text:
Chinese

2021,  3 (1):   55 - 64   Published Date：2021-2-20

DOI: 10.1016/j.vrih.2020.11.005

Abstract

Background
Continuous emotion recognition as a function of time assigns emotional values to every frame in a sequence. Incorporating long-term temporal context information is essential for continuous emotion recognition tasks.
Methods
For this purpose, we employ a window of feature frames in place of a single frame as inputs to strengthen the temporal modeling at the feature level. The ideas of frame skipping and temporal pooling are utilized to alleviate the resulting redundancy. At the model level, we leverage the skip recurrent neural network to model the long-term temporal variability by skipping trivial information for continuous emotion recognition.
Results
The experimental results using the AVEC 2017 database demonstrate that our proposed methods are beneficial to a performance improvement. Further, the skip long short-term memory (LSTM) model can focus on the critical emotional state when training the models, thereby achieving a better performance than the LSTM model and other methods.

Content

1 Introduction
Emotion recognition has become an increasingly important research topic owing to its essential role in human communication[1]. A person's emotional state can be described using a dimension emotion model, which uses numerical values to indicate the type and degree of emotions continuously[2], such as arousal (calm versus active) and valence (negative versus positive). The dimension emotion model has the advantages of representing multiple subtle and complex emotional states.
Prior studies have investigated various discriminative emotional features and effective emotion recognition methods[3,4]. However, an emotional state is a gradual and smooth process with temporal changes. Therefore, it is essential to incorporate long-term temporal context information when making a dense frame-level prediction for continuous emotion recognition. Two types of approaches exploiting long-term temporal contexts use feature-level representations and can learn the long-term dependencies based on short-term feature representations.
Many emotional feature-level representations have been designed to contain temporal context information. Valstar et al. showed that it was necessary to consider a larger window to obtain segment-level features, e.g., 4s for arousal and 6s for valence[5]. Le et al. and Cardinal et al. found that increasing the number of contextual frames with deep neural networks can improve the performance[6,7]. In addition, Povolny et al. applied simple frame stacking and temporal content summarization to consider contextual information[8]. Moreover, Huang et al. utilized temporal pooling subsampling to obtain wider contexts and achieve emotional temporal modeling[3] and Han et al. extracted segment-level BoAW(bag-of-audio-words) features to encapsulate the contexts with neighboring frames[9]. In speech recognition, incorporating long-term temporal information by splicing the input sequences is a common practice to improve the performance[10,11]. In addition, Zeyer et al. used max-pooling in the time dimension to downscale the original inputs for speech recognition[12].
The above methods are operated at the feature level to obtain temporal context information. However, these methods still depend on the temporal modeling ability of the emotion recognition model. Researchers have explored many models to learn emotional dynamic information from sequence signals. For example, Wöllmer et al. first employed a long short-term memory (LSTM) network to achieve a continuous emotion recognition[13] and Khorram et al. investigated two convolutional network architectures, dilated convolutional networks and downsampling–upsampling networks, to learn emotional long-term temporal dependencies[14]. In addition, Zhang et al. utilized a recurrent neural network (RNN) to learn spatial and temporal dependencies for emotion recognition[15], whereas Li et al. proposed a convolution neural network (CNN) with two different groups of filters to achieve a similar purpose[16]. Finally, Trigeorgis et al. trained a convolutional recurrent network using time-domain signals directly[17].
Most results have indicated that the performance of an RNN is superior to that of a CNN in terms of continuous emotion recognition. At the model level, LSTM-RNNs have become common deep neural network structures for continuous emotion recognition because they naturally learn emotional dynamic information between emotional features. However, the training of basic RNNs on long sequences often faces numerous challenges such as a slow inference, vanishing gradients, and difficulty in capturing long-term dependencies. Simple solutions are to reduce the length of continuous input signals, such as segmenting the training samples[18] and frame skipping[19] to shorten the length of the inputs, which are based on heuristics and can be suboptimal. In addition, these approaches are based on the feature level.
Fortunately, researchers have proposed many improved RNNs for enhancing the ability to model long sequences. A more recent thought is to skip the input information at certain steps, such as LSTM-Jump[20], skim RNN[21], and skip RNN[22]. Each of these methods can skip unimportant parts to reduce the length of the processing of the input signals. In particular, a skip RNN can select important parts of the input signals adaptively by determining whether the state needs to be updated or copied from the previous time step. This thought is similar to the attention mechanism that accommodates active selection from the most emotionally salient frames[23]. In this study, we take advantage of a skip RNN with emotional long-term feature representations to learn temporal contexts of continuous emotion recognition. We focus on dimension emotion recognition; in addition, for the emotion recognition model, the inputs are audio or video features, and the outputs are arousal and valence.
In the following, Section 2 briefly introduces the proposed methods. Section 3 presents the database and feature sets. Section 4 describes the experimental results and analysis. Finally, Section 5 provides some concluding remarks regarding this research.
2 Methods
In this study, the models learn long-term temporal context information to improve the performance of continuous emotion recognition. At the feature level, we preprocess the features with a large window of consecutive frames to cover the temporal contexts. Meanwhile, the ideas of frame skipping and temporal pooling release the impact of redundant information. In addition, a skip RNN is utilized to focus on more valuable information for modeling long sequences of continuous emotion recognition.
2.1 Data processing
We leverage four simple data processing methods to incorporate temporal contexts at the feature level. A straightforward method is to take a window of feature frames from the past and the future. Specifically, consecutive emotional features within the window are concatenated as the inputs, shown in Figure 1a and Figure 1b. The difference is that the features are stacked with overlapping frames in Figure 1a, as represented by "Context1," whereas those in Figure 1b are based on frame skipping without overlapping frames, as represented by "Context2." However, these methods will result in some redundancy of the emotional features. Thus, we introduce temporal pooling to average the features instead of applying a concatenation within the window in Figure 1c, as represented by "Context3." The operation of "Context1" followed by "Context3" is shown in Figure 1d and is represented by "Context4." Note that we average the labels within the same window to match the shortened length of the emotional features for "Context2," "Context3," and "Context4" when training their models. The window length is regarded as a parameter to be optimized.
2.2 Skip RNN
We introduce a skip RNN model to conduct a continuous emotion recognition. A skip RNN, a variant of an RNN, conducts fewer updating time steps to decrease the number of sequential operations and provide faster convergence in involving long sequences. As a result, trivial information can be skipped and a superior ability to model long-term dependencies for continuous emotion recognition is achieved.
A standard RNN takes an input sequence
and generates a state sequence
by iteratively applying a parametric state transition model
$S$
from
to
$T$
:
A skip RNN has a binary state update gate,
$u t ∈ 0,1$
, selecting whether the state of the RNN will be updated (
$u t = 1$
) or copied from the previous time step (
$u t = 0$
). At every time step
$t$
, the probability
of applying a state update at
$t + 1$
is emitted. The resulting architecture is depicted in Figure 2a and can be characterized as follows:
$u t = f b i n a r i z e u ˜ t$
$Δ u ˜ t = σ W p s t + b p$
$u ˜ t + 1 = u t ∙ Δ u ˜ t + 1 - u t ∙ u ˜ t + m i n Δ u ˜ t , 1 - u ˜ t$
where
$W p$
is a weight vector,
$b p$
is a scalar bias, σ is the sigmoid function, and
$f b i n a r i z e : 0,1$
binarizes the input value. In addition,
$f b i n a r i z e$
is a deterministic step function,
$u t = r o u n d u ˜ t$
, and its gradient is set to
$∂ f b i n a r i z e / ∂ x = 1$
during a backward pass[22]. Thus, the model can be trained to minimize the target loss function with standard backpropagation without defining any additional parameters.
Whenever a state update is omitted, the pre-activation of the state update gate for the following time step
$u ˜ t + 1$
is incremented by
$Δ u ˜ t$
. When a state update is conducted,
$u t = 1$
, as shown in Figure 2b, the accumulated value is flushed and
$u ˜ t + 1 = Δ u ˜ t$
. When the model skips the state update,
$u t = 0$
, as shown in Figure 2c,
$s t = s t - 1$
, which enables more efficient implementations in which no computations are conducted at all. Thus, redundant computations are avoided because
$Δ u ˜ t = σ W p s t + b p = σ W p s t - 1 + b p = Δ u ˜ t - 1$
, and there is no need to evaluate it again, as shown in Figure 2(d). This enables the skip RNN to shield the hidden state over a longer period of time.
3 Database and feature sets
3.1 Database
We use the Audio/Visual Emotion Challenge and Workshop (AVEC 2017) database to show the benefits of our proposed methods[24]. This database collects spontaneous and naturalistic human-human interactions in the wild, consisting of audio, visual, and textual modalities. The recordings are annotated time-continuously in terms of the emotional dimensions, including arousal for the emotion activation and valence for the emotion positiveness. All emotional dimensions are annotated every 100ms and scaled into [-1, +1]. There are 64 German subjects in the dataset, which is divided into a training set with 34 subjects, a development set with 14 subjects, and a testing set with 16 subjects. We focused on an estimation of the arousal and valence in this study.
3.2 Feature sets
In this study, we utilized audio and visual modalities for continuous emotion recognition. The extended Geneva minimalistic acoustic parameter set (eGeMAPS)[25] is regarded as the audio feature. The window segment-level acoustic features are computed over a 4-s period, resulting in 88 dimensional features. The extraction of the LLDs and the computations of the functionals are conducted using openSMILE[26]. The geometric features are regarded as visual features, including facial landmark locations, facial action units, head pose features, and eye gaze features extracted by OpenFace[27]. We then apply a principal component analysis to reduce the dimensionality and maintain 90% variance, resulting in 55 dimensions.
4 Experiment and analysis
4.1 Experimental setup
We utilized an LSTM as a basic continuous emotion recognition model. The number of hidden layers is 1 with 64 hidden nodes for the LSTM and skip LSTM models. We use the Adam optimization algorithm[28] with a dropout rate of 0.5 and a batch size of 3. The maximum number of training epochs is 100. The evaluation measure is the concordance correlation coefficient (CCC)[29]. Owing to a lack of availability of the testing set, we utilized the training set to train the models and the development set to evaluate the performance.
4.2 Effect of data processing
To incorporate temporal contexts, we use several novel methods to process the input features at the feature level. In this section, we describe the experiments conducted using a basic LSTM model.
First, we explore the effect of the window length for these methods. Figure 3 shows the experimental results of the audio features in terms of arousal and valence against the length of the window, which denotes the number of frames. The performance is improved with the increase in window length, showing that a longer window is necessary to model the emotional temporal context information. In particular, Context3 requires a longer window compared with the other methods because it compresses the features to obtain extremely long span contexts. The results indicate that the window length of the valence is larger than that the arousal; thus, valence accords longer temporal contexts, which is same as in [5]. The conclusions drawn from the visual features are similar to those from the audio features, except that visual features require longer windows, which are not displayed in this study owing to limited space.
The performance of different data processing methods is shown in Table 1, and the corresponding window lengths are from the optimal results of Figure 3. The Original models were trained without data processing. The results indicate that the performance of Context1 is worse than that of the Original models. We speculate that the LSTM is able to store past information internally; thus, the concatenation of consecutive frames increases the information redundancy, which results in a degraded performance. To address this problem, we borrow the ideas of frame skipping with concatenation for Context2 and mean temporal pooling for Context3, which achieved a better performance than Context1 and the Original models. This indicates that it is important to alleviate the redundancy of the emotional features, which effectively improves the performance. In addition, the ideas of frame skipping and temporal pooling shorten the input signals, which makes LSTM more accessible to longer sequence information.
Performance of different methods with LSTM model using audio and visual features in terms of arousal and valence
Arousal Valence
Audio Visual Audio Visual
Original 0.473 0.589 0.423 0.650
Context1 0.431 0.571 0.431 0.655
Context2 0.484 0.584 0.469 0.695
Context3 0.512 0.602 0.455 0.736
Context4 0.464 0.611 0.436 0.701
The optimal performance of arousal and valence are both from the visual features, specifically 0.611 for Context4 under arousal and 0.736 for Context3 under valence. The similarity between Context3 and Context4 is the operation of temporal pooling to compress the features. Thus, temporal pooling is essential to an emotion recognition model based on visual features. When using the audio features, the optimal performance is 0.512 for Context3 under arousal and 0.469 for Context2 under valence. The similarity between Context3 and Context2 is the operation of frame skipping to cover wider temporal contexts. Thus, frame skipping is more effective for emotion recognition models based on audio features. The results reveal the effectiveness of our proposed data processing methods to incorporate long-term context information, which is beneficial to a performance improvement. In addition, the performance of valence is better than that of arousal, which is similar to previous studies[3,4].
4.3 Continuous emotion recognition with skip RNN
It is important to consider long-term temporal dependencies to improve the performance of continuous emotion recognition. Based on the outstanding temporal modeling capability of the LSTM model, we further investigate the skip LSTM model to capture as much longer temporal information as possible by performing fewer state updates.
The experimental results are presented in Table 2. Context2 and Context3 achieve a better performance than other methods when using the skip LSTM model. Context2 obtains a better performance of 0.622 in terms of arousal with the visual features and 0.447 in terms of valence with the audio features. Context3 obtains a better performance of 0.531 in terms of arousal with the audio features and 0.698 in terms of valence with the visual features. This shows that frame skipping plays an important role in the emotion recognition model based on the skip LSTM model.
Performance of different methods with skip LSTM model using audio and visual features in terms of arousal and valence
Arousal Valence
Audio Visual Audio Visual
Original 0.487 0.483 0.420 0.634
Context1 0.473 0.472 0.415 0.641
Context2 0.465 0.622 0.447 0.690
Context3 0.531 0.603 0.427 0.698
Context4 0.492 0.562 0.441 0.666
Comparing Table 2 with Table 1, we can observe that the skip LSTM model achieves a better performance in arousal with the audio features of 0.531 versus 0.512 and visual features of 0.622 versus 0.611, whereas the LSTM model achieves a better performance in terms of valence with the audio features of 0.469 versus 0.447 and visual features of 0.736 versus 0.698. The skip LSTM does not show advantages over LSTM in all cases. This is perhaps because the skip LSTM model can access a longer temporal context in Context2 and Context3, which is beneficial to the emotion recognition model for arousal. This may be limited by the insufficiency of the training data. In total, incorporating long-term temporal context information on the feature level and model level together can improve the performance.
4.4 Analysis
As the true benefit of skip LSTM, fewer state updates focus on more valuable information for modeling longer sequences. We calculate the state update rates based on the feature frames when training the skip LSTM models. Table 3 compares the average state update rates with the standard deviation over five trials for different methods as a comparison. It is obvious that the Original models and Context1 have low update state rates, which implies that many sequence frames of the input signals are trivial and can be skipped. Context2 still has some redundancy after frame skipping with concatenation. Context3 and Context4 have high state update rates because temporal pooling has relieved the redundancy to a large extent. The visual features have more useless information than the audio features because of their lower update state rates. The results provide evidence that skip LSTM can effectively alleviate redundant information that exists in continuous emotion recognition tasks.
State update rates (mean±std %) of skip LSTM model using audio and visual features in terms of arousal and valence
Arousal Valence
Audio Visual Audio Visual
Original 59.17±8.89 52.91±5.90 84.43±1.56 59.09±0.92
Context1 65.00±2.74 50.02±0.73 59.02±3.78 45.27±3.53
Context2 89.02±1.93 69.06±1.91 88.39±3.10 81.56±3.03
Context3 93.01±1.93 91.54±2.04 95.29±1.66 89.03±1.07
Context4 99.29±1.83 76.92±3.95 99.99±0.01 82.55±0.94
We take an example of visualization with the Original models in terms of arousal, as shown in Figure 4. The blue line is the ground truth, the black line indicates the predictions, and the red dots denote the updated frames. Figure 4 illustrates that the updated frames concentrate on critical fluctuant and low-value areas, which helps to model complicated and subtle emotional states well. We compared our results with the baseline results[24] and the winner[4] of AVEC 2017. Ringeval et al. utilized a static SVM model as an emotional model[24]. Our models can learn long-term temporal dependencies; thus, the performance significantly outperforms the baseline results. Our proposed methods achieve a better performance than Chen's approach for a single modality[4], which utilized a basic LSTM model for emotion recognition. The skip LSTM model focuses on the modeling of crucial emotional states based on context information, which significantly promotes the ability of continuous emotional temporal modeling compared with the LSTM model.
CCC comparison between our methods and other approaches.
Arousal Valence
Audio Visual Audio Visual
Ringeval et al.[24] 0.344 0.466 0.351 0.400
Chen et al.[4] 0.526 0.623 0.447 0.708
Our methods 0.531 0.622 0.469 0.736
5 Conclusion
In this study, techniques for incorporating long-term temporal context information for continuous emotion recognition were considered. We widened a larger view of the feature frames with frame skipping and temporal pooling on the feature level and employed a skip RNN on the model level. The experimental results reveal that a longer window is necessary to model the temporal emotional contexts. It is also important to alleviate the redundancy of the emotional features. We illustrated the benefit of skip LSTM in the modeling of long-term temporal dependencies, which skips the trivial information to focus on critical emotional values. Therefore, the skip LSTM can model subtle and complicated emotional states based on the context information for continuous emotion recognition. Our methods improve the performance of the LSTM model and achieve a better performance than the other approaches. These promising results suggest that skip LSTM can lead to important improvements in continuous emotion recognition. In the future, we will further improve the performance of skip LSTM and explore multimodal emotion fusion of temporal modeling.

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text:
# Reference

The reference is the exhaustive description of the whole library's API, with a description of every module. All you need to know in order to make your project is here.

Overview

Carbide

Core, clocks and power management

Input/output

Communication ports

Timer/counters

Miscellanous

Utils modules

Modules below are not based on a peripheral in the microcontroller but are instead common-use classes that can be used to speed up development. They are compiled using the UTILS_MODULES option in the Makefile.

# Under the hood

This is a more advanced read for those who want to know how the library works and how to customize it. It is by no means necessary to make something great with libtungsten.

format: plain


text:
## anonymous 4 years ago HElp me Out? 1. What is the degree measure of angle A? __________ degrees http://www.flickr.com/photos/63839927 @N04/8219196862/in/photostream/lightbox/ 2. What did you use to find the measure of angle A? (1 point) A. 60 + x + A = 180 B. 180 – 150 = A C. Interior angles are equal D. Vertical angles are equal Have to copy and paste the whole link.

1. anonymous

i don't know...I can upload it somewhere else.

2. anonymous

Thanks.

3. anonymous

A should equal 60, intersection and all that

4. anonymous

|dw:1353897846442:dw|

5. anonymous

So your saying it equals 60 and number 2 is ???

6. anonymous

I'd go with the interior angle answer,...if I remember correctly, it means the same thing as saying " opposit angles are equal"

7. anonymous

that;s not right saw my error

8. anonymous

vertical angles are equal

9. anonymous

That's true^

10. anonymous

meaning that A= 60 because it is an angle of intersecting lines, aka A=60 because they are vertical angles

11. anonymous

Ohh..so did you just solve those equations then subtracted it?? I'm trying to understand I'm just not good with Geometry.

12. anonymous

What I did was make an inference based off of vertical angle theorem. If there are two lines that intersect, then they will create four angles. Each angle has an opposing angle, vertical from its perspective. The two opposite or vertical angles are equal

13. anonymous

Jazy seems to have a more current handle on geometry than me, so correct me if any of that is innacurate

14. anonymous

Vertical angles are equal. ex: |dw:1353898300187:dw|

15. anonymous

Nope. You're completely right @karama (:

16. anonymous

^^^ I understand that with the pic..thanks guys. :))

17. anonymous

you're welcome...let us know if you have any other questions! :)

18. anonymous

Sure thing.

19. anonymous

Can I have help with finding the degree of x? and Which of the following can you use to find the degree measure of angle x? 180 – (60 + A) = x 150 – 60 = x x = 60 + A x = 150 – 50 180 – 90 = x These are my last questions that I need help with.

20. anonymous

The same triangle?

21. anonymous

Yes.

22. anonymous

let's see, it might be 90...because y=30, since 180-150=30...then 60+30=90...triangles must have angles that sum to 180...180-90=90

23. anonymous

ohhh..Okay I understand that.

24. anonymous

$180=\alpha+\beta +\gamma$ each symbol represents an interior (meaning inside of) angle of a triangle

25. anonymous

|dw:1353899048358:dw| as well, in case you need any reference for how I got the value of Y

26. anonymous

Yeah.. I'm understanding it..Thanks again.


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Rogers, J. Nanoribbons on the edge. Nature Nanotech 5, 698–699 (2010). https://doi.org/10.1038/nnano.2010.200


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## College Algebra (6th Edition)

Fill the blanks with $-6$ and $(x+6)(2x^{2}-x-1)=0$
When dividing a polynomial with (x-c) using synthetic division, the last value of the last row is $f(c)$. We have a polynomial equation $f(x)=0$. from the synthetic division we see that $f(-6)=0$. So, $-6$ is a zero of f, which means it is a root of the equation. Since $-6$ is a zero of f, $(x-(-6))=(x+6)$ is a factor of $f(x)$ The first three values of the last row give coefficients of the quotient $f(x)\div(x+6)$ So we can write $f(x)=(x+6)(2x^{2}-x-1)$, and the equation as $(x+6)(2x^{2}-x-1)=0$


text:
## ranyai12 3 years ago Use Laplace transforms to solve the initial value problem x''+4x'+8x=2e^(−t) x(0)=0 x'(0)=4

1. ranyai12

@richyw

2. richyw

are you allowed to use a table?

3. ranyai12

this is my work and im not sure what i did wrong

4. ranyai12

and yes i am

5. richyw

i'm already confused lol. where did you get the 18 from?

6. ranyai12

ooops thats supposed to be an 8

7. richyw

yeah if you put an 8 it looks pretty straightforward as you clearly know what to do. I never finished it though as I am cramming for calculus haha!

8. ranyai12

id you get x=(4s+6)/(s=1)(s^2+4a+8)

9. ranyai12

that part always confuses me so if you can please check that id be fine

10. ranyai12

(s-1)*

11. richyw

yes I got there that's where I quit. I'm thinking about it again now

12. ranyai12

ok thanks

13. richyw

hmm after my partial fractions I'm getting one that is hard to fix into a nice way to inverse laplace

14. ranyai12

what did you end up getting?

15. richyw

I can't figure out the answer. I really do apologize but I have a big final coming up and rushed through it. I don't mean to leave you hanging!

16. ranyai12

its ok thanks i just wanted what x= to anyway not the whole thing but I totally understand!!! Finals suck!

17. ranyai12

@jagan

18. phi

x'' + 4x' +8x= 2*exp(-t), x(0)=0, x'(0)= 4 $(s^2+4s+8)L[x] -(s+4)*0 - (1)*4 = \frac{2}{s+1}$ $(s^2+4s+8)L[x] = \frac{2}{s+1} +4$ $L[x]= \frac{4s+6}{(s+1)(s^2+4s+8)}$ partial fraction expansion $L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s-7}{s^2+4s+8}$ complete the square on the denominator of the 2nd term to match the laplace tables rewrite the numerator $L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s+2-9}{(s+2)^2+2^2}$ re-write the last term as 2 fractions that match the Laplace tables. $L[x]= \frac{2}{5}\frac{1}{(s+1)} -\frac{2}{5}\frac{s+2}{(s+2)^2+2^2}+\frac{9}{5}\frac{2}{(s+2)^2+2^2}$ $x= \frac{2}{5}e^{-t}- \frac{2}{5}e^{-2t}cos(2t)+\frac{9}{5}e^{-2t}sin(2t)$ see http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

19. phi

I just gave the highlights, as this takes quite a bit of work. Ask if you have any questions.


text:
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# Thickness-independent capacitance of vertically aligned liquid-crystalline MXenes

## Abstract

The scalable and sustainable manufacture of thick electrode films with high energy and power densities is critical for the large-scale storage of electrochemical energy for application in transportation and stationary electric grids. Two-dimensional nanomaterials have become the predominant choice of electrode material in the pursuit of high energy and power densities owing to their large surface-area-to-volume ratios and lack of solid-state diffusion1,2. However, traditional electrode fabrication methods often lead to restacking of two-dimensional nanomaterials, which limits ion transport in thick films and results in systems in which the electrochemical performance is highly dependent on the thickness of the film1,2,3,4. Strategies for facilitating ion transport—such as increasing the interlayer spacing by intercalation5,6,7,8 or introducing film porosity by designing nanoarchitectures9,10—result in materials with low volumetric energy storage as well as complex and lengthy ion transport paths that impede performance at high charge–discharge rates. Vertical alignment of two-dimensional flakes enables directional ion transport that can lead to thickness-independent electrochemical performances in thick films11,12,13. However, so far only limited success11,12 has been reported, and the mitigation of performance losses remains a major challenge when working with films of two-dimensional nanomaterials with thicknesses that are near to or exceed the industrial standard of 100 micrometres. Here we demonstrate electrochemical energy storage that is independent of film thickness for vertically aligned two-dimensional titanium carbide (Ti3C2Tx), a material from the MXene family (two-dimensional carbides and nitrides of transition metals (M), where X stands for carbon or nitrogen). The vertical alignment was achieved by mechanical shearing of a discotic lamellar liquid-crystal phase of Ti3C2Tx. The resulting electrode films show excellent performance that is nearly independent of film thickness up to 200 micrometres, which makes them highly attractive for energy storage applications. Furthermore, the self-assembly approach presented here is scalable and can be extended to other systems that involve directional transport, such as catalysis and filtration.

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## Relevant articles

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## Acknowledgements

We thank W.-S. Wei and Z. Davison for providing insights into the assembly of discotic liquid crystals in our experiments. We acknowledge support from the National Science Foundation Materials Science and Engineering Center Grant to the University of Pennsylvania, DMR-1120901 and DMR-1720530 (to S.Y.). Work on MXene synthesis and electrochemical characterization at Drexel University was supported by the Fluid Interface Reactions, Structures & Transport Center, an Energy Frontier Research Center funded by the US Department of Energy, Office of Science, Office of Basic Energy Sciences (to Y.G.).

## Author information

Authors

### Contributions

Y.X., T.S.M., M.-Q.Z., Y.G. and S.Y. conceived the idea and designed the experiments. Y.X., T.S.M. and M.-Q.Z. performed the experiments. Z.Z., A.D., H.C. and B.A. helped with the experiments. Y.G. and S.Y. supervised the work. Y.X. and T.S.M. drafted the manuscript, and all the authors contributed to the editing of the manuscript.

### Corresponding authors

Correspondence to Yury Gogotsi or Shu Yang.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Extended data figures and tables

### Extended Data Fig. 1 Size statistics of MXene nanosheets.

a, SEM image of the nanosheets. The flake dimensions used to report the length and width of the sheets are marked. b, Size distribution of MXene nanosheets, measured from SEM over 250 sheets. c, Average sizes of the nanosheets.

### Extended Data Fig. 2 POM images of MXene aqueous solutions.

a, At a concentration of 50 mg ml−1, the MXene solution shows a nearly isotropic phase with low birefringence under crossed polarizers. b, At a concentration of 250 mg ml−1, higher birefringence starts to appear. A nematic phase is formed, as clearly indicated from the Schlieren texture shown in the inset. c, d, POM images of the slow-appearing MXene liquid-crystal phase on top of one-dimensional microchannels at two polarizer angles: 45° (c) and 0° (d). Higher birefringence starts to appear when water evaporates, and the resulting nematic liquid-crystal phase of MXene can be well aligned with microchannels. The microchannels used here have the following dimensions: diameter 2 µm, spacing 2 µm and depth 1.5 µm. The Onsager theory of liquid crystals predicts the formation of the liquid-crystal phase as a function of the volume fraction of molecules in a media: low volume fraction gives an isotropic phase, and high volume fraction gives a liquid-crystal phase. The empirical value of the critical volume fraction of liquid-crystal phase formation can be estimated by $$\varphi \approx \frac{4T}{W}$$, where Φ is the critical volume fraction, W and T are the width and thickness of the nanosheet, respectively. In our MXene system, Φ is estimated to be around 2 vol%, which is equivalent to about 80 mg ml−1 of MXene nanosheets in aqueous solution. However, because the MXene nanosheets are highly polydisperse in terms of size, and the surface charges differ from system to system, the critical value of Φ could vary in experiments. In this work, we demonstrated the isotropic phase of MXene liquid crystal at around 50 mg ml−1, in good agreement with theory, and we showed the nematic phase of MXene at around 250 mg ml−1. This concentration was chosen to be sufficiently high above the critical value of Φ such that a liquid-crystal phase could be ensured.

### Extended Data Fig. 3 SEM image of nematic MXene liquid crystal after mechanical shear.

Horizontally aligned MXene nanosheets are obtained.

### Extended Data Fig. 4 Schematic of the preparation of MXLLC films.

a, Preparation of the MXLLC slurry. b, Fabrication of the liquid-crystal cell. c, Separation of the MXLLC layer from the liquid-crystal cell. d, The free-standing MXLLC film obtained after supercritical drying.

### Extended Data Fig. 5 POM and SEM images of MXLLC films.

a, b, POM images of MXLLC before (a) and after (b) shear. The inset of b shows the POM image of the MXLLC with the shear direction parallel to the polarizer. ch, Views of the MXLLC film from SEM: random alignment before shear (c), vertical alignment after shear (d), lateral views (e, f), front views (g, h). Dashed red lines indicate the bending directions of the MXene layers.

### Extended Data Fig. 6 AFM images of the spin-coated flat MXene film.

a, Height map. b, Film thickness measured across the boundary of the film and the glass. The thickness of the flat MXene film is estimated to be about 30 nm.

### Extended Data Fig. 7 SEM images of a vacuum-filtered MXene–SWCNT paper with a thickness of around 35 µm.

a, The full cross-section of the freestanding film. b, A higher-magnification image of the top portion of a. Both images show horizontally stacked layers of MXene nanosheets interpenetrated with SWCNTs, a configuration that has been reported to facilitate ion diffusion1.

### Extended Data Fig. 8 Plots of the anodic peak current against scan rate and peak separation against scan rate.

a, Plot of the peak current of the second pair of redox peaks seen at −0.4 V and – 0.6 V versus Hg/HgSO4 in the voltammograms for the MXLCC electrodes in Fig. 3a. b, Plot of the peak separation (ΔEp) for the second pair of redox peaks; the dashed line corresponds to the expected trend for a quasi-electrochemical process2. From a it can be seen that the trend of the current against scan rate of the anodic peak of the second pair of redox peaks is similar (Fig. 3c) to that of the main peaks at – 0.8 V versus Hg/HgSO4 (Fig. 3a), which are characteristic of Ti3C2Tx. The main difference in behaviour for this second pair of peaks compared with previous reports on Ti3C2Tx can be seen in b. Previous characterization of the peak separation against the scan rate for the redox peaks in thin (around 90 nm) Ti3C2Tx electrodes showed a region at low scan rates that corresponds to quasi-equilibrium behaviour3; it is clear from b that this is not the case for the second pair of redox peaks of the MXLLC electrodes. The primary reason for this is probably the large difference in thickness of the MXLLC electrodes (200 μm) compared with previous reports of thin Ti3C2Tx electrodes (90 nm).

### Extended Data Fig. 9 Cyclic voltammograms of pure Ti3C2Tx vacuum-filtered paper and MXLLC film.

a, For the second pair of redox peaks at 0.6 V versus the reference, the current grows slightly during cycling and then stabilizes during continuous cycling. b, The current will fade if the cell is allowed to rest. This pair of peaks is thought to originate from changes occurring in the transition-metal surface of the MXene during cycling and will be the subject of further studies. It is thought that these peaks are more pronounced in the cyclic voltammograms of the MXLLC samples owing to the large amount of active material surface that is exposed to the electrolyte in the MXLLC samples relative to the vacuum-filtered papers. c, d, Cyclic voltammograms of a 35-µm-thick vacuum-filtered MXene paper (c) and a 40-µm-thick MXLLC film (d). For similar film thicknesses, MXLLC films have much better rate-handling ability compared to the vacuum-filtered papers, as only a small decay of the cyclic voltammogram is observed until high scan rates are used.

### Extended Data Fig. 10 Capacitance retention as a function of scan rate for both vacuum-filtered films and MXLLC films.

All data points are normalized to the capacitance value at 10 mV s−1 for each sample curve, respectively. At 2,000 mV s−1, vacuum-filtered MXene papers retain around 14% (35 µm thick) and 30% (6 µm thick) capacitance, whereas MXLLC films maintain more than 75% capacitance over a wide range of film thickness from 40 µm to 200 µm. However, in thicker (320 µm) MXLLC films, the retention curve starts to behave similarly to that of thin vacuum-filtered paper (6 µm). These data again suggest that the optimal thickness of the working-electrode film of MXLLC is 200 µm.

## Supplementary information

### Video 1: Polarized optical microscope (POM) images demonstrating MXene nanosheets aligned by microchannels during water evaporation.

MXene aqueous solution is sandwiched between a patterned epoxy substrate and a flat glass. As water gradually evaporated from the right side of the sample, the MXene LC phase began to form and aligned. Video is accelerated by 4x.

## Rights and permissions

Reprints and Permissions

Xia, Y., Mathis, T.S., Zhao, MQ. et al. Thickness-independent capacitance of vertically aligned liquid-crystalline MXenes. Nature 557, 409–412 (2018). https://doi.org/10.1038/s41586-018-0109-z

• Accepted:

• Published:

• Issue Date:

• DOI: https://doi.org/10.1038/s41586-018-0109-z

• ### 2D materials: increscent quantum flatland with immense potential for applications

• Pranay Ranjan
• Snehraj Gaur
• Prashant Kumar

Nano Convergence (2022)

• ### Ultra-compact MXene fibers by continuous and controllable synergy of interfacial interactions and thermal drawing-induced stresses

• Tianzhu Zhou
• Yangzhe Yu
• Lei Wei

Nature Communications (2022)

• ### Two-dimensional quantum-sheet films with sub-1.2 nm channels for ultrahigh-rate electrochemical capacitance

• Wenshu Chen
• Jiajun Gu
• Y. Morris Wang

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• ### MXene chemistry, electrochemistry and energy storage applications

• Xinliang Li
• Zhaodong Huang
• Chunyi Zhi

Nature Reviews Chemistry (2022)

• ### Field-induced orientational switching produces vertically aligned Ti3C2Tx MXene nanosheets

• Changjae Lee
• Soon Mo Park
• Dong Ki Yoon

Nature Communications (2022)


text:
What are $\limsup{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}$ and $\;\;\liminf{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}\;?$

What are $\limsup{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}$ and $\;\;\liminf{\left(\dfrac{3}{2^{2k+1}}\right)^{1/k}}\;?\;\;$ I find both $1/4$. Thank you.

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See my answer on your other question and think it through. The hint (and properties of exponents) will get you where you need to go. –  Cameron Buie Dec 27 '12 at 16:41

$$\left(\frac{3}{2^{2k+1}}\right)^{1/k}=\left(\frac{3}{2}\right)^{1/k}\frac{1}{2^2}$$
Well, now you only need to know/remember that $\,\sqrt [n] a\xrightarrow [n\to\infty]{}1\,\,\,,\,\,\forall\,a>0$...and you're right: they both are $\,\frac14\,$ (which btw happens iff the sequence converges to the common supremum and infimum limit)
This may be a useful hint: $${\left(\dfrac{3}{2^{2k+1}}\right)^\frac{1}{k}}=3^\frac{1}{k}(2^{-2k-1})^{\frac{1}{k}}=3^\frac{1}{k}(2^{-2-\frac{1}{k}})=...$$


text:
# Intercept Form of a Plane – Equation and Example

Here you will learn intercept form of a plane equation with examples.

Let’s begin –

## Intercept Form of a Plane

The equation of a plane intercepting lengths a, b and c with x-axis, y-axis and z-axis respectively is

$$x\over a$$ + $$y\over b$$ + $$z\over c$$ = 1

Note :

1). The above equation is known as the intercept form of the plane, because the plane intercepts length a, b and c with x, y and z-axes respectively.

2). To determine the intercepts made by a plane with the coordinate axes we proceed as follows :

For x-intercept : Put y = 0, z = 0 in the equation of the plane and obtain the value of x. The value of x is the intercept on x-axis.

For y-intercept : Put x = 0, z = 0 in the equation of the plane and obtain the value of y. The value of y is the intercept on y-axis.

For z-intercept : Put x = 0, y = 0 in the equation of the plane and obtain the value of z. The value of z is the intercept on z-axis.

Example : Write the equation of the plane whose intercepts on the coordinates axes are -4, 2 and 3 respectively.

Solution : We know that the equation of a plane having a, b and c intercept on the coordinate axes is given by $$x\over a$$ + $$y\over b$$ + $$z\over c$$ = 1

Here a = -4, b = 2, and c = 3. So ,the equation of the required plane is

$$x\over -4$$ + $$y\over 2$$ + $$z\over 3$$ = 1

$$\implies$$  -3x + 6y + 4z =12


text:
# Monotone circuit complexity of computing functions on sparse inputs

The weight $|x|$ of a binary string $x\in\{0,1\}^n$ is the number of ones in the string. What happens if we are interested in computing a monotone function on inputs with few ones?

We know that deciding if a graph has a $k$-clique is hard for monotone circuits (see among others Alon Boppana, 1987), but if a graph has for example at most $k^3$ edges it possible to find a monotone bounded depth circuit of size $f(k)\cdot n^{O(1)}$ which decides $k$-clique.

My question: is there any function which is hard to compute by a monotone circuit even on inputs of weight less than $k$? Here hard means circuit size $n^{{k}^{\Omega(1)}}$.

Even better: is there an explicit monotone function which is hard to compute even if we only care about inputs of weight $k_1$ and $k_2$?

Emil Jeřábek already observed that known lower bounds hold for monotone circuits that separate two classes of inputs ($a$-cliques vs maximal $(a-1)$-colorable graphs), thus at cost of some independence in the probabilistic argument it is possible to make it work for two classes of input of fixed weight. This would cause $k_2$ to be a function of $n$ which I want to avoid.

What is would really like is an explicit hard function for $k_1$ and $k_2$ much smaller than $n$ (as in the parameterized complexity framework). Even better if $k_1=k_2+1$.

Notice that a positive answer for $k_1=k_2$ would imply an exponential lower bound for arbitrary circuits.

Update: This question may be partially relevant.

-
To your very first (general) question (not about Clique). I think, even the case of inputs with at most $2$ ones is very difficult. Take a bipartite $n\times m$ graph $G$ with $m=o(n)$. Assign to each vertex $u$ a boolean variable $x_u$. Let $f_G(x)$ be a monotone boolean function whose minterms are $x_u\land x_v$ for the edges $uv$ of $G$. Let $s(G)$ be the minimal size of a monotone circuit which correctly computes $f_G$ on inputs with $\leq 2$ ones. Then any lower bound $s(G)\geq (2+c)n$ for a constant $c>0$ would imply an exponential lower bound for nonmonotone circuits. –  Stasys Dec 8 '11 at 15:46
Existing arguments for monotone circuits need that many inputs with many ($\gg n/2$) ones must be rejected. The best we can do so far is to prove $\exp(\min\{a,n/b\}^{1/4})$ lower bound when circuit must accept all $b$-cliques, and reject all complete $a$-partite graphs ($a<b$). B.t.w. important is that you deal with sparse, not with dense inputs. Say, $k$-Clique requires monotone circuits of size about $n^k$ for every constant $k\geq 3$, but $(n-k)$-Clique has monotone circuits of size $O(n^2\log n)$ for every constant $k$. –  Stasys Dec 8 '11 at 17:06
I should clarify that I care about sparse inputs in the sense of sparse graph. Looking for a $k$-clique in a very sparse graph (with say $k^{10}$ edges) can be done in FPT monotone circuit size. –  MassimoLauria Dec 8 '11 at 18:25
Your example in the first comment is very nice. If I understand correctly this is a similar issue with monotone functions which are hard on a fixed weight $k$. Using pseudo complement functions to simulate negated inputs, the circuit complexity does not differ between monotone and non-monotone case. For constant (or small) $k$ this pseudo complement can be implemented efficiently by a monotone circuit. –  MassimoLauria Dec 8 '11 at 19:48
my first comment relied on graph complexity. The "$(2+c)n$" phenomenon can be found on page 13 of this draft. B.t.w. I haven't quite understood what you mean by being "hard for k and k+1"? (My fault, of course.) –  Stasys Dec 8 '11 at 20:22
specifically considering one part of the question (eg for $k_1$=1,$k_2$=2), Lokam studied "2-slice" functions in this paper & proves that strong lower bounds on them can be generalized, therefore this is a very hard open problem related to basic complexity class separation & any such construction/explicit function would be a breakthrough; from the abstract:


text:
# Linear Algebra - Change of Bases

Mumba
Hi, again another problem:

Let B1 = {( $$\stackrel{1}{3}$$),($$\stackrel{1}{2}$$)} and

$$B_{2} = [ \frac{1}{\sqrt{2}}( \stackrel{1}{1}), \frac{1}{\sqrt{2}} (\stackrel{-1}{1}) ]$$

Determine the representing matrix $$T = K_{B_{2},B_{1}} \in \Re^{2\times2}$$ for the change from B1 coordinates to B2 coordinates.

I have no idea what i should do here. I ve found how to calculate the representing matrix from a domain to a codomain.
Is this the same way? Can you give me atleast a hint, please ^^.

Thx Mumba

PS. Sorry, it looks really strange. I dont know how to formate this better.

dacruick
This reminds me of quantum mechanics 1, a course that crushed m average and of which I remember very little.

I tried a couple things and the answer i got was 1/sqrt(2) ( 2 5 )
( 0 1 )
for the matrix T. But as i said again, thats just ag uess

Homework Helper
Gold Member
If S denotes the standard basis for $\mathbb{R}^2$, do you know how to find

$$K_{S,B_1}$$ and $$K_{S,B_2}$$?

If so, then observe that

$$K_{B_2,B_1} = K_{B_2,S} K_{S,B_1} = K_{S,B_2}^{-1}K_{S,B_1}$$

Mumba
No, i dont even know what K is supposed to be...

Homework Helper
Gold Member
No, i dont even know what K is supposed to be...

$$K_{S,B_1}$$

is the change-of-basis matrix that transforms $B_1$ coordinates to $S$ coordinates. I suggested doing it this way because you already know how to express $B_1$ and $B_2$ in $S$ coordinates: that is what you are given.

Mumba
So i should calculate the change of base matrix for B1 to S and the inverse of the change of base matrix for B2 to S coordinates?
And multiply this to get my result?

Ok i ll try to find out how to calculate the change of base matrix ^^
Thx

Mumba
Is [text]K_{S,B1}[/tex] not the same as B1?
And the same for B2? So i calculate the inverste and multply them and then im finished?

Mumba
the i get as matrix:

4/sqrt(2) 3/sqrt(2)
2/sqrt(2) 1/sqrt(2)

correct?

Homework Helper
Gold Member
Is [text]K_{S,B1}[/tex] not the same as B1?
And the same for B2? So i calculate the inverste and multply them and then im finished?

Well, $B_1$ is a basis (set of vectors), not a matrix, so what you said is not exactly correct. However, I think what you are trying to say is this:

$$K_{S,B_1}$$ is the matrix whose columns are the basis vectors from $B_1$ expressed in $S$ coordinates, namely

$$K_{S,B_1} = \left[\begin{array}{cc}1 & 1 \\ 3 & 2\end{array}\right]$$

and similarly for $$K_{S,B_2}$$.

So yes, now you can find $$K_{B_2,B_1}$$ as I described earlier.

Mumba
Coool thx a lot, easy this way.
:D

Homework Helper
Gold Member
the i get as matrix:

4/sqrt(2) 3/sqrt(2)
2/sqrt(2) 1/sqrt(2)

correct?


text:
Proportional Relationships

Objective

Solve ratio and rate problems by setting up a proportion, including part-part-whole problems.

Common Core Standards

Core Standards

?

• 7.RP.A.1 — Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2/1/4 miles per hour, equivalently 2 miles per hour.

• 7.RP.A.3 — Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

?

• 6.RP.A.3

Criteria for Success

?

1. Write a proportion of two equivalent ratios, either part:part or part:whole.
2. Identify situations when a part:whole ratio is required, compared to a part:part.
3. Attend to precision with units when setting up a proportion (MP.6).
4. Solve ratio and rate problems using proportions or other strategies such as tape diagrams.

Tips for Teachers

?

Remote Learning Guidance

If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Problem 1 (benefits from worked example). Find more guidance on adapting our math curriculum for remote learning here.

Fishtank Plus

• Problem Set
• Student Handout Editor
• Vocabulary Package

Anchor Problems

?

Problem 1

Macaroni and cheese is made by combining noodles and cheese in a ratio of 4:1. A pan must be big enough to hold the total amount of food. The table below shows 3 different variations of the recipe and the size pan that is needed.

 Noodles (cups) Cheese (cups) Pan Size (cups) $4$ $1$ $5$ $\frac{2}{3}$ $3\frac{3}{4}$
1.  If you have ${{2\over3}}$ cups of noodles, how many cups of cheese should you add? What size pan do you need?
2. If you have a pan that holds ${3{3\over4}}$ cups, how many cups of noodles and cheese should you use to fill this pan?

References

EngageNY Mathematics Grade 7 Mathematics > Module 1 > Topic C > Lesson 13Exercise

Grade 7 Mathematics > Module 1 > Topic C > Lesson 13 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by The Match Foundation, Inc.

Problem 2

Gus is painting a large bookshelf to match a desk in his office. The desk is painted a color orange, in the ratio of 1 part red to 2 parts yellow. If Gus thinks he will need 2 gallons of the orange paint, how many gallons of each color paint should he mix to get the right shade of orange?

Problem Set

?

The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.

• Include a variety of problems from Lessons 13-16

 Flour (cups) Salt (cups) Total (cups) $1$ $2$ $\frac{3}{4}$ $4\frac{1}{2}$ $7\frac{1}{3}$


text:
# A random sample of 10 boys...

• Dec 30th 2009, 02:45 AM
zorro
A random sample of 10 boys...
Question : A random sample of 10 boys had a following IQ's 70, 120,110,101, 88,83, 95,98,107, 100. Whether data supports the assumption of a population mean =100?
--------------------------------------------------------------------------------------------------------

I actual didn't understand what to do in this
• Dec 31st 2009, 04:50 PM
matheagle
This seems to be a two sided test
$H_0:\mu=100$ vs. $H_a:\mu\ne 100$

The test statistic is a t with 9 degrees of freedom

$t^*_9={\bar X-100\over s/\sqrt{10}}$
• Dec 31st 2009, 08:17 PM
zorro


text:
# LaTeX for JIRA

LaTeX for JIRA is the only plugin for rendering LaTeX inside JIRA Server. The latest version is 2.0.0. You can install this JIRA plugin at https://marketplace.atlassian.com/plugins/com.fulstech.jira-latex.

## Usage

This plugin renders LaTeX anywhere inside JIRA – even on Issue’s Summary. The syntax is very simple.

• For fields that use the Atlassian Wiki editor like Description or Comment, inline LaTex content must be wrapped inside the {latex-inline}…{latex-inline} macro and LaTex block content must be wrapped between {latex-block}…{latex-block}. For JIRA 7+, the macros have to be used in Text mode and not Visual mode.
• For simple text fields like Summary, inline LaTex must be wrapped inside //(..//)  and LaTex block must be wrapped inside //[..//].

## Example

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Solve equation //(ax^2 + bx + c = 0//)

And for Issue’s Description, please use the following one:

Solution for equation {latex-inline}ax^2 + bx + c = 0{latex-inline} should be {latex-block}x = {-b \pm \sqrt{b^2-4ac} \over 2a}{latex-block}.

For JIRA 7+, the macros have to be used in Text mode and not Visual mode.

All LaTeX content will be rendered as below.

## Support forum

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text:
# B Box in gravitational field

1. Jul 18, 2016

### Einstein's Cat

Let's say there's a massive and uniform sphere of radius, r, and there's a box in its gravitational field and the sphere is so massive that the box is accelerated to a velocity near the speed of light; could the classical coordinate system still be used or is the Schwarzschild coordinate system needed?

Furthermore say there's an observers further from the gravitational field than the box. Is the affect of time dilation on different aspects of the box negotiable from the observers perspective and what is the radial coordinate of the observer? Wikipedia says that it is analogous to the classical distance; could the classical distance and classical coordinate system be used instead where the classical coordinate system is a three dimensional version of Cartesian coordinates?

I've researched this but do not understand.

Also is it suitible to you use classical physics to describe the acceleration of the box?

Last edited by a moderator: Jul 20, 2016
2. Jul 18, 2016

### Mr-R

What do you mean by "Classical coordinate system"? If you mean the Cartesian $(x,y,z)$ then it would not work because this is a relativistic situation(Acceleration $\approx c$ !!) . So you need Minkowski Coordinates $(t,x,y,z)$. But since we are dealing with gravitational fields and noninertial frames we need to take account of curvature in space-time. So we use the Schwarzchild metric.

Last edited by a moderator: Jul 18, 2016
3. Jul 18, 2016

### Staff: Mentor

Relative to what? As far as the box is concerned it is floating in empty space, although if you're in the box and look out in the right direction, you'll see the massive spherical body rushing towards you at relativistic velocities.
In any sufficiently small (and singularity-free) region every spacetime is locally flat, so you can use the Minkowski metric with ordinary Cartesian coordinates if you wish. You need to use the Schwarzschild metric when you're considering a region of space large enough that the curvature effects matter. There's an analogy with the curved surface of the earth here: If you're laying out the foundations of a house you can use ordinary x/y Cartesian coordinates as if the earth is flat; but if you're plotting a course across the ocean you'll have to pay attention to the curvature to calculate the correct great-circle route. Same thing around a gravitating object - you can ignore the curvature in small regions but not large regions. So if the box is whizzing by just under your nose, you can use ordinary coordinates at least for the few moments before it's moved far away again.

Again, that depends on the size of the box relative to the strength of the gravitational field. If the box is small enough that the effects of curvature from one side to the other is negligible, then the difference in time dilation effects from one side to the other will be also be negligible. The time dilation may be very large if the observer is very far from the box, but it will be about the same for both sides.

Only within a region small enough that curvature effects can be ignored (and even then you'll find classical polar coordinates more convenient). Consider two spheres around the object, one with Schwarzschild $r$ coordinate equal to $r_1$ and the second with Schwarzschild $r$ coordinate equal to $r_2$. the area of the two spheres will be $4\pi{r}_1^2$ and $4\pi{r}_1^2$ - but the distance between them will not be $r_2-r_1$. There is no way of making this work using flat-space Cartesian coordinates (or polar coordinates).
Not really. The box is accelerating, and classical physics says that means there a force on it (in GR, there's no force) but there's no way of calculating that force to see what the acceleration is - the classical $F=Gm_1m_2/r^2$ doesn't work under relativistic conditions.

4. Jul 18, 2016

### pervect

Staff Emeritus

If you are using coordinates in which the sphere is at rest, and you also assume the box was initially at rest in these coordinates and "dropped", allowing it to accelerate towards the sphere, aquiring near light-speed velocity in it's fall due to the gravity of the sphere, , then yes, it's a problem that needs the Schwarzschild coordinate system. (Actually, of course, you can use other coordinates, but they'd all be related to the Schwarzwschild coordinate by a 1:1 map, i.e. a diffeomorphism).

5. Jul 18, 2016

### Yukterez

With Newton the particle can be accelerated beyond c relative to the central mass, while under Einstein it can't, see Plot 3 in this post.

The distance from r1 to r2 is not r2-r1 but ∫(1/√(1-rs/r), r=r1..r2), see Flamm's paraboloid in this post.

Yes if you differentiate coordinates by proper time (then you get the rapidity instead of the velocity) and no if you differentiate by coordinate time.

6. Jul 18, 2016

### Einstein's Cat

Could the spacetime be treated flat if it was say 4m in length and 3m from a sphere of 1m in radius? And also how would one calculate the acceleration of the box and thus it's displacement

7. Jul 18, 2016

### Staff: Mentor

That depends on how strong the gravitational field is. There's an easy way to find out: look for tidal effects. The box is in free fall, so if we release a weight inside the box it will just float. So we release six weights in the box - one just above the floor, one just below the ceiling, and one against each wall. In a flat-enough spacetime they will stay put. Any detectable curvature will show up as follows:
- The weights near the walls will drift closer to one another, because their paths aren't quite parallel - they point towards the center of the gravitating mass.
- The weight at the top and the weight at the bottom will move apart and towards the ceiling and the floor, because the bottom weight is a bit closer to the mass so is feeling a slightly stronger force.

The smaller the box and the weaker the gravitational field, the smaller these effects will be. When they're small enough to be irrelevant we say that the spacetime is flat enough to ignore the curvature. You'll notice that when that happens all of Newton's laws work perfectly; the objects started at rest (as far we can tell from inside the box) and because there is no force on them they stay that way.

Last edited: Jul 18, 2016
8. Jul 19, 2016

### Einstein's Cat

Does that mean when spacetime is flat enough Newtonian mechanics could be used to describe the box at relativistic velocities?

9. Jul 19, 2016

### A.T.

Newton or SR, depending on what happens inside the box and which frame you choose.

10. Jul 19, 2016

### Staff: Mentor

At relativistic velocities relative to what? If you're inside the box and free-falling with it, the only deviation from Newtonian mechanics you'll see will be the tidal effects I mentioned above, and these can become arbitrarily small as the box becomes smaller. If you're outside the box and it is moving at relativistic speeds relative to you, you'll need to use relativity. SR will work if you're only considering a very short period of time during which the box is right under your nose; as soon as the distances and times become large enough that curvature effects appear you'll have to use GR.

In all of these discussions, "large enough" and "small enough" depends on the strength of the gravitational field; the more the curvature the more effect it has and the shorter the distances over which it can be ignored.

11. Jul 19, 2016

### Einstein's Cat

The relativistic velocities would be relative to an observer further from the gravitational field than the box; and if I was to use GR then how would the box be affected by a sphere of g of x, and where it's accelerated to near the speed of light, relative to the observer?

12. Jul 19, 2016

### Einstein's Cat

Say there is a box of height, h, width, w, and length, l, with a distance of d from the centre of a sphere of radius, r, and mass,m. The box is accelerated by the gravitational field to near the speed of light.

How could this box be described in the gravitational field?
My knowledge of GR is very limited so I am entirely ignorant to this.

13. Jul 19, 2016

### A.T.

14. Jul 19, 2016

### Einstein's Cat

15. Jul 20, 2016

### Staff: Mentor

It's not clear what you mean by "a description". Do you want to know its trajectory? What it's like to be inside the box?

16. Jul 20, 2016

### Einstein's Cat

By description I mean what would happen to the box and the box would be stationary when time (t) is zero. Would it be compressed? To what extent?

17. Jul 20, 2016

### Staff: Mentor

Start by solving that problem using ordinary Newtonian gravity. You can model the box as eight point masses, one at each corner of the box, connected together by twelve rigid and massless sticks. Assuming just ordinary classical physics, what is the tension or compression in each stick when the box is free-falling in Newtonian $F=Gm_1m_2/r^2$ gravity?

That will give you some understanding into the various factors that influence what happens to the box. After you thoroughly understand that problem, you can start to think about the effect of corrections due to GR.

18. Jul 20, 2016

### Einstein's Cat

I've done the first part of the task; it is just the GR aspect that I am confused with


text:
Electromagnetic tensor use in QFT

(This question may lose me some physics knowledge points, but please be patient with me, my training is mostly in chemistry)

What is the purpose of using the electromagnetic tensor over simply directly working with electric and magnetic fields? Specifically in the quantum electrodynamic Lagrangian we use the tensor. 1 However, in my work with classical electromagnetism I have found that simply working with vector fields is much more profitable and the EM tensor is usually not needed. (If nothing else vector fields are simply easier to use.)

So in brief, why does no quantum Lagrangian exist in terms of the vector fields we know from classical EM?

• We require a relativistically invariant Lagrangian which is easier to construct in terms of relativistically covariant objects such as $A_\mu, J_\mu, F_{\mu\nu}$ rather than in terms of ${\bf E}, {\bf B}, \rho, {\bf J}$. All you have to look for terms where the Lorentz indices are contracted. If you wish, you can express the Lagrangian in terms of Lorentz noncovariant objects ${\bf E}, {\bf B}, \rho, {\bf J}$. – SRS Jun 28 '20 at 2:52

2 Answers

It does. If you don't want to use $$F_{\mu\nu}$$ you can always write the pure QED Langrangian as

$$\mathcal L = (\vec E^2 - \vec B^2)$$

which is equivalent. However it's not so easy to see that this quantity is Lorentz invariant, while it is easy to see that $$F_{\mu\nu}F^{\mu\nu}$$ is. For the purposes of QED, the use of $$F_{\mu\nu}F^{\mu\nu}$$ is more common than the above (though the above can just as easily be quantized, it's just notation after all) because it is more convenient to work in a way that makes the symmetries of the theory manifest.

This may not be what you're looking for, but electric and magnetic fields are not Lorentz invariant, i.e. well-defined in the special theory of relativity. A point particle will (classically) emit only an electric field if it's at rest, but it will emit both electric and magnetic fields if you boost out of its rest frame. Quantum field theory is designed to be Lorentz invariant, so its fundamental objects should be Lorentz invariant objects like four-vectors (e.g. the electromagnetic four potential) and tensors (e.g. the EM tensor, which just consists of derivatives acting on the four potential). This Lorentz-invariant formulation of EM in terms of four-potentials and tensors is actually entirely classical, as you can see here: https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Covariant_formulation_in_vacuum. The quantum part is just interpreting the four-potential as a quantum field operator, rather than a classical field with a definite value.

Actually, it turns out that the four potential is somewhat more fundamental than the EM tensor: certain quantum phenomena, like the Aharonov-Bohm effect, rely on the four potential directly, producing measurable effects even if there are no electric or magnetic forces present.


text:
# I Commutator Properties

1. Dec 16, 2016

### Kyle Nemeth

Given the property,

[A,BC] = [A,B]C+B[A,C],

is it true that, if B=C, then

[A,B]C+B[A,C]=[A,B]C+B[A,B]=[A,B](C+B)?

I apologize if I have posted in the wrong forum.

2. Dec 16, 2016

### Staff: Mentor

The commutator (in the assumed context above) is defined as $[A,B]=AB-BA$.
Now you have $[A,B]C+B[A,C]=[A,B]B+B[A,B]=AB^2-B^2A$ on the left and a factor $2$ on the right.
The (presumably) operators (or linear mappings) $A,B,C$ do not commutate, so you cannot switch orders.

3. Dec 16, 2016

### Kyle Nemeth

Thank you for your help. I understand now.


text:
# proper

proper is one of the 1000 most common headwords.
 proper

## Contents

### Pronunciation

• (Australia) IPA: /ˈprɔp.ə/, SAMPA: /"prOp.@/
• (UK) IPA: /ˈprɒp.ə/, SAMPA: /"prQp.@/
• (US) enPR: präpʹər, IPA: /ˈprɑp.ɚ/, SAMPA: /"prAp.@`/
•  Audio (US) (file)

 Positive proper Comparative more proper Superlative most proper
1. (usually before the noun) The proper way, place, use, etc. is the right one.
What is the proper way to write a letter?
2. If something is proper, it is polite or acceptable.
It's not proper for you to live alone like this.
She was a very proper young lady.
Why don't you get a proper job at a good company?
3. (only after the noun) The ${\displaystyle x}$ proper is the main or important part of ${\displaystyle x}$, not the parts near it.
We stayed in a village and only went into the city proper during the day.


text:
## Stream: new members

### Topic: translation on reals

#### Damiano Testa (Dec 23 2020 at 15:04):

Dear All,

I am struggling to find a proof of the lemma below: I am convinced that it should be in mathlib, I have found continuous_add, but I know so little od the topology side of mathlib, that I am unable to get it to work. Can anyone give me a pointer, please?

Thanks!

import analysis.convex.topology

open real

lemma translate_open (α : ℝ) (U : set ℝ) (oU : is_open U) :
is_open ((λ (y : ℝ), y - α) '' U) :=
begin
sorry,
end

#xy

#### Damiano Testa (Dec 23 2020 at 15:22):

I have a set A of reals and a real x ∈ A. I want to show that A \ {x} ≠ ∅ by showing that A contains a neighbourhood of x. The neighbourhood in question is the translate of an intersection of sets that I already know are open and nonempty, hence the question!

#### Johan Commelin (Dec 23 2020 at 15:23):

@Damiano Testa there is continuous_add_left

#### Johan Commelin (Dec 23 2020 at 15:23):

the first thing I would do is change your goal to preimage under addition, instead of images under subtraction.

#### Damiano Testa (Dec 23 2020 at 15:24):

Hmm, maybe I should import more? I get

unknown identifier 'has_continuous_add_left'

#### Damiano Testa (Dec 23 2020 at 15:24):

Let me try with replacing preimages!

#### Johan Commelin (Dec 23 2020 at 15:24):

without the has_

#### Damiano Testa (Dec 23 2020 at 15:27):

I will try to make this work, thanks!

#### Damiano Testa (Dec 23 2020 at 15:29):

This works!

lemma translate_open (α : ℝ) (U : set ℝ) (oU : is_open U) :
is_open ((λ (y : ℝ), y + α) ⁻¹' U) :=

Now I will adjust the rest to what I have!

#### Damiano Testa (Dec 23 2020 at 15:36):

There is even preimage_add_const_Ioo! Found by simp!

#### Adam Topaz (Dec 23 2020 at 15:52):

Is there no homeomorph_add_left in mathlib?

#### Heather Macbeth (Dec 23 2020 at 15:54):

Ah, thanks for the pointer to homeomorph.add_left! I am doing some very simple-minded argument, so I may not be needing it, but I am relieved to see that it is in mathlib!


text:
# Transfer function - get rid of zero

I have a close loop transfer function, consisting of a system and a controller (PID). For some reason does it create 2 zeroes which creates an undershoot which should not appear at all.

I don't understand how zeroes placed at the LHP should cause an overshoot. the poles are placed on the line thereby create a critical damped system.

the system is

$$G(s) =\frac{10.95 s + 0.9574}{s^2 + 0.09149 s + 6.263*10^{-6}}$$

With P = 0.1, I= 0.617746, d = 0.0147173 I get a close loop system which is $$G_cl(s) = \frac{0.1612 s^4 + 1.109 s^3 + 6.86 s^2 + 0.5914 s}{ 0.1612 s^4 + 2.109 s^3 + 6.952 s^2 + 0.5914s}$$

poles =

0,
-7.0000,
-6.0000,
-0.0874,

zeroes =

0.0000 + 0.0000i
-3.3974 + 5.5165i
-3.3974 - 5.5165i
-0.0874 + 0.0000i

I don't know how, but it seems like it 2 zeroes which causes the overshoot. Could someone explain to me why it causes overshoot, and how i can get rid of it using a PID controller.

Calculation:

x1 := -7
x2 := -6

Solve[x1 (x1^2 + 0.09149 x1 + 6.263*10^-6) +
kp*x1 (10.95 x1^2 + 0.9574 x1) + ki (10.95 x1^2 + 0.9574 x1) +
kd*x1^2 (10.95 x1^2 + 0.9574 x1) == 0 &&
x2 (x2^2 + 0.09149 x2 + 6.263*10^-6) +
kp*x2 (10.95 x2^2 + 0.9574 x2) + ki (10.95 x2^2 + 0.9574 x2) +
kd*x2^2 (10.95 x2^2 + 0.9574 x2) == 0, {kp, ki, kd}]
{{ki -> 0.294669 + 3.23077 kp, kd -> 0.007025 + 0.0769231 kp}}

I am pratically solving the denum for kp,kd,ki as if x1 and x2 where roots.


text:
zbMATH — the first resource for mathematics

Ordre, convergence et sommabilité de produits de séries de Dirichlet. (Order, convergence and summability of products of Dirichlet series.). (French) Zbl 0977.11037
Summary: Optimal or almost optimal answers are given to the following questions, going back to Stieltjes, Landau and Bohr, about Dirichlet series $$A_j= \sum^{\infty}_{ n=1}a(j,n)n^{-s}$$ $$(j=1,2,\cdots ,k)$$ and their product $$C= \sum^{\infty}_{n=1}c(n)n^{-s}.$$
1) Assuming that the $$A_j$$ converge at points $$\rho_ j$$ and converge absolutely at points $$\rho_j+\tau_j,$$ at which points $$s$$ does it follow that $$C$$ converges?
2) Assuming that the $$A_j$$ converge at points $$\rho_j,$$ at which points $$s$$ does it follow that $$C$$ converges?
3) Assuming that the $$A_j$$ are $$\alpha_j$$-summable at points $$\rho_j,$$ at which points $$s$$ does it follow that $$C$$ is $$\beta$$-summable ?
The answers involve convex functions which enjoy another extremal property: they are the largest order (= Lindelöf) functions compatible with the data.

MSC:
 11M41 Other Dirichlet series and zeta functions 30B50 Dirichlet series, exponential series and other series in one complex variable 40G10 Abel, Borel and power series methods
Full Text:
References:
 [1] H. BOHR, On the convergence problem for Dirichlet series, Dan. Mat. Fys. Medd., 25, 6 (1946), 1-18 (I A 17 in Collected Math. Works). [2] H. BOHR, On multiplication of summable Dirichlet series, Mat. Tidsskr., (1950), 71-75 (I A 19 in Collected Math. Works). · Zbl 0039.08203 [3] H. BOHR, On the summability function and the order function of Dirichlet series, Dan. Mat. Fys. Medd., 27, no. 4 (1952), 3-38 (manuscript prepared by E. Følner) (I A 22 in Collected Math. Works). · Zbl 0046.30203 [4] N. BOURBAKI, Utilisation des nombres réels en topologie générale, Chapitre 9, Paris, Hermann, 1958. · Zbl 0085.37103 [5] H. DELANGE & G. TENENBAUM, Un théorème sur LES séries de Dirichlet, Monatshefte Mat., 113 (1992), 99-105. · Zbl 0765.30002 [6] G. H. HARDY & M. RIESZ, The general theory of Dirichlet series, Cambridge tracts in Math. and Math. Physics, 18 (1915). · JFM 45.0387.03 [7] J.-P. KAHANE, The last problem of harald Bohr, J. Austr. Math. Soc., A 47 (1989), 133-152. · Zbl 0734.30003 [8] J.-P. KAHANE, Sur trois notes de Stieltjes relatives aux séries de Dirichlet. Numéro spécial “100 ans après Stieltjes” des Annales de la Faculté des Sciences de Toulouse, (1996), 33-56. · Zbl 0877.01023 [9] E. LANDAU, Über die multiplication Dirichlet’schen reihen, Rendiconti di Palermo, 24 (1907), 81-160. · JFM 38.0322.01 [10] E. LANDAU, Über das konvergenzproblem der Dirichlet’schen reihen, Rendiconti di Palermo, 28 (1909), 113-151. · JFM 40.0311.03 [11] H. QUEFFELEC, Propriétés presque-sûres et quasi-sûres des séries de Dirichlet et des produits d’Euler, Canad. J. Math., 32 (1980), 531-558. · Zbl 0475.30006 [12] M. RIESZ, Sur l’équivalence de certaines méthodes de sommation, Proc. London Math. Soc., (2) 22 (1924), 412-419. · JFM 50.0154.01 [13] W. RUDIN, Real and complex analysis, 3rd ed. Mc-Graw Hill, 1987, p. 118. · Zbl 0925.00005
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Cohomology of automorphism groups of free groups with twisted coefficients (Englisch)

Freier Zugriff
arXiv ; 2016

We compute the groups $H^*(\mathrm{Aut}(F_n); M)$ and $H^*(\mathrm{Out}(F_n); M)$ in a stable range, where $M$ is obtained by applying a Schur functor to $H_\mathbb{Q}$ or $H^*_\mathbb{Q}$, respectively the first rational homology and cohomology of $F_n$. For reasons which are not conceptually clear, taking coefficients in $H_\mathbb{Q}$ and its related modules behaves in a far less trivial way than taking coefficients in $H^*_\mathbb{Q}$ and its related modules. The answer may be described in terms of stable multiplicities of irreducibles in the plethysm $\mathrm{Sym}^k \circ \mathrm{Sym}^l$ of symmetric powers. We also compute the stable integral cohomology groups of $\mathrm{Aut}(F_n)$ with coefficients in $H$ or $H^*$, respectively the first integral homology and cohomology of $F_n$, and compute the stable cohomology with coefficients in Schur functors of $H$ or $H^*$ modulo small primes.

## Dokumentinformationen

• Titel:
Cohomology of automorphism groups of free groups with twisted coefficients
• Autor / Urheber:
• Verlag:
arXiv
• Erscheinungsjahr:
2016
• Anmerkungen:
20 pages. Supersedes arXiv:1012.1433. v2 21 pages, to appear in Selecta Mathematica
• Medientyp:
Preprint
• Format:
Elektronische Ressource
• Sprache:
Englisch
• Klassifikation:
 MSC: 20F28 / 20J06 / 57R20
• Lizenzbestimmungen:

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text:
## tex4ht - Bugs: bug #560, problem using tex4ht with xr-hyper...

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## bug #560: problem using tex4ht with xr-hyper when other documents are in separate folders

 Submitted by: Nasser M. Abbasi Submitted on: Sat Apr 23 05:23:31 2022 Category: None Priority: 5 - Normal Severity: 5 - Normal Status: Fixed Privacy: Public Assigned to: None Open/Closed: Closed

Tue Apr 26 08:21:28 2022, comment #2:

I've also fixed it for the Xr package. I think we can fix this now.

Michal Hoftich <michal_h21>
Mon Apr 25 13:32:32 2022, comment #1:

Hi Nasser, thanks for the report. I've fixed this issue in the TeX4ht sources and posted answer on TeX.sx.

Michal Hoftich <michal_h21>
Sat Apr 23 05:23:31 2022, original submission:

reference and screen shots at

I am splitting my large document into smaller separate documents, where now each chapter will be compiled separately, as separate PDF file (not using \include or input, but as a separate document with its own title and \begin{document})

But I still need to have reference from a main document to the other documents to build tables and references.

So I am using xr-hyper. Which worked well. I have this MWE tree to show my current set

main.tex
|
+------ chapter_1/ch1.tex
+------ chapter_2/ch2.tex

I put each chapter in separate folder to reduce clutter. This is what my main.tex, ch1.tex and ch2.tex look like

%main.tex
\documentclass{book}
\usepackage{xr-hyper}
\usepackage{hyperref}
\externaldocument{chapter_1/ch1} %notice folder name included
\externaldocument{chapter_2/ch2}
\begin{document}
See problem \hyperref[1]{1} below and problem \hyperref[2]{2}.
\end{document}

and ch1.tex which is inside folder chapter_1/

\documentclass{book}
\usepackage{xr-hyper}
\usepackage{hyperref}

\begin{document}
\chapter{chapter 1 in document ch1.tex}
\section{some section name}
\subsection{problem 1 from some book}
\label{1}
This is problem 1
\end{document}

and ch2.tex

\documentclass{book}
\usepackage{xr-hyper}
\usepackage{hyperref}
\begin{document}

\chapter{chapter 2 in document ch2.tex}
\section{some section name}
\subsection{problem 2 from some book}
\label{2}
This is problem 2
\end{document}

Next, compiled ch1.tex,ch2.tex and finally main.tex in this order using lualatex. Now I opened main.pdf and the links there correct and work.

Next, I did the same for tex4ht. Compiled ch1.tex,ch2.tex then main.tex in this order, all using the commands

make4ht ch1.tex 'mathjax,htm'
make4ht ch2.tex 'mathjax,htm'
make4ht main.tex 'mathjax,htm'

Then opened main.htm and it shows correct output

Now here is the problem, when I click on the link, it drops the folder name from the link. The link for 1 says

ch1.htm#x1-30001.1.1

chapter_1/ch1.htm#x1-30001.1.1

So the link did not work.

By manually editing the link and adding the folder name chapter_1/ to it, then it works and opens chapter 1 page correctly.

I can fix all of this by having all separate documents (main.tex, ch1.tex and ch2.tex) in same top level folder.

But to reduce clutter, and since I have many chapters that I want to make now completely separate PDF files for each chapter (but still have cross references from main document to them), it will be much better to have each in separate folder to make it easier to manage and reduce clutter in same folder, since I need to compile each one separately now.

This looks like a bug in tex4ht, in that it drops the folder name from the reference. Or Am I doing something wrong?

TL 2021

Nasser M. Abbasi <nma123>

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• -unavailable- added by michal_h21 (Posted a comment)
• -unavailable- added by nma123 (Submitted the item)
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