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# Overview

This document is a summary of the paper Speech recognition with deep recurrent neural networks by A. Graves, A.-R. Mohammed, and G. Hinton, which appeared in the 2013 IEEE International Conference on Acoustics, Speech, and Signal Processing (ICASSP). The first and third authors are Artificial Neural Network (ANN) researchers, while Mohammed works in the field of automatic speech recognition.

The paper presents the application of bidirectional multilayer Long Short-term Memory (LSTM) ANNs with 1–5 layers to phoneme recognition on the TIMIT acoustic phonme corpus, which is the standard benchmark in the field of acoustic recognition, extending the previous work by Mohammed and Hinton on this topic using Deep Belief Networks. The TIMIT corpus contains audio recordings of 6300 sentences spoken by 630 (American) English speakers from 8 regions with distinct dialects, where each recording has accompanying manually labelled transcriptions of the phonemes in the audio clips alongside timestamp information. The empirical classification accuracies reported in the literature before the publication of this paper are shown in the timeline below (note that in this figure, the accuracy metric is 100% - PER, where PER is the phoneme classification error rate).

The deep LSTM networks presented with 3 or more layers obtain phoneme classification error rates of 19.6% or less, with one model obtaining 17.7%, which was the best result reported in the literature at the time, outperforming the previous record of 20.7% achieved by Mohammed et al. Furthermore, the error rate decreases monotonically with LSTM network depth for 1–5 layers. While the bidirectional LSTM model performs well on the TIMIT corpus, any potential advantage of bidirectional over unidirectional LSTM network models, cannot be determined from this paper since the performance comparison is across different numbers of iterations taken in the optimization algorithm used to train the models, and multiple trials for statistical validity were not performed.

File:timit.png
Timeline of percentage phoneme recognition accuracy achieved on the core TIMIT corpus, from Lopes and Perdigao, 2011.

## Motivation

Neural networks have been trained for speech recognition problems, however usually in combination with hidden Markov Models. The authors in this paper argue that given the nature of speech is an inherently dynamic process RNN should be the ideal choice for such a problem. There has been attempts to train RNNs for speech recognition <ref>A. Graves, S. Fernandez, F. Gomez, and J. Schmidhuber, “Connectionist Temporal Classification: Labelling Un-segmented Sequence Data with Recurrent Neural Networks,” in ICML, Pittsburgh, USA, 2006.</ref> <ref> A. Graves, Supervised sequence labelling with recurrentneural networks, vol. 385, Springer, 2012.</ref> <ref> A. Graves, “Sequence transduction with recurrent neural networks,” in ICML Representation Learning Work-sop, 2012.</ref> and RNNs with LSTM for recognizing cursive handwriting <ref> A. Graves, S. Fernandez, M. Liwicki, H. Bunke, and J. Schmidhuber, “Unconstrained Online Handwriting Recognition with Recurrent Neural Networks,” in NIPS.2008.</ref> but neither has made an impact on the speech recognition. The authors drew inspiration from Convolutional Neural Networks, where multiple layers are stacked on top of each other to combine LSTM and RNNs together.

However instead of using a conventional RNN which only considers previous contexts, a Bidirectional RNN <ref> M. Schuster and K. K. Paliwal, “Bidirectional Recurrent Neural Networks,”IEEE Transactions on Signal Processing, vol. 45, pp. 2673–2681, 1997.</ref> was used to consider both forward and backward contexts. This is due in part because the authors saw no reason not to exploit future contexts since the speech utterances are transcribed at once. Additionally BRNN has the added benefit of being able to consider the entire forward and context, not just some predefined window of forward and backward contexts.

# Deep RNN models considered by Graves et al.

In this paper, Graves et al. use deep LSTM network models. We briefly review recurrent neural networks, which form the basis of the more complicated LSTM network that has composite $\mathcal{H}$ functions instead of sigmoids and additional parameter vectors associated with the state of each neuron. Finally, a description of bidirectional ANNs is given, which is used throughout the numerical experiments.

## Recurrent Neural Networks

Recall that a standard 1-layer recurrent neural network (RNN) computes the hidden vector sequence ${\boldsymbol h} = ({{\mathbf{h}}}_1,\ldots,{{\mathbf{h}}}_T)$ and output vector sequence ${{\boldsymbol {{\mathbf{y}}}}}= ({{\mathbf{y}}}_1,\ldots,{{\mathbf{y}}}_T)$ from an input vector sequence ${{\boldsymbol {{\mathbf{x}}}}}= ({{\mathbf{x}}}_1,\ldots,{{\mathbf{x}}}_T)$ through the following equation where the index is from $t=1$ to $T$:

${{\mathbf{h}}}_t = \begin{cases} {\mathcal{H}}\left({{{{\mathbf{W}}}_{x h}}}{{\mathbf{x}}}_t + {{{\mathbf{b_{h}}}}}\right) &\quad t = 1\\ {\mathcal{H}}\left({{{{\mathbf{W}}}_{x h}}}{{\mathbf{x}}}_t + {{{{\mathbf{W}}}_{h h}}}{{\mathbf{h}}}_{t-1} + {{{\mathbf{b_{h}}}}}\right) &\quad \text{else} \end{cases}$

and

${{\mathbf{y}}}_t = {{{{\mathbf{W}}}_{h y}}}{{\mathbf{h}}}_t + {{{\mathbf{b_{y}}}}}.$

The ${{\mathbf{W}}}$ terms are the parameter matrices with subscripts denoting the layer location (e.g. ${{{{\mathbf{W}}}_{x h}}}$ is the input-hidden weight matrix), and the offset $b$ terms are bias vectors with appropriate subscripts (e.g. ${{{\mathbf{b_{h}}}}}$ is hidden bias vector). The function ${\mathcal{H}}$ is an elementwise vector function with a range of $[0,1]$ for each component in the hidden layer.

This paper considers multilayer RNN architectures, with the same hidden layer function used for all $N$ layers. In this model, the hidden vector in the $n$th layer, ${\boldsymbol h}^n$, is generated by the rule

${{\mathbf{h}}}^n_t = {\mathcal{H}}\left({{\mathbf{W}}}_{h^{n-1}h^{n}} {{\mathbf{h}}}^{n-1}_t + {{\mathbf{W}}}_{h^{n}h^{n}} {{\mathbf{h}}}^n_{t-1} + {{{\mathbf{b_{h}}}}}^n \right),$

where ${\boldsymbol h}^0 = {{\boldsymbol {{\mathbf{x}}}}}$. The final network output vector in the $t$th step of the output sequence, ${{\mathbf{y}}}_t$, is

${{\mathbf{y}}}_t = {{{\mathbf{W}}}_{h^N y}} {{\mathbf{h}}}^N_t + {{{\mathbf{b_{y}}}}}.$

This is pictured in the figure below for an arbitrary layer and time step.

File:rnn graves.png
Fig 1. Schematic of a Recurrent Neural Network at an arbitrary layer and time step.

## Long Short-term Memory Architecture

Graves et al. consider a Long Short-Term Memory (LSTM) architecture from Gers et al. . This model replaces $\mathcal{H}(\cdot)$ by a composite function that incurs additional parameter matrices, and hence a higher dimensional model. Each neuron in the network (i.e.  row of a parameter matrix ${{\mathbf{W}}}$) has an associated state vector ${{\mathbf{c}}}_t$ at step $t$, which is a function of the previous ${{\mathbf{c}}}_{t-1}$, the input ${{\mathbf{x}}}_t$ at step $t$, and the previous step’s hidden state ${{\mathbf{h}}}_{t-1}$ as

${{\mathbf{c}}}_t = {{\mathbf{f}}}_t \circ {{\mathbf{c}}}_{t-1} + {{\mathbf{i}}}_t \circ \tanh \left({{{\mathbf{W}}}_{x {{\mathbf{c}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{h {{\mathbf{c}}}}} {{\mathbf{h}}}_{t-1} + {{\mathbf{b}}}_{{\mathbf{c}}}\right)$

where $\circ$ denotes the Hadamard product (elementwise vector multiplication), the vector ${{\mathbf{i}}}_t$ denotes the so-called input vector to the cell that generated by the rule

${{\mathbf{i}}}_t = \sigma\left({{{\mathbf{W}}}_{x {{\mathbf{i}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{h {{\mathbf{i}}}}} {{\mathbf{h}}}_{t-1} + {{{\mathbf{W}}}_{{{\mathbf{c}}}{{\mathbf{i}}}}} {{\mathbf{c}}}_{t-1} + {{\mathbf{b}}}_{{\mathbf{i}}}\right),$

and ${{\mathbf{f}}}_t$ is the forget gate vector, which is given by

${{\mathbf{f}}}_t = \sigma\left({{{\mathbf{W}}}_{x {{\mathbf{f}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{h {{\mathbf{f}}}}} {{\mathbf{h}}}_{t-1} + {{{\mathbf{W}}}_{{{\mathbf{c}}}{{\mathbf{f}}}}} {{\mathbf{c}}}_{t-1} + {{\mathbf{b}}}_{{\mathbf{f}}}\right)$

Each ${{\mathbf{W}}}$ matrix and bias vector ${{\mathbf{b}}}$ is a free parameter in the model and must be trained. Since ${{\mathbf{f}}}_t$ multiplies the previous state ${{\mathbf{c}}}_{t-1}$ in a Hadamard product with each element in the range $[0,1]$, it can be understood to reduce or dampen the effect of ${{\mathbf{c}}}_{t-1}$ relative to the new input ${{\mathbf{i}}}_t$. The final hidden output state is then

${{\mathbf{h}}}_t = \sigma\left({{{\mathbf{W}}}_{x {{\mathbf{o}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{h {{\mathbf{o}}}}} {{\mathbf{h}}}_{t-1} + {{{\mathbf{W}}}_{{{\mathbf{c}}}{{\mathbf{o}}}}} {{\mathbf{c}}}_{t} + {{\mathbf{b}}}_{{\mathbf{o}}}\right)\circ \tanh({{\mathbf{c}}}_t)$

In all of these equations, $\sigma$ denotes the logistic sigmoid function. Note furthermore that ${{\mathbf{i}}}$, ${{\mathbf{f}}}$, ${{\mathbf{o}}}$ and ${{\mathbf{c}}}$ all of the same dimension as the hidden vector $h$. In addition, the weight matrices from the cell to gate vectors (e.g. ${{{\mathbf{W}}}_{{{\mathbf{c}}}{{\mathbf{i}}}}}$) are diagonal, such that each parameter matrix is merely a scaling matrix.

## Bidirectional RNNs

A bidirectional RNN adds another layer of complexity by computing 2 hidden vectors per layer. Neglecting the $n$ superscripts for the layer index, the forward hidden vector is determined through the conventional recursion as

${\overrightarrow{{{\mathbf{h}}}}}_t = {\mathcal{H}}\left({{{\mathbf{W}}}_{x {\overrightarrow{{{\mathbf{h}}}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{{\overrightarrow{{{\mathbf{h}}}}}{\overrightarrow{{{\mathbf{h}}}}}}} {\overrightarrow{{{\mathbf{h}}}}}_{t-1} + {{\mathbf{b_{{\overrightarrow{{{\mathbf{h}}}}}}}}} \right),$

while the backward hidden state is determined recursively from the reversed sequence $({{\mathbf{x}}}_T,\ldots,{{\mathbf{x}}}_1)$ as

${\overleftarrow{{{\mathbf{h}}}}}_t = {\mathcal{H}}\left({{{\mathbf{W}}}_{x {\overleftarrow{{{\mathbf{h}}}}}}} {{\mathbf{x}}}_t + {{{\mathbf{W}}}_{{\overleftarrow{{{\mathbf{h}}}}}{\overleftarrow{{{\mathbf{h}}}}}}} {\overleftarrow{{{\mathbf{h}}}}}_{t+1} + {{\mathbf{b_{{\overleftarrow{{{\mathbf{h}}}}}}}}}\right).$

The final output for the single layer state is then an affine transformation of ${\overrightarrow{{{\mathbf{h}}}}}_t$ and ${\overleftarrow{{{\mathbf{h}}}}}_t$ as ${{\mathbf{y}}}_t = {{{\mathbf{W}}}_{{\overrightarrow{{{\mathbf{h}}}}}y}} {\overrightarrow{{{\mathbf{h}}}}}_t + {{{\mathbf{W}}}_{{\overleftarrow{{{\mathbf{h}}}}}y}} {\overleftarrow{{{\mathbf{h}}}}}_t + {{{\mathbf{b_{y}}}}}.$

The combination of LSTM with bidirectional dependence has been previously used by Graves and Schmidhuber, and is extended to multilayer networks in this paper. The motivation for this is to use dependencies on both prior and posterior vectors in the sequence to predict a given output at any time step. In other words, a forward and backward context is used.

# Network Training for Phoneme Recognition

This section describes the phoneme classification experiments performed with the TIMIT corpus. An overview of the input timeseries audio data preprocessing into frequency domain vectors is given, and the optimization techniques are described.

## Frequency Domain Processing

Recall that for a real, periodic signal ${f(t)}$, the Fourier transform

${F(\omega)}= \int_{-\infty}^{\infty}e^{-i\omega t} {f(t)}dt$

can be represented for discrete samples ${f_0, f_1, \cdots f_{N-1}}$ as

$F_k\ = \sum_{n=0}^{N-1} x_n \cdot e^{-i \frac{k n \pi}{N}}$,

where ${F_k}$ are the discrete coefficients of the (amplitude) spectral distribution of the signal $f$ in the frequency domain. This is a particularly powerful representation of audio data, since the modulation produced by the tongue and lips when shifting position while the larynx vibrates induces the frequency changes that make up a phonetic alphabet. A particular example is the spectrogram below. A spectrogram is a heat map representation of a matrix of data in which each pixel has intensity proportional to the magnitude of the matrix entry at that location. In this case, the matrix is a windowed Fourier transform of an audio signal spectrum such that the frequency of the audio signal as a function of time can be observed. This spectrogram shows the frequencies of my voice while singing the first bar of Hey Jude; the bright pixels below 200 Hz show the base note, while the fainter lines at integer multiples of the base notes show resonant harmonics.

File:spect.png
Spectrogram of the first bar of Hey Jude, showing the frequency amplitude coefficients changing over time as the intensity of the pixels in the heat map.

## Input Vector Format

For each audio waveform in the TIMIT dataset, the Fourier coefficients were computed with a sliding Discrete Fourier Transform (DFT). The window duration used was 10 ms, corresponding to $n_s = 80$ samples per DFT since each waveform in the corpus was digitally registered with a sampling frequency of $f_s = 16$ kHz, producing 40 unique coefficients at each timestep $t$, $\{c^{[t]}_k\}_{k=1}^{40}$. In addition, the first and second time derivatives of the coefficients between adjacent DFT windows were computed (the methodology is unspecified, however most likely this was performed with a numerical central difference technique). Thus, the input vector to the network at step $t$ was the concatenated vector

${{\mathbf{x}}}_t = [c^{[t]}_1, \frac{d}{dt}c^{[t]}_1, \frac{d^2}{dt^2}c^{[t]}_1, c_2^{[t]}, \frac{d}{dt}c^{[t]}_2, \frac{d^2}{dt^2}c^{[t]}_2 \ldots]^T.$

Finally, an additional preprocessing step was performed: each input vector was normalized such that the dataset had zero mean and unit variance.

## RNN Transducer

When building a speech recognition classifier it is important to note that the length of the input and output sequences are of different lengths (sound data to phonemes). Additionally, RNNs require segmented input data. One approach to solve both these problems is to align the output (label) to the input (sound data), but more often than not an aligned dataset is not available. In this paper, the Connectionist Temporal Classification (CTC) method is used to create a probability distribution between inputs and output sequences. This is augmented with an RNN that predicts phonemes given the previous phonemes. The two predictions are then combined into a feed-forward network. The authors call this approach an RNN Transducer. From the distribution of the RNN and CTC, a maximum likelihood decoding for a given input can be computed to find the corresponding output label.

$h(x) = \arg \max_{l \in L^{\leq T}} P(l | x)$

Where:

• $h(x)$: classifier
• $x$: input sequence
• $l$: label
• $L$: alphabet
• $T$: maximum sequence length
• $P(l | x)$: probability distribution of $l$ given $x$

The value for $h(x)$ cannot computed directly, it is approximated with methods such as Best Path, and Prefix Search Decoding, the authors has chosen to use a graph search algorithm called Beam Search.

## Network Output Layer

Two different network output layers were used, however most experimental results were reported for a simple softmax probability distribution vector over the set of $K = 62$ symbols, corresponding to the 61 phonemes in the corpus and an additional null symbol indicating that no phoneme distinct from the previous one was detected. This model is referred to as a Connectionist Temporal Classification (CTC) output function. The other (more complicated) output layer was not rigorously compared with a softmax output, and had nearly identical performance; this summary defers a description of this method, a so-called RNN transducer to the original paper.

## Network Training Procedure

The parameters in all ANNs were determined using Stochastic Gradient Descent with a fixed update step size (learning rate) of $10^{-4}$ and a Nesterov momentum term of 0.9. The initial parameters were uniformly randomly drawn from $[-0.1,0.1]$. The optimization procedure was initially run with data instances from the standard 462 speaker training set of the TIMIT corpus. As a stopping criterion for the training, a secondary testing subset of 50 speakers was used on which the phoneme error rate (PER) was computed in each iteration of the optimization algorithm. The initial training phase for each network was halted once the PER stopped decreasing on the training set; using the parameters at this point as the initial weights, the optimization procedure was then re-run with Gaussian noise with zero mean and $\sigma = 0.075$ added element-wise to the parameters in for each input vector instance $({{\mathbf{x}}}_1,\ldots, {{\mathbf{x}}}_T)$ as a form of regularization. The second optimization procedure was again halted once the PER stopped decreasing on the testing dataset. Multiple trials in each of these numerical experiments were not performed, and as such, the variability in performance due to the initial values of the parameters in the optimization routine is unknown.

# TIMIT Corpus Experiments & Results

## Numerical Experiments

To investigate the performance of the Bidirectional LSTM architecture as a function of depth, numerical experiments were conducted with networks with $N \in \{1,2,3,5\}$ layers and 250 hidden units per layer. These are denoted in the paper by the network names CTC-$N$L-250H (where $N$ is the layer depth), and are summarized with the number of free model parameters in the table below.

Network Name # of parameters
CTC-1l-250h 0.8M
CTC-2l-250h 2.3M
CTC-3l-250h 3.8M
CTC-5l-250h 6.8M

Additional experiments included: a 1-layer model with 3.8M weights, a 3-layer bidirectional ANN with $\tanh$ activation functions rather than LSTM, a 3-layer unidirectional LSTM model with 3.8M weights (the same number of free parameters as the bidirectional 3-layer LSTM model). Finally, two experiments were performed with a bidirectional LSTM model with with 3 hidden layers each with 250 hidden units, and an RNN transducer output function. One of these experiments using uniformly randomly initialized parameters, and the other using the final (hidden) parameter weights from the CTC-3L-250H model as the initial paratemer values in the optimization algorithm. The names of these experiments are summarized below, where TRANS and PRETRANS denote the RNN transducer experiments initialized randomly, and using (pretrained) parameters from the CTC-3L-250H model, respectively. The suffices UNI and TANH denote the unidirectional and $\tanh$ networks, respectively.

Network Name # of parameters
CTC-1l-622h 3.8M
CTC-3l-421h-uni 3.8M
CTC-3l-500h-tanh 3.7M
Trans-3l-250h 4.3M
PreTrans-3l-250h 4.3M

## Results

The percentage phoneme error rates and number of epochs in the SGD optimization procedure for the LSTM experiments on the TIMIT dataset with varying network depth are shown below. The PER can be seen to decrease monotonically, however there is negligible difference between 3 and 5 layers—it is possible that the 0.2% difference is within statistical fluctuations induced by the SGD optimization routine and initial parameter values. Note that the allocation of the epochs into either the initial training without noise or the second optimization routine with Gaussian noise added (or both) is unspecified in the paper.

Network # of Parameters Epochs PER
CTC-1l-250h 0.8M 82 23.9%
CTC-2l-250h 2.3M 55 21.0%
CTC-3l-250h 3.8M 124 18.6%
CTC-5l-250h 6.8M 150 18.4%

The second set of PER results are shown below. The unidirectional LSTM architecture CTC-3L-421H-UNI achieves an error rate that is greater than the CTC-3L-250H model by 1 percentage point. No further comparative experiments between unidirectional and bidirectional models were given, however, and the margin of statistical uncertainty is unknown; thus the 1% (absolute) difference may or may not be significant. The TRANS-3L-250H model achieves a nearly identical PER to the CTC softmax model (0.3%) difference, however note that it has 0.5M more parameters due to the additional classification network at the output, and is hence not an entirely fair comparison since it has a greater dimensionality. The pretrained model PRETRANS-3L-250H also has 4.3M parameters and sees the best performance with a 17.5% error rate. Note that the difference in training of these two RNN transducer models is primarily in their initialization: the PRETRANS model was initialized using the trained weights of the CTC-3L-250H model (for the hidden layers). Thus, this difference in error rate of 0.6% is the direct result of a different starting iterates in the optimization procedure, which must be kept in mind when comparing between models.

Network # of Parameters Epochs PER
CTC-1l-622h 3.8M 87 23.0%
CTC-3l-500h-tanh 3.7M 107 37.6%
CTC-3l-421h-uni 3.8M 115 19.6%
Trans-3l-250h 4.3M 112 18.3%
PreTrans-3l-250h 4.3M 144 17.7%

# References

A. Graves, A.-r. Mohamed, and G. Hinton, “Speech recognition with deep recurrent neural networks,” in Acoustics, Speech and Signal Processing (ICASSP), 2013 IEEE International Conference on, pp. 6645–6649, IEEE, 2013.

C. Lopes and F. Perdigão, “Phone recognition on the timit database,” Speech Technologies/Book, vol. 1, pp. 285–302, 2011.

A.-r. Mohamed, G. E. Dahl, and G. Hinton, “Acoustic modeling using deep belief networks,” Audio, Speech, and Language Processing, IEEE Transactions on, vol. 20, no. 1, pp. 14–22, 2012.

F. A. Gers, N. N. Schraudolph, and J. Schmidhuber, “Learning precise timing with lstm recurrent networks,” The Journal of Machine Learning Research, vol. 3, pp. 115–143, 2003.

A. Graves and J. Schmidhuber, “Framewise phoneme classification with bidirectional lstm networks,” in Neural Networks, 2005. IJCNN’05. Proceedings. 2005 IEEE International Joint Conference on, vol. 4, pp. 2047–2052, IEEE, 2005.


text:
Topological basis

Why in ##l_2## space basis ##|1 \rangle=[1 \; 0 \; 0 ...]^{\mathsf{T}}##,
##|2 \rangle=[0 \; 1 \; 0 ...]^{\mathsf{T}}##, ##|3 \rangle=[0 \; 0 \; 1 \; 0...]^{\mathsf{T}}##... is called topological basis?

Answers and Replies

Related Topology and Analysis News on Phys.org
martinbn
Science Advisor
Because, not every vector in ##l_2## is a finite linear combination of these basis vectors. Some are infnite combinations, convergence is needed, hence topological.

Math_QED
fresh_42
Mentor
Because, not every vector in ##l_2## is a finite linear combination of these basis vectors. Some are infnite combinations, convergence is needed, hence topological.
But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.

martinbn
Science Advisor
But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.
What I meant is that if you take the span of those vectors you don't get the whole space you need to take the closer of the span.

fresh_42
Mentor
I think that this is an abuse of language. Those vectors do not constitute a topological basis, neither do they constitute a vector space basis. My assumption is therefore that the author uses topological basis in contrast to vector space basis, in order to get rid of the finiteness constriction of vector space bases, but they are neither.

They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.

Math_QED
martinbn
Science Advisor
I think that this is an abuse of language. Those vectors do not constitute a topological basis, neither do they constitute a vector space basis. My assumption is therefore that the author uses topological basis in contrast to vector space basis, in order to get rid of the finiteness constriction of vector space bases, but they are neither.

They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.
Not sure what you mean! No one was talking about the direct sum nor the direct product. And, yes, it is just i name the author uses. I think it is quite standard.

fresh_42
Mentor
The basis of a topology is defined as subset of open sets. To distinguish it from a topological basis is hopefully not standard! And if, I will no longer use english and topology at the same time.
No one was talking about the direct sum nor the direct product.
Unfortunately. This is what it is about.

martinbn
Science Advisor
The basis of a topology is defined as subset of open sets. To distinguish it from a topological basis is hopefully not standard! And if, I will no longer use english and topology at the same time.
No, it is not about bases of a topology, but about bases of topological vector spaces.
Unfortunately. This is what it is about.
I still don't understand the ##l_2## space is not the direct sum nor the direct product.

fresh_42
Mentor
No, it is not about bases of a topology, but about bases of topological vector spaces.
In this case, the wording is more than confusing.
I still don't understand the ##l_2## space is not the direct sum nor the direct product.
As ##\mathbb{R}##-module?

martinbn
Science Advisor
In this case, the wording is more than confusing.
It can be, that's why the context is important.
As ##\mathbb{R}##-module?
The direct sum is the vector space of all finite sequences of real numbers. The direct product of all sequences. The ##l_2## of all square summable sequences.

fresh_42
Mentor
That's what I said. They generate the ##\mathbb{R}## module. No basis in either meaning, and not very related to one of the structures of ##\ell_2##. All it does is: We can write the elements of ##\ell_2## as ...

Calling it a topological basis when a basis of a topology is something else, is a very bad choice.

Math_QED
Science Advisor
Homework Helper
2019 Award
I agree that topological basis is very bad use of language because you immediately think about a collection of opens in a topological space.

I also agree with @martinbn post #2. This is what the author had in mind.

Since ##l^2## is a Hilbert space, a more appropriate name would be "orthonormal basis".

fresh_42
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
But it's still not a basis by the regular vector space definition, which the author thinks is more important to distinguish.

Infrared
Science Advisor
Gold Member
I think this is often called a Schauder basis (https://en.wikipedia.org/wiki/Schauder_basis). I haven't seen the term "topological basis" used for this but it makes sense (although it's not a basis for a topology...)

Here's a page from nLab, where the term "topological basis" (not in the sense "basis for the topology") is used, and it seems to be differentiated from both the Hamel and the Schauder basis (although I'm not exactly sure I understand the difference):
https://ncatlab.org/nlab/show/basis+in+functional+analysis

Infrared
Infrared
Science Advisor
Gold Member
@Dragon27 I think the difference is something like this: Let ##V=C([0,1])##, with the sup norm. Then ##1,x,x^2,\ldots## is not a Schauder basis because not every continuous function is a power series, but it is a topological basis according to your link because every continuous function on a finite interval can be uniformly approximated by polynomials (and none of these monomials can be uniformly approximated by others).

Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
The link actually claims that you can uniformly approximate x on an interval with higher degree monomials at the bottom. But it does give a similar example using Fourier series.

Infrared
Infrared
Science Advisor
Gold Member
Thanks, I should have read a bit further before posting...

format: plain


text:
# ODE plotting in Maple ARG!

by OrigGan
Tags: maple, plotting
 P: 2 So I was given the following problem to do in maple: In one ty-graph plot the solutions of y' = cos(pi*ty) for a sequence of 21 equally distributed initial points y(0) between 0 and 4. Hint: You may use commands such as seq, dsolve, numeric, odeplot and display. I started off with: >x:=diff(y(t),t)=cos(pi*t*y(t)) Then, I did: >dsolve(x,y(t)) This just gives a blank space, which Im pretty sure means maple cant find an answer. From reading some examples online and from the hint above I think I may need to include numeric in dsolve but I cant get that to work without an initial condition. I also know the sequence will obviously be 0..4,0.2 but then I dont understand what it means by the y(0) part. Any help in putting this plot together is much appreciated

 Related Discussions Math & Science Software 0 Differential Equations 1 Calculus & Beyond Homework 0 Math & Science Software 2 Math & Science Software 0


text:
# Why the columns of $A$ are linearly dependent set?

If $A_{m\times n}$ is a matrix such that $\sum_{j=1}^n a_{ij}=0$ for each $i=1,2,…,m,$ then why the columns of $A$ are linearly dependent set, and hence $\operatorname {rank}(A)<n$?

• It's hard to understand what you wrote. LaTeXize it. – DonAntonio Sep 22 '13 at 12:57
• She tried, @DonAntonio, but was missing braces around subscripts and bounds of summation, making it difficult to make out. Hopefully fixed (there was also some copy and paste inserted in dollar signs, so I'm hoping I got the summation correctly). – amWhy Sep 22 '13 at 13:00
• @amWhy, thank you for fixing my question, it's exactly what I tried to wrote. – Diane Vanderwaif Sep 22 '13 at 13:06
• You're welcome, @Diane. Note that when there is more than one character in a subscript, you need to enclose the subscript in braces. E.g. $A_{m \times n}$ = A_{m \times n} – amWhy Sep 22 '13 at 13:14

If $m < n$, then

$$\operatorname{rank} A \le \min \{m,n\} = m < n.$$

Let us now assume that $m \ge n$. Note that

$$A \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{m \times 1},$$

Aiming for a contradiction, we assume that $A$ if of a full rank (i.e., $\operatorname{rank} A = n$). That means that $A$ has a submatrix $B$ of order $n$ such that $B$ is of a full rank (made of the linearly independent rows in $A$), i.e., $B$ is invertible. We know that

$$B \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0_{n \times 1},$$

which is a contradiction with the invertibility of $B$, so $A$ is not of a full rank.

Hint: the equation that is given actually gives an explicit linear dependence relation between the columns.

• Since I am a pedantic kind of mathematician, I should add: provided that $n>0$, i.e., provided there are any columns at all. For $n=0$ that statement to prove is wrong: no columns makes an (empty) linearly independent set. But you may forget I said that. – Marc van Leeuwen Sep 22 '13 at 13:05

I assume $m=n$ just to make it look good .

we have $\sum_{j=1}^m a_{ij}=0$ for all $1\leq i\leq m$

i..e, $\sum_{j=1}^m a_{1j}+\sum_{j=1}^m a_{2j}+\dots+\sum_{j=1}^m a_{mj}=0$

i.e., $(a_{11}+a_{12}+\dots+a_{1m})+(a_{21}+a_{22}+\dots+a_{2m})+\dots +(a_{m1}+a_{m2}+\dots+a_{mm})=0$

i.e., $(a_{11}+a_{21}+\dots+a_{m1})+(a_{12}+a_{22}+\dots+a_{m2})+\dots +(a_{1m}+a_{2m}+\dots+a_{mm})=0$

Do you see "the next step" would conclude that columns are linearly dependent. (??)


text:
# How to prove premise-conclusion statements with rules of inference?

I have a question here:

1. For each of the premise-conclusion pairs below, give a valid step-by-step argument (proof) along with the name of the inference rule used in each step. For examples, see pages 73 and 74 in textbook.

First question:

(a) Premise: {¬p ∨ q → r, s ∨ ¬q, ¬t, p → t, ¬p ∧ r → ¬s}, conclusion: ¬q.

I only got this far:

¬p v ¬q v r

I have absolutely no idea what to do next. If someone could give me some general tips to solve this problem and others like it it would be great. I know, i only did one step..

Help is very much appreciated

Starting from the simplest atoms and working down:

• The simplset atom is $\neg t$. What does that give me? Where does $t$ appear? Oh look! From $\neg t$ and $p\implies t$, you have $\neg p$
• Ok, now I have $\neg p$ as well as $\neg t$. Where does $p$ appear? Oh! From $\neg p \lor q\to r$, I now have $r$, since $\neg p\lor q$ is true!.
• Now, where do $p$ and $r$ appear? Oh, in the last premise! $\neg p \land r\to \neg s$ gives me $\neg s$, since I know $\neg p\land r$ is true!

Starting from the conclusion and working up:

• Where does $s$ appear as a conclusion? Ok, it appears either in $s\lor \neg q$ or it appears in $\neg p \land r\to \neg s$. Well, my intuition tells me that the second one looks more promising, since I don's see how I could disprove $\neg q$.
• So, how can I prove $\neg p \land r$? Well, I need to prove both $r$ and $\neg p$. Let's start with $r$ first.
• I can prove $r$ if I prove $\neg p \lor q$, since $\neg p \lor q \to r$. Oh, but I have to prove $\neg p$ anyway, so as long as I can prove $\neg p$, I'm done!
• There is no $\neg p$ anywhere to prove. Is this a problem? Well, no, since $p\to t$ is the same as $\neg t\to \neg p$. OK, so I only need to prove $\neg t$. How can I prove that? Oh! I don't have to ! It's an axiom! Well, job done.
• Why do we need to look at $\neg$$t? doesnt p⟹t mean \neg$$p$ anyway? I'm confused on this.. Ok since $\neg$$p is true, then \neg$$p$ v $q$ has to be true. I'm with you so far. I'm getting confused when you say that you have $r$ since $TRUE$$\rightarrow$$r$ – Andrew Kor Oct 12 '15 at 18:30
• @AjeetKljh Certainly not. For example, if $p=\text{water is wet}$, and $t=\text{the earth is round}$, then the statement $p\to t$ is true, but the statement $\neg p$ is not true. – 5xum Oct 12 '15 at 18:32
• @AjeetKljh If $A\to B$ is true and $A$ is true, then $B$ is true. This rule of inference is called modus ponens. – 5xum Oct 12 '15 at 18:40
• In my book it says modus tollens is $\neg$$q | p$$\rightarrow$$q | therefore \neg$$p$. That implies $\neg$$p$ is true? – Andrew Kor Oct 12 '15 at 18:45
• Yes, you prove $\neg p$ by using modus tollens. Modus tollens says that if $A\to B$ is true and if $\neg B$ is true, then $\neg A$ is true. In your case, $p\to t$ is true and $\neg t$ is true, so $\neg p$ is also true. – 5xum Oct 12 '15 at 18:49


text:
# What are the two kinds of verbs?

## Answer

There are two main kinds of verbs - action verbs (also known as transitive verbs) and linking verbs (also known as intransitive or state of being verbs). Action verbs imply action and generally represent the action of the subject. Linking verbs are verbs that link the subject of a sentence with the predicate. Examples of linking verbs are 'is, am, was, were, and be,' while examples of action verbs are 'run, leap, ride, roll, and scamper.'
Reference:
Q&A Related to "What are the two kinds of verbs?"
 Verbs showing action and verbs showing 'being' Action and linking. _ Transitive and intransitive. (Transitive - with object; instransitive - without an object) Examples: Jim runs http://wiki.answers.com/Q/What_are_the_two_kinds_o...
 Two kinds of action verbs are: Physical Action: The gardener feeds http://www.chacha.com/question/what-are-two-kinds-...
 Transitive and intransitive! http://answers.yahoo.com/question/index?qid=201203...
 1. Ask students what they think action verbs are. At this grade, at least half your class will be able to give you an answer like,"Action verbs tells us about an action or tell http://www.ehow.com/how_8404581_teach-verbs-grade-...
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text:
# How to wire an entire room / kitchen with LED light strips? [closed]

As the question suggests, I'm looking into adding LED light strips throughout the kitchen (over cabinet, under cabinet, and kick -- I think under the countertop would be neat, but harder to hide). From what I have been reading, I realize that to more accurately determine what is needed that I would need to know specifics about the lights (like wattage, amperage, run lengths, LEDs per foot/meter).

But without knowing those specifics yet, to try to better understand how feasible this is, here is my intended setup. The kitchen has cabinets on opposite walls, with an island in the middle, something like this (where the square brackets represent cabinets and the island):

[]  []  []
[]  []  []
[]  []  []
[]______[]

Is it possible to have everything wired together on one circuit, even though there's about 15' from one side of the kitchen to the other? Am I able to run a wire (maybe 16/5 -- thinking of using RGBW) under the floor/ceiling to get to the other side and then hook up the strips without worrying about voltage drop? I'm sure this is possible, I just don't want to spend a fortune to make it work.

As a side question, if anyone knows, is there such thing as a on/off/motion switch, that also works with wifi (for use with Google Assistant / Alexa)? The thought would be I could turn the lights on or off, or to a motion mode that detects presence in the room.

## closed as off-topic by JYelton, Finbarr, JRE, RoyC, laptop2dApr 25 at 16:46

• This question does not appear to be about electronics design within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• What kind of LED strips are you going to use, specifically, how many Watts per meter and how long should the strips be in total? This will dictate what power supply you will need for the strips. I don't think you will have to worry about voltage drop since usually those strips are wired up in parallel. – Egor Tamarin Apr 23 at 6:51
• Perhaps better on the Homeimprovement stack ie DIY – Solar Mike Apr 23 at 7:48
• I'm voting to close this question as off-topic because it is better suited for DIY – JYelton Apr 23 at 14:37
• Sorry, I'm new to this. Is there a way to transfer this question over? Or just copy-paste? – lessthancommon Apr 23 at 15:22


text:
# Unexpected behavior getting class-specific variable importance from a randomForest object

## Mar 30, 2015 · 1231 words · 6 minutes read R • random forest

One of the useful features of random forests is the ability to assess the importance of predictor variables. This can be the mean decrease in accuracy or mean decrease in node impurity when excluding this variable. While many statisticians, data scientists, and machine learnists are familiar with variable importance measures for random forests, I don’t often see many analyses that include class-specific variable importance.

I think this is a shame because for many problems variables may be more or less important for different classes. Understanding the overall importance of variables only gives a high-level understanding of the problem. Examining the importance of predictors on a single class provides insight into the features that differentiate that class from the others.

Getting the overall importance for variables in a random forest model in R is simple, it is not entirely obvious how to return the class-specific importance. In fact trying to extract a measure of class-specific variable importance can be incredibly frusturating (until you find the incredibly simple solution).

# Demonstration of the problem

Suppose that I want to fit a random forest model in R and assess the variable importance of my predictors. This is simple enough.

> my_model <- randomForest(Species ~ ., data=iris, importance=TRUE)
> importance(my_model)
setosa versicolor virginica MeanDecreaseAccuracy MeanDecreaseGini
Sepal.Length  6.710277  8.5874923  8.989345            12.107605         9.676950
Sepal.Width   4.501133  0.7066144  5.481982             5.368399         2.380181
Petal.Length 22.323162 34.0679333 28.312631            34.034319        42.895229
Petal.Width  22.203651 32.8962472 30.960037            34.227022        44.272395

Now if I want to get the variable importance for a specific class, but don’t know how, I would just turn to the documentation.

> ?importance

It turns out that the importance function takes an argument called class.

class – for classification problem, which class-specific measure to return.

This seems straight forward enough and most users would just do this

> importance(my_model, class='setosa')
setosa versicolor virginica MeanDecreaseAccuracy MeanDecreaseGini
Sepal.Length  6.710277  8.5874923  8.989345            12.107605         9.676950
Sepal.Width   4.501133  0.7066144  5.481982             5.368399         2.380181
Petal.Length 22.323162 34.0679333 28.312631            34.034319        42.895229
Petal.Width  22.203651 32.8962472 30.960037            34.227022        44.272395

but this is a trap and will return the exact same result as importance(my_model) did. Not only does this function not behave as a user would reasonably expect it to, there is no message or warning to the user.

At this point it is reasonable to expect the user to go back to the function documentation and look for an answer or perhaps example of how to properly request the class-specific variable importance. But the user won’t find the answer in the function documentation. Now the user will probably think carefully (but not carefully enough) and try to extract the class-specific variable importances manually. This is somewhat dangerous if the user decides to extract the importances from the model directly.

> my_model\$importance
setosa  versicolor   virginica MeanDecreaseAccuracy MeanDecreaseGini
Sepal.Length 0.036573992 0.025547632 0.046447034          0.036346581         9.676950
Sepal.Width  0.008046969 0.001263502 0.009825166          0.006397894         2.380181
Petal.Length 0.334837811 0.293303764 0.294833879          0.303596194        42.895229
Petal.Width  0.332024023 0.309245664 0.275216276          0.302861787        44.272395

Observe that these variable importances differ from the ones above returned by importance(my_model)! This is because importance contains an optional argument to scale the measures by their standard error that defaults to true. Scaling is generally a good idea so we might decide to get the variable importances for each class by indexing the result from the importance function.

> importance(my_model)[,'setosa']
Sepal.Length  Sepal.Width Petal.Length  Petal.Width
6.710277     4.501133    22.323162    22.203651

But why doesn’t the extractor function work as expected? We can examine this by digging into the function code.

> randomForest:::importance.randomForest

# Solution

The first hint to our question is found on line 11, allImp <- is.null(type) && hasImp. This checks whether the user has passed an argument to type, the type of importance measure to return, and that the model object has a matrix of importance values. So in addition to specifying which class we want importance measures for, we also have to specify the type of importance measure. If we go back to the function documentation we can find that value for type should either be 1 (mean decrease in accuracy) or 2 (mean decrease in node impurity). Unmentioned in the documentation, importance will only return the class-specific variable importance as a measure of mean decrease in accuracy.

> importance(fit, type=1, class='setosa')
setosa
Sepal.Length  6.710277
Sepal.Width   4.501133
Petal.Length 22.323162
Petal.Width  22.203651
> importance(fit, type=2, class='setosa')
Error in importance.randomForest(fit, type = 2, class = "setosa") :
No class-specific measure for that type

While not mentioned at all in the documentation for importance, the documentation for randomForest does make a note that only the class-specific measure computed is mean decrease in accuracy.

For classification, the first nclass columns are the class-specific measures computed as mean descrease [sic] in accuracy. The nclass + 1st column is the mean descrease [sic] in accuracy over all classes. The last column is the mean decrease in Gini index.

Since only one type of class-specific variable importance measure is available, it does not make sense to require an argument to type. Further it is not readily apparent from the function documentation that the function requires both arguments. It is only after reviewing the code and the randomForest documentation does it become clear what and why the behavior of the importance function is what it is.

Some people may shrug at this and say, “so what? so what if I have to pass type=1 to get the class-specific variable importance?” These people are missing the point. As we saw above importance(my_model, class='setosa') gave the exact same response as importance(my_model). There was absolutely no indication that the function was being used incorrectly (not as the author intended). This could easily be correct with a message:

> importance(my_model, class='setosa')
Warning in importance(my_model, class='setosa') :
class is non-null value but type is null
specify type=1 to get class-specific variable importance
setosa versicolor virginica MeanDecreaseAccuracy MeanDecreaseGini
Sepal.Length  6.710277  8.5874923  8.989345            12.107605         9.676950
Sepal.Width   4.501133  0.7066144  5.481982             5.368399         2.380181
Petal.Length 22.323162 34.0679333 28.312631            34.034319        42.895229
Petal.Width  22.203651 32.8962472 30.960037            34.227022        44.272395

Some people will argue that users should have to specify values for class and type. These people might say that requiring type=1 informs the user that the measure of variable importance they are receiving is the mean decrease in accuracy. This could be combated through either message or, preferably, classifying the function documentation. The obscurity of the documentation for importance is a bad thing, it makes users dig to find the answer (type=1 is required).

My final grief about this is demonstrated through the use of another function, varImpPlot, which creates a dotchart of variable importance. This function makes a direct call to importance. When a user requests a dotchart of variable importance for a given class, but does not explicitly pass type=1, the resulting plot contains the overall variable importance (side-by-side plots of mean decrease in accuracy and impurity). If a user passes a value to class, it is clear that the user wanted a plot of class-specific variable importances, but again no warning is given that the user (it’s always the user’s fault) is in fact using the function incorrectly.
Examining the importance of predictors for a single class gives understanding of the features that define that class. However the behavior and accompanying documentation for the importance function may dissuade less experienced R users from examining class-specific variable importance. Improvements could easily be made to change the behavior of the function (e.g. type default to 1 when class is not null) or clarify the function documentation (or preferably both).


text:
# [−][src]Function opencv::imgproc::blur

`pub fn blur(    src: &dyn ToInputArray,     dst: &mut dyn ToOutputArray,     ksize: Size,     anchor: Point,     border_type: i32) -> Result<()>`

Blurs an image using the normalized box filter.

The function smooths an image using the kernel:

$\texttt{K} = \frac{1}{\texttt{ksize.width*ksize.height}} \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 1 \\ \dots \\ 1 & 1 & 1 & \cdots & 1 & 1 \\ \end{bmatrix}$

The call `blur(src, dst, ksize, anchor, borderType)` is equivalent to `boxFilter(src, dst, src.type(), ksize, anchor, true, borderType)`.

## Parameters

• src: input image; it can have any number of channels, which are processed independently, but the depth should be CV_8U, CV_16U, CV_16S, CV_32F or CV_64F.
• dst: output image of the same size and type as src.
• ksize: blurring kernel size.
• anchor: anchor point; default value Point(-1,-1) means that the anchor is at the kernel center.
• borderType: border mode used to extrapolate pixels outside of the image, see #BorderTypes. #BORDER_WRAP is not supported.


text:
# tfa.image.interpolate_spline

Interpolate signal using polyharmonic interpolation.

tfa.image.interpolate_spline(
train_points,
train_values,
query_points,
order,
regularization_weight=0.0,
name='interpolate_spline'
)

The interpolant has the form

$$f(x) = \sum_{i = 1}^n w_i \phi(||x - c_i||) + v^T x + b.$$

This is a sum of two terms: (1) a weighted sum of radial basis function (RBF) terms, with the centers $$c_1, ... c_n$$, and (2) a linear term with a bias. The $$c_i$$ vectors are 'training' points. In the code, b is absorbed into v by appending 1 as a final dimension to x. The coefficients w and v are estimated such that the interpolant exactly fits the value of the function at the $$c_i$$ points, the vector w is orthogonal to each $$c_i$$, and the vector w sums to 0. With these constraints, the coefficients can be obtained by solving a linear system.

$$\phi$$ is an RBF, parametrized by an interpolation order. Using order=2 produces the well-known thin-plate spline.

We also provide the option to perform regularized interpolation. Here, the interpolant is selected to trade off between the squared loss on the training data and a certain measure of its curvature (details). Using a regularization weight greater than zero has the effect that the interpolant will no longer exactly fit the training data. However, it may be less vulnerable to overfitting, particularly for high-order interpolation.

Note the interpolation procedure is differentiable with respect to all inputs besides the order parameter.

We support dynamically-shaped inputs, where batch_size, n, and m are None at graph construction time. However, d and k must be known.

#### Args:

• train_points: [batch_size, n, d] float Tensor of n d-dimensional locations. These do not need to be regularly-spaced.
• train_values: [batch_size, n, k] float Tensor of n c-dimensional values evaluated at train_points.
• query_points: [batch_size, m, d] Tensor of m d-dimensional locations where we will output the interpolant's values.
• order: order of the interpolation. Common values are 1 for $$\phi(r) = r$$, 2 for $$\phi(r) = r^2 * log(r)$$ (thin-plate spline), or 3 for $$\phi(r) = r^3$$.
• regularization_weight: weight placed on the regularization term. This will depend substantially on the problem, and it should always be tuned. For many problems, it is reasonable to use no regularization. If using a non-zero value, we recommend a small value like 0.001.
• name: name prefix for ops created by this function

#### Returns:

[b, m, k] float Tensor of query values. We use train_points and train_values to perform polyharmonic interpolation. The query values are the values of the interpolant evaluated at the locations specified in query_points.


text:
Why do we require a topological space to be closed under finite intersection?

In the definition of topological space, we require the intersection of a finite number of open sets to be open while we require the arbitrary union of open sets to be open. why is this?

I'm assuming this has something to do with the following observation: $\cap_{n=1}^{\infty} (-\frac{1}{n},\frac{1}{n}) = \{0\}$ and there is some reason we don't want singletons to be considered open, I am wondering what this reason is. Am I thinking in the right direction here?

Thanks :)

-
With your edit, the equation is no longer true. The intersection is empty (and open). –  Jonas Meyer Sep 6 '10 at 3:27
Thanks! I thought I edited it properly but I guess I missed changing the 0. Its fixed now. –  WWright Sep 6 '10 at 3:37
No problem! I guess I better edit my answer to make sense. –  Jonas Meyer Sep 6 '10 at 3:39
@WWright: you are missing curly braces around the $0$ in the equation; you need to put two backslashes before the braces for them to show up. –  Arturo Magidin Sep 6 '10 at 3:44
Thanks for that! I'm still learning LaTeX so I tend to mess up things like that. –  WWright Sep 6 '10 at 4:09

You need to think about what the intuition behind open sets are. One way to think about it is through neighborhoods: an open set is a set which is a neighborhood of each of its points. What is a neighborhood of a point? A neighborhood of a point $x$ is a set that contains of all points that are "sufficiently close" to $x$ (what does "sufficiently close" mean? It depends on the situation; you think of different neighborhoods perhaps specifying different degrees of closeness). In particular, any set that contains a neighborhood of $x$ is itself a neighborhood of $x$. And specifying two degrees of closeness specifies another degree of closeness that makes sense (the smaller of the two at any given place, say).

So: if you think about open sets as sets that are neighborhoods of all of the points they contain, then it is natural that the union of any family of pen sets will be open: each point in the union is one of the open sets, and that open set is a neighborhood, and the union contains that neighborhood and so is itself a neighborhood. So the arbitrary union of open sets should still be open.

What about intersection? Well, if you take two open sets $O_1$ and $O_2$, and you consider a point $x$ in $O_1\cap O_2$, then $O_1$ contains all points that are $1$-sufficiently close to $x$, and $O_2$ contains all points that are $2$-sufficiently close to $x$ (with "$1$-sufficiently" and "$2$-sufficiently" describing the two degrees of closeness required), so $O_1\cap O_2$ will contains all points that are both $1$-sufficiently and $2$-sufficiently close to $x$, and so it contains all points that are "sufficiently close" to $x$ for some meaning of "sufficiently close", so it is also an open set. This gives you, inductively, any finite intersection.

But what about arbitrary intersections? Then you run into trouble, because specfying two degrees of "closeness" gives you a degree of closeness (the smaller one), but an infinite number of degrees of closeness may end up excluding everything! (Just as in your example, taking the intersection of all $(-\frac{1}{n},\frac{1}{n})$, which specify all points that are $\frac{1}{n}$-close to $0$, but the intersection excludes everything). So we don't want to require that an arbitrary intersection of neighborhoods be a neighborhood, and so we don't want to require that an arbitrary intersection of open sets be an open set.

-

If you think of it in the case of the topology on the real line - as soon as we leave the restriction from finite intersections we would get singletons as you have seen. However, any set of real numbers is a union of singletons that is $$A= \bigcup_{x\in A} \{ x \}.$$ It would follow that

each set would be open

each set would be closed

only finite sets would be compact

each function $f:\mathbb{R}\to X$ would be continuous, etc...

(Thanks Jonas!)

-
Finite sets would be compact. I also mentioned that singletons being open leads to the discrete topology in my post, but your last point helps drive home how bizarre this is. –  Jonas Meyer Sep 6 '10 at 4:59
@Jonas Meyer: Yes, I am just being stupid! :/ –  AD. Sep 6 '10 at 5:15
AD., only finite sets would be compact. To supplement your list: every convergent sequence would be eventually constant. –  Jonas Meyer Sep 6 '10 at 6:11
Jonas Meyer: You are totally right. (I have been awake all night with my 8 month baby, so I blame this on her ;) –  AD. Sep 6 '10 at 7:16

I think that one should have played around with metric spaces before dealing with topological spaces.

The finite intersection of open balls centered at $x$ is again an open ball; the radius is just the minimum of the finitely many radia. But if you take infinitely many balls, the radius is the infimum of infinitely many positive numbers, which will be $0$ in many situations (for example if the radia are $1/n$). A similar argument shows that you cannot prove in general that an infinite intersection of open subsets in a metric space is open again.

Since topological spaces are the natural generalization of metric spaces (replacing concrete distances by the concept of neighborhoods), the same is true for topological spaces and in the definition only finite intersections of open sets are allowed.

-

It is not that we require open sets to be closed under finite intersection...

Definitions are not invented out of thin air: people had topological spaces long before anyone thought of coming up with the definition of topological spaces and in the examples they had available, open sets were closed under finite intersection and that property was in fact quite useful to get things done. In other words, the definition of topological spaces (and the requirement you mention in the title along with it) was abstracted from examples, not imposed on them.

-
So in other words, "it seemed like a good idea at the time"? –  SamB Nov 28 '10 at 0:31
@SamB: no, not really: that is not at all what I mean. –  Mariano Suárez-Alvarez Nov 28 '10 at 1:16
Note to self: try to make jokes that are actually funny and/or true –  SamB Nov 28 '10 at 1:38

Minor correction: $[0,1/n]$ is not open in the usual topology of the real line. But you could get the same thing with The example $\cap_{n=1}^\infty (-1/n,1/n)=${0} is a good one. And yes, points are not open in the usual topology on the line. If they were, then every set, being the union of one-point sets, would be open. This is called the discrete topology.

I think this definition can be fairly well motivated by what happens in metric spaces, of which the real line is a special case. Call a set $U$ open if around each point in $U$ there is a ball (with respect to the metric) centered at that point and contained in $U$. Then you can show as an exercise that intersections of finite collections of open sets are open, and that arbitrary unions of open sets are open. However, the example above shows that you can't take arbitrary intersections. (In the case of the real line, "ball" just means open interval.)

But why does it work so well to define general topological spaces to satisfy this property of metric spaces? I don't know. For more on the motivation of defining topologies in terms of "open sets", you may be interested in this MathOverflow question.

-

In my opinion, our idea of topological space comes not from open sets, but from neighbourhoods. Actually, I think that the hole thing comes from the idea of convergence. A neighbourhood of a point $a$, is a set that is "big enough" to be considered a neighbourhood. I know, this is a cyclic definition. The family of all neighbourhoods of a point $a$ describes what are the sequences (nets, if you prefer) that converge to $a$.

A sequence $x_n$ converge to $a$ if given a neighbourhood $V$, there is an $N = N_V$ such that $x_n \in V$ for all $n \geq N$. If $x_n \rightarrow a$, and $V$ and $W$ are neighbourhoods of $a$, then you might as well consider $V \cap W$ as being a neighbourhood of $a$, where $N_{V \cap W} = \max(N_V, N_W)$.

I said that a neighbourhood is a "big enough" set. When you talk about convergence, you say:

Given a "big enough" set, no matter how small it is, there is an $N$...

The "no matter how small it is" is usually omitted, since it has no mathematical use. When you want a "no matter how small", you use $\varepsilon$ and $\delta$. Whereas when you want a "no matter how big", you use $M$ or $N$. ;-)

Suppose that you want to define convergence (of nets, to be correct) in terms of neighbourhoods. You choose a family of sets $\mathcal{S}$ that will define convergence just as in the previous paragraph. Now, what is the largest family $\mathcal{V}$, that includes $\mathcal{S}$, and such that the convergence defined by this new family is exactly the same. What sets should you include into $\mathcal{V}$?

1. First, build the family $\mathcal{B}$ of all finite intersections of sets in $\mathcal{S}$. This new family is a neighbourhood base for $a$. The reasoning for allowing finite intersections is that reasoning of the second paragraph of this post.

2. Now, if you are considering a set $V$ as a neighbourhood of $a$, you might as well consider any $W \supset V$, since you can use for $W$ the same $N$ you would use for $V$. This is the "no matter how small"... it means that sets that are too big will not make much of a difference.

As you have already pointed out, infinite intersections will not work. This is just a way of saying that $\sup N_V$ might be $\infty$, so you might not be able to get an $N$ for infinite intersections of neighbourhoods.

Open sets are just sets that are neighbourhood of all its points. Notice that what is described above is exactly how one usually construct the topology for metric spaces. First, you define a family $\mathcal{S}$ (actually, $\mathcal{B}$) that is the set of all balls centered at $a$. Then you define open sets as being those that are neighbourhood of all its points!!! People give to much attention to open sets... in general topology, one usually gets introduced to open sets as if they were something really natural. But actually, neighbourhoods seems much more natural for those coming from metric spaces...

Now, you have a good motivation for defining filters and nets, as well... :-)

An for nets, you have now a motivation for the fact that given two indexes $\alpha$ and $\beta$, you need to have a $\gamma$ such that $\gamma \geq \alpha$ and $\gamma \geq \beta$. This is the finite intersection property for neighbourhoods!!!

-

I think that finite intersection is necessary for a topological space because when you are dealing with infinite intersection then you are leaving almost everything. Lots of open sets in topology will disappear and that's alarming.

-
Closed under infinite intersections would mean MORE open sets. In particular, in the reals every singleton would become open, making the space discrete. –  Forever Mozart Jul 15 '13 at 18:13


text:
#### 3.OA.B.6

Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8.


text:
# Pokemon Crossing: Gijinka World [SU]

Discussion in 'Profile Threads' started by Elegante, Dec 16, 2012.

1. You weren't meant to be, yet you found your way here.

This is Koyoto, the supposed region that gijinkas are supposed to be safe in.

~~~~~~~~~~~~~~~~~~~~~~
The Form of Filling Out and Filing

Code:
[b]Name:[/b] The obvious part. First Middle Last
[b]Age:[/b] Herp-a-derp.
[b]Gijinka Pokemon:[/b] What Pokemon are you combined with?
[b]Apperance:[/b] I prefer to see than read. Please include an image within the spoiler ( [spoiler] ) and image ( [img] ) tags. Only in a worst case scenario will I accept words.
[b]Gender & Sexuality:[/b] Yes, I am allowing sexuality, but this will all remain PG-13.
[b]Biography:[/b] Give me a bio. Please include a hometown. Somewhere between 4 and 6 sentences will sufice. More is good. Fewer sentences will be taken into account.
[b]Personality:[/b] What does your character act like? Same number of sentences as bio.
[b]Boosted Stats:[/b] This is the main part to be approved, thus the reason this form is messaged to yours truly.
[b]Other:[/b] Something different about your gijinka that can't fit above? Here ya go.
RULES: NO SHINIES. NO LEGENDS OUTSIDE OF TOWN LEADERS. ONE TOWN LEADER PER PERSON. THIS IS A LITERATE ROLEPLAY, MEANING AT LEAST ONE PARAGRAPH (3~5 sentences) IS EXPECTED PER POST. NO GOD MODDING. I AM ALLOWING UP TO 2 CHARACTERS PER PERSON, FOR NOW.

Private message character profiles to me, with the subject line reading "PC: GW". You may post them here once accepted.

#1
2. MAP RESERVED

City descriptions to come.

#2
3. My Characters:

Name: Nicholas Jaie Redfearn
Age: Ageless (Legendary)
Gijinka Pokemon: Dialga
Apperance:
Gender & Sexuality: Gender & Sexuality Unknown
Biography: Nicholas was raised in a small town that played home to much of nothing. This town was New Bark Town. Even though he looked completly human, from a small age he was able to control time for short bursts, sending his parents into shock at first. Once he grew older, he began to develop more and more into his true form. Around what his parents thought was his thirteenth birthday, he began to grow horns on both sides of his head. These horns continued to grow from then on. His parents didn't know what to do with him, so they sent him out at the age of seemingly sixteen. He eventually wound up in the Whirl Islands, alone and sad. Not much is known of what happened to him, or how long he was there. He eventually left and found the area where Ziemas is now located. Alongside several other legendaries, he founded the town as a Gijinka Safehaven.
Personality: Nicholas is nuetral territory. Depending on who you are and what you act like will decide what he treats you like. He enjoys music, and thus is constantly singing or messing with an instrument. He is neither loud nor quiet and thus enjoys the presence of others in a small group. He welcomes anyone with open arms, until they prove themselves otherwise. A lot gets under his skin, but some things get under there quicker. His pet peeves include: rudeness, uncleanleness, and stupidity.
Boosted Stats: Minor time control ability that lasts up to thirty minutes without hurting himself.
Other: Allergy to Peanuts.

#3
4. If one person is confused, I realize others might be too. So, here is some things crazE and I came up with.

crazE tell me of your Gijinka RP
16:35 crazE what is it about? I was only able to quickly scan over it when I had time to get on
16:36 RaivisGalante Well, all I have released is that you are a gijinka, supposed to be dead, that has made it to a safehaven.
16:36 RaivisGalante I have the entire plot already done, but will release it as the roleplay goes on.
16:37 crazE ok well uhm
16:37 crazE how am I to make a background and a hometown if I know nothing of them? Are they regions from the pokemon series? Do I have to explain why my character is a gijinka or are they common?
16:38 RaivisGalante Hrm. Hometown can be any region within the actual games/anime
16:38 RaivisGalante You can explain why they are gijinka if you'd like, but you don't have to.
16:39 crazE Well, can you tell me how gijinka are made? I'm not familiar with that dealio
16:45 RaivisGalante Really, gijinka aren't made. (the best i can come up with)
16:45 RaivisGalante Being a gijinka, from what I can deceiver, is sort of a genetic mutation.
16:46 crazE digi evolution?
16:46 crazE *shot*
16:46 RaivisGalante XD
16:46 RaivisGalante But, um, no. Do you know how real life genetic mutations work?
16:48 crazE cancer?
16:48 crazE *shotagain*
16:49 RaivisGalante *bricks crazE*
16:49 crazE explodes
16:49 crazE I haven't familiarized myself with that level of biology
16:50 RaivisGalante Darn.
16:50 crazE why do you ask?
16:51 RaivisGalante Well, basically, for a gijinka: The genes within them, that cause them to be gijinkas, may remain dormant for many years, or may show themselves at birth.
16:51 crazE soooo
16:51 crazE our character's ancestors might've been **** by a pokemon and then this?
16:52 RaivisGalante That can be one way to put it, I guess.
16:52 crazE well
16:52 crazE I find that half disturbing
16:53 crazE and half interesting
16:53 RaivisGalante LoL
16:53 crazE I would've thought that we could go by some kind of digimon thing where a person is able to "absorb" (with a lack of any better description) a pokemon and their attributes/abilities/other and therefore make gijinka
16:54 crazE or they're pokemon spirits in general recycled into human beings
16:54 RaivisGalante That could be another way.
16:54 crazE and when a certain power ignites their inner pokemon spirit
16:54 crazE they gain abilities
16:54 crazE or something
16:54 crazE IDK
16:54 RaivisGalante This is the reason I am having people to PM the forms, so I can make sure nothing is too drastic.
16:55 RaivisGalante ALL OF THIS SOUNDS GREAT, HOLY SH!T
16:55 crazE do stats count as abilities or what?
16:56 RaivisGalante Stats?
16:57 crazE "Boosted Stats"
16:58 RaivisGalante Oh, on the form?
16:58 crazE yesserie
16:58 RaivisGalante Hang on a sec
17:00 RaivisGalante What I was meaning there, and I will go back and word it better, is the part of the human that is changed.
17:00 RaivisGalante Boosted speed, psychic abilites, etc
17:00 crazE o
17:00 crazE well
17:01 crazE do the gijinkas get any special powers such as performing moves like a pokemon?
17:01 crazE because Hyper Beam yes.
17:01 RaivisGalante I was thinking about that. Hang on.
17:03 crazE XD otay
17:03 crazE also, maybe not just leave it to one move at a time. Humans are more innovative and gijinkas may be able to mix and match moves
17:03 crazE as in combos
17:03 crazE kinda like KH
17:03 crazE (Kingdom Hearts)
17:04 RaivisGalante I know KH
17:04 crazE I didn't know if you knew the abbreviation
17:04 crazE XD
17:04 RaivisGalante i did
17:04 RaivisGalante I think I'll cut it down to where they can only know up to 4 moves.
17:04 RaivisGalante No OP MAry Sue's though
17:05 crazE well of course not
17:05 crazE and four moves? Must be a mind limit or something then
17:05 RaivisGalante Mhmm
17:05 RaivisGalante And Maybe a recharge time.
17:05 crazE so are they mentally incapable of keeping knowledge of more moves or no?
17:06 RaivisGalante YOU ARE MAKING ME THINK
17:06 RaivisGalante STAHP
17:06 RaivisGalante lol, jk
17:06 crazE I'm just trying to figure this ou- Oh XD
17:06 RaivisGalante I would say the Pokemon side prevents that, yes
17:07 crazE pokemon's brains are more powerful than ours
17:07 crazE I mean, you never see pokemon get drunk and beat their spouses
17:07 RaivisGalante i have -_-
17:07 crazE . . .
17:07 RaivisGalante she **** him afterwards
17:07 crazE I see
17:07 crazE well that's unfortunate
17:07 RaivisGalante and his seed created the first gijinka
17:07 crazE but you should probably close the hentai tab
17:08 RaivisGalante SEE WHAT I DID THERE!?
17:08 crazE Nope!!!!!
17:08 crazE I don't see anything!
17:08 crazE I can only read!
17:08 RaivisGalante -_-
17:08 RaivisGalante read what i did there!?
17:08 crazE yes
17:08 crazE and my mind is now on the wall
17:08 RaivisGalante see, it all comes back
17:09 RaivisGalante I BLEW CRAZE'S MIND~
17:09 crazE can a gijinka be a delta species?
17:09 crazE now I want to see your mind on the wall
17:09 crazE :>
17:10 RaivisGalante Delta Species?
17:10 crazE where a pokemon is a different element than what their species is supposed to be
17:10 crazE
17:11 RaivisGalante No.
17:11 crazE
17:12 RaivisGalante Negative
17:12 crazE http://pokegym.net/gallery/displayimage.php?imageid=40011
17:14 RaivisGalante I SAID NO

#4


text:
# How can one be sure that type-Ia supernovae have constant properties in time?

This type Ia category of supernovae produces consistent peak luminosity because of the uniform mass of white dwarfs that explode via the accretion mechanism. The stability of this value allows these explosions to be used as standard candles to measure the distance to their host galaxies because the visual magnitude of the supernovae depends primarily on the distance.

Could it be, however, that this (hypothesized) stability of the peak luminosity value is not constant in time? Let's assume the laws of physics didn't change since the big bang (this seems pretty obvious, but some physicists say these laws vary from place to place and from time to time).
This would imply that these "standard candles" can't be used for measuring the expansion rate of the Universe in the past? This, on its turn, could mean that dark energy may not exist after all (contrary to dark matter, the effects of which are certain, contrary to its Nature).

• My guts feeling tell me something to do with Chandrasekhar mass limit Jan 31, 2020 at 12:59
• @user6760 That's indeed a constant in time. Given that the laws of physics don't change in time, which seems pretty obvious to me. But maybe the white dwarfs in earlier times were composed out of different material as they are later on. Jan 31, 2020 at 14:09
• you of course assume that physics does not change with time, otherwise you can make any predictions you want
– user65081
Jan 31, 2020 at 22:12
• and candles have a range of values, when you far two far away and supernovas are less trustworthy, you can use other methods, see here en.wikipedia.org/wiki/Cosmic_distance_ladder
– user65081
Jan 31, 2020 at 22:15
• @Wolphramjonny If you know how they change in time you can make them. Feb 2, 2020 at 23:00

• @descheleschilder Anysotropies in CMB give a value of the curvature of the universe and parameters of cosmological models: $\Omega_0$,$\Omega_{dm,0}$,$\Omega_{b,0}$,$\Omega_{\Lambda,0}$ (curvature, darkmatter, barionic and dark energy). Feb 3, 2020 at 9:57


text:
How to find constrains on variables that make a system of inequalities not have a solution

My problem is as follows:

How do I ask Mathematica the following question. Let $$f(x,y,z,a,b)$$ be a 5 variable polynomial. I want to find all values of $$a,b$$ for which $$f$$ has no zeros in the region $$h_1(x,y,z)\leq g_1(a,b)$$ and $$h_2(x,y,z)\leq g_2(a,b)$$ ewe an assume that all the functions are smooth.

I think it is appropriate so I will attach what my particular problem is.

$$\left({a_1} {b_2}+{a_2} {b_1}+\frac{{xy}}{2}-z\right)^2-\frac{1}{4} \left(-4 {a_1} {a_2}-4 e+x^2\right) \left(-4 {b_1} {b_2}-4 f+y^2\right)=0$$ While: $$a_1a_2\leq -e+(\frac{x}{2})^2\quad\text{and}\quad b_1b_2\leq -f+(\frac{y}{2})^2$$

I want to figure out for what $$x,y,z,e,f$$ the polynomial has no zero's given the above inequalities. I am working over reals, so every variable should be considered real. Ideally I would like some list of inequalities involving $$x,y,z,e,f$$ that tell me exactly when a solution does not exist. However, I doubt that is an easy problem. Hence, perhaps it is possible to find inequalities which are at least sufficient for non-existance of a zero.

Decided to add my clunky attempt that I am not sure even works:

Reduce[Reduce[(a2 b1 + a1 b2 + x* y/2 - z)^2 - (1/ 2 Sqrt[-4 a1 a2 - 4 e + x^2] Sqrt[-4 b1 b2 - 4 f + y^2])^2 == 0 && a1*a2 <= -e + (x/2)^2 && b1*b2 <= -f + (y/2)^2, {a1, a2, b1, b2}] == {}, {x, y, z, e, f}]

I am not sure if I have full understand the problem. Here just provide a thinking.

conditions =
ForAll[{a1, a2, b1, b2},
a1*a2 <= -e + (x/2)^2 &&
b1*b2 <= -f + (y/2)^2, (a2 b1 + a1 b2 + x*y/2 -
z)^2 - (1/
2 Sqrt[-4 a1 a2 - 4 e + x^2] Sqrt[-4 b1 b2 - 4 f + y^2])^2 !=
0]

Resolve[conditions, Reals]
Reduce[conditions, Reals]

$$\left(y<0\land e>\frac{x^2}{4}\land \left(\left(z<\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\lor \left(z=\frac{x y}{2}\land f>\frac{y^2}{4}\right)\lor \left(z>\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\right)\right)\lor \left(y=0\land e>\frac{x^2}{4}\land f>\frac{z^2}{4 e-x^2}\right)\lor \left(y>0\land e>\frac{x^2}{4}\land \left(\left(z<\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\lor \left(z=\frac{x y}{2}\land f>\frac{y^2}{4}\right)\lor \left(z>\frac{x y}{2}\land f>\frac{e y^2-x y z+z^2}{4 e-x^2}\right)\right)\right)$$

• I think this is it! Thank you so much. Since this is Reduce this actually solved exactly correct? If my $x,y,z,e,f$ satisfy any of these conditions, then the polynomial will have a zero in the region defined in the conditions. This was very interesting. I did not know ForAll can be put into Reduce. Jan 18 at 23:36


text:
Question

The weight w of a person in kilograms and the person’s height h in meters are related by the equation In (10w) = 4.6 +3lnh If In 2 = 0.6 In 3 = 1.1 and In 5 = 1.6 then what is the height of a person whose weight is 60kg?

1. thachthao

The equation states that In (10w) = 4.6 + 3lnh, so we can rewrite it as ln (10w) = 4.6 + 3lnh.

We can then use the property of logarithms that states that ln (a^b) = b ln a to rewrite the equation as ln (10^(10w)) = 4.6 + 3lnh.

This simplifies to 10w = 4.6 + 3lnh, or w = (4.6 + 3lnh)/10.

Substituting the given value of w = 60, we get 60 = (4.6 + 3lnh)/10.

Solving for h, we find that h = e^((60 – 4.6)/3).

To find the value of h, we need to use the given information about the logarithms of 2, 3, and 5. We know that In 2 = 0.6, In 3 = 1.1, and In 5 = 1.6, so we can rewrite the expression for h as:

h = e^((60 – 4.6)/3) = e^(55.4/3) = e^(18.466666666666665)

We can use the fact that e^(In x) = x to rewrite this expression as:

h = 2^(18.466666666666665/0.6) = 3^(18.466666666666665/1.1) = 5^(18.466666666666665/1.6)

We can then use the given values of In 2, In 3, and In 5 to find that h = 2^30.777 = 3^16.727 = 5^11.541.

Since the height h is in meters, the final answer is h = 30.777 meters.


text:
# The integrability of fundamental solution of laplace equation follows from integrability of f ?

Hi, I am really struggling with this question.

The question is :

Let $f:R^3\to R$ and $f\in L^2(R^3)$. $f$ is supported on a ball of radius 1/2 centred at origin. Let $u$ be the solution to $\Delta u=f$ , where $u$ is given by $u(x)= \frac{1}{4\pi}\int_{R^3}\frac{1}{|x-y|}f(y)\,dy$.

1. Show that $L^2$ norm of u in the unit ball of radius 1, centred at origin, is bounded by C$||f||_{L^2}$, where C is a constant independent of f.
2. Show that $u$ is $C^\infty$ outside the unit ball centred at origin.
3. Suppose that $\int_{R^3}f(y)dy = 0$ , show $u\in L^2(R^3)$. (Consider how an good approximation it is to replace $\frac{1}{|x-y|}$ by $\frac{1}{|x|}$ for $|x|$ large.

$(1)$ seems straightforward since $f$ is compactly supported. For (2), I managed to show $\frac{1}{|x-y|}$ can be approximated by a sequence of smooth function say $\delta_n$ such that $u(x)= \int \delta_n(x-y)f(y)dy$ . But no success.

Could anyone here help me with this problem? Thanks in advance.

-
This question was asked here: math.stackexchange.com/q/188416/5773 – B R Aug 30 '12 at 22:49
Are you both sarah and chris? – Steven Gubkin Aug 30 '12 at 23:34
I'm surprised you're not able to get more help from math.stackexchange.com. You could also try the appropriate forum on The Art of Problem Solving web site. Or get help from your classmates or professor? But you should also keep struggling with it yourself. You appear to be making progress, and even if it takes you a long time, it's always best to somehow do it yourself. Don't worry about how long it takes, even if it all seems too easy and obvious afterwards. Even some of us now successful research mathematicians suffered like that as undergraduates. – Deane Yang Aug 30 '12 at 23:34
You don't need to know anything. Just use your bare hands and struggle. – Deane Yang Aug 30 '12 at 23:59
This is obviously homework, voting to close. – Igor Rivin Aug 31 '12 at 15:49

Your question text suggests that yo know how to do (1). (Interchange the order of integration). Now (2) is easy: outside the support of $f$ we have $\Delta u=0$, that is $u$ is harmonic outside the support of $u$. Harmonic functions are surely $C^\infty$. For (3), expand the kernel $1/|x-y|$ in a series of spherical harmonics outside of some ball, and integrate this series term by term (it is absolutely and uniformly convergent). You obtain $u(x)=O(1/|x|^2)$ which is more than enough for $L^2$.
Let $E$ be the fundamental solution of the Laplace equation in $\mathbb R^n$ for $n\ge 3$: $E=c_n\vert x\vert^{2-n}$ so that $E$ is $L^1_{loc}$. For $f$ compactly supported, $\chi\in C^\infty_c$, $$\chi u=\chi (E\ast f)=\chi(x)\int_{y\in supp f} E(x-y) f(y) dy.$$ Since $x$ and $y$ are moving in compact sets, it is also the case of $x-y\in K=supp \chi-supp f$: as a result $$\chi u=\chi \bigl((1_KE)\ast f\bigr)\Longrightarrow \Vert \chi u\Vert_{L^2}\le \Vert 1_K E \Vert_{L^1}\Vert f\Vert_{L^2},\quad\text{proving (1)}.$$ Moreover since $\Delta u=0$ on the open set $(supp f)^c$, the function $u$ is harmonic there, thus in fact real-analytic on that open set, proving more than (2). Going back to $u(x)=\int_{y\in supp f} E(x-y) f(y) dy,$ we already know that $u\in L^2_{loc}$. We have also from the vanishing of the integral of $f$, $$u(x)=\int_{y\in supp f} \bigl(E(x-y)-E(x)\bigr) f(y) dy,$$ so that for $\vert x\vert\gg diam supp f$, $y\in suppf$, $\vert E(x-y)-E(x)\vert\lesssim \vert x\vert^{1-n}\in L^2(\mathbb R^n\cap \vert x\vert \ge 1)$ since we have $$2n-2>n,\quad \text{from n>2.}$$


text:
zbMATH — the first resource for mathematics

A convergence result for stochastic partial differential equations. (English) Zbl 0653.60049
It is shown that for regular approximations $$w_ n$$ of the Brownian motion w, the solutions of $du_ n-A u_ n dt+B u_ n dw_ n$ converge to the solution of the abstract Stratonovich stochastic differential equation $$du=A u dt+B u dw$$. The assumptions on the operators A and B permit physically reasonable applications, and some PDE examples are given.
Reviewer: T.C.Gard

MSC:
 60H15 Stochastic partial differential equations (aspects of stochastic analysis)
Full Text:
References:
 [1] Acquistapace P., An approach to Ito linear equations in Hilbert spaces by approximations of white noise with coloured noise, Preprint 42 · Zbl 0547.60066 [2] Batchelor G. K., The Theory of Homogenous Turbulence (1959) [3] DOI: 10.1512/iumj.1976.25.25049 · Zbl 0319.60038 · doi:10.1512/iumj.1976.25.25049 [4] Chow P. L., Probabilistic Analysis and Related Problems 1 (1978) [5] DOI: 10.1080/07362998308809004 · Zbl 0511.60055 · doi:10.1080/07362998308809004 [6] Daprato G., Stochastics 6 pp 105– (1982) [7] DOI: 10.1016/0022-247X(82)90041-5 · Zbl 0497.93055 · doi:10.1016/0022-247X(82)90041-5 [8] Ikeda N., Stochastic Differential Equations and Diffusion Processes (1981) · Zbl 0495.60005 [9] Lions J. L., Optimal Control of Systems Governed by Partial Differential Equations (1971) · Zbl 0203.09001 [10] Lions J. L., Non-homogenous Boundary Value Problems and Applications (1972) [11] Mcshane E. I., Stochastic Calculus and Stochastic Models (1974) · Zbl 0292.60090 [12] Pardoux E., C.R. Acad. Sc. Paris, t. 21 pp 101– (1972) [13] Pazy A. J., Semigroups of Bounded Linear Operators (1984) [14] DOI: 10.1214/aop/1176995608 · Zbl 0391.60056 · doi:10.1214/aop/1176995608 [15] Szafirski B., Bull. Acad. Polon. Sci. Ser. Math.Astron. Phys 19 pp 785– (1971) [16] DOI: 10.1214/aoms/1177699916 · Zbl 0138.11201 · doi:10.1214/aoms/1177699916 [17] Yosida K., Functional Analysis (1980) · Zbl 0435.46002
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:
+0

# Floor function

0
152
1

How many values of r are there such that $\lfloor r/4 \rfloor + r = 15.5?$

May 10, 2022

#1
+9462
0

Moving r to the other side gives $$\left\lfloor\dfrac r4\right\rfloor = 15.5 -r$$.

Then 15.5 - r is an integer. Then r = k + 0.5 for some integer k.

Then $$\left\lfloor\dfrac {k + \frac12}4\right\rfloor = 15 - k$$.

That means $$15-k\leq \dfrac{k + \frac12}4 < 16 - k$$ by definition of floor function. Now it becomes a compound inequality. To proceed, solve the compound inequality and list its integer solutions.

May 10, 2022


text:
Our tech stack at Freetrade is 99% serverless. This means that for the vast majority of our applications, the physical infrastructure, deployment and other operational concerns are largely abstracted away by the cloud provider (Google Cloud in our case).

We focus on building what our users need, and our cloud provider focuses on what we need to keep it running smoothly at scale.

Serverless computing has huge benefits in terms of allowing more resources to be devoted to solving user problems, but it also presents some different challenges. One of those is how you go about applying automated testing to your software systems when you’re in a serverless environment.

Automated testing is essential to produce and maintain high-quality software. With traditional infrastructure where you’re responsible for the actual server instances and processes that run your applications, it can be easier to implement automated testing because you have direct control over more of the stack. That extra responsibility is exactly what you’re trying to abstract away with serverless computing, though, so you have to adapt your testing strategy.

## Unit testing serverless applications

Unit testing is the bedrock of an automated testing strategy. Unit tests are the most numerous and lowest-level tests, which isolate tiny parts of application logic and test them on their own.

With a well-architected application, the infrastructure that will ultimately run the application shouldn’t impact unit testing at all. Each individual unit is oblivious to what server environment it might be running in. Serverless computing facilitates this, as units of code cannot rely on particular environment details being available, such as disk i/o or some particular application state being persisted across requests. Good abstractions have to be used for these instead.

In other words, serverless encourages you to write code that depends on abstractions (see dependency inversion). It also encourages stateless applications.

Both of these make it easier to write comprehensive unit tests, as you can isolate those parts out.

As an example, requests from Freetrade’s apps to the platform are handled by HTTP handlers in Google Cloud Functions. There is little to no logic in these handlers; they merely pass the request on to services that don’t need to know about how the request came in or even about HTTP at all.

Those services and their components are then easy to unit test because they’re not tied to particular details of the environment.

## Integration testing serverless applications in Google Cloud Functions

Integration testing could be described as the next level up from unit testing. Unit testing mocks out interactions between components to isolate logic, but what if you’ve got that interaction wrong?

Integration tests fill this gap in a testing strategy by confirming that components also play nicely with each other.

While serverless does not impact unit testing, it can present a couple of challenges for integration testing. You need to test the interactions between HTTP requests, databases, queues and other systems, so you need those to be available for the integration tests. With a traditional server infrastructure, you might run instances of all those things locally on a single machine and run the integration tests that way.

There are options to try and emulate serverless components locally for this purpose, but that kind of defeats the point of the integration testing — what if the emulation is wrong? You’re reliant on the emulation being accurate and up to date with the serverless platform, and that’s the kind of work you want to avoid with serverless.

It is possible to spin up a fresh environment in Google Cloud for each integration test run, and that’s certainly an option. However, it takes a bit of time for each setup, and you want the integration tests to be applied frequently and rapidly, so this approach might not be ideal.

So far we’ve taken a middle-ground approach by having a dedicated CI (continuous integration) environment in Google Cloud. It isn’t spun up and torn down between test runs. Instead, our integration tests aim to achieve that robustness by creating the specific state to be tested within each test, and avoiding dependency on any other state (as that would be a bad practice).

As an imaginary example:

// Given I have a user account;

(code to create a fresh user account)

// When I make a valid sign-in request;

(code to send a real sign-in request over the Internet to the test environment)

// Then I should get a valid auth token;

(code to check received token works in the test environment)

// And a sign-in event should have been persisted.

(code to check a sign-in event was persisted)

We have various helper code to set up and assert on the expected state within a test, and a test client to mimic user app requests against the system. Individual tests or the entire test suite can be run as needed without needing to refresh the whole environment.

This is a pragmatic approach that is working well for us so far. Tests can be run continuously, and each test run completes in reasonable time with the individual tests executed in parallel.

The serverless cloud functions are spun up and down as in production use, and interact realistically with other components.

Because serverless functions are spun up to meet demand, they can have cold starts. As the name suggests, this is when a function that had no “hot” instances running is spun up.

Inevitably this will happen during integration testing, which can add several seconds to the time taken by the test. However, we actually want this to occur, as it mimics behaviour in production. We want our integration tests to help us identify potential issues due to cold function starts just as we would with any other integration issue.

## End-to-end testing serverless applications

The top tier of a testing strategy is end-to-end (E2E) testing. This applies user-level interactions via the user interface, and also checks that the application behaviour is correct via the user interface. These tests are the most time-consuming to run, but also provide the strongest confirmation that the entire stack works as expected for our users.

As with unit tests, the fact that serverless infrastructure is being used should not impact end-to-end testing. They only care about what they can interact with in the user interface, with the platform behind that being a black box.

That said, serverless computing does offer some benefits to end-to-end testing. You don’t need to maintain dedicated infrastructure to handle E2E tests: the infrastructure spins up automatically to meet the demand. This can reduce costs as you don’t pay for idle infrastructure that might sit waiting for a test run.

Getting more specific, Google Cloud also makes it easier to manage E2E testing by grouping components within specific environments, with consistent URLs based on that.

Managing test suites and their environments is then straightforward.

## To sum up

We’ve had a great experience applying different tiers of automated testing to our serverless infrastructure.

Having strong abstractions provided by the cloud platform lets us devote more of our time to building things for our users, and that carries through to our approach to testing.

This allows a lot of our test code to be highly readable and meaningful from a user perspective, keeping us on track to bring free investing to more and more people.


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# Show that arbitrary $A$ and $A^T$ have same eigenvalue, algebraic and geometric multiplicity

Show that an arbitrary $n \times n$ matrix $A$ and its transpose $A^T$ have the same eigenvalues, algebraic multiplicity and geometric multiplicity.

I'm not sure if I did it correctly and especially how to show that they have same geometric multiplicity?

same eigen values

Assume $A$ and $A^T$ have same eigenvalues, then they have the same chracteristic polynomial. So we need to show that $p_A(\lambda)=\det(A-\lambda I)$ is same as $p_{A^T}(\lambda)=\det(A^T-\lambda I)$.

So we have $$p_{A^T}(\lambda)=\det(A^T-\lambda I) = \det(A^T-\lambda I^T) = \det\left((A-\lambda I)^T\right) = \det(A-\lambda I)=p_A(\lambda)$$

We see their characteristic polynomials are same so their eigenvalues are same as well.

same algebraic multiplicity

I'm not sure if this is a correct reason proof but: Because the characteristic polynomials are same, we have that the algebraic multiplicities of the eigenvalues of $A$ nd $A^T$ sre the same.

same geometric multiplicity

I don't know? :/

• Hint: Use the fact that $rank(A) = rank(A^T)$ and then use rank-nullity to deduce that $dim(ker(A-\lambda I)) = dim(ker(A^T-\lambda I))$ – Osama Ghani May 28 '18 at 8:08

Geometric Multiplicity of $\lambda$ in $A=\text{rank} (A-\lambda I)=\text{rank }(A-\lambda I)^T=\text{rank } (A^T-\lambda I)=$Geometric multiplicity of $\lambda$ in $A^T$

• The geometric multiplicity of $\lambda$ is measured by the dimension of the kernel, not the rank. A line referencing rank-nullify to show that the dimensions of the kernels are equal would complete the proof. – Osama Ghani May 28 '18 at 8:09


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User quetzalcube - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:56:09Z http://mathoverflow.net/feeds/user/12204 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130689/reference-for-cech-cohomology-for-nisnevich-topology Reference for Cech cohomology for Nisnevich topology Quetzalcube 2013-05-15T10:06:32Z 2013-05-15T10:06:32Z
I need the theorems that prove that Nisnevich cohomology can be computed by the Cech complex similar to what happens in the étale topology.

I know this has to be true since (Morel-Voevodsky: $\mathbb{A}^1$-homotopy theory of schemes. Publ. Math. IHES (1999)) say it at the beggining of sectio 3, page 95... but they don't give the reference!

Thank you very much for your help!
 http://mathoverflow.net/questions/91312/why-are-kan-fibrations-and-serre-fibrations-so-important Why are Kan fibrations and Serre fibrations so important? Quetzalcube 2012-03-15T17:23:34Z 2012-03-16T09:05:28Z
I know so far that Kan fibrations are the abstract analog of Serre fibrations but i don't have any expierience with neither one of them so

Could someone give a motivation why Kan fibrations (or if easier, Serre fibrations) are so important in homotopy theory of simplicial sets?

Thank you
 http://mathoverflow.net/questions/78314/alexander-duality-theorem-for-cw-complexes-and-stable-homotopy-theory Alexander duality theorem for CW-complexes and stable homotopy theory Quetzalcube 2011-10-17T07:40:31Z 2011-10-31T03:54:43Z
In Adams, J.F. Infinite Loop Spaces Princ. Univ. Press. page 9 he states Alexander duality theorem

Theorem:[Alexander Duality] $$H^r(X,G)=H_{n-r+1}(S^n-X,G)$$

for finite CW-complexes with a "nice embedding". That is to say, $S^n-X$ has a CW-complex $Y$ as deformation retract and $X$ is a deformation retract of $S^n-Y$.

My first question is the following: Do you know if every CW-complex admits a nice embedding into some $S^n$? if so, could you give a reference?

Thanks in advance.
 http://mathoverflow.net/questions/54186/reference-learning-noncommutative-geometry-and-c-algebras Reference: Learning noncommutative geometry and C^* algebras Quetzalcube 2011-02-03T11:27:41Z 2011-02-05T02:55:33Z
I am starting to study noncommutative geometry and C^* algebras so my question is

Does anyone knows a good reference on this subject?

I would like a basic book with intuitions for definitions and this kind of things. I come from algebraic geometry so if it talks a bit about the relation with it would be highly appreciated.

I have already taken a look to Conne's book but I find it too hard and I'm currently studying Landi's but it lacks lots of proofs (he refers to several papers)

Thanks in advance
 http://mathoverflow.net/questions/51991/are-finite-correspondances-flat Are finite correspondances flat? Quetzalcube 2011-01-13T19:04:21Z 2011-01-14T14:10:01Z
In Voevodsky&Mazza&Weibel's book on motivic cohomology they define "an elementary correspondance from X (Smooth connected scheme over $k$) to Y (separated scheme over $k$) as an irreducible closed subset $W$ of $X\times Y$ whose associated integral subscheme is finite and surjective over $X$."

As far as I know $W$ would be flat over $X$ if it was Cohen-Macaulay so...

1.- ¿Is $W$ flat over $X$?

If not,

2.- why isn't this a common sense assumption? Could anyone give an example of why nonflat elementary correspondaces should be allowed?

Thanks in advance
 http://mathoverflow.net/questions/130689/reference-for-cech-cohomology-for-nisnevich-topology Comment by Quetzalcube Quetzalcube 2013-05-17T07:49:59Z 2013-05-17T07:49:59Z Oups, you are right. Thank you very much. Prop. 3.1.9 is actually what i was looking for. Sorry for my mistake! http://mathoverflow.net/questions/461/understanding-steenrod-squares/19713#19713 Comment by Quetzalcube Quetzalcube 2011-06-14T14:15:13Z 2011-06-14T14:15:13Z $v$ is the normal bundle sorry ( i don't how to érase the comment) but still: what is $w_i$? thanks http://mathoverflow.net/questions/461/understanding-steenrod-squares/19713#19713 Comment by Quetzalcube Quetzalcube 2011-06-14T14:10:09Z 2011-06-14T14:10:09Z please, could you define what is $w_i$ and $v$ in the formula $Sq^i(x)=f_*(w_i(v))$? Thanks a lot in advance