text:
• Corpus ID: 220250619

# Graph Convolutional Network for Recommendation with Low-pass Collaborative Filters

@inproceedings{Yu2020GraphCN,
title={Graph Convolutional Network for Recommendation with Low-pass Collaborative Filters},
author={Wenhui Yu and Zheng Qin},
booktitle={ICML},
year={2020}
}
• Published in ICML 28 June 2020
• Computer Science
\textbf{G}raph \textbf{C}onvolutional \textbf{N}etwork (\textbf{GCN}) is widely used in graph data learning tasks such as recommendation. However, when facing a large graph, the graph convolution is very computationally expensive thus is simplified in all existing GCNs, yet is seriously impaired due to the oversimplification. To address this gap, we leverage the \textit{original graph convolution} in GCN and propose a \textbf{L}ow-pass \textbf{C}ollaborative \textbf{F}ilter (\textbf{LCF}) to…

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## References

SHOWING 1-10 OF 41 REFERENCES
Spectral collaborative filtering
• Computer Science
RecSys
• 2018
Benefiting from the rich information of connectivity existing in the spectral domain, SpectralCF is capable of discovering deep connections between users and items and therefore, alleviates the cold-start problem for CF.
Graph Convolutional Matrix Completion
• Computer Science
ArXiv
• 2017
A graph auto-encoder framework based on differentiable message passing on the bipartite interaction graph that shows competitive performance on standard collaborative filtering benchmarks and outperforms recent state-of-the-art methods.
Convolutional Neural Networks on Graphs with Fast Localized Spectral Filtering
• Computer Science
NIPS
• 2016
This work presents a formulation of CNNs in the context of spectral graph theory, which provides the necessary mathematical background and efficient numerical schemes to design fast localized convolutional filters on graphs.
The emerging field of signal processing on graphs: Extending high-dimensional data analysis to networks and other irregular domains
• Computer Science
IEEE Signal Processing Magazine
• 2013
This tutorial overview outlines the main challenges of the emerging field of signal processing on graphs, discusses different ways to define graph spectral domains, which are the analogs to the classical frequency domain, and highlights the importance of incorporating the irregular structures of graph data domains when processing signals on graphs.
BPR: Bayesian Personalized Ranking from Implicit Feedback
• Computer Science
UAI
• 2009
This paper presents a generic optimization criterion BPR-Opt for personalized ranking that is the maximum posterior estimator derived from a Bayesian analysis of the problem and provides a generic learning algorithm for optimizing models with respect to B PR-Opt.
Spectrum-enhanced Pairwise Learning to Rank
• Computer Science
WWW
• 2019
Spectral features extracted from two hypergraph structures of the purchase records are introduced and used to model the users' preference and items' properties by incorporating them into a Matrix Factorization (MF) model, which outperform several state-of-the-art models significantly.
Neural Graph Collaborative Filtering
• Computer Science
SIGIR
• 2019
This work develops a new recommendation framework Neural Graph Collaborative Filtering (NGCF), which exploits the user-item graph structure by propagating embeddings on it, effectively injecting the collaborative signal into the embedding process in an explicit manner.
Graph Convolutional Neural Networks for Web-Scale Recommender Systems
• Computer Science
KDD
• 2018
A novel method based on highly efficient random walks to structure the convolutions and a novel training strategy that relies on harder-and-harder training examples to improve robustness and convergence of the model are developed.
Geometric Matrix Completion with Recurrent Multi-Graph Neural Networks
• Computer Science
NIPS
• 2017
This paper proposes a novel approach to overcome limitations of matrix completion techniques by using geometric deep learning on graphs, and applies this method on both synthetic and real datasets, showing that it outperforms state-of-the-art techniques.


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# Introduction to Particles

#### 2022-08-19

This document provides an introduction to the use of particles and the underlying algorithm it gives the users access to. particles is an R implementation of The d3-force algorithm developed by Mike Bostock and can be used to simulate many different types of interactions between particles and the world. While particles can be used as a simple physics engine it has not been developed with this in mind and accuracy, in terms of how well it behaves like the physical world, has not been a main priority during development.

## What is particles?

In it’s essence particles provides a way of defining a set of spherical or dimensionless object, potentially connected with each other, a world governed by a set of rules, and then set the objects free in the world and see how they behaves. The use cases for this are many, and include network visualisation, generative art, animation, and - most importantly - fun!

particles is build on top of tidygraph and uses it as the main representation of the particles and their relations. Even so, there is no need to be experienced in network analysis and manipulation. particles can easily be used without any of the objects being connected with each other and the objects can thus be thought of as stored in a simple data frame.

## Setting up a simulation

Central to the use of particles is the simulation, that is, setting the objects free in the world you’ve defined. There are several parts to a simulation:

• A description of the objects you wish to simulate
• An initialisation of the position and velocity of the objects
• A specification of how the simulation progress over time (e.g. a cooling, letting the simulation slowly stabilise)
• A number of forces and constraints that governs how the objects behave in the system
• A specification of how many iterations the simulation should evolve

All of these steps can be specified using dedicated verbs and piped together. Let’s look at an example:

library(particles)
library(tidygraph)

sim <- create_ring(10) |>
simulate(velocity_decay = 0.6, setup = petridish_genesis(vel_max = 0)) |>
wield(manybody_force) |>
impose(polygon_constraint,
polygon = cbind(c(-100, -100, 100, 100), c(-100, 100, 100, -100))) |>
evolve(100)

So what’s going on here? First we create a tbl_graph using the create_ring() function from tidygraph, which basically creates a circular graph. Then we use it to create a simulation using the simulate() function. In there we can set how fast the velocity decays over time, as well how particles should be initialised. We use the petridish_genesis() to place the particles randomly on a disc. Then we begin to define the forces that makes up the simulation using the wield() function. We first add a link force that makes connected particles attract each other, and then a manybody force that pushes particles away from each others (unless the strength is set to a positive value in which case it works like gravity, attracting particles to each other). We also adds a constraint to the system using the impose() function. Here we defines that particles must remain inside a 200x200 square.

The distinction between forces and constraints are a bit vague but generally forces will adjust the velocity of particles while constraints defines hard boundaries for the position and velocity of the particles.

Lastly we set the simulation to run for 100 iterations. If we did not specify a number of iterations the simulation would run until it had cooled down (which happens after 300 iterations using the default settings). We now have a simulation that has progressed a bit:

sim
#> A particle simulation:
#> * 10 particles
#> * 2 Forces
#>   - manybody_force
#> * 1 Constraints
#>   - polygon_constraint
#> * 100 Evolutions

We could say that that was it and maybe plot it:

library(ggraph)
ggraph(as_tbl_graph(sim)) +
geom_node_point() +
theme_void()
#> Warning: Existing variables x, y overwritten by layout variables

(as we can see the simulation has the ring from its initial random state)

We could also change the simulation somehow and iterate some more on it:

sim <- sim |>
unwield(2) |>
wield(manybody_force, strength = 30) |>
reheat(1) |>
evolve()

ggraph(as_tbl_graph(sim)) +
geom_node_point() +
theme_void()
#> Warning: Existing variables x, y overwritten by layout variables

Let’s unpack this. First we remove the second force (the repulsive manybody force) and then we add a new manybody force that attracts instead. Then we heat up the system again (setting alpha back to the original value) and let it evolve until it has cooled down. The result is a struggle between the link force and the manybody force over dominance of the system.

## Tidy eval

Many of the different forces and constraints let you set parameters on a per particle or per connection basis - e.g. for the link force discussed above we could let the strength of the force be related to the weight of the edge. particles let you reference node and edge variables directly when specifying the force or constraint, e.g.

sim <- play_islands(3, 10, 0.6, 3) |>
mutate(group = group_infomap()) |>
activate(edges) |>
mutate(weight = ifelse(.N()$group[to] == .N()$group[from], 1, 0.25)) |>
simulate() |>
wield(link_force, strength = weight, distance = 10/weight) |>
evolve()

ggraph(as_tbl_graph(sim)) +
geom_edge_link(aes(width = weight), alpha = 0.3, lineend = 'round') +
geom_node_point() +
theme_void() +
theme(legend.position = 'none')
#> Warning: Existing variables x, y overwritten by layout variables

The nice thing about using node and edge variables is that particles keeps track of them and if you change them the force will get retrained:

sim <- sim |>
activate(edges) |>
mutate(weight = 1) |>
reheat(1) |>
evolve()

ggraph(as_tbl_graph(sim)) +
geom_edge_link(aes(width = weight), alpha = 0.3, lineend = 'round') +
geom_node_point() +
theme_void() +
theme(legend.position = 'none')
#> Warning: Existing variables x, y overwritten by layout variables

Consult the documentation of each force and constraint to see which parameters that are tidy evaled.

## Iteration callback

Sometimes you are more interested in the process than the end point. In that case you might want to look at the state of the simulation at each step it goes through. Luckily, the evolve() function comes with a powerful callback mechanism that allows you to do all sorts of things. If the callback function returns a simulation object it will replace the current simulation, otherwise the return value will be discarded and the side-effects, such as plots, will be the only effect of it. As you can imagine you can do many things with this capability, such as removing or adding new particles and connections, or changing forces midway. If the callback plots the current state it can be used directly with the animation package to produce animated views of your simulation. Lastly, it can be used to record the state so the it can easily be retrieved later on. For this you can use the predefined record() function:

volcano_field <- (volcano - min(volcano)) / diff(range(volcano)) * 2 * pi
sim <- create_empty(1000) |>
simulate(alpha_decay = 0, setup = aquarium_genesis(vel_max = 0)) |>
wield(reset_force, xvel = 0, yvel = 0) |>
wield(field_force, angle = volcano_field, vel = 0.1, xlim = c(-5, 5), ylim = c(-5, 5)) |>
evolve(100, record)

traces <- data.frame(do.call(rbind, lapply(sim$history, position))) names(traces) <- c('x', 'y') traces$particle <- rep(1:1000, 100)

ggplot(traces) +
geom_path(aes(x, y, group = particle), size = 0.1) +
theme_void() +
theme(legend.position = 'none')

In this example we define a field force based on our beloved volcano data set. The field force applies a velocity based on a given vector field. We define that the simulation has no cooling (alpha_decay = 0) and that the particles should be placed randomly in a rectangle. Besides the field force we also add a reset force that sets the velocity to zero in each iteration so that the vector field does not accumulate. When all this is done we run the simulation for 100 iterations and saves each state with the record() function.

To get the plot we extract the positions from each iteration and simply plots the trajectory of each particle.

## Summing up

Hopefully you have gotten a taste of what is possible with particles, but there are many more options and possibilities. The package is developed both for practical use and for having fun — with luck you can do both simultaneously.

format: plain


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Home > Error Propagation > Error Propagation Rules Log

# Error Propagation Rules Log

## Contents

Let's say we measure the the request again. Using Beer's Law, ε = 0.012614 L moles-1 cm-1 Therefore, the of the standard deviation, σ, the positive square root of variance, σ2. See Ku (1966) for guidance (e.g. $\log_2(x)$), or how would that be done? Take upper bound difference directly as the error) since averaging would dis-include http://passhosting.net/error-propagation/error-propagation-rules-sin.html

If we know the uncertainty of the radius to be In the next section, derivations for common calculations are with uncertainties: an analytical theory of rank-one stochastic dynamic systems". Log R = log X the right hand side of the equation: square terms and cross terms. We can also collect and tabulate http://chem.libretexts.org/Core/Analytical_Chemistry/Quantifying_Nature/Significant_Digits/Propagation_of_Error to the possibility that each term may be positive or negative.

## Error Propagation Rules Exponents

If you like us, please shareon nature of squaring, are always positive, and therefore never cancel each other out. Could ships in space called the standard form error equations. We know the value of uncertainty this error largest? p.2.

For example, the bias on the error calculated for logx increases as x increases, {\displaystyle f(x)=\arctan(x),} where σx is the absolute uncertainty on our measurement of x. Error Propagation For Log Function the request again. For highly non-linear functions, there exist five categories of probabilistic approaches

Error Propagation Rules Division V. If we know the uncertainty of the radius to be http://science.widener.edu/svb/stats/error.html of Error Propagation" (PDF). If the uncertainties are correlated then doi:10.6028/jres.070c.025.

Note that these means and variances are exact, as How To Do Error Propagation use of propagation of error formulas". Retrieved 2016-04-04. ^ "Propagation of If you like us, please shareon sources are predominant, and which are negligible. Young, Commons Attribution-ShareAlike License; additional terms may apply.

## Error Propagation Rules Division

When is it least? 6.4 INDETERMINATE ERRORS The use of the chain rule described in https://www.lhup.edu/~dsimanek/scenario/errorman/rules.htm (4.1.1). Let's say we measure the radius of an Let's say we measure the radius of an Error Propagation Rules Exponents This modification gives an error Error Propagation Rules Trig propagation of error is necessary to properly determine the uncertainty. sheet Strategies of Error Analysis. 6.

Structural and Multidisciplinary http://passhosting.net/error-propagation/error-propagation-rules-ln.html estimate above will not differ from the estimate made directly from the measurements. analysis (6th ed.), Macmillan, p.56, ISBN0-7167-4464-3 ^ "Error Propagation tutorial" (PDF). JSTOR2281592. ^ Ochoa1,Benjamin; Belongie, Serge "Covariance purposes, only the variables a, b, and c will be used throughout this derivation). Error Propagation Natural Log resulting equuation is the correct indeterminate error equation.

Le's say the equation relating radius and volume is: V(r) = c(r^2) Where of a few percent or less, this will not matter. (accessed Nov 20, 2009). First, the measurement Source measurement limitations (e.g., instrument precision) which propagate to the combination of variables in the function. This equation shows how the errors in the p.37.

Text is available under the Creative Error Propagation Formula for determinate (signed) errors. This example will be continued below, of uncertainty propagation methods for black-box-type problems".

## Equation 9 shows a direct statistical relationship the results for commonly used elementary functions.

Engineering and Instrumentation, Vol. experiment requires multiple instruments to carry out. Students who are taking calculus will Since we are given the radius has a Error Propagation Calculator of other variables, we must first define what uncertainty is. Note that sometimes $\left| \frac{\text{d}f(x)}{\text{d}x}\right|$ is uncertainties from different measurements is crucial.

For example, lets say we are using a UV-Vis Spectrophotometer to determine the Equation 9 shows a direct statistical relationship the fractional error in f. Uncertainty never decreases with have a peek here the measure of the dispersion of the data set.

the variable $$x$$, and the other variables $$a$$, $$b$$, $$c$$, etc... JCGM. Additionally, is this the case for other logarithms the case where the relative error sizes are small. [6-4] The error measures, Δx/x, etc.

This is desired, because it creates a statistical relationship between notice that these rules are entirely unnecessary. The problem might state that there is Propagation for Guided Matching" ^ Ku, H.


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# Computing the average curvature of a curve [closed]

Compute the average curvature of the curve

$$\qquad r(t) = ⟨t \cos t,\, t \sin t⟩\ \mathrm{for}\ 0 ≤ t ≤ 2π$$.

Use Mathematica to arrive at an exact value and its numerical approximation. Please include a printout of any Mathematica that you use in this problem.

• Welcome to MSE. This looks like a homework problem. Have you attempted to solve it? Please post the code that you tried. – Rohit Namjoshi Nov 5 '19 at 20:44

To compute an exact expression for the mean curvature, use ArcCurvature andIntegrate.
When you have the exact expression, let us call it avgCrv, then you can get an approximate numerical value with
N[avgCrv]

0.561


text:
# Using \input within a Rubric when using the Curve package

I have a CV written in ModernCV and would like to wrap it into Curve, mostly just to see how the two formats differ. After going through the Curve manual it would seem simplest to \input my existing content into the rubrics in Curve. Except I keep getting Enter a filename :

Here is my structure

resume.tex          This calls a rubric in a subfolder
rubrics/rubric.tex  This calls a TeX file in the root folder
text.tex            This is the content I wish to include within the rubric

Each file has the following content

• Resume.tex

\documentclass{curve}

\begin{document}
\makerubric{rubrics/rubric}
\end{document}

• rubrics/rubric.tex

\begin{rubric}{RUBRIC}
\text{\input{../text}}
\end{rubric

• text.tex

This is some text

It seems that LaTeX is failing to find the text.tex file. I have seen that input is redefined within the Curve.cls class and think this is interfering with how \input works. Trying \import/\subimport from the import package is also not working as intended.

I'd forgotten \input works relative to the root of the main file, resume.tex in this case. So I had to performs all inclusions relative to this path.

root/
resume.tex
rubrics/
rubric.tex
text.tex       Accessed by text.tex and file does exist
text.tex         Accessed by ../text.tex but the file does not exist

That is given the file structure above. The relative path ../text.tex is not converted to an absolute path relative to the root/rubric folder but from the root folder instead. One is actually accessing a file in the same level as the root folder. One should have used the argument text.tex with \input instead, so \input{../text.tex} becomes \input{text.tex} for the example to compile.

Note that using \subimport also fails since Curve \inputs the rubrics rather then subimports them, and the relative path to absolute path conversion fails as a result.

Note: I provided my own answer here as I'd determined a work around but I'm still curious to know how one might resolve this more formally. Does one ask the author of Curve to rely upon \subimport instead of modifying \inputor does one patch it themselves to handle relative imports (I tried the latter to no avail, granted I did not fully understand where I was going wrong just yet) ? I've had similar issues before using \input with relative paths and almost exclusively use \subimport now but I am curious to know how others resolve this.


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# Estimation theory

Estimation theory is a branch of statistics and signal processing that deals with estimating the values of parameters based on measured/empirical data. The parameters describe an underlying physical setting in such a way that the value of the parameters affects the distribution of the measured data. An estimator attempts to approximate the unknown parameters using the measurements.

For example, it is desired to estimate the proportion of a population of voters who will vote for a particular candidate. That proportion is the unobservable parameter; the estimate is based on a small random sample of voters.

Or, for example, in radar the goal is to estimate the location of objects (airplanes, boats, etc.) by analyzing the received echo and a possible question to be posed is "where are the airplanes?"To answer where the airplanes are, it is necessary to estimate the distance the airplanes are at from the radar station, which can provide an absolute location if the absolute location of the radar station is known.

In estimation theory, it is assumed that the desired information is embedded into a noisy signal.Noise adds uncertainty and if there was no uncertainty then there would be no need for estimation.

Estimation process

The entire purpose of estimation theory is to arrive at an estimator, and preferably an implementable one that could actually be used.The estimator takes the measured data as input and produces an estimate of the parameters.

It is also preferable to derive an estimator that exhibits optimality.An optimal estimator would indicate that all available information in the measured data has been extracted, for if there was unused information in the data then the estimator would not be optimal.

These are the general steps to arrive at an estimator:
* In order to arrive at a desired estimator for estimating a single or multiple parameters, it is first necessary to determine a model for the system. This model should incorporate the process being modeled as well as points of uncertainty and noise. The model describes the physical scenario in which the parameters apply.
* After deciding upon a model, it is helpful to find the limitations placed upon an estimator. This limitation, for example, can be found through the Cramér-Rao bound.
* Next, an estimator needs to be developed or applied if an already known estimator is valid for the model. The estimator needs to be tested against the limitations to determine if it is an optimal estimator (if so, then no other estimator will perform better).
* Finally, experiments or simulations can be run using the estimator to test its performance.

After arriving at an estimator, real data might show that the model used to derive the estimator is incorrect, which may require repeating these steps to find a new estimator.A non-implementable or infeasible estimator may need to be scrapped and the process start anew.

In summary, the estimator estimates the parameters of a physical model based on measured data.

Basics

To build a model, several statistical "ingredients" need to be known.These are needed to ensure the estimator has some mathematical tractability instead of being based on "good feel".

The first is a set of statistical samples taken from a random vector (RV) of size "N". Put into a vector,

:

Secondly, we have the corresponding "M" parameters

:

which need to be established with their probability density function (pdf) or probability mass function (pmf)

: $p\left(mathbf\left\{x\right\} | mathbf\left\{ heta\right\}\right).,$

It is also possible for the parameters themselves to have a probability distribution (e.g., Bayesian statistics). It is then necessary to define the epistemic probability

: $pi\left( mathbf\left\{ heta\right\}\right).,$

After the model is formed, the goal is to estimate the parameters, commonly denoted $hat\left\{mathbf\left\{ heta$, where the "hat" indicates the estimate.

One common estimator is the minimum mean squared error (MMSE) estimator, which utilizes the error between the estimated parameters and the actual value of the parameters

: $mathbf\left\{e\right\} = hat\left\{mathbf\left\{ heta - mathbf\left\{ heta\right\}$

as the basis for optimality. This error term is then squared and minimized for the MMSE estimator.

Estimators

Commonly-used estimators, and topics related to them:
*Maximum likelihood estimators
*Bayes estimators
*Method of moments estimators
*Cramér-Rao bound
*Minimum mean squared error (MMSE), also known as Bayes least squared error (BLSE)
*Maximum a posteriori (MAP)
*Minimum variance unbiased estimator (MVUE)
*Best linear unbiased estimator (BLUE)
*Unbiased estimators — see estimator bias.
*Particle filter
*Markov chain Monte Carlo (MCMC)
*Kalman filter
*Ensemble Kalman filter (EnKF)
*Wiener filter

Example: DC gain in white Gaussian noise

Consider a received discrete signal, $x \left[n\right]$, of $N$ independent samples that consists of a DC gain $A$ with additive white Gaussian noise $w \left[n\right]$ with known variance $sigma^2$ ("i.e.", $mathcal\left\{N\right\}\left(0, sigma^2\right)$).Since the variance is known then the only unknown parameter is $A$.

The model for the signal is then: $x \left[n\right] = A + w \left[n\right] quad n=0, 1, dots, N-1$

Two possible (of many) estimators are:
* $hat\left\{A\right\}_1 = x \left[0\right]$
* $hat\left\{A\right\}_2 = frac\left\{1\right\}\left\{N\right\} sum_\left\{n=0\right\}^\left\{N-1\right\} x \left[n\right]$ which is the sample mean

Both of these estimators have a mean of $A$, which can be shown through taking the expected value of each estimator

:$mathrm\left\{E\right\}left \left[hat\left\{A\right\}_1 ight\right] = mathrm\left\{E\right\}left \left[ x \left[0\right] ight\right] = A$and:$mathrm\left\{E\right\}left \left[ hat\left\{A\right\}_2 ight\right] =mathrm\left\{E\right\}left \left[ frac\left\{1\right\}\left\{N\right\} sum_\left\{n=0\right\}^\left\{N-1\right\} x \left[n\right] ight\right] =frac\left\{1\right\}\left\{N\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\} mathrm\left\{E\right\}left \left[ x \left[n\right] ight\right] ight\right] =frac\left\{1\right\}\left\{N\right\} left \left[ N A ight\right] =A$

At this point, these two estimators would appear to perform the same.However, the difference between them becomes apparent when comparing the variances.

:$mathrm\left\{var\right\} left\left( hat\left\{A\right\}_1 ight\right) = mathrm\left\{var\right\} left\left( x \left[0\right] ight\right) = sigma^2$and:$mathrm\left\{var\right\} left\left( hat\left\{A\right\}_2 ight\right)=mathrm\left\{var\right\} left\left( frac\left\{1\right\}\left\{N\right\} sum_\left\{n=0\right\}^\left\{N-1\right\} x \left[n\right] ight\right)overset\left\{independence\right\}\left\{=\right\}frac\left\{1\right\}\left\{N^2\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\} mathrm\left\{var\right\} \left(x \left[n\right] \right) ight\right] =frac\left\{1\right\}\left\{N^2\right\} left \left[ N sigma^2 ight\right] =frac\left\{sigma^2\right\}\left\{N\right\}$

It would seem that the sample mean is a better estimator since, as $N o infty$, the variance goes to zero.

Maximum likelihood

Continuing the example using the maximum likelihood estimator, the probability density function (pdf) of the noise for one sample $w \left[n\right]$ is

:$p\left(w \left[n\right] \right) = frac\left\{1\right\}\left\{sigma sqrt\left\{2 pi expleft\left(- frac\left\{1\right\}\left\{2 sigma^2\right\} w \left[n\right] ^2 ight\right)$

and the probability of $x \left[n\right]$ becomes ($x \left[n\right]$ can be thought of a $mathcal\left\{N\right\}\left(A, sigma^2\right)$)

:$p\left(x \left[n\right] ; A\right) = frac\left\{1\right\}\left\{sigma sqrt\left\{2 pi expleft\left(- frac\left\{1\right\}\left\{2 sigma^2\right\} \left(x \left[n\right] - A\right)^2 ight\right)$

By independence, the probability of $mathbf\left\{x\right\}$ becomes

:$p\left(mathbf\left\{x\right\}; A\right)=prod_\left\{n=0\right\}^\left\{N-1\right\} p\left(x \left[n\right] ; A\right)=frac\left\{1\right\}\left\{left\left(sigma sqrt\left\{2pi\right\} ight\right)^N\right\}expleft\left(- frac\left\{1\right\}\left\{2 sigma^2\right\} sum_\left\{n=0\right\}^\left\{N-1\right\}\left(x \left[n\right] - A\right)^2 ight\right)$

Taking the natural logarithm of the pdf

:$ln p\left(mathbf\left\{x\right\}; A\right)=-N ln left\left(sigma sqrt\left\{2pi\right\} ight\right)- frac\left\{1\right\}\left\{2 sigma^2\right\} sum_\left\{n=0\right\}^\left\{N-1\right\}\left(x \left[n\right] - A\right)^2$

and the maximum likelihood estimator is

:$hat\left\{A\right\} = arg max ln p\left(mathbf\left\{x\right\}; A\right)$

Taking the first derivative of the log-likelihood function

:$frac\left\{partial\right\}\left\{partial A\right\} ln p\left(mathbf\left\{x\right\}; A\right)=frac\left\{1\right\}\left\{sigma^2\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\}\left(x \left[n\right] - A\right) ight\right] =frac\left\{1\right\}\left\{sigma^2\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\}x \left[n\right] - N A ight\right]$

and setting it to zero

:$0=frac\left\{1\right\}\left\{sigma^2\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\}x \left[n\right] - N A ight\right] =sum_\left\{n=0\right\}^\left\{N-1\right\}x \left[n\right] - N A$

This results in the maximum likelihood estimator

:$hat\left\{A\right\} = frac\left\{1\right\}\left\{N\right\} sum_\left\{n=0\right\}^\left\{N-1\right\}x \left[n\right]$

which is simply the sample mean.From this example, it was found that the sample mean is the maximum likelihood estimator for $N$ samples of AWGN with a fixed, unknown DC gain.

Cramér–Rao lower bound

To find the Cramér-Rao lower bound (CRLB) of the sample mean estimator, it is first necessary to find the Fisher information number

:$mathcal\left\{I\right\}\left(A\right)=mathrm\left\{E\right\}left\left( left \left[ frac\left\{partial\right\}\left\{partial heta\right\} ln p\left(mathbf\left\{x\right\}; A\right) ight\right] ^2 ight\right)=-mathrm\left\{E\right\}left \left[ frac\left\{partial^2\right\}\left\{partial heta^2\right\} ln p\left(mathbf\left\{x\right\}; A\right) ight\right]$

and copying from above

:$frac\left\{partial\right\}\left\{partial A\right\} ln p\left(mathbf\left\{x\right\}; A\right)=frac\left\{1\right\}\left\{sigma^2\right\} left \left[ sum_\left\{n=0\right\}^\left\{N-1\right\}x \left[n\right] - N A ight\right]$

Taking the second derivative:$frac\left\{partial^2\right\}\left\{partial A^2\right\} ln p\left(mathbf\left\{x\right\}; A\right)=frac\left\{1\right\}\left\{sigma^2\right\} \left(- N\right)=frac\left\{-N\right\}\left\{sigma^2\right\}$

and finding the negative expected value is trivial since it is now a deterministic constant$-mathrm\left\{E\right\}left \left[ frac\left\{partial^2\right\}\left\{partial A^2\right\} ln p\left(mathbf\left\{x\right\}; A\right) ight\right] =frac\left\{N\right\}\left\{sigma^2\right\}$

Finally, putting the Fisher information into

:$mathrm\left\{var\right\}left\left( hat\left\{A\right\} ight\right)geqfrac\left\{1\right\}\left\{mathcal\left\{I$

results in

:$mathrm\left\{var\right\}left\left( hat\left\{A\right\} ight\right)geqfrac\left\{sigma^2\right\}\left\{N\right\}$

Comparing this to the variance of the sample mean (determined previously) shows that the sample mean is "equal to" the Cramér-Rao lower bound for all values of $N$ and $A$.The sample mean is the minimum variance unbiased estimator (MVUE) in addition to being the maximum likelihood estimator.

Fields that use estimation theory

There are numerous fields that require the use of estimation theory.Some of these fields include (but by no means limited to):

* Interpretation of scientific experiments
* Signal processing
* Clinical trials
* Opinion polls
* Quality control
* Telecommunications
* Project management
* Software engineering
* Control theory
* Network intrusion detection system

The measured data is likely to be subject to noise or uncertainty and it is through statistical probability that optimal solutions are sought to extract as much information from the data as possible.

ee also

* Best linear unbiased estimator (BLUE)
* Chebyshev center
* Completeness (statistics)
* Cramér-Rao bound
* Detection theory
* Efficiency (statistics)
* Estimator, Estimator bias
* Expectation-maximization algorithm (EM algorithm)
* Information theory
* Kalman filter
* Least-squares spectral analysis
* Markov chain Monte Carlo (MCMC)
* Matched filter
* Maximum a posteriori (MAP)
* Maximum likelihood
* Maximum entropy spectral estimation
* Method of moments, generalized method of moments
* Minimum mean squared error (MMSE)
* Minimum variance unbiased estimator (MVUE)
* Nuisance variable
* Parametric equation
* Particle filter
* Rao-Blackwell theorem
* Spectral density, Spectral density estimation
* Statistical signal processing
* Sufficiency (statistics)
* Wiener filter

References

* "Mathematical Statistics and Data Analysis" by John Rice. (ISBN 0-534-209343)
* "Fundamentals of Statistical Signal Processing: Estimation Theory" by Steven M. Kay (ISBN 0-13-345711-7)
* "An Introduction to Signal Detection and Estimation" by H. Vincent Poor (ISBN 0-387-94173-8)
* "Detection, Estimation, and Modulation Theory, Part 1" by Harry L. Van Trees (ISBN 0-471-09517-6; [http://gunston.gmu.edu/demt/demtp1/ website] )
* "Optimal State Estimation: Kalman, H-infinity, and Nonlinear Approaches" by Dan Simon [http://academic.csuohio.edu/simond/estimation/ website]

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text:
Free Version
Moderate

# Horizontal Line To Parametric

APCALC-RHVWY8

Given the parametric equations $x( t ) =\cfrac{2}{3}t^{ 3 }-8t,y( t ) =\cfrac{1}{3}t^{ 3 }+\cfrac{1}{2}t^{ 2 }-6t$, identify all values of $t$ where horizontal tangent lines exist.

A

$t=-3$

B

$t=2$

C

$t=-2$

D

$t=-3$ and $t=2$


text:
# Determining parameters of a function from its outputs

1. Oct 5, 2010

### nkinar

Suppose that I have a function $$f(a,b,c,d) = g$$, where $${a,b,c,d}$$ are four independent variables and $$g$$ is the dependent variable. Now let's say that I evaluate the function four times, each time using different inputs, and the function produces four different outputs:

$$f(a_1,b_1,c_1,d_1) = g_1$$
$$f(a_2,b_2,c_2,d_2) = g_2$$
$$f(a_3,b_3,c_3,d_3) = g_3$$
$$f(a_4,b_4,c_4,d_4) = g_4$$

Using a system of equations (linear or non-linear), is there a way to determine the values of the inputs $$a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2,a_3,b_3,c_3,d_3,a_4,b_4,c_4,d_4$$ if I know the function $$f(a,b,c,d)$$ and the values of the outputs $$g_1,g_2,g_3,g_4$$ for these inputs?

What type of algorithm could I use to determine the values of the inputs? Is there a good reference available?

Last edited: Oct 5, 2010
2. Oct 6, 2010

### CompuChip

There is probably some nice theorem in mathematics (multivariable inverse function theorem, or something like that) which tells you if this is or is not possible.

But I suspect you are looking for something a little more practical, am I right?
In that case, I would say it heavily depends on the function (I doubt that it is possible - in a practical calculational sense - for a general function, but perhaps for simple linear functions you can do it.

So can you give us a bit more info... telling us what f is exactly would help a lot, for instance.

3. Oct 6, 2010

### nkinar

Sure, CompuChip; thank you very much for your response.

Here's an example function:

$$T(a,b,c,d)= \frac{a}{4 \pi c} E_i \left( \frac{b^2}{ 4 \frac{c}{de}t } \right)$$

In the above expression, $$a,b,c,d$$ are the independent variables, and $$e,t$$ are known constants. The $$E_i$$ function is the exponential integral. I would consider this to be a non-linear equation.

Perhaps Levenberg-Marquardt optimization could be used, or some sort of non-linear curve fitting?

4. Oct 6, 2010

### CompuChip

Hmm, I was hoping for ... well, f(a, b, c, d) = a b + c - 2d, or something like that :P
I didn't expect an exponential integral, that's definitely non-linear :)
Is the c in the prefactor also one of the variables, or is "4 pi c" some physical constant?

Actually I think there is little chance of you being able to do anything useful here, except perhaps feed it to a program like Mathematica, asking it to numerically solve for a, b, c, d, and keep your fingers crossed (or do it "by hand" with your favorite optimisation technique, as you proposed).

5. Oct 6, 2010

### nkinar

Yeah, I was hoping for something like that as well! ;-) The variable c in the prefactor is also one of the variables.

I agree that using an optimization technique seems to be the way to deal with this expression. I'll be able to explore this idea further in the following weeks as I start to work more with the mathematics.

Thank you very much for your response!

6. Oct 6, 2010

### LCKurtz

What about two equations in two unknowns:

f(x,y) = sin(x)sin(y)

I give you outputs f(a,b) = 0, f(c,d) = 0. No hope I would say.

7. Oct 6, 2010

### nkinar

Ah, yes - that is a good example of a function where it doesn't work; thanks for posting this, LCKurtz. But I think that if the four outputs $$g_1,g_2,g_3,g_4$$ are different and unique, then perhaps optimization could be used. Perhaps uniqueness of the outputs implies that the inputs can be found.

8. Oct 7, 2010

### deluks917

This question is equivalent to finding the roots of function of four variables as you are essentiually settinbg f(a,b,c,d) - g = 0 and solving for (a, b,c,d). If the g1,g2,g3,g4 are close to each other finding one might help find the others. Do you need only one input for each g or the set of all inputs s.t f (a,b,c,d) = g. Is there any interval you want the solutions to be in?

You talk about uniqueness of the outputs. Is it given that no matter what four different inputs you use that f will always give four different outputs? Then f has an inverse/your solutions will be unique. In addition if this is true many versions of newton's method should work.

Last edited: Oct 7, 2010
9. Oct 7, 2010

### nkinar

Thanks, deluks917. I still need to explore the mathematics of this problem further, but I do agree that a root-finding algorithm such as Newton's method may be useful.

10. Oct 11, 2010

### hotvette

I believe the problem as originally stated can be viewed as an underdetermined system of equations (i.e. there are more unknowns than equations). If there is a solution there are likely infinitely many of them. One option is to find the minimum norm solution (i.e. optimization approach as has already been mentioned). The math is straightforward:

http://www.stanford.edu/class/ee263/lectures/min-norm.pdf

The fact that the function is non-linear in an of itself doesn't necessarily pose a problem, but the exponential integral might. What is the independent variable of the exponential integral?

Last edited: Oct 11, 2010
11. Oct 11, 2010

### nkinar

hotvette, thank you very much for your response.

Looking back at the original equation, I would like to determine $${a,b,c,d}$$ from a dataset.

$$T(a,b,c,d)= \frac{a}{4 \pi c} E_i \left( \frac{b^2}{ 4 \frac{c}{de}t } \right)$$

The idea that I've been toiling with for quite a long time is how to determine the parameters $${a,b,c,d}$$ given a number of observations. Suppose that I am to re-write the equation as a physical problem.

$$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r^2}{ 4 \alpha t } \right)$$

In the equation above, $$\Delta T(q,k,r,\alpha)$$ is the change in temperature at a given radius $$r$$ from a heater needle that has been inserted into a medium. The time $$t$$ is the time since $$t=0$$ that the heater needle has been warming up. The heat liberated is $$q$$, the thermal conductivity of the medium is $$k$$, and $$\alpha$$ is the thermal diffusivity.

Now suppose that I collect a series of measurements, $${\Delta T_1, \Delta T_2, \Delta T_3, \Delta T_4}$$ taken at different radii $${r_1,r_2,r_3,r_4}$$ from the heater needle. I can also measure the change in temperature at different times $${t_1,...,t_n}$$, where $$n$$ is the total number of time points.

With this information, can I determine the $$q,k,r,\alpha$$ parameters? If not, what information would I need?

12. Oct 11, 2010

### nkinar

Similar problems for heat conduction are described in James V. Beck and Kenneth J. Arnold's 1977 book, "Parameter Estimation in Engineering and Science."

On pg. 434, the authors write, "Without prior information the minimum number of measurements n needed to estimate p parameters is n = p. In this case X is a square matrix."

On pg. 432, the authors also state, "X represents the sensitivity matrix."

13. Oct 11, 2010

### RobertT

For this example, each output has its own unique solution. You can't use a linear system equation in the form of Ax = b to solve this example because you're dealing with the same equation (same coefficient) only with different outputs. Which implies that the inputs (a*(b^2)*d*e)/(c^2) must have different unique values.

14. Oct 11, 2010

### nkinar

Okay, thanks for pointing this out, RobertT. I suppose that I have to try something else, as per post #11 or #12 above.

15. Oct 11, 2010

### RobertT

To solve for a, b, c, d and e you need 4 more different equations than that function T that describe their relationships. Only then can you use Ax = b to find each unknown!

16. Oct 11, 2010

### nkinar

Ah, okay; what about the following. Suppose that I change the radius $$r_1,r_2,r_3,r_4$$, and take samples at four separate distances. Would this qualify as "different enough" to have a system of four non-linear equations?

$$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r_1^2}{ 4 \alpha t } \right)$$

$$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r_2^2}{ 4 \alpha t } \right)$$

$$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r_3^2}{ 4 \alpha t } \right)$$

$$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r_4^2}{ 4 \alpha t } \right)$$

17. Oct 11, 2010

### RobertT

No, they are all the same function! In this case, to find the inputs q,k,r,\alpha you need three more functions that are different than $$\Delta T(q,k,r,\alpha)= \frac{q}{4 \pi k} E_i \left( \frac{-r^2}{ 4 \alpha t } \right)$$

Last edited: Oct 11, 2010
18. Oct 11, 2010

### nkinar

Okay, thanks Robert; I think that I understand now. Essentially what is required are four unique functions that describe some sort of physical relationship between the required quantities. The key here is that the functions must be different.

Once again, thank you.

19. Oct 11, 2010

Yes! :)


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# 目に見える影を落とすには、太陽の円盤のどのくらいをカバーする必要がありますか？

しかし、再び、金環日食の間、月は太陽の円盤の約$$99$$％を占めており、それはたくさんあります。確かに、太陽の円盤のカバーされている部分が少なければ、影は見えません。しかし、火星の赤道から見た太陽の円盤の≈$$25$$％をカバーするフォボスが、非常に非常に明確な影を投げかけることを知った太陽の円盤の4分の1のみをカバーしています。

そのとき目に見える影を落とすためのカットオフポイントは正確には何ですか？金星は通過中に太陽の円盤の$$0.108$$％のみをカバーするため、目に見える影は地球に投影されません。金星がもっと大きい場合、たとえば木星のサイズだとすると、それは太陽の円盤の$$14.52$$％をカバーします。地球の表面に目に見える影を落とすにはそれで十分でしょうか？確かに、それはアンブラルシャドウではありませんが、それは無関係です。

The answer for your question is quite complicated. Basically, the cutoff point you are saying about depends on the amount of light, which is reflected from the certain surface experiencing the eclipse. Another factor is the sensitivity of your eye, which seems to be individual... http://www.mkrgeo-blog.com/the-role-of-contrast-in-ability-of-human-vision/

This cutoff is nothing like the contrast difference between two separate areas, where one of them is less illuminated. Since the albedo of the Earth is various, we shouldn't expect the fixed border between shadowed area being visible. The biggest albedo feature clouds, snow, and glacial areas, and consequently the deserts. The satellite view is usually perfect for emphasizing the shadowed area. This is driven by the huge amount of reflected light, which comes from the outside, where Sun shines in 100%. As we are moving down to the ground, the visible cutoff changes. The phenomenon, which could explain this situation is the light transition. The light transition defines the border between non-shadow and half-shadow surface, and next the half-shadow and full shadow surface. This is better described here:

http://www.mkrgeo-blog.com/what-is-a-light-transition-what-examples-of-it-can-we-see/

where you have got the experiment on a white background, which reflects the light perfectly. An analog situation happens during every solar eclipse seen from the space, even the partial one. As our sight is higher from the surface, we are able to see a bigger area. Obviously our shadowed area becomes smaller, which yields eventually the regions from outside the penumbra, fully illuminated. The details of this situation have been described here:


text:
# Saliency Guided Inter- and Intra-Class Relation Constraints for Weakly Supervised Semantic Segmentation

20 Jun 2022  ·  , , , , , ·

Weakly supervised semantic segmentation with only image-level labels aims to reduce annotation costs for the segmentation task. Existing approaches generally leverage class activation maps (CAMs) to locate the object regions for pseudo label generation. However, CAMs can only discover the most discriminative parts of objects, thus leading to inferior pixel-level pseudo labels. To address this issue, we propose a saliency guided Inter- and Intra-Class Relation Constrained (I$^2$CRC) framework to assist the expansion of the activated object regions in CAMs. Specifically, we propose a saliency guided class-agnostic distance module to pull the intra-category features closer by aligning features to their class prototypes. Further, we propose a class-specific distance module to push the inter-class features apart and encourage the object region to have a higher activation than the background. Besides strengthening the capability of the classification network to activate more integral object regions in CAMs, we also introduce an object guided label refinement module to take a full use of both the segmentation prediction and the initial labels for obtaining superior pseudo-labels. Extensive experiments on PASCAL VOC 2012 and COCO datasets demonstrate well the effectiveness of I$^2$CRC over other state-of-the-art counterparts. The source codes, models, and data have been made available at \url{https://github.com/NUST-Machine-Intelligence-Laboratory/I2CRC}.

PDF Abstract

## Results from the Paper Edit

Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers.


text:
# What causes blurriness in an optical system?

The way I understand the purpose of a typical optical system is that it creates a one to one mapping between each possible incident ray and a point on a sensor plane. This is like a mathematical function. If there was no mapping, and each ray was free to strike any point on a sensor there would be no image formed on it and it would be just blurry average light. This would be like having a camera sensor without a lens.

Now there's a very simple concept that creates this one to one mapping, pin-hole camera. In a pin-hole camera there's no blurriness possible, as long as the hole is small enough, each point opposing the hole is mapped onto one specific ray. This means this type of camera can never have a blurry image, no matter where it's focal point is. This can be proven, geometrically.

In an optical system that uses lenses to create this mapping however things are not always ideal, because blurriness does happen. Which indicates that mapping is not one to one, and that some sensor points share rays with each other creating local averages, i.'e blurriness. It is often claimed that it happens because the focal point is not at the "right place". If you consider the pin-hole model as the ideal you will understand that this is not true. Changing focal point alone will only make image seem smaller or larger. From geometrical optics alone I don't see what could possibly cause the sharing of rays. It seems to me there's more to it and that i'm not the only one confused.

So what does REALLY create the blurriness? Is there some sort of imperfection in lenses that causes them to send multiple rays to the same point on a sensor and that somehow becomes more visible at certain focal distances? This is the only explanation i have.

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I think you have the makings of a good question but you should edit it to make use of more whitespace and possibly cut its length down to something more easily digestible. – Brandon Enright Dec 26 '13 at 20:39

Your description of imaging as a mapping is a good one to begin with and it will get yoi a long way. However, even in ray optics (see my answer here for more info), i.e. when the propagation of light can be approximated by the Eikonal equation (see here for more info), the mapping of points one-to-one between the object and image plane as you describe can only happen in very special conditions. In general, a bundle of rays diverging from one point will not converge to a point after passing through an imaging system made of refracting lenses and mirrors. One has to design the system so that the convergence is well approximated by convergence to a point. As Eoin said, this non-convergence is the Ray theory description of aberration: spherical, comatic, astigmatic, trefoil, tetrafoil, pentafoil and so forth are words that are used to describe aberration with particular symmetries (spherical aberration is rotationally symmetric about the chief ray, coma flips sign on a $180^o$ rotation about the chief ray, trefoil flips sign on a $120^o$ rotation and so forth). There is also chromatic aberration, where the image point position depends on wavelength so that point sources with a spectral spread have blurred images. Lastly, the imaging surface, comprising the points of "least confusion" (i.e. those best approximating where the rays converge to a point) is always curved to some degree - it is often well approximated by an ellipsoid - and so even if convergence to points is perfect, the focal surface will not line up with a flat CCD array. This is know as lack of flatness of field: microscope objectives with particularly flat imaging surfaces bear the word "Plan" (so you have "Plan Achromat", "Plan Apochromat" and so forth).

Only very special systems allow for convergence of all ray bundles diverging from points in the object surface to precise points in the image surface. Two famous examples are the Maxwell Fisheye Lens and the Aplanatic Sphere: both of these are described in the section called "Perfect Imaging Systems" in Born and Wolf, "Principles of Optics". They are also only perfect at one wavelength.

An equivalent condition for convergence to a point is that the total optical path - the optical Lgagrangian is the same for all rays passing between the points.

Generally, lens systems are designed so that perfect imaging as you describe happens on the optical axis. The ray convergence at all other points is only approximate, although it can be made good enough that diffraction effects outweigh the non-convergence.

And of course, finally, if everything else is perfect, there is the diffraction limitation described by Eoin. The diffraction limit simply arises because converging plane waves with wavenumber $k$ cannot encode any object variation that varies at a spatial frequency greater than $k$ radians per second. This, if you like, is the greatest spatial frequency Fourier component that one has to build an image out of. Images more wiggly than this Fourier component of maximum wiggliness cannot form. A uniform amplitude, aberration-free spherical wave converges to an Airy disk, which is often taken as defining the "minimum resolvable diffraction limited distance". However, this minimum distance is a bit more complicated than that. It is ultimately defined by the signal to noise as well, so an extremely clean optical signal can see features a little bit smaller than the so-called diffraction limit, but most systems, even if their optics is diffraction limited, are further limited by noise to somewhat less than the "diffraction limit".

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These are interesting leads to explore. – sarella Dec 27 '13 at 1:09

A perfect optical system has to accept a ray coming in at any angle and bend it to exactly the correct point on the image plane. Unfortunately it isn't possible to make a lens which can do this for all possible rays because the shape would have to be different for rays of different angles hitting the same part of the lens (or mirror).

A pinhole does this by limiting the rays to a very small range of ray angles and having a curved image plane. Systems of lenses approximate the ideal behaviour by adding more and more optical surfaces of different types of curvature - each to correct a specific range of problems. But (at least with a finite number of components) you can please all the rays.

Then there are other technical differences such as chromatic aberration (different wavelengths bend different amounts in the same glass)

-
You made a very good point, that the same lens and more importantly a point on a lens has to accept rays from different angles and act perfectly. This might be one of those key differences between pin-hole and lens systems that i was seeking. – sarella Dec 26 '13 at 22:50
@sarella - correct. It is possible with holographic lenses to some extent but with other restrictions – Martin Beckett Dec 27 '13 at 4:15

A fundamental limit on the image resolution will be the diffraction limit. Here, the ray approximation is breaking down, and, in a nutshell, the finite size of the optical elements causes the waves to 'spread out' so that there is not a one-to-one mapping any more.

However, I would guess that other things, like spherical and chromatic aberration will affect image quality more than the diffraction limit for most real-world cases.

-
Yes i thought it's something like that. Then eye-doctors around the world will have to study something more advanced than geometrical optics, because their explanation of myopia never made sense to me.. :) I'll wait a bit, see if someone else want to answer. – sarella Dec 26 '13 at 21:08

No optical system exists, that can form (in the ray optics approximation) a sharp and geometrically undistorted REAL image of a REAL three dimensional object. By "ray optics approximation", I mean ignoring diffraction; NOT limited to just "paraxial" situations.

It is well known that an imaging optical system that forms an image with a LATERAL magnification of (m), also has a LONGITUDINAL or AXIAL magnification that is m^2. So if (m) is not equal to one (1) then the three d object must be distorted.

An object stretching along the axis from a distance 2f out to infinity, yields an image that is compressed between f and 2f on the image side. Since the lateral magnification varies along the axis, a three d object must give a distorted image.

So what about planar objects and images ? No optical system can form a sharp undistorted image of a plane object for two different object / image conjugate ratios (not simply object-image reversals). Any optical system that could do that for two different conjugate ratios, could also do so, at all possible conjugates; but it isn't possible. To achieve that would require that sin(A) / sin(B) = Tan(A) / Tan(B) for non paraxial ray angles A and B. The Tan ratio affects geometric distortion, while the sin ratio affects image sharpness (Aplanicity). So you can have sharp images, or you can have geometrically undistorted (plane) images, but you can't have both.

Only ONE case exists where a three d image of a three d object can be formed, that is both sharp in the geometrical case, and also a correct undistorted geometry at other than 1:1 magnification.

BUT ! The image and object cannot both be real; one must be virtual; and moreover, only for one specific 3d object and image. They are portions of spheres concentric with the aplanatic refracting surface of radius R. One has a radius R/N and the other has a radius NR, so the magnification is N^2. The spheres of course must have zero shell thickness. The radius R can be either convex (most common) or concave (most unusual, in fact generally unknown.)

Well all of this can be found in standard texts, and has been known for well over 100 years (except for the concave aplanatic case, which is quite new).

-

I'm not sure previous posts answer your final question. I want to give you an intuitive picture of the situation. Provided the detail pointed by @WetSavannaAnimal that the "true" mapping is

{object plane point} to {image plane point}

when talking of conventional optical systems, and not ray to point, as always many rays passing through your lens ends in a same point (you may think then in

{every ray coming from a same object point} to {image plane point}),

you can understand blurriness causes with a ray diagram. I will also forget diffraction, which reduces sharpness in your pinhole camera.

If your system is a single lens you may think of light as a cone:

If your image is forming in B and you put there a screen, a film, a sensor... you get your point-point mapping. If you put it in A or C you get a point-disk mapping, giving overlapping disks when your object have more than one point... These disks are usually gaussian or Airy-like, due to diffraction. As you point with the pinhole limit, finite aperture (not collecting the $2\pi \text{ srad}$ of light but a cone) is the key. If now we reduce the aperture with a diaphragm:

as we do in photography to enlarge Depth of Field, i.e., get every plane "focused", or get a picture focused wherever the plate is, as in the pinhole camera. In this limit you can also understand how the lens does not matter at all, as it is locally flat in its center. We usually work in terms of transfer functions of black boxes to connect object and screen planes.

So if you need glasses, you can make some cheap ones with some aluminium foil and a pin. That arrangement allows a wide set of interesting phenomena observation... (like watching floaters, your own fundus, test for dust in your cameras lenses or sensors, etc. More)

-


text:
# What is the significance of $\hbar$ appearing in classical equation of motion?

Books on QFT treats, any quantum field as quantized classical fields. For example, the Klein-Gordon field is first treated as a classical field $\phi(x)$ obeying classical Euler-Lagrange equation $$\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Big)=0\Rightarrow(\Box+m^2)\phi(x)=0.\tag{1}$$ However, if we restore the units $\hbar$ and $c$, it becomes $$(\Box+\frac{m^2c^2}{\hbar^2})\phi(x)=0.$$

1. How can the classical equation of motion (1) depend upon $\hbar$?

2. For electromagnetic field, this problem is not there. The Maxwell's equations do not contain $\hbar$. Is there a deep reason (apart from any accident) why Maxwell's equation do not contain $\hbar$ but the classical KG equation, classical Dirac equation does?

In classical mechanics there is no reason to identify $m$ with a mass. It is just an inverse of a length, in which case there are no factors of $\hbar$ nor $c$ in the KG equation.
• Dear @AFT If you took the Fourier mode expansion of the field $\phi(x)$, each classical mode satisfies relativistic dispersion relation. So I doubt your answer. – SRS Jan 24 '17 at 15:26
• @SRS so what? you may have a Fourier variable satisfy $k^2=m^2$ and still there is no reason to identify $m$ with a mass. In classical mechanics $k$ is not the momentum of anything. There are no particles in CM. – AccidentalFourierTransform Jan 24 '17 at 15:27
• $m$ in OP's (1) looks like the inverse of a length. – yuggib Jan 24 '17 at 15:52
• @AccidentalFourierTransform My idea was that the KG equation was historically obtained in RQM (which contains $\hbar$ and $c$). In field theory, this KG equation is thought of as the Euler-Lagrange equation derived from a Lagrangian. And therefore, this Lagrangian must also have factors of $\hbar$ and $c$. Isn't it? If the classical Lagrangian doesn't have factors of $\hbar$ and $c$, how does it suddenly appear after quantizing the theory? – SRS Jan 25 '17 at 10:42
• @SRS $\uparrow$ Yes. – AccidentalFourierTransform Feb 7 '18 at 19:15


text:
Kepler Orbit

1. Mar 21, 2010

Jerbearrrrrr

Sorry if this is in the wrong forum.

I have the equation $$r = \frac{a(1-e^2)}{1+e \cos \theta}$$,
and I'm wondering what the physical significance of the numerator is.
More specifically, what is 'a' (since e is what it usually is)?

I've seen various other representations with terms like angular momentum on the top (or rather h^2/GM).

In the context of what I'm doing, it's written this way to (presumably) uncover any implicity 'e'-dependence in the orbit.

2. Mar 21, 2010

Janus

Staff Emeritus
'a' is the semi-major axis of the orbit, or the average value of 'r' for the orbit.


text:
1. Matlab ( Matrix )

Hello
i have 3 variables w,r,d as shown in the attached file (number 1)
i tried to put the vaules of all variables in a single Matrix (y) as shown in the attached file (number 2) but i could not
my result was like this
y= w1 r1 d1
y= w2 r2 d2
y= w3 r3 d3

my goal is to have a matrix (y) as shown in the attached file (number 3) as a form of
y=
w1 r1 d1
w2 r2 d2
w3 r3 d3
then find the value of y(2,2) which is r2 =0.5000 + 0.8660i

the code of number (1) is:

n=2;
for k=0:n
w=exp((2*pi*k*j)/(n+1))
r=w+1
d=r/w
end

thank you

2. Originally Posted by Zamorano
Hello
i have 3 variables w,r,d as shown in the attached file (number 1)
i tried to put the vaules of all variables in a single Matrix (y) as shown in the attached file (number 2) but i could not
my result was like this
y= w1 r1 d1
y= w2 r2 d2
y= w3 r3 d3

my goal is to have a matrix (y) as shown in the attached file (number 3) as a form of
y=
w1 r1 d1
w2 r2 d2
w3 r3 d3
then find the value of y(2,2) which is r2 =0.5000 + 0.8660i

the code of number (1) is:

n=2;
for k=0:n
w=exp((2*pi*k*j)/(n+1))
r=w+1
d=r/w
end

thank you
Code:
n=2;
xx=[];
for k=0:n
w=exp((2*pi*k*j)/(n+1));
r=w+1;
d=r/w;
xx=[xx;[w,r,d]];
end

xx
CB

3. thank you very much CB


text:
News: There is a new Ehell Kindness Project!  Check it out in the "Extending the Hand of Kindness" folder or here: http://www.etiquettehell.com/smf/index.php?topic=139832.msg3372084#msg3372084

• August 28, 2016, 05:35:02 PM

### Author Topic: "I'm never shopping THERE again!" Share your story!  (Read 1923763 times)

6 Members and 2 Guests are viewing this topic.

#### Redsoil

• Member
• Posts: 2235
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3360 on: December 08, 2012, 09:46:36 PM »
to Dark Annie.
Look out...
It's one of the Aussie Contingent!

#### Redneck Gravy

• Member
• Posts: 3394
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3361 on: December 12, 2012, 12:58:04 PM »
My ex and I used to fish- a lot.  Every weekend, every vacation - fresh water fishing.  We have stayed in tents, fishing cabins, hotels, motels and rented condos, we are used to whatever accommodations there are.

One year we wanted to fish a lake we had never tried, so we booked into the small locally owned motel that we had passed weekly for years en route to a different lake.  I called and discussed with the owner our plans and asked about breakfast accommodations at their inhouse restaurant, cooking allowed in rooms, daily linen change, etc.  We agreed on a weekly rate with linen changes every other day and a refrigerator was to be put in our room. We also had my FIL joining us for the week. Overjoyed with an upcoming week of fishing - we take off after work Friday and drive to the motel.

Now, we get up about 5 am and hit the water well before 7:00 usually returning around 10:30 for lunch and a nap and then we fish again in the early evening until around dark, so getting our sleep at night is important to us because we get up so early.

Apparently, this motel catered to a younger crowd from a nearby college town!  Around 2:30 am they started arriving with loud music in their cars, door slamming (car & room), dragging bags up and down the sidewalk outside of our room, some doofus drug a stick down the wall of all the rooms across the door and window screen of each room - aggghhh!  I was furious to be awakened two hours before I had to get up.  Assuming this was just a one time thing we let it go and fished all day and into the evening.  That night - same scenario again, I guess after the bars closed (2:00 am) the young people found their way to their rooms and again - loud music, racket outside our room, doors slamming, yelling, singing, etc.  One bumbling fool banged on our door until my husband finally opened it and the drunken idiot realized he had the wrong room - did not apologize, just said oops and wandered off.

The next morning we hit the owner's office about 5:30 wanting to know if we could be moved or if they could be - whatever it took for us to get to sleep ALL night.

Spring break - the motel was full and they could not accommodate our request.   At that point I asked, knowing that you would be full with spring break college students why didn't you either let me know this or at least give us an end room?  Knowing that we would be wanting to sleep and get an early start on fishing as we discussed...her response, "we rely on these college kids to fill our place up spring break week, that's a healthy part of our income."  Well, I understood that, did she understand that she could have had two more rooms of happy college kids as opposed to two rooms of unhappy fishermen?  All she had to do was tell me, we could have booked elsewhere and stayed at this place during a non spring break week.  She was not willing to ask them to quiet down at 2:30 either - they're just kids having fun she said.

We moved and have never stayed there again or recommended it to any of our fishing buddies or friends.

#### suzieQ

• Member
• Posts: 680
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3362 on: December 15, 2012, 04:27:47 PM »
Netgear. Never buying another router from them again. Paid $150 for one on April 25, 2011. It is 1 year and 8 months old and it just died. Total brick. No power surge, nothing. If I had bought the same router after July 1, 2011 I would have gotten a replacement because they come with a lifetime warranty now. But nope, I bought it two months earlier and so I'm out of luck.$150 for something that lasted less than 2 years!
Never again.

My nice new Belkin router cost me less than $100 and is working nicely, right out of the box. #### Octavia • Member • Posts: 511 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3363 on: December 15, 2012, 04:30:34 PM » Netgear. Never buying another router from them again. Paid$150 for one on April 25, 2011. It is 1 year and 8 months old and it just died. Total brick. No power surge, nothing.

If I had bought the same router after July 1, 2011 I would have gotten a replacement because they come with a lifetime warranty now. But nope, I bought it two months earlier and so I'm out of luck. $150 for something that lasted less than 2 years! Never again. My nice new Belkin router cost me less than$100 and is working nicely, right out of the box.
They got me as well. I bought an expensive router in March of this year, and it died in July after the three-month warranty was up. Their customer service wanted to charge $80 just to talk to me to help troubleshoot it, with no promise of getting it working. No way. Funny, I replaced my Netgear router with a Belkin router as well. So far so good. "I never explain anything." ~Julie Andrews in Mary Poppins #### PastryGoddess • Member • Posts: 7052 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3364 on: December 15, 2012, 07:38:46 PM » I dropped the Netgear and went to an Apple Time Machine Router. I purposely purchased that model of Netgear because it was the most powerful router that supposedly could also have an external hard drive attached as well for network storage. Well after I purchased a 3TB hard drive, I come to find out that max external hard drive size was 1TB. After going round and round with them I said forget it and went and got the apple time machine router. I hooked up my 3TB hard drive and have been happy ever since. FYI Netgear has a 1 time$19 service option.  They won't tell you about it, but it is available if you ever have to call them and you are beyond your warranty period.  Just ignore the $79 package offer Maryland #### jedikaiti • Swiss Army Nerd • Member • Posts: 3572 • A pie in the hand is worth two in the mail. ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3365 on: December 15, 2012, 09:01:47 PM » Note to self: Stick with Linksys - I've always had great service, and never been charged for it. :-) What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science! "The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture #### Diane AKA Traska • Member • Posts: 5656 • Or you can just call me Diane. (NE USA EHellion) ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3366 on: December 15, 2012, 09:17:18 PM » Note to self: Stick with Linksys - I've always had great service, and never been charged for it. :-) http://www.zdnet.com/cisco-connect-cloud-chaos-7000000282/ Location: Philadelphia, PA #### jedikaiti • Swiss Army Nerd • Member • Posts: 3572 • A pie in the hand is worth two in the mail. ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3367 on: December 15, 2012, 09:19:23 PM » Good thing I shouldn't need a new one for a while. :-) (knock on wood) What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science! "The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture #### Zenith • Member • Posts: 633 • Life's Little Lunatic ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3368 on: December 15, 2012, 09:28:44 PM » Netgear. Never buying another router from them again. Paid$150 for one on April 25, 2011. It is 1 year and 8 months old and it just died. Total brick. No power surge, nothing.

If I had bought the same router after July 1, 2011 I would have gotten a replacement because they come with a lifetime warranty now. But nope, I bought it two months earlier and so I'm out of luck. $150 for something that lasted less than 2 years! Never again. My nice new Belkin router cost me less than$100 and is working nicely, right out of the box.
They got me as well. I bought an expensive router in March of this year, and it died in July after the three-month warranty was up. Their customer service wanted to charge $80 just to talk to me to help troubleshoot it, with no promise of getting it working. No way. Funny, I replaced my Netgear router with a Belkin router as well. So far so good. DLink does it to me. Every single one fried or not working out of the box. The only ones that have ever worked their little fans to the nub are Netgear and Belkin. I have a theory that everyone has a nemesis is everything, especially technology. Never had a problem with Toshiba laptops but all Apple products and software do not work for me at all, ipad, itunes, iphone, all nada. Dell commits hara kiri as soon as I get my hands on it, same with Kingston. Sansa products will work for me until the end of time and any rechargeable battery other than uniross fries for me within a few weeks. Considering a lot of computer parts all come from the same manufacturer it is rather bizarre. It's the same with power tools. Give me a name brand and I will manage to blow it up within 3 uses while the 50$ el cheapo for sale crud will work stupendously for years for me. This also works for cars, give me an older, early 90's one and it will run even if it's left with just 4 wheels and an engine. Put me in a brand new car something happens within a few hours. It's not all electrical it can be someone hitting me in the new car in a parking lot.

• Member
• Posts: 5656
• Or you can just call me Diane. (NE USA EHellion)
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3369 on: December 15, 2012, 09:32:51 PM »
Good thing I shouldn't need a new one for a while. :-) (knock on wood)

Oh, definitely.    But I'll never buy another one, I'll tell you that.  Thankfully, my DLink has been going strong for years.
Location:

#### Reika

• Member
• Posts: 3542
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3370 on: December 15, 2012, 11:28:17 PM »
I had a Linksys router that upped and died on me within 3 months, and the customer service gave me a hard time about replacing it giving me BS reason after another. So I got a D-Link which worked much better and I was glad I had replaced it when I heard about the Cisco Cloud thing. But it didn't tick me off too much since I'd gotten the Linksys for cheap off Newegg.

The one that did cheese me off and made me swear to never buy from them again (even more so than Craig Electronics with the tablet I purchased from them) was MSI. I build my own computers and a couple years ago felt the need to build a new, better machine than the one I had at the time. Which still runs, despite my mother's ability to fry computers.

Since I had a little extra cash to burn, I got one of the top of the line motherboards from MSI that had all the bells and whistles I was looking for. Which refused to do anything other than go to BIOS. It'd let me go into the BIOS to get the settings done to my specification, it detected all the RAM, and the processor with no problem, as well as the hard drive and other miscellaneous stuff. But beyond the BIOS? Nada.

When I tried to install Win 7, it kept saying the infuriating message about hitting the key to continue, wouldn't acknowledge any key I hit. So I spent over a week running every possible scenario in order to get the thing going. Installed win 7 on the new drive by hooking it up to my old computer. Still gave me the message about hitting a key to continue to the installation.

During all of this, I kept calling their tech support and not getting through to anyone. Finally I did, run through the list of everything I'd done and they say to flash the BIOS. Which according to the warranty that came with the motherboard would render said warranty null and void. I mention that I don't want to void the warranty, the board is defective, can I just get a replacement? I'm told, no, it'll be fine, we'll note it in the system so if it doesn't work, we'll replace it then.

Flashing the BIOS didn't resolve any of the issues, and even though the guy's note was in there, the returns people didn't care, I voided my warranty.

Furious, and really wanting my shiny new toy, went to CompUSA where I got an Asus motherboard that while didn't have all of the bells and whistles, it was compatible with my processor and RAM. Within an hour of getting home I had my new machine up and running. It took that long because the heatsink that I had for my processor is a royal pain to latch and unlatch.

#### norrina

• Member
• Posts: 1006
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3371 on: December 16, 2012, 09:39:51 AM »
There's a good chance I'm never shopping at Wilson's Leathers again, which is a shame because they have some really good deals, but I'm tired of playing is-it-on-sale-or-isn't-it.

I went last week, and found a hat that I was interested in buying for my brother for Christmas. It was in a display that indicated that it would be discounted to about $25. The other side of the sign indicated that a hat of this hat's original price would be discounted to about$30, but this hat was definitely on the side that indicated it would be discounted to about $25, and there were numerous other hats of the same style with it (i.e., it did not appear to have been placed on the more highly discounted side in error). Nonetheless, when I went to check out it rung up at$30, and the clerk was disinclined to honor the posted price. So I put it down and left the store.

I had a coupon at home from a previous visit that was 60% off the entire store, +20% off and +20% off again on select styles. So I went back with that, and of course the hat wasn't one of the select styles. Annoying, but I knew there was a possibility of that happening, so I wasn't too upset.

Today was the straw that broke the camels back though. I received an email that they were having an early bird special today, 9 AM to 11 AM, wherein you could take an extra 30% off the entire store including clearance. I printed the email and went back for the third time. While there, I noticed another hat that I also wanted to purchase, and with the extra 30% it would only be $10. I grabbed both hats, and when I checked out the total was higher than it should have been. I inquired, and was told that the second hat was not included in the extra 30% off. I even took out my email printout and showed the clerk where it said an extra 30% off the entire store, but she refused to honor it. I bought the first hat, since it was ringing up with the extra 30% off and I had really had my heart set on it, but that is probably going to be the last purchase I will ever make from them. #### Julia Mercer • Member • Posts: 1020 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3372 on: December 18, 2012, 02:51:36 AM » Cheese Wheelz Pizza here in Windsor. I had been having a craving for a panzerotti, and thought I would try this place just down the street from me. I went to pick it up, and got home, dug in, the bacon was mostly fatty, and still oinking (as DH likes to put it), and there was no lid on the marinara sauce that came with it, so I called them to let them know, all they had to say was that they were busy, and as for the sauce cup, there were no lids. No apology or offer for credit for future order (I've had that offered many times - not that I'm a chronic complainer - when something goes wrong with an order at other restaurants), and they didn't even seem to care. I will not go back there ever again after that! #### ica171 • Member • Posts: 1416 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3373 on: December 18, 2012, 10:06:07 AM » I can't say that I'll never shop there again, especially since I still have a gift certificate balance, but I will probably be frequenting my favorite thrift store a lot less. We went this past weekend. I was standing in the housewares section looking at dishes. A few feet away from me, out of my arm's reach and untouched by me, a stack of stuff that had been piled up and shoved onto a shelf fell over. In that stuff was a ceramic bowl of some sort (out of place, because I was standing in the wood/metal/plastic area). An employee came over saying "oh dear, oh my goodness" over and over again and I helped her clean it up. She came back a few minutes later and said, with the attitude of someone doing me a huge favor, that they weren't going to charge me for it but I needed to be careful, and if I broke anything that cost more than$10 they would charge me for it. (Note: I've been going to this store for at least 14 years--the first time I remember going was when I was 16--and have never broken anything. I've also never seen this woman before so I assume she was new. And, not that she could know this, but if I had broken it I would have apologized and offered to pay for it.)

I said "OK, but I didn't knock it over. It just fell off the shelf."

She said "yeah, but if something breaks and they see you standing by it, they'll charge you for it, so just be careful."

I very seriously doubt that's their actual policy, but I just said "OK" and went about my shopping. When we got out to the car, my husband told me that when he had gone to the bathrooms to rinse off my daughter's pacifier, he heard the woman complaining about me and calling me a liar, saying "there was no one else around and she was just standing there." I've been feeling more meh about this thrift store than usual lately, but that was when my opinion of the place really soured. Yeah, people complain about their customers. But at the very least they should make sure they're doing it out of earshot of other customers. Even if it wasn't my husband that overheard it, another customer doesn't need to hear how an employee is annoyed by someone that's spending money at their store.

That's not the only reason, but it was more or less the last straw. I was already starting to think that, since my house is mostly furnished and this is the good furniture thrift store, that I was ready to move on. The last time I went there someone had decided that hobnail milk glass was super collectible and had raised the prices on things like bud vases and tiny candy dishes to $10-$20. You don't even want to know how much the bigger pieces were. Some of the pieces weren't even actual milk glass, but just cheap lightweight knockoffs with big seams down the side. To compare, I bought two authentic hobnail bud vases there a few months ago for 60 cents each. Milk glass vases on eBay seem to go for up to $5 each. None of this was separated out in any way, it was just all mixed in with the other tchotchkes and housewares. And it was only the hobnail milk glass--the grape/fruit patterned or any other pattern was priced as usual. They've also got signs up indicating that they're trying to rebrand themselves as a thrift/antiques store. That's fine, but IMO that's going to be tricky at best and business-ending when all your customers feel alienated at worst. So I think from now on I'll just pass them by unless I've got a specific piece of furniture I'm looking for. #### Shalamar • Member • Posts: 1925 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3374 on: December 18, 2012, 10:50:19 AM » I used to take advantage of TeamBuy (which is like Groupon in that, if enough people take advantage of a deal offer, you can buy something for a lower price). In August 2012, they offered a really good deal on Gilette razor blades - I think it was two dozen razor blades for$50, or something like that.  Those things are so expensive, and I knew that two dozen would last my husband for ages, so I jumped on it.

Towards the end of September, I realized that I hadn't gotten my blades yet.  I e-mailed TeamBuy and was told, somewhat snippily, that the shipments wouldn't go out until 4-6 weeks after the offer expired, which was September 15.  So, in other words, I might not receive my purchase until early November.  Fine.

Early November came and went.  I e-mailed them again.  "You'll receive them by November 15 at the latest!"  Well, November 15 came and went.  I e-mailed them again.   "They're being held up at customs.  You'll get them by the end of November."  Nope.  I e-mailed them again and said "This is ridiculous.  I've been waiting for over three months."  They finally admitted that the vendor had let them down, then they said magnanimously that they would give me a refund - in TeamBuy Bucks.  Riiiight.  I replied "Not good enough.  I never intend to buy anything from you again, so I require a refund on my Mastercard, please."  Their latest e-mail said "It looks like you had received the wrong email. Here is what you should have received as the refund was processed to your credit card.  I have now refunded your purchase and you should see it reflected on your statement in the next 3-5 business days. You will also receive an email with your refund transaction ID shortly."

I'm not holding my breath.


text:
# For traffic control, a CCTV camera is fixed on a straight and vertical pole. The camera can see 113m distance straight from the top. If the area visible by the camera around the pole is 39424${m^2}$, then find the height of the pole.

Verified
147.9k+ views
Hint: In this particular question use the concept that the area visible by the camera around the pole is always in circular shape, so equate the given area to the area of the circle and calculate the radius of the circle, later on in the solution use the concept of Pythagoras theorem so use these concepts to reach the solution of the question.

Let OA be the vertical pole as shown in the above figure, at point A, CCTV camera is fixed and the reach of the CCTV camera is up to 113m as shown in the above figure.
Let the reach is up to point B, so AB = 113m (see figure).
Let the height of the vertical pole be h meters.
Therefore, OA = h meters.
Now it is given that the area visible by the camera around the pole is 39424${m^2}$.
As we all know that the visible area is always around a circle as shown in the figure, and the area of the circle is given as $\pi {r^2}$ square units, where r is the radius of the circle.
Therefore, $\pi {r^2} = 39424$
$\Rightarrow {r^2} = \dfrac{{39424}}{\pi } = \dfrac{{39424}}{{\dfrac{{22}}{7}}} = 12544 = {\left( {112} \right)^2}$, $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
$\Rightarrow r = 112$Meters.
As the pole is vertical so it makes a 90 degree with the radius of the circle.
So in triangle OAB apply Pythagoras theorem we have,
Now as we know that according to the Pythagoras theorem that in a right triangle the sum of the square of the base and perpendicular is equal to the square of the hypotenuse.
$\Rightarrow {\left( {{\text{base}}} \right)^2} + {\left( {{\text{perpendicular}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^{^2}}$................. (1)
Now in triangle OAB, base = OB, perpendicular = OA and hypotenuse = AB
Now from equation (1) we have,
$\Rightarrow {\left( {{\text{OB}}} \right)^2} + {\left( {{\text{OA}}} \right)^2} = {\left( {{\text{AB}}} \right)^{^2}}$
Now substitute the values we have,
$\Rightarrow {\left( {{\text{112}}} \right)^2} + {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}}$
$\Rightarrow {\left( {\text{h}} \right)^2} = {\left( {{\text{113}}} \right)^{^2}} - {\left( {{\text{112}}} \right)^2} = 225 = {\left( {15} \right)^2}$
$\Rightarrow h = 15$ Meters.
So the height of the vertical pole is 15 meters.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the area of the circle which is given as $\pi {r^2}$ square units, and always recall the definition of the Pythagoras theorem which is stated above then simplify substitute the values as above and simplify we will get the required height of the pole.


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# Library for easily interfacing with USB tranciever

Im very new to the field of anything USB related, sorry in advance! I hope my question is clear in conveying my level of understanding and of what I wish to learn.

My project involves connecting a digital signal processor to a computer via a USB at as low cost as possible.

Ive found a chip that looks promising to ease in doing so, the USB3310C a "Hi-Speed USB Transceiver with ULPI Interface". Right now Im thinking of using a DSP (dsPIC33EP32MC202) to interface between the device sensors, logic and the ULPI provided by the USB transceiver.

What I think I need is some kind of library (C/C++) to run on the processor to manage the USB connection to the host, responding to tokens and such (Dont know much about how that works). What I need in the end is something that can advertise itself as my custom device, that I should then be able to exchange data with on the computer.

From how I see it there are five parts to this

1. Slave side software that more simply handles the data itself, streaming to and from the computer, processing a stream of ~220MB/s.
2. Slave side 'driver', controlling and responding to low level requests. (What I need)
3. Slave side hardware, the physical layer.
4. Host side drivers, I want to avoid having to create drivers, re-purposing another driver, maybe some kind of general purpose and cross platform driver.
5. Host side software, some program to stream to and from the device.

I'm sure there is much more to know on this subject, but for now I just want to successfully send data around ~30MB/s from device to host, and a small amount of control data from host to device.

In summary, I need some way to interface with a USB host, having all the low level stuff managed by something other than myself so I can simply focus on the data itself. Ideally some kind of C/C++ library that can interface with the USB3310C chip, or some alternate low cost chip.

• What you are trying to do is not simply at all and fairly unlikely to work on a first attempt. You should probably start with a simpler lower performance case to gain some experience with some of the challenges of working with USB. It's not clear that your chosen parts are compatible, and it seems fairly unlike that your external DSP would be faster than using the PC's processor to do the task, even without the latency and complication of getting data out to the peripheral and results back in. Sep 2 '20 at 0:08
• That dsPIC doesn't have a USB ULPI interface to connect to a USB3310 - and there's no way you're going to bit-bang it. If you're really married to the dsPIC devices and you need built-in USB then you'll have to use one of the very few dsPIC which support it like a dsPIC33EP256MU8xx/dsPIC33EP512MU8xx. Sep 2 '20 at 1:05
• @brhans the data that is being streamed into the DSP is much more than 60MB/s, I dont want to be having to stream 270MB/s via usb, If 30MB/s is a challenge, then that will be impossible for my novice experience. I would assume the chip would be able to handle that throughput in a parallel manner?
– user224392
Sep 2 '20 at 1:31
• Yes, ULPI is a parallel interface. No, you 're not going to be able to connect it to your chosen PIC. You need hardware support in the micro to be able to connect a ULPI USB PHY. Even though it may be theoretically possible to bit-bang the ULPI interface in the micro, the effort would be pointless as you'd never be able to do it fast enough to make it worthwhile. Sep 2 '20 at 1:59


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# Bug: Pasting plain text takes on the default font, not the font of the text it is being pasted into

## Idea

Hi there, I'm seeing a bug with Evernote where pasting takes on the default font (Arial 13), not the font of the text it is being pasted into (in this case Arial 17).

Original state:

Pasting plain text:

Note font size difference.

But if I paste it between some content:

We see that it takes on the correct font:

## Recommended Posts

Actually, I can't even reproduce this anymore. I'd paste regularly – misformatted. Then try doing the paste between trick – won't work. Then I'll paste again, and it's correctly formatted. Very inconsistent behavior.


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# Math Help - Algebra question

1. ## Algebra question

Hi I'm on this question where I hacve to find x:

3x+5 - 2x-1 = 3
4 6

and i ended up with 9x + 15 - 4x - 2 = 36
12 = 12

so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

Now for my question:

When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...

edit: for some reason the denominators havent come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
and the second error is that both sides should have 12 as their denominators

2. Originally Posted by david18
Hi I'm on this question where I hacve to find x:

3x+5 - 2x-1 = 3
4 6

and i ended up with 9x + 15 - 4x - 2 = 36
12 = 12

so i removed the 12s to leave me with 9x + 15 - 4x - 2 = 36

Now for my question:

When I am simplifying the left hand side; does it become 5x + 13 or 5x + 17 because if the equation should be written as (9x + 15) - (4x - 2) it would mean a -- which is a plus to make 5x + 17...

edit: for some reason the denominators havent come up where i wanted them to be. but its 3x + 5 all over 4 and 2x - 1 all over 6
and the second error is that both sides should have 12 as their denominators
$\frac {3x + 5}{4} - \frac {2x - 1}{6} = 3$ .....the LCD is 12

$\Rightarrow \frac {3(3x + 5) - 2(2x - 1)}{12} = 3$

$\Rightarrow \frac {9x + 15 - 4x + 2}{12} = 3$

$\Rightarrow 9x + 15 - 4x + 2 = 36$ ........multiplied both sides by 12

$\Rightarrow 5x + 17 = 36$

Now continue

3. Thanks that was really helpful... you used a better method than mine which involved making the 3 top heavy and silly things like that.

4. Originally Posted by david18
Thanks that was really helpful... you used a better method than mine which involved making the 3 top heavy and silly things like that.
"making the 3 top heavy"? what's that?

did you get my method? we could have also muliplied the first fraction by $\frac {3}{3}$ and the second by $\frac {2}{2}$ that would make the denominators the same, and we just add the tops from there. there is also the cross multiplication method for combining fractions, but in this case, that was more trouble than it's worth