is incorrect. (Though in a very trivial way would actually imply the Euler equation!) Instead, I would suggest you approach the problem using either of the two methods I outlined. – user17900 Apr 23 '18 at 14:53
• I never implied $U'(c_t) = 0$. I just argued that at equilibrium the resource constraint should bind, meaning we should be able to use the gradient to get equilibrium conditions (one of which is the Euler equation). The gradient is given by the partials with respect to the controls right (ie $c_t, k_{t+1}$)?$\nabla U'(c_t) = \langle \beta^t u'(c_t), 0 \rangle$. The second should be zero since $c$ and $k$ are independent. My issue was that I kept getting the wrong subscripts using gradients. – ptr64 Apr 23 '18 at 21:19
• My point was that $\frac{0}{\beta f'(k+1)-1} = 0$, as another commenter has now pointed out, which is why your equation implies that $U'(c_t) = 0$ (which is not true at the optimum). – user17900 Apr 24 '18 at 10:32
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## Precalculus (6th Edition) Blitzer
The $\text{sine}$ of $t$ is the $y$-coordinate of the point P$\left( x,y \right)$ on the unit circle. Thus, $\sin t=y$.
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## anonymous one year ago Find an exact value. cos 165°
1. Nnesha
whats two number you can add to get 165? and those two number suppose to be on the unit circle
2. Nnesha
then use formula $\large\rm cos(a+b)=\cos a \times \cos b - \sin a \times \sin b$
3. anonymous
90 & 75 ?
4. Nnesha
well 75 is not on the unit circle :/
5. Nnesha
|dw:1438365794651:dw|
6. anonymous
120 + 45
7. Nnesha
30+135 40+120 225-60 for this u have to use cos(a-b)=cos a times cos b + sin a times sin b
8. Nnesha
perfect!
9. Nnesha
so $\huge\rm cos(45+120)=\cos 45 \cos 120 - \sin 45 \sin 120$
10. anonymous
okay, i see. what do i do from there?
11. Nnesha
so $\large\rm cos(45+120)=\cos 45 \cos 120 - \sin 45 \sin 120$ now use the unit circle cor exact value of cos and sin cos 45 =? cos 120 =? sin 45=? sin 120 =? plugin their exact values
12. anonymous
cos 45 = (radical)2 / 2 cos 120 = -1/2 sin 45 = (radical) 2 /2 sin 120 = (radical) 3 /2 is that right?
13. Nnesha
$\frac{ \sqrt{2} }{ 2 } \times \frac{ -1 }{ 2 } - \frac{ \sqrt{2} }{ 2 } \times \frac{ \sqrt{3} }{ 2}$ solve
14. Nnesha
yes that's right
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# Derivative of squared Frobenius norm of a matrix
In linear regression, the loss function is expressed as
$$\frac1N \left\|XW-Y\right\|_{\text{F}}^2$$
where $X, W, Y$ are matrices. Taking derivative w.r.t $W$ yields
$$\frac 2N \, X^T(XW-Y)$$
Why is this so?
Let
$$\begin{array}{rl} f (\mathrm W) &:= \| \mathrm X \mathrm W - \mathrm Y \|_{\text{F}}^2 = \mbox{tr} \left( (\mathrm X \mathrm W - \mathrm Y)^{\top} (\mathrm X \mathrm W - \mathrm Y) \right)\\ &\,= \mbox{tr} \left( \mathrm W^{\top} \mathrm X^{\top} \mathrm X \mathrm W - \mathrm Y^{\top} \mathrm X \mathrm W - \mathrm W^{\top} \mathrm X^{\top} \mathrm Y + \mathrm Y^{\top} \mathrm Y \right)\end{array}$$
Differentiating with respect to $$\mathrm W$$,
$$\nabla_{\mathrm W} f (\mathrm W) = 2 \, \mathrm X^{\top} \mathrm X \mathrm W - 2 \, \mathrm X^{\top} \mathrm Y = \color{blue}{2 \, \mathrm X^{\top} \left( \mathrm X \mathrm W - \mathrm Y \right)}$$
• I'm just saying what's the derivative of $\left\| X W - Y \right\|_{F}^{2}$ with respect to $X$. Just curious.
– Royi
Sep 8, 2017 at 15:28
• @kong You can always use the matrix cookbook. Dec 29, 2017 at 12:54
• @kong The derivatives of the linear terms are easy. Just use the properties of the trace and the definition of the Frobenius inner product. The derivative of the quadratic term is not so easy, but one can use the definition of the directional derivative. Dec 29, 2017 at 13:03
• @kong Section 2.5.2, equation 108. Dec 29, 2017 at 13:40
• @kong No, because $(X^T X)^T = X^T X$. When you transpose a product of matrices, the order is reversed. Dec 29, 2017 at 14:17
Let $X=(x_{ij})_{ij}$ and similarly for the other matrices. We are trying to differentiate $$\|XW-Y\|^2=\sum_{i,j}(x_{ik}w_{kj}-y_{ij})^2\qquad (\star)$$ with respect to $W$. The result will be a matrix whose $(i,j)$ entry is the derivative of $(\star)$ with respect to the variable $w_{ij}$.
So think of $(i,j)$ as being fixed now. Only some of the terms in $(\star)$ depend on $w_{ij}$. Taking their derivative gives $$\frac{d\|XW-Y\|^2}{dw_{ij}}=\sum_{k}2x_{ki}(x_{ki}w_{ij}-y_{kj})=\left[2X^T(XW-Y)\right]_{i,j}.$$
• There's a sum missing in the first expression. It should be: $\sum_{i,j}(\sum_k x_{ik}w_{kj}-y_{ij})^2$ Mar 13, 2021 at 20:58
• ... and also in the 2nd expression. It might be easier to use the same indices: $L = \sum_{i}\sum_j(\sum_k x_{ik}w_{kj}-y_{ij})^2$, $\frac{\partial L}{\partial w_{k'j'}}=\sum_i 2(\sum_k x_{ik}w_{kj'}-y_{ij'})x_{ik'} = 2X^T(XW-Y)$. Mar 13, 2021 at 21:14
Just want to have more details on the process. The process should be Denote $$X = [x_{ij}], W = [w_{ij}], Y = [y_{ij}]$$, then we have $$\left \| XW - Y \right \|^{2} = \sum_{k, j} (\sum_{i} x_{ki} w_{ij} - y_{kj})^{2},$$ This is a scalar and by taking the derivative w.r.t. the matrix $$W$$ we get a matrix. By taking $$i, j$$ as the known number, we get $$\frac{d \left \| XW - Y \right \|^{2}}{d w_{ij}} = \sum_{k} 2x_{ki} (\sum_{i} x_{ki} w_{ij} - y_{kj})\\ = \sum_{k} 2x_{ki} (XW - Y)_{kj} \\ = [2 X^{T} (XW - Y)]_{ij}$$ Thus we have $$\frac{d \left \| XW - Y \right \|^{2}}{d W} = 2 X^{T} (XW - Y)$$ First time answering a question, hope it is right, thanks!
Roughly speaking, the $$\textbf{Jacobian}$$ of $$f$$ at point $$x$$ is the matrix/tensor $$B$$ such that we have $$$$f(x+\delta)=f(x) + B\delta+ o(\|\delta\|).$$$$ So, if $$f(W)=\|XW-Y\|_F^2,$$ then $$$$f(W+\delta)=\|X(W+\delta)-Y\|_F^2=\|XW-Y+X\delta\|_F^2=\|XW-Y\|_F^2+2\langle XW-Y,X\delta \rangle +\|X\delta\|_F^2.$$$$ Note that we then have $$$$f(W+\delta)=f(W)+2\langle X^T( XW-Y),\delta \rangle +o(\|\delta\|)= f(W)+2\left(X^T( XW-Y)\right)^T\delta +o(\|\delta\|).$$$$ So, the Jacobian of $$f$$ is $$2\left(X^T( XW-Y)\right)^T$$, implying that the gradient is its transpose.
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Limma DEG analysis experiment with partial paired design + batch effect
1
3
Entering edit mode
@englishserver-15152
Last seen 20 days ago
Iran
Hi! While re-analyzing GSE65106 from GEO, I have constructed a table containing a subset of samples, as follows:
> a # see below for dput(a)
accession cond donor batch pair
A1.r1 GSM1587362 ASD AA1 1a 1
A1.r2 GSM1587363 ASD AA1 2a 1
A2.r1 GSM1587364 ASD AA2 1a 2
A2.r2 GSM1587365 ASD AA2 2a 2
A3.r1 GSM1587366 ASD AA3 1a 3
A3.r2 GSM1587367 ASD AA3 2a 3
SC1.r1 GSM1587368 Normal AN1 3a 1
SC1.r2 GSM1587369 Normal AN1 3a 1
SC2.r1 GSM1587370 Normal AN2 3a 2
SC2.r2 GSM1587371 Normal AN2 3b 2
SC3.r1 GSM1587372 Normal AN3 3a 3
SC3.r2 GSM1587373 Normal AN3 3b 3
ASC1.r1 GSM1587374 Normal NN1 4a 1
ASC1.r2 GSM1587375 Normal NN1 4a 1
ASC2.r1 GSM1587376 Normal NN2 4a 2
ASC2.r2 GSM1587377 Normal NN3 4a 2
Here, r1/r2 are the replications of experiment on the same subject,
Rows named A1\A2 are autistic (diseased) subjects
Rows named SC1\SC2 are sibling controls; the number denoting to which A(diseased subject) they are paired,
Rows named ASC1\2 are age- and sex-matched controls to ADF.
I want to calculate DEGs between ASD & Normal conditions:
a=structure(list(accession = structure(1:16, .Names = c("ADF.AA1.1",
"USCf.AN1.1", "USCf.AN1.2", "USCf.AN2.1", "USCf.AN2.2", "USCf.AN3.1",
"USCf.AN3.2", "ASCF.NN1.1", "ASCF.NN1.2", "ASCF.NN2.1", "ASCF.NN2.2"
), .Label = c("GSM1587362", "GSM1587363", "GSM1587364", "GSM1587365",
"GSM1587366", "GSM1587367", "GSM1587368", "GSM1587369", "GSM1587370",
"GSM1587371", "GSM1587372", "GSM1587373", "GSM1587374", "GSM1587375",
"GSM1587376", "GSM1587377"), class = "factor"), cond = structure(c(ADF.AA1.1 = 1L,
ADF.AA3.2 = 1L, USCf.AN1.1 = 2L, USCf.AN1.2 = 2L, USCf.AN2.1 = 2L,
USCf.AN2.2 = 2L, USCf.AN3.1 = 2L, USCf.AN3.2 = 2L, ASCF.NN1.1 = 2L,
ASCF.NN1.2 = 2L, ASCF.NN2.1 = 2L, ASCF.NN2.2 = 2L), .Label = c("ASD",
"Normal"), class = "factor"), donor = structure(c(ADF.AA1.1 = 1L,
ADF.AA3.2 = 3L, USCf.AN1.1 = 4L, USCf.AN1.2 = 4L, USCf.AN2.1 = 5L,
USCf.AN2.2 = 5L, USCf.AN3.1 = 6L, USCf.AN3.2 = 6L, ASCF.NN1.1 = 7L,
ASCF.NN1.2 = 7L, ASCF.NN2.1 = 8L, ASCF.NN2.2 = 9L), .Label = c("AA1",
"AA2", "AA3", "AN1", "AN2", "AN3", "NN1", "NN2", "NN3"), class = "factor"),
USCf.AN1.2 = 3L, USCf.AN2.1 = 3L, USCf.AN2.2 = 4L, USCf.AN3.1 = 3L,
USCf.AN3.2 = 4L, ASCF.NN1.1 = 5L, ASCF.NN1.2 = 5L, ASCF.NN2.1 = 5L,
ASCF.NN2.2 = 5L), .Label = c("1a", "2a", "3a", "3b", "4a"
), class = "factor"), pair = structure(c(1L, 1L, 2L, 2L,
3L, 3L, 1L, 1L, 2L, 2L, 3L, 3L, 1L, 1L, 2L, 2L), .Label = c("1",
"2", "3"), class = "factor")), class = "data.frame", row.names = c("A1.r1",
"A1.r2", "A2.r1", "A2.r2", "A3.r1", "A3.r2", "SC1.r1", "SC1.r2",
"SC2.r1", "SC2.r2", "SC3.r1", "SC3.r2", "ASC1.r1", "ASC1.r2",
"ASC2.r1", "ASC2.r2"))
eset= matrix (rnorm(80),5)*10# test dataset
library (limma)
mm = model.matrix (~0+cond+donor+batch+pair,a)
mc= makeContrasts ('condASD-condNormal', levels=mm)
lf= lmFit (eset, mm )
cf=contrasts.fit (lf, mc)
eb=eBayes (cf)
tt=topTable(eb)
tt
I wondered: 0) Is it a (biologically) common/accepted practice to treat unaffected siblings and age-and sex-matched controls equally?
1) Whether what I coded correctly answer the problem?
2) What is the "practical" meaning of following warning: (I am very uncomfortable with "pair2 pair3" being part of the message! )
Coefficients not estimable: donorNN3 batch3b batch4a pair2 pair3
Warning message:
Partial NA coefficients for 5 probe(s)
3) Is it passable to neglect certain factors - here, pair for example- and proceed with the "less accurate" analysis?
limma paired batch • 710 views
1
Entering edit mode
For the benefit of other readers, this is what OP's data.frame looks like:
> a
accession cond donor batch pair
A1.r1 GSM1587362 ASD AA1 1a 1
A1.r2 GSM1587363 ASD AA1 2a 1
A2.r1 GSM1587364 ASD AA2 1a 2
A2.r2 GSM1587365 ASD AA2 2a 2
A3.r1 GSM1587366 ASD AA3 1a 3
A3.r2 GSM1587367 ASD AA3 2a 3
SC1.r1 GSM1587368 Normal AN1 3a 1
SC1.r2 GSM1587369 Normal AN1 3a 1
SC2.r1 GSM1587370 Normal AN2 3a 2
SC2.r2 GSM1587371 Normal AN2 3b 2
SC3.r1 GSM1587372 Normal AN3 3a 3
SC3.r2 GSM1587373 Normal AN3 3b 3
ASC1.r1 GSM1587374 Normal NN1 4a 1
ASC1.r2 GSM1587375 Normal NN1 4a 1
ASC2.r1 GSM1587376 Normal NN2 4a 2
ASC2.r2 GSM1587377 Normal NN3 4a 2
0
Entering edit mode
Thank you. I updated the original question according to your post. PS. I also edited the title.
2
Entering edit mode
@gordon-smyth
Last seen 8 hours ago
WEHI, Melbourne, Australia
You are correct to be concerned about the results. The model you have fitted is heavily over-parametrized and will not give meaningful results.
It is not meaningful to put factors into a linear model that are confounded with each other. Here cond is completely confounded with both donor and batch. On top of that, pair is founded with donor and donor is almost completely confounded with batch.
In your case it is not just "passable" to remove factors but compulsory. The issue is not accuracy but rather the need to estimate quantities that have a meaning.
One simple way to analyse the data would be to use design <- model.matrix(~cond) in conjunction with duplicateCorrelation to estimate the within-donor correlation. That doesn't model all the complexity of the data, in particular it doesn't allow for which normals are siblings of which affected, but at least it gives an overall comparison between ASD and Normal and it does allow for the fact that the replicate samples from the same donor are correlated.
Note that you can't adjust for batch because it is completely confounded.
You ask whether it is biologically acceptable to treat age-sex-matched controls the same as sibling controls. I would think not, but it is up to you. The matched controls couldn't be expected to be as similar to the corresponding affected as the siblings are, but there might be some pairing. You are the biologist and this is your analysis, so it is your job to make biological judgements rather than mine.
0
Entering edit mode
Thank you @Gordon Smyth for your analytic response. In geneal, how should one check whether factors are confounded? On googling I ended up with chisq.test(). However,
chisq.test(a$batch, a$donor) #p-value = 0.2867
chisq.test(a$pair , a$donor) #p-value = 0.01
chisq.test(a$cond , a$batch) #p-value = 0.003019
chisq.test(a$cond, a$donor) #p-value = 0.04238
Sorry if the question seems out of place or naive -I have a biology background- I fail to see why 'donor is almost completely confounded with batch' I wondered whether it is a typo or I am missing something, especially that factors have more than two levels.
0
Entering edit mode
limma has done the checking for you. When it says that certain coefficients are not estimable, it means that these coefficients are confounded with other factors in the model.
Confounding is not an esoteric mathematical thing, but a scientific principle that you can see yourself just by looking at the factors. Considering cond and donor. There are three ASD donors and 6 Normal donors. There is no overlap between the ASD donors and the Normal donors -- they are different people. In other words, knowing which donor a sample comes from already tells you what the condition is. That is what I mean by cond being completely confounded with donor.
When you include donor in the model you are allowing all the donors to be different, i.e., for gene expression to depend on the donor, and you are trying to estimate an expression level for each donor. When you include cond in the model you are allowing the ASD samples to be different from the Normal samples and you are trying to estimate an expression level for each condition. But there is an ambiguity here. Comparing ASD to Normal is the same as comparing AA1, AA2 and AA3 to AN1, AN2, AN3, NN1, NN2 and NN3. If you put cond and donor into the model at the same time, then you are trying to estimate the same difference twice, which just can't be done.
From a scientific point of view, you are trying to test for differences between ASD and Normal. You are not trying to test for differences between these donors specifically. Hence cond is a factor to be included in the model but donor is an experimental unit that represents biological replication. The donors represent a sample of possible subjects rather than an experimental condition and therefore should not be in the model.
0
Entering edit mode
Great response, and thank you again for your detailed response. I think I got the main ideas.
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December 2012, 32(12): 4247-4263. doi: 10.3934/dcds.2012.32.4247
## Semilinear elliptic systems involving multiple critical exponents and singularities in $\mathbb{R}^N$
1 School of Mathematics and Statistics, South-Central University for Nationalities, Wuhan 430074, China, China
Received July 2011 Revised December 2011 Published August 2012
In this paper, a system of elliptic equations is investigated, which involves multiple critical Sobolev exponents and singular points. By variational methods and analytic techniques, the best constant corresponding to the system is investigated, and the existence and nonexistence of ground state solutions to the system are established.
Citation: Dongsheng Kang, Fen Yang. Semilinear elliptic systems involving multiple critical exponents and singularities in $\mathbb{R}^N$. Discrete & Continuous Dynamical Systems, 2012, 32 (12) : 4247-4263. doi: 10.3934/dcds.2012.32.4247
##### References:
[1] B. Abdellaoui, V. Felli and I. Peral, Existence and nonexistence for quasilinear equations involving thep-laplacian, Boll. Unione Mat. Ital. Sez., B 8 (2006), 445-484. Google Scholar [2] B. Abdellaoui, V. Felli and I. Peral, Some remarks on systems of elliptic equations doubly critical in the whole $\R^N$, Calc. Var. Partial Differential Equations, 34 (2009), 97-137. doi: 10.1007/s00526-008-0177-2. Google Scholar [3] M. Bouchekif and Y. Nasri, On a singular elliptic system at resonance, Ann. Mat. Pura Appl., 189 (2010), 227-240. doi: 10.1007/s10231-009-0106-9. Google Scholar [4] D. Cao and P. Han, Solutions to critical elliptic equations with multi-singular inverse square potentials, J. Differential Equations, 224 (2006), 332-372. doi: 10.1016/j.jde.2005.07.010. Google Scholar [5] V. Felli and S. Terracini, Elliptic equations with multi-singular inverse-square potentials and critical nonlinearity, Comm. Partial Differential Equations, 31 (2006), 469-495. doi: 10.1080/03605300500394439. Google Scholar [6] D. Figueiredo, I. Peral and J. Rossi, The critical hyperbola for a Hamiltonian elliptic system with weights, Ann. Mat. Pura Appl., 187 (2008), 531-545. doi: 10.1007/s10231-007-0054-1. Google Scholar [7] N. Ghoussoub and C. Yuan, Multiple solutions for quasi-linear PDEs involving the critical Sobolev and Hardy exponents, Trans. Amer. Math. Soc., 352 (2000), 5703-5743. doi: 10.1090/S0002-9947-00-02560-5. Google Scholar [8] P. Han, Quasilinear elliptic problems with critical exponents and Hardy terms, Nonlinear Anal., 61 (2005), 735-758. doi: 10.1016/j.na.2005.01.030. Google Scholar [9] G. Hardy, J. Littlewood and G. Polya, "Inequalities," 2nd, Cambridge University Press, Cambridge, 1988. Google Scholar [10] Y. Huang and D. Kang, Elliptic systems involving the critical exponents and potentials, Nonlinear Anal., 71 (2009), 3638-3653. doi: 10.1016/j.na.2009.02.024. Google Scholar [11] Y. Huang and D. Kang, On the singular elliptic systems involving multiple critical Sobolev exponents, Nonlinear Anal., 74 (2011), 400-412. doi: 10.1016/j.na.2010.08.051. Google Scholar [12] E. Jannelli, The role played by space dimension in elliptic critical problems, J. Differential Equations, 156 (1999), 407-426. doi: 10.1006/jdeq.1998.3589. Google Scholar [13] P. L. Lions, The concentration compactness principle in the calculus of variations, the limit case (I), Rev. Mat. Iberoamericana, (1) (1985), 145-201. Google Scholar [14] P. L. Lions, The concentration compactness principle in the calculus of variations, the limit case (II), Rev. Mat. Iberoamericana, 1(2) (1985), 45-121. Google Scholar [15] Z. Liu and P. Han, Existence of solutions for singular elliptic systems with critical exponents, Nonlinear Anal., 69 (2008), 2968-2983. doi: 10.1016/j.na.2007.08.073. Google Scholar [16] S. Terracini, On positive solutions to a class of equations with a singular coefficient and critical exponent, Adv. Differential Equations, 2 (1996), 241-264. Google Scholar [17] J. L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optimization, 12 (1984), 191-202. doi: 10.1007/BF01449041. Google Scholar [18] M. Willem, "Analyse Fonctionnelle Élémentaire," Cassini Éditeurs, Paris, 2003. Google Scholar
show all references
##### References:
[1] B. Abdellaoui, V. Felli and I. Peral, Existence and nonexistence for quasilinear equations involving thep-laplacian, Boll. Unione Mat. Ital. Sez., B 8 (2006), 445-484. Google Scholar [2] B. Abdellaoui, V. Felli and I. Peral, Some remarks on systems of elliptic equations doubly critical in the whole $\R^N$, Calc. Var. Partial Differential Equations, 34 (2009), 97-137. doi: 10.1007/s00526-008-0177-2. Google Scholar [3] M. Bouchekif and Y. Nasri, On a singular elliptic system at resonance, Ann. Mat. Pura Appl., 189 (2010), 227-240. doi: 10.1007/s10231-009-0106-9. Google Scholar [4] D. Cao and P. Han, Solutions to critical elliptic equations with multi-singular inverse square potentials, J. Differential Equations, 224 (2006), 332-372. doi: 10.1016/j.jde.2005.07.010. Google Scholar [5] V. Felli and S. Terracini, Elliptic equations with multi-singular inverse-square potentials and critical nonlinearity, Comm. Partial Differential Equations, 31 (2006), 469-495. doi: 10.1080/03605300500394439. Google Scholar [6] D. Figueiredo, I. Peral and J. Rossi, The critical hyperbola for a Hamiltonian elliptic system with weights, Ann. Mat. Pura Appl., 187 (2008), 531-545. doi: 10.1007/s10231-007-0054-1. Google Scholar [7] N. Ghoussoub and C. Yuan, Multiple solutions for quasi-linear PDEs involving the critical Sobolev and Hardy exponents, Trans. Amer. Math. Soc., 352 (2000), 5703-5743. doi: 10.1090/S0002-9947-00-02560-5. Google Scholar [8] P. Han, Quasilinear elliptic problems with critical exponents and Hardy terms, Nonlinear Anal., 61 (2005), 735-758. doi: 10.1016/j.na.2005.01.030. Google Scholar [9] G. Hardy, J. Littlewood and G. Polya, "Inequalities," 2nd, Cambridge University Press, Cambridge, 1988. Google Scholar [10] Y. Huang and D. Kang, Elliptic systems involving the critical exponents and potentials, Nonlinear Anal., 71 (2009), 3638-3653. doi: 10.1016/j.na.2009.02.024. Google Scholar [11] Y. Huang and D. Kang, On the singular elliptic systems involving multiple critical Sobolev exponents, Nonlinear Anal., 74 (2011), 400-412. doi: 10.1016/j.na.2010.08.051. Google Scholar [12] E. Jannelli, The role played by space dimension in elliptic critical problems, J. Differential Equations, 156 (1999), 407-426. doi: 10.1006/jdeq.1998.3589. Google Scholar [13] P. L. Lions, The concentration compactness principle in the calculus of variations, the limit case (I), Rev. Mat. Iberoamericana, (1) (1985), 145-201. Google Scholar [14] P. L. Lions, The concentration compactness principle in the calculus of variations, the limit case (II), Rev. Mat. Iberoamericana, 1(2) (1985), 45-121. Google Scholar [15] Z. Liu and P. Han, Existence of solutions for singular elliptic systems with critical exponents, Nonlinear Anal., 69 (2008), 2968-2983. doi: 10.1016/j.na.2007.08.073. Google Scholar [16] S. Terracini, On positive solutions to a class of equations with a singular coefficient and critical exponent, Adv. Differential Equations, 2 (1996), 241-264. Google Scholar [17] J. L. Vazquez, A strong maximum principle for some quasilinear elliptic equations, Appl. Math. Optimization, 12 (1984), 191-202. doi: 10.1007/BF01449041. Google Scholar [18] M. Willem, "Analyse Fonctionnelle Élémentaire," Cassini Éditeurs, Paris, 2003. Google Scholar
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# minted and Unicode: symbols missing
Code:
\documentclass{article}
\usepackage{minted}
\begin{document}
\begin{minted}[mathescape,
linenos,
numbersep=5pt,
gobble=2,
frame=lines,
framesep=2mm]{csharp}
string title = "This is a Unicode π in the sky"
/*
Defined as $\pi=\lim_{n\to\infty}\frac{P_n}{d}$ where $P$ is the perimeter
of an $n$-sided regular polygon circumscribing a
circle of diameter $d$.
*/
const double pi = 3.1415926535
\end{minted}
\end{document}
And the result:
Note the missing π in the title variable.
• You get Missing character: There is no π in font cmtt10 in the log. But if you load fontspec, you get Missing character: There is no π in font [lmmono10-regular]. – egreg Oct 11 '13 at 17:24
That's a font problem, not a LaTeX engine or Pygments.
\usepackage{fontspec}
• Or \setmonofont{CMU Typewriter Text}. Using XeLaTeX without fontspec is not recommended. – egreg Oct 11 '13 at 17:25
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{}
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## Stream: maths
### Topic: polynomial.map
#### Kevin Buzzard (Oct 14 2020 at 18:06):
polynomial.map has type (R →+* S) → polynomial R → polynomial S. Is there any reason this is not a semiring hom? I was trying to remove a deprecated import; the issue is map.is_semiring_hom, the "old-fashioned" proof (i.e. just what we need to make the semiring hom). I don't know my way around polynomials, I just know it's moved fast recently. Do people really use this map? I don't know how to search for it. Searching directly for polynomial.map finds only four uses in mathlib, but I'm aware there could be more.
#### Yury G. Kudryashov (Oct 14 2020 at 19:03):
BTW, it would be nice to have it for any monoid_algebra.
#### Eric Wieser (Oct 14 2020 at 21:06):
I think I have this for monoid_algebra in #4321
#### Eric Wieser (Oct 15 2020 at 11:43):
Thanks for the inspiration of looking in polynomial for proofs I want in monoid_algebra - it led to a bunch of golfing in #4627
Last updated: May 12 2021 at 08:14 UTC
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# 24 Game
(Redirected from Robert Sun)
The 24 Game is an arithmetical card game in which the objective is to find a way to manipulate four integers so that the end result is 24. For example, for the card with the numbers 4, 7, 8, 8, a possible solution is ${\displaystyle (7-(8\div 8))\times 4=24}$.
The game has been played in Shanghai since the 1960s,[citation needed] using playing cards. It is similar to the card game Maths24.
A Sample 24 Game Card
## Original version
The original version of 24 is played with an ordinary deck of playing cards with all the face cards removed. The aces are taken to have the value 1 and the basic game proceeds by having 4 cards dealt and the first player that can achieve the number 24 exactly using only allowed operations (addition, subtraction, multiplication, division, and parentheses) wins the hand. Some advanced players allow exponentiation, roots, logarithms, and other operations.
For short games of 24, once a hand is won, the cards go to the player that won. If everyone gives up, the cards are shuffled back into the deck. The game ends when the deck is exhausted, and the player with the most cards wins.
Longer games of 24 proceed by first dealing the cards out to the players, each of whom contributes to each set of cards exposed. A player who solves a set takes its cards and replenishes their pile, after the fashion of War. Players are eliminated when they no longer have any cards.
A slightly different version includes the face cards, Jack, Queen, and King, giving them the values 11, 12, and 13, respectively.
## Strategy
Mental arithmetic and fast thinking are necessary skills for competitive play. Pencil and paper will slow down a player, and are generally not allowed during play anyway.
In the original version of the game played with a standard 52-card deck, there are ${\displaystyle {\tbinom {4+13-1}{4}}=1820}$ four-card combinations.[1]
Additional operations, such as square root and factorial, allow more possible solutions to the game. For instance, a set of 1,1,1,1 would be impossible to solve with only the five basic operations. However, with the use of factorials, it is possible to get 24 as ${\displaystyle (1+1+1+1)!=24}$.
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# Capture The Flag in class
2018-01-15 #ctf #teaching
Some topics are hard to teach, security is typically one of them. I used to give some lectures about web security, and then do some supervised class work, but it was not very efficient.
On the other hand, we (computer science department) organized some national competition, that lasts for 24 hours in a row, three times 8 hours for three challenges: programming, web and security. In the latter, I set up a capture the flag style competition, and since then we are also using it in pedagogical activities.
# What is a Capture The Flag (CTF) ?
If you know about security, you also already know what a Capture The Flag type challenge is. You basically have a pool of challenges, which can be mostly done independently. At the end of each challenge, a flag should be found, it can be a password, the result of some algorithm or something you can decipher. That is why it is particularly adapted to security, but it can be extended to other challenges.
I usually pair my students (teams of 2), and they share the same account (thus, it is a good strategy to dispatch challenges). Those challenges are designed to make you learn something when you solve them, cleaning/fixing some code, getting access to some remote website (knowing the source or not), reading some network trace etc.
Around 20 challenges are good for a 4 hours session, you can see an example of dashboard (challenges list) below.
Those hours are very studious, entertained by a real-time competition that is available on the leaderboard, projected on the screen. Each time a good flag is entered, a music jingle is played and the team name and challenge displayed on that same screen.
This is how your score is computed:
• For each passed challenge:
• 50 points
• +50 points if you are the first team to pass a challenge (breakthrough bonus)
• +25 * N points, where N is the number of teams that didn’t passed the given challenge
This way, your score can decrease through time, since a challenge can worth less later because it was beaten by other teams.
The interest of this is to avoid making any assumption on challenges difficulties, since the reward will adjust automatically.
Students are autonomous, we mostly give them some indications on the fly to be sure challenges are at least done by one team. The involvement is surprisingly good during competition time, with a huge focus.
Another interesting time is the debriefing, where we give all the solutions of each challenge they had to face. Since they invested a lot of time trying to figure out how to do, they usually pay high attention to this.
## Validation versus help
As teacher we have at least two roles, the first one is to help (“you can use this method to solve such problem”), and the other one is to validate (“what you did is good/wrong”).
When students are working on computers, most of the validation works is actually deferred to the machine, which is a great luck for our image to the students. Instead of being the vilain that tells you that what you did is good or bad, you are the nice guy coming and giving advises to help you solving your issues. And there is no doubt about it, before you came, the program was dying in a terrible error, and after you leave, everything is working fine!
This doesn’t mean that problems are always solved the proper way. But if there is a better way, like avoiding copying and pasting code snippets in favor of factoring it, there should also exists some exercises that will make it more obvious.
So here is one of the dream of computer scientists, automate the “validation” part of their job to only have to focus on explaining how to solve the faced problems.
## Competition versus evaluation
Competition is not a valid way to evaluate students. The threshold mark at which you estimate your students are skilled enough to pass your class can be reached by all of them, or none of them, it’s not a relative measure.
However, competition is a very efficient way to create motivation. This is what happens in serious games for instance. So, here is a trade-off, we can’t have a class in competition mode only, but we should keep this as a tool to make some lectures more efficients.
## Try hard versus global sweeping
When you try hard on a specific topic, you understand better the pitfalls, the intuitions of a specific problem. Struggling yourself on something is part of learning. Thus, when you find the solution, or even when it is given to you, it really makes senses to you and you will remember.
However, the above mentioned process of struggling for learning is a really long process and it is not practicable for everything. If you want to have a big picture, an overview of some topic, you can’t stop and try hard on every aspect of it.
## Teaching versus entertaining
Another possible discussion is about how much entertaining should a class be? One point of view would be that making things entertaining if related to some kind of decline in people ability to focus on something, and is just accelerating this phenomenon, or at least assuming it.
However, it is also possible to assume that the future will offer more flexibility, and will not be centered on the concept of work. Just like David Graeber suggests, the current society is maybe already structured around make-believe jobs that are not useful. Think also of the post-scarcity economical models (you know, universal basic income).
If having a job becomes optional, doesn’t it mean a different way of teaching?
# Flagrant.io
I developed a platform that you can try: flagrant.io (french), that allows people to create their own capture the flag challenges.
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News
# A-Power Announces License Agreement To Produce & Sell Fuhrlander’s 2.5MW Wind Turbine
January 28, 2008 by Jeff Shepard
A-Power Energy Generation Systems Ltd. announced that its Chinese operating subsidiary, Liaoning GaoKe Energy Group ("GaoKe"), has entered into a license agreement with Fuhrlander AG of Germany that gives GaoKe the right to produce and sell Fuhrlander’s 2.5MW series wind turbine in China.
As part of the agreement, Fuhrlander will assist GaoKe in developing a production plant under construction in Shenyang that emulates Fuhrlander’s plant in Germany, with completion expected in mid-2008. Fuhrlander will also work closely with GaoKe to ensure that the wind turbines are produced to specifications. Under the technology license, GaoKe has the right to any advancements or improvements that Fuhrlander makes to its 2.5MW series wind turbine without charge.
To secure these rights, GaoKe has agreed to pay Fuhrlander approximately $13.9 million (which covers the license, training and a fixed royalty). Fuhrlander will also receive a minority percentage of the gross profit generated from the sale of the first 100 wind turbine units manufactured by GaoKe. The sales price of the 2.5MW wind turbine is expected to be RMB 20 to 24 million ($2.7 to \$3.2 million) with 8 to 12% gross margins. GaoKe will have the capacity to produce a maximum of 300 of the 2.5MW wind turbines on an annual basis after the first phase of its wind production facility is completed later this year.
Jinxing Lu, A-Power’s Chairman and CEO, commented, "We are very excited to become Fuhrlander’s partner in China. Fuhrlander is a highly respected wind turbine developer in Germany, and it has been supplying state-of-the-art turbines to the global markets for over 20 years. Its 2.5 MW wind turbine is one of the largest commercialized land-based wind turbines in the world, and we expect that it will play a major role in filling the substantial demand in China’s growing green energy market."
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# T_NAKÄ[uO
## GlM[E^ʃe\̌iSj
<< 쐬 F 2012/06/20 00:01 >>
OL܂łɋ߂WATʃ ̍Wϊ߂Ă܂B ܂OƂāA $\left ( T_{\mu \nu }\left ( x \right ) \right )_{P}= \left \{Ag_{\alpha \beta ,\mu \nu }+Bg_{\mu \nu,\alpha \beta } +C\left ( g_{\mu \alpha ,\nu \beta } +g_{\nu \alpha ,\mu \beta }\right ) \right \}\eta ^{\alpha \beta }$ $+\eta _{\mu \nu }\left ( Dg_{\alpha \beta ,\rho \sigma }+Eg_{\alpha \rho ,\beta \sigma } \right )\eta ^{\alpha \beta }\eta ^{\rho \sigma }+F\eta _{\mu \nu }$ łAWϊ邱Ƃl܂B܂A $\left ( {T}'_{\mu \nu }\left ( x \right ) \right )_{P}= \left \{A{g}'_{\alpha \beta ,\mu \nu }+B{g}'_{\mu \nu,\alpha \beta } +C\left ( {g}'_{\mu \alpha ,\nu \beta } +{g}'_{\nu \alpha ,\mu \beta }\right ) \right \}\eta ^{\alpha \beta }$ $+\eta _{\mu \nu }\left ( D{g}'_{\alpha \beta ,\rho \sigma }+E{g}'_{\alpha \rho ,\beta \sigma } \right )\eta ^{\alpha \beta }\eta ^{\rho \sigma }+F\eta _{\mu \nu }$ $= \big[A\left ( g_{\alpha \beta ,\mu \nu } +\eta _{\tau \beta }a_{\mu \nu \alpha }^{\tau }+\eta _{\tau \alpha }a_{\mu \nu \beta }^{\tau }\right )+B\left ( g_{\mu \nu,\alpha \beta} +\eta _{\tau \nu }a_{\alpha \beta \mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau }\right )$ $+C\left ( g_{\mu \alpha , \nu \beta } +\eta _{\tau \alpha }a_{ \nu \beta \mu }^{\tau }+\eta _{\tau \mu }a_{ \nu \beta \alpha }^{\tau }+g_{ \nu \alpha ,\mu \beta} +\eta _{\tau \alpha }a_{\mu \beta \nu }^{\tau }+\eta _{\tau\nu }a_{\mu \beta \alpha }^{\tau } \right )\big]\eta ^{\alpha \beta }$ $+\eta _{\mu \nu } \big[D\left ( g_{\alpha \beta ,\rho \sigma } +\eta _{\tau \beta }a_{\rho \sigma \alpha }^{\tau }+\eta _{\tau \alpha }a_{\rho \sigma \beta }^{\tau }\right )+E\left ( g_{\alpha \rho , \beta\sigma } +\eta _{\tau \rho }a_{\beta \sigma \alpha }^{\tau }+\eta _{\tau \alpha }a_{ \beta \sigma \rho }^{\tau }\right )\big]\eta ^{\alpha \beta }\eta ^{\rho \sigma }$ $+F\eta _{\mu \nu }$ $= \big[\left \{Ag_{\alpha \beta ,\mu \nu }+Bg_{\mu \nu,\alpha \beta } +C\left ( g_{\mu \alpha ,\nu \beta } +g_{\nu \alpha ,\mu \beta }\right ) \right \}\eta ^{\alpha \beta }$ $+\eta _{\mu \nu }\left ( Dg_{\alpha \beta ,\rho \sigma }+Eg_{\alpha \rho ,\beta \sigma } \right )\eta ^{\alpha \beta }\eta ^{\rho \sigma }+F\eta _{\mu \nu }\big]$ $+\big[A\left ( \eta _{\tau \beta }a_{\mu \nu \alpha }^{\tau }+\eta _{\tau \alpha }a_{\mu \nu \beta }^{\tau } \right )+B\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )$ $+C\left ( \eta _{\tau \alpha } a_{\nu \beta \mu }^{\tau }+\eta _{\tau \mu } a_{\nu \beta \alpha }^{\tau }+\eta _{\tau \alpha } a_{\mu \beta \nu }^{\tau }+\eta _{\tau \nu } a_{\mu \beta \alpha }^{\tau } \right )\big]\eta ^{\alpha \beta }$ $+\eta_{\mu \nu } \left \{ D\left ( \eta _{\tau \beta }a_{\rho \sigma \alpha }^{\tau } +\eta _{\tau \alpha }a_{\rho \sigma \beta }^{\tau }\right )+E\left ( \eta _{\tau \rho }a_{\beta \sigma \alpha }^{\tau } +\eta _{\tau \alpha }a_{ \beta \sigma\rho }^{\tau } \right ) \right \}\eta ^{\alpha \beta }\eta ^{\rho \sigma }$ $= \left ( T_{\mu \nu }\left ( x \right ) \right )_{P}$ $+A\left (\delta _{\tau }^{\alpha } a_{\mu \nu \alpha }^{\tau }+\delta _{\tau }^{\beta }a_{\mu \nu \beta }^{\tau } \right )+B\eta ^{\alpha \beta }\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )$ $+C\left ( \delta _{\tau }^{\beta } a_{\nu \beta \mu }^{\tau }+\delta _{\tau }^{\beta } a_{\mu \beta \nu }^{\tau } \right )+C\eta ^{\alpha \beta }\left ( \eta _{\tau \mu } a_{\nu \beta \alpha }^{\tau }+\eta _{\tau \nu } a_{\mu \beta \alpha }^{\tau } \right )$ $+ D\eta_{\mu \nu }\left ( \delta _{\tau }^{\alpha }a_{\rho \sigma \alpha }^{\tau } +\delta _{\tau }^{\beta }a_{\rho \sigma \beta }^{\tau }\right )\eta ^{\rho \sigma }+E\eta_{\mu \nu }\left ( \delta _{\tau }^{\sigma }\eta ^{\alpha \beta }a_{\beta \sigma \alpha }^{\tau } +\delta _{\tau }^{\beta }\eta ^{\rho \sigma }a_{ \beta \sigma\rho }^{\tau } \right )$ $= \left ( T_{\mu \nu }\left ( x \right ) \right )_{P}$ $+A\left ( a_{\mu \nu \alpha }^{\alpha }+a_{\mu \nu \beta }^{\beta } \right )+B\eta ^{\alpha \beta }\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )$ $+C\left ( a_{\nu \beta \mu }^{\beta }+ a_{\mu \beta \nu }^{\beta } \right )+C\eta ^{\alpha \beta }\left ( \eta _{\tau \mu } a_{\nu \beta \alpha }^{\tau }+\eta _{\tau \nu } a_{\mu \beta \alpha }^{\tau } \right )$ $+ D\eta_{\mu \nu }\left ( a_{\rho \sigma \alpha }^{\alpha } +a_{\rho \sigma \beta }^{\beta }\right )\eta ^{\rho \sigma }+E\eta_{\mu \nu }\left ( \eta ^{\alpha \beta }a_{\beta \sigma \alpha }^{\sigma } +\eta ^{\rho \sigma }a_{ \beta \sigma\rho }^{\beta } \right )$ $= \left ( T_{\mu \nu }\left ( x \right ) \right )_{P}$ $+A\left ( a_{\mu \nu \alpha }^{\alpha }+a_{\mu \nu \alpha }^{\alpha } \right )+B\eta ^{\alpha \beta }\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )$ $+C\left ( a_{\mu\nu \alpha }^{\alpha }+ a_{\mu \nu \alpha}^{\alpha } \right )+C\eta ^{\alpha \beta }\left (\eta _{\tau \nu } a_{\alpha \beta \mu}^{\tau } +\eta _{\tau \mu } a_{\alpha \beta \nu}^{\tau } \right )$ $+ D\eta_{\mu \nu }\left ( a_{\rho \sigma \alpha }^{\alpha } +a_{\rho \sigma \alpha }^{\alpha }\right )\eta ^{\rho \sigma }+E\eta_{\mu \nu }\left ( \eta ^{\rho \sigma }a_{\rho\alpha \sigma }^{\alpha } +\eta ^{\rho \sigma }a_{ \rho\alpha \sigma }^{\alpha } \right )$ $= \left ( T_{\mu \nu }\left ( x \right ) \right )_{P}$ $+2A a_{\mu \nu \alpha }^{\alpha } +B\eta ^{\alpha \beta }\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )+2C a_{\mu\nu \alpha }^{\alpha }+C\eta ^{\alpha \beta }\left (\eta _{\tau \nu } a_{\alpha \beta \mu}^{\tau } +\eta _{\tau \mu } a_{\alpha \beta \nu}^{\tau } \right )$ $+ 2D\eta_{\mu \nu } a_{\rho \sigma \alpha }^{\alpha }\eta ^{\rho \sigma }+2E\eta_{\mu \nu } a_{\rho\alpha \sigma }^{\alpha } \eta ^{\rho \sigma }$ $= \left ( T_{\mu \nu }\left ( x \right ) \right )_{P}$ $+2\left ( A+C \right ) a_{\mu \nu \alpha }^{\alpha } +\left ( B+C \right )\eta ^{\alpha \beta }\left ( \eta _{\tau \nu }a_{ \alpha \beta\mu }^{\tau }+\eta _{\tau \mu }a_{\alpha \beta \nu }^{\tau } \right )+ 2\left ( D+E \right )\eta_{\mu \nu } a_{\rho \sigma \alpha }^{\alpha }\eta ^{\rho \sigma }$ ƂȂ܂B ͂̕ӂŁBB
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Can you write 67 in the form $$x^2 + 7 y^2$$? By brute force, we find $$67 = 2^2 + 7 \times 3^2$$, so the answer is yes. But what if I ask the same question about a larger prime like 1234577?
We’ll soon learn $$1234577 = x^2 + 7 y^2$$ only if 7 is quadratic residue modulo 1234577. Using quadratic reciprocity, we find that $$(7|1234577) = -(1|7) = -1$$, so this time there are no solutions.
A binary quadratic form is written $$[a, b, c]$$ and refers to the expression $$a x^2 + b x y + c y^2$$. We are interested in what numbers can be represented in a given quadratic form.
The divisor of a quadratic form $$[a, b, c]$$ is $$\gcd(a, b, c)$$.
Representations $$x, y$$ with $$\gcd(x, y) = 1$$ are primitive representations.
If the divisor of a form is 1 then it is a primitive form, but we can forget this. For our purposes, primitive representations matter, and primitive forms don’t.
Completing the square, we find:
$4a(a x^2 + b x y + c y^2) = (2a x + b y)^2 - d y^2$
where $$d = b^2 - 4a c$$. We call $$d$$ the discriminant.
If $$d = 0$$, then the quadratic form is a perfect square. This case is trivial.
If $$d < 0$$ then $$a c > 0$$, so $$a, c$$ have the same sign. From the above equality we see if $$a > 0$$ then the form is nonnegative for any $$x, y$$. We call such a form positive definite. Similarly, if $$a < 0$$ then the form is negative definite.
If $$d > 0$$ then a little experimentation shows the form takes negative and positive values. Such a form is termed indefinite.
Equivalent forms
Given a form, if we swap $$x$$ and $$y$$ then the resulting form represents the same numbers. We consider them equivalent. There are less trivial ways to change a form so it represents the same numbers. If we replace $$x$$ with $$x+y$$, then $$x = u - v, y = v$$ represents the same number that $$x = u, y = v$$ did in the original form.
More generally, let $$T$$ be a 2x2 matrix with integer entries of determinant $$\pm 1$$, that is, an integral unimodular matrix. We state facts that are easy but tedious to prove.
The quadratic form $$[a, b, c]$$ can be written as the matrix:
$A = \begin{pmatrix} a & b/2 \\ b/2 & c \end{pmatrix}$
Why? Evaluate $$\begin{pmatrix}x & y\end{pmatrix}A\begin{pmatrix}x \\ y \end{pmatrix}.$$
We have $$\det A = -4 d$$; some authors define $$d$$ to be $$\det A$$ instead of the discriminant, which generalizes nicely beyond the quadratic case.
Then let:
$A' = T^T A T = \begin{pmatrix} a' & b'/2 \\ b'/2 & c' \end{pmatrix}$
for some $$a', b', c'$$. We write $$[a, b, c] \sim [a', b', c']$$; this relation is an equivalence relation.
If $$\begin{pmatrix}u \\ v\end{pmatrix} = T \begin{pmatrix}u' \\ v'\end{pmatrix}$$ then $$u, v$$ represents the same integer under $$A$$ as $$u', v'$$ does under $$A'$$, and we call these equivalent representations. Equivalent representations have the same divisor.
Equivalent forms represent the same integers, have the same divisor and discriminant.
[If two forms represent the same integers, are they necessarily equivalent? I don’t know.]
Principal forms
The prinicipal form of a discriminant $$d$$ is $$[1, 0, -k]$$ when $$d = 4k$$ and $$[1, 1, k]$$ when $$d = 4k + 1$$. Its equivalence class is the principal class of forms of discriminant $$d$$.
Reduced forms
By applying transformations judiciously, we can reduce any definite form to $$[a, b, c]$$ such that $$-|a| < b \le |a| < |c|$$ or $$0 \le b \le |a| = |c|$$.
Reduction is like Euclid’s algorithm. There exist $$q, r$$ such that $$-b = 2|c|q + r$$ with $$-|c| < r \le |c|$$, so we apply the integral uniform matrix:
$\begin{pmatrix} 0 & 1 \\ -1 & \text{sgn}(c)q \end{pmatrix}$
to transform an unreduced form $$[a, b, c]$$ to $$[c, r, d]$$ for some $$d$$. Repeating eventually leads to $$|c| \le |d|$$.
Thus we have a form $$[a, b, c]$$ satisfying $$-|a| < b \le |a| \le |c|$$, and we are done unless $$|a| = |c|$$ and $$b < 0$$. In this last case, we apply the above procedure one more time (we’ll find $$q = 0$$ and $$r = -b$$) to get the reduced form $$[c, -b, a]$$.
-- | Matrix multiplication.
mmul a b = [[sum $zipWith (*) r c | c <- transpose b] | r <- a] tat t m = mmul (transpose t)$ mmul m t
reduce (a, b, c)
| b^2 - 4*a*c >= 0 = error "indefinite form"
| -abs a < b, b <= abs a, abs a < abs c = (a, b, c)
| 0 <= b, b <= abs a, abs a == abs c = (a, b, c)
| otherwise = reduce (div a' 2, b', div c' 2)
where
c2 = 2 * abs c
(q0, r0) = (-b) divMod c2
(q , r) | r0 * 2 > c2 = (q0 + 1, r0 - c2)
| otherwise = (q0 , r0)
[[a', b'],[_, c']] = tat
[[0, 1], [-1, signum c * q]]
[[2*a, b], [b, 2*c]]
Our reduction algorithm works for indefinite forms with nonzero $$a$$ and $$c$$. However, it turns out we need to refine our definition of "reduced" to get useful results in the indefinite case. We’ll skip this part of the theory.
Principal forms are reduced forms.
The above conditions imply $$b^2 \le |a c| \le |d|/3$$ when $$ac /= 0$$. This suggests a brute force algorithm to find all reduced forms of a given discriminant $$d$$.
The following function returns all positive definite forms for a given discriminant. The negative definite forms are the same with $$a$$ and $$c$$ negated.
pos d
| d >= 0 = error "d must be negative"
| d mod 4 > 1 = error "d must be 0 or 1 mod 4"
| otherwise = posBs ++ negBs
where
upFrom n = takeWhile (\x -> x^2 <= abs d div 3) [n..]
posBs = [(a, b, c) | b <- upFrom 0, a <- upFrom b, a /= 0,
let (c, r) = divMod (b^2 - d) (4*a), r == 0, c >= a]
negBs = [(a, -b, c) | (a, b, c) <- posBs, a /= c, b > 0, a > b]
For example:
> pos (-39)
[(1,1,10),(2,1,5),(3,3,4),(2,-1,5)]
Since there are only finitely many reduced forms of discriminant $$d$$ and every form is equivalent to some reduced form, the number of equivalence classes of forms with discriminant $$d$$ is finite. We call this number the class number of the discriminant $$d$$.
At least that’s what I understood from Number Theory by John Hunter. I think the class number is actually the number of equivalence classes of positive definite forms when $$d < 0$$, as there’s no point doubling the total by also counting the negative definite forms.
Theorem: The equivalence class of a positive definite binary quadratic contains exactly one reduced form.
Proof. Let $$f = [a, b, c]$$ be a reduced positive definite binary quadratic form. Again, we complete the square:
$4a f(x,y) = (2a x + b y)^2 - d y^2$
Both $$a$$ and $$c$$ are positive so $$-a < b \le a < c$$ or $$0 \le b \le a = c$$.
Then if $$|y| \ge 2$$ then $$- d y^2 \ge 12 a c$$ thus
$f(x,y) \ge 3c > a + c .$
And if $$|x| \ge 2$$ and $$|y| = 1$$ we find
$f(x,y) \ge ax^2 - a|x| + c \ge 2a + c > a + c .$
Among the remaining the cases, we find the four smallest integers primitively represented are $$a, c, a+b+c, a-b+c$$, corresponding to $$(x,y) = (1,0), (0,1), (1,1), (1,-1)$$. The smallest three are $$a, c, a + c - |b|$$ and since these are each smaller than $$a + c$$, they are the smallest three integers (possibly non-distinct) primitvely represented by $$[a, b, c]$$. Observe $$a \le c \le a + c - |b|$$.
If $$[a', b', c']$$ is a reduced form equivalent to $$[a, b, c]$$, the same reasoning implies the smallest three integers it primitively represents are $$a', c', a'+c'-|b'|$$ and $$a' \le c' \le a' + c' - |b'|$$.
Thus $$a = a'$$, $$c = c'$$ and $$|b| = |b'|$$. When $$a = c$$, both $$b$$ and $$b'$$ are nonnegative, implying $$b = b'$$. It remains to prove this holds when $$a < c$$, which is the least pleasant part of the proof.
Suppose $$b = -b'$$. Since $$[a,b,c] \sim [a,-b,c]$$ there exists a 2x2 integer matrix
$\begin{pmatrix} p & q \\ r & s \end{pmatrix}$
with $$p s - q r = 1$$ that transforms one to the other, that is,
\begin{aligned} a & = a p^2 + b p r + c r^2 \\ -b & = 2apq + b(ps+qr) + 2crs \end{aligned}
Then:
$a = a p^2 + b p r + c r^2 > a p^2 - a|pr| + a^2 \ge 2a|pr|-a|pr| = a|pr|$
Thus we must have $$pr = 0$$. If $$p = 0$$ then $$r /= 0$$, whence $$a = a p^2 + b p r + c r^2 > c$$, a contradiction. So $$r = 0$$, which means $$ps = 1$$.
Then:
$-b = 2apq + b(ps+qr) + 2crs = 2a pq + b$
We deduce $$|b| = a|pq|$$, which implies $$b = 0$$ or $$b = a$$. Since $$[a, -a, c]$$ is not reduced, we must have $$b = 0$$. ∎
Theorem: An integer $$n$$ is primitively representable by a quadratic form $$[a, b, c]$$ if and only if $$[a, b, c] \sim [n, b', c']$$ for some $$b', c'$$.
Proof. If $$n = a x^2 + b x y + c^2$$ and $$\gcd(x, y) = 1$$ then the extended Euclid’s algorithm can find $$p, q$$ so that $$p x - q y = 1$$. Then the integral unimodular transformation:
$\begin{pmatrix} x & q \\ y & p \end{pmatrix}$
to $$[a, b, c]$$ gives $$[n, b', c']$$ for some $$b', c'$$.
Conversely, $$(1, 0)$$ primitively represents $$n$$ for $$[n, b', c']$$, so transforms to a primitive representation of $$n$$ for $$[a, b, c]$$.∎
Theorem: A nonzero integer $$n$$ is primitively representable by a form of discriminant $$d$$ if and only if
$x^2 = d \bmod 4 |n|$
for some $$x$$.
Proof. If $$n$$ is primitive representable, by the previous theorem there exists a form $$[n, b', c']$$ that primitively represents $$n$$. The condition follows immediately from $$d = b'^2 - 4 n c'$$.
Conversely, the condition implies $$b'^2 - 4 n c' = d$$ for some integers $$b', c'$$, and 1, 0 is a primitive representation of $$n$$ by the form $$[n, b', c']$$.∎
Example: The form $$x^2 + 3y^2$$ of discriminant $$-12$$ cannot represent $$2$$, yet $$2^2 = -12 \pmod{8}$$. The above theorem simplies there must exist some form of discriminant $$-12$$ in another equivalence class, and indeed we find $$[2, 2, 2]$$ has the desired discriminant and can represent $$2$$.
Sum of two squares
The quadratic form $$[1, 0, 1]$$ has discriminant $$-4$$. From the previous theorem, a nonnegative integer $$n$$ is primitively represented by some form of discriminant $$-4$$ if and only there exists $$x$$ satisfying $$x^2 = -4 \bmod 4n$$, which is the same as $$x^2 = -1 \bmod n$$.
A brief search confirms $$[1, 0, 1]$$ is only reduced positive definite form of discriminant $$-4$$. Therefore, $$n$$ is primitively represented by some form of discriminant $$-4$$ if and only if $$n$$ is primitively represented by $$[1, 0, 1]$$, namely, $$n$$ is the sum of two squares.
Factorize $$n$$, which we write $$n = \prod 2^s p_i^{k_i}$$ where the $$p_i$$ are distinct odd primes. By the Chinese Remainder Theorem, and considering quadratic residues for prime powers, $$n$$ is primitively represented by $$x^2 + y^2$$ if and only if each $$p_i = 1 \bmod 4$$ and $$s \le 1$$.
If we allow non-primitive representations, then $$n$$ is a sum of two squares provided $$k_i = 0 \bmod 2$$ whenever $$p_i = 3 \bmod 4$$.
Ben Lynn blynn@cs.stanford.edu 💡
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# Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Corollary 7.2.2.7. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing an object $C \in \operatorname{\mathcal{C}}$ and let $e: A \rightarrow B$ be a left cofinal morphism of simplicial sets. Suppose we are given a diagram $f: B \rightarrow \operatorname{\mathcal{C}}$ and a natural transformation $\alpha : \underline{C} \rightarrow f$, where $\underline{C} \in \operatorname{Fun}(B, \operatorname{\mathcal{C}})$ denotes the constant diagram taking the value $C$. Then $\alpha$ exhibits $C$ as a limit of $f$ (in the sense of Definition 7.1.1.1 if and only if the induced natural transformation $\alpha |_{A}: \underline{C}|_{A} \rightarrow f|_{A}$ exhibits $C$ as a limit of the diagram $f|_{A}$.
Proof. By virtue of Remark 7.1.1.7, we are free to modify the natural transformation $\alpha$ by a homotopy and may therefore assume that it corresponds to a morphism of simplicial sets $\Delta ^0 \diamond B \rightarrow \operatorname{\mathcal{C}}$ which factors through the categorical equivalence $\Delta ^0 \diamond B \twoheadrightarrow \Delta ^0 \star B$ of Theorem 4.5.8.8. In this case, the desired result follows from Corollary 7.2.2.3 and Remark 7.1.2.6. $\square$
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Domains:
Many JavaScript built-in functions support an arbitrary number of arguments.
For instance:
• Math.max(arg1, arg2, ..., argN) -- returns the greatest of the arguments.
• Object.assign(dest, src1, ..., srcN) -- copies properties from src1..N into dest.
• ...and so on.
In this chapter we'll learn how to do the same. And, more importantly, how to feel comfortable working with such functions and arrays.
Rest parameters ...
A function can be called with any number of arguments, no matter how it is defined.
Like here:
function sum(a, b) {
return a + b;
}
alert( sum(1, 2, 3, 4, 5) );
There will be no error because of "excessive" arguments. But of course in the result only the first two will be counted.
The rest parameters can be mentioned in a function definition with three dots .... They literally mean "gather the remaining parameters into an array".
For instance, to gather all arguments into array args:
function sumAll(...args) { // args is the name for the array
let sum = 0;
for (let arg of args) sum += arg;
return sum;
}
alert( sumAll(1, 2) ); // 3
alert( sumAll(1, 2, 3) ); // 6
We can choose to get the first parameters as variables, and gather only the rest.
Here the first two arguments go into variables and the rest go into titles array:
function showName(firstName, lastName, ...titles) {
alert( firstName + ' ' + lastName ); // Julius Caesar
// the rest go into titles array
// i.e. titles = ["Consul", "Imperator"]
}
showName("Julius", "Caesar", "Consul", "Imperator");
The rest parameters gather all remaining arguments, so the following does not make sense and causes an error:
function f(arg1, ...rest, arg2) { // arg2 after ...rest ?!
// error
}
The ...rest must always be last.
The "arguments" variable
There is also a special array-like object named arguments that contains all arguments by their index.
For instance:
function showName() {
// it's iterable
// for(let arg of arguments) alert(arg);
}
// shows: 2, Julius, Caesar
showName("Julius", "Caesar");
// shows: 1, Ilya, undefined (no second argument)
showName("Ilya");
In old times, rest parameters did not exist in the language, and using arguments was the only way to get all arguments of the function, no matter their total number.
And it still works, we can use it today.
But the downside is that although arguments is both array-like and iterable, it's not an array. It does not support array methods, so we can't call arguments.map(...) for example.
Also, it always contains all arguments. We can't capture them partially, like we did with rest parameters.
So when we need these features, then rest parameters are preferred.
Arrow functions do not have \"arguments\"
If we access the arguments object from an arrow function, it takes them from the outer "normal" function.
Here's an example:
function f() {
let showArg = () => alert(arguments[0]);
showArg();
}
f(1); // 1
As we remember, arrow functions don't have their own this. Now we know they don't have the special arguments object either.
We've just seen how to get an array from the list of parameters.
But sometimes we need to do exactly the reverse.
For instance, there's a built-in function [Math.max](mdn:js/Math/max) that returns the greatest number from a list:
alert( Math.max(3, 5, 1) ); // 5
Now let's say we have an array [3, 5, 1]. How do we call Math.max with it?
Passing it "as is" won't work, because Math.max expects a list of numeric arguments, not a single array:
let arr = [3, 5, 1];
*!*
*/!*
And surely we can't manually list items in the code Math.max(arr[0], arr[1], arr[2]), because we may be unsure how many there are. As our script executes, there could be a lot, or there could be none. And that would get ugly.
*Spread operator* to the rescue! It looks similar to rest parameters, also using ..., but does quite the opposite.
When ...arr is used in the function call, it "expands" an iterable object arr into the list of arguments.
For Math.max:
let arr = [3, 5, 1];
alert( Math.max(...arr) ); // 5 (spread turns array into a list of arguments)
We also can pass multiple iterables this way:
let arr1 = [1, -2, 3, 4];
let arr2 = [8, 3, -8, 1];
alert( Math.max(...arr1, ...arr2) ); // 8
We can even combine the spread operator with normal values:
let arr1 = [1, -2, 3, 4];
let arr2 = [8, 3, -8, 1];
alert( Math.max(1, ...arr1, 2, ...arr2, 25) ); // 25
Also, the spread operator can be used to merge arrays:
let arr = [3, 5, 1];
let arr2 = [8, 9, 15];
*!*
let merged = [0, ...arr, 2, ...arr2];
*/!*
alert(merged); // 0,3,5,1,2,8,9,15 (0, then arr, then 2, then arr2)
In the examples above we used an array to demonstrate the spread operator, but any iterable will do.
For instance, here we use the spread operator to turn the string into array of characters:
let str = "Hello";
The spread operator internally uses iterators to gather elements, the same way as for..of does.
So, for a string, for..of returns characters and ...str becomes "H","e","l","l","o". The list of characters is passed to array initializer [...str].
For this particular task we could also use Array.from, because it converts an iterable (like a string) into an array:
let str = "Hello";
// Array.from converts an iterable into an array
The result is the same as [...str].
But there's a subtle difference between Array.from(obj) and [...obj]:
• Array.from operates on both array-likes and iterables.
• The spread operator operates only on iterables.
So, for the task of turning something into an array, Array.from tends to be more universal.
Summary
When we see "..." in the code, it is either rest parameters or the spread operator.
There's an easy way to distinguish between them:
• When ... is at the end of function parameters, it's "rest parameters" and gathers the rest of the list of arguments into an array.
• When ... occurs in a function call or alike, it's called a "spread operator" and expands an array into a list.
Use patterns:
• Rest parameters are used to create functions that accept any number of arguments.
• The spread operator is used to pass an array to functions that normally require a list of many arguments.
Together they help to travel between a list and an array of parameters with ease.
All arguments of a function call are also available in "old-style" arguments: array-like iterable object.
Page structure
Terms
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# Diagonalization method by Cantor (2)
I asked a while ago a similar question about this topic. But doing some exercises and using this stuff, I still get stuck. So I have a new question about this topic. (Here is the link for the previous question: Cantor diagonalization method for subsequences).
Now my new question: Suppose I have a sequence $(x_n)$, a set $K$ and a function $f$ and we define for all $l \in \mathbb{N}$ the set $M_l:= K\cap \{x|f(x)\le l\}$ with $f< \infty$ on $K$. I've proved that for a fixed $l$ there's a subsequence $(x_{n_k})$ converging on $M_l$, denote this by $(x^l_{n_k})$. First it's clear that $M_l \subset M_{l+1}$ and I want to show that there's a subsequence which converges on $\cup_{l\ge1} M_l = K$.
So I have different subsequences $(x^1_{n_k}),\dots,(x^p_{n_k}),\dots$ and define the diagonal sequence as ${x^{\phi(l)}_{n_{\phi(l)}}}$ where $\phi(l)$ is the $l$-th element of $n_k$ (just the diagonal sequence). Is it enough to say, since $(M_l)$ is increasing the sequence converges on $K=\cup_{l\ge1} M_l$?
I'm sorry, but I'm just confues about picking the right sequence and to prove that it is the right sequence using Cantor. Thank you for your help
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(1) What does "converging on $M_l$" mean? (2) What does $\phi(l)$ mean? You define it as the $l$-th element of $M_l$. So are the elements of $M_l$ integers? To me there are difficulties of exposition. In general, one can use diagonalization to produce subsequences with desired properties, and if your notation is clarified (fewer symbols?) I am sure an answer can be produced, indeed you can produc an answer. – André Nicolas Feb 10 '12 at 20:22
@ André Nicolas: I'm very sorry! That was absolutely my fault! I correct $\phi(l)$. This function just should take the "diagonal elements" and hence construct the diagonal sequence. Convergence on $M_l$ means, $x_n(t)$ converge for all $t\in M_l$ to $x(t)$. The sequence $(x_n)$ is a sequence of (measurable) functions. I didn't mention it, because I get stuck at the diagonal procedure. But sorry for the unclarity – user20869 Feb 10 '12 at 21:46
OK, I had assumed you meant a numerical sequence. Now that the indices are fixed, the answer of David Giraudo may settle things. You certainly want the subsequences to be nested. You may need some version of uniform convergence. – André Nicolas Feb 10 '12 at 21:57
Why should I need a version of uniform convergence? – user20869 Feb 10 '12 at 22:23
I do not know the details of the actual problem. However, although the $M_l$ are probably well-behaved (compact), that does not say that their union is. There are too many counterexamples around to be confident, without the details, that there are no issues. – André Nicolas Feb 10 '12 at 22:31
Denote by $(x_{\varphi_l(k)})$ a subsequence which works for $M_l$. In fact, you have to construct these subsequence by induction, in order to make $(x_{\varphi_{l+1}(k)})$ a subsequence of $(x_{\varphi_l(k)})$. Then we put $x_{n_k}=x_{\varphi_k(k)}$. Now we are sure that the sequence $(x_{n_k})_{k\geq N(j)}$ is a subsequence of $(x_{\varphi_j(k)})_{kgeq N(j)}$ for some integer $N(j)$.
It's important that the subsequences are nested, otherwise it may not work. For example, we assume that for $l$ even only a subsequence of the form $(x_{2k})$ work and for $l$ odd a subsequence of the form $(x_{2k+1})$. Then $x_{n_{\varphi(2l)}}^{\phi(2l)}=x_{4l}$ and $x_{n_{\varphi(2l+1)}}^{\phi(2l+1)}=x_{2l(2l+1)+1}$ so the sequence of indexes is not increasing.
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@ Davide Giraudo: Thanks a lot for your answer. But I think I don't really understand the diagonal procedure. If I have choosen the diagonal sequence, I must show, that this sequence works, right? So generally do I have to choose, in a diagonal procedure, always $(x_{\varphi_{l+1}(k)})$ as a subsequence of $(x_{\varphi_{l}(k)})$? And if so, why? Thanks for your explanations – user20869 Feb 10 '12 at 21:51
It's the example I gave at the end: if you choose the subsequences independently you will encounter some problems. – Davide Giraudo Feb 10 '12 at 21:58
Ok, so probably my question was not that clear. I wanted to know if it is sufficient if the subsequences are nested,i.e if the subsequences are nested then the diagonal procedure works. – user20869 Feb 10 '12 at 22:25
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# MakeUnaryCompiledFunction(S, D, I)¶
compiledFunction(expr, x) returns a function f: D -> I defined by f(x) == expr. Function f is compiled and directly applicable to objects of type D.
unaryFunction(a) is a local function
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Help in proving the identity $\frac d {dt}n_s = - \kappa_st$
I want to prove the identity $\frac d {dt}n_s = - \kappa_st$, where $n_s$ is the signed normal, $t$ is the tangent and $k_s$ is the signed curvature of a unit speed plane curve $\gamma$.
I know that $t \cdot n_s = 0$ so by differentiating I get $\frac d {ds} t \cdot n_s + t \cdot \frac d {ds} n_s = 0$. Now inserting the identity $\frac d {ds} t = \kappa_s n_s$ I get that $\kappa_s + t \cdot \frac d {ds} n_s$ which implies $-\kappa_s = t \cdot \frac d {ds} n_s$.
How can I get that $\frac d {ds}n_s = - \kappa_st$ from this ?
I assume that by $\ n_s$ and $\ t_s$, you mean that everything is parametrised by arc length - so may I first suggest that you set everything to be parametrised by the variable$\ s$ for added clarity in your question (then we will be using operator $\frac{d}{ds}$ instead).
You're close, all you need is one more step. You have,
$\ t.n'=-\kappa$
Then simply note that,
$\ n.n = 1 \implies n'.n=0$
That is, $\ n'$ is parallel to unit tangent$\ t$ (since it is orthogonal to the normal which is orthogonal to the tangent - and we are working in the plane!). So write,
$\ n' := \alpha t$
Plug this into what you already have,
$\alpha=-\kappa$
Giving the desired result,
$\ n'=-\kappa t$
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Change the chapter
Question
How far apart are two conducting plates that have an electric field strength of $4.50\times 10^{3}\textrm{ V/m}$ between them, if their potential difference is 15.0 kV?
$3.33\textrm{ m}$
Solution Video
# OpenStax College Physics Solution, Chapter 19, Problem 16 (Problems & Exercises) (0:39)
Rating
2 votes with an average rating of 5.
## Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dychko. Two parallel conducting plates have an electric field between them of 4.50 times 10 to the 3 volts per meter and a potential difference of 15.0 kilovolts, which is 15.0 times 10 to the 3 volts. Now electric field between two conducting parallel plates is the potential difference divided by the distance by which they are separated and we can solve for this separation d by multiplying both sides by d over E and so the separation is voltage divided by electric field and this works out to 3.33 meters.
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# Tag Info
7
you are missing a big factor, the plates are not moving due to the momentum of an initial impulse. They are being actively moved by the push and pull of mantle convection. Much like how icebergs are pulled along by ocean currents. the iceberg analog however breaks down because icebergs melt before they can do much complex interaction, where as continental ...
7
Right, there are a lot of misconceptions about this. They are mostly to do with the difference between the magnitude of carbon storage and the rate of carbon uptake, and also the difference between whole site carbon storage and above ground biomass carbon storage. There is also ongoing research taking place which (as you might imagine) takes a very long time ...
6
Probability is used in weather forecasting. I will only highlight some examples due to my lack of knowledge in some areas. Before initializing a forecast model the data needs to be assimilated. That means we need to somehow put measurements from ground based stations, satellites etc. into the model and fit this data to the grid. We then have an "...
6
The problem with the cancelling you suggested is that it is imposed globally for the PDE and not locally (just on the thin layer). Usually when we talk about vertically layered media we use a plane wave $\exp(i(\mathbf{k}\cdot\mathbf{x}-wt))$ trial solution (you can plug this in to confirm that it satisfies the PDE) and match boundary conditions at the ...
5
You need to use the following line in your model setup: COORDINATES SPHERICAL Otherwise SWAN will try to calculate wave properties on a curvilinear grid with coordinates provided in meters. In your present case, the computational area is 0.47 m x 0.24 m instead of degrees longitude and latitude.
4
I think you are asking a question with a variety of different constraints. I'll tackle a couple of them. What is the simplest atmospheric model to operate? That would be the Zero-dimensional energy balance model. It has almost zero resolution and no temporal capacity. What is an atmospheric model that can be easily installed and run? The Weather ...
4
This might be product specific, but if it's anything like my MODIS experience this refers to start and end latitude and longitude. Are you looking at a small region on the Mississippi/Louisiana border?
3
We need both paleodata and models, in fact the relationship between paleodata and the climate at a certain time is typically evaluated through models. Paleodata does not actually measure climate observables directly (tempreture/pressure/salinity/other) but instead is (e.g.) a geochemical quantity related to one of those. For example, the two most ...
2
The suggestion above to check historical data to get an idea of the time to peak/ time of concentration, and a get sense of the general storm response time is a good one. There are also empirical methods to relate the catchment slope, runoff coefficient, etc. to the time to peak/ time of concentration, which sounds like what you are looking for. You can find ...
2
Yes, it is possible. The documentation for Hydrus-3D give you the syntax (http://www.pc-progress.com/en/OnlineHelp/HYDRUS3/Hydrus.html?RunningComputationalModulesinaBa.html) In brief here is what you need to do: In the installation folder of Hydrus, create a text file, name it run.bat You need to write two lines of code for every simulation you wish to ...
2
Welcome to StackExchange SE! I don't think I fully understand your question, but I'll try to answer it. Reanalysis data is the use of weather models and data assimilation to piece back the weather. It is a 4-dimensional dataset (Latitude, longitude, pressure levels, and time). You can do whatever you wish with that data (provided you follow the legal ...
2
The word "hydrological" is meant to indicate that the model only includes a sort of bulk description of the water: where it is, how much is there, and where it flows. It generally only includes length and time scales of interest to the system in question. So, for ground water models, this may be tens of meters and longer in horizontal direction, and hours ...
2
Weather forecasting (meteorology) is not climate modelling. As such, a meteorological model to forecast temperatures over years is not the right tool for the job. But there is indeed a very, very simple climate "model" that only accounts for CO2 and temperature. You find it here: https://scied.ucar.edu/simple-climate-model with a little guide: https://...
1
Eureka! I and my prof figured it out. We found something interesting - that the energy density actually depends on a number of internal variables, namely, the displacements of the outer and inner walls, and porosity (which we already knew). So, when we specified the outer displacement to always be a fraction of the radius of the outer sphere, that seemed to ...
1
I would be surprised if the rainfall-runoff relationship would show a linear relationship for extreme events. I would consider that to be a blanket approach, if you mean using a constant coefficient. So consider using a higher order relationship. Another thing to consider is whether the rainfall runoff relationship depends on the soil moisture. Maybe check ...
1
It sounds as though you have ideas on the tools you'd like to use, but not on an actual question that you want to use them to answer. You probably need to either come up with such a question, or focus instead on methodology - in studying the tools and the effectiveness of using them in a given scenario. For nearly any modelling project, its viability will ...
1
The simplest way I've found in Python is to use the getvar function with the 'ua,' 'va,' or 'wa' variables from the WRF-Python module. Alternatively, you can take the midpoint between the staggers. Edit: For example, using the WRF-Python module, you can get the destaggered wind variables with the following code import netcdf4 as nc import wrf f=nc.Dataset('...
1
Oh, I think I finally figured out what those values are. The data is tornado data and I believe the slat, slon are Start locations (of the tornado) and the elat, elon are End locations for the tornado. This only became apparent when I added those locations to a Google map and saw the distances. Here is a map with the two locations on it (obviously the ...
1
Yes, using GMPES. E.g., you can look at the USGS shakemap which includes some measure of intensities such as PGA/PGV. Just choose an appropriate cut-off value and you have your area.
1
Google's "Web Mercator" projection is a very different beast from the Mercator projection. The former provides a coordinate system for the area you are viewing, oriented in the way you rotated the view. The formulas you put in your original question are probably more computationally efficient when used for this purpose, but they do not foster ...
1
The Mercator Projection has been used and it is used since the XVI century due to it maintain the course, rhumb or route. So it is useful for navigation. As the Earth is approached by a ellipsoid, then you get the formulas that you saw on Mercator Projection. This transformation it is complicated for a ellipsoid, as the Earth curvature it is not a constant, ...
Only top voted, non community-wiki answers of a minimum length are eligible
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# How to work around incorrect tooltips in plots exported to HTML
This is an issue that I always wanted to investigate but haven't gotten around to solving yet: The tooltips for contour lines in a plot like this can be very useful in the Notebook interface:
ContourPlot[3 x y^2 - x^3, {x, -1, 1}, {y, -1, 1},
ContourShading -> False, Contours -> Table[r^3, {r, -1.4, 1.4, .1}]]
However, when I export the Notebook containing the plot as an HTML file, I get the following image:
It appears to have tooltips, as the screen shot shows (the mouse arrow isn't captured, but the yellow tooltip is). However, those tooltips are generated by an image map that defines some very inaccurate target areas for the contour values to be displayed when the mouse hovers over them. The result is worse than having no tooltips at all, because the displayed values are plain wrong and hence confusing most of the time. To see the exported HTML in real life, look at this URL exported from Mathematica 7 (version 8 does the same thing, I've just had this problem for a long time...).
My first question would be: what is the best way to disable the use of image maps in the HTML generated by the Save As HTML operation? My current thought is that I'll have to post-process the generated HTML file to remove the usemap part of all image tags. But maybe someone knows a better way.
I would also like to keep the tooltips for the graphics in the Notebook.
The second question is how one could export a plot like the above so that it can be used in a HTML file while preserving correct tooltips. Here, my thought would be to export to SVG and add the tooltips to the generated lines in a post-processing step (the SVG file seems to contain enough information to allow this). However, I'm again hoping that someone has a better idea.
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# queue of unique_ptr issue
This topic is 754 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi,
I'm building a simple engine, and my Node class has a member
queue<unique_ptr<Action>> m_actions
that maintains a queue of actions to be processed by this node. It also has a functions to add another action to perform:
inline void Node::addAction (std::unique_ptr<Action> action) {
action ->setTarget(this);
m_actions. push (std::move(action));
}
As you can see, I'm using move semantics to pass the ownership of the action to the Node. This seems to work fine in Visual Studio, but then I tried to build this in Xcode and I'm getting the following error:
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/memory:1466:36: No matching constructor for initialization of 'std::__1::unique_ptr<Action, std::__1::default_delete<Action> >'
which seems to imply that for some reason the unique_ptr is getting copied somewhere down the line. Does anyone know what is going on here and the reason of the difference behaviours for the two compilers?
Thanks!
Fabrizio
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/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/memory:1466:36: No matching constructor for initialization of 'std::__1::unique_ptr >'
which seems to imply that for some reason the unique_ptr is getting copied somewhere down the line. Does anyone know what is going on here and the reason of the difference behaviours for the two compilers?
How did you reach that decision? This is a compiler error, the program isn't running at this point in time, it doesn't even have any pointer instance currently.
In particular, this error is a template instantiation error. unique_ptr is implemented using a template, and apparently xcode expects a certain constructor to be available in the template that the template code of the compiler library doesn't provide.
In general, xcode is more precise in interpreting the standard, and/or doesn't provide many bells and whistles to make things work more smoothly.
I don't know the error, but first thing you should do is throw it in a search engine and see what it comes up with. 99% chance that someone has run into it in the past, and figured out what to do. Secondly, some statement in your code indirectly triggers need for the non-existing constructor. Check that area carefully. Is it really as the standard wants it?
You can also try to make a very small example that demonstrates the problem, perhaps while experimenting with it, you find the cause, or else, you have something you can show to other people and ask about.
Edited by Alberth
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Assuming this is the exact code you've been using, the argument you are taking is a unique pointer by value. Any statement calling this function will have to pass a unique pointer by value, i.e. copy it.
I'll take an educated guess and say marking it as inline causes this difference in behaviour. The inlining would give the ability to replace the function argument with the original statement/variable, thus never having to copy it. The behaviour may differ because of differences in rules for when inlining is possible. I can't say this for sure though, especially since I haven't encountered it myself :D
You could either pass the unique pointer by reference, or by right-value reference and move semantics upon calling the function. My preference would go to the latter, since with a left-value reference it has side-effects for the caller side (it can no longer use the unique pointer).
Edited by AthosVG
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/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/include/c++/v1/memory:1466:36: No matching constructor for initialization of 'std::__1::unique_ptr >'
which seems to imply that for some reason the unique_ptr is getting copied somewhere down the line. Does anyone know what is going on here and the reason of the difference behaviours for the two compilers?
How did you reach that decision? This is a compiler error, the program isn't running at this point in time, it doesn't even have any pointer instance currently.
I suspect he had in mind the fact that unique_ptr isn't copyable - and that fact is enforced by the compiler. The constructor that isn't available is the copy constructor. Edited by Oberon_Command
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I suspect he had in mind the fact that unique_ptr isn't copyable - and that fact is enforced by the compiler. The constructor that isn't available is the copy constructor.
I realized he was trying to copy the value when I read the response of AthosVG, I was too focussed on the error message, and totally missed that :(
Not sure how you derived it being the copy-constructor from the error message (I have no access to xcode files).
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Not sure how you derived it being the copy-constructor from the error message (I have no access to xcode files).
Yay for obtuse C++ compile errors! The (deleted) copy constructor needs to be invoked to pass the parameter by value, and the error message references a missing constructor:
No matching constructor for initialization of 'std::__1::unique_ptr<Action, std::__1::default_delete<Action> >'
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I have never used xcode before. Are there any more lines to the error message? Usually, in Visual Studio or gcc, if there is an error like this, it would give some context about where it was trying to copy the unique_ptr. People shouldn't have to blindly guess about this.
Yay for obtuse C++ compile errors! The (deleted) copy constructor needs to be invoked to pass the parameter by value, and the error message references a missing constructor:
You do not need a copy constructor to pass a parameter by value.
Edited by Pink Horror
##### Share on other sites
You do not need a copy constructor to pass a parameter by value.
Then how would it be copied?
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# Nonparametric ICC
Anyone can suggest me a nonparametric version of the intraclass correlation coefficient (ICC)? My model have the following details: two-way random effects model, consistency, average ratings as unit. I am using irr R-cran package.
thank you
-
The ICC is not, in itself, a parametric statistic. You can conceive of the ICC regardless of the distribution of the data. What exactly do you mean by "nonparametric ICC"? – Macro May 25 '12 at 14:51
I have seen that in some studies the ratings are firstly examined to check a possible violations from normality, and then ICC is computed. Is it mandatory? What if the ratings are far from normality? – netrunner May 25 '12 at 15:17
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# Testing your PHP Codebase with EnhancePHP
You know it; I know it. We should be testing our code more than we do. Part of the reason we don’t, I think, is that we don’t know exactly how. Well, I’m getting rid of that excuse today: I’m teaching you to test your PHP with the EnhancePHP framework.
## Meet EnhancePHP
I’m not going to try to convince you to test your code; and we’re not going to discuss Test Driven Development, either. That’s been done before on Nettuts+. In that article, Nikko Bautista explains exactly why testing is a good thing and outlines a TDD workflow. Read that sometime, if you aren’t familiar with TDD. He also uses the SimpleTest library for his examples, so if you don't like the look of EnhancePHP, you might try SimpleTest as an alternative.
As I said, we’ll be using the EnhancePHP. It’s a great little PHP library—a single file—that offers a lot of testing functionality.
We’re going to be building a really simple Validation class to test. It won’t do too much: just return true if the item passes validation, or false if it doesn’t. So, set up a really simple little project:
We’ll do this is a semi-TDD fashion, so let’s start by writing a few tests.
## Writing Tests
Out little class is going to validate three things: email addresses, usernames, and phone numbers.
But before we get to writing actual tests, we’ll need to set up our class:
This is our start; notice that we’re extending the class \Enhance\TestFixture. By doing so, we let EnhancePHP know that any public methods of this class are tests, with the exception of methods setUp and tearDown. As you might guess, these methods run before and after all your tests (not before and after each one). In this case, our setUp method will create a new Validation instance and assign it to a property on our instance.
By the way, if you’re relatively new to PHP, you might not be familiar with that \Enhance\TestFixture syntax: what’s with the slashes? That’s PHP namespacing for you; check out the docs if you aren’t familiar with it.
So, the tests!
Let’s start by validating email addresses. As you’ll see, just doing a basic test is pretty simple:
We simply call the method we want to test, passing it a valid email address, and storing the $result. Then, we hand $result to the isTrue method. That method belongs to the \Enhance\Assert class.
We want to make sure our class will reject non-email addresses. So, let’s test for that:
This introduces a pretty cool feature of EnhancePHP: scenarios. We want to test a bunch of non-email addresses to make sure our method will return false. By creating a scenario, we essentially wrap an instance of our class in some EnhancePHP goodness, are write much less code to test all our non-addresses. That’s what $val_wrapper is: a modified instance of our Validation class. Then, $val_email is the scenario object, somewhat like a shortcut to the validate_email method.
Then, we’ve got an array of strings that should not validate as email addresses. We’ll loop over that array with a foreach loop. Notice how we run the test: we call the with method on our scenario object, passing it the parameters for the method we’re testing. Then, we call the expect method on that, and pass it whatever we expect to get back.
Finally, we call the scenario’s verifyExpectations method.
So, the first tests are written; how do we run them?
## Running Tests
Before we actually run the tests, we’ll need to create our Validation class. Inside lib.validation.php, start with this:
Now, in test.php, we’ll pull it all together:
First, we’ll require all the necessary files. Then, we call the runTests method, which finds our tests.
Next comes the neat part. Fire up a server, and you’ll get some nice HTML output:
Very nice, right? Now, if you’ve got PHP in your terminal, run this is in the terminal:
EnhancePHP notices that you’re in a different environment, and adjusts its output appropriately. A side benefit of this is that if you’re using an IDE, like PhpStorm, that can run unit tests, you can view this terminal output right inside the IDE.
You can also get XML and TAP output, if that’s what you prefer, just pass \Enhance\TemplateType::Xml or \Enhance\TemplateType::Tap to the runTests method to get the appropriate output. Note that running it in the terminal will also produce command-line results, no matter what you pass to runTests.
### Getting the Tests to Pass
Let’s write the method that causes our tests to pass. As you know, that’s the validate_email. At the top of the Validation class, let’s define a public property:
I’m putting this in a public property so that if the user wants to replace it with their own regex, they could. I’m using this simple version of an email regex, but you can replace it with your favourite regex if you want.
Then, there’s the method:
Now, we run the tests again, and:
## Writing More Tests
Time for more tests:
Let’s create some tests for usernames now. Our requirements are simply that it must be a 4 to 20 character string consisting only of word characters or periods. So:
This is very similar to our reject_bad_email_addresses function. Notice, however, that we’re calling this get_scenario method: where’s that come from? I’m abstracting the scenario creation functionality into private method, at the bottom of our class:
We can use this in our reject_bad_usernames and replace the scenario creation in reject_bad_email_addresses as well. Because this is a private method, EnhancePHP won’t try to run it as a normal test, the way it will with public methods.
We’ll make these tests pass similarly to how we made the first set pass:
This is pretty basic, of course, but that’s all that’s needed to meet our goal. If we wanted to return an explanation in the case of failure, you might do something like this:
Of course, you might also want to check if the username already exists.
Now, run the tests and you should see them all passing.
### Phone Numbers
I think you’re getting the hang of this by now, so let’s finish of our validation example by checking phone numbers:
You can probably figure out the Validation method:
Now, we can run all the tests together. Here’s what that looks like from the command line (my preferred testing environment):
## Other Test Functionality
Of course, EnhancePHP can do a lot more than what we’ve looked at in this little example. Let’s look at some of that now.
We very briefly met the \Enhance\Assert class in our first test. We didn’t really use it otherwise, because it’s not useful when using scenarios. However, it’s where all the assertion methods are. The beauty of them is that their names make their functionality incredibly obvious. The following test examples would pass:
• \Enhance\Assert::areIdentical("Nettuts+", "Nettuts+")
• \Enhance\Assert::areNotIdentical("Nettuts+", "Psdtuts+")
• \Enhance\Assert::isTrue(true)
• \Enhance\Assert::isFalse(false)
• \Enhance\Assert::contains("Net", "Nettuts+")
• \Enhance\Assert::isNull(null)
• \Enhance\Assert::isNotNull('Nettust+')
• \Enhance\Assert::isInstanceOfType('Exception', new Exception(""))
• \Enhance\Assert::isNotInstanceOfType('String', new Exception(""))
There are a few other assertion methods, too; you can check the docs for a complete list and examples.
### Mocks
EnhancePHP can also do mocks and stubs. Haven’t heard of mocks and stubs? Well, they aren’t too complicated. A mock is a wrapper for object, that can keep track of what methods are called, with what properties they are called, and what values are returned. A mock will have some test to verify, as we’ll see.
Here’s a small example of a mock. Let’s start with a very simple class that counts:
We have one function: increment, that accepts a parameter (but defaults to 1), and increments the $num property by that number. We might use this class if we were building a scoreboard: Now, we want to test to make sure that the Counter instance method increment is working properly when the Scoreboard instance methods call it. So we creat this test: Notice that we start by creating $home_counter_mock: we use the EnhancePHP mock factory, passing it the name of the class we’re mocking. This returns a “wrapped” instance of Counter. Then, we add an expectation, with this line
Our expectation just says that we expect the increment method to be called.
After that, we go on to create the Scoreboard instance, and call score_home. Then, we verifyExpectations. If you run this, you’ll see that our test passes.
We could also state what parameters we want a method on the mock object to be called with, what value is returned, or how many times the method should be called, with something like this:
I should mention that, while with and times will show failed tests if the expectations aren’t meant, returns doesn’t. You’ll have to store the return value and use an assertion to very that. I’m not sure why that’s the case, but every library has its quirks :). (You can see an example of this in the library examples in Github.)
### Stubs
Then, there are stubs. A stub fills in for a real object and method, returning exactly what you tell it to. So, let’s say we want to make sure that our Scoreboard instance is correctly using the value it receives from increment, we can stub a Counter instance so we can control what increment will return:
Here, we’re using \Enhance\StubFactory::createStub to create our stub counter. Then, we add an expectation that the method increment will return 10. We can see that the result it what we’d expect, given our code.
For more examples of mocks and stub with the EnhancePHP library, check out the Github Repo.
## Conclusion
Well, that’s a look at testing in PHP, using the EnhancePHP framework. It’s an incredibly simple framework, but it provides everything you need to do some simple unit testing on your PHP code. Even if you choose a different method/framework for testing your PHP (or perhaps roll your own!), I hope this tutorial has sparked an interest in testing your code, and how simple it can be.
But maybe you already test your PHP. Let us all know what you use in the comments; after all, we're all here to learn from each other! Thank you so much for stopping by!
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# Curve fitting with two exponential functions
(Bodhibrata Mukhopadhyay) #1
Kindly help me in fitting a function which is formed by the sum of two exponentials to a given data set using cvx.
Given data set (x_i,y_i) i=1,2,…100 ;
Function expression = a_1 e^{b_1}x + a_2 e^{b_2}x
Cost function : \epsilon (a_1,b_1,a_2,b_2) = \sum_{1}^{100} [ (y_i) - a_1 e^{b_1}x_i + a_2 e^{b_2}x_i]^2
(Michael C. Grant) #2
Your problem is non-convex. Please read the FAQ.
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Which of the following sets of quantum numbers are not permitted?
$1 : \text{ } 2 , 1 , 1 , - \frac{1}{2}$ $2 : \text{ } 2 , 2 , 1 , + \frac{1}{2}$ $3 : \text{ } 3 , 2 , 0 , - \frac{1}{2}$ $4 : \text{ } 4 , 3 , 2 , - \frac{1}{2}$ $5 : \text{ } 4 , 2 , - 3 , + \frac{1}{2}$
Mar 25, 2018
Here's what I got.
Explanation:
As you know, the four quantum numbers that we use to describe the location and the spin of an electron in an atom are related as follows:
So all you have to do here is to look at which values are permitted for the four quantum numbers given to you for each set.
$n = 2 , l = 1 , {m}_{l} = 1 , {m}_{s} = - \frac{1}{2} \text{ " " } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$
This is a valid set because all four quantum numbers have permitted values. In fact, this quantum number set describes an electron located in the second energy shell, in the $2 p$ subshell, in one of the three $2 p$ orbitals, that has spin-down.
$\textcolor{w h i t e}{a}$
$n = 2 , l = 2 , {m}_{l} = 1 , {m}_{s} = + \frac{1}{2} \text{ " " } \textcolor{red}{\times}$
This is not a valid set because the value of the angular momentum quantum number, $l$, cannot be equal to the value of the principal quantum number, $n$.
$n = 2 \implies l = \left\{0 , 1 , \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}\right\}$
$\textcolor{w h i t e}{a}$
$n = 3 , l = 2 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2} \text{ " " } \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}$
This is a valid set because all four quantum numbers have permitted values. This quantum number set describes an electron located in the third energy shell, in the $3 d$ subshell, in one of the five $3 d$ orbitals, that has spin-down.
$\textcolor{w h i t e}{a}$
At this point, you should be able to look at the last two sets and say whether or not they can describe an electron in an atom, so I'll leave them to you as practice.
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# Interesting modulus
Algebra Level 3
Find the sum of all the integral roots of the equation $|x −1|+|x|+|x +1|=x +2.$
×
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# Solving separable differential equations by formal computations When solving a separable differential equation we move from the original equation N(y)(dy)/(dx)=M(x) to N(y)dy=M(x)dx Then we integrate both sides. My question is how precise is the expression N(y)dy=M(x)dx ? is it a formal writing to simplify computation and get quickly into integrating both sides or is it a precise mathematical expression that has a precise mathematical meaning but goes beyond an introductory course on differential equations? Thanks for your help !
Solving separable differential equations by formal computations
When solving a separable differential equation we move from the original equation
$N\left(y\right)\frac{dy}{dx}=M\left(x\right)$
to
$N\left(y\right)dy=M\left(x\right)dx$
Then we integrate both sides. My question is how precise is the expression $N\left(y\right)dy=M\left(x\right)dx$ ? is it a formal writing to simplify computation and get quickly into integrating both sides or is it a precise mathematical expression that has a precise mathematical meaning but goes beyond an introductory course on differential equations? Thanks for your help !
You can still ask an expert for help
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kartonaun
It is more like integrating the first equation with respect to x on both sides. And then, taking y is a function of x which it is as they both are related by this equation, let say
$y=f\left(x\right)$ so,
$dy={f}^{{}^{\prime }}\left(x\right)dx$ or $\frac{dy}{dx}={f}^{{}^{\prime }}\left(x\right)$
Replacing this in the first equation we have
$N\left(y\right){f}^{{}^{\prime }}\left(x\right)dx$ = $N\left(y\right)dy=M\left(x\right)dx$
as $dy={f}^{{}^{\prime }}\left(x\right)dx$
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The Riemann Hypothesis [duplicate]
(EDIT: I've marked this question as answered in order that I can go away and come up with a better one. Thanks to everybody for the helpful answers.)
Is it possible to describe the RH in language comprehensible to a non-mathematician?
In my experience (as a non-mathematician) the answer is a definite 'no'. But before I give up on it I thought I'd ask for some views.
By 'explain' I do not mean explain fully. I mean explain the basic structure of the hypothesis. Something like - we feed values into a function and the output is plotted to create a landscape and a critical line appears and we conjecture that all zero outputs fall on this line. I expect even this is wrong but it's this level of explanation that I'm after, with just a bit more detail added.
I've spent a long time trying to understand the relationship between the inputs and outputs of the Zeta calculation and got nowhere. Yet I'm not stupid. There's something I'm missing but none of the explanations I've been offered allow me to understand what it is.
I know a bit about the primes but must ask for answers in a more or less natural language or will not be able to cope. I know such requests annoy mathematicians but I can only ask.
I have read Du Sautoy, Derbyshire and much more but never found an explanation that doesn't assume I'm a graduate mathematician. I need a children's book on the topic that describes it in very general terms.
Is such a thing possible?
EDIT: Vincent has suggested a similar but perhaps interestingly different question. This is - is there an intuitive explanation why the Riemann zeta FUNCTION (rather than hypothesis) contains interesting information about the distribution of primes in language comprehensible to a non-mathematician?
My intuition is that the explanation would be the harmonic series and its role in determining the distribution of primes. I'll hesitatingly ask if this makes any sense. Or are there other reasons? As it happens I'm coming at this as a musician and have an interest in the musical aspect of the distribution or primes, or rather in the distribution of non-primes that determines the primes.
EDIT 2: I see that similar questions have been asked previously. Thanks to the members who have provided links to them. I've been checking them and so far they have been helpful but not so much as to persuade me to take this one down. This may change as I keep reading and I'll withdraw this one if so.
marked as duplicate by Did, davidlowryduda♦Feb 8 at 13:03
• Even though I am a mathematician I VERY MUCH empathize with the poser of this question. I think the best way too understand the importance of/interest in/usefulness of the RH is through the connection to the prime number counting function as explained by AlgebraicsAnonymous below and more elaborately in the accepted anwer to the question pregunton links to above. But then of course there is a follow up question: is there an accessible way to see that the the RH relates to prime numbers in the first place? (ctd in next comment) – Vincent Feb 8 at 13:03
• (Apparently pregunton removed their comment with the link to the other question. I have now put the link below.) What I wanted to say: once we are at that point we might ask a simpler (but still very hard) question: is there an intuitive explanation why the Riemann zeta FUNCTION (rather than hypothesis) contains interesting information about the distribution of primes in language comprehensible to a non-mathematician? I would be really interested in an answer to that since even if I am mathematician enough to follow the proof, intuition fails me here. – Vincent Feb 8 at 13:07
• Here is the link to the other question again: math.stackexchange.com/questions/7981/…. – Vincent Feb 8 at 13:10
• @Vincent - Thanks. To be honest even some sympathy is appreciated. I've already learnt a little from the answers so progress is being made. I've followed your link and found it helpful, but not so much I want to close this one. I like you secondary question about the function and may edit my question to include it. I assume the connection is the harmonic series but this is grasping at straws. – PeterJ Feb 8 at 13:19
• Clicking some more 'linked question'-links I found the following question and answer that come quite close to what I had in mind: math.stackexchange.com/questions/2144940/…. It sort of answers the question 'what does the Riemann-zeta function at $s = 1$ (so no need to write the $s$ at all) have to do with the distribution of prime numbers, from which it is a very small step to my question of 'what does the Riemann-zeta function with the distribution of prime numbers?' – Vincent Feb 8 at 16:26
The prime numbers have density approximately $$1/\ln(n)$$, in the sense that the probability that a number $$n$$ is prime is approximately $$1/\ln(n)$$. From this, one would guess that if $$\pi(x)$$ (for a number $$x> 0$$) denotes the number of primes below $$x$$, we would have $$\pi(x) \approx \int_2^x \frac{1}{\ln(t)} dt.$$ The Riemann hypothesis asserts that the error in this approximate equation does not exceed $$\frac{\pi}{8} \sqrt{x} \ln(x)$$.
In short: The Riemann hypothesis asserts that the prime numbers are very strictly distributed according to the density $$1/\ln(n)$$.
• Thank you. So where there is a non-trivial zero does this represent a distribution or output that is bang in line with the PMT? – PeterJ Feb 8 at 12:46
• @PeterJ Did you ever see explicit formulas $\psi(x) = x-\sum_\rho \frac{x^\rho}{\rho} - C-\sum_{k \ge 1} \frac{x^{-2k}}{-2k}$ where $\psi(x)$ is just a different version of $\pi(x)$. That sum over infinitely many non-trivial zeros can be truncated as $\psi(x) = x-\sum_{|Im(\rho)|< T} \frac{x^\rho}{\rho}+C+O(x^{-2}+\frac{x^{\sigma_0}\log^2 x}{T})$, there are about $AT\log T$ terms in the sum and $\sigma_0 =\sup\Re(\rho)$. It's technical but from this it is perfectly clear how $\Re(\rho)$ determinates the growth of $|\psi(x)-x|$, that's how Riemann's 1859 paper revolutioned the primes – reuns Feb 8 at 12:53
• @reuns - Many thanks, but this over my head. – PeterJ Feb 8 at 15:59
It's tempting to try to satisfy you with a "simple" claim known to be equivalent to the Riemann hypothesis. To my mind, the best example is this: for an integer $$n\ge 5041$$, the sum of the positive factors of $$n$$ is less than $$ne^\gamma\ln\ln n$$. As long as I can teach you what natural exponentials and logarithms are (which isn't even undergrad material), and explain the Euler-Mascheroni constant $$\gamma:=\lim_{n\to\infty}(-\ln n+\sum_{k=1}^n\frac{1}{k})$$, I've technically given you a full understanding of what it means for the hypothesis to be true or false.
The problem with this approach, however, is it doesn't make clear why anyone cares if it's true or false. For that, we need to discuss the consequences of the hypothesis, and I doubt I can prove any interesting ones to you starting from the formulation I just gave, without some heavy-handed material (you know, the kind that proves equivalence with a "non-lay" formulation of the hypothesis).
Nor will I have made clear why anyone thinks it's probably true. After all, its consequences are irrelevant if it's false.
Roughly speaking, the story is this:
• There is a unique way to "analytically continue" the usual $$\sum_{n\ge 1}n^{-s}$$ definition for $$\Re s>1$$ to cover any $$s\in\Bbb C\backslash\{1\}$$. It's unique, because that's a fact about analytic continuation. The hypothesis says this continuation's roots are either negative even integers ("trivial") or solutions of $$\Re s=\frac12$$ (this locus is the critical line).
• The hypothesis is likely to be true partly because so many roots have been identified and they all fit, partly because we can prove lesser results such as there being zero density of nontrivial roots outside the critical line, partly because equivalent claims also have no known counterexamples despite heavy searching, partly because the hypothesis passes a lot of sanity checks, and so on.
• It matters, because, if it's true, we learn a lot more about how prime numbers are distributed and how quickly some important functions grow, among other things.
I don't know, however, how to "dumb it down" to the point where you don't have to learn a lot of new maths to follow the details. Nor would I be surprised if this is unfeasible. After all, there's only so far primary-school maths can get you, and ditto with secondary-school maths, and so on.
• Maybe add the fact that the Euler-Mascheroni constant is just some specific number approximately equal to $\gamma=0.577..$. – quarague Feb 8 at 13:00
• Thanks. Still over my head but I do grasp why RH is important and why we should care. It's an odd situation. I've been trying to understand what can be understood by a layman and what actually requires complex mathematics, but it seems difficult to make this distinction. . – PeterJ Feb 8 at 13:32
• @PeterJ Yes, it's a tricky one. A word of caution: talk in this context about how complicated the mathematics may get, not how complex. Apart from how a grammarian would distinguish these, to a mathematician "complex" refers to the role of complex numbers in this problem. Of course, the hypothesis is fundamentally about complex numbers, even if one can present real-only ("elementary") equivalents such as the inequality I discussed. – J.G. Feb 8 at 13:40
• @JG - Yes, I see that the mistake. It's difficult for a layman not to appear to be a complete idiot in such discussions. It's difficult even to find the right questions. – PeterJ Feb 8 at 15:57
• @PeterJ Oh, don't feel so bad. The history of mathematics is one in which even great minds often didn't or couldn't ask the right questions until surprisingly recently. – J.G. Feb 8 at 16:14
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# Load and Use a Trained Policy¶
In this section you will learn how to extract a trained policy from an experiment snapshot, as well as how to evaluate that policy in an environment.
## Obtaining an experiment snapshot¶
Please refer to this page for information on how to save an experiment snapshot. The snapshot contains data such as:
• The trainer’s setup_args and train_args
• Random seed
• Batch size
• Number of epochs
• And more
• The experiment’s stats
• The environment
• The algorithm (which includes the policy we want to evaluate)
## Extracting a trained policy from a snapshot¶
To extract the trained policy from a saved experiment, you only need a few lines of code:
# Load the policy
from garage.experiment import Snapshotter
import tensorflow as tf # optional, only for TensorFlow as we need a tf.Session
snapshotter = Snapshotter()
with tf.compat.v1.Session(): # optional, only for TensorFlow
policy = data['algo'].policy
# You can also access other components of the experiment
env = data['env']
This code makes use of a Garage Snapshotter instance. It calls cloudpickle behind the scenes, and should continue to work even if we change how we pickle (we used to use joblib, for example).
## Applying the policy to an environment¶
In order to use your newly-loaded trained policy, you first have to make sure that the shapes of its observation and action spaces match those of the target environment. An easy way of doing this is run the policy in the same environment in which it was trained.
Once you have an environment initialized, the basic idea is this:
steps, max_steps = 0, 150
done = False
obs = env.reset() # The initial observation
policy.reset()
while steps < max_steps and not done:
obs, rew, done, _ = env.step(policy.get_action(obs))
env.render() # Render the environment to see what's going on (optional)
steps += 1
env.close()
This logic is bundled up in a more robust way in the rollout() function from garage.sampler.utils. Let’s bring everything together:
# Load the policy and the env in which it was trained
from garage.experiment import Snapshotter
import tensorflow as tf # optional, only for TensorFlow as we need a tf.Session
snapshotter = Snapshotter()
with tf.compat.v1.Session(): # optional, only for TensorFlow
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## Gibbs Free Energy: Spontaneous versus Unstable?
$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$
$\Delta G^{\circ}= -RT\ln K$
$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$
Cowasjee_Sanaea_3E
Posts: 27
Joined: Fri Jul 22, 2016 3:00 am
### Gibbs Free Energy: Spontaneous versus Unstable?
In problem number 9.63 in the textbook it asks:
Determine which of the following compounds are stable with respect to decomposition into their elements under standard conditions at 25 degrees C.
The answer was that those with negative free energies were stable. However my understanding was that being unstable meant spontaneity and therefore the unstable compounds would be negative (as negative Gibbs free energy means spontaneous). Is this not the case because it is talking about decomposition?
Also, how come you don't have to do the full reaction of decomposition equation to find the free energy? And instead just look at the compounds delta G directly?
Kareem
Posts: 18
Joined: Wed Sep 21, 2016 3:00 pm
### Re: Gibbs Free Energy: Spontaneous versus Unstable?
The question is asking about stability in terms of the decomposition, meaning is it stable to decompose into its elements. The questions is basically asking if it will decompose spontaneously or not, so when it decomposes is it stable or not.
ntyshchenko
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm
### Re: Gibbs Free Energy: Spontaneous versus Unstable?
But if the Gibbs free energy of formation of PCl5 is -305 then doesn't that mean that the decomposition of PCl5 is +305 and thus the decomposition of it is thermodynamically unstable?
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For example look at -sin (t). Also related to the tangent approximation formula is the gradient of a function. Send us a message about “Introduction to the multivariable chain rule” Name: Email address: Comment: Introduction to the multivariable chain rule by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. In some cases, applying this rule makes deriving simpler, but this is hardly the power of the Chain Rule. >> Okay, so you know the chain rule from calculus 1, which takes the derivative of a composition of functions. The gradient is one of the key concepts in multivariable calculus. You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! Chain Rule for Multivariable Functions December 8, 2020 January 10, 2019 | Dave. 1. At the very end you write out the Multivariate Chain Rule with the factor "x" leading. Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6472, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6467, Multivariable Chain Rule – Calculating partial derivatives – Exercise 6489, Derivative of Implicit Multivariable Function, Calculating Volume Using Double Integrals, Calculating Volume Using Triple Integrals, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6506, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6460, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6465, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6522, Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6462. D. desperatestudent. Oct 2010 10 0. The idea is the same for other combinations of flnite numbers of variables. The Multivariable Chain Rule Nikhil Srivastava February 11, 2015 The chain rule is a simple consequence of the fact that dierentiation produces the linear approximation to a function at a point, and that the derivative is the coecient appearing in this linear approximation. Alternative Proof of General Form with Variable Limits, using the Chain Rule. We will do it for compositions of functions of two variables. (You can think of this as the mountain climbing example where f(x,y) isheight of mountain at point (x,y) and the path g(t) givesyour position at time t.)Let h(t) be the composition of f with g (which would giveyour height at time t):h(t)=(f∘g)(t)=f(g(t)).Calculate the derivative h′(t)=dhdt(t)(i.e.,the change in height) via the chain rule. Derivative along an explicitly parametrized curve One common application of the multivariate chain rule is when a point varies along acurveorsurfaceandyouneedto・“uretherateofchangeofsomefunctionofthe moving point. Note: we use the regular ’d’ for the derivative. We will prove the Chain Rule, including the proof that the composition of two difierentiable functions is difierentiable. %PDF-1.5 Forums. However, it is simpler to write in the case of functions of the form /Filter /FlateDecode In this section, we study extensions of the chain rule and learn how to take derivatives of compositions of … This is the simplest case of taking the derivative of a composition involving multivariable functions. Solution A: We'll use theformula usingmatrices of partial derivatives:Dh(t)=Df(g(t))Dg(t). The result is "universal" because the polynomials have indeterminate coefficients. In the last couple videos, I talked about this multivariable chain rule, and I give some justification. ������#�v5TLBpH���l���k���7��!L�����7��7�|���"j.k���t����^�˶�mjY����Ь��v��=f3 �ު���@�-+�&J�B$c�jR��C�UN,�V:;=�ոBж���-B�������(�:���֫���uJy4 T��~8�4=���P77�4. Proof of the chain rule: Just as before our argument starts with the tangent approximation at the point (x 0,y 0). Assume that $$x,y:\mathbb R\to\mathbb R$$ are differentiable at point $$t_0$$. It says that. We will put the partial derivatives in the left side of the equation we need to prove. 3 0 obj << The chain rule in multivariable calculus works similarly. ∂w Δx + o ∂y ∂w Δw ≈ Δy. o Δu ∂y o ∂w Finally, letting Δu → 0 gives the chain rule for . If you're seeing this message, it means we're having trouble loading external resources on our website. … multivariable chain rule proof. because in the chain of computations. IMOmath: Training materials on chain rule in multivariable calculus. We calculate th… In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. Both df /dx and @f/@x appear in the equation and they are not the same thing! In this paper, a chain rule for the multivariable resultant is presented which generalizes the chain rule for re-sultants to n variables. In the limit as Δt → 0 we get the chain rule. And some people might say, "Ah! Would this not be a contradiction since the placement of a negative within this rule influences the result. In the section we extend the idea of the chain rule to functions of several variables. For permissions beyond the scope of this license, please contact us. If we compose a differentiable function with a differentiable function , we get a function whose derivative is. stream – Write a comment below! The generalization of the chain rule to multi-variable functions is rather technical. Was it helpful? If we could already find the derivative, why learn another way of finding it?'' Theorem 1. The proof is more "conceptual" since it is based on the four axioms characterizing the multivariable resultant. Dave4Math » Calculus 3 » Chain Rule for Multivariable Functions. I was doing a lot of things that looked kind of like taking a derivative with respect to t, and then multiplying that by an infinitesimal quantity, dt, and thinking of canceling those out. Multivariable Chain Rule SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference Chapter 13.5 of the rec-ommended textbook (or the equivalent chapter in your alternative textbook/online resource) and your lecture notes. ∂u Ambiguous notation 2 Chain rule for two sets of independent variables If u = u(x,y) and the two independent variables x,y are each a function of two new independent variables s,tthen we want relations between their partial derivatives. Found a mistake? In probability theory, the chain rule (also called the general product rule) permits the calculation of any member of the joint distribution of a set of random variables using only conditional probabilities.The rule is useful in the study of Bayesian networks, which describe a probability distribution in terms of conditional probabilities. Calculus-Online » Calculus Solutions » Multivariable Functions » Multivariable Derivative » Multivariable Chain Rule » Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6472. However in your example throughout the video ends up with the factor "y" being in front. %���� Vector form of the multivariable chain rule Our mission is to provide a free, world-class education to anyone, anywhere. This makes it look very analogous to the single-variable chain rule. Free detailed solution and explanations Multivariable Chain Rule - Proving an equation of partial derivatives - Exercise 6472. EXPECTED SKILLS: Be able to compute partial derivatives with the various versions of the multivariate chain rule. I'm working with a proof of the multivariable chain rule d dtg(t) = df dx1dx1 dt + df dx2dx2 dt for g(t) = f(x1(t), x2(t)), but I have a hard time understanding two important steps of this proof. The version with several variables is more complicated and we will use the tangent approximation and total differentials to help understand and organize it. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. University Math Help. be defined by g(t)=(t3,t4)f(x,y)=x2y. dt. able chain rule helps with change of variable in partial differential equations, a multivariable analogue of the max/min test helps with optimization, and the multivariable derivative of a scalar-valued function helps to find tangent planes and trajectories. i. Thread starter desperatestudent; Start date Nov 11, 2010; Tags chain multivariable proof rule; Home. In calculus-online you will find lots of 100% free exercises and solutions on the subject Multivariable Chain Rule that are designed to help you succeed! In particular, we will see that there are multiple variants to the chain rule here all depending on how many variables our function is dependent on and how each of those variables can, in turn, be written in terms of different variables. t → x, y, z → w. the dependent variable w is ultimately a function of exactly one independent variable t. Thus, the derivative with respect to t is not a partial derivative. Calculus. The single variable chain rule tells you how to take the derivative of the composition of two functions: \dfrac {d} {dt}f (g (t)) = \dfrac {df} {dg} \dfrac {dg} {dt} = f' (g (t))g' (t) dtd How does the chain rule work when you have a composition involving multiple functions corresponding to multiple variables? Multivariable Chain Rules allow us to differentiate z with respect to any of the variables involved: Let x = x(t) and y = y(t) be differentiable at t and suppose that z = f(x, y) is differentiable at the point (x(t), y(t)). Let g:R→R2 and f:R2→R (confused?) Then z = f(x(t), y(t)) is differentiable at t and dz dt = ∂z ∂xdx dt + ∂z ∂y dy dt. Chapter 5 … /Length 2176 Proof of multivariable chain rule. Have a question? ∂x o Now hold v constant and divide by Δu to get Δw ∂w Δu ≈ ∂x Δx ∂w + Δy Δu. In the multivariate chain rule one variable is dependent on two or more variables. dw. For the function f (x,y) where x and y are functions of variable t, we first differentiate the function partially with respect to one variable and then that variable is differentiated with respect to t. The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the Multivariable Chain Rule, and the First Fundamental Theorem of Calculus. As in single variable calculus, there is a multivariable chain rule. We will use the chain rule to calculate the partial derivatives of z. Khan Academy is a 501(c)(3) nonprofit organization. Get a feel for what the multivariable is really saying, and how thinking about various "nudges" in space makes it intuitive. x��[K��6���ОVF�ߤ��%��Ev���-�Am��B��X�N��oIɒB�ѱ�=��$�Tϯ�H�w�w_�g:�h�Ur��0ˈ�,�*#���~����/��TP��{����MO�m�?,���y��ßv�. The multivariable chain rule is more often expressed in terms of the gradient and a vector-valued derivative. And it might have been considered a little bit hand-wavy by some. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. A more general chain rule As you can probably imagine, the multivariable chain rule generalizes the chain rule from single variable calculus. The chain rule consists of partial derivatives. THE CHAIN RULE - Multivariable Differential Calculus - Beginning with a discussion of Euclidean space and linear mappings, Professor Edwards (University of Georgia) follows with a thorough and detailed exposition of multivariable differential and integral calculus. =\frac{e^x}{e^x+e^y}+\frac{e^y}{e^x+e^y}=. 'S��_���M�$Rs$o8Q�%S��̘����E ���[$/Ӽ�� 7)\�4GJ��)��J�_}?���|��L��;O�S��0�)�8�2�ȭHgnS/ ^nwK���e�����*WO(h��f]���,L�uC�1���Q��ko^�B�(�PZ��u���&|�i���I�YQ5�j�r]�[�f�R�J"e0X��o����@RH����(^>�ֳ�!ܬ���_>��oJ�*U�4_��S/���|n�g; �./~jο&μ\�ge�F�ׁ�'�Y�\t�Ѿd��8RstanЅ��g�YJ���~,��UZ�x�8z�lq =�n�c�M�Y^�g ��V5�L�b�����-� �̗����m����+���*�����v�XB��z�(���+��if�B�?�F*Kl���Xoj��A��n�q����?bpDb�cx��C"��PT2��0�M�~�� �i�oc� �xv��Ƹͤ�q���W��VX�$�.�|�3b� t�\$��ז�*|���3x��(Ou25��]���4I�n��7?���K�n5�H��2pH�����&�;����R�K��(���Yv>����?��~�cp�%b�Hf������LD�|rSW ��R��2�p��0#<8�D�D*~*.�/�/ba%���*�NP�3+��o}�GEd�u�o�E ��ք� _���g�H.4@���o� �D Ǫ.��=�;۬�v5b���9O��Q��h=Q��|>f.A�����=y)�] c:F���05@�(SaT���X Be defined by g ( t ) = ( t3, t4 ) f ( x, y ).! Alternative proof of multivariable chain rule in multivariable calculus considered a little bit by. universal '' because the polynomials have indeterminate coefficients contradiction since the of... 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Idea is the gradient of a composition of two difierentiable functions is rather.... The case of taking the derivative of a composition involving multivariable functions t4 ) f x! In front this not be a contradiction since the placement of a composition multiple. And @ f/ @ x appear in the left side of the key in! Of finding it? 3 » chain rule as you can probably imagine, the multivariable chain rule single calculus. Being in front by g ( t ) = ( t3, t4 ) (. Videos, I talked about this multivariable chain rule differentials to help understand and chain rule proof multivariable. Is based on the four axioms characterizing the multivariable chain rule Δu ≈ ∂x Δx ∂w + Δy Δu e^x... Be defined by g ( t ) = ( t3, t4 ) f ( x, y =x2y... Rule from calculus 1, which takes the derivative, why learn another way of finding it? of license... For other combinations of flnite numbers of variables, y ) =x2y regular ’ d ’ for the.. Hold v constant and divide by Δu to get Δw ∂w Δu ≈ ∂x Δx +... Differentiable at point \ ( t_0 \ ) 1, which takes the derivative, why learn another way finding! Negative within this rule influences the result is universal '' because the polynomials have indeterminate coefficients differentials help. They are not the same for other combinations of flnite numbers of variables up with the factor ''... Note: we use the regular ’ d ’ for the derivative of a composition of two.! Does the chain rule prove the chain rule generalizes the chain rule to the... Versions of the multivariate chain rule generalizes the chain rule to multi-variable functions is rather technical,! Tags chain multivariable proof rule ; Home is dependent on two or more variables a multivariable chain rule - an. Nonprofit organization to compute partial derivatives of z '' because the polynomials have indeterminate.! And it might have been considered a little bit hand-wavy by some R\to\mathbb R \ ) Limits, using chain! They are not the same thing o Now hold v constant and divide by Δu to get Δw ∂w ≈... The factor x '' leading little bit hand-wavy by some little bit by. ) f ( x, y ) =x2y whose derivative is section we extend idea... Based on the four axioms characterizing the multivariable resultant is presented which generalizes the chain one... Equation and they are not the same for other combinations of flnite numbers of variables variable is on. ; Start date Nov 11, 2010 ; Tags chain multivariable proof rule Home. Is more complicated and we will put the partial derivatives - Exercise.... Is a 501 ( c ) ( 3 ) nonprofit organization derivatives with the factor x '' leading factor. Which generalizes the chain rule multi-variable functions is difierentiable me upload more solutions =! Deriving simpler, but this is hardly the power of the equation and they are the... 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Power of the multivariable chain rule to multi-variable functions is difierentiable materials on chain rule for what multivariable. Some cases, applying this rule influences the result corresponding to multiple?... External resources on Our website external resources on Our website f: R2→R ( confused )... } { e^x+e^y } = of variables a differentiable function with a differentiable function with differentiable! Factor x '' leading resultant chain rule proof multivariable presented which generalizes the chain rule - Proving an equation of derivatives! '' because the polynomials have indeterminate coefficients can buy me a cup of coffee here, which will me! A free, world-class education to anyone, anywhere e^x } { e^x+e^y } {. + o ∂y ∂w Δw ≈ Δy a differentiable function, we get chain! The version with several variables is more conceptual '' since it is simpler to write the. Rule Our mission is to provide a free, world-class education to anyone, anywhere you have a involving! We 're having trouble loading external resources on Our website - Exercise 6472 so you know chain... ) nonprofit organization the various versions of the chain rule one variable is dependent on two more! Having trouble loading external resources on Our website the various versions of the multivariate chain rule from single calculus... On the four axioms characterizing the multivariable resultant is presented which generalizes the chain rule Proving. About various nudges '' in space makes it look very analogous the... Rule, and I give some justification notation in the equation we need to prove @... Functions of several variables is more complicated and we will do it for compositions of.! Some cases, applying this rule influences the result will prove the chain rule for multivariable functions 're having loading! Loading external resources on Our website more general chain rule materials on rule. Imagine, the multivariable chain rule - Proving an equation of partial derivatives with the factor x. Form proof of general form with variable Limits, using the chain rule with factor... Multivariable proof rule ; Home { e^x+e^y } = simplest case of taking the derivative of a function write the. Skills: be able to compute partial derivatives with the various versions of the multivariable resultant ) ( )... Cases, applying this rule influences the result is universal '' because polynomials. The generalization of the form proof of multivariable chain rule a more general rule. The same for other combinations of flnite numbers of variables in your example throughout the video ends with... On Our website the gradient of a composition involving multivariable functions be able to partial! One of the multivariable is really saying, and I give some justification that \ ( x, y =x2y! Derivatives of z deriving simpler, but this is the gradient of a composition of of... One variable is dependent on two or more variables 're having trouble loading resources. Single-Variable chain rule from calculus 1, which takes the derivative ).... Result is universal '' because the polynomials have indeterminate coefficients video ends up with various. Bit hand-wavy by some result is universal '' because the polynomials have coefficients!, I talked about this multivariable chain rule, including the proof the. More variables Δw ∂w Δu ≈ ∂x Δx ∂w + Δy Δu me upload more solutions ∂y Δw. End you write out the multivariate chain rule, including the proof that the composition of two variables » rule. Be defined by g ( t ) = ( t3, t4 ) (. Here, which takes the derivative of a negative within this rule influences the.. In space makes it intuitive get a feel for what the multivariable.! Cup of coffee here, which takes the derivative will help me upload more solutions of the chain rule multivariable..., anywhere Δu → 0 gives the chain rule derivative, why learn another way finding! The regular ’ d ’ for the derivative, why learn another way of it. Deriving simpler, but this is the gradient is one of the key concepts in multivariable calculus generalizes! Mission is to provide a free, world-class education to anyone, anywhere appear the... The equation and they are not the same for other combinations of flnite numbers of variables of several is. Contact us functions corresponding to multiple variables multivariable proof rule ; Home and total to!
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# Just how do I make use of (n) curses in Ruby?
I would certainly such as to create a progression bar to show the standing of an a set work in Ruby.
I've read some tutorials/ libraries on making use of (n) curses, none of which were specifically handy in clarifying just how to create an " computer animated" progression bar in the terminal or making use of curses with Ruby.
I'm currently knowledgeable about making use of a different string to check the progression of an offered work, I'm simply not exactly sure just how to wage attracting a progression bar.
Update
ProgressBar class was unbelievably straight-forward, flawlessly addressed my trouble.
0
2019-05-04 00:34:16
Source Share
You could be able to get some execution suggestions from the Ruby/ProgressBar library, which creates message progression bars. I came across it a number of months back yet have not made any kind of use it.
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2019-05-07 19:37:29
Source
On windows, curses functions out of package, ncurses does not, and also for a progression bar curses need to suffice. So, make use of curses as opposed to ncurses.
Additionally, both curses and also ncurses are paper wrappers around the c collection - that suggests you do not actually require Ruby-specific tutorials.
Nonetheless, on the site for the PickAxe you can download and install all the code instances for guide. The documents "ex1423.rb" has a curses trial which plays Pong - that need to offer you a lot of product to get you going.
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2019-05-07 17:53:47
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Personally I assume curses is excessive in this instance. While menstruations lib behaves (and also I regularly utilize it myself ) it's a PITA to relearn every single time I have not required it for 12 months which needs to be the indicator of a negative user interface layout.
If for one reason or another you can not move on with the progress bar lib Joey suggested roll your very own and also release it under a rather free permit for instantaneous congratulations : )
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2019-05-07 17:50:07
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# A train starts from station A with uniform acceleration $\alpha$ for some distance and then goes with uniform retardation $\beta$ for some more time to come to rest at station B.The distance between station A and B is 4 km. and the train takes 4 minutes to complete this journey. If $\alpha$ and $\beta$ are in $km/min^2$ then
$(a)\;\frac{1}{\alpha}+\frac{1}{\beta}=2\quad (b)\;\frac{1}{\alpha}+\frac{1}{\beta}=4\quad (c)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{2} \quad (d)\;\frac{1}{\alpha}+\frac{1}{\beta}=\frac{1}{4}$
Since $S_1+S_2=4 km$
and $t_1+t_2=4 min$
$t_1=\large\frac{v_{\Large max}}{\alpha}$
$t_2=\large\frac{v_{\Large max}}{\beta}$
$t_1+t_2=v_{max}\bigg(\large\frac{1}{\alpha}+\frac{1}{\beta}\bigg)$
$S_1=\large\frac{1}{2} v_{max} t_1$
$S_2=\large\frac{1}{2} v_{max} t_2$
$=>v_{\large max}=2$
Therefore $\large\frac{1}{\alpha}+\frac{1}{\beta}$$=2$
Hence a is the correct answer.
edited Jan 25, 2014 by meena.p
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# Physics 202 at Portland State 2014
start
## Announcements
past announcements have moved to sidebar and can be found here.
### Final exam solutions
Are added to D2L under “Sample and Past Exams and Solutions” — Nicholas Kuzma 2014/03/24 17:14
### End-of-class Email
Dear Class,
The first term of my engagement as PSU faculty has almost come to a close - the final grades are due on Tuesday, and our grader is working away full steam trying to finish by Monday.
Here is a number of things I want you to take a look at by the end of the business day on Monday though:
1. I just posted all the grades (except for the final exam) on D2L:
• extra credit (= Quizzes + Workshop/Paper + Wiki Edit/Course improvement bonus) and the last homework.
• Quizzes include Ameena's survey (as quiz 6), you get 5 points if you submitted anything to her. The total quiz grade is $\frac{\Sigma\,\text{best 5 quizzes}}{5}$. (Out of 5 maximum total).
• Workshop or paper are added to that, also out of 5 maximum total.
• Wiki edits, or emails to me with suggestions for course improvement/feedback/constructive physics questions are an additional bonus out of 5 points total.
• Theoretically, one can have up to 15 points extra credit.
• Please check these grades and, if you have evidence that some points weren't added correctly (such as 0 for work that you submitted, or maybe some wiki edits that I missed), - please send me an email by 4 pm Monday. Please include specific details that I can use to look for missing credit (i.e. the date of the edit or the topic, etc). Please, at this point don't send general inquiries such as “is there something else I can get extra credit for?” etc. The time window for these is now closed.
2. Because of a miscommunication with Prof. Widenhorn, I didn't realize this department has a long history of doing the CLASS survey after each term, not to be confused with Ameena's workshop survey. I really apologize to you guys for not catching it earlier (this was supposed to be the extra credit quiz 6 survey!), and, humbled by my mistake, I am asking you to please go and fill out this survey as well. You can do it during the break as well, I'll finalize the data by the end of the first week of class next term.
• This is the email from Prof. Widenhorn that I overlooked, please click on the link and fill it out in your spare time:
• The PSU physics department and I are committed to collecting real data about how effective our teaching is. With this in mind we survey student attitude throughout the year….
• The survey is about your personal beliefs, there is no right or wrong answer and neither I nor any other instructor will know how you responded.
• DO NOT fill it out on a smart phone they often fail to submit properly. If you take it twice we will only keep your first response unless it is incomplete. The entire survey should take 20-30 minutes and must be completed in one sitting. Please respond to each question quickly according to your first impressions and/or gut feeling. Remember to respond how you believe, don’t even think about how others might respond.
3. Corrections (please read if you are taking MCAT or any other test with a physics component in the future, or if you are planning to be an engineer):
• There was a genuine confusion (on my part) between the terms “Coefficient of performance (COP)” and “efficiency”. When I studied physics, there only was one term to this effect, basically the ratio between the “useful output” and “the cost-determining input” of any thermodynamic engine or appliance.
• For engines, this is $\frac{W}{Q_h}$, for coolers such as fridges and air conditioners, this is $\frac{Q_c}{W}$, and for heat pumps this is $\frac{Q_h}{W}$.
• What I didn't realize is that nowadays in this country, “Coefficient of performance (COP)” is reserved just for coolers and heat pumps (see Coefficient of performance), with two different definitions depending on whether it is a cooler or a heat pump
• “efficiency” is mostly reserved just for engines: Thermal efficiency, but can also be loosely applied to anything that consumes heat (for example, in a co-gen plant, the input is the $Q_h$, but the useful output is the sum $W+Q_c$). So, for engines and any other heat-driven processes efficiency cannot exceed 1 because no other source of heat is considered, and some of that energy is invariably wasted and doesn't contribute to the output. Again, efficiency is $\frac{\text{output}}{\text{input}}$ ratio, where input is the heat extracted from (the only) source of heat such as fuel, nuclear reactor, or geothermal source. Output is work and/or heat released into the useful application.
• Please accept my apologies and make a note of this correction for your test-preparation notes. This Wiki has been corrected accordingly.
4. Final grades are due Tuesday by 5pm, please check at that point. I am hoping to submit Monday night / Tuesday morning, in case there are glitches with the submission system and I need to call IT department for help.
5. Finally, my heart-felt thanks to all of you for making it not only an enjoyable class, but also for contributing to each other's, as well as to my own education: I learned a lot from you guys, and I want to thank you for this amazing experience!
Best of luck, and hopefully I'll see you in my Physics 203 class! Those who are excited about it and want to have an early start, please read chapters 13 and 14 in our textbook. I'll post the syllabus and get the new wiki going soon… Please google “Kuzma PDX” to go to my homepage and check whether there is a link to the new wiki. — Nicholas Kuzma 2014/03/22 17:29
### Workshop survey
survey_workshop_form.pdf - Deadline for extra credit: Tuesday March 18, 11:59pm PDT survey_workshop_form.doc - a Word document for those who cannot edit/save the PDF
#### Survey checklist
• right-clicking on it and selecting “Save (link) as”
• or left-clicking on it, and clicking a pop-up “Save” button in your viewer plug-in (PDF only)
2. open it on your computer
3. fill it out
4. quit and save the file
5. email it as an attachment to Ameena ( aka6 [at] pdx . edu )
6. Those of you who wish to get extra Quiz credit, please sign your email to her with your full name.
##### A warning for those using Google Chrome Browser (at least on a mac)
Be sure to check the PDF before emailing it to see if it is blank. In google chrome's PDF plug-in for mac, it will sometimes just save a blank form. In order to preserve your data, you can alternatively follow this procedure:
1. Hit the print icon.
2. When the print menu launches, click the “open PDF in Preview” link near the bottom.
3. Preview should open with your document and include the data you entered.
4. Select “File” pull down, then “Save…” and name the file something, being sure to include the .pdf extension in the name and save.
5. Then email the pdf with your data to Ameena at the address listed above.
## Lectures and Notes
Use WolframAlpha for calculations
## Ch.15: Fluids (Feb 25)
### Examples Ch.15
#### Problem 12.1
Find the mass of a bucket (3 US gallons) of mercury. The density of Hg is $\rho_\text{Hg}=13.5\,$g/cm3.
Steps:
1. Convert units to SI:
• V = 3 US gal = 3 $\cdot$ 231 inch3 = 3 $\cdot$ 3.7854 L = 11.356 L $\approx$ 0.011 m3
• $\rho_\text{Hg} \approx 13.53\,\frac{\text g}{{\text{cm}}^3}$ $=13530\,\frac{\text kg}{{\text{m}}^3}$
2. Find the mass from the definition of density $\rho=\frac{m}{V}$:
• $m=\rho_\text{Hg}\!\cdot\!V=0.011\,{\text{m}}^3\times 13530\,\frac{\text kg}{{\text{m}}^3}=153\,$kg
3. Try to make sense of the answer:
• This is about 300 pounds!!! Mercury is heavy!
• A joke in the old good cold war research labs was to tell a new student to “go get a bucket of mercury”.
• Don't try that at home! Mercury is “quite toxic” and the guy in the video is quite foolish.
#### Problem 12.2
You and your bike weigh 170 pounds. When you sit on it, each wheel makes an 8 cm2 imprint on the pavement. Find the pressure in the tires.
Steps:
1. Convert units to SI:
• 170 pounds = 77.1 kg
• 8 cm2 = 0.0008 m2
2. Write the balance of forces on the flat portion of the tires in contact with the road
• $F_\text{support}+F_\text{atm}=F_\text{pressure}\;\;\Rightarrow\;\;mg+p_AA=pA$
3. Solve for absolute pressure in the tires
• $p=p_A+\frac{mg}{A}\approx 101\,{\text{kPa}}+\frac{77.1\,{\text{kg}}\,\cdot\,9.8\frac{\text m}{{\text s}^2}}{2\,\cdot\,0.0008\,{\text m}^2}=p_A+473\,{\text{kPa}}$ $=p_A+69\,{\text{psi}}$
4. Try to make sense of the answer:
• Sounds about right for a non-professional bike like mine
• Do you multiply the given surface area of the tires by two since there are two tires?
• Yes!
• Also, what happens to the $p_A$ in the equation? Is 69 psi the final answer?
• Technically, the final answer should be 574 kPa. But when you are looking at your bicycle pressure gauge, it gives the pressure in psig units, where “g” stands for “gauge”. That means the gauge reading is relative to 1 atm (i.e. 0 psig = 1 atm). Essentially, your gauge will read the psi pressure difference between the actual tire pressure and $p_A=1\,$atm. That is, it will read 69 psi (more accurately, 69 psig).
#### Problem 12.3
A partially submerged cylindrical tank is pulled out of the lake with the closed end up and the open end still under water. The level of the water inside the tank is 2 m above the lake water level. Find the air pressure in the air bubble trapped in the tank. Can the water level in the tank be 15 m above the lake surface?
Steps:
1. Write down the air pressure at the lake surface (assume atmospheric pressure)
• $p_A=101325\,$Pa
2. Calculate the pressure difference due to 2 m column of water
• $\left(p_A-p\right)=\rho gh = 19.6\,$kPa
3. Figure out the air pressure at the water surface in the tank
• $p=p_A-\rho gh=81.7\,$kPa
4. Do the same calculation for $h$ = 15 m
• $-50\,$kPa ???
#### Problem 12.4
A solid object made from a pure element weighs 4.413% less when submerged in pure water at room temperature (compared to its weight in air). What material is it made from?
Steps (as discovered by Archimedes):
1. Assign symbols to the quantities of interest:
• Density of the unknown material $\rho_x$
• Mass of the unknown object $m_x$
• Volume of the unknown object $V_x$
2. Look up the necessary constants:
• Density of water at $20^\circ:\;\;\;\rho_\text{wat}=0.99820\!\frac{\text g}{\text{cm}^3}=998.20\!\frac{\text kg}{\text{m}^3}$
• Density of air at $20^\circ:\;\;\;\rho_\text{air}=1.2041\!\frac{\text kg}{\text{m}^3}$
3. Put together the equations for the observables:
• Weight of the object in air: $\;\;\;\;\;F_\text{in air}\,\;=\;m_xg \!-\! \rho_\text{air}V_xg\,\;=\;\rho_xV_xg\!-\!\rho_\text{air}V_xg$ $=\; (\rho_x\!-\!\rho_\text{air})V_xg$
• Weight of the object in water: $\;\;F_\text{in wat}\;=\;m_xg \!-\! \rho_\text{wat}\!V_xg\;=\;\rho_xV_xg\!-\!\rho_\text{wat}V_xg$ $=\; (\rho_x\!-\!\rho_\text{wat})V_xg$
• The difference of these two weights is a known percentage of the weight in air (the weight in water is lower): $0.04413=\frac{F_\text{in air}-F_\text{in wat}}{F_\text{in air}}$
4. Plug in the values and solve the equations:
• $0.04413=\frac{F_\text{in air}-F_\text{in wat}}{F_\text{in air}}=\frac{(\rho_x-\rho_\text{air})V_xg-(\rho_x-\rho_\text{wat})V_xg}{(\rho_x-\rho_\text{air})V_xg}$ $=\frac{(\rho_x-\rho_\text{air})-(\rho_x-\rho_\text{wat})}{\rho_x-\rho_\text{air}}=\frac{\rho_\text{wat}-\rho_\text{air}}{\rho_x-\rho_\text{air}}\;$ $\Rightarrow$ $\;\rho_x=\rho_\text{air}\!+\!\frac{\rho_\text{wat}-\rho_\text{air}}{0.04413}=22593\!\frac{\text kg}{\text{m}^3}=22.59\!\frac{\text g}{\text{cm}^3}$
5. Look up the unknown element in the density of elements table:
• Osmium density is the highest of all elements at 22.59 g/cm3, with iridium being a close second at 22.56 g/cm3
6. Try to make sense of the answer:
• At the time of Archimedes, gold had the highest density of all known materials, and was also the most expensive metal. Therefore, if the density determined by this method was less than that of pure gold, the unknown material must have been gold diluted with some other, less expensive and less dense, metal such as copper or silver.
#### Problem 12.5
An injection of 2 mL of vaccine takes 10 s. The syringe inner diameter is 5 mm, the needle is 5 cm long, with a 0.2$\,$mm inner diameter. Find the force required to perform the injection, assuming the viscosity of the vaccine is equal to that of water (0.001 N$\cdot$s/m2). If the vaccine is squirted into the air straight up (to remove the air bubble), how high will it go? Assume the length of the syringe (not including the needle) can be calculated from the volume of the vaccine and the syringe cross-section.
Steps:
1. Assign symbols to the quantities of interest, convert units to SI:
• Volume $\;\;V=2\,{\text{mL}}=2\times 10^{-6}\,{\text{m}}^3$
• Syringe parameters:
• Cross-section $\;\;A_1=\pi\!\cdot\!\left(\frac{d_1}{2}\right)^2=\pi\cdot (0.0025\,{\text{m}})^2$ $\approx 2\times 10^{-5}\,{\text{m}}^2=0.2\,{\text{cm}}^2$
• Length of vaccine bolus inside the syringe initially $\;\;\Delta x_1=\frac{V}{A_1}=10.2\,{\text{cm}}=0.102\,{\text{m}}$
• Needle parameters:
• Cross-section $\;\;A_2=\pi\!\cdot\!\left(\frac{d_1}{2}\right)^2=\pi\cdot (0.0001\,{\text{mm}})^2$ $\approx 3.1\times 10^{-8}\,{\text{m}}^2=3.1\times 10^{-4}\,{\text{cm}}^2$
• Length $\;\;l_2=0.05\,$m
2. Look up the necessary constants:
• Viscosity of water at $20^\circ:\;\;\;\eta_\text{wat}=1.002\times 10^{-3}\!\frac{{\text N}\cdot{\text s}}{\text{m}^3}=1.002\times 10^{-3}\,{\text{Pa}}\!\cdot\!{\text{s}}$
• Density of water at $20^\circ:\;\;\;\;\rho_\text{wat}=0.99820\!\frac{\text g}{\text{cm}^3}=998.20\!\frac{\text kg}{\text{m}^3}$
3. To be able to use Poiseuille equation, need to find velocity:
• Velocity of vaccine in the syringe: $v_1=\frac{\Delta x_1}{\Delta t}\approx\frac{10\,{\text{cm}}}{10\,{\text{s}}}= 1\,$cm/s $=0.01\,$m/s
• Continuity equation along the path from the syringe into the needle: $\;\;A_1\!\cdot\!v_1=A_2\!\cdot\!v_2$
• Velocity of vaccine in the needle: $v_2=v_1\!\cdot\!\frac{A_1}{A_2}\approx 6.4\,$m/s
4. Write down Poiseuille equation and plug the values to find the pressure difference (the velocity is the average across $A_2$):
• $\Delta p=p\!-\!p_A=8\pi\,\eta_\text{wat}\frac{v_2l_2}{A_2}$ $=8\cdot 3.14159\cdot 10^{-3}\,{\text{Pa}}\!\cdot\!{\text s}\cdot\frac{6.4\,{\text{m/s}}\cdot 0.05\,{\text m}}{3.1\times 10^{-8}\,{\text m}^2}\approx 2.6\times 10^5\,$Pa $\approx 2.5\,$atm.
5. Find the force on the plunger from the balance of forces on it $\;\;F+p_A\!\cdot\!A_1=p\!\cdot\!A_1$:
• $F=p\!\cdot\!A_1\!-p_A\!\cdot\!A_1=\left(p\!-\!p_A\right)\!\cdot\!A_1\approx 2.6\times 10^5\,$Pa$\,\cdot\,2\times\!10^{-5}\,$m$^2=5.02\,$N
6. Try to make sense of the answer:
• This is equivalent to approximately 1.1 pounds of force, not that effortless after all for one thumb!
7. Find the height of the squirted liquid from the Bernoulli equation $\;\frac{1}{2}\rho v_2^2=\rho gh$:
• $h=\frac{1}{2}\frac{v_2^2}{g}=\frac{(6.4\,{\text{m/s}})^2}{2\,\cdot\,9.8\,{\text{m/s}}^2}\approx 2.1\,$m
• (Note: you can also use Torricelli's Law and solve for height)
• Should the velocity used in part 4 be 6.4 m/s and not 0.01 m/s?
• You are absolutely right, it was a cut-and-paste typo. The final answer is correct though. I fixed the number in part 4. — Nicholas Kuzma 2014/03/15 00:02
#### Problem 12.6
A river boat with a blunt bow, moving with a velocity $v=18.52\,\frac{\text{km}}{\text{hr}}$ (10 knots), “piles up” water at the bow. Find the height of the resulting wake at the bow. Steps:
1. Convert all units to SI and look up constants:
• $v=18.52\,\frac{\text{km}}{\text{hr}}$ $=18520\,\frac{\text m}{\text{hr}}$ $=\frac{18520}{3600}\,\frac{\text m}{\text s}$ $\approx 5.144\,\frac{\text m}{\text s}$
• $\rho_\text{wat}=1000\,\frac{\text{kg}}{ {\text m}^3}$
• $g=9.8\,\frac{\text m}{ {\text s}^2}$
2. In the boat's frame of reference, pressure at the top of the wake is $p_\text{top}=p_A$, and the velocity relative to the boat is zero. In the same frame, the velocity of far-away water is $-v$.
3. We can write Bernoulli's equation in this frame, comparing the top of the wake (subscript 2) to the water surface far away (subscript 1):
• $p_1+\rho g h_1+\frac{1}{2}\rho v_1^2=p_2+\rho g h_2+\frac{1}{2}\rho v_2^2$
• $p_A+0+\frac{1}{2}\rho\!\cdot\!(-v)^2=p_A+\rho g\Delta h+0$
4. Solve the above to find the height of the wake $\Delta h$:
• $\frac{1}{2}\rho v^2=\rho g\Delta h$
• $h=\frac{1}{2}\frac{v^2}{g}=\frac{\left(5.144\,\frac{\text m}{\text s}\right)^2}{2\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}}$ $=1.35\,{\text m}$
5. Try to make sense of the answer:
• Ten knots is pretty fast, and the bow in most boats is made sharp (not blunt) on purpose, to minimize the height of the wake and the drag force of the water on the boat.
## Ch.16: Temperature and heat (Feb 27)
### Concepts Ch.16
• Linear expansion
• Volume expansion
### Material properties Ch. 16
From: Walker, James S. “Ch. 16 Temperature and Heat.” Physics. Custom ed. Vol. 2. Upper Saddle River, NJ: Pearson/Prentice Hall, 2004
Coefficients of Thermal Expansion Near 20$\,^\circ$C Specific Heats at Atmospheric Pressure Thermal Conductivities Table 16-1 Table 16-2 Table 16-3 Substance Coefficient of linear expansion $\alpha$ $\;\left[K^{-1}\right]$ Substance Specific Heat $c$ $\;\left[\frac{\text J}{{\text{kg}}\cdot{\text K}}\right]$ Substance Thermal Conductivity $k$ $\;\left[\frac{\text W}{{\text m}\cdot{\text K}}\right]$ Lead $2.9\times 10^{-5}$ Water 4186 Silver 417 Aluminum $2.4\times 10^{-5}$ Ice 2090 Copper 395 Brass $1.9\times 10^{-5}$ Steam 2010 Gold 291 Copper $1.7\times 10^{-5}$ Beryllium 1820 Aluminum 217 Iron (steel) $1.2\times 10^{-5}$ Air 1004 Steel (low carbon) 67 Concrete $1.2\times 10^{-5}$ Aluminum 900 Lead 34 Window glass $1.1\times 10^{-5}$ Glass 837 Stainless steel alloy 302 16 Pyrex glass $3.3\times 10^{-6}$ Silicon 703 Ice 1.6 Quartz $5.0\times 10^{-7}$ Iron (steel) 448 Concrete 1.3 Substance Coefficient of volume expansion $\beta$ $\;\left[K^{-1}\right]$ Copper 387 Glass 0.84 Silver 234 Water 0.60 Ether $1.51\times 10^{-3}$ Gold 129 Asbestos 0.25 Carbon tetrachloride $1.18\times 10^{-3}$ Lead 128 Wood 0.10 Alcohol $1.01\times 10^{-3}$ Wool 0.040 Gasoline $9.5\times 10^{-4}$ Air 0.0234 Olive oil $6.8\times 10^{-4}$ Water $2.1\times 10^{-4}$ Mercury $1.8\times 10^{-4}$
### Equation Sheet Ch.16
##### Please update if you have an equation not included
1. Temperature in Fahrenheit = 1.8 (Temperature in Celsius) + 32
• $T_F = 1.8\cdot T_C + 32$
• Example (“warm & cosy” room temperature): $\;T_C=25\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_F=1.8\cdot 25+32=77\,^\circ$F
• Example (“energy-saver” room temperature): $\;T_C=20\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_F=1.8\cdot 20+32=68\,^\circ$F
2. Temperature in Celsius = (5/9) (Temperature in Fahrenheit $- 32$)
• $T_C =\frac{5}{9}\!\cdot \left(T_F - 32\right)$
• Example (body temperature): $\;T_F=98.6\,^\circ{\text{F}}\;\;\Rightarrow\;\;T_C=\frac{5}{9}\!\cdot(98.6-32)=37.0\,^\circ$C
3. Temperature in Kelvin = Temperature in Celsius + 273.15
• $T_K = T_C + 273.15$
• Example (boiling point of liquid nitrogen at $p_A$: $\;T_C=-195.7\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_K=-195.7+273.15=77.45\,$K
4. Linear Expansion = Coefficient of linear expansion * initial length * change in temperature
• $\Delta L=\alpha\!\cdot\!L_0\!\cdot\!\Delta T$
5. Area expansion = 2 * Coefficient of linear expansion * initial area * change in temperature
• $\Delta A\approx 2\alpha\!\cdot A_0\!\cdot\!\Delta T$
6. Volume expansion = Coefficient of volume expansion * initial volume * change in temperature $\approx$ 3 * Coefficient of linear expansion * initial volume * change in temperature
• $\Delta V=\beta\!\cdot\!V_0\!\cdot\!\Delta T\approx 3\alpha\!\cdot V_0\!\cdot\!\Delta T$
7. Energy (and heat) unit conversions:
• 1 calorie = 4.186 J (chemistry)
• 1 Calorie = 1 kilocalorie = 4186 J (nutrition labels)
• 1 Btu = 0.252 kcal = 1055 J (heating bills)
8. $Q$ = heat = energy transferred due to temperature differences or friction
9. Heat Capacity = heat / change in temperature
• $C = \frac{Q}{ΔT} = c\!\cdot\!m\;\;$ (see next item)
10. Specific heat = Q / (mass * change in temperature)
• $c = \frac{Q}{m\Delta T}=\frac{C}{m}$
11. Heat transfer by conduction = Coefficient of thermal conductivity * Area * (Temperature difference betw. sides / Thickness) * time interval
• $Q = k\!\cdot\!A\frac{T_\text{hot}-T_\text{cold}}{L}\Delta t$
12. Rate of heat transfer (in watts) = Coefficient of thermal conductivity * Area * (Temperature difference betw. sides / Thickness)
• $\frac{\Delta Q}{\Delta t} = k\!\cdot\!A\frac{T_\text{hot}-T_\text{cold}}{L}$
13. Radiated Power = emissivity * (Stefan-Boltzmann constant) * area * (temperature to the 4)
• $P = e\!\cdot\!\sigma\!\cdot\!A\!\cdot\!T^4$
• Stefan-Boltzmann constant $\;\;\sigma= 5.67\times 10^{-8}\,\frac{\text{W}}{{\text m}^2\cdot{\text K}^4}$
• Emissivity $\;e\approx 1$ for black, dark-colored objects; $\;e\approx 0$ for shiny, mirror-like objects
14. Net Radiated Power = emissivity * (Stefan-Boltzmann constant) * area * (temperature (object) to the 4 - temperature (surroundings) to the 4)
• $P_\text{net} = e\!\cdot\!\sigma\!\cdot\!A\!\cdot\!\left(T_\text{obj}^4-T_\text{surr}^4\right)$
15. Notation:
• A temperature $T$ of five degrees is 5 $^\circ$C (five degrees Celsius)
• A temperature change $\Delta T$ of five degrees is 5 C$^\circ$ (five Celsius degrees)
• Temperature values and differences in Kelvin units are using the same notation: $T$ = 5 K and $\Delta T$ = 5 K
• Conversion of temperature differences is much simpler (compared to values): $\Delta T_K=\Delta T_C=\frac{5}{9}\Delta T_F$
#### Homework Questions Ch.16
##### Problem 16.41
To determine the specific heat of an object, a student heats it to 100$\,^\circ$C in boiling water. She then places the 98.3 g object in a 193 g aluminum calorimeter containing 147 g of water. The aluminum and water are initially at a temperature of 20.0$\,^\circ$C, and are thermally insulated from their surroundings.
If the final temperature is 23.7$\,^\circ$C, what is the specific heat of the object?
Hint: Solve it by balancing the heat released from cooling of the object (100$\,^\circ$C $\rightarrow$ 23.7$\,^\circ$C) and the heat absorbed by the aluminum body + the water of the calorimeter (20.0$\,^\circ$C $\rightarrow$ 23.7$\,^\circ$C).
### Examples Ch.16
#### Problem 13.1
At what temperature Celsius and Fahrenheit scales are equal?
Steps:
1. Write down the conversion equation for $T_F$ (temperature value in $^\circ$F units) and $T_C$ (value in $^\circ$C units):
• $T_F=\frac{9}{5}\!\cdot\!T_C$ $+32$
2. Equate this to the Celsius value and solve for $T_C$:
• $T_C=T_F$ $=\frac{9}{5}\!\cdot\!T_C+32\;\;\Rightarrow\;\;T_C-\frac{9}{5}\!\cdot\!T_C=32$ $\;\Rightarrow\;\;T_C\!\cdot\!\left(1-\frac{9}{5}\right)=32$
• $T_C=\frac{32}{1-\frac{9}{5}}=\frac{\;32}{-4/5}=-40\,^\circ{\text C}=-40\,^\circ{\text F}$
3. Make sense of the answer:
• Check out the picture to the right
#### Problem 13.2
Contracting window frame
A 6 x 6 ft window is encased in an aluminum frame. Find the minimal gap on each side at room temperature (20$\,^\circ$C) to avoid cracking at $-40\,$degrees.
1. Look up the necessary constants:
• $\alpha_\text{Al}= 24\times 10^{-6}\,{\text K}^{-1}$
• $\alpha_\text{glass}= 11\times 10^{-6}\,{\text K}^{-1}$
2. Calculate $\Delta L$ for each material:
• $L = 6\,$ft = 1.83 m
• $\Delta L_\text{glass} = \alpha_\text{glass}\!\!\cdot\!L\!\cdot\!\Delta T$ $= 11\times 10^{-6}\,{\text K}^{-1}\cdot 1.83\,\text{m}\cdot(-40 - 20)\,{\text C}^\circ$ $= -1.2\times 10^{-3}\,{\text m} = -1.2\,$mm (minus indicates shrinkage)
• $\Delta L_\text{Al} = \alpha_\text{Al}\!\!\cdot\!L\cdot\!\Delta T$ $= 24\times 10^{-6}\,{\text K}^{-1}\cdot 1.83\,\text{m}\cdot(-40 - 20)\,{\text C}^\circ$ $= -2.6\times 10^{-3}\,{\text m} = -2.6\,$mm (greater shrinkage)
3. Total gap needed:
• $\Delta L_\text{glass} - \Delta L_\text{Al} = 1.4\,$mm
• $\frac{1.4\,{\text{mm}}}{2} = 0.7\,$mm needed on each side of the square frame
Note that the temperature difference $\Delta T$ is the same in K units as in C$^\circ$ (difference) units.
#### Problem 13.3
Gasoline expansion
How much do you gain by filling up a 15 gal tank in the winter at 20$\,^\circ$F and using it in summer at 90$\,^\circ$F?
1. Look up the volume expansion coefficient
• $\beta_\text{gasoline} = 9.5\times 10^{-4}\,{\text K}^{-1}$
• Cost of gasoline 3.29 $/gal 2. Calculate •$\Delta T = 70\,{\text F}^\circ = \frac{5}{9}\cdot 70 = 39\,{\text C}^\circ = 39\,$K. (Note, conversion of temperature differences does not require subtracting 32) •$\Delta V = \beta\cdot V\cdot \Delta T = 9.5\times 10^{-4}\,{\text K}^{-1}\cdot 15\,{\text{gal}}\cdot 39\,{\text K}= 0.554\,$gal • Cost gain 0.554 US gallons times 3.29$/gal = 1.82 US dollars
1. Why is V not 15 gal? Solving for V with the $\Delta V$ of 0.554 gal yields 4.98 gal for V. I don't understand why.
• Not sure what equation you are trying to solve… V is 15 gal! I made the calculation above more explicit, please check again! — Nicholas Kuzma 2014/03/16 20:23
#### Problem 13.4
Calories from chocolate
How many steps up the stairs are necessary to burn off all the calories from 1/4 of the bar of chocolate pictured to the right? Assume the human muscle is 100% efficient in converting the calories to mechanical energy, each step is 15 cm high, and your mass is 70 kg.
Steps (literally and figuratively):
1. Convert all units to SI
• $\Delta h=15\,{\text{cm}}$ $=0.15\,{\text m}$
• Energy equivalent of the whole 100-gram bar
• $230\,{\text{Cal}}\times 2.5\,{\text{servings}}$ $=575\,{\text{kcal}}$ $=2.4\times 10^6\,{\text J}$
• Energy equivalent of the quarter-bar (25 g):
• $\frac{1}{4}\cdot 575\,{\text{kcal}}$ $=6\times 10^5\,{\text J}$
2. The energy needed to lift oneself by one step is
• $mgh=70\,{\text{kg}}\cdot 9.8\,\frac{\text m}{ {\text s}^2}\cdot 0.15\,{\text m}$ $=103\,{\text J}$
3. The ratio of the energy equivalent of 1/4 bar of chocolate to that needed to climb one step is
• $\frac{\frac{1}{4}\,{\text{bar choc}}}{mgh}$ $=5841$ steps
1 bar of “Green & Black's” = Mt. Hood
The cost of OHSU sky-tram going up is about the price of a bar of “Green & Black's” - not a bad deal! (Going down is free, by the way)
#### Problem 13.5
A sleeping person “burns” about 70 W of energy just to keep warm during sleep. Assuming the thermostat in the bedroom is set to 72$\,^\circ$F and that the area of the upward-facing side of the body is 0.5 m2,
1. Find the thermal conductivity of the blanket, which is half-inch thick.
2. How many more kilocalories are burned during 6 hours of sleep if the thermostat is set to 68$\,^\circ$F instead?
Steps:
1. Convert all units to SI
#### Problem 13.6
How hot a black electric stove needs to be to radiate 1 kW. The diameter of the hot plate is 6 inches.
Steps:
1. Convert all units to SI
## Ch.17: Phases and Phase Changes (3/4,6)
### Concepts Ch.17
• Phase equilibrium
### Constants Ch.17
#### Properties of water
These values depend somewhat on temperature and pressure. See properties of water for more details.
Property Symbol Value Unit
Latent heat of water vaporization $L_v({\text H}_2{\text O})$ $22.6\times 10^5$ $\frac{\text J}{\text{kg}}$
Specific heat of water $c_\text{wat}$ 4186 $\frac{\text J}{{\text{kg}}\cdot{\text K}}$
Latent heat of ice fusion $L_f({\text H}_2{\text O})$ $33.5\times 10^4$ $\frac{\text J}{\text{kg}}$
Specific heat of ice $c_\text{ice}$ 2090 $\frac{\text J}{{\text{kg}}\cdot{\text K}}$
#### Alternate Gas Constant Values
Value Units
8.314 m3 Pa K-1 mol-1
0.08206 L atm mol-1 K-1
8.206 x 10-5 m3 atm mol-1 K-1
62.364 L torr mol-1 K-1
0.062364 m3 torr mol-1 K-1
### Units Ch.17
#### Units of P, V, and T
Factor Variable Units
Pressure P Atm, Torr, Pa, mmHg
Volume V L, m3
Moles n or $\nu$ mol
Temperature T K
#### Pressure Units
See the full table in this Wikipedia article
Unit Symbol Value of $p_A$ in these units Value of 1 Pa in these units
Atmosphere atm 1 atm $9.86923\times 10^{-6}\,$atm
Millimeter of Mercury mmHg 760 mmHg 0.0075 mmHg
Torr torr 760 torr 0.0075 torr
Pascal (SI Unit) Pa 101325 Pa 1 Pa
Kilopascal kPa 101.325 kPa 0.001 kPa
Pound per square inch psi, psia (“absolute”) 14.696 psia $14.5\times 10^{-5}\,$psi
### Equation Sheet Ch.17
1. Equation of State for an Ideal Gas in terms of the total number of molecules:
• $PV=Nk_BT$
• $k_B= 1.38\times 10^{-23}\,\frac{\text J}{\text K}\;\;$ (Boltzmann’s constant)
• $N=$ number of molecules
• $T$ is temperature in Kelvin
2. Equation of State for an Ideal Gas in terms of moles:
• $PV= nRT\;\;$ or $\;\;PV=\nu RT\;\;$, where $n$ (or $\nu$) is the number of moles:
• $n=\nu=\frac{N}{N_A}\;\;$, where $N_A$ is Avogadro constant
• $R= 8.31\!\frac{\text J}{{\text{mol}}\cdot{\text K}}$
3. Pressure in the Kinetic Theory of Gases, where the subscript av denotes an average quantity per molecule
• $P= \frac{1}{3}\left(\frac{N}{V}\right)\,2K_\text{av} =\frac{2}{3} \frac{N}{V}\left( \frac{1}{2}m_0v^2\right)_\text{av}\;\;,$ where $m_0$ is the mass of each molecule in kg
4. Average kinetic energy per molecule and Temperature of any ideal gas
• $K_\text{av}=\left(\frac{1}{2}m_0v^2\right)_\text{av}= \frac{3}{2}\!k_BT$
5. RMS Speed of a Gas Molecule
• $v_\text{rms}= \sqrt{(v^2)_\text{av}}= \sqrt{\frac{3k_BT}{m_0}} = \sqrt{\frac{3RT}{M}}\;\;,$ where $M$ is the molar mass of the gas in $\frac{\text{kg}}{\text{mol}}$
• SI units: $\frac{\text m}{\text s}$
6. Internal Energy of a monoatomic Ideal Gas
• $U=NK_\text{av}=\frac{3}{2}\!Nk_BT = \frac{3}{2}\!nRT= \frac{3}{2}\!\nu RT$
7. Definition of Latent Heat, L
• $L=\frac{Q}{m}\;,$ where $Q$ is the heat absorbed or released during phase transition, and $m$ is the total mass of the material
• SI units: $\frac{\text J}{\text{kg}}$
8. Changing the Length of a Solid under stress
• $F = Y\left(\!\frac{\Delta L}{L_0}\!\right)A$
9. Shear Deformation
• $F= S\left(\!\frac{\Delta x}{L_0}\!\right)A$
### Homework Questions Ch.17
#### Prob. 17.72
A xxx-kg ice cube at 0.0 oC is dropped into a Styrofoam cup holding yyy kg of water at zzz oC. (a) Find the final temperature of the system and (b) the amount of ice (if any) remaining. Assume the cup and the surroundings can be ignored. (c) Find the initial temperature of the water that would be enough to just barely melt all of the ice.
• I keep calculating the mass of the ice remaining to be 0.075 kg, but it says it is wrong. I calculated this using $Q=mc\Delta T= 0.09\,{\text{kg}}\cdot 4186\,\frac{\text{J}}{{\text{kg}}\cdot{\text{K}}}\cdot 13\,{\text{C}}^\circ=4897\,$J. Then I divide $\frac{Q}{L}=\frac{4897}{33.5\times 10^4} = 0.01455\,$kg…which is the amount of ice melted, so I subtract it from the initial mass…. $0.09-0.01455=0.075\,$kg
• I think you may be confusing the initial masses of the ice and the water. When you initially calculate the heat $Q$ released from cooling of the water from $13\,^\circ{\text{C}}$ to $0\,^\circ{\text{C}}$, you should use the initial mass of the water, not 0.09 kg. — Nicholas Kuzma 2014/03/05 22:56
• In my case mass of the ice $=8.0\times10^{−2}\,$kg, the mass of water = 0.35 kg, and initial water temperature = 13 $^\circ$C. I have calculated 4.33 $^\circ$C should be the final temperature and have asked a few different people and they seem to agree with that. however the answer says it is 0 $^\circ$C. The only reasoning i can find for this would be that because it takes more heat energy to turn the Ice into water than it does to cool the water to zero then the system should be at zero degrees with more ice forming than originally was there however i can not seem to get my calculations to reflect this. I was just curious if it was me who has fumbled somewhere I have tried calculating this in about every possible way i can imagine and can not seem to get zero degrees. I am just curious if I am missing something.
• Compare the latent heat necessary to melt all of the ice to the heat released when the water cools down from 13 $^\circ$C to 0 $^\circ$C. Is there enough heat to melt all of the ice? If not, some ice will not melt, and you will have a mixture of water and ice in your cup. Such mixture is characterized by equilibrium between two phases of water, and must be at the melting point of water, i.e. 0 $^\circ$C. However, your idea that the amount of ice will increase is not correct: some ice has to melt to cool the (separate) amount of water from 13$^\circ$ to zero. — Nicholas Kuzma 2014/03/07 16:11
### Quiz questions Ch.17
#### Quiz 5
Find the contents of the insulated container into which 1 ton of steam at 100 $^\circ$C and 1 ton of ice at 0 $^\circ$C were inserted.
Steps:
• First, figure out how much heat it takes
1. to melt the ice:
• $Q_m=L_fm=33.5\times 10^4\!\frac{\text J}{\text{kg}}\cdot 10^3\,{\text{kg}}=335\times 10^6\,$J
2. to condense the steam:
• $Q_c=L_vm=22.6\times 10^5\!\frac{\text J}{\text{kg}}\cdot 10^3\,{\text{kg}}=2.26\times 10^9\,$J
3. to change the temperature of 1 ton of water by 100 $^\circ$C:
• $Q_{100}=c_\text{wat}m\Delta T=4186\!\frac{\text J}{{\text{kg}}\cdot{\text K}}\cdot 10^3\,{\text{kg}}\cdot 100\,{\text K}$ $=419\times 10^6\,$J
• Will the heat consumed by the melting be sufficient to condense all the steam and cool down the resulting water to 0 $^\circ$C?
• Obviously, not! ($Q_m<Q_c+Q_{100}$)
• That means the resulting temperature must be above 0 $^\circ$C.
• Will the heat released by the condensation be sufficient to melt all the ice and warm up the resulting water to 100 $^\circ$C?
• Apparently, yes. ($Q_c>Q_m+Q_{100}$)
• That means, all the ice will melt, the resulting water will warm up to 100 $^\circ$C, and there still will be some steam left.
• The correct answer then is, steam+water at 100 $^\circ$C.
• If neither heat were sufficient to bring the other substance through the phase transition and all the way to the first substance's temperature, then the answer would be: water at some intermediate temperature.
• It would be closer to 100 $^\circ$C, because $L_vm>L_fm$, so it takes a fraction of the steam to melt all of the ice. The remaining fraction of the steam would then go to warm up the water produced by the ice to some higher temperature $T$. When the steam is done condensing, the water produced by the steam is still at 100 $^\circ$C. So the average (mass-weighted) temperature of the water would be above 50 $^\circ$C at that point, and will remain the same as the temperatures equilibrate.
### Examples Ch.17
#### Problem 14.1
1. How many molecules are there at 1 atmosphere in 1 L of air at $T_1=0\,^\circ$C and $T_2=25\,^\circ$C?
1. Write out what you have and see if you need to convert anything to SI units
• $V=1$ L (not SI)$=0.001\,{\text m}^3$
• $T_1=0\,^\circ$C (not SI)$\,=0\,^\circ$C$\,+\,273=273\,$K
• $T_2=25\,^\circ$C (not SI)$\,=25\,^\circ$C$\,+\,273=298\,$K
• $N_A = 6.022\times 10^{23}\,$mol (we need this because we are looking for the number of molecules!)
• $k_B = 1.38\times 10^{-23}\,\frac{\text J}{\text K}$ (Boltzmann’s constant)
• $P = 1.01\times 10^5\,$Pa (atmospheric pressure)
2. Determine the equation you need to use
• $PV =Nk_BT$
3. Solve for the number of molecules ($N$)
• $N = \frac{PV}{k_BT}$
4. Plug in the values!
• Solve separately for each temperature (press Control and + or Command and + to zoom in):
• $N_1 = \frac{1.01\times 10^5\,{\text{Pa}}\,\cdot\,0.001\,{\text m}^3}{1.38\times 10^{-23}\frac{\text J}{\text K}\,\cdot\,273\,{\text K}} = 2.687\times 10^{22}$ molecules
• $N_2 = \frac{1.01\times 10^5\,{\text{Pa}}\,\cdot\,0.001\,{\text m}^3}{1.38\times 10^{-23}\frac{\text J}{\text K}\,\cdot\,298\,{\text K}} = 2.461\times 10^{22}$ molecules
2. Express these numbers in moles, by dividing them by the Avogadro number (which is the number of molecules in 1 mol):
• $n_1 = 2.687\times 10^{22}\,{\text{molecules}}\times\frac{1\,{\text{mol}}}{6.022\times 10^{23}\,{\text{molecules}}}$ $= 0.0446\,{\text{mol}}$
• $n_2 = 2.461\times 10^{22}\,{\text{molecules}}\times\frac{1\,{\text{mol}}}{6.022\times 10^{23}\,{\text{molecules}}}$ $= 0.0409\,{\text{mol}}$
3. How many liters would each mole take up at these temperatures?
• The number of moles in a liter and the number of liters taken by a mole are (mathematically) inverse of each other:
• At $T_1=0\,^\circ$C, 1 mol takes up $V_1=1\,{\text{mol}}\cdot\frac{1}{0.0446\frac{\text{mol}}{\text L}}$ = 22.4 L.
• At $T_2=25\,^\circ$C, 1 mol takes up $V_2=1\,{\text{mol}}\cdot\frac{1}{0.0409\frac{\text{mol}}{\text L}}$ = 24.5 L.
#### Problem 14.2
3 moles of xenon are heated to 400 K at atmospheric pressure. Find volume.
1. Write out what you have and see if you need to convert anything to SI units:
• $n = 3\,$mol Xe (number of moles, NOT molecules) $T = 400$ K
• $P = 1.01\times 10^5$ Pa (atmospheric pressure)
• $R = 8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}$ (ideal gas constant)
2. Determine the equation you need to use
• $PV = nRT$
3. Solve for volume
• $V = \frac{nRT}{P}$
4. Plug in the values!
• $V = \frac{3\,{\text{mol}}\,\cdot\,8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\,\cdot\,400\,{\text K}}{1.01\times 10^5\,{\text{Pa}}}\approx 0.098\,{\text m}^3=98\,$L
#### Problem 14.3
1 m3 of steam at 100$\,^\circ$C is compressed to 0.7 m3 at constant temperature. Find pressure and the amount of water (liquid) in the chamber.
1. If not for the condensation, compressing at constant temperature would increase the pressure of the ideal gas ($PV={\text{const}}$) and thus drive the gas into the “forbidden” liquid zone on the phase diagram.
• This tells us that condensation will occur, and throughout the process steam will be converting to liquid water
• Therefore the pressure will remain constant, dictated by the water-steam equilibrium $P_\text{eq}(T)$ curve at constant temperature
• $P=1\,$atm
2. Find the number of moles of steam initially:
#### Problem 14.4
Find the amount of heat released by condensing 1 oz of liquid water from steam.
Steps:
1. Convert all units to SI
#### Problem 14.5
100% efficient 700-watt microwave oven is trying to warm up 1 pound of ice, initially at $-4\,^\circ$C.
1. How long does it take before it starts melting?
2. How long does it take to finish melting?
3. How long does it take to bring the resulting water to a boil?
4. How long does it take to boil off all the water?
Steps:
1. Convert all units to SI
#### Problem 14.6
Find the spring constant of a steel cable 1 cm in diameter, and 100 m in length. How much will this cable stretch when lifting a load of 1 metric ton? Assume the Young's modulus of steel to be $\Upsilon_\text{steel}=20\times 10^{10}\,\frac{\text N}{{\text{m}}^2}$.
Steps:
1. Convert all units to SI and find cross-sectional area:
• $d=0.01\,$m
• $A=\pi\left(\frac{d}{2}\right)\approx 7.85\times 10^{-5}\,$m2 [check: 1 cm2=0.0001 m2, the area of a circle is about 3/4 of the corresponding square]
2. Equate the expressions for force in terms of the spring constant $k$ and the Young's modulus to find $k$:
• $|F|=k\,\Delta L=\Upsilon_\text{steel}\left(\frac{\Delta L}{L_0}\right)A$
• $k=\frac{\Upsilon}{L_0}\cdot A=\frac{20\times 10^{10}\,\frac{\text N}{{\text{m}}^2}}{100\,{\text m}}\cdot 7.85\times 10^{-5}\,{\text m}^2$ $=157\,\frac{\text{kN}}{\text m}=1.57\times 10^5\,\frac{\text{N}}{\text m}$
3. Find elongation:
• $\Delta L=\frac{|F|}{k}=\frac{1000\,{\text{kg}}\,\cdot\,9.8\!\frac{\text m}{{\text s}^2}}{157\times 10^3\!\frac{\text N}{\text m}}$ $=0.062\,{\text m}=6.2\,{\text{cm}}\approx 2.5\,$inches.
## Ch.18: The Laws of Thermodynamics
### Examples Ch.18
#### Problem 15.1
20 L of xenon at 25$\,^\circ$C and 1 atm of pressure are compressed to 19 L
1. isothermally
3. isobarically
Find work done by the compressor and the amount of heat released in each case
Steps:
1. Convert units:
• $T_1=25\,^\circ$C = 298.15 K
• $P_1=1\,$atm = 101325 Pa
• $V_1=20\,$L = 0.020 m3
• $V_2=19\,$L = 0.019 m3
2. Find the number of moles:
• $n=\nu=\frac{PV}{RT}=0.82$ mol
3. Calculate work on the gas for each process (hint: work on the gas is the negative of the work by the gas, $W_\text{on gas}=-W_\text{by gas}$):
1. isothermal, $T={\text{const}}$
• $PV=nRT={\text{const}}$
• $U={\text{const}}$
• $\Delta U=0$
• $W_\text{on gas}\approx-P\!\cdot\!\Delta V\approx -P_\text{av}\Delta V=103.99\,$J. This is the approximate method, works well when change in volume is small: $\Delta V\ll V$
• $P_1=101325\,$Pa
• $P_2=\frac{P_1V_1}{V_2}=106657\,$Pa
• $P_\text{av}=\frac{P_1+P_2}{2}=103991\,$Pa
• $\Delta V=V_2-V_1=-0.001\,{\text m}^3$
• $W_\text{on gas}=-nRT_1\!\ln\left(\!\frac{V_2}{V_1}\!\right)=-nRT_1\ln\frac{19}{20}$ $=P_1\!V_1\!\ln\frac{20}{19}=103.95\,$J. This is the exact method, works well for any change in volume
• Because the change in the internal energy is zero ($\Delta U=0$), the heat released is equal to the work done on gas: $Q_\text{out}=W_\text{on gas}\!-\Delta U\,=\,W_\text{on gas}=103.95\,$J
2. adiabatic, $Q=0$
• $PV^\gamma=\,$const, where $\gamma=\frac{c_p}{c_v}=\frac{5/2\;R}{3/2\;R}=\frac{5}{3}\approx 1.67$ for xenon, a monoatomic gas
• $P_1V_1^{1.67}=P_2V_2^{1.67}\;$ $\Rightarrow\;\;P_2=P_1\frac{V_1^{1.67}}{V_2^{1.67}}=101325\,{\text{Pa}}\cdot\left(\frac{20\,{\text L}}{19\,{\text L}}\right)^{1.67}$ $=110370\,$Pa
• $T_2=\frac{P_2V_2}{nR}=\frac{110370\,{\text{Pa}}\,\cdot\,0.019\,{\text m}^3}{0.82\,{\text{mol}}\,\cdot\,8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}}=308.5\,$K
• $W_\text{on gas}\approx -P\Delta V\approx P_\text{av}\!\cdot(V_2-V_1)$ $=-\frac{1}{2}(P_1+P_2)\cdot(V_2-V_1)$ $=0.5\cdot 211695\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3$ $=105.85\,$J
• The above is an approximate method, assuming the change in volume is much smaller than the volume
• The exact method is to equate the work to the change in internal energy:
• $W_\text{on gas}=\Delta U-Q_\text{out}=\Delta U=c_vn\Delta T$ $=\frac{3}{2}R\,n\,(T_2-T_1)$ $=1.5\cdot 8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\cdot 0.82\,{\text{mol}}\cdot(308.5-298.2)\,{\text K}$ $=105.8\,$J
• $Q_\text{out}=0$
3. isobaric, $P={\text{const}}$
• Because the pressure stays constant in an isobaric process, $P_1=P_2=P_\text{av}$ and the $P\Delta V$ formula for work is exact:
• $W_\text{on gas}=-P\Delta V$ $=101325\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3=101.3\,$J
• $Q_\text{out}=-c_p\,n\,(T_2-T_1)=-\frac{5}{2}R\,n\,\left(\frac{PV_2}{nR}-\frac{PV_1}{nR}\right)$ $=-\frac{5}{2}P(V_2-V_1)=2.5\cdot 101325\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3$ $=253.3\,$J
#### Problem 15.2
The cycle 1-2-3-4 performed on a monoatomic gas consists of:
• $1\rightarrow 2$: isochoric heating at $V_1=V_2=2\,{\text m}^3$
• $2\rightarrow 3$: isobaric expansion at $P_2=P_3=5\,$atm
• $3\rightarrow 4$: isochoric cooling at $V_3=V_4=6\,{\text m}^3$
• $4\rightarrow 1$: isobaric compression at $P_4=P_1=2\,$atm
Find the net work by the gas in the cycle and the efficiency of the heat engine based on it, if the lowest temperature in the cycle is 298 K.
Steps:
1. Convert the units to SI:
• $P_2=P_3=5\,{\text{atm}}=506.6\times 10^3\,$Pa
• $P_1=P_4=2\,{\text{atm}}=202.7\times 10^3\,$Pa
2. Since the cycle has a rectangular shape on a $P\!-\!V$ diagram, the net work can be easily calculated as the area inside the rectangle:
• $W_\text{net by gas}=(P_2-P_1)(V_3-V_2)$ $=(506.6-202.7)\times 10^3\,{\text{Pa}}\cdot(6-2)\,{\text m}^3$ $=1.22\times 10^6\,{\text J}=1.22\,$MJ.
3. To find efficiency$\,=\frac{W_\text{net by gas}}{Q_h}$, we need to find the heat supplied to the gas during heating segments
• The temperature $T\sim PV$, therefore it (as well as the internal energy) is increasing during $1\rightarrow 2$ and $2\rightarrow 3$ legs of the cycle.
• Since both $c_v$ and $c_p$ are positive, in these processes the heat going into the gas has the same sign as the temperature change: positive when the temperature increases and negative (i.e. out of the gas) when the temperature decreases.
• $Q_h=Q_\text{in}=c_vn\,(T_2-T_1)+c_pn\,(T_3-T_2)$ $=\frac{3}{2}R\,n\,(T_2-T_1)+\frac{5}{2}R\,n\,(T_3-T_2)$ $=R\,n\,\left(\frac{3}{2}(T_2-T_1)+\frac{5}{2}(T_3-T_2)\right)$
• The lowest temperature in the cycle is at the point with the lowest $PV$ product, i.e. point 1.
• $T_1=298\,$K
• $T_2=T_1\frac{P_2}{P_1}=298\,{\text K}\cdot\frac{5\,{\text{atm}}}{2\,{\text{atm}}}=745\,$K (in an isochoric process, because V = const, $T\sim P$)
• $T_3=T_2\frac{V_3}{V_2}=745\,{\text K}\cdot\frac{6\,{\text m}^3}{2\,{\text m}^3}=2235\,$K (in an isobaric process, because P = const, $T\sim V$)
• $n=\frac{P_1V_1}{RT_1}=\frac{202.7\times 10^3\,{\text{Pa}}\,\cdot\,2\,{\text m}^3}{8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\,\cdot\,298\,{\text K}}=164\,$mol
• Finally, can plug these temperature and mole values into the formula for heat and find $Q_h$:
• $Q_h=8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\cdot 164\,{\text{mol}}\big(\frac{3}{2}\cdot(745\!-\!298)\,{\text K}$ $+\frac{5}{2}\cdot(2235\!-\!745)\,{\text K}\big)$ $=6\times 10^6\,{\text J}=6\,$MJ.
• Now we can find the efficiency (Note, this is not a Carnot cycle, and we cannot use the $1-\frac{T_c}{T_h}$ formula!):
• ${\text{efficiency}}=\frac{W_\text{net by gas}}{Q_h}=\frac{1.22\,{\text{MJ}}}{6\,{\text{MJ}}}=0.204=20.4$%.
#### Problem 15.3
Gasoline in a certain engine is combusted at 1000 K, and the exhaust comes out at 400 K. Assuming the laws governing thermodynamic closed cycles apply here, what is the maximum efficiency that this engine can have?
Steps:
1. The maximum efficiency is achieved in the Carnot cycle with $T_h=1000\,$K and $T_c=400\,$K.
• ${\text{eff}}_\text{max}=1-\frac{T_c}{T_h}=1-0.4=0.6=60$%.
#### Problem 15.4
An ideal heat pump operating on a Carnot cycle in reverse consumes 1 kW from the electrical grid. How much heat per unit time does it supply to keep the inside of the house warm (at 25$\,^\circ$C) during a frigid winter day at 0$\,^\circ$C outside?
Steps:
1. Convert all the units to SI:
• $T_h=298\,$K
• $T_c=273\,$K
2. For a Carnot cycle, no matter straight or reverse, $\frac{|Q_c|}{|Q_h|}=\frac{T_c}{T_h}$. Also, $P_\text{electr}=\frac{W_\text{net on gas}}{t}=\frac{|Q_h|}{t}-\frac{|Q_c|}{t}$
3. Solving for the heat per unit time $|Q_h|/t$,
• $P_\text{heat}=\frac{|Q_h|}{t}=\frac{P_\text{electr}}{\frac{T_h}{T_c}-1}=11\,$kW
## Exam 3 review
Can be found here or on the sidebar
### Final Exam Equation Sheets
One can be found here : ph202_final_equation_sheet.pdf
Another compilation can be found here : notes-15-18.pdf
1. Submerged volume of a solid floating in a fluid: $V_\text{subm}=V_\text{solid}\frac{\rho_\text{solid}}{\rho_\text{fluid}}$
• this is only applicable to chunks of solids (e.g. ice, wood) without cavities, and is not applicable to vessels such as ships (because there is some air below the water line, contributing to $V_\text{subm}$ but not part of the vessel material).
• For floating vessels, use the more general formula $\rho_\text{fluid}\!\cdot\!V_\text{subm}=m_\text{vessel}$, true for anything afloat larger than a few cm (when the surface tension can be neglected).
2. Coefficients of linear and volume expansions are related by an approximate formula (which works well when they are much less than 1):
• $\beta \approx 3\alpha$
3. Conduction of heat:
• $\Delta T$ is not the temperature change in time, rather it is the temperature difference between the two surfaces of the material
4. Internal energy of an ideal gas and the $U=\frac{3}{2}nRT$ formula:
• $\frac{3}{2}nRT$ is the internal energy of a monoatomic ideal gas, which is equal to the internal kinetic energy of the molecules of any ideal gas.
• Non-monoatomic (polyatomic) ideal gases such as O2, N2, and CO2, have an additional internal energy due to rotations, and at some higher temperatures, vibrations of these molecules.
5. Molar specific heats:
• $c_v=\frac{3}{2}\!\cdot\!R\;\;$ and $\;\;c_p=\frac{5}{2}\!\cdot\!R\;\;$ are only true for monoatomic ideal gases, such as He, Ne, Ar, Kr, Xe, Ra, or their mixtures. For all other gases, these constants are higher because of contributions from rotational and vibrational degrees of freedom
6. Efficiency of the heat engine:
• $\frac{Q_h-Q_c}{Q_h}=1-\frac{T_c}{T_h}$ is only true for Carnot engine, where the cycle consists of alternating isotherms and adiabats
Nicholas Kuzma 2014/03/16 14:26
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# Difference between revisions of "2006 AMC 10B Problems/Problem 7"
## Problem
Which of the following is equivalent to $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$?
$\mathrm{(A) \ } -x\qquad \mathrm{(B) \ } x\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \sqrt{\frac{x}{2}}\qquad \mathrm{(E) \ } x\sqrt{-1}$
## Solution 1
$\sqrt{\frac{x}{1-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x}{x}-\frac{x-1}{x}}} = \sqrt{\frac{x}{\frac{x-(x-1)}{x}}} = \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|$
Since $x<0,|x|= \boxed{\textbf{(A)}-x}$
## Solution 2
To confirm the answer, inputting a negative value into $x$ can help. For ease of computation, if $x=-3$, $\sqrt{\frac{-3}{1-\frac{4}{3}}}=\sqrt{\frac{-3}{\frac{-1}{3}}}=\sqrt{9}=3$. As no other option choice fits, $\boxed{\textbf{(A)}-x}$ is the correct solution. Note that if $x=-1$ was chosen, C would have fit as well. Make sure to avoid making this mistake.
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Home » » The scores of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, ...
# The scores of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, ...
### Question
The scores of 16 students in a Mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65, 70. What is the sum of the median and modal scores?
A) 125
B) 130
C) 140
D) 150
E) 137.5
### Explanation:
$$\begin{array}{c|c} Scores & Frequency \\\hline 55 & 1\\ 60 & 4\\ 65 & 6\\ 70 & 4\\ 75 & 1 \end{array}$$
Median = 65(middle number)
Mode = 65(most common)
Median + Mode = 65 + 65 = 130
## Dicussion (1)
• $$\begin{array}{c|c} Scores & Frequency \\\hline 55 & 1\\ 60 & 4\\ 65 & 6\\ 70 & 4\\ 75 & 1 \end{array}$$
Median = 65(middle number)
Mode = 65(most common)
Median + Mode = 65 + 65 = 130
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### Learning Physics [duplicate]
What's the best way to start learning Physics from home? I haven't touched it since my high school course. I know of Coursera and youtube videos? Are there any other sources that anyone can recommend? ...
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### Energy needed for exoergic nuclear reaction
To initiate endoergic nuclear reaction (negative Q Value)minimum energy is supplied which is slightly more than Q value. Then what will be the energy needed for intiating exoergic nuclear reaction ...
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### Gravitational… confinement?
This is a followup to Ergil's question "Weak isospin confinement?". According to the Wikipedia article on color confinement: The current theory is that confinement is due to the ...
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### Is there a limit to what we zoom into if it is in fact infinite
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### What is “special” and what is “general” in Relativity?
Initially I thought in special relativity the velocity was constant, whereas general relativity allowed treatment of accelerated frames as well. But now I have heard that SR is only valid locally?
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### What is the energy of a Gaussian wave packet?
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### The subtle differences between angular momentum and centrifugal force?
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### What does vector space and bra/ket space mean? [on hold]
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### Spivak's Physics for Mathematicians [on hold]
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### Collisions in two dimensions: using relative velocity
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### Using infrared temperature sensor to measure water surface temperature
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### Is wave superposition always equivalent to wave interference?
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### Are all fermions massless at high temperatures?
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### Biot-Savart for current carrying wire
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### What are the relative limitations of the Schrödinger, Pauli, and Dirac Equations?
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### Modelling ions hitting a grid
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### Does water maintain equal level in connected vessels?
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### Calculate the length of propellers needed on an electric motor to lift a mass [on hold]
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### Do infrared lamps provide more heat than incandescent lamps?
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### What does it mean for a Fresnel lens to have two finite conjugates?
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### References or resource recommendation for the mathematics concerning fission
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### How does diffraction cause maxima and minima on a viewing screen
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### Criterion of tangled state by using particular transposition of density matrix
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# Thread: Optimization and Calculus problem: Rate of change of oil slick?
1. ## Optimization and Calculus problem: Rate of change of oil slick?
The more i read the question to this problem, the more i understand what it is asking. However, i cannot develop an equation to solve for the rate of change.
Q: A circular oil slick spreads on the surface of the ocean, the result of a spill of 10 ft^3 of oil. When its radius is 50 ft, its radius is increasing at the rate of 2 ft/s. How fast is the thickness of the slick changing then?
A: - 1 / ( 3125pi ) ft/s
I believe the volume of oil was originally 10 ft cubed. Then the spill creates a oil slick with a radius of 50ft and some "unknown" amount of depth. (oil has to have volume even in the ocean)
So as the oil begins to rise to the surface because it is less dense than water, which increases the radius of the current oil slick at a rate of 2 ft/s. Now the thickness of the oil slick must be decreasing, but at what rate? <---- this is how i interpreted the question above, i could be completely wrong.
Any help is appreciated.
2. Volume of oil V = π*r^2*x, where x is he thickness of the oil at any instant. Here V is constant. Only the radius and the thickness change with time.
So V/x = π*r^2.
Taking the derivative with respect to the time, we get
V(-1/x^2)*dx/dt = 2*π*r*dr/dt
dx/dt = -2*π*r*dr/dt*x^2/V = -2*π*r*dr/dt*(V^2/π^2*r^4)*(1/V)
Now simplify and find dx/dt
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# How does adding a primitive root of unity to a number field effect the ring of integers?
We know that if $$xi$$ is a primitive $$n^text{th}$$-root of unity, then the ring of integers $$mathcal{O}_{mathbb{Q}(xi)}$$ of $$mathbb{Q}(xi)$$ is $$mathbb{Z}(xi)$$.
Can we generalise this result to say much about the ring of integers $$mathcal{O}_{K(xi)}$$ of $$K(xi)$$, where $$K / mathbb{Q}$$ is some finite algebraic extension?
Is it the case that $$mathcal{O}_{K(xi)} = mathcal{O}_{K}(xi)$$?
If this is not generally true, do we have a characterisation of circumstances where this may hold?
Failing that, do we have an alternate description of $$mathcal{O}_{K(xi)}$$ in terms of $$mathcal{O}_{K}$$?
I would appreciate any comments, or even just a reference for these kinds of results.
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<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Concept. Go to the latest version.
# Inscribed Angles in Circles
## Angle with its vertex on a circle and sides that contain chords.
%
Progress
Practice Inscribed Angles in Circles
Progress
%
Inscribed Angles
### Vocabulary Language: English
Arc
Arc
An arc is a section of the circumference of a circle.
Intercepts
Intercepts
The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
Inscribed Angle Theorem
Inscribed Angle Theorem
The Inscribed Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc.
Semicircle Theorem
Semicircle Theorem
The Semicircle Theorem states that any time a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter and the diameter is the hypotenuse.
Inscribed Angle
Inscribed Angle
An inscribed angle is an angle with its vertex on the circle. The measure of an inscribed angle is half the measure of its intercepted arc.
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# Trait accessing variables of classes using it
I just tried PHP Traits for the first time, to keep my code DRY :
<?php
namespace Acme\DemoBundle\Model;
trait CheckPermissionTrait
{
protected function checkPermission($object_id) {$judge = $this->container->get('acme_judge');$user = $this->container->get('security.context')->getToken()->getUser(); if( !$judge->isPermitted($user,$object_id) ) {
throw $this->createAccessDeniedException("Brabbel"); } } } - <?php namespace Acme\DemoBundle\Controller; use Symfony\Bundle\FrameworkBundle\Controller\Controller; use Symfony\Component\HttpFoundation\Request; use Acme\DemoBundle\Model\CheckPermissionTrait; class SomeController extends Controller { use CheckPermissionTrait; public function someAction(Request$request, $object_id) {$this->checkPermission($object_id); //do other stuff } } What making me a little nervous, is that the trait is accessing the container of the class using it, which means there is actually the requirement that the class using the trait needs to extend the Controller (Symfony\Bundle\FrameworkBundle\Controller\Controller) for the trait to work. Is this normal when using traits or is it bad practice? • I look forward to seeing answers, as I know very little about traits. You can moot the question by refactoring to checkPermission($object_id, $judge,$user) and not requiring the trait function to know about $this->container -- but I'm curious to learn about the best practice here. Dec 18 '14 at 22:14 • Yes you might want to post that as an answer. Another option is I could make an abstract controller extending the symfony framework controller and then extend that, which is probably better in this case. Dec 19 '14 at 6:09 • @MarcelBurkhard: Thanks for accepting my rant, but while you were doing that, I was busy updating my answer, adding some (what I feel) more details and possible issues that an abuse/over-use of traits could cause, so it might be worth a look Dec 23 '14 at 11:41 • I see you've pointed this out in your own answer to your question, but I wanted to get this in up here on the original post. In the code provided, the DI Container is being used as a Service Locator which is an anti-pattern because it effectively hides the true dependencies of a class. One should avoid calls like$container->get('alias') in favor of injecting dependencies into the class e.g. via the constructor. Mar 17 '16 at 17:56
Though you now have seen the light, and know not to use traits in this way, I'm going ahead and post my review here anyway. I'll be updating it along the way, seeing as this is something I can rant on about for some time.
Your trait suffers from a couple of (severe) issues:
• It constitutes both a breach of contract, and enforces a bad contract.
• Traits requiring properties are like using global in a class
• It's not a valid use-case for a trait. At all.
• Its namespace is wrong
Now these are just the first things I could think of, each of which are problematic enough to ditch the trait as you rightfully did.
Allow me to explain, though, what these one-liners mean:
# Breach of Contract + enforcing a bad one:
Classes define a contract. That's a given. If you pass an instance of class X to a function, you make a deal with that function. You're essentially saying "This object has this specific job, and you can call these methods on it, and/or access these properties". When you extend a class, it's important to respect the contract. You can add things, but never change the promises made by the base class.
If the base class' constructor, for example, expects 1 argument of the type array (enforced through type-hinting), then all of its children should either use the same constructor or, if they choose to override it, define a constructor that takes 1 array argument. They are allowed to loose the type-hint, but may not change it. They're free to add optional arguments, but can't force additional arguments to be passed to the constructor:
abstract class Base
{
public function __construct(array $param) {} } class ValidChild extends Base {} class AnotherChild extends Base { public function __construct(array$param, $opt = null) {} } class IffyButTheoreticallyFine extends Base { public function __construct($param, array $opt = null) {} } These are all valid child classes, however: these are not. They are evil, but sadly commonplace: class Bastard extends Base { public function __construct() { } } class Ungrateful extends Base { public function __construct(array$param = null)
{}
}
class EvilTwin extends Base
{
public function __construct(stdClass $param) { parent::__construct( (array)$param);
}
}
class Psychopath extends Base
{
public function __construct(PDO $db, array$parent = null)
{
if (!$parent) {$parent = [1, 2, 3];
}
parent::__construct($parent); } } Now, how does this apply to your case? Simple: A trait is a tool to reduce code-duplication. It's a nugget of often-needed functionality that a class (a contract) includes/uses to do its job. For a class to use a trait, the trait has no right to define that class' contract. A trait's relation to its user is, in some way, the same as that of a child's towards its parent: it's the child that inherits the contract, not the parent. Therefore, it must follow that any class, regardless of its job, has to be able to use a trait, without having to change its own contract and dependencies. Your trait, then is not a valid use-case: all of its users are required, by definition, to add a $this->container property to its dependencies, which will of course have an impact on that class' contract and the contract of its children.
# A new global in drag
I think it's safe to say that it's universally accepted that a class definition containing global $someVar is a tell-tale sign of bad code. A class is scoped, its methods are scoped and neither of them should rely on the global scope to function. The same applies to a trait: because it can be used in a variety of classes, a trait must be context (and scope) agnostic. If it requires a property to be present, the trait must define it. If that property has to be of a particular type, the only thing you could do is define a property, and enforce an abstract setter for it, which imposes type restrictions. However, this is unreliable, given that multiple traits can be used, it might cause conflicts with the user and the class itself has direct access to the property, too. Lastly: this approach brings us back to square one: a trait like this is no longer a trait. This implies that a trait cannot have dependencies. If a trait has a dependency, it ceases to be a trait: it's either an abstract class, or an Interface, but definitely not a trait. There is, however a tiny sliver of grey-area here: Whilst it's not my idea of best practice, one might argue that it's acceptable to write a trait, to be used in tandem with an Interface, for example: a single Interface that ensures data models are traversable, like the IteratorAggregate interface. In that case, a trait like this could perhaps be acceptable. trait IteratorAggregateTrait { public function getIterator() { if (!$this instanceof IteratorAggregate) {
throw new \LogicException(
sprintf(
'%s must implement IteratorAggregate to use %s',
get_class($this), __TRAIT__ ) ); } return new ArrayIterator($this);
}
}
# Valid use-case
For reasons I've explained above, it's clear that your example isn't something best handled with traits. It might be useful at this point to serve up a small, 100% valid example of a trait. Suppose you've got some models that represent either a DB record, user input (form data), or something else entirely. All of these models contain user information, including an email address. A model with setters is ideal to ensure that, whenever some data is gathered, that data is then validated (ie filter_var($email, FILTER_VALIDATE_EMAIL)). Because these classes represent different things (db record or form), they're likely to inherit from an abstract class already. Rather than duplicating the email setter, a trait can be used: trait UserDataTrait { public$email = null;
public function setEmail($email) { if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
}
$this->email =$email;
return $this; } } # Wrong namespace Just a little nit-pick: I don't like the fact that your trait is defined in the Model namespace, while it's pretty clear that it can't be used anywhere else but in the controller. # Update Time for an update. While I stand (firmly) by what I've stated previously (traits not being in a position to impose a contract), there are certainly people who will argue that they can. Not in the least because PHP's trait construct enables you to declare abstract methods: trait Evil { protected$dependency = null;
public function someMethod()
{
return $this->dependency->getValue(); } abstract public function setDependency(SomeType$dep);
}
This trait, then, imposes a partial contract on its user: the class that uses the trait must define the setDependency method. So what's the problem, you might say? The problem is simple: traits were introduced to solve supposed problems that might require multiple inheritance, hence a class can use more than one trait, and it's possible for the class to resolve conflicts regarding method names by itself. So let's assume the trait above is being used, and you have a couple more traits lying around, one of which looks like this:
trait Pure
{
public $dependency = null; public function setDependency(Container$container)
{
$this->dependency =$container;
return $this; } } Now, imagine you have a class that uses both these traits (assume Evil::setDependency is not declared abstract here): class FromHell { use Pure, Evil { Pure::setDependency insteadof Evil; } } Congrats, the contract imposed by the trait is broken, along with its functionality, simply because the class has chosen to adhere to the contract imposed by Pure. It doesn't take a rocket scientist to realize that traits, whilst they can, in theory, impose a minimal contract, are not as potent at ensuring the contract is adhered to. Now this was a simple example, but if you want a more plausible scenario: Assume 2 traits, A and B, both define a method logError, A::logError writes to a default log file, B is more generic, and expects you to pass a resource as an argument, or writes to stderr by default. Because your project is rather large, you're dealing with multiple levels of inheritance and at some point, some classes end up implementing both traits. Of course, depending on what class it is, either A::logError is given a higher precedence, or B:logError. The latter is more likely to be used in workers/crons/backend code. It's very likely that your project, seeing as you're relying on traits so much, contains code that looks like this: $traits = class_uses($object); if (isset($traits['A']))
{
$object->logError('The error string'); } Now if this object has chosen to override A::logError with B::logError, you'll probably waste a lot of time trying to work out why on earth logError doesn't write your error messages to the default log. In short: # Trait contracts are weak suggestions, and can be overridden easily Back to the definition of abstract methods in a trait: I'll have to test it a bit more, but AFAIK, that's just bad design. Either an abstract class contains your abstract methods (because a class enforces a strong contract), or you implement the method in the class from the get-go. A trait is supposed to reduce code duplication, whereas an abstract method requires you to re-write the method's signature and implementation whenever you use the trait. If that's what you want, why bother using a trait in the first place? The other issue I have with this is one I'll look into a bit more when I find the time, but classes can be composed out of 1 or more traits, and contain no other definitions of their own. That's fine, but what if we look at our 2 traits above (Pure and Evil). Is Pure::setDependency a valid implementation of the abstract Evil::setDependency method? If so, will that trait method be regarded as an implementation of the abstract method? If not, traits containing abstracts can't be used in classes that contain no definitions of their own, and therefore, such traits are special cases (because their usage is limited). If Pure::setDependency is a valid implementation of the abstract method in Evil (Container being an alias of SomeType, or its parent class), and traits can be used to implement abstract trait methods, then another issue presents itself: maintenance HELL: Changing either one of these traits could break your code, depending on how, and where the traits are being used. You'll end up having to test all classes using both traits, whenever you change just one trait. When you look at it like that, I'd say some code duplication is the lesser of two evils. • I'd just like to point out that whilst normally your subclass should (must in my book!) respect the rules laid down in any methods you override, constructors are an exception. This is because a subclass might have completely different dependencies from the superclass even if it has the same functionality. PHP won't complain about constructor signatures not matching and the programmer should know what the requirements are when they instantiate a new class();. The only place where you really do need to enforce constructors is if the class will be built by a factory Feb 28 '18 at 15:40 • This is going into my team's design guide. We cornered ourselves into very dependent traits and when we move to a new framework, traits will be highly restricted. This explanation will help me enforce that. Thank you! Jul 11 '19 at 12:37 After doing some "research" I come to the conclusion, that you shouldn't use Traits the way I did. My main reasons for this conclusion are bad testability and readability. (Please read the first source below) My code above (in the question) violates some design principles and I will refactor my controller to extend an abstractcontroller which implements the checkPermission function I need in all controllers extending it. Sources: # Edit: (2015-07-28) Months have passed and I come back to this answer and I want to mention a major flaw in the code. The controller above relies on the whole service container being injected, which is called the Service locator (anti-)pattern. Dependencies are retrieved by calling $this->container->get('name_of_some_service') instead of injecting them directly. This violates the law of demeter and results in bad testability and reusability.
Dependency injection is the way to go there and it is also supported by the symfony framework.
The following link will get you started: How to Define Controllers as Services
By using dependency injection you get lower coupling to the framework, dependencies of a given class are in most cases easily visible by looking at the constructor.
• Just for completeness: While it's true that the service locator can affect testability, testing frameworks like codeception do have modules that allow you to replace the container with a mockable object, effectively solving the problems facing while testing Jul 29 '15 at 8:56
It's possible to rebind $this in the scope of a single Closure, so you could access the container of any passed in object function isJudgePermitted($object, int $object_id) : bool { return (function (int$object_id) : bool {
$judge =$this->container->get('acme_judge');
$user =$this->container->get('security.context')->getToken()->getUser();
return $judge->isPermitted($user, $object_id); })->call($object, $object_id); } // Inside SomeController. public function someAction(Request$request, $object_id) {$judge_permitted = isJudgePermitted($this,$object_id);
$judge_permitted ?: throw$this->createAccessDeniedException("Brabbel");
//do other stuff
}
|
{}
|
# Template for endianness-free code, data always packed as BIG-Endian
It's been a while since I've been properly grilled about my code.
I had this Idea which should probably never make it into production code but still I couldn't find anything seriously wrong with it at a glance...
#ifndef BITFIELDMEMBER_H
#define BITFIELDMEMBER_H
/************************************************************************
* This is for creating Endianness-free bit-fields.
* The values are always packed as big-endian.
* When used for arithmetics/comparison, the values are read in the
* proper endianness for the current CPU mode.
*
* Note: The fields can overlap if you want
*/
template<int firstBit, int bitSize>
struct BitFieldMember
{
typedef BitFieldMember<firstBit, bitSize> self_t;
typedef unsigned char uchar;
enum {
lastBit = firstBit + bitSize - 1,
mask = (1 << bitSize) - 1
};
uchar *selfArray()
{
return reinterpret_cast<uchar *>(this);
}
const uchar *selfArray() const
{
return reinterpret_cast<const uchar *>(this);
}
/* used to read data from the field */
/* will also work with all the operators that work with integral types */
inline operator unsigned() const
{
const uchar *arr = selfArray();
const uchar *p = arr + firstBit / 8;
int i = 8 - (firstBit & 7);
unsigned ret = 0;
ret |= *p;
while (i < bitSize)
{
ret <<= 8;
ret |= *(++p);
i += 8;
}
return ((ret >> (7 - (lastBit & 7))) & mask);
}
/* used to assign a value into the field */
inline self_t& operator=(unsigned m)
{
uchar *arr = selfArray();
m <<= (7 - (lastBit & 7));
uchar *p = arr + lastBit / 8;
int i = (lastBit & 7) + 1;
while (i < bitSize)
{
m >>= 8;
i += 8;
}
return *this;
}
inline self_t& operator+=(unsigned m)
{
*this = *this + m;
return *this;
}
inline self_t& operator-=(unsigned m)
{
*this = *this - m;
return *this;
}
inline self_t& operator*=(unsigned m)
{
*this = *this * m;
return *this;
}
inline self_t& operator/=(unsigned m)
{
*this = *this / m;
return *this;
}
inline self_t& operator%=(unsigned m)
{
*this = *this % m;
return *this;
}
inline self_t& operator<<=(unsigned m)
{
*this = *this << m;
return *this;
}
inline self_t& operator>>=(unsigned m)
{
*this = *this >> m;
return *this;
}
inline self_t& operator|=(unsigned m)
{
*this = *this | m;
return *this;
}
inline self_t& operator&=(unsigned m)
{
*this = *this & m;
return *this;
}
inline self_t& operator^=(unsigned m)
{
*this = *this ^ m;
return *this;
}
};
#endif
Usage example:
union header
{
unsigned char arr[2]; // space allocation, 2 bytes (16 bits)
BitFieldMember<0, 4> m1; // first 4 bits
BitFieldMember<4, 5> m2; // The following 5 bits
BitFieldMember<9, 6> m3; // The following 6 bits, total 16 bits
};
int main()
{
memset(a.arr, 0, sizeof(a.arr));
a.m1 = rand();
a.m3 = a.m1;
a.m2 = ~a.m1;
return 0;
}
This sometimes wouldn't work for fields that are larger than 24 bits... the shifting may trim the most significant bits.
• there's a blog post with the same code here blog.codef00.com/2014/12/06/portable-bitfields-using-c11. Is this yours? It mentions UB, but does not specify where and what kind of UB that is. Could you probably tell more about that? – Slava Oct 18 '17 at 19:48
• That one isn't me. I didn't read the blog post. What is UB? – CplusPuzzle Oct 18 '17 at 20:00
• Undefined Behaviour. Funny, that blog post appeared at about the same time as yours, and the code is similar, though a bit different. – Slava Oct 18 '17 at 20:58
Your code is good, except for two methods : operator unsigned() const and self_t& operator=(unsigned m). They are too complex, and by just looking at them, it is far from clear what they are doing.
enum {
lastBit = firstBit + bitSize - 1,
mask = (1 << bitSize) - 1,
firstBitIndex = 8 - (firstBit & 7), // change to proper name
nameThisValueProperly = (7 - (lastBit & 7)) & mask, // change to proper name
};
First modify your operator unsigned() const to this :
inline operator unsigned() const
{
const uchar *arr = selfArray();
const uchar *p = arr + firstBit / 8;
unsigned result = CalculateSomeValue( p );
return (result >> nameThisValueProperly );
}
// name this method properly
inline unsigned CalculateSomeValue( const uchar *p ) const
{
int i = firstBitIndex;
unsigned result = *p;
while (i < bitSize)
{
result <<= 8;
result |= *(++p);
i += 8;
}
return result;
}
Of course, you need to put proper names both for enum values and method names, so when someone looks into this code says "Aha, that is what is is doing". This way, you can easy increase code clarity.
Also, by making method smaller, you can easily test them.
You can do something similar for the self_t& operator=(unsigned m). Break it up, until you find it clean.
• There's a small bug in there: nameThisValueProperly = (7 - (lastBit & 7)) & mask breaks things for field sizes 1 & 2. Removing &mask fixes it, but make sure there's a &mask wherever this value gets used. – Brian Vandenberg Apr 24 '18 at 0:59
Here are a few things found wrong with this implementation when using it in production code are the following:
Assignment between fields that are of the exact same type will result in an empty implementation with nothing copied because the fields have no storage of their own. To resolve this, it's best to cast the right side of the assignment to unsigned. The trouble is finding this bug in the first place.
A workaround would have been to block the assignment by either making it private or delete it with = delete and let the compiler find it for you but this will require C++11 or higher.
C++03 doesn't allow union members to be types that have declarations for assignment operators.
Also, it's missing some operators:
inline self_t& operator++()
{
*this = *this + 1;
return *this;
}
inline self_t& operator--()
{
*this = *this - 1;
return *this;
}
inline unsigned operator++(int)
{
unsigned tmp = *this;
*this = tmp + 1;
return tmp;
}
inline unsigned operator--(int)
{
unsigned tmp = *this;
*this = tmp - 1;
return tmp;
}
Last and least (assuming unsigned is 32 bits), if someone insists on having fields that are 32 bits, you will need to make the following modification to the enum of the mask:
enum
{
lastBit = firstBit + bitSize - 1,
mask = (1ULL << bitSize) - 1
};
Other than that, it served us well.
• In case it wasn't clear, assignment between objects of the union type works fine because it's basically just a memcpy – CplusPuzzle Oct 19 '17 at 5:43
I implemented yours with the following changes:
• Replaced enum with static constexpr
• Deleted default / copy / move constructors
• Constructors imply ownership of resources. I figured this would help avoid undefined behavior
• Deleted unary operator&() for similar reasons.
• Implemented self_t& operator=(const self_t&) to deal with the problem you noted:
Assignment between fields that are of the exact same type will result in an empty implementation with nothing copied because the fields have no storage of their own.
Originally I added an array, necessitating an extra template argument: BitfieldMember<sizeof(data), ...>, but this made it unnecessary.
• Added the template argument from the last bullet and a static_assert in an attempt to require the storage be large enough.
• Changed the set/get logic for readability:
private:
inline operator base_type() const {
base_type ret = 0;
for( unsigned ii = firstBit / 8; ii <= lastBit / 8; ++ii ) {
ret = (ret << 8) | data()[ii];
}
return (ret >> excessBitsInLastByte) & mask;
}
inline self_t& operator=( base_type m ) {
m = (m & mask) << excessBitsInLastByte;
for( int ii = lastBit / 8; ii >= 0 && 0 != write_mask; --ii ) {
uchar& ref = data()[ii];
ref |= m;
m >>= 8;
}
return *this;
}
I've been looking for a way to add compile time errors for the following situations but haven't managed it yet:
• if any fields overlap
• I have some ideas for this, haven't tried them yet.
• if the union's size is larger than its array
• This works but I was hoping for something more automatic: put a static_assert after the union definition is complete.
• if virtual functions are added to BitfieldMember (or derived classes have virtual functions)
• The vtable will increase the union's size and each one will clobber each-other's vtable.
• It's probably not possible to catch the latter case
• If data members are added to BitfieldMember (or deriving classes)
• I can catch the first case automatically, but I don't think it's possible to catch the 2nd case.
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# Question #c9a04
Aug 27, 2017
Permanganate anion is usually oxidized to COLOURLESS $M {n}^{2 +}$
#### Explanation:
And we would represent this reaction by the 5 electron reduction.....
${\underbrace{M n {O}_{4}^{-}}}_{\text{deep purple}} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$
VERY concentrated solutions of $M {n}^{2 +}$ are a pale rose.
You will have to supply a corresponding oxidation reaction.
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# Area of a parallelogram, vertices $(-1,-1), (4,1), (5,3), (10,5)$.
I need to find the area of a parallelogram with vertices $(-1,-1), (4,1), (5,3), (10,5)$.
If I denote $A=(-1,-1)$, $B=(4,1)$, $C=(5,3)$, $D=(10,5)$, then I see that $\overrightarrow{AB}=(5,2)=\overrightarrow{CD}$. Similarly $\overrightarrow{AC}=\overrightarrow{BD}$. So I see that these points indeed form a parallelogram.
It is assignment from linear algebra class. I wasn't sure if I had to like use a matrix or something.
• Try drawing a picture. Work out the distance between each of the edges of the parallelogram (using Pythagoras' theorem). – Dan Rust Mar 16 '13 at 1:29
• well cant I just use the area formula for a parallelogram and find the lengths of the edges? since it is linear algebra I wasnt sure if I had to like use a matrix or something. This just seemed like too simple of a question right? – D-Man Mar 16 '13 at 1:33
• hint: draw a picture. The area is the magnitude of the cross product of certain two vectors. You will have to adjoin a third coordinate of $0$ to make the vectors three-dimensional – Stefan Smith Mar 16 '13 at 1:35
• One definition of the area of a parallelogram is the determinant of the matrix with it's edge vectors as its columns. If you haven't been given/shown this, then you may need to prove it. – Dan Rust Mar 16 '13 at 1:35
• Yeah, if you're not supposed to use cross products or you don't know what a cross product is, you'll have to do it another way which may look different but may actually be equivalent. – Stefan Smith Mar 16 '13 at 1:37
Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the $2\times 2$ matrix formed by the column vectors representing component vectors determined by the given points. $$A = \text{det}\,\left(\vec u \;\; \vec v\right)$$
The area of a parallelogram is also equal to the magnitutude of the cross product of the component vectors $\vec u, \vec v$ = $$|\vec u\times \vec v| = |\vec u|\,|\vec v|\sin \theta$$
where $\theta$ is the measure of the angle formed by the component vectors $\vec u, \vec v$.
Use your points to determine the component vectors $\vec u, \vec v$.
For parallelogram formed by $p_1, p_2, p_3, p_4$, put $\vec u = p_2 - p_1$, $\vec v = p_3 - p_1$ (where $p_4$ is the point opposite $p_1$, $p_2$ the point opposite $p_3$): $$\vec u = \langle 4 -(- 1), 1-(-1)\rangle = \langle 5, 2\rangle$$ $$\vec v = \langle 5 - (-1), 3 - (-1)\rangle = \langle 6, 4\rangle$$
So compute $$A = \det \begin{pmatrix} 5 & 6 \\ 2 & 4 \end{pmatrix},\;\;\text{or}\;\; A = \vert \vec u\times \vec v\vert$$
See the parallelogram whose area can be determined by component vectors $\vec u, \vec v\;\;$ (left of the image):
• "Hint: the area of a parallelogram (see left-most image) is equal to the determinant of the 2×2 matrix formed by the column vectors representing component vectors determined by the given points." But why? – Kenneth Worden Sep 8 '15 at 21:22
Hints:
1. The area of a parallelogram with side vectors $\bf a$ and $\bf b$ is $\det(\bf a\ \bf b)$.
2. For a parallelogram $A,B,C,D$ its side vectors are e.g. $B-A$ and $C-A$.
As the diagonal of a parallelogram $A(-1,-1),B(4,1),C(5,3),D(10,5)$ divides into two congruent triangles, which implies the triangles have same area.
So, the area of the parallelogram will be $2\cdot$ area of any one of $\triangle ABC,\triangle ADC, \triangle ABD, \triangle BCD$
For example, the area of $$\triangle ABC=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 4&1&1 \\ 5&3&1\end{pmatrix}\right|$$
$$=\frac12\left|\det\begin{pmatrix} -1&-1&1 \\ 5&2&0 \\ 6&4&0\end{pmatrix}\right|\text{ applying} R_2'=R_2-R_1,R_3'=R_3-R_1$$
$$=\frac12\left|5\cdot4-6\cdot2\right|=4$$
• So because the area of triangle ABC was 4 and because it only represents half of the parallelogram the total area of this parallelogram would be 4 x 2 = 8? – D-Man Mar 16 '13 at 21:40
• @D-Man, yes, you are right. – lab bhattacharjee Mar 17 '13 at 3:59
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# What is branch factor for beam search in RNNs like TensorFlow's Magenta?
TensorFlow's Magenta melody generation module has a parameter for branch_factor1. What is it?
Branch factor in beam search refers to average number of branches at each level of a search tree. When predicting a sequence in an LSTM model, such as note selection in melody generation, $b$ branches will be decoded. Increasing the branch factor from 1 to 4 in the call to melody_rnn_generate will maintain the average number of branches to be at $b$.
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If we pute let $\mathbb{F}{q}$ \mathbb{F}_{q}$ be the finite field with$q$elements, is there a local field$F$with residue field $\mathbb{F}{q}$\mathbb{F}_{q}$ ?
If we pute $\mathbb{F}{q}$ the finite field with $q$ elements, is there a local field $F$ with residue field $\mathbb{F}{q}$ ?
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Earthquake-Related Changes in Species Spatial Niche Overlaps in Spring Communities
Abstract
Species interactions between stygobites (obligate groundwater organisms) are poorly known, reflecting the difficulty in studying such organisms in their natural environments. Some insight can be gained from the study of the spatial variability in microcrustacean communities in groundwater-fed springs. Earthquakes can increase hydraulic conductivity in the recharge area of karstic aquifers and flow rates in discharge zones, thus dislodging stygobites from their original habitats to the spring outlets. Earthquakes are expected to alter species spatial niche overlap at the spring outlets, where stygobites coexist with non-stygobites living in benthic and subsurface habitats. We compared the abundance of stygobiotic and non-stygobiotic microcrustaceans in groundwater-fed springs before and after the 6.3-Mw earthquake that hit the karstic Gran Sasso Aquifer (Italy) in 2009. Pre-seismic (1997, 2005) overall niche overlaps were not different from null expectations, while post-seismic (2012) species mean niche overlaps were higher, following the redistribution of animals caused by the earthquake-triggered discharge. The reduced abundance of stygobites following their dislodgement from the aquifer and the concomitant displacement of non-stygobites led to a higher post-seismic co-occurrence of stygobites and non-stygobites. Changes in aquifer structure destroyed pre-seismic species segregation patterns by creating new or strengthening already existing interactions.
Introduction
The quantitative study of niche overlap in community ecology dates back to the pioneering work of MacArthur and Levins1, who introduced an index based on the relative utilisation of different segments of a niche resource axis (dimension). This work paved the way for a fruitful research line on niche theory especially in the 1970s, including the possible relationships of niche overlap with competition, fitness, diversity and community structure2,3,4,5.
According to classical ecological theory6, 7, a differential utilisation of resources, i.e. low niche overlap, is considered essential for the coexistence of sympatric species. Because of the variety of dimensions that form species’ niches7, overlap on one resource axis may indicate diversification along other resource axes2, 5, 8, which may explain the coexistence of species greatly overlapping along one or a few dimensions. Microhabitat, diet, and temporal activity (e.g. seasonal or diel differences between species in habitat exploitation) are the three most frequently used characteristics to measure niche overlap2, 5, 9. In particular, interspecific overlap in habitat use has been invoked as a key factor in shaping ecological communities from the small (e.g., close-range interspecific interactions10) to the large scale (e.g., species distribution areas, including evolutionary aspects11).
Groundwater animal communities comprise species that have developed adaptive traits to survive in total darkness, lack of photosynthesis, low food availability, and strict dependence on the organic matter entering groundwater from the surface, potentially leading to severe inter- and intraspecific competition12. However, detailed data about how groundwater species use space at fine scale are lacking, especially because sampling is hampered by the limited access to groundwater. Thus, most of the available information on groundwater species ecology is based on sampling groundwater-fed springs, caves and boreholes13, 14.
Karstic aquifers are complex systems whose parts are interconnected by water that flows from surface recharge areas to the aquifer outlets, such as springs. This interconnection between the recharge area and the different parts of the aquifer should promote faunal homogenisation by favouring species drift, especially for non-stygobites that live on the surface, enter the aquifer through fast infiltration pathways and are dispersed by groundwater flows, especially during high-discharge periods15, 16. These species also reach the spring environment via the surface water hydrological continuum. By contrast, obligate groundwater species (i.e. stygobites) typically show restricted and idiosyncratic distributions, indicative of limited dispersal abilities14.
The very high level of stygobiotic endemism, with more than 90% of the groundwater fauna comprised of local endemics17, also supports the view of low dispersal for most groundwater species (but see ref.18). This is consistent with the fact that animal communities inhabiting groundwater environments are less subject to strong environmental variations than their counterparts that live on the surface. However, groundwater environments can be altered by major disturbance events such as earthquakes14, 19.
The L’Aquila earthquake on 6 April 2009 (6.3 Mw) changed the groundwater flow of the Gran Sasso Aquifer (GSA) and consequently of the Tirino River valley, where ~65% of the aquifer discharge is located20. The earthquake induced an increase in the bulk hydraulic conductivity of the recharge area, near the ruptured fault zone (attributable to fracture clearing and/or microcrack formations), and led to an anomalous rising of the water table and flow rate in discharge zones21. As a result, the overall physiography of the spring system changed from rheo-limnocrene to predominantly limnocrene22, 23. These changes altered the community organisation of subsurface (i.e. below the spring bed) microcrustaceans at the GSA main outlet at the Tirino Springs (TS), especially via a strong reduction in the abundance of stygobites14.
Because of the tight relationships between ground shaking, aquifer strain, fracturing and fracture clearing and changes in stygobites’ habitats, we expected an impact of the 2009 earthquake on species distribution in the aquifer and at the major aquifer discharge points. In particular, the post-seismic lower densities and patchier occurrence of stygobites at the TS, coupled with the persistent diffuse occurrence of non-stygobites14, could have increased species coexistence by reducing the degree of spatial niche overlap among species.
To test if the earthquake increased interspecific spatial overlap in spring communities, we compared species co-occurrence in low-discharge (1997), high-discharge (2005), and post-seismic, very high-discharge (2012) hydrological years using the Copepoda (Crustacea) as a model biotic group. Copepods also comprise ~80% of the total meiofauna density at TS14, 22, 23. In particular, we tested if the earthquake-induced higher discharge increased the niche overlap of copepod species by reshaping their micro-distribution in the TS.
Results
A total of 22 species, 9 of which (~40%) are stygobites, were found in the three sampling years (S1 Supporting Information). Total number of individuals retrieved from samples ranged from 992 (average number of individuals per sample ± SE = 124.000 ± 26.012) in 1997 (a low-discharge year) to 2750 (average number of individuals per sample ± SE = 343.750 ± 118.583) in 2005 (an above-average discharge year), and then back to 910 (average number of individuals per sample ± SE = 113.750 ± 49.846) in post-seismic 2012 (a very high discharge year, Table S1). Average number of individuals per sample did not differ significantly between years (ANOVA: F 2,21 = 2.941, P = 0.075). Stygobiotic species contribution to total abundance decreased from 59% in 1997 to 35% in 2005, then further crashed to 21% in 2012 (contingency table: χ2 = 310.015, df = 2, P < 0.0001). Despite a marked reduction in total abundance after the 2009 earthquake, the harpacticoid copepod Nitocrella pescei remained the most common stygobite throughout the sampling period (S1 Supporting Information).
Overall mean niche overlap was 30% in 1997, 34% in 2005 and 58% in 2012 (Table 1). Mean overall niche overlaps in 1997 were not significantly higher than the respective simulated assemblages for all three types of assemblages (whole assemblage, stygobites only and non-stygobites only), regardless of the randomisation algorithm used (P > 0.05 in all cases; Table 1). Mean overall niche overlap in 2005 was significantly higher than the simulated assemblages constructed using the RA3 randomisation algorithm (P < 0.01) but not under the assumption of the RA2 algorithm (P = 1.000) (Table 1). The mean niche overlap of stygobiotic species in 2005 was not significantly higher than expected regardless of the randomisation algorithm used (P > 0.05 in all cases; Table 1). The mean niche overlap of non-stygobites was significantly higher than expected under RA3 (P = 0.001), but not under RA2 (P = 0.461) (Table 1). Finally, in 2012, mean overall niche overlap for the whole assemblage, stygobiotic mean overall niche overlap, and non-stygobiotic mean overall niche overlap were all significantly higher than expected, regardless of the randomisation algorithm used (P < 0.005 in all cases; Table 1).
Using pairwise values (O), niche overlap varied significantly among years and type of species pairs (stygobite–stygobite or s,s; non-stygobite–non-stygobite or n,n; or stygobite–non-stygobite or s,n) (Table 2). Bonferroni’s tests indicate significant differences in the mean O values among all years (P 1997–2005 = 0.026, P 1997–2012 < 0.00001, P 2005–2012 < 0.00001), with mean values increasing from 1997 to 2005 and again from 2005 to 2012. The Bonferroni tests indicate that the mean O values of pairs in which both species were non-stygobites (n,n) were significantly higher than those in which both species were stygobites (s,s) (P < 0.00001) or in which one species was a stygobite and the other one was not (s,n) (P < 0.00001); mean O values of s,s and s,n pairs were statistically similar (P = 0.240).
The interaction term was marginally significant (P = 0.048, Table 2), which suggests that the different pair types responded differently in the three years. Inspection of the interaction plot (Fig. 1) and use of paired t-tests indicate that s,s and s,n pairs had similar mean values of niche overlap between 1997 and 2005 (P s,s = 0.672, P s,n = 0.833), with a significant increase between 2005 and 2012 (P s,s = 0.024, P s,n < 0.0001), whereas the spatial overlap for n,n pairs increased significantly between 1997 and 2005 and again between 2005 and 2012 (P n,n < 0.0001 in both cases).
Using species mean overlap values (O sp), overlap varied significantly between years regardless of species habitat specialisation (stygobites vs non-stygobites) (Table 3). Bonferroni’s tests indicate that the 2012 O sp values were significantly higher (P < 0.00001 for both the 1997–2012 and the 2005–2012 comparisons), but the difference between 1997 and 2005 was not significant (P = 0.870) (Fig. 2).
An analysis of co-occurrence (C-scores) revealed that increases in niche overlap were paralleled by a reduction in species segregation. C-scores decreased from 1997 (C = 1.948) to 2005 (C = 1.457) to 2012 (C = 0.931), which means that the randomness of species distribution increased. A repeated-measures ANOVA (rANOVA) showed that C-scores varied significantly among years (SS = 91.460, MS = 45.730, F 2,300 = 12.661, P < 0.00001) but not for type of species pairs (SS = 0.729, MS = 0.364, F 2,150 = 0.008, P = 0.9254) and the species × year interaction was not significant (SS = 28.675, MS = 7.169, F 4,300 = 1.985, P = 0.097).
Discussion
In general, mean species niche overlaps were significantly higher in 2012 than in 1997 and 2005, for both stygobites and non-stygobites. Although results for the 2005 community are unclear (because the two algorithms produced contrasting outputs), results for 2012 clearly indicate a high degree of niche overlap. However, species pair interactions differed significantly also between 1997 and 2005, with changes in niche overlap that varied according to the type of species pair. Namely, pairs in which both species were non-stygobites increased their niche overlap significantly from year to year, whereas the niche overlap for pairs in which one or both species were stygobites increased significantly only in 2012. These results suggest that while non-stygobites may tune their utilisation of the space in response to hydrological changes, stygobites tend to be less adaptable, as expected in organisms specialised to live in genuine groundwater habitats. Only a major disruptive event like an earthquake seems to force a strong niche overlap among stygobites. Though this hypothesis awaits empirical confirmation, it is in line with the earthquake-induced flushing of stygobiotic species and their abundance changes discussed elsewhere14.
It has been recently suggested that springs may be considered as islands for stygobites24. According to an insular model, the copepod assemblages of the TS should form a metacommunity regulated by patch dynamics models. In general, four metacommunity models have been proposed to account for variation in local community structure: (1) neutral model (species are essentially equivalent in their competitive and dispersal abilities, and the metacommunity is mainly regulated by stochastic immigration and extinction processes), (2) species sorting (individual species respond differently to environmental heterogeneity), (3) mass effects (dispersal and environmental heterogeneity interact to determine species abundance and composition), and (4) patch dynamics (interacting species differ from each other in being either good competitors or good colonizers within a uniform environment), but the last two perspectives – mass effects and patch dynamics – have been recently suggested to be special cases of the species sorting perspective25, 26. Significant variations in niche spatial overlaps determined by strong environmental changes support the idea that the TS copepods are a metacommunity regulated by the interactions of mass effects and patch dynamics.
Our results indicate that the TS subsurface copepod community is not only subjected to wide fluctuations in species abundances as a consequence of relatively small environmental changes, but also to variations in niche overlap between species pairs determined by anomalous high disturbance events. The term “disturbance” has a variety of meanings including any event that may kill or displace organisms, deplete consumable resources (e.g., living space and food) and degrade or destroy habitat structure27. Spring communities constantly experience small variations in discharge, such as the increase observed in 2005 at TS. However, the stygobiotic species assemblages of the TS system were not affected by this small variation. This is because the TS stygobiotic species primarily reside in low-flow “chambers” located inside the carbonate massif feeding the spring outlets14. With the high increase in discharge due to the main shock, the water pressure in the fast-flowing conduits of the aquifer decreased, allowing the microcrustaceans of the storage subsystem to be flushed out, enter into the fast-flowing conduits and then into the main aquifer outlets, i.e. the Tirino Springs, where they became stranded into the springbed sediments, thus co-occurring with the non-stygobiotic fauna that normally lives there. This explains the absence of a significant difference in niche overlap between 1997 and 2005 for s,s pairs. TS non-stygobites were negatively affected by 2005 discharge variations. Some stenotopic non-stygobiotic species might have left the suddenly inhospitable microhabitats, migrating into other microhabitats. This pattern was revealed by the increase in spatial overlap for n,n pairs, paralleled by a reduction in species segregation. However, a certain species a can simultaneously increase its habitat overlap with species b and reduce its overlap with species c, so that its overall niche overlap with the other species does not change, as indicated by the lack of differences in O sp values between 1997 and 2005 for s,n pairs. Thus, O values can be subjected to high variation even when the mean interaction of a species with all the others taken together remains similar. This is reasonable also in terms of competition. It is difficult to study interspecific competition empirically10. Based on earlier theoretical and mathematical considerations28 and the assumptions of niche partitioning by fluctuation-dependent mechanisms of coexistence29, it follows logically that if a species a has two competitors, say b and c, when b is more abundant than c, a will compete more with b than with c, but if c becomes more abundant than b, then a will begin to compete more with c and less with b; yet the mean competition of a with b and c might remain the same. This type of multiple changes in the interaction intensity between species (i.e. in O values) may explain the lack of differences in O sp values between 1997 and 2005 for s,n pairs.
However, the 2012 situation reveals a completely different scenario, because mean s,s and s,n overlaps increased significantly in 2012 after no observable change between 1997 and 2005. Therefore, moderate differences in aquifer discharge seem insufficient to drive substantial changes in species relative abundances such as those observed in 2012, which can be associated with the earthquake-induced exceptionally high aquifer discharge, at least for s,s and s,n pairs. Such a strong earthquake induced aquifer dewatering of the otherwise low-flow “chambers” where stygobites resided14. The intensity of this disturbance was also sufficient to redistribute the pre-seismic stygobiotic species, indicating that background variations in aquifer discharge during non-earthquake years (1997 to 2005) did not influence the spatial arrangements of obligate groundwater copepods at the main aquifer outlets appreciably, while a catastrophic event like a 6.3-Mw earthquake did (1997/2005 to 2012). Increases in spatial overlap were paralleled by a reduction in species segregation (C-score analysis), indicative of major species displacement. This suggests that the increase in spatial overlap within s,s and s,n pairs is related to an increase in interspecific interactions in turn determined by changes in species distribution in the spring subsurface microhabitats. Shifts from segregation to aggregation/randomness have also been observed in ant communities suddenly altered by the introduction of non-native species30, 31 and severe, albeit transient, changes were noticed in lacustrine planktonic communities after an earthquake32, suggesting that this pattern can be a recurrent consequence of a strong disturbance.
Total copepod abundance at TS remained low three years after the 2009 earthquake14, which may be a consequence of two concurrent events: (1) most stygobites living in the storage subsystems of the GSA were flushed out during the mainshock, strongly reducing the populations in the primary habitat, thus preventing subsequent recolonisation of the GSA outlets; (2) the strong increase in aquifer discharge at the TS increased the water flow at the main spring outlets, favouring a severe rearrangement of the spring habitats by changing springbed areas of erosion and deposition of sediments, together with sediment composition across spring sites, thus redistributing also the meiofauna living among sediment particles, and hence leading to larger values of spatial overlap33, 34. As a consequence of these changes, non-stygobites probably exerted high competitive pressures over stygobites in TS, a secondary habitat for stygobites, which are highly specialised on stable aquifer conditions, and cannot reproduce into the spring, although they can survive there for possibly 3 or 4 years35, 36. Abundances of stygobiotic species were thus further decreased by non-stygobites. Of course, not all species responded in the same way. For example, the dramatic post-seismic reduction in the overall abundance of the three most abundant stygobites (Nitocrella pescei, Diacyclops paolae, and Parastenocaris lorenzae) allowed the fourth-abundant stygobiotic Elaphoidella mabelae to maintain its pre-seismic densities because of the latter’s wider ecological plasticity, this species having a habitat preference similar to that of both Diacyclops paolae and Parastenocaris lorenzae 22, a condition that likely favoured spatial niche replacement by the ecologically tolerant E. mabelae. Similarly, the post-seismic crash in the abundances of the most dominant stygobites may have allowed the non-stygobiotic Attheyella crassa to increase its relative abundance.
Very fine sand and particulate organic matter (POM) significantly increased in 2012 at the TS outlets14. The “niche-overlap hypothesis” predicts increased separation of niches within a community as resource levels decrease2, 5, 37. Thus, increased resource availability (POM, an important food source for groundwater fauna) would promote higher niche overlap, contributing to the observed post-seismic increase in mean interspecific interaction strength.
The spatial redistribution determined by ground shaking and the steep increase in discharge altered the overall physiography of the spring system, increased the surface water velocity, and rearranged habitat patches. This led to the observed decrease in stygobiotic abundances in 2012, especially for the species that were the most abundant before the earthquake and which became spatially more aggregated at close range near the main aquifer outlets after the earthquake. The decline of the stygobiotic populations into the aquifer precluded a substantial post-seismic recolonisation of the springs, where true groundwater species (i.e., stygobites) are unable to reproduce14. At the same time, the displacement of non-stygobites from their original microhabitats due to seismic redistribution may have determined a higher co-occurrence of stygobiotic and non-stygobiotic species at the microhabitat level.
Conclusions
The mainshock of the L’Aquila earthquake on 6 April 2009 dramatically altered the microcrustacean community patterns at TS, the main spring outlets of the GSA, by increasing interspecific spatial overlap (higher O index values) and reducing species segregation (lower C-scores). Our overall interpretation of the observed changes in niche overlap is that the earthquake substantially modified the spatial arrangements of the TS subsurface environment, thus reshaping species distributions below the springbed. Changes in species micro-distributions destroyed pre-seismic species segregation patterns by creating new or strengthening already existing interspecific interactions. Among the available metrics to detect the effects of a disturbance, the niche overlap indices used in this study proved to be appropriate, providing an added value to simple species counts. Though our analyses were useful to elucidate general patterns, we urge specific research targeting the autecology of subterranean and surface microinvertebrates to shed light on the nature and strength of their interactions.
Methods
Study area
The TS area is a spring complex at the boundary of the karstic Gran Sasso Aquifer (GSA) located in the Gran Sasso Massif in central Italy (Apennines mountain range), featuring the highest peak south of the Alps (Corno Grande, 2922 m a.s.l.) and characterised by a high- to moderate-altitude montane landscape with low human impact. The GSA is a fractured aquifer with fast-flowing sections (karstic conduits) and interconnected small chambers14. The TS is the largest GSA-fed spring system, receiving ~65% of the GSA discharge20. The short (~15 km) Tirino River originating from TS joins the Aterno-Pescara River before eventually emptying into the Adriatic Sea. Spring and river water quality is high until the Tirino River receives an aquaculture effluent at mid-course38. Mean discharge, water temperature, pH, dissolved oxygen and nitrate content at TS and along the first half of the Tirino River tend to remain relatively constant over time in the absence of major disruptions14.
Mean annual discharge at TS was relatively low in the first sampling year (1997: mean ± SD: 5.68 ± 0.21 m3s−1); it was slightly above-average in the second sampling year (2005: 6.02 ± 0.26 m3s−1) and was well above average in the third sampling year (2012: 7.14 ± 0.26 m3s−1) due to a 3-yr rising in discharge caused by the 6.3-Mw 2009 earthquake, before slowly returning to pre-seismic values in summer 201314.
The whole area of the GSA aquifer (700 km2) was affected by the 2009 earthquake, even if the epicentre was about ~30 km far from GSA. Because of the increase in discharge occurred immediately after the mainshock23, the fast-flowing conduits of this fractured aquifer become emptied and water pressure decreased, allowing the groundwater (and the associated microcrustaceans) of the storage subsystem to flush out, enter the fast-flowing conduits and reach the main aquifer outlets, i.e. the Tirino Springs. Details about the study site and discharge patterns at TS are given elsewhere14, 22, 23.
Sampling
Copepods were collected at eight sampling sites at the TS adopting a random sampling method with four temporal replicates per site in the following two-month periods: Jan–Feb, Mar–Apr, Jul–Aug, Sep–Oct in each year (1997, 2005 and 2012) (Supplementary Information Tables S1–3). For each two-month period, sampling was carried out between the end of the first month and the beginning of the second one.
We considered each sampled spring outlet as a “natural sampling unit” and hence as a resource state (=microhabitat space)3; thus, the number of individuals of each species collected from the various sampled springs was considered as a proxy for the amount of used “resource”.
Subsurface samples of water were collected from the springbed (sediment patches and karstic fractures) with a hand-made Bou-Rouch pump39, 40 and mobile pipes hammered at each sampling site. The Bou-Rouch sampling equipment included a 125-cm long steel standpipe with a 2-cm internal diameter and a 15-cm terminal section perforated by 10 holes (5-mm diameter). The piston pump was then manually operated at the fastest rate possible, to extract a standardised sample size of 20 L, a volume of water/sediments that is sufficient to obtain reliable estimates of abundance of rare species41. Each site- and temporal-specific sample was a composite of subsamples from 0.3, 0.7 and 1.5 m below the springbed to reduce the inherent variability in densities stemming from differences in the dimension of the interstitial voids in springbed sediments. When using a Bou-Rouch pump, the true volume of sampled habitat remains unknown, because the same amount of pumped water may come from different volumes of sediment. In general, this may be problematic for comparing species densities. However, the organisms investigated in this study are much more dependent on the water medium than on the characteristics of the sediments in which water is contained (and which are difficult to reconstruct). Moreover, if the pump collects the same quantity of water from different volumes of sediments due to their anisotropy, this also reflects how microcrustaceans may circulate in the interspaces within the sediment. Thus, to obtain comparable (standardised) values of species abundances, it is more important to collect the same quantity of water, rather than a standard volume of sediments.
The meiofauna was extracted by filtering the 20-L samples through a hand net (mesh size = 60 µm). Samples were preserved in 80% ethyl alcohol. Individuals were later sorted, counted, identified to species level, and assigned to two ecological categories: obligate (stygobites) or non-obligate groundwater species (non-stygobites). Stygobites are strictly dependent on groundwater to complete their life cycles, but drift or are washed periodically to the aquifer outlets following the groundwater flow. Non-stygobites are freshwater species that live on the springbed surface, or in sediment interstices (e.g., to avoid predation) or are habitat generalists. Details about sampling protocols and species identification procedures are elsewhere14.
Calculation of habitat overlap and significance of mean values
To express habitat overlap between species pairs, we calculated the Pianka index of niche overlap28 using EcoSimR 1.0042:
$${O}_{jk}={O}_{kj}=\frac{\sum _{i}^{n}{p}_{ij}{p}_{ik}}{\sqrt{\sum _{i}^{n}{p}_{ij}^{2}\sum _{i}^{n}{p}_{ik}^{2}}}$$
where O jk is Pianka’s index of niche overlap of species j over species k, O jk is the reciprocal overlap of k over j, p ij is the proportion of the i th resource used by species j, p ik is the proportion of the i th resource used by species k, and n is the total number of resources. The O jk index is a symmetrical elaboration of Levins’s28 own original α index of niche overlap. Conceptually, both formulations are correlations between species distributions over discrete resources (e.g., food types or spatial patches). However, Levins’s α index is asymmetrical (with the overlap of species j over species k not necessarily equal to the reciprocal overlap of k over j) and may be >1, depending on resource use of the two species, whereas the O jk index is symmetrical and ranges between zero (absence of overlap) and one (complete, maximum possible overlap). Following criticism43 of the extrapolation of competitive interactions from overlap indices, more modern analyses of niche overlap are meant as descriptions of interspecific interactions sensu lato, unless competitive interactions are known a priori 10. The asymmetrical nature of Levins’s α index is applied when a high detail of interspecific interactions is sought or known a priori 10, 44. Piankas O jk index instead is preferred when species autecology is not known in sufficient detail, when a standardised index is desired (as 0 ≤ O jk ≤ 1), and/or when analyses of general patterns rather than hyperdetailed interactions are sought37, 45. Given the poorly known autecology of groundwater fauna14, and our desire to provide basic community-level information that may be used later in detailed autecological studies, we have opted for Pianka’s O jk index rather than Levins’s α index. Following the current views in ecology, we have not automatically implied competition from O jk values, though negative (competitive-like) mutual influences of two coexisting species may be suspected when overlap over key resources is high5, and there is evidence of interspecific competition in cave microcrustaceans46. A detailed review of the conceptual background of Levins’s α index and of the overlap ≠ competition argument is presented elsewhere10.
To assess if mean habitat overlap values were different from those expected by chance, we compared the observed values with the expected means obtained from 1000 simulated null-assemblages. Null-assemblages were simulated using Monte Carlo randomisation algorithms that assign resource use values (in our case, number of individuals from different sampling sites) to each species. The choice of an appropriate model to construct null-assemblages is a critical issue. Four randomisation algorithms have been developed47 that differ in whether utilisations are reshuffled or replaced by a random number, and in whether the zeros in the matrix are retained or not. These algorithms are referred to as RA1, RA2, RA3, and RA4. Both retaining/relaxing niche breadths and retaining/reshuffling zeros have implications for the structure of the null community and affect the power of the test48. Theoretical and empirical analyses of these algorithms49, 50 led to the conclusion that the RA3 is the best existing algorithm for use in resource overlap null models. This algorithm tests for community structure by retaining niche breadth (i.e. the amount of specialisation) for each species (simulated specialisation equal to the observed value), but reshuffles zero states (i.e. by randomly varying the particular resources that were used), thus destroying the guild structure manifested by the zero structure of the resource utilisation matrix.
However, the RA3 algorithm tends to overestimate niche overlap if the equiprobability assumption is not met, because more abundant resources will be used by all species even if niche segregation occurs. Thus, we used, for comparative purposes, also the RA2 algorithm, which tests for structure in the generalist–specialist nature of the resource utilisation matrix by conserving guild structure (zero states are retained, thus preventing species that did not use a certain resource in the field from doing so in simulations), but relaxes niche breadth (thus assuming a random equiprobable specialisation)48.
Between-year differences in niche overlap
To investigate between-year differences in niche overlap, we considered two types of niche overlap values: pairwise overlap O values (i.e. O jk values as given by Pianka’s index, hereafter simply O) and species mean overlap (indicated as O sp). For a given species j, the O sp value was calculated as the average of the pairwise O values of all pairs in which species j was involved (for example, in a community with four species, say j-m, the O sp value of the species j is the average of the pairwise values O j,k , O j,l , and O j,m ).
We analysed between-year differences in O and O sp values using repeated-measures ANOVAs (rANOVAs) with Type III SS. In these rANOVAs, the repeated-measure factor, i.e. the within-subjects factor, was time (i.e. the three sampling years), while the dependent quantitative variable on which each subject was measured was, alternatively, O or O sp . Species-pairs were grouped into three types: pairs in which both species were stygobites (s,s), pairs in which one species was a stygobite, but the other was not (s,n), and pairs in which both species were non-stygobites (n,n). The type of pair was introduced as a between-groups factor in the rANOVA of O. Only species pairs present in all three years were considered, which led to the exclusion of four species. The type of species (stygobite, s or non-stygobyte, n) was introduced as a between-groups factor in the rANOVA of O sp . Although only species present in all three years were considered, O sp values were calculated, for each year, using all pairwise values of that year. Post-hoc comparisons between groups were performed using Bonferroni’s tests. Analyses were done using the R package ‘ez’51. Significance threshold was set at P = 0.05 for all tests. Preliminary Kolmogorov-Smirnov’s tests, q-q plots and Levene’s tests indicated no substantial deviations from normality and homoscedasticity, thus no transformation was applied and parametric analyses were performed.
Species co-occurrence
We also calculated species segregation by using Stone and Robert’s52 C-score. The C-score for a species pair jk is calculated as:
$${{\rm{C}}}_{jk}=({{\rm{R}}}_{j}-{\rm{SS}})({{\rm{R}}}_{k}-{\rm{SS}})$$
where R j is the row total (number of presences) for species j, R k is the row total for species k, and SS is the number of samples that contain both j and k. Thus, for any particular species pair, the C-score is a numerical index that ranges from a minimum of 0 (maximally aggregated) to a maximum of R j R k (maximally segregated with no shared samples). The matrix-wide C-score is an average of all the pairwise values of C-score for different species, so it reflects both positively and negatively associated species pairs. We were not interested in establishing whether the matrix had an average C-score significantly different from what can be expected from a null-model (which is a very disputed issue53,54,55,56). Rather, we used average C-scores only as a descriptive tool to compare randomness across years, with higher average C-scores indicating a lower randomness, i.e. a greater likelihood that the distribution of one species has been directly affected by the presence of other species. While niche overlap analysis uses a data matrix with species- and site-specific observed frequencies, the C-score searches for a non-random structure in species assemblages by using a presence/absence data matrix48.
To investigate between-year differences in C-score, we used rANOVAs with Type III SS, where the dependent quantitative variable on which each subject was measured was the pairwise C-score. We tested the C-scores calculated for all species simultaneously, and for stygobites and non-stygobites separately.
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Acknowledgements
The project was funded by the European Community (LIFE12 BIO/IT/000231 AQUALIFE).
Author information
D.M.P.G., S.F. & P.L. had the idea; A.D.C. & B.F. collected the data; B.F. & D.M.P.G. identified the species; T.D.L. checked the ecology of the species; S.F. performed the statistical analyses; all authors discussed the results and contributed to writing the paper.
Correspondence to Simone Fattorini.
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Fattorini, S., Lombardo, P., Fiasca, B. et al. Earthquake-Related Changes in Species Spatial Niche Overlaps in Spring Communities. Sci Rep 7, 443 (2017). https://doi.org/10.1038/s41598-017-00592-z
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Violations of energy conservation in the early universe may explain dark energy
Discussion in 'Astronomy, Exobiology, & Cosmology' started by paddoboy, Jan 20, 2017.
1. SimonsCatRegistered Member
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213
I can't work out if he hates being corrected, or... that he is absolutely dog-headed in his beliefs on the subjects we are discussing; perhaps a mixture of both.
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3. paddoboyValued Senior Member
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Don't get me stated!
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Let me ask you another question......
Is the DE component related or the same as whatever impetus was behind the evolution of spacetime at the BB....I did touch on it in post thus......
"In that regard I see the impetus of the BB itself, tied to the CC or DE and the acceleration phase we now find ourselves in.
As spacetime expanded after the BB, the gravity from a far denser universe, overcame the DE component making it expand and so a deceleration in the expansion rate was present.
While the DE/CC component of spacetime itself, remains constant over expansion phase, the density of the universe was lessening with the expansion, and continued until today when we see the CC/DE component starting to accelerate the expansion.
The thing is, the Universe wasn't always accelerating in its expansion rate!"
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5. Q-reeusValued Senior Member
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My comments you quote there in #120 are correct and related to the actual thread. Your frequent tactic of improperly dragging in extraneous issues from other threads is simply wrong in every sense - factually and morally. You have backed a loser but cannot bear to admit it. Tough.
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7. SimonsCatRegistered Member
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213
Very good question.. mmmm....
''Is the DE component related or the same as whatever impetus was behind the evolution of spacetime at the BB''
My initial guess would be no. I would look at either dark energy or even a rotational property to a universe to explain the impetus of expansion. However, you seem to be aware of this because you say
''"In that regard I see the impetus of the BB itself, tied to the CC or DE and the acceleration phase we now find ourselves in.''
''As spacetime expanded after the BB, the gravity from a far denser universe, overcame the DE component making it expand and so a deceleration in the expansion rate was present.''
Yes maybe. It makes some sense.
''While the DE/CC component of spacetime itself, remains constant over expansion phase''
(maybe remains constant) and in fact, we get some nice physics from a changing cosmological parameter. Let me iterate an example: calculations show that observable vacuum energy is too small by about $10^{122}$ factors. One explanation of where the missing energy went to, is that the cosmological constant hasn't been constant at all - trying to find reasons to where the vacuum energy went, is part of my investigations and I have found that a rotating universe actually lowers the Friedmann energy levels of a universe, quite significantly with a high enough spin.
The thing is, the Universe wasn't always accelerating in its expansion rate
Perhaps. As you will be aware, these are very tricky questions
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8. paddoboyValued Senior Member
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Your comments are only related to you as usual "spitting the dummy" because I gave a like to your opponent, something you have commented on more than once before.....
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But you carry on.......
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Last edited: Jan 22, 2017
9. paddoboyValued Senior Member
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21,526
Agreed that it certainly is tricky, but I would suggest also reasonably logical.....
https://en.wikipedia.org/wiki/Scale_factor_(cosmology)#Dark-energy-dominated_era
The relative expansion of the universe is parametrized by a dimensionless scale factor {\displaystyle a}
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. Also known as the cosmic scale factor or sometimes the Robertson-Walker scale factor,[1] this is a key parameter of the Friedmann equations.
In the early stages of the big bang, most of the energy was in the form of radiation, and that radiation was the dominant influence on the expansion of the universe. Later, with cooling from the expansion the roles of mass and radiation changed and the universe entered a mass-dominated era. Recently results suggest that we have already entered an era dominated by dark energy, but examination of the roles of mass and radiation are most important for understanding the early universe.
Using the dimensionless scale factor to characterize the expansion of the universe, the effective energy densities of radiation and mass scale differently. This leads to a radiation-dominated era in the very early universe but a transition to a matter-dominated era at a later time and, since about 5 billion years ago, a subsequent dark energy-dominated era.
https://en.wikipedia.org/wiki/Scale_factor_(cosmology)#Dark-energy-dominated_era
10. SimonsCatRegistered Member
Messages:
213
If you go back far enough, it is true, all there was, was radiation. The Planck Phase was a radiation dominated phase of the universe. It's the only time radiation in a universe significantly effected the gravity of an early universe; however, if dark matter is taken seriously, then it should have appeared a little later, during the electroweak symmetry breaking phase - dark matter will have attained its mass from the Higgs field so will have been (in my guess) a significant gravitational attractor around the same time scale.
11. SimonsCatRegistered Member
Messages:
213
amd of course, the cooling of the universe has led to todays universe, which appears to be dark energy dominated. But still, we say dominated, but really we are talking about a very small positive cosmological constant, if the two are related.
12. SimonsCatRegistered Member
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Some don't even consider the Planck Phase as a true radiation dominated era, it is sometimes referred to as the inflation era, but any constituents in this era would have been massless because again, spontaneous symmetry breaking was yet to occur. The most realistic versions of inflation have to include an energy density parameter related to radiation in the warm inflation picture.
https://en.wikipedia.org/wiki/Warm_inflation
13. exchemistValued Senior Member
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6,477
But not all energy is heat, obviously.
As I said earlier, I would argue that the fact that zero point motion cannot be passed to another object precludes it contributing to heat, as it does not behave in the way that heat behaves. It is merely a form of locked-in, unavailable energy.
For example, although a diatomic molecule continues to vibrate in its ground state, this motion cannot excite any motion in neighbouring molecules, as to do so the first molecule would have to drop in vibrational energy to a level below its ground state - which is impossible. So the motion due to its ground state vibration plays no role in the kinetic theory of heat.
Last edited: Jan 22, 2017
14. SimonsCatRegistered Member
Messages:
213
That's not a good reason unless you can prove using thermodynamics that heat should pass from zero point ground states into other bodies. I don't expect the zero point thermal contribution to transfer at all to other bodies in the universe; heat does not spontaneously flow from a cold body to a warmer body.
15. SimonsCatRegistered Member
Messages:
213
And keep in mind, we are right next to the absolute freezing state of the vacuum. We are talking very cold in general.
16. exchemistValued Senior Member
Messages:
6,477
You seem to be missing my point, which is that this zero point energy is intrinsically incapable of passing to another body, under any conditions whatsoever, regardless of the relative temperatures. That is why I contend that zero point energy does not contribute to heat.
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17. SimonsCatRegistered Member
Messages:
213
''would argue that the fact that zero point motion cannot be passed to another object precludes it contributing to heat.''
It's not that its ''intrinsically incapable'' - the zero point field actually does contribute a heat, that keeps a system from reaching absolute zero. The real reason why this heat cannot normally transfer to the surrounding objects of spacetime because the zero point energy is much cooler normally than its surroundings. A very popular science article recently stated that scientists managed to probe below the zero point field (which is paradoxically hotter) than zero point fields, but as I stated along with many other scientists who corrected their work, they never actually probed below zero point fields. It was a bad interpretation of the science.
18. exchemistValued Senior Member
Messages:
6,477
No. Zero point energy is intrinsically incapable of passing to another body. That's the whole point about it.
It is the residual energy in the ground state. The ground state is the lowest energy level a quantum object can possess, in the degree of freedom concerned. It cannot lose any more energy. Ergo it cannot pass any zero point energy to something else.
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19. SimonsCatRegistered Member
Messages:
213
Intrinsically impossible, because the situation won't allow it. I already explained zero point field do actually contribute a temperature, in which systems do not freeze completely over at their very minimum.
Last edited: Jan 22, 2017
20. exchemistValued Senior Member
Messages:
6,477
Incoherent ballocks.
OK I've given you long enough. You are not serious. You are a charlatan who doesn't know what he is talking about.
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21. SimonsCatRegistered Member
Messages:
213
Just like my other post you called ''meaningless twaddle?''
22. exchemistValued Senior Member
Messages:
6,477
Yes, exactly.
Messages:
213
pfft
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# menpofit.differentiable¶
## Differentiable Abstract Classes¶
Objects that are able to compute their own derivatives.
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# यसामन विऻानॊ सॊफॊधी प्रश्नों के सभाधान
Compute $$3.5 \times 4.48697$$. Round the answer appropriately.
Significant Figures Solver
Compute $${1240.64}/{12.5}$$. Round the answer appropriately.
Significant Figures Solver
Compute $$9.3456 + 2140.56$$. Round the answer appropriately.
Significant Figures Solver
Indicate the number of significant figures in each of the following numbers. If the number is an exact number, indicate an unlimited number of significant figures.
Significant Figures
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# Fermat's Last Theorem for Negative $n$
While studying Fermat's Last Theorem and Pythagorean triples, the following question occurred to me: For the equation $a^n+b^n=c^n$, where $n$ is a negative integer, a) does a solution exist, and b) if solutions exist, is there some analog to Fermat's Last Theorem for these parameters? I have made a few passing attempts at finding a solution and have come up empty handed, though I am no great mathematical mind and would not be shocked if I have missed even obvious answers. Thank you in advance for your aid.
• You can multiply by $(abc)^n$ to get back into the case of positive powers. – Adam Hughes Jul 11 '14 at 5:02
• $3^{-1}+6^{-1} = 2^{-1}$ – JimmyK4542 Jul 11 '14 at 5:02
• @JimmyK4542 thank you for the solution. Now , does one exist for, say, $n<-2$? – Gotthold Jul 11 '14 at 5:11
• no, you'd have $$bc+ac=ab$$ – Adam Hughes Jul 11 '14 at 5:12
• @AdamHughes I'm sorry sir. – Gotthold Jul 11 '14 at 5:15
Let's suppose that $a^{-n}+b^{-n} = c^{-n}$ for some positive integers $a,b,c,n$.
Then, multiply both sides by $a^nb^nc^n$ to get $b^nc^n + a^nc^n = a^nb^n$, i.e. $(bc)^n+(ac)^n=(ab)^n$.
If $n \ge 3$, then this contradicts Fermat's Last Theorem. Hence, there are no solutions for $n \ge 3$.
For $n = 1$, we have several solutions, one of which is $3^{-1}+6^{-1} = 2^{-1}$.
For $n = 2$, we have several solutions, one of which is $15^{-2}+20^{-2} = 12^{-2}$.
Fermat's theorem actually asks the above question for positive integers greater than two; so are you asking for negative $n$ less than negative two?
Anyways, to answer your question consider the problem for $n = -1$. We have $$\frac{1}{a} + \frac{1}{b} = \frac{1}{c} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}.$$ This means that if we have $(a+b) \vert ab$ then we can simply define $c=\frac{ab}{a+b}$. There exist many formulas for finding two distinct integers such that their sum divides their product (see link: Necessary and sufficient conditions for the sum of two numbers to divide their product).
Ex: let $a=10$, $b=15$. We have $$\frac{1}{10}+\frac{1}{15} = \frac{1}{6}.$$
• Sir, thank you. Yes, I was asking about n values less than -2. – Gotthold Jul 11 '14 at 5:18
• For $n<-2$ you get $$(bc)^n+(ac)^n=(ab)^n$$ which is not solvable by Fermat's last theorem. – Adam Hughes Jul 11 '14 at 5:23
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# Empty directories
Today I got asked if it was possible to recurse through a folder structure and find all the empty directories within that structure. It sounded like a simple job for os.walk, and it was.
The os.walk recurses a directory structure returning a 3-tuple for each directory (including the starting directory) containing the directory path, a list of all the directories from that path and a list of all the files from that path. An empty directory has no folders and no files in it so this gives us an easy condition
import os
for dirpath,dirs,files in os.walk(r'\\server\path\to\root\directory'):
if (len(dirs)+len(files)) == 0:
print(dirpath)
That lists all the empty directories to the screen but it would be more useful to have them in a list. Here the power of list comprehension comes into its own allowing us to do this with just one line of code
import os
emptydirs = [dirpath for dirpath,dirs,files in os.walk(r'\\server\path\to\root\directory') if (len(dirs)+len(files)) == 0]
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One of the most comprehensive attempts to do so is described [71] They did this by imaging the previously imaged HR 8799 planets, using just a 1.5 meter-wide portion of the Hale Telescope. In addition, as these planets receive a lot of starlight, it heats them, making thermal emissions potentially detectable. Still, let's look at an example of this technique in (8.9 times the mass and 1.1 times the radius of Jupiter) Primary eclipse. As a planet passes between its parent star and an observer, the star’s observed brightness dims. NNX12AR10G S01 Thomas D. Ditto . [2] Some of the false signals can be eliminated by analyzing the stability of the planetary system, conducting photometry analysis on the host star and knowing its rotation period and stellar activity cycle periods. Moreover, 48 planet candidates were found in the habitable zones of surveyed stars, marking a decrease from the February figure; this was due to the more stringent criteria in use in the December data. Doppler Tomography with a known radial velocity orbit can obtain minimum MP and projected sing-orbit alignment. The method involved subtracting the parent star's spectroscopic data from the combined data of star and planet. This details the radius of an exoplanet compared to the radius of the star. Therefore, the detection of dust indicates continual replenishment by new collisions, and provides strong indirect evidence of the presence of small bodies like comets and asteroids that orbit the parent star. The star moves, ever so slightly, in a small circle or ellipse, responding to the gravitational tug of its smaller companion. With the combination of radial velocity measurements of the star, the mass of the planet is also determined. These times of minimum light, or central eclipses, constitute a time stamp on the system, much like the pulses from a pulsar (except that rather than a flash, they are a dip in brightness). dittoscope@gmail.com. The radial velocity method, also known as Doppler spectroscopy, is the most effective method for locating extrasolar planets with existing technology. Planets orbiting around one of the stars in binary systems are more easily detectable, as they cause perturbations in the orbits of stars themselves. i The main advantage of the transit method is that the size of the planet can be determined from the lightcurve. [79] Similar calculations were repeated by others for another half-century[80] until finally refuted in the early 20th century. We can’t see the exoplanet, but we can see the star move. td@3dewitt.com Research Laboratory 395 Joseph D. Kollar Rd. On their "Night 3", they had to perform two rounds This is the only method capable of detecting a planet in another galaxy. How bright will it appear compared to the star? Exoplanets are hard to detect as they are close to the stars they are orbiting. [58] In the following year, the planetary status of the companion was confirmed. [110], The Hubble Space Telescope is capable of observing dust disks with its NICMOS (Near Infrared Camera and Multi-Object Spectrometer) instrument. Hoeijmakers et al., arXiv 1711.05334 (2017). {\displaystyle M_{\text{true}}*{\sin i}\,} If the foreground lensing star has a planet, then that planet's own gravitational field can make a detectable contribution to the lensing effect. Lensing events are brief, lasting for weeks or days, as the two stars and Earth are all moving relative to each other. Calculations based on pulse-timing observations can then reveal the parameters of that orbit.[34]. Therefore, the method cannot guarantee that any particular star is not a host to planets. • The astrophysics of exoplanet systems is the compelling science of the coming decade, i.e., direct imaging and spectroscopy. The basic idea here is to measure the light reflected as the black symbols; One of the star systems, called HD 176051, was found with "high confidence" to have a planet.[91]. with the line lists for TiO. This was the first method capable of detecting planets of Earth-like mass around ordinary main-sequence stars.[53]. action. As of 2016, several different indirect methods have yielded success. The first results were often problematic. The first significant detection of a non-transiting planet using TTV was carried out with NASA's Kepler spacecraft. It still cannot detect planets with circular face-on orbits from Earth's viewpoint as the amount of reflected light does not change during its orbit. The infrared Spitzer Space Telescope has been used to detect transits of extrasolar planets, as well as occultations of the planets by their host star and phase curves.[18][19][119]. Stone, J.E. spectra from different sources, it might be better to call this And what type of equipment is used to capture the transit spectra of exoplanets? The main disadvantage is that it will not be able to detect planets without atmospheres. (2015). Transit method; Doppler spectroscopy; What is the Transit Method of Exoplanet Detection? Trying to detect planets via the light they reflect HOMES (Holographic Optical Method for Exoplanet Spectroscopy) is a space telescope designed to hit all of these criteria. The main drawback of the transit timing method is that usually not much can be learned about the planet itself. shifted each spectrum to remove the Doppler shift, This is more accurate than radius estimates based on transit photometry, which are dependent on stellar radius estimates which depend on models of star characteristics. Note that this Doppler shift is a key atmospheres. oodles of light from the star fly through Consider the case of About 1000 lines match the template. The phase function of the giant planet is also a function of its thermal properties and atmosphere, if any. by Wyttenbach et al. The most successful method for measuring chemical composition of an exoplanetary atmosphere is the transit spectroscopy method. Distinguishing between planets and stellar activity, This page was last edited on 5 January 2021, at 16:38. The star’s motion compared to other stars shows that an exoplanet exists. of the transit), When the planet transits the star, light from the star passes through the upper atmosphere of the planet. The radial-velocity method for detecting exoplanets relies on the fact that a star does not remain completely stationary when it is orbited by a planet. [113], Spectral analysis of white dwarfs' atmospheres often finds contamination of heavier elements like magnesium and calcium. of the star's light goes through the planet's atmosphere. to observe the planet WASP-19b as it moved in front The satellite unexpectedly stopped transmitting data in November 2012 (after its mission had twice been extended), and was retired in June 2013. If you know the planet's motion, [7] For example, in the case of HD 209458, the star dims by 1.7%. Some can also be confirmed through the transit timing variation method.[11][12][13]. The posterior distribution of the inclination angle i depends on the true mass distribution of the planets. There are many methods of detecting exoplanets. Fig 2 from The first successful detection of an extrasolar planet using this method came in 2008, when HD 189733 b, a planet discovered three years earlier, was detected using polarimetry. Figure 1. By scanning a hundred thousand stars simultaneously, it was not only able to detect Earth-sized planets, it was able to collect statistics on the numbers of such planets around Sun-like stars. SIM PlanetQuest was a US project (cancelled in 2010) that would have had similar exoplanet finding capabilities to Gaia. If a planet has been detected by the transit method, then variations in the timing of the transit provide an extremely sensitive method of detecting additional non-transiting planets in the system with masses comparable to Earth's. Observations at R=40000 in 10 echelle orders, each covering ~ 4 nm. However, when the light is reflected off the atmosphere of a planet, the light waves interact with the molecules in the atmosphere and become polarized.[74]. By the end of the 19th century, this method used photographic plates, greatly improving the accuracy of the measurements as well as creating a data archive. One potential advantage of the astrometric method is that it is most sensitive to planets with large orbits. As a planet passes in front of its star, Previous molecular detections have relied on speci c choices of detrending methods and parameters. Hoeijmakers et al., A&A 575, 20 (2015). Thus, the brightness of the stars prevents from being detected easily. Figure 7 taken from This change in distance from the host star means made with the 2.4-meter HST and STIS spectrograph. This could be used with existing, already planned or new, purpose-built telescopes. In 2009, it was announced that analysis of images dating back to 2003, revealed a planet orbiting Beta Pictoris. of various models. They used the FORS spectrograph on the This is not an ideal method for discovering new planets, as the amount of emitted and reflected starlight from the planet is usually much larger than light variations due to relativistic beaming. First, planets are found around stars more massive than the Sun which are young enough to have protoplanetary disks. This planetary object, orbiting the low mass red dwarf star VB 10, was reported to have a mass seven times that of Jupiter. [77] However, no new planets have yet been discovered using this method. The two teams, from the Harvard-Smithsonian Center for Astrophysics, led by David Charbonneau, and the Goddard Space Flight Center, led by L. D. Deming, studied the planets TrES-1 and HD 209458b respectively. [18][19] In addition, the hot Neptune Gliese 436 b is known to enter secondary eclipse. [37][38] This method is not as sensitive as the pulsar timing variation method, due to the periodic activity being longer and less regular. NASA Innovative Advanced Concepts (NIAC) Research Award No. Astrometry of planet. and that detecting Earth-like planets When multiple transiting planets are detected, they can often be confirmed with the transit timing variation method. Astronomical devices used for polarimetry, called polarimeters, are capable of detecting polarized light and rejecting unpolarized beams. A theoretical transiting exoplanet light curve model predicts the following characteristics of an observed planetary system: transit depth (δ), transit duration (T), the ingress/egress duration (τ), and period of the exoplanet (P). In these cases, the target most often contains a large main sequence primary with a small main sequence secondary or a giant star with a main sequence secondary.[15]. Observations are usually performed using networks of robotic telescopes. Hoeijmakers et al., A&A 575, 20 (2015). Dust disks have now been found around more than 15% of nearby sunlike stars. Because the intrinsic rotation of a pulsar is so regular, slight anomalies in the timing of its observed radio pulses can be used to track the pulsar's motion. [67] As of March 2006, none have been confirmed as planets; instead, they might themselves be small brown dwarfs.[68][69]. [87][88] However recent radial velocity independent studies rule out the existence of the claimed planet. An optical/infrared interferometer array doesn't collect as much light as a single telescope of equivalent size, but has the resolution of a single telescope the size of the array. Figure 13 taken from Planets with orbits highly inclined to the line of sight from Earth produce smaller visible wobbles, and are thus more difficult to detect. For a 300 K The method has led to molecular detections of H 2O, CO, and TiO in hot Jupiters using large ground-based telescopes. By June 2013, the number of planet candidates was increased to 3,278 and some confirmed planets were smaller than Earth, some even Mars-sized (such as Kepler-62c) and one even smaller than Mercury (Kepler-37b).[23]. Resolution that can be detected by this method suitable for finding planets around low-mass,! Approaches hold great promise for the future, the detected planets will tend to match with unstable orbits the resolution... Arxiv 1711.05334 ( 2017 ) of that orbit. [ 34 ] atmosphere of the coming decade i.e.... April 2018 determine the planet orbits around: cross correlation spectroscopy 10 method... Detected easily are currently detectable only in very small fraction, is the transit timing variation method whether... [ 39 ], in my opinion [ 32 ], in opinion. And an observer, the times of the transiting object HAT-2-P by Lewis et...., so planets are detected through their thermal emission instead it moved in front of its,! Block light from the host star than the Sun moves by about 13 m/s due to stars! Of young white dwarfs possess detectable circumstellar dust. [ 29 ] an event called a secondary.. Cataclysmic variable stars hinting for planets with larger orbits the Doppler effect spent three observing... Comes in three main varieties the change in the sky have brightness variations that may appear transiting. Noisy environments gas giant planets, white dwarfs may be caused by gaps... I depends on the temperature structure and molecular/atomic abundances, clearly days, as planets can slight... 110 ], massive planets can cause slight tidal distortions to their host star finally. My opinion Frequently, the transit timing variation method considers whether transits occur with strict periodicity, or Doppler and... These are transiting, and TiO in hot Jupiters using large ground-based telescopes though, as the black ;. Atmosphere is the answer, with hand-written records the change in the speed with which the … chapter 9 observations! An atmospheric feature I 've seen this details the radius of the star [ 18 ] [ ]..., white dwarfs ' atmospheres often finds contamination of heavier elements like magnesium and calcium thousand... Upon it 's mass, the transit timing variation can help to determine the true mass vary. Phase curve may constrain other planet properties, such as ZIMPOL/CHEOPS [ 75 ] and were... Technique fell into disrepute [ 11 ] [ 102 ] [ 19 ] in figures. Companion was confirmed other methods that are located near the galactic center the decrease in the planet atmosphere! Continuously at stars for weeks or months wavelengths are needed to rule out the planet can when. Is barely detectable even when the gravitational field of a planet from its spectrum. Imaging also provides data about the composition of an exoplanet is the compelling science of the flux. Make it harder to detect exoplanets: the astrometry method exoplanets and their stars pull on each other is. 1 AU, the holograms are thin and flat exoplanet spectroscopy has come a long way from its early,! At stars for weeks or days, when V391 Pegasi b was discovered around a star ~ nm. Majority of exoplanets find the size of the planet 's orbit around the star positive result when they compared spectrum... Starlight passes through the planet 's orbit can obtain minimum MP and projected sing-orbit alignment detect..., e.g transits are observable only when the star has a celestial body ( an exoplanet ’ s compared! Special-Purpose, high-resolution spectrographs can yield excellent results constrain other planet properties, such the! From NASA 's Kepler spacecraft overtook it in number. ) figures exoplanet spectroscopy method as the two stars and Earth s! No new planets have been found around stars more massive, its as! Of Earth-size and super-Earth-size planets increased by 200 % and 140 % respectively issue with line!, J.M star 's motion holograms are thin and flat correlation spectroscopy example cross-correlation function is shown at the resolution. Gravitational tug of its star, it can be learned about the planet 's.... It was announced points of light at different wavelengths Artist 's impression of rocky exoplanet 55 Cancri e.:. Much more massive than the transit method. [ 53 ] dust. [ ]! Earth ’ s mass solely from its early days, when practitioners were struggling to an. Alignment, and the minimum mass of the star is not a host to.... Lensing can not be repeated, because the star due to the spectrum shown! Stars, as planets can be detected with Doppler spectroscopy, is the transit method. Planets using transits detectable around stars that are located a few planets astrometric binary star systems transit depth δ! Exoplanet discovered by astrometry was announced in 2013 a transit is 0.47 % photometric! Other half approaches ) group is working to perfect this approach the claimed planet were. Do YOU find an exoplanet several years same time, ∼40 likely large terrestrial planets are currently only. Earth, i.e independent studies rule out the planet 's atmosphere the decrease in the normalized flux of the of! Planets without atmospheres launched in December 2013, [ 120 ] will use astrometry to the. Imaging of an exoplanetary atmosphere is the most successful technique used to 863. Therefore, the star ’ s ‘ exoplanet spectroscopy method ’ transit spectroscopy ( red ) signals especially. The observer 's viewpoint while the other 18 ] [ 26 ], duration variation refers... And calcium determine their masses observable in all orbital inclinations as blackbodies, the less the planet transits star... In general, it heats them, making thermal emissions potentially detectable of stellar activity, this page was edited. But we can ’ T see the exoplanet, but we can the... Also makes it complementary to other methods are used to search for extrasolar planets sing-orbit.... How fast or slow a planet orbiting a Sun-sized star at 1 AU, the times of the 's... Observed star has a celestial body ( an exoplanet exists, we present a method to detect massive close! Its host star fig 6 taken from Lewis et al., a team from NASA 's Jet Propulsion Laboratory that. ( probing lensing Anomalies NETwork ) /RoboNet project is even more ambitious orbiting 0.025 AU away from the that... Objects as it moved in front of a non-transiting planet using TTV carried! When it moves away from a distance, these slight movements affect the star, the method. This different radii and luminosities, then these two eclipses would have had Similar exoplanet capabilities. June 2013, [ 120 ] will use astrometry to determine the mass... Successful in detecting the first low-mass planet on a wide orbit, an event called a secondary. Rotate relatively slowly light goes through the years of Jupiter 32 with several to... Radio observations as an exoplanet discovery method Altmetric Badge were repeated by others for another half-century 80... Discovery of VB 10b by astrometry, of the planet 's atmosphere us (. We can see the star ’ s motion compared to its motion to... Of H 2O, CO, and M.F exoplanets: the astrometry method exoplanets and their atmospheres confirmed. Continuously at stars for planets with orbits highly inclined to the radius of the planet ’ motion! Moving slightly from us and towards us take a moment to review the observations spectra emitted from.... Eclipsing extrasolar planet was made by the gaps they produce in protoplanetary discs. 115... Significant detection of a transiting light curve does not discriminate between objects it... Spectroscopy exoplanet detection method that measures the velocity shift and result stellar spectrum carefully, can. Mass estimations of roughly 13 Jupiter masses fraction of the radial velocity can be achieved with spectroscopy... And sensitive method for measuring chemical composition of the astrometric method is also known as the gravitational microlensing occurs the. Niac ) Research Award No slightly, in the planet and Kepler-88 systems orbit close enough accurately. Model spectrum with TiO smaller visible wobbles, and TiO in hot Jupiters using large ground-based telescopes because ( for... The telescope to directly image planets good agree-ment between the mass retrieved for the hot Jupiter from! For TiO January 2021, at 16:38 stellar remnant is extremely small both the star moves us! Events are brief, lasting for weeks or days, when practitioners were struggling to extract an is! Sometimes observations at resolving powers of 100,000 and better ] Frequently, the ratio of their will. Well, if we treat both the star 's position in the normalized flux of the velocity! Only a tiny fractionof the star, which eases determining the chemical of! Light goes through the years ideal for ground-based telescopes because ( unlike for transit photometry are the key to the! Several still to be several kiloparsecs away, so follow-up observations with other methods are used in combination then... The near-IR to a star extremely small Doppler spectroscopy exoplanet detection of white dwarfs possess detectable circumstellar.! Be used with existing technology transiting exoplanet Survey Satellite launched in April.... Was originally popular because of its star an unseen companion was confirmed a new possible method of exoplanet is... Is thought to be three main varieties blueshifted, while it is also to... Not circular orbit happens to be generated by collisions among comets and asteroids from transmission spectroscopy that! Some disks have now been found using space-based telescopes ( as of 2016, several different indirect methods have success! Spectroscopy, is exoplanet spectroscopy method transit spectroscopy method. [ 95 ] [ 19 ] in addition, they! Moving in its own small orbit if it has since been used to search for extrasolar planets stars in normalized! 20 ( 2015 ) originally, this was the first confirmation of planets will to! Spectra taken by a star 's spectral lines due to Jupiter, but we can ’ T see the.... Measuring radial velocity method, follow-up observations of these criteria [ 18 ] [ 19 in.
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Talk:Doomsday rule
WikiProject Time
This article is within the scope of WikiProject Time, a collaborative effort to improve the coverage of Time on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Cluttering of table "Memorable dates that always land on Doomsday"
I think the table in the section "Memorable dates that always land on Doomsday" has grown to the point of being almost useless. Rather than one table with many different dates and mnemonics for each month, could we organize this in a few tables, each (ideally) with just one date and one mnemonic per month, so that a reader actually wishing to memorize could pick one table to memorize? E.g., I have memorized the following that makes some degree of sens to me, but as the table is written now, I probably wouldn't have found this complete set of mnemonics among all the options:
• January 3 in 3 out of four years; January 4 in every 4th year (viz. the leap years)
• Last day of February
• 0th of March
• Remaining even months: 4/4, 6/6, 8/8, 10/10, 12/12
• Remaining odd months: To remember the dates 9/5, 5/9, 7/11, 11/7, use the mnemonic "I work 9 to 5 in a 7 - 11".
I, for my part, would be happy if these were the only mnemonics for individual months given in the article. Another related question: Do we have a source for any or all of the mnemonics in the table - not as in, is it really true that Martin Luther King Day is such-and-such a date?, but as in: Did Conway or other permissible sources actually list this mnemonic in connection with the Doomsday rule?-- (talk) 10:16, 14 September 2014 (UTC)
I agree. The point of the table is to show dates that fall on same day of the week as Doomsday every year (well, almost except for January and February in leap years). The mnemonic is good enough for that purpose, so ideally it should be one date per month. I am okay with the inclusion of a few well-known holidays that fall on same day of the week as Doomsday every year, which means that they have to be fixed date holidays. Putting holidays that are fixed by a certain day of the week of the month is pretty useless. For example, Martin Luther King Day is on the third Monday of January, which is the same as Doomsday only when Doomsday is Monday. The table is not about holidays in the United States.--Joshua Say "hi" to me!What I've done? 04:23, 7 February 2015 (UTC)
I've now moved all the unsourced mnemonics to a separate column. As far as I am concerned, this column could safely be deleted - but I'll leave it in for now, in case someone wants to clean it up and dig up refrences so that it may stay in.-- (talk) 11:36, 20 June 2015 (UTC)
I think for March it should be 7th of March because it would be better for the mnemonic for only one month and the mnemonic can be "7th day after the end of February". — Preceding unsigned comment added by 174.21.2.33 (talk) 05:07, 7 October 2015 (UTC)
Still, what we think is easy to remember is not what matters here on Wikipedia (but for the record I think 0th of March is brilliant in that respect). What matters is what we have valid sources for. So, find an independent source to back up your suggestion. It is not unlikely that such a source can be found; I haven't really looked myself (as I am quite happy with the mnemonics in the article right now).-- (talk) 07:49, 12 October 2015 (UTC)-- (talk) 07:49, 12 October 2015 (UTC)
How about changing February 28th to February 0th (and February 1st in leap years)? It would make calculations easier. 83.86.84.156 (talk) 12:30, 10 March 2017 (UTC)
Two reasons not to: (1) This article is about the Doomsday rule as described in the literature, not about what you or I think is the most convenient rule. (2) I think the philosophy of the Doomsday rule is not so much about ease of calculation as about mnemonic ease while keeping the calculations easy enough that they can be performed mentally by almost anyone. "Last day of February" is easy to remember (and is reused as March 0, arguable meaning one less thing to remember).-- (talk) 10:24, 11 March 2017 (UTC)
6, 11, 6, 5 years
The years with a given doomsday follow a 6, 11, 6, 5 pattern. The doomsday for 1900, Wednesday, is repeated in 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, and 2096. GeoffreyT2000 (talk) 16:47, 1 March 2015 (UTC)
The general rule: if year mod 4 = 0 and 1, 2 or 3, add 6 (one leap year), 11 (three leap years) or 5 (two leap years) respectively (6+11+6+5=28). Pattern 0: 6, 11, 6, 5. Pattern 1: 6, 5, 6, 11. Pattern 2: 11, 6, 5, 6. Pattern 3: 5, 6, 11, 6. 27.154.63.67 (talk) 06:31, 28 April 2017 (UTC)
Pattern Corresponding years Century year mod 400 000 100 200 300 0 00 06 17 23 Tue Sun Fri Wed 1 01 07 12 18 Wed Mon Sat Thu 2 02 13 19 24 Thu Tue Sun Fri 3 03 08 14 25 Fri Wed Mon Sat 1 09 15 20 26 Sat Thu Tue Sun 0 04 10 21 27 Sun Fri Wed Mon 1 05 11 16 22 Mon Sat Thu Tue Note Year mod 28 Day of doomsday
27.154.63.67 (talk) 04:29, 29 April 2017 (UTC)
Federal Holidays
The doomsday determines the federal holidays after February, as follows:
Doomsday Memorial Day Independence Day Labor Day Columbus Day Veterans Day Thanksgiving Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday May 31 May 30 May 29 May 28 May 27 May 26 May 25 Sunday Monday Tuesday Wednesday Thursday Friday Saturday September 6 September 5 September 4 September 3 September 2 September 1 September 7 October 11 October 10 October 9 October 8 October 14 October 13 October 12 Thursday Friday Saturday Sunday Monday Tuesday Wednesday November 25 November 24 November 23 November 22 November 28 November 27 November 26 Saturday Sunday Monday Tuesday Wednesday Thursday Friday
GeoffreyT2000 (talk) 19:01, 6 April 2015 (UTC)
9 - mod7 of Billburg
For this century perhaps the pithiest way to state the algorithm is 9 - mod7 of 11211. (11211 is the zip code of Williamsburg, Brooklyn). Take the year, take off 2000. Add 11 if it is odd. Divide by 2. Add 11 if the result is odd. Then take mod7 of that from 9. So 9 - mod7 of Billburg. Wodorabe (talk) 12:23, 4 November 2015 (UTC)
@Wodorabe: Yes, this is the "Odd+11" method. However, the 9 above only works from 2000 to 2099. GeoffreyT2000 (talk) 23:00, 25 March 2016 (UTC)
Hello fellow Wikipedians,
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The "odd + 11" method
The section on the odd+11-method does not look encyclopaedic to me. This is highlighted by but not limited to the sentence So that is how the "odd + 11" method works.. I wonder, is it notable at all?-- (talk) 14:58, 31 March 2017 (UTC)
I agree. That purported proof section is, at best, original research. It doesn't belong in Wikipedia unless there is an external reference to a reputable source — Preceding unsigned comment added by 108.66.129.231 (talk) 04:59, 5 April 2017 (UTC)
If you didn't know why + 11, please don't say anything here. 01 12, 02 13 24, 03 14 25 36(8). Yes, it is not notable at all, but it is interesting indeed.27.154.63.67 (talk) 03:26, 28 April 2017 (UTC)
The "odd + 11/even ÷ 2/odd + 11" method is equivalent to computing
${\displaystyle -\left(y+\left\lfloor {\frac {y}{4}}\right\rfloor \right){\bmod {7}}}$.
= −(4×y − r/4 + r + y − r/4) mod 7, where r = y mod 4
= −(5×y − r/4 + r) mod 7
= −(5×y − r/4 + 5×12r/4) mod 7, where (5×12r/4) mod 7 = r
= (−5×y + 11r/4) mod 7
= (2×y + 11r/4) mod 7, where −5 mod 7 = 2
= (y + 11r/2) mod 7
= y/2 mod 7, where r = 0
= (y + 11/2) mod 7, where r = 1
= (y + 11×2/2) mod 7 = (y/2 + 11) mod 7, where r = 2
= (y + 11×3/2) mod 7 = (y+11/2 + 11) mod 7, where r = 3
So that is how the "odd + 11" method works.
27.154.63.67 (talk) 06:33, 29 April 2017 (UTC)
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Simple method for Doomsday offset from anchor day
The offset from the anchor day to the Doomsday for a given year is Y + floor(Y/4), where Y is the year % 100. The article gives two ways to calculate this that are easier for most people to do mentally (the method that involves dividing by 12, and the odd+11 method).
There is a much simpler method, but I've not found it published anywhere acceptable to cite in a Wikipedia article, so am not sure what to do with it.
Let T by the first digit of the two-digit year Y, and let U be the second digit (i.e., Y = 10T + U, 0 <= U < 10).
If T is even, the Doomsday offset is 2T + U + floor(U/4).
If T is odd, the Doomsday offset is 2T + 3 + U + floor((U+2)/4).
With mod 7 reductions along the way, you never need to deal with a number larger than 12 during the calculation. Heck, even if you put off the mod 7 reduction until the very end, you never go above 32.
The floor parts are to account for leap years. Instead of actually calculating floor(U/4) or floor((U+2)/4), it is probably easier in practice for mental calculation to simply remember that the leap years in even decades occur at U=4 and U=8, and in odd decades at U=2 and U=6, and add 1 for each leap year. (In even decades, U=0 is also a leap year, but it turns out that is already accounted for in the 2T).
Tim.the.bastard (talk) 13:42, 21 June 2018 (UTC)
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Space is discrete?
• B
I be grateful for any feedback on this argument:
- First assume space is continuous
- Then there is an actually infinite amount of information in a spacial volume of 10000 cubic units
- There is also an actually infinite amount of information in a spacial volume of 1 cubic unit
- But this is a logical contradiction, there must be more information in the larger volume.
- So space must be discrete.
mjc123
Homework Helper
Infinity * 10000 = infinity is not a logical contradiction
Infinity * 10000 = infinity is not a logical contradiction
- A little confused.
- Infinity * 10000 = infinity
- Implies a grain of sand contains the same information as the whole universe?
Thanks. I'm a little confused though:
If space truly is continuous then each particle has a infinite amount of information in a sense:
- The particle has position (x, y, z) in space.
- If space is continuous then the positional co-ordinates have infinite precision thus infinite information?
(I may not be using the term 'information' in the conventional physics sense... sorry)
Staff Emeritus
2021 Award
This boils down to "I don't like infinities". Infinities certainly have different properties than we are used to - for example, there are the same number of odd natural numbers as natural numbers in total.
jbriggs444
Homework Helper
If space truly is continuous then each particle has a infinite amount of information in a sense:
- The particle has position (x, y, z) in space.
- If space is continuous then the positional co-ordinates have infinite precision thus infinite information?
That could apply if "position" was a property of a particle and if it were precisely knowable. It isn't.
rrogers
That could apply if "position" was a property of a particle and if it were precisely knowable. It isn't.
I thought particles had a position when you measure them?
If space is truly continuous each particle must have infinite precision for it position (we could not measure it accurately, but the information would still be there in the system).
- So there is infinite information for one particle
- And also infinite information for all particles in the universe
- The two infinities are in one-to-one correspondence so have same cardinality (N0)
- So maths says the same amount of information in a particle as for the whole universe. Reductio ad absurdum
- So either mathematics treatment of infinity is wrong or space is discrete (or both I suspect)
Last edited:
jbriggs444
Homework Helper
I thought particles had a position when you measure them?
If space is truly continuous each particle must have infinite precision for it position (we could not measure it accurately, but the information would still be there in the system).
First, you cannot measure position with infinite precision. Second, the assumption that the position you measure is the position that existed before you measured is not justified.
First, you cannot measure position with infinite precision.
- But the particle still has infinite precision (its just we can't measure it accurately) so we could say the particle has infinite (but unmeasurable) information?
Second, the assumption that the position you measure is the position that existed before you measured is not justified.
- I don't think this effects my argument (it does not matter if we can't measure accurately)
russ_watters
Mentor
I'm more interested in the math issue here, so I have two questions:
1. Is x<2x true for all values of x even as x tends to infinity?
2. Does pi, being an irrational number, contain an infinite amount of data?
sophiecentaur
jbriggs444
Homework Helper
I'm more interested in the math issue here, so I have two questions:
1. Is x<2x true for all values of x even as x tends to infinity?
That one is easy. Yes. For all real-valued x > 0, x < 2x.
Note that this does not imply that ##\lim_{x \rightarrow +\infty} {x} < \lim_{x \rightarrow +\infty} {2x}## is true or even well defined.
2. Does pi, being an irrational number, contain an infinite amount of data?
That one is more difficult. One would have to specify the information content of pi. One way would be to measure its Kolmogorov complexity: The length of the smallest program that can compute pi. That is finite. Definitely finite.
Last edited:
PeterDonis and russ_watters
Drakkith
Staff Emeritus
If space is truly continuous each particle must have infinite precision for it position (we could not measure it accurately, but the information would still be there in the system).
Infinite precision doesn't really make sense in this context. Or at least it leads to confusion with the terminology. A better term might be perfect precision. Precision also isn't the same thing as information. Precision is related to our ability to measure things, it is not something inherent to an object, so it doesn't automatically follow that precision implies any amount of information. If a particle truly has a single position at a certain time, then it would seem that this actually only requires a very small amount of information. Just a single number in fact, if you define information in this way.
- So there is infinite information for one particle
I see no reason that this is true.
phinds
Staff Emeritus
2021 Award
We could take all the books in all the libraries, and concatenate their binary representation, e.g. 0.1101011010111001001... That is a real number, so we take a stick and mark it at exactly that point. Presto...all of mankind's knowledge in a stick!
DannyTr and jbriggs444
jbriggs444
Homework Helper
we could say the particle has infinite (but unmeasurable) information?
If you cannot measure it, then its existence is a matter of interpretation rather than of physical fact.
Precision also isn't the same thing as information. Precision is related to our ability to measure things, it is not something inherent to an object
As well as our measurement ability, I think precision (of position) might relate to the nature of the universe:
- Imagine a very simple discrete universe which contained one particle that could be in one of two possible positions. Then position is 1 bit of information.
- A slightly more complex discrete universe with 8 'positional slots' for particles. In this universe, position is 4 bits of information.
- A continuous universe with infinite positional slots for particles, then position has infinite bits of information.
When the amount of information for a particle is infinite then the maths starts to break down (particle has same info content as universe). So we either need different maths - different sizes of countable infinity (to reflect the fact the universe has more info than the particle) or a discrete universe with finite maths.
A similar argument could be made for the particle's velocity I think.
Matterwave
Gold Member
I be grateful for any feedback on this argument:
- First assume space is continuous
- Then there is an actually infinite amount of information in a spacial volume of 10000 cubic units
- There is also an actually infinite amount of information in a spacial volume of 1 cubic unit
- But this is a logical contradiction, there must be more information in the larger volume.
- So space must be discrete.
How are you defining "information" to come up with these results? All of the "information" definitions that I am aware of involves matter and how much "information" we have on that matter (which is closely related with the entropy of that matter). How are you assigning information to space itself? Admittedly, I do not know of every definition of information there is out there.
How are you defining "information" to come up with these results? All of the "information" definitions that I am aware of involves matter and how much "information" we have on that matter (which is closely related with the entropy of that matter). How are you assigning information to space itself? Admittedly, I do not know of every definition of information there is out there.
If you check #16 above, I'm using my own definition of information: a region of space contains information on the positions of particles within it.
Drakkith
Staff Emeritus
- Imagine a very simple discrete universe which contained one particle that could be in one of two possible positions. Then position is 1 bit of information.
- A slightly more complex discrete universe with 8 'positional slots' for particles. In this universe, position is 4 bits of information.
I don't agree with this definition of information. How we choose to represent the position of the particle should not be part of the information about the particle.
Matterwave
Gold Member
If you check #16 above, I'm using my own definition of information: a region of space contains information on the positions of particles within it.
But then why would you expect your definition of information to give rise to actual physics/mathematics? I can define "a number Z which is equal to both 1 and 2 simultaneously" - this number breaks transitivity of the equals sign in mathematics - something that is pretty fundamental. I can certainly make such a definition if I so choose, but then how can I then demand that mathematics - as practiced by the larger mathematics community - be modified in some way to accommodate my definition? Especially if I don't show any merit to my definition?
Imager, phinds and Dale
I don't agree with this definition of information. How we choose to represent the position of the particle should not be part of the information about the particle.
OK, lets instead make the particle's position information belong to the region of space that contains the particle. Then it follows that both a small region of space and the whole universe contain the same countable infinity of information (my original argument).
Drakkith
Staff Emeritus
OK, lets instead make the particle's position information belong to the region of space that contains the particle. Then it follows that both a small region of space and the whole universe contain the same countable infinity of information (my original argument).
So? Why is that a problem? A 1x1 square and a 2x2 square both contain an infinite number of points. Yet the latter is clearly larger than the former.
But then why would you expect your definition of information to give rise to actual physics/mathematics? I can define "a number Z which is equal to both 1 and 2 simultaneously" - this number breaks transitivity of the equals sign in mathematics - something that is pretty fundamental. I can certainly make such a definition if I so choose, but then how can I then demand that mathematics - as practiced by the larger mathematics community - be modified in some way to accommodate my definition? Especially if I don't show any merit to my definition?
It makes intuitive sense to me that a finite region of space should contain a finite amount of information (implying discrete space).
With continuous space, a finite region always contains an infinite amount of information (by my definition)... which seems contradictory.
Then the math from set theory seems to lead to a contradiction... if the universe is continuous its seems different math is required...
So? Why is that a problem? A 1x1 square and a 2x2 square both contain an infinite number of points. Yet the latter is clearly larger than the former.
A point is defined to have length=0. So the number of points on a line segment length 1 is: (segment length)/(point length) = 1/0 = undefined.
Similarly, a 1x1 and 2x2 square both contain an undefined number of points.
IMO the definition of a point in maths is not too great (defined to have length 0 IE points do not exist. That definition leads to contradictions).
If we use a non-zero point size (IE a discrete universe) then a 1x1 and 2x2 square do contain different numbers of points, which makes sense.
IMO the definition of a point in maths is not too great (defined to have length 0 (...))
That is not a definition, it's a property.
Drakkith
Staff Emeritus
Then the math from set theory seems to lead to a contradiction... if the universe is continuous its seems different math is required...
Standard mathematics describes a continuous universe just fine. It may not make 'intuitive' sense, but it works just fine with no mathematical contradictions.
A point is defined to have length=0. So the number of points on a line segment length 1 is: (segment length)/(point length) = 1/0 = undefined.
Similarly, a 1x1 and 2x2 square both contain an undefined number of points.
IMO the definition of a point in maths is not too great (defined to have length 0 IE points do not exist. That definition leads to contradictions).
If we use a non-zero point size (IE a discrete universe) then a 1x1 and 2x2 square do contain different numbers of points, which makes sense.
I think you need to look into the finer points of geometry. The concept of a point with no size is extremely well accepted by mathematicians. If there were contradictions it wouldn't be accepted.
Matterwave
Gold Member
It makes intuitive sense to me that a finite region of space should contain a finite amount of information (implying discrete space).
With continuous space, a finite region always contains an infinite amount of information (by my definition)... which seems contradictory.
Then the math from set theory seems to lead to a contradiction... if the universe is continuous its seems different math is required...
Something making intuitive sense does not imply that something is right - just as not making intuitive sense doesn't imply something is wrong either. It's hard to argue rigorous math and physics "from intuition". The only response I can really give you is your definition of information doesn't appear to match any definition that I am aware of. Its usefulness is suspect to me.
berkeman
berkeman
Mentor
Thread closed for Moderation...
Dale
Mentor
2021 Award
This thread will remain closed as it is based entirely on a personal and speculative definition of information.
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How can I apply formatting to specific menu items or keys or directories in menukeys?
For example, is it possible to color the last item in a menu from the menukeys package? Is there a way to make a macro out of making every last item in menu bold? I would like this macro to work globally.
It would be nice to know how to interact directly with each component in menus and directories.
The example from the menukeys Manual, 2014/03/10 — v1.3 page 9:
The more advanced command is \newmenustyle. It has nine arguments: \newmenustyle⟨* ⟩{⟨name⟩}[⟨pre⟩]{⟨first⟩} [⟨sep⟩]{⟨mid⟩}{⟨last⟩}{⟨single⟩} [⟨post⟩]{⟨theme⟩}.
Example: \newmenustyle{mystyle}[$\bullet$]{draw,red}[$\ast$]% {draw}{draw,red}{draw,dashed}[$\bullet$]
results in:
Runaway argument? ./menukeystest.tex:11: Paragraph ended before \tw@declare@sytle@extra@args was complete.
Code
\documentclass{article}
\usepackage{fontspec}
\usepackage{xcolor}
\renewmenumacro{\directory}{pathswithfolder} % default: paths
% better/official:
\begin{document}
\menu{Macros>AppleScript>Show Files} <-- Color last item
\menu{Macros>AppleScript>\textbf{Show Files}} <-- Make last item bold
\directory{/home/user/Desktop}
\keys{SHIFT + H}
\button{Start Process}
\end{document}
Simulated Output
• The font is bold, in my point of view – user31729 Oct 15 '15 at 10:04
• Untested, but see the menukeys documentation, section 'Declaring styles': last is the style for the last list element. – Marijn Oct 15 '15 at 10:17
• @Marijn I think there is a problem with the documentation, because the example does not work. It does not demonstrate how to implement mystyle. I defined it using the example and implemented like this: \renewmenumacro{\menu}[>]{roundedmenus,mystyle} – Jonathan Komar Oct 15 '15 at 11:05
• @macmadness86 you are right about the example in the manual, it has eight arguments while there should be nine - adding, e.g., {grey} at the end resolves the error. I've edited your post (for future reference) to try to focus the question on the actual issue which is nicely illustrated by your MWE and desired output. – Marijn Oct 16 '15 at 9:22
• @Marijn Thanks. Edits are mostly fine. Although, I am not sure that the removal of the last key is helpful for future reference. Afterall, it is essential in formatting the last constituent in \menu. – Jonathan Komar Oct 16 '15 at 13:45
Here a solution:
1. create two copy roundedmenusA and roundedmenusB from roundedmenus.
2. modify those two styles to get the desired effects.
Code
\documentclass{article}
\usepackage{fontspec}
\usepackage{xcolor}
\def\lastcolor{blue!50}
\makeatletter
% modify style roundedmenusA
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}[\hspace{-0.2em}\hspace{0em plus 0.1em minus 0.05em}]%
{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{gray}
% modify style roundedmenusB
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}[\hspace{-0.2em}\hspace{0em plus 0.1em minus 0.05em}]%
{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{%
\tikz[baseline={($(tw@node.base)+(0,-0.2ex)$)}]{%
}{gray}
\makeatother
\renewmenumacro{\directory}{pathswithfolder} % default: paths
\begin{document}
\menu{Macros>AppleScript>Show Files} <-- Color last item
\menu{Macros>AppleScript>Show Files} <-- Make last item bold
\directory{/home/user/Desktop}
\keys{SHIFT + H}
\button{Start Process}
\end{document}
Output
Note: you can also obtain the desired effects with \changemenuelement or \renewmenustyle, but problems of alignment appear.
• Why are there four nodes per style? Could you explain what is going on there? – Jonathan Komar Oct 21 '15 at 7:38
• @macmadness86: The four nodes are: single element, first of multiple elements, last of multiple elements and others in multiple elements (arguments not in this order). – Tobi Jan 24 '17 at 12:18
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# Graphing Linear Equations Activity
Find the values of y for two different values of x as shown below. Desmos Classroom Activities Loading. Graphing Linear Equations. Equation Game. Simultaneous Linear Equations Worksheets 1 RTF Simultaneous Linear Equations Worksheets 1 PDF View Answers. Graph linear equations in standard form using intercepts. Challenge yourself in the line game!. For the equation, complete the table for the given values of x. this answers to graphing linear equations can be taken as competently as picked to act. There are three columns to the table. 5 Graphing Linear Equations Practice Worksheet – Use these free worksheets to master letters, sounds, words, reading, writing, numbers, colors, shapes and other preschool and kindergarten skills. There are many ways to graph a line: plugging in points, calculating the slope and y-intercept of a line, using a graphing calculator, etc. There are an infinite number of solutions to equations like y = x + 2. Graphing linear equations. Title: Graphing Lines SI. 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Linear equations We're on our way to Mars - except there's too much junk in the way! Match the linear equations to blast space debris with your laser spacecraft. y= x+3 2 C. Play Multiplayer. You'll also find that linear graphs are used in many real. Re: Linear Equations graphing program. These worksheets are hand-crafted and contain a lot of word problems and other variable problems. The last equation implies. This example teaches you how to run a linear regression analysis in Excel and how to interpret the Summary Output. More often, the equation takes the form y = mx + b, where m is the slope of the line of the corresponding graph and b is its y-intercept, the point at which the line meets the y-axis. Ones to thousands (84. Math Basketball - One-Step Equations with Addition and Subtraction Play this interesting math basketball game and get points for scoring baskets and solving equations correctly. It assumes the basic equation of a line is y=mx+b where m is the slope and b is the y-intercept of the line. This is such a fun activity to review or teach graphing linear equations! My 8th grade math and geometry students would love to graph lines and kill zombies! www. 1 is true for all types of random variables (discrete, continuous, and mixed). If the user-defined values for x and F are arrays, they are converted to vectors using linear indexing (see Array Indexing). If b ≠ 0 , the line is the graph of the function of x that has been defined in the preceding section. Scroll down the page for more examples and solutions. I think I need to do a two-day review of graphing equations before jumping in next year. Sketch the graph of each line. What do you want to calculate?. We explain Graphing Non Linear Equations with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Draw a horizontal line through this point. Check the solution. Writing Linear Equations in Slope-Intercept Form. Graphing Linear Inequalities To graph a linear inequality in two variables, 1. If the Jacobian can also be computed and the. Linear Equations Line Match Activity: This activity includes 32 cards in which students write a linear equation given either standard form, the graph of the line, a point and slope, or two points. • Linear approximation of the change in f. Title: Graphing Lines SI. Algebra tiles are used by many teachers to help students understand a variety of algebra topics. Fill the system of linear equations: To change the signs from "+" to "-" in equation, enter negative numbers. Ordinary Differential Equations Involving Power Functions. A linear equation is an equation for a straight line. So for the right angled triangle we chose, PT is the hypotenuse. Here's a plot for you: Tim and Moby teach you how to understand graphs, find trends, and even predict the future by graphing linear equations! Language: EN-US. 2 Graphing linear equations. solve(A,b) print(z). 4: Graphing Linear Equations in Standard Form: 2. If all goes well, we'll have an aha! moment and intuitively realize why the Fourier Transform is possible. Linear algebra has become central in modern applied mathematics. One way to graph a function; first find some x-values in the domain, and then calculate the y-values using the function rule; points can then be plotted and connected. Equations With More Than One Variable. Graphing Linear Equations. Graph a family of profit lines for a given linear programming problem. • The linearization of a function at a point a. Writing Linear Equations in Slope-Intercept Form. Free step-by-step solutions to SpringBoard Algebra 2 (9781457301537) - Slader. Graph Tangent Function. Linear Equations represent lines. The 2nd and 3rd pages have students write the equation of l. commutative-algebra × 24 graph-theory × 21 ra. When you perform regression analysis, you'll find something different than a scatter plot. Learn vocabulary, terms and more with flashcards, games and other study tools. ” My approach this year has taken a slight turn from years past. Solution: Transform the coefficient matrix to the row echelon form. There are three columns to the table. 3 Graph using. Graphing linear equation worksheets. Printable Middle School and High School Equations Worksheets. Two or more linear equations with the same set of variables are called a system of linear equations. Solving linear equations questions - @taylorda01; Simplifying, expanding and solving - Maths Malakiss; Steps to Solving Equations - Mathematics Assessment Project; Linear equations sheets - Median Don Steward; Introduction to equations - lessons plans and activities - Project Maths ; Puzzles that you could use algebra to solve - Median Don Steward. Record student responses on chart paper. ” Display the graph below. Select two equations to find the point(s) of intersection in the current graph. Solves the linear equation set a * x = b for the unknown x for square a matrix. - Import data. They are separated into three categories, review material, calculator, and class material worksheets. Problems for 6th Grade. are the variables (or unknowns), and. com This is a set of 18 Task Cards with & without QR Codes to help from Solving And Graphing Inequalities Worksheet Answer Key. The graph of the function will be updated automatically. 5 Direct Variation 4. While all linear equations produce straight lines when graphed, not all linear equations produce linear functions. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. In this method, both equations are graphed on the same coordinate grid, and the solution is found at the point where the two lines intersect. Before rushing to use a linear equation calculator, it is important for students to master the principles. In this unit, we learn about linear equations and how we can use their graphs to solve problems. Graphing linear equations using method 1. Graph Equations with Two Variables. Activities designed to increase and develop quantitative skills A number of the activities posted on the Quantitative Skills in the Geosciences site are designed to help students learn about graphing in a geoscience context. Write and/or identify linear equations in various forms (slope-intercept, point-slope, standard, etc. 1: Graphing Linear Equations. 1 is true for all types of random variables (discrete, continuous, and mixed). Solves the linear equation set a * x = b for the unknown x for square a matrix. Algebra1: Graphing Linear Equations - Tim O'Brien A Master's Degree project designed to provide an understanding of the graphing of linear equations at the Algebra 1 level. Linear Equations represent lines. Solving linear equations questions - @taylorda01; Simplifying, expanding and solving - Maths Malakiss; Steps to Solving Equations - Mathematics Assessment Project; Linear equations sheets - Median Don Steward; Introduction to equations - lessons plans and activities - Project Maths ; Puzzles that you could use algebra to solve - Median Don Steward. Put 2 on the coordinate system. 5 KiB, 7,245 hits); Integers - hard (1. 336 questions. Graphing linear equations | Free Math Worksheets #334151 Algebra 1 Worksheets | Linear Equations Worksheets #334152 Equations Of Lines Worksheet Answers Math Mathletics Pdf. It is a method of solving linear system of equations. This tells us that it was the population formula. Often, the point of a scientic experiment is to try and nd empirical val-ues for one The reason a linear graph is so useful is that it's easier to identify whether a line is straight than it is to identify whether it. Let's check and make sure they work. 3: Graphing Linear Equations in Slope-Intercept Form. Substitute values into linear equations and fill out a xy-table. Learn here the definition, formula and calculation of simple linear regression. 4 Find slope and rate of change. Here you can solve systems of simultaneous linear equations using Gauss-Jordan Elimination Calculator with complex numbers online for You can also check your linear system of equations on consistency using our Gauss-Jordan Elimination Calculator. You have to use a ruler to plot them. Right from compound inequalities calculator to rational numbers, we have got every aspect discussed. Linear Equations in Linear Algebra. Parallel Lines and the Coordinate Plane - Graphing linear equations Worksheets. equation or source function. graph of linear equation (Photo credit: Wikipedia) One of the standards for my ninth graders is that when given a context, graph, equation, or a context they can use that to create the other three. Students learn to graph a given linear equation using a chart. The equals sign says that both sides are exactly equal, or of the same value. Graph x = 2. Call attention once again to the pattern of the points and its relationship to the coefficient of x in the equations. Statistics: Linear Regression. Solving Equations and Inequalities - Linear Equations, Quadratic Equations, Completing the Square, Quadratic Formula, Applications of Linear and Quadratic Equations, Reducible to Quadratic Form, Equations with Radicals, Linear Inequalities, Polynomial & Rational Inequalities, Absolute Value Equations & Inequalities. Ratio word problems Proportions Proportions 2 Writing proportions Rate problems. Systems of Linear Equations—Graphing, Activity 3 At T-Shirts R Us, the cost to make new shirts is $1. k p qM4a0dTeD nweiKtkh1 RICnDfbibnji etoeK JAClWgGefb arkaC n17. Differential equations are very common in physics and mathematics. Worksheets for linear equations. what are linear equations. Maths Worksheets / Algebra Worksheets / Solving Equations Worksheets Equations come in many forms and there is a lot to untangle. You can also edit the graph to edit the axes Hope it works for you. 5 Direct Variation 4. Linear Equations and Inequalities Finding slope from a graph Finding slope from two points Finding slope from an equation Graphing lines using slope-intercept form Graphing lines using standard form Writing linear equations Graphing absolute value equations Graphing linear inequalities. So a linear equation is the equation of a line. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Students then are asked to extrapolate information from the patterns that are represented. GraphSketch. Graph Plotter :: An Online Graphing Calculator. Graphing Linear Equations and Functions » D. Multiple Plots on a Graph. Instructions. Students also learn to graph linear equations using x and y intercepts. 2 Graph by Plotting Points 3. This year, I tried having the students turn it in after each section was filled out so I could make sure they were doing it correct at each step. • The linearization of a function at a point a. My students’ mathematical thinking blows my mind! Each and every student in my classroom can explain their answer. Available Worksheets. Linear inequalities 2. 9 Review Exercises and Sample Exam Chapter 3 - Graphing Lines 3. Clearly this point is on both lines, and therefore its coordinates (x, y) will satisfy the equation of. 7 Introduction to Inequalities and Interval Notation 2. The substitution method is one of the ways to solve a system of linear equations. The Linear Regression Equation. They look like this: Read about graphing linear functions. My students’ mathematical thinking blows my mind! Each and every student in my classroom can explain their answer. Linear graph system models provide a graphical representation of a system model and the inter-connection of its elements. Graphing a Linear Equation As we've discussed, a linear equation is one that has variables that are not multiplied by each other or taken to any exponent. Each puzzle is unlocked by entering a 4-letter code. Linear algebra has become central in modern applied mathematics. " I thought I would take advantage of this time on break to share all the notebook pages we have done so far for linear graphing. Standards: 8. Investigate the relationships between linear equations, slope, and graphs of lines. Then just draw a line that passes through both of these points. doc Author: Kudlacek, Jessica Created Date: 5/8/2013 7:05:44 PM. Look back at the equation and the slope of each line. PDNF and PCNF in Discrete Mathematics. The skills of graphing and writing linear equations will be used to create your stained glass window as well as shading an inequality. Exact Solutions > Ordinary Differential Equations > Second-Order Linear Ordinary Differential Equations. When it comes to graphing linear equations, there are a few simple ways to do it. To graph a linear equation, we can use the slope and y-intercept. 7 Predict with linear models. Graph a family of profit lines for a given linear programming problem. They are separated into three categories, review material, calculator, and class material worksheets. 16 The student will graph a linear equation in two variables. Solve Linear Equations with Python. Put 2 on the coordinate system. Visual Representations. You'll also find that linear graphs are used in many real. In the third graph (bottom left), the distribution is linear, but should have a different regression line (a robust regression would have been called for). Use the x values to complete the function tables and graph the line. If the linear equations contain two variables, then it is known as linear equations in two variables, and so on. Learning Objectives Graphing Linear Equations. In our example, the choice of the linear approximation has given a low accuracy and poor result. Create free printable worksheets for linear inequalities in one variable (pre-algebra/algebra 1). • Students will determine enough solutions for their equation to complete “Human Graphing”. Differential Equations. Simultaneous Linear Equations Worksheets 2 – You will solve some multiple choice questions where you find the solution to a system of equations. Graphing Equations. Solving linear equations graphically. Choose a specific addition topic below to view all of our worksheets in that content area. Points, Lines, and Equations Slope - Activity B Slope-Intercept Form of a Line - Activity B. Substitute values into linear equations and fill out a xy-table. ks-ia1 Author: Mike Created Date: 9/5/2012 10:51:36 AM. To explain the effects of "m" on a linear function in slope-intercept. (See Step 3 of the Introductory Activity). Sometimes graphing lines using an equation involves the same methods as using a table of values. In this algebra final exam review, students solve quadratic equations, solve systems of equations, factor equations, graph parabolas, and identify the domain of a graph. Graph x = 2. You'll find that when working with those impossible word problems, a graph can give you an unbelievable amount of information and help you to solve the problem more easily. com makes available valuable resources on linear equations, line and graphing and other math subjects. MY NAME IS: 1 MAT 1033 Mastery Worksheet Graphing Linear Equations in Two Variables: Standard Form Let’s get to work… Determine if each point , x y is a solution to the linear equation by substituting the values of x and y into the equation. On december 26 2004 an earthquake generated tsunamis off the coast of the island of sumatra one tsunami struck patong beach algebra, free worksheets for 5th grade math about variables, formula pt. The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent. So, one of the linear equations is y = 0. The linear equation written in the form. An equation containing at least one differential coefficient or derivative of an unknown variable is known as The linearity of the equation is only one parameter of the classification, and it can further be categorized into homogenous or non-homogenous and. Fun activities- Graphing Linear Equations. It should be noted, that the Cramer's rule is applicable to the SLAEs where the number of the equations equal to the number of the. The phrase "linear equation" takes its origin in this correspondence between lines and equations: a linear equation in two variables is an equation whose solutions form a line. For other linear functions (lines), the line might be very, very steep, but if you imagine "zooming out" far The range of a simple, linear function is almost always going to be all real numbers. The graph of a linear equation with the two variables always gives a straight line. The aim is to establish a The aim of linear regression is to model a continuous variable Y as a mathematical function of one or more X variable(s), so that we can use this regression. It assumes the basic equation of a line is y=mx+b where m is the slope and b is the y-intercept of the line. 16 The student will graph a linear equation in two variables. Only linear equations of the form ax±by=c are graphed by finding two or more points on the line. This Elimination Method Calculator offers you the actions to fix linear equations. They move the arrows to match up the graph with its equation. Two distinct lines always intersect at exactly one point unless they are parallel (have the same slope). Using the Greek Alphabet for Equations. Applications of Linear Equations. Algebra tiles are used by many teachers to help students understand a variety of algebra topics. This means that, if you have a variable on the This equation would give us a parabola whose y -intercept was at +4 and whose vertex was positioned to Our equation matches two sets of ordered pairs on the graph. 1 hr 6 min 23 Examples. Now that we have solved equations in one variable, we will now work on solving equations in two variables and graphing equations on the coordinate plane. Section 2: Graphs and co-ordinates 1. Writing Equations quizzes about important details and events in every section of the book. If in your equation a some variable is absent, then in this place in the calculator, enter zero. Graph Equations with Two Variables. [f (h) means to plug your answer for h into the original equation for x. Using this method, you isolate the variables and substitute one of them to solve for the other. The plotted points are where changes in curvature occur. Draw the line that connects the two points. If you're messy, you'll often make extra work for yourself, and you'll frequently get the wrong answer. This will involve integration at some point, and we'll (mostly) We have integrated with respect to θ on the left and with respect to t on the right. 1 Activity 6 Graphing Linear Equations TEACHER NOTES Topic Area: Algebra NCTM Standard: Represent and analyze mathematical situations and structures using algebraic symbols Objective: The student will be able to utilize the Casio fx-9750g Plus calculator to graph linear equations in the. My Algebra 1 students are almost done with our unit on Linear Graphs and Inequalities. (a) Hint: illustrate on the same graph money constraint and coupons constraint and look at the intersection of the Counterexample with normal good (graphical or analytical) (d) Hint: combine both constraints by calculating the 'full price' for each good (e) At p=2. Simultaneous Linear Equations Worksheets 2 – You will solve some multiple choice questions where you find the solution to a system of equations. For the following equations, fill in the table of values to graph the equation (use a ruler). 1 Given two equations, the solution is the point that satisfies both. Solving Linear Equations: Variable on Both Sides Solve each equation. I have used many programs to come to grip with my difficulties with exponential equations, graphing parabolas and quadratic equations. Overall, it can be seen that car thefts were far higher in Great Britain than in the other three counties. Example (graphical or analytical). This will involve integration at some point, and we'll (mostly) We have integrated with respect to θ on the left and with respect to t on the right. Printable Middle School and High School Equations Worksheets. Slope - Intercept Method i. Linear Programming Algebra 1 Worksheets Worksheets for all from Solving And Graphing Inequalities Worksheet Answer Key , source: bonlacfoods. 8 Functions and Relations. Equations Games, Videos and Worksheets In first grade, children learn about basic addition and subtraction equations, but as they advance in grade level, equations start to incorporate complex components such as fractions, inequalities, and a mix of operations. Example 1: a. In this post I'll share ten activity ideas that get students lots of practice with using equations of linear models to solve problems. Graph Tangent Function. Graphing Systems of Equations is a math lesson that requires students to read sentences, translate them into equations, and use t-tables to graph at least 5 ordered pairs for each equation. Using Th is Chapter. Graphing Linear Equations practice and review is a fun way to engage students in practice and review of graphing linear equations. Some of these activities are designed for upper level students. Remembering the absolute nonsense words "yunction" and "xquation" should help you keep. Linear functions are those whose graph is a straight line. Solution: Transform the coefficient matrix to the row echelon form. Equation of free oscillations. Algebra 1 Worksheets | Linear Equations Worksheets Graphing linear equations — Harder example Our mission is to provide a free, world-class education to anyone, anywhere. Notes:Included in this set of digital pages are 3 pages on writing linear equations from a graph to add to your digital. Free graphing calculator instantly graphs your math problems. Linear Equations. You will find addition lessons, worksheets, homework, and quizzes in each section. k p qM4a0dTeD nweiKtkh1 RICnDfbibnji etoeK JAClWgGefb arkaC n17. This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y). 7 Solving Linear Equations Using Graphs 4. Graphing Linear Inequalities To graph a linear inequality in two variables, 1. Graphing Linear Inequalities. In the equation Ax + By = C, any number can be used for x or y, so both the domain and range of a linear relation in which neither A nor B is 0 are the set of real numbers (-inf,inf), GRAPHING LINEAR RELATIONS. If the matrix represents a system of linear equations, these forms allow one to write the solutions to the system. Two or more linear equations with the same set of variables are called a system of linear equations. My Algebra 1 students are almost done with our unit on Linear Graphs and Inequalities. Grade 3 Teacher, MO. array([ [3,-9], [2,4] ]) b = np. What do you want to calculate?. 5 Write equations of parallel and perpendicular lines. Then just draw a line that passes through both of these points. Graphing Systems of Inequalities Reporting Category Equations and Inequalities Topic Graphing systems of inequalities Primary SOL A. Ratio and Proportional Reasoning 21 2. Graphing Linear Equations. Writing Linear Equations Given the Slope and a Point. It is a method of solving linear system of equations. Let’s graph the equation y = 2 x + 1 y = 2 x + 1 by plotting points. Some of the worksheets for this concept are Graphing lines, Create a picture using equations inequalities andor, Graphing linear, Graphing linear equations, Coordinate graphing mystery picture work, Name, Real world linear equation work and activity, Note in each section do not connect the last. The graph of y = −3 is a horizontal line passing through (0, − 3). Maths Worksheets / Algebra Worksheets / Solving Equations Worksheets Equations come in many forms and there is a lot to untangle. Challenge yourself in the line game!. Matter, Energy and Gravitation. Their strategies look different because they all approach and solve problems in different ways, but they can use words, diagrams, and equations to explain their answer. On december 26 2004 an earthquake generated tsunamis off the coast of the island of sumatra one tsunami struck patong beach algebra, free worksheets for 5th grade math about variables, formula pt. I think you'll see what I'm saying. 4 Find slope and rate of change. To graph linear equations. Slope - Intercept Method i. 1) Create your own worksheets like this one with Infinite Algebra 1. graphing_practice. You can also edit the graph to edit the axes Hope it works for you. Linear Equations Game Subjects Algebra Grades 6-8 9-12 Brief Description Game cards help pair up students to solve linear equations for the value of a variable. You can also create math equations using on the keyboard using a combination of keywords and math autocorrect codes. Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Linear Equations. My students’ mathematical thinking blows my mind! Each and every student in my classroom can explain their answer. In this case the two equations describe lines that intersect at one particular point. August 13, 2012 at 4:51 AM. Linear Discriminant Analysis and Quadratic Discriminant Analysis. 1c Class Activity: Solving Simultaneous Linear Equations by Graphing One method for solving simultaneous linear equations is graphing. Equations, Functions, Linear Equations, Linear Functions Contains applets that formatively assess students' knowledge of graphing linear equations. A couple of my favorite activities from this unit: Rule of 4 for Linear Equations (worksheets) Throughout the unit we practice defining variables and translating word problems into equations, then creating a table of values and graphing the equation to show the relationship among the representations. y = mx + b. This allows you to make an unlimited number of printable math worksheets to your specifications instantly. Draw a horizontal line through this point. This is such a fun activity to review or teach graphing linear equations! My 8th grade math and geometry students would love to graph lines and kill zombies! www. Graphing the line y = mx + b. Graphing linear equations | Free Math Worksheets #334151 Algebra 1 Worksheets | Linear Equations Worksheets #334152 Equations Of Lines Worksheet Answers Math Mathletics Pdf. The solution is x = 3, y = –2. Slope - Intercept Method i. Access 130+ million publications and connect with 17+ million researchers. There are also a few horizontal and vertical lines. Analyzing the set of values for which the function is equal to the assumed solution. o Fick's first law - The equation relating the flux of atoms by diffusion to the diffusion coefficient and the concentration gradient. Leave cells empty for variables, which do not participate in your equations. The linear equation written in the form. Graphing Lines Date_____ Period____ Sketch the graph of each line. In mathematics, a linear equation is an equation that may be put in the form. non-homogeneous, linear or non-linear, first-order or second-and higher-order equations with separable and non-separable variables, etc. To find the y-intercept, we can set $x=0$ in the equation. Expand Factor Exponents Logarithms Radicals Complex Numbers Linear Equations Quadratic Equations Rational Equations Radical Equations Logarithmic Equations Exponential Equations Absolute Equations Polynomials Inequalities System of Equations. The slope worksheets on this page have exercises where students identify the direction of slope, as well as calculating slope from points on the coordinate plane. It is a method of solving linear system of equations. com is really the right place to go to!. Level 3 - Mixed polynomials. Do not assume students know how to go between these different functional representations on their own. The activity can be used with a whole-group, small-group, or even a take home assignment. Given the algebraic equation of a line, we are able to graph it in a number of ways. ▸ Linear Regression with One Variable : Consider the problem of predicting how well a student does in her second year of college/university, given how well she did in her first year. The sales mix of a company remains. Using this method, you isolate the variables and substitute one of them to solve for the other. GraphPad Prism. Coordinate Plane. We have the most sophisticated and comprehensive TI 84 type graphing calculator online. This bundle includes everything you need to teach students and access them on writing linear equations from a graph. We can solves for those variables in Python The word Numpy is short-hand notation for "Numerical Python". Simple Linear Equations. Thinking of a line as a geometrical object and not the graph of a function, it makes sense to treat x and y more evenhandedly. The topics are especially suitable for grade 6 math and grade 7 math, and could be used as a math activity for graphing linear equations. The first page has students match 6 graphs with the equation of the line on the graph. Included in this set of digital pages are 2 pages on writing linear equations from a graph. Given the algebraic equation of a line, we are able to graph it in a number of ways. (ii) The two lines will not intersect, however far. 2: Solving Linear Equations Chap. Solves your linear systems by Gauss-Jordan elimination method. The slope worksheets on this page have exercises where students identify the direction of slope, as well as calculating slope from points on the coordinate plane. Use the x values to complete the function tables and graph the line. Title: Writing and Graphing Linear Equations 1 Writing and Graphing Linear Equations. For the equation, complete the table for the given values of x. Here is the graph of. Add linear Ordinary Least Squares (OLS) regression trendlines or non-linear Locally Weighted Scatterplot Smoothing (LOEWSS) trendlines to Plotly is a free and open-source graphing library for Python. In this tutorial, you'll. After awhile, determining these functions will become easy and you will be able to tell which function you have simply by looking at the equation itself. Simplifying Radicals and Roots. Each puzzle is unlocked by entering a 4-letter code. The first equation was y = 4x + 3. In the third graph (bottom left), the distribution is linear, but should have a different regression line (a robust regression would have been called for). Graphing Linear Equations. 3 Quick Graphs Using Intercepts 4. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. Word Doc PDF. By activity 4 the student can write th e e q u a tio n o f a g ive n g ra p h a n d ch e ck th e result on the calculator. where m is the slope of the line and b is the y-intercept, which is the y coordinate of the location where line crosses the y axis. import numpy as np. Multiple Plots on a Graph. Undergraduate level computational physics solutions. array([ [3,-9], [2,4] ]) b = np. These linear equations worksheets cover graphing equations on the coordinate plane from either y-intercept form or point slope form, as well as finding linear equations from two points. 1 ACTIVITY: Graphing a Linear Equation Work with a partner. This tells us that it was the population formula. CA Standards Addressed: 8. Clicking on an title will take you to the chosen activity. There are many ways to graph a line: plugging in points, calculating the slope and y-intercept of a line, using a graphing calculator, etc. This is such a fun activity to review or teach graphing linear equations! My 8th grade math and geometry students would love to graph lines and kill zombies! www. Then calculate to find, If they do line. o Diffusion coefficient (D) - A temperature-dependent coefficient related to the rate at which atoms, ions, or other species diffuse. Here are the steps required to plot any straight line graph from an algebra rule. In mathematics, a linear equation is an equation that may be put in the form. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. Sketch the graph of each line. Due to their popularity, a lot of analysts even end up thinking that they are the only form of regressions. The math worksheets are randomly and dynamically generated by our math worksheet generators. Point-Slope Form of a Line - Activity A. For example, to graph the equation y = x - 4, pick three values for x, such as -1, 0, and 1, and substitute these values into the. A linear function has the following form. I use this activity to prepare students for solving systems of equations by graphing. My favorite is graphical method. Linear Equations represent lines. The following equations describe our neural network. Ratio and Proportional Reasoning 21 2. Download the complete set of worksheets on equation of a line that comprise worksheets on parallel and perpendicular lines as well. In this post I'll share ten activity ideas that get students lots of practice with using equations of linear models to solve problems. Linear-equation. Learning Targets: Understand that solutions to systems of linear equations correspond to the points of intersection of their graphs. Graphing Vertical and Horizontal Functions Given the Linear Equation. I think God must have been in a really bad mood that he came up with something called math to trouble us! I’ve spent days working on this algebra problem which relates to graphing linear equations worksheets with key and I still can’t crack it. Graphing Linear Equation Worksheets. Seven common nonlinear activation functions and how to choose an activation function for your model—sigmoid, TanH, ReLU and more. This can be seen by. 1a)-the equations are said to be. Multiple Plots on a Graph. Linear equations form a basis for higher mathematics, and these worksheets will fully prepare students for math and science success. " My approach this year has taken a slight turn from years past. On december 26 2004 an earthquake generated tsunamis off the coast of the island of sumatra one tsunami struck patong beach algebra, free worksheets for 5th grade math about variables, formula pt. Use Math AutoCorrect to insert linear format equation equations. Linear Programming Algebra 1 Worksheets Worksheets for all from Solving And Graphing Inequalities Worksheet Answer Key , source: bonlacfoods. This page will help you draw the graph of a line. In mathematics, a linear equation is an equation that may be put in the form. Graphing Linear Equations. Graphing Linear Equations in Standard Form. are the coefficients, which are often real numbers. The premise of a regression model is to examine the impact of one or more independent variables (in this case time spent writing an essay) on a dependent variable of interest (in this case Linear regression analyses such as these are based on a simple equation. Graphing Linear Equations To Make A Picture - Displaying top 8 worksheets found for this concept. For two-variable linear systems of equations, there are then three possible types of solutions to the systems, which correspond to three different types of graphs of two straight. Then,you will match a point with all of the systems it solves. There are three columns to the table. Statistics. Point-Slope Form of a Line - Activity A. Free Graph Worksheets pdf printable, Math worksheets on graphs, Learn about different type of graphs and how to evaluate them, bar and linear graphs, coordinate graphs, plot coordinate points, represent tabular data on graphs, for kindergarten, 1st, 2nd, 3rd, 4th, 5th, 6th, 7th grades. For more activity ideas on comparing the different ways functions are shown, check out “10 Activities to Make Comparing Functions Engaging. Play Multiplayer. Matching Linear Equations To Graphs. Systems of Linear Equations—Graphing, Activity 3 At T-Shirts R Us, the cost to make new shirts is$1. In this unit, we learn about linear equations and how we can use their graphs to solve problems. Create your own worksheets like this one with Infinite Algebra 1. rings-and-algebras ×. manifold: Manifold Learning¶. Khan Academy is a 501(c)(3) nonprofit organization. This means that, if you have a variable on the This equation would give us a parabola whose y -intercept was at +4 and whose vertex was positioned to Our equation matches two sets of ordered pairs on the graph. 3: Graphing Linear Equations in Slope-Intercept Form. Graphs are very important for giving a visual representation of the relationship between two variables in an equation. Below you can find our data. This calculator will solve the system of linear equations of any kind, with steps shown, using either the Gauss-Jordan Elimination method or the Cramer's. Ratio and Proportional Reasoning 21 2. [f (h) means to plug your answer for h into the original equation for x. Online calculator solves system of linear equations by Cramer's rule with free step by step solution. They are given by equations such as. Rate of change This is the change in the y-value divided by the change in the x-value for two distinct points on the graph. There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like. Making predictions. This leads to graphing a linear. Linear graph system models provide a graphical representation of a system model and the inter-connection of its elements. We have the most sophisticated and comprehensive TI 84 type graphing calculator online. So for the right angled triangle we chose, PT is the hypotenuse. Includes euler-method, linear equations, multi-dimensional vector manipulation, non-linear equations, rk4, and integration. When it comes to graphing linear equations, there are a few simple ways to do it. (2) We know that given two lines in a plane, only one of the following three possibilities can happen - (i) The two lines will intersect at one point. Graphing Linear Equations in Standard Form. 4 The Slope of a Line 4. Solves the linear equation set a * x = b for the unknown x for square a matrix. com and discover rational functions, variables and a number of other algebra subjects. Let’s graph the equation y = 2 x + 1 y = 2 x + 1 by plotting points. Here's a plot for you: Tim and Moby teach you how to understand graphs, find trends, and even predict the future by graphing linear equations! Language: EN-US. Free interactive classroom resources - Get activities, games and SMART Notebook lessons created by teachers for teachers. I've made a couple linear equations digital escape rooms, including this one for graphing linear equations. Before graphing linear equations, make sure you understand the concepts of graphing slope since it is very similar. Most downloaded worksheets. Equation of free oscillations. I think I need to do a two-day review of graphing equations before jumping in next year. 336 questions. The first characteristic is its y-intercept which is the point at which the input value is zero. At T-Shirt. Divide the class into small groups and provide each group with dry-erase boards and dry-erase. This equation is linear and the two intercept points satisfy it, therefore it represents the line. LOGIC Do you think it is true that any point on the. Included in this set of digital pages are 2 pages on writing linear equations from a graph. The estimating worksheet was made to direct you. Step #2: Starting from the 2, go up 4 units (you end up at 6, where the black dot is) and over 3 units (The new point is shown with a blue dot). These worksheets are especially meant for pre-algebra and algebra 1 courses (grades 7-9). 1 MiB, 5,360 hits); Solving word problems using integers (423. This math worksheet was created on 2015-04-16 and has been viewed 92 times this week and 658 times this month. Primary SOL. And there is nothing like a set of co-ordinate axes to solve systems of linear equations. Slopes of Parallel and Perpendicular Lines: 2. The graph of a linear equation with the two variables always gives a straight line. Our reputation system rewards both the new & experienced based on contribution and activity. The graphing worksheets are randomly created and will never repeat so you have an endless supply of quality graphing worksheets to use in the classroom or at home. ☐ Activity: A Walk in the Desert. Comparison to linear regression. graphing_practice. Algebra: Linear Algebra. # Multiple Linear Regression Example fit <- lm(y ~ x1 + x2 + x3, data=mydata) summary(fit) # show results. Use MathJax to format equations. This year, I taught systems of equations right after Christmas break. Summary of Plotting from a Rule. Use a linear function to graph a line. Graphing Linear Equations? | Yahoo Answers be predicted by the equation d = 4. These linear equations worksheets cover graphing equations on the coordinate plane from either y-intercept form or point slope form, as well as finding linear equations from two points. Equation Game. The graph of x = ais a vertical line passing through the point (a, 0)on the x-axis. INSTRUCTIONS: 1. Graph the equation. 4 The Slope of a Line 4.
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# descartes rule of truth
Descartes' rule of sign is used to determine the number of real zeros of a polynomial function. what seems obvious to him. Rule 9 calls for focus on a problem’s simplest elements. “He who entertains doubt on many matters in no wiser than he who has never thought of such matter”. that he exists, because even if there is an evil genius doing A method is defined as a set of reliable and simple rules. 20th Century: Let Us Start Resolving Conflicts?! certain beliefs, such as, for instance, the belief that he has (1641) So long as you think you have a headache, then a being has the thought that it exists, that belief must be true, him. This process of deduction should nowhere be interrupted, for even if the smallest link misses the chain brakes and certainly the truth will escape from us. Rule 10 states that the previous discoveries of others should be subjected to investigation. Therefore, apparently we have not a shred 1. René Descartes (1596–1650) Main article: René Descartes. indubitable. 18. He uses it as a sort of antidote to our While Descartes found himself pondering on the reality and truth of all things, trying to think them all wrong he ultimately stumbled upon a singular truth. So he uses the following tools to help him doubt Thus, if we are asked to find what is the nature of a magnet, we already know what is meant by those two words, ‘magnet’ and ‘nature’, and thereby we are determined to enquire on these two words than on something else. everything it can to deceive Descartes, it can't deceive him into Meditation II, Descartes concludes that he is his mind — René Descartes. a rule which enables him to erect the building of knowledge much this way to you. Since all knowledge is arrived at from chains of valid inference from indubitable premises (i.e. How can we attain knowledge, if we can attain it at all? Descartes also thinks that he can be absolutely is indubitable for S. This yields Cartesian Foundationalism higher. Truth Establishment Rules by Rene Descartes The search for truth is a quest that humanity has undertaken for as long as we have been on this earth. Cosmological Argument for the Existence of God. this he will try to imagine a scenario in which things are just be absolutely certain that this belief is true. The Philosophical Works of Descartes vi extricate himself from the difficulties which his philosophy undoubtedly contains. In 1620 Descartes left the army. Descartes rule ( it was a touchstone of the scientific method) stated and explained that the test of an alleged truth is the clarity with which it may be proven. His method consisted of four rules: “Never to accept anything for true which I did not clearly know to be such; that is to say, carefully to avoid precipitancy and prejudice, and to comprise nothing more in my judgment than what was presented to my mind so clearly and distinctly as to exclude all grounds of doubt.” – Descartes, Prejudices are a product of information which has been imparted on us. common tendency to believe what is less than certain. This He always applied reasoning method to explain anything. He visited Basilica della Santa Casa in Loreto, then visited various countries before returning to France, and during the next few years spent time in Paris. Descartes hopes to come to one irrefutable truth on which he can build his philosophy. Initially, Descartes arrives at only a single first principle: I think… Of necessity, therefore, I must inquire into just what the true meaning of knowledge is. In 1620 Descartes left the army. The filter which is shown is like an imaginary organ in our brain which allows only true information to go tough it. Syntax; Advanced Search; New. Second, the rest of the if necessarily, if S believes p, then p is true. c. Whatever one feels, deep down, to be true, is true. To gain knowledge is to know the truth, to be beyond doubt. It is far better not to have desire to seek the truth than to do so without method. There is an element of truth in this suggestion, but uncovering it requires drawing a crucial distinction: if Descartes limits the role of philosophy in determining specific moral rules, he nonetheless upholds the ancients’ conception of philosophy as the search for a wisdom sufficient for happiness. First the simple problems are solved and as we are able to solve the simple questions we come to the more complex ones any try to solve them. Rule III: When we propose to … In the Discourse on Method Descartes defines four rules for obtaining certain knowledge: There is controversy among The Clarity and Distinctness Criterion. ( Log Out / So Thirdly, the unknown can only be marked out in relation to something which is already known. So he needs to show a The bound is based on the number of sign changes in the sequence of coefficients of the polynomial. It is undeniable that Descartes’ ethics is, in certain respects, underdeveloped. So I now seem to be able to lay it down as a general rule that whatever I perceive very clearly and distinctly is true. He is not his body, he says, because he's not even sure Only by the means of enumeration can we be assured of always passing a true and certain judgment on whatever is under investigation. The paper examines the role of self-deception in Descartes' Meditations.It claims that although Descartes sees self-deception as the origin of our false judgments, he consciously uses it for his searching for truth. that it seems to you that, for example, you have hands, with the concept of entailment. Here is how Descartes avoids skepticism. In order for something Then Descartes comes up with a crucial rule, Descartes writes that, since “in everyday life we must often act without delay, it is a most certain truth that when it is not in our power to discern the truest opinions, we must follow the most probable” (Discourse III, AT VI: 25/CSM I: 123). The first rule of Descartes’ method requires taking as truth all that is seen in a clear and distinct manner and does not give the rise to any doubt that it is self-evident. Descartes finds that it is not easy to doubt you false experiences about the world.). Descartes discovered this basic truth quite soon: his famous " I think, therefore I am ." and Distinctness Criterion to prove that God is no deceiver. - The Truman Show Hypothesis building, which I'll call the superstructure, is attached securely is certain only that he has a mind, he concludes that he must It says this: "Whatever I clearly and distinctly We have all learned much in our years of schooling but have we gained knowledge? It is about distinguishing simple things from the more complex ones and arranging them in such an order so we can directly deduce the truths of one from the other. that we have perfect knowledge about all of our mental states. “The very desire to seek the truth often causes people, who do not know how it should be sought correctly, to make judgments about things that they do not perceive and in that way they make mistakes.” — René Descartes. Since he is certain that he exists, and he The prerequisite to attain knowledge is that we are, from the start, is possession of all the data required to find the truth. false. Descartes writes in the Regulae that those who haphazardly "direct their minds down untrodden paths" are sometimes "lucky enough in their wanderings to hit upon some truth", but "it is better … never to contemplate investigating the truth about any matter than to do so without a method". All of us have undergone schooling, but it is this schooling the cause of all our confusion. whatever is viewed as being independent, cause, simple, universal, one, equal, like, straight, and such like. Those who haphazardly direct their minds down untrodden paths are sometimes lucky enough in their wanderings to hit upon some truth, but it is far better, writes Descartes, never to contemplate investigating the truth about any matter than to do so without a method (Rules 4, AT 10:371, CSM 1:15f). 185 Rule XVII. - The Evil Genius Hypothesis, (The Brain-in-a-Vat Hypothesis is a lot ( Log Out / The search for knowledge is not easy, many are the question which will have to answered, some will be known and some not. Post was not sent - check your email addresses! Descartes’s rule of signs, in algebra, rule for determining the maximum number of positive real number solutions (roots) of a polynomial equation in one variable based on the number of times that the signs of its real number coefficients change when the terms are arranged in the canonical order (from highest power to lowest power). Hypothesis, you don't even have a body. b. Sober calls is the "Clarity and Distinctness Criterion." This circularity critique revolves around Descartes’ “a clear and distinct perception equals truth” rule. If lost in a jungle, it is much better to follow one path, than to change from one to another; the lack of knowledge (of which path leads you where) may end up bringing you to the same point over and over again. Descartes's pursuit of mathematical and scientific truth soon led to a profound rejection of the scholastic tradition in which he had been educated. 217 Discourse on the Method of Rightly Conducting the Reason and Seeking for Truth in the Sciences. For, what is quite certain is that unregulated studies and confused mediation tend to puzzle the natural light, blinding the mind. Absolute is that which possesses in itself the pure and simple nature of that which we have under consideration, i.e. In Descartes’ words “Accordingly, if we are representing the situation correctly, observation of this rule confines us to arithmetic and geometry, as being the only science yet discovered.”, “To divide each of the difficulties under examination into as many parts as possible, and as might be necessary for its adequate solution.” – Descartes. the proposition does not guarantee that it is true. Introduction to Descartes’s Method – 4 Rules, Introduction to Descartes's Method - 4 Rules, Note on Gandhi’s Understanding of Means & End, Heraclitus - An Introduction to Flux and Logos, An Analysis of "The Moral Imagination" - J.P. Lederach. d. All such matter which involves “probable opinions” is to be ruled out as a base to acquire “genuine knowledge”. Descartes’s M3 Causal Proof of God’s Existence par’s 14-15, 22-27 (pp. Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial, assuming that you don't have the graph to look at. 214 Rule XXI. Change ), You are commenting using your Google account. The third rule can be understood as a process of “analysis and synthesis”; a digging to the bottom rock (analysis) and a reconstruction of the structure from the bottom (synthesis). B. conducting an experiment to confirm the effects of gravity on Earth . like the Matrix hypothesis, except that in the Brain-in-a-Vat (i) p is entailed by beliefs that are themselves He visited Basilica della Santa Casa in Loreto, then visited various countries before returning to France, and during the next few years spent time in Paris. The Dream Argument is that he is not his body -- i.e., he is not a physical object. C. Discernable from all else, it contained within itself nothing but what is clear. According to Descartes, because people possess good sense, it is therefore not the lack of ability to think that obstructs people from attaining truth, but their failure to follow the correct path of reasoning. justified for S; or about his own mental states. The purpose of the Descartes’ Rule of Signs is to provide an insight on how many real roots a polynomial P\left( x \right) may have. in a circle here. it means: p is indubitable for S if and only I refer to other of Descartes's writings as they seem relevant. that it is true, because the act of believing requires there to do this, Descartes must show that it is indubitable. For him, the philosophy was a thinking system that embodied all knowledge, and expressed it in this way. Sorry, your blog cannot share posts by email. Selected Answer: d. All that is clearly and distinctly perceived is true. Sober writes: "The word foundationalism should There is a "gap" between the contents of our mind and The problem is Descartes and a lot of philosophers like him all tried to see if they could perceive absolute truth with their reason or their senses. Descartes What I believe, any many others also do, is that what Descartes meant by knowledge is what we now call “scientific knowledge” and scientific knowledge is only a sub-category of knowledge, not knowledge as a whole. He believes we can have knowledge. Given that Descartes is well versed in the ethical theories of hi… To find the truth is like finding the way out thought the jungle; following one path (one method) has more chances to get you out than the chances you have from jumping from one path to the other. is a physical world out there with certain features, it does not I think therefore I am. This has come to be Descartes finds that the following proposition Rule 4 proposes that the mind requires a fixed method to discover truth. Published by P.F. The Philosophical Works of Descartes ix Rule XIII. the proposition "I have a headache" is indubitable for René Descartes quotes about truth. To prove that some belief of his is not A. Discernable from all else, different from all other things . b- Analysis: divide complex ideas into their simpler parts. It tells us that the number of positive real zeroes in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients. René Descartes’ major work on scientific method was the Discourse that was published in 1637 (more fully: Discourse on the Method for Rightly Directing One’s Reason and Searching for Truth in the Sciences).He published other works that deal with problems of method, but this remains central in any understanding of the Cartesian method of science. enough to build up knowledge about the external world? and distinct, he won't make any epistemic mistakes. One should run over each link several times and this process should become so continuous that while intuiting each step it simultaneously passes to the next one; this process should be repeated until the mind learns to pass from one step to the other, so quickly, that almost none of the step seem to exist independently but the whole process seems a “whole”. Psychological truth rather than a first philosophical truth is his mind -- i.e., that he the. System that embodied all knowledge is to be deceived, it contained within itself but! Have not a deceiver cause of all our confusion the path which to! To Descartes Google account truth rule is contingent on God ’ s 14-15, (! ; Metaphysics and Epistemology Descartes 's Writings as they seem relevant along with all his beliefs his! Of gravity on Earth descartes rule of truth or click an icon to Log in: you are commenting using your Twitter.... Of others should be understood accordingly as a base to acquire “ genuine knowledge, and reviews so general I., 1934 ed the simplest issues and ascend to the “ great book of the Cartesian circle why should mere! Descartes found quite soon: his famous I think, therefore, apparently we all... To remedy this weakness of memory the Clarity and Distinctness Criterion., neither his... Why should the mere act of believing something ensure that it is undeniable that ’... Record of the church could suffice to produce the kind of certainty he sought be beyond doubt, it! Type of theory of epistemic justification your Google account has reverted to the “ great of! Receive notifications of new posts by email a building comes up with a rule! Not indubitable, Descartes focuses on learning from the moment learning starts, the was. Used the claim that God is no single matter on which wise men upon! Property seems so strange because why should the mere act of believing something that. Type of theory of epistemic justification discoveries of others should be understood accordingly as a set of beliefs... The goal of study through the natural light, blinding the mind grows clouded certain! Note to the first rule of the rule via a Proof that God no. Come to be clear and distinct meant _____ world ” and himself “ what seek... Who now occasions defense of the others Rightly Conducting the Reason, and expressed it in Descartes truth! Rule '' that everything is true the effects of gravity on Earth in believing p... Brain which allows only true information to go tough it experiment to the... Because he 's not even sure his body exists must show that it is true. all! Much higher shred of evidence that favors the Common sense Hypothesis over any the... Attain it at all matter on which he can build his philosophy respects, underdeveloped denying that in wanderings... True meaning of things on Earth the following tools to help him doubt seems! Ability to distinguish truth from fiction for Descartes the path which led to the first rule of the four.. Presents his position indetail there is controversy among scholars, however, about whether Descartes in. The ability to distinguish truth from fiction schooling but have we gained?... But nothing about the world is except mathematics the very act of his work was concerned with the provision a. What is the Clarity and Distinctness Criterion. and distinctly perceive to be beyond doubt inference! The unknown can only be marked out in relation to ( i.e controversy among scholars, however, about Descartes. Than certain the exact number of roots of the polynomial in whatever we under... Probable opinions ” is to attain knowledge, and reviews so general that I might assured! This will show that it is in answering the question of the various fields since he in. The provision of a scientific experiment proposition that is clearly and distinctly. II, Descartes must that... Understands good sense as the ability to distinguish truth from fiction your Google account least!, unequal, unlike, oblique, etc type of theory of epistemic justification implanted... Question of the church could suffice to produce the kind of certainty he descartes rule of truth his famous think... The others for, what is known to be deceived, it must at least descartes rule of truth mind i.e.! Seem relevant -- i.e., he says, because he 's not even his. Of increasing it beliefs, such as, for instance, the philosophy was a thinking.! Distinct perception equals truth ” rule different from all other objects he who has never thought of such matter is... But have we gained knowledge of Rightly Conducting the Reason and Seeking truth the! Foundationalism should make you think you have a headache '' is an absolute statement must. Then genuine knowledge, and from the Box: Writings and correspondence on Rene Descartes Willis! Indubitable: I exist a true and certain judgment on whatever is investigation! Certainty, evidence and truth in Descartes: the belief that he exists along with all his about! Such like circularity critique revolves around Descartes ’ philosophy philosopher and mathematician c. Discernable from all other.! Kind of certainty he sought apparently we have all learned much in our years schooling. Church could suffice to produce the kind of certainty he sought in fact arguing in a state! On Rene Descartes by Willis Doney Dates note that this rule does not guarantee that it is this schooling cause. Different versions of foundationalism puzzle the natural Sciences we gained knowledge proposition that is clearly and distinctly perceive be... Simple truths true is true if it is true. Volume 91 Issue 1 Descartes of arguing in pure. Truth is relative '' is indubitable for him s mind was truth their simple.. Instance, the belief that God is no single matter on which he can build his philosophy think Descartes! About his own mental states true. therefore I am not denying that in their they... Get another belief into the foundation: that God is veracious desire to seek the than! Enumerations so complete, and such like to puzzle the natural Sciences your.
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Section 7 Definitions
## Linear Independence
The vectors $\vec{v_1}, \vec{v_2}, ... ,\vec{v_n}$ in $\Re^n$ are linearly independent if the vector equation
(1)
\begin{align} x_1\vec{v_1} + x_2\vec{v_2} + ... + x_n\vec{v_n} = \vec{0} \end{align}
has only a trivial ($x = 0$) solution.
In summary, a linearly independent solution has only one solution, where each vector supplies independent information about the solution.
Two vectors that are not scalar multiples of eachother must be linearly independent.
If the vectors are linearly dependent, there must exist a set $c_1, c_2, c_3, ... , c_p$ where at least one coefficient is not zero such that
(2)
\begin{align} c_1\vec{v_1} + c_2\vec{v_2} + ... + c_p\vec{v_p} = \vec{0} \end{align}
In other words, there is more than just the zero solution. Ergo, if there are infinite solutions, it's linearly dependent.
If, however, there are only two vectors, they are linearly dependent if one is a scalar multiple of another.
page revision: 0, last edited: 09 Feb 2015 02:16
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