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http://electronics.stackexchange.com/questions/121363/why-do-current-transformers-have-a-minimum-frequency-rating | # Why do current transformers have a minimum frequency rating?
Datasheets of current transformers usually specify a working frequency range. In other words, they specify a minimum and maximum frequency for normal operation.
I understand that the core losses grow exponentially with frequency. So the datasheets give an upper limit for frequency. But, why is there a lower frequency limit? What happens when we apply an AC current to the input of a current transformer with a frequency lower than the rated minimum frequency?
An example product:
EPCOS B82801B Series
Frequency range: 50 kHz ... 1 MHz
-
For a voltage transformer: -
If the applied voltage were very low in frequency i.e. 0.1 Hz, the primary current taken (if the voltage were 120VAC or 230VAC) would be several tens of amps and the winding would fry and burn. Nonetheless, what about trying it at 1VAC? Now the current (remember we're talking 0.1 Hz) is a couple of hundred milliamps (that's OK) and the flux is just about avoiding saturation (phew) but, a decent flux rate of change is needed to produce the requisite voltage in the secondary and at 0.1 Hz it aint changing that quick so you get a really small output. No big deal output voltage and input voltage are still approximately related by the turns ratio but you can't connect to live and neutral anymore because it'll burn.
At some point above (maybe) 35 Hz, you can run full voltage.
Current transformer
Primary current is always defined by the "external circuit" - that's good because now we don't need to worry about saturation or the current getting bigger at lower frequencies. Normally the current transformer has a burden resistor of 1 to 100 ohms and this totally swamps the magnetization impedance; consider, at 50 Hz the mag inductance is 10 uH - this has a reactance of 3.14 milliohms. Now the burden may be 10 ohms and the winding ratio maybe 1:200 - referred to the primary the 10 ohms is 40,000 times smaller at 250 micro ohms. The mag inductive reactance (3 milliohms) is swamped by the primary-referred burden of 250 micro ohms.
At 5 Hz the burden is still 250 micro ohms and the mag reactance is 300 micro ohms - see how things are changing - what this means is that the CT at low frequencies can no-longer sustain the maths that underlies it. At 0.5 Hz the output will be negligible because the primary referred burden resistance is totally swamped by the extremely small magnetization impedance.
Equivalent circuits of VT and CT
Comparing CT with VT, you can forget about the primary DC resistance, leakage inductance and core losses - all you have a fat wire going thru a hole and it generates magnetism hence Xm is still present. If the secondary leakage impedances are small compared to the burden, then you can usually forget these as well. What remains is the burden (redirected to be on the primary) in parallel with Xm and, the burden normally dominates i.e. it is the lowest impedance because of turns-ratio-squared.
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Thanks for correcting me and teaching! +1 – horta Jul 15 '14 at 21:33
Ha! Me too :) I missed the "current" part of the current transformer. Thanks! – bitsmack Jul 15 '14 at 21:46
Could the turns ratio be 1:200 in your example? – hkBattousai Jul 16 '14 at 0:18
@hkBattousai oops - well spotted on the typo!!!! – Andy aka Jul 16 '14 at 7:18
Will the reflected burden resistor be in series or parallel with $X_m$ in your equivalent model? – hkBattousai Jul 16 '14 at 10:15
Non-superconducting current transformers have a rolloff below a certain frequency, so the output for a given current simply goes down as the frequency decreases. It's not core saturation nor is it directly related to the inductance- this is a linear effect.
The -3dB rolloff frequency is determined from L/R where L is the inductance of the transformer secondary and R is the resistance of the secondary circuit.
Take the case of the B82801B0504A050, which has a recommended 50R load and inductance of 500uH looking into the secondary. The L/R constant is 10usec, corresponding to a -3dB frequency of 16kHz. So the error at the rated 50kHz should be minimal.
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Its the primary magnetization inductance being low impedance at low frequencies that spoils the game. – Andy aka Jul 15 '14 at 21:43
@Andyaka Hmm.. amperes * turns should be the same regardless of frequency when it's driven with a constant current, so I don't see how the primary inductance enters into it directly. What am I missing? – Spehro Pefhany Jul 15 '14 at 22:12
The reflected impedance of the burden to the primary is normally at least ten times lower than the ct primary mag Reactance hence, it dominates and the device acts as one expects. At low frequency, the mag inductive Reactance is lower than the reflected burden and the burden does not receive the current it deserves but that's life! – Andy aka Jul 15 '14 at 22:29
How did you find 16kHz frequency from the 10$\mu$s time constant? – hkBattousai Jul 15 '14 at 23:52
@hkBattousai $\omega$ = 2$\pi$f – Spehro Pefhany Jul 16 '14 at 0:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8530458211898804, "perplexity": 1932.1753788840788}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246636255.43/warc/CC-MAIN-20150417045716-00085-ip-10-235-10-82.ec2.internal.warc.gz"} |
http://mathhelpforum.com/statistics/176878-maximum-likelihood-estimate.html | 1. ## Maximum Likelihood Estimate
Hey guys, I'm having difficulty with this problem. I don't even really know how to start it, or what formulas to use.
Find the maximum likelihood estimate for the parameter mu of the normal distribution with known variance sigma^2 = sigma_0_^2
_0_ meaning subscript.
Any help on this problem would be greatly appreciated!
2. it should still be the sample mean
Just write the joint density, which is called the likelihood function
Then take the logarithm and differentiate (twice) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9874400496482849, "perplexity": 490.24910185346164}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171664.76/warc/CC-MAIN-20170219104611-00078-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathhelpforum.com/algebra/148622-evaluate-9-3-2-a.html | # Math Help - Evaluate 9^(-3/2)
1. ## Evaluate 9^(-3/2)
Evaluate:
9^(-3/2)
Got a mental blank, please guys help me quickly!
How is it 1/27? Please show complete working out.
2. $
9^{1/2}\;=\;3
$
4. 9^(-3/2)
9^(-3/2) = 1/(square root of 9^3 )
= 1/27
5. For any number $a\ne 0$, the number $a^{-1}$ is defined as such a number that $a\cdot{a^{-1}} = a^{-1}\cdot{a} = 1.$ If we for example take $3$, then the number $3^{-1}$ such that $3\cdot{3^{-1}} = 3^{-1}\cdot{3} = 1$ is clearly $\dfrac{1}{3}$, as $3\cdot\left(\dfrac{1}{3}\right) = \left(\dfrac{1}{3}\right)\cdot{3} = 1.$ It can be easily seen that $a^{-1}$ is always $\dfrac{1}{a}$ (with the exception of $a = 0$, of course). So, for your case, we have $9^{-\frac{3}{2}} = \dfrac{1}{9^\frac{3}{2}}$ Remember that $X^{\frac{Y}{Z}} = (X^{\frac{1}{Z}})^Y$, so you have $\dfrac{1}{9^\frac{3}{2}} = \dfrac{1}{(9^{\frac{1}{2}})^3}.$ Since $9^{\frac{1}{2}} = 3$, we get $\dfrac{1}{3^3},$ which is of course $27$.
6. Originally Posted by Joker37
Evaluate:
9^(-3/2)
Got a mental blank, please guys help me quickly!
How is it 1/27? Please show complete working out.
You need to thoroughly review your basic index laws (and that's you job, not ours).
$9^{-3/2} = \frac{1}{9^{3/2}} = \frac{1}{(9^{1/2})^3}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9367200136184692, "perplexity": 649.1249242082105}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507444582.16/warc/CC-MAIN-20141017005724-00191-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/80118?sort=oldest | ## Stable Law with Rates
If $X_{i}$ are a bunch of iid random variables with mean 0 and finite second moments, we know that $\sum_{i=1}^{n} \frac{X_{i}}{\sqrt{n}}$ converges in law to a Gaussian. Furthermore, by the Berry-Esseen theorem, we have some bounds on the rates of this convergence. Similar results hold, even if $X_{i}$ are only mixing in some sense, rather than actually iid.
If the $X_{i}$ are now instead regularly varying with exponent 1 (and satisfying some other conditions), and are 'very strongly mixing' but not iid, then I know that their suitably scaled partial sums converge in law to an appropriate Levy function with exponent 1.
My question is, what is a reasonable rate for this convergence? Of course, it will have something to do with the mixing rate, and with the constants associated with regular variation, and maybe other things. Are there any standard results in this direction?
I am not much of an expert, and have so far tried only some very standard tricks (e.g. concentration of regeneration times) which work in very specialized circumstances, and often not very well. Any pointers would be appreciated.
Edited: In response to Igor's very nice answer below, to state explicitly that I did not wish to assume the $X_{i}$ were iid.
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## 1 Answer
Check out this paper:
Harmonic mean, random polynomials and stochastic matrices Natalia L. Komarova, , Igor Rivin
There are some results of the sort you are asking about there (and since the authors were not then and are not now probabilists, the proofs start from essentially nothing).
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Hello Igor, thank you for the response - the paper is quite interesting (and the application is even somewhat related to what I was thinking of studying). I think that you are studying the iid case, rather than the mixing case. On the other hand, the rates are much better than I would have expected. I may try to use the calculations anyway... – QAMS Nov 7 2011 at 4:44 Also, I just realized I was being a little rude there with the edit - I think my initial question was quite unclear, with one mention of iid and then another mention of a mixing rate. Perhaps the new version is better. – QAMS Nov 7 2011 at 4:45 No problem. I don't know much about the mixing case, but am guessing that some of the techniques work there too... – Igor Rivin Nov 7 2011 at 20:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9732118844985962, "perplexity": 417.49096058066925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368706933615/warc/CC-MAIN-20130516122213-00004-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://socratic.org/questions/what-is-the-discriminant-of-x-2-9-0-and-what-does-that-mean | Algebra
Topics
# What is the discriminant of x^2-9=0 and what does that mean?
Jul 30, 2015
In your case, $\Delta = 36$, which means that your equation has two distinct, rational solutions.
#### Explanation:
The general form of a quadratic equation is
$a {x}^{2} + b x + c = 0$
for which the discriminant is defined as
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
In your case, $a = 1$, $b = 0$, and $c = - 9$, so the discriminant becomes
$\Delta = {0}^{2} - 4 \cdot 1 \cdot \left(- 9\right) = \textcolor{g r e e n}{36}$
A quadratic equation that has $\Delta > 0$ has two distinct, real solutions. Moreover, since $\Delta$ is a perfect square, two two solutions will be rational numbers.
The general form for the solutions of a quadratic equation is
color(blue)(x_(1,2) = (-b +- sqrt(Delta))/(2a)
In your case, those two solutions will be
${x}_{1 , 2} = \frac{0 \pm 6}{2} = \left\{\begin{matrix}{x}_{1} = 3 \\ {x}_{2} = - 3\end{matrix}\right.$
Note that these solutions could have easily been determined by
${x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} = 9$
$\sqrt{{x}^{2}} = \sqrt{9} \implies {x}_{1 , 2} = \pm 3$
##### Impact of this question
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http://monolix.lixoft.com/tasks/population-parameter-estimation-using-saem/ | Select Page
# Population parameter estimation using SAEM
### Purpose
The estimation of the population parameters is the key task in non-linear mixed effect modeling. In Monolix, it is performed using the Stochastic Approximation Expectation-Maximization (SAEM) algorithm [1]. SAEM has been shown to be extremely efficient for both simple and a wide variety of complex models: categorical data [2], count data [3], time-to-event data [4], mixture models [56], differential equation based models, censored data [7], … The convergence of SAEM has been rigorously proven [1] and its implementation in Monolix is particularly efficient. No other algorithms are available in Monolix.
### Running the population parameter estimation task
#### Overview
The pop-up window which permits to follow the progress of the task is shown below. The algorithm starts with a small number (5 by default) of burn-in iterations for initialization which are displayed in the following way: (note that this step can be so fast that it is not visible by the user)
Afterwards, the evolution of the value for each population parameter over the iterations of the algorithm is displayed. The red line marks the switch from the exploratory phase to the smoothing phase. The exact value at each iterations can be followed by hovering over the curve (as for Cl_pop below). The convergence indicator (in purple) helps to detect that convergence has been reached (see below for more details).
The “Population parameter” estimation task must be launched before running any other task. To skip this task, the user can fix all population parameters. If all population parameters have been set to “fixed”, the estimation will stop after a single iteration and allow the user to continue with the other tasks.
#### The convergence indicator
The convergence indicator (also sometimes called complete likelihood) is defined as the joint probability distribution of the data and the individual parameters and can be decomposed using Bayes law:
$$\sum_{i=1}^{N_{\text{ind}}}\log\left(p(y_i, \phi_i; \theta)\right)=p(y_i| \psi_i; \theta)p(\psi_i; \theta)$$
Those two terms have an analytical expression and are easy to calculate, using as $$\phi_i$$ the individual parameters drawn by MCMC for the current iteration of SAEM. This quantity is calculated at each SAEM step and is useful to assess the convergence of the SAEM algorithm.
The convergence indicator aggregates the information from all parameters and can serve to detect if the SAEM algorithm has already converged or not. When the indicator is stable, that is it oscillates around the same value without drifting, then we can be pretty confident that the maximum likelihood has been achieved. The convergence indicator is used, among other measures, in the auto-stop criteria to switch from the exploratory phase to the smoothing phase.
Note that the likelihood (i.e the objective function) $$\sum_{i=1}^{N_{\text{ind}}}\log\left(p(y_i; \theta)\right)$$ cannot be computed in closed form because the individual parameters $$\phi_i$$ are unobserved. It requires to integrate over all possible values of the individual parameters. Thus, to estimate the log-likelihood an importance sampling Monte Carlo method is used in a separate task (or an approximation is calculated via linearization of the model).
#### The simulated annealing
The simulated annealing option (setting enabled by default) permits to keep the explored parameter space large for a longer time (compared to without simulated annealing). This allows to escape local maximums and improve the convergence towards the global maximum.
In practice, the simulated annealing option constrains the variance of the random effects and the residual error parameters to decrease by maximum 5% (by default – the setting “Decreasing rate” can be changed) from one iteration to the next one. As a consequence, the variances decrease more slowly:
The size of the parameter space explored by the SAEM algorithm depends on individual parameters sampled from their conditional distribution via Markov Chain Monte Carlo. If the standard deviation of the conditional distributions is large, the individual parameters sampled at iteration k can be quite far away from those at iteration (k-1), meaning a large exploration of the parameter space. The standard deviation of the conditional distribution depends on the standard deviation of the random effects (population parameters ‘omega’). Indeed, the conditional distribution is $$p(\psi_i|y_i;\hat{\theta})$$ with $$\psi_i$$ the individual parameters for individual $$i$$, $$\hat{\theta}$$ the estimated population parameters, and $$y_i$$ the data (observations) for individual $$i$$. The conditional distribution thus depends on the population parameters, and the larger the population parameters ‘omega’, the larger the standard deviation of the conditional distribution. That’s why we want to keep large ‘omega’ values during the first iterations.
#### Methods for the parameters without variability
Parameters without variability are not estimated in the same way as parameters with variability. Indeed, the SAEM algorithm requires to draw parameter values from their marginal distribution, which exists only for parameters with variability.
Several methods can be used to estimate the parameters without variability. By default, these parameters are optimized using the Nelder-Mead simplex algorithm (Matlab’s fminsearch method). Other options are also available in the SAEM settings:
• No variability (default): optimization via Nelder-Mead simplex algorithm
• Add decreasing variability: an artificial variability (i.e random effects) is added for these parameters, allowing estimation via SAEM. The variability starts at omega=1 and is progressively decreased such that at the end of the estimation process, the parameter has a variability of 1e-5. The decrease in variability is exponential with a rate based on the maximum number of iterations for both the exploratory and smoothing phases. Note that if the autostop is triggered, the resulting variability might me higher.
• Variability at the first stage: during the exploratory phase of SAEM, an artificial variability is added and progressively forced to 1e-5 (same as above). In the smoothing phase, the Nelder-Mead simplex algorithm is used.
Depending on the specific project, one or the other method may lead to a better convergence. If the default method does not provide satisfying results, it is worth trying the other methods. In terms of computing time, if all parameters are without variability, the first option will be faster because only the Nelder-Mead simplex algorithm will be used to estimate all the fixed effects. If some parameters have random effects, the first option will be slower because the Nelder-Mead and the SAEM algorithm are computed at each step. In that case the second or third option will be faster because only the SAEM algorithm will be required when the artificial variability is added.
Alternatively, the standard deviation of the random effects can be fixed to a small value, for instance 5% for log-normally distributed parameters. To enforce a fixed value, click on the wheel next to the initial omega value and select “Fixed”.
With this method, the SAEM algorithm can be used, and the variability is kept small.
### Outputs
#### In the graphical user interface
The estimated population parameters are displayed in the Pop. Param section of the Results tab. Fixed effects are named “*_pop”, the standard deviation of the random effects “omega_*”, parameters of the error model “a”, “b”, “c”, the correlation between random effects “corr_*_*” and parameters associated to covariates “beta_*_*”.
When the “Standard errors” task has also been run, the standard error (s.e), relative standard error (r.s.e) and p-values for covariate betas are also displayed in this result section. The total elapsed time for this task is shown at the bottom.
Notice that there is a “Copy table” icon on the top of each table to copy them in Excel, Word, … The table format and display will be kept.
#### In the output folder
After having run the estimation of the population parameters, the following files are available:
• summary.txt: contains the estimated population parameters, in a format easily readable by a human (but not easy to parse for a computer)
• populationParameters.txt: contains the estimated population parameters (by default in csv format)
• predictions.txt: contains for each individual and each observation time, the observed data (y), the prediction using the population parameters and population median covariates value from the data set (popPred_medianCOV), the prediction using the population parameters and individual covariates value (popPred), the prediction using the individual approximate conditional mean calculated from the last iterations of SAEM (indivPred_SAEM) and the corresponding weighted residual (indWRes_SAEM).
• IndividualParameters/estimatedIndividualParameters.txt: individual parameters corresponding to the approximate conditional mean, calculated using the last estimations of SAEM (*_SAEM)
• IndividualParameters/estimatedRandomEffects.txt: individual random effects corresponding to the approximate conditional mean, calculated using the last estimations of SAEM (*_SAEM)
More details about the content of the output files can be found here.
### Settings
The settings are accessible through the interface via the button next to the parameter estimation task.
Burn-in phase:
The burn-in phase corresponds to an initialization of SAEM: individual parameters are sampled from their conditional distribution using MCMC using the initial values for the population parameters (no update of the population parameter estimates).
Note: the meaning of the brun-in phase in Monolix is different to what is called burn-in in Nonmem algorithms.
• Number of iterations (default: 5): number of iterations of the burn-in phase
Exploratory phase
• Auto-stop criteria (default: yes): if ticked, auto-stop criteria are used to automatically detect convergence during the exploratory phase. If convergence is detected, the algorithm switches to the smoothing phase before the maximum number of iterations. The criteria take into account the stability of the convergence indicator, omega parameters and error model parameters.
• Maximum number of iterations (default: 500, if auto-stop ticked): maximum number of iterations for the exploratory phase. Even if the auto-stop criteria are not fulfilled, the algorithm switches to the smoothing phase after this maximum number of iterations. A warning message will be displayed in the GUI if the maximum number of iterations is reached while the auto-stop criteria are not fulfilled.
• Minimum number of iterations (default: 150, if auto-stop ticked): minimum number of iterations for the exploratory phase. This value also corresponds to the interval length over which the auto-stop criteria are tested. A larger minimum number of iterations means that the auto-stop criteria are harder to reach.
• Number of iterations (default: 500, if auto-stop unticked): fixed number of iterations for the exploratory phase.
• Step-size exponent (default: 0): The value, comprised between 0 and 1, represents memory of the stochastic process, i.e how much weight is given at iteration k to the value of the previous iteration compared to the new information collected. A value 0 means no memory, i.e the parameter value at iteration k is build based on the information collected at that iteration only, and does not take into account the value of the parameter at the previous iteration.
• Simulated annealing (default: enabled): the Simulated Annealing version of SAEM permits to better explore the parameter space by constraining the standard deviation of the random effects to decrease slowly.
• Decreasing rate for the variance of the residual errors (default: 0.95, if simulated annealing enabled): the residual error variance (parameter “a” for a constant error model for instance) at iteration k is constrained to be larger than the decreasing rate times the variance at the previous iteration.
• Decreasing rate for the variance of the individual parameters (default: 0.95, if simulated annealing enabled): the variance of the random effects at iteration k is constrained to be larger than the decreasing rate times the variance at the previous iteration.
Smoothing phase
• Auto-stop criteria (default: yes): if ticked, auto-stop criteria are used to automatically detect convergence during the smoothing phase. If convergence is detected, the algorithm stops before the maximum number of iterations.
• Maximum number of iterations (default: 200, if auto-stop ticked): maximum number of iterations for the smoothing phase. Even if the auto-stop criteria are not fulfilled, the algorithm stops after this maximum number of iterations.
• Minimum number of iterations (default: 50, if auto-stop ticked): minimum number of iterations for the smoothing phase. This value also corresponds to the interval length over which the auto-stop criteria is tested. A larger minimum number of iterations means that the auto-stop criteria is harder to reach.
• Number of iterations (default: 200, if auto-stop unticked): fixed number of iterations for the smoothing phase.
• Step-size exponent (default: 0.7): The value, comprised between 0 and 1, represents memory of the stochastic process, i.e how much weight is given at iteration k to the value of the previous iteration compared to the new information collected. The value must be strictly larger than 0.5 for the smoothing phase to converge. Large values (close to 1) will result in a smoother parameter trajectory during the smoothing phase, but may take longer to converge to the maximum likelihood estimate.
Methodology for parameters without variability (if parameters without variability are present in the model):
The SAEM algorithm requires to draw parameter values from their marginal distribution, which does not exist for parameters without variability. These parameters are thus estimated via another method, which can be chosen among:
• No variability (default choice): After each SAEM iteration, the parameter without variability are optimized using the Nelder-Mead simplex algorithm. The absolute tolerance (stopping criteria) is 1e-4 and the maximum number of iterations 20 times the number of parameters to calculate via this algorithm.
• Add decreasing variability: an artificial variability is added for these parameters, allowing estimation via SAEM. The variability is progressively decreased such that at the end of the estimation process, the parameter has a variability of 1e-5.
• Variability in the first stage: during the exploratory phase, an artificial variability is added and progressively forced to 1e-5 (same as above). In the smoothing phase, the Nelder-Mead simplex optimization algorithm is used.
Handling parameters without variability is also discussed here.
### Good practice, troubleshooting and tips
#### Choosing to enable or disable the simulated annealing
As the simulated annealing option permits to more surely find the global maximum, it should be used during the first runs, when the initial values may be quite far from the final estimates.
On the other side, the simulated annealing option may keep the omega values artificially high, even after a large number of SAEM iterations. This may prevent the identification of parameters for which the variability is in fact zero and lead to NaN in the standard errors. So once good initial values have been found and there is no risk to fall in a local maximum, the simulated annealing option can be disabled. Below we show an example where removing the simulated annealing permits to identify parameters for which the inter-individual variability can be removed.
Example: The dataset used in the tobramycin case study is quite sparse. In these conditions, we expect that estimating the inter-individual variability for all parameters will be difficult. In this case, the estimation can be done in two steps, as shown below for a two-compartments model on this dataset:
• First, we run SAEM with the simulated annealing option (default setting), which facilitates the convergence towards the global maximum. All four parameters V, k, k12 and k21 have random effects. The estimated parameters are shown below:
The parameters omega_k12 and omega_k21 have high standard errors, suggesting that the variability is difficult to estimate. The omega_k12 and omega_k21 values themselves are also high (100% inter-individual variability), suggesting that they may have been kept too high due to the simulated annealing.
• As a second step, we use the last estimates as new initial values (as shown here), and run SAEM again after disabling the simulated annealing option. On the plot showing the convergence of SAEM, we can see omega_V, omega_k12 and omega_k21 decreasing to very low values. The data is too sparse to correctly identify the inter-individual variability for V, k12 and k21. Thus, their random effects can be removed, but the random effect of k can be kept.
Note that because the omega_V, omega_k12 and omega_k21 parameters decrease without stabilizing, the convergence indicator does the same. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8835046291351318, "perplexity": 1038.9900530553948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203168.70/warc/CC-MAIN-20190324022143-20190324044143-00350.warc.gz"} |
https://crypto.stackexchange.com/questions/87332/doubt-on-elliptic-curve-over-a-finite-field-and-binary-representation | # Doubt on elliptic curve over a finite field and binary representation
I'm a programmer, i.e. agnostic to the mathemathics behind most of cryptographic scheme, but I'm trying to remediate. I'm writing this premise for any possible error or imprecision that I probably put in this question.
However, I'm studying Elliptic Curve and I got to Elliptic Curves over Finite fields topic. For what I understood an elliptic over a finite field $$F$$ are the points that fulfill the Weierstrass equation where coefficients and coordinates belongs to $$F$$.
So far I always thought of F as the integer modulo $$p$$ (with $$p$$ prime). Point addition, doubling must be thus performed modulo p and all the points that pop out along with the point at infinity form a cyclic group. Ok, I got this.
Now problems begin: reading some handouts that a professor of mine gave to me, I read two things that I couldn't figure out.
First, they say that the order of a finite field is always a prime power (and such a power is named extension degree that he denotes with $$m$$): of course I trust this result but I couldn't figure out a field of order, say, $$4$$, $$8$$, $$9$$ or $$16$$. What are examples of such fields?
Secondly, if $$p=2$$, $$m>1$$ and $$q=p^m$$ (binary field), they say that (I'm citing):
The elements of the a binary field of order $$q=2^m$$ cannot be represented as integers modulo $$2^m$$. A convenient way to represent them is by means of binary polynomials of degree less than m.
Why can't they be represented as integers modulo $$2^m$$? Any answers and/or reference is appreciated.
• There are prime field-based curves, that is $F_p$ What you are asking is the finite field question. This might help. The hint is that $F_{p^m}$ is considered as a vector space over $F_p$ with $m$ dimension. Could your share your slides? Jan 4, 2021 at 16:45
• The wikipedia article has an explicit example of a/the finite field with 4 elements: en.wikipedia.org/wiki/Finite_field#Non-prime_fields Jan 4, 2021 at 17:39
• Welcome to Crypto Stack Exchange. I've added some paragraph breaks and math formatting to your question to make it easier readable – please check the changes, and feel free to roll back, or edit again if I made something wrong. Jan 5, 2021 at 1:00
I couldn't figure out a field of order, say, 4, 8, 9 or 16. What are examples of such fields?
Let's do that with $$8=2^3$$.
• Elements of that field $$\mathbb F_{2^3}$$ will be assimilated to $$3$$-bit quantities, that is the set $$\{\mathtt0,\mathtt1\}^3$$, or equivalently polynomials of degree less than $$3$$ with binary coefficients, where e.g. $$\mathtt{110}$$ is the polynomial $$x^2+x$$, and $$\mathtt{101}$$ is the polynomial $$x^2+1$$.
• Our addition is bitwise exclusive-OR, or equivalently addition of polynomials, so that $$\mathtt{110}\oplus\mathtt{101}=\mathtt{011}$$, or equivalently $$(x^2+x)+(x^2+1)=x+1$$.
• For our multiplication we choose an irreducible_polynomial $$P(x)$$ of degree $$3$$ with binary coefficients (among two such polynomials, see this list of irreducible polynomials over GF(2) up to degree 11), e.g. $$P(x)=x^3+x+1$$; and we define multiplication as polynomial multiplication followed by reduction modulo $$P(x)$$. This simply tells that when in the product we get a term of degree $$d\ge3$$, we can get rid of it by adding the polynomial $$x^{d-3}\,P(x)\ =\ x^d+x^{d-2}+x^{d-3}$$. So for example $$\begin{array}{lll}(x^2+x)\,(x^2+1)&=(x^2+x)\,x^2+(x^2+x)\\ &=x^4+x^3+x^2+x\\ &\equiv(x^4+x^3+x^2+x)+(x^4+x^2+x)&\pmod{x^3+x+1}\\ &\equiv x^3&\pmod{x^3+x+1}\\ &\equiv x^3+(x^3+x+1)&\pmod{x^3+x+1}\\ &\equiv x+1&\pmod{x^3+x+1}\\ \end{array}$$ thus $$\mathtt{110}\otimes\mathtt{101}=\mathtt{011}$$.
The full multiplication table goes $$\begin{array}{c|cccccccc} \otimes &\mathtt{000}&\mathtt{001}&\mathtt{010}&\mathtt{011}&\mathtt{100}&\mathtt{101}&\mathtt{110}&\mathtt{111}\\ \hline \mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}&\mathtt{000}\\ \mathtt{001}&\mathtt{000}&\mathtt{001}&\mathtt{010}&\mathtt{011}&\mathtt{100}&\mathtt{101}&\mathtt{110}&\mathtt{111}\\ \mathtt{010}&\mathtt{000}&\mathtt{010}&\mathtt{100}&\mathtt{110}&\mathtt{011}&\mathtt{001}&\mathtt{111}&\mathtt{101}\\ \mathtt{011}&\mathtt{000}&\mathtt{011}&\mathtt{110}&\mathtt{101}&\mathtt{111}&\mathtt{100}&\mathtt{001}&\mathtt{010}\\ \mathtt{100}&\mathtt{000}&\mathtt{100}&\mathtt{011}&\mathtt{111}&\mathtt{110}&\mathtt{010}&\mathtt{101}&\mathtt{001}\\ \mathtt{101}&\mathtt{000}&\mathtt{101}&\mathtt{001}&\mathtt{100}&\mathtt{010}&\mathtt{111}&\mathtt{011}&\mathtt{110}\\ \mathtt{110}&\mathtt{000}&\mathtt{110}&\mathtt{111}&\mathtt{001}&\mathtt{101}&\mathtt{011}&\mathtt{010}&\mathtt{100}\\ \mathtt{111}&\mathtt{000}&\mathtt{111}&\mathtt{101}&\mathtt{010}&\mathtt{001}&\mathtt{110}&\mathtt{100}&\mathtt{011}\\ \end{array}$$ The neutral for $$\otimes$$ is $$\mathtt{001}$$ that is the polynomial $$1$$. The distributive property and other commutative field properties follow from that for polynomials.
The elements of the a binary field of order $$q=2^m$$ cannot be represented as integers modulo $$2^m$$.
Actually it's OK to represent them as integers, and even convenient in some computer languages (perhaps our $$\oplus$$ is just the bitwise-XOR operator ^). But when $$m>1$$, addition and multiplication modulo $$q=2^m$$ give the ring $$\mathbb Z_q$$, which is essentially useless to build the field $$\mathbb F_q$$, for $$\mathbb Z_q$$'s addition and multiplication bear no relation with $$\mathbb F_q$$'s $$\oplus$$ and $$\otimes$$.
A convenient way to represent elements of the a binary field of order $$q=2^m$$ is by means of binary polynomials of degree less than $$m$$.
Indeed. That's what we did above.
Following comment
If $$m=1$$ then coordinates over elliptic curve are just scalars, whereas if $$m>1$$ then a coordinate is in its turn a "set of coordinate".
Yes, that's a useful way of seeing it. An element of the field $$\mathbb F_{p^k}$$ is most naturally expressed as $$k$$ "coordinates" each in $$\{0,1\ldots,p-1\}$$ when devising a general computer implementation of arithmetic in $$\mathbb F_{p^k}$$. The usual mathematical statement of the same thing is that such element is a polynomial of degree less than $$k$$, with coefficients in $$\mathbb F_p==\mathbb Z_p$$.
In the first part of the answer I have specialized to $$p=2$$, since the question mentioned binary in the title, but we can do the same for any prime $$p$$, and that makes polynomial notation shine: it implies the definition of addition, and of multiplication with the help of an irreducible polynomial.
• Ok, you've been clear but just to be sure can you confirm the following? If m=1 then coordinates over elliptic curve are just scalars whereas if m>1 then a coordinate is in its turn a "set of coordinate" (in your example belonging to {0,1}^3). And about the integer representation: what are you trying to tell me is that it is not convenient to represent an element of a F when m>1 because in that case I would have, for example, 110 XOR 111 = 101 becoming 6 XOR 7 = 5 that may result meaningless? Jan 4, 2021 at 22:42
• @user1108 "convenience" is relative. What the author meant here is that the "natural" ($+$ and $·$) operations on the integers modulo $p^m$ are not giving a field (they are a ring, but you have divisors of 0, so division is not unique), but the "normal" operations on polynomials modulo an irreducible polynomial to form a field (and up to isomorphism the only field of order $p^m$), so we don't need to define special operations here. XOR on integers mod $2^m$ is convenient enough, the multiplication is a bit more complicated to implement. Jan 5, 2021 at 0:58
• @user1108: I tried to clarify in updated answer (and changed primitive to irreducible as it should be).
– fgrieu
Jan 5, 2021 at 7:38
• Thanks @fgrieu. Now it's much clearer to me. I still would have a lot of questions about this marvelous branch of mathematics so I ask one more thing: can you suggest me a set (forgive me the joke) of reference (textbook and/or online handouts) that gather elliptic curves theory for cryptographer? I would enjoy them. Jan 5, 2021 at 10:58
• Thanks also to @PaŭloEbermann (apparently I cannot cite more than a user in a comment) Jan 5, 2021 at 11:00
It's more like a problem within abstract algebra rather than a problem within elliptic curves.
Integer modulo primepower $$p^k$$ would contain a zero divisor for k>1. Therefore, $$\mathbb{Z}/p^k$$ cannot be a field because you can't find, say, the multiplicative inverse of $$p$$. You could always consider $$\mathbb{F}_{p^k}$$ as $$\mathbb{F_p}[x]/f(x)$$ where $$f(x)$$ is some irreducible polynomial of degree $$k$$. Furthermore, whichever $$f(x)$$ you chose, they are always isomorphic.
There are many such polynomials. Exactly how many? Since $$\mathbb{F}_{p^k}$$ have order $$p^k$$ but they could possibly falls into lower extension degree. Write the prime factorization $$k=\ell_1^{e_1}\dots \ell_r^{e_r}$$. Say $$c_d$$ be the number of degree-d monic irreducible polynomial. Then we have $$p^k-p=\#\mathbb{F}_{p^k}\setminus\bigcup_{d|k}\mathbb{F_{p^d}}=\sum_{d|k}d\cdot c_d$$ Using Mobius inversion, we easily obtain $$k\cdot c_k=\sum_{d|k}\mu(d) (p^{k/d}-p).$$ You could therefore randomly pick one polynomial and then use algorithms such as Berlekamp's or Cantor-Zassenhaus to check that it is irreducible and resample if otherwise. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 75, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8970480561256409, "perplexity": 334.6111409878514}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00703.warc.gz"} |
http://mathoverflow.net/questions/75078/does-obstruction-class-for-deformations-of-a-pair-x-d-lie-in-rm-ext2-o?answertab=oldest | # Does obstruction class for deformations of a pair $(X,D)$ lie in ${\rm Ext}^2(\Omega^1(\log D), \mathcal{O}_X)$ when $X$ is singular?
This question is related to this one and this one.
Let $X$ be a normal variety over algebraically closed field $k$ and $D$ be an effective Cartier divisor. Let $\zeta := (\mathcal{D} \subset \mathcal{X} \rightarrow {\rm Spec} A)$ be a deformation of the pair $(X,D)$ over Artin local $k$-algebra $A$. Let $e := (0 \rightarrow (t) \rightarrow \tilde{A} \rightarrow A \rightarrow 0)$ be a small extension where $(t) \simeq k$.
Question Can we define an obstruction element
$o_{\zeta}(e) \in {\rm Ext}^2( \Omega^1_X( \log D), \mathcal{O}_X)$
with the property that, if $o_{\zeta}(e) =0$, then there exists a deformation $\tilde{\zeta} = (\tilde{\mathcal{D}} \subset \tilde{\mathcal{X}}\rightarrow {\rm Spec} \tilde{A})$ which is a lifting of $\zeta$?
If $X$ is smooth, the answer is in this post. If $X$ has only l.c.i. singularities and $D =0$, this is Proposition 2.4.8 in Sernesi's book. Is there a similar interpretation? If you know about some reference around this topic, please let me know about it.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9756194353103638, "perplexity": 112.84981963398437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701169261.41/warc/CC-MAIN-20160205193929-00260-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://fiction.eveonline.com/lore/stargates | # Stargates
Stargates, also known as jump gates, are the primary means of interstellar travel in New Eden.
The theory and technology behind stargates opened up a whole new era in the history of mankind and is readily accepted as being one of the most important discoveries of all times. Jump gates have now been in usage for centuries and new versions appear regularly that make them more sophisticated and safe.
## Operational uncertainty
Even if the functions of jump gates are well known from a theoretical point of view, there still remain a lot of unanswered questions about the fundamentals of dimensional inter-connections. Naturally, many theories exist on the subject, but none are comprehensive enough to fully explain how the universe is divided into many dimensions and the connections between them, some also touch upon the subject of hyperspace, an alternative plain in another dimension. About the only statement these theories agree upon is that these issues are definitely not as simple as they seem on the surface. However, these theories are still a topic of research, and stargate technology is being continually refined, with new jump gate designs appearing regularly to improve safety and efficiency.
Part of a stargate's operation involves the use of superheated plasma, which can be vented into space. If vented improperly, it is possible that the plasma may be attracted away from the jump gate's boson sphere toward an approaching ship, where it may react with the ship's shields to produce a harmless visual phenomenon, such as the Lutins observed near the Perimeter gate in Iyen-Oursta. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8334569334983826, "perplexity": 1092.5007646401316}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00028.warc.gz"} |
https://www.varsitytutors.com/calculus_2-help/maclaurin-series | # Calculus 2 : Maclaurin Series
## Example Questions
← Previous 1
### Example Question #1 : Maclaurin Series
Suppose that . Calculate
Explanation:
Let's find the power series of centered at to find . We have
This series is much easier to differentiate than the expression . We must look at term , which is the only constant term left after differentiating 48 times. This is the only important term, because when we plug in , all of the non-constant terms are zero. So we must have
### Example Question #55 : Taylor And Maclaurin Series
What is the value of the following infinite series?
Explanation:
We can recognize this series as since the power series is
with the value plugged into since
.
So then we have
.
### Example Question #301 : Series In Calculus
What is the value of the following infinite series?
The infinite series diverges.
Explanation:
The infinite series can be computed easily by splitting up the two components of the numerator:
Now we recall the MacLaurin series for the exponential function , which is
which converges for all . We can see that the two infinite series are with , respectively. So we have
### Example Question #57 : Taylor And Maclaurin Series
Find the value of the infinite series.
The series does not converge.
Explanation:
We can evaluate the series
by recognizing it as a power series of a known function with a value plugged in for . In particular, it looks similar to :
After manipulating the series, we get
.
Now it suffices to evalute , which is .
So the infinite series has value
.
### Example Question #58 : Taylor And Maclaurin Series
Find the value of the following infinite series:
Explanation:
After doing the following manipulation:
We can see that this is the power series
with plugged in.
So we have
### Example Question #53 : Taylor And Maclaurin Series
Find the value of the following series.
Divergent.
Explanation:
We can split up the sum to get
.
We know that the power series for is
and that each sum,
and
are simply with plugged in, respectively.
Thus,
.
### Example Question #60 : Taylor And Maclaurin Series
Find the value of the infinite series.
Infinite series does not converge.
Explanation:
The series
looks similar to the series for , which is
but the series we want to simplify starts at , so we can fix this by adding a and subtracting a , to leave the value unchanged, i.e.,
.
So now we have with , which gives us .
So then we have:
### Example Question #61 : Taylor And Maclaurin Series
Write out the first two terms of the Maclaurin series of the following function:
Explanation:
The Maclaurin series of a function is simply the Taylor series of a function, but about x=0 (so a=0 in the formula):
To write out the first two terms (n=0 and n=1), we must find the first derivative of the function (because the zeroth derivative is the function itself):
The derivative was found using the following rule:
Next, use the general form, plugging in n=0 for the first term and n=1 for the second term:
### Example Question #1 : Maclaurin Series
Find the Maclaurin series for the function:
Explanation:
Write Maclaurin series generated by a function f. The Maclaurin series is centered at for the Taylor series.
Evaluate the function and the derivatives of at .
Substitute the values into the power series. The series pattern can be seen as alternating and increasing order.
### Example Question #3092 : Calculus Ii
Find the first three terms of the Maclaurin series for the following function: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9887606501579285, "perplexity": 931.9889475305837}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813818.15/warc/CC-MAIN-20180221222354-20180222002354-00096.warc.gz"} |
http://math.stackexchange.com/questions/113999/is-the-conclusionlast-line-correct-if-not-whats-wrong-with-this-proof-pr?answertab=votes | # Is the conclusion(last line) correct? If not, what's wrong with this proof ? (Probability Density Estimation, Bayesian inference)
(Probability Density Estimation, Bayesian inference)
$x_1$ and $x_2$ are independent random variables, so $p\left(x_1,x_2\right)=p\left(x_1\right)\cdot p\left(x_2\right)$ and $p\left(x_1,\left.x_2\right|\theta \right)=p\left(\left.x_1\right|\theta \right)\cdot p\left(\left.x_2\right|\theta \right)$,
from $p\left(x_1,x_2\right)=p\left(x_1\right)\cdot p\left(x_2\right)$,
I get $\int p\left(x_1,\left.x_2\right|\theta \right)p(\theta )d\theta =\int p\left(\left.x_1\right|\theta \right)p(\theta )d\theta \cdot \int p\left(\left.x_2\right|\theta \right)p(\theta )d\theta$
And because $p\left(x_1,\left.x_2\right|\theta \right)=p\left(\left.x_1\right|\theta \right)p\left(\left.x_2\right|\theta \right)$
$\therefore \int p\left(\left.x_1\right|\theta \right)p\left(\left.x_2\right|\theta \right)p(\theta )d\theta =\int p\left(\left.x_1\right|\theta \right)p(\theta )d\theta \cdot \int p\left(\left.x_2\right|\theta \right)p(\theta )d\theta$
-
Could you perhaps be persuaded to use English prose to connect your formulas rather than those little constellation of dots that nobody remembers what are supposed to mean? You're not taking dictation here; you have time to write things out. – Henning Makholm Feb 27 '12 at 14:31
just one thing remember for ever,that integral of product does not equal to product of integrals – dato datuashvili Feb 27 '12 at 14:35
What do you think is wrong? If $x_1$ and $x_2$ are independent random variables, does that imply that they are also conditionally independent given $\theta$? If not, the last part of your first sentence is not necessarily true. – Dilip Sarwate Feb 27 '12 at 14:41
1. Independence per se, which translates as $p(x_1,x_2)=p(x_1)p(x_2)$.
2. Conditional independence, which translates as $p(x_1,x_2\mid\theta)=p(x_1\mid\theta)p(x_2\mid\theta)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9471224546432495, "perplexity": 601.177007081395}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115868812.73/warc/CC-MAIN-20150124161108-00039-ip-10-180-212-252.ec2.internal.warc.gz"} |
https://brilliant.org/problems/permutation-and-combination/ | # PERMUTATION AND COMBINATION
Find the number of ways of permuting the numbers 1,2,3,....n so that they are first increasing and then decreasing.
× | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9880788922309875, "perplexity": 1113.567421305279}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560282926.64/warc/CC-MAIN-20170116095122-00532-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://workflow.arts.ac.uk/artefact/artefact.php?view=279429&artefact=3245267 | Note (15) | Research
Finger traps are often used as practical jokes, involving the wearer struggling to remove their digits from either end of the tube, which is usually woven from bamboo.
Both ends of the trap have to be pushed inward to relax the mesh and release the fingers. "The finger trap is often used as a metaphor for a problem that can be overcome by relaxing," says Piatt. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8630274534225464, "perplexity": 1262.3366632584025}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221209755.32/warc/CC-MAIN-20180815004637-20180815024637-00030.warc.gz"} |
https://infoscience.epfl.ch/record/153693 | ## Signification of the phase function parameter gamma explored with a fractal distribution of scatterers
The optical properties within limited volumes of diffusive media can be probed by carrying spatially-resolved measurements of diffused light at short source-detector separation (typically one scattering mean free path). At such distance, analytical models only relying on the absorption and reduced scattering coefficients fail at correctly predicting reflectance and it was demonstrated that adding a third optical coefficient γ improves the description of light propagation conditions near the source. In an attempt to relate the γ coefficient to physical properties of turbid media, this paper uses a fractal distribution law for modeling scatterers’ sizes distributions and investigates numerically and experimentally how γ is related to the fractal power α. The results indicate that within the range of γ typically encountered in biological samples, this coefficient is approximately linearly correlated with α.
Published in:
Optics Express, 18, 23, 23664-23675
Year:
2010
Publisher:
Optical Society of America
ISSN:
1094-4087
Keywords:
Laboratories: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8282579779624939, "perplexity": 2177.5023947396094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00522.warc.gz"} |
https://www.physicsforums.com/threads/interesting-physics-question.84888/ | # Interesting physics question
1. Aug 11, 2005
### asdf1
! question...
Why does lnX!=XlnX-X?
2. Aug 11, 2005
### VietDao29
???
$$\ln{1!} = \ln1 = 0$$
$$1\ln1 - 1 = 1 \times 0 - 1 = -1$$
So 0 = -1??
Viet Dao,
3. Aug 11, 2005
### Timbuqtu
I think you're referring to the Stirling approximation:
$$\ln n! = \ln 1 + \ln 2 + \ldots + \ln n = \sum_{k=1}^{n}\ln k \approx \int_1^n \ln x dx = n \ln n - n + 1 \approx n \ln n - n$$
where the approximation gets relatively better when n becomes larger.
4. Aug 11, 2005
### HallsofIvy
Staff Emeritus
What you said was "Why does lnX!=XlnX-X?". Now you tell us not only did you NOT mean "ln X!= X ln X- X, but you also tell us that you already KNOW the answer to your (unstated) question. What was your purpose in posting that?
5. Aug 11, 2005
### mathwonk
he means antiderivative maybe. answer: check it.
6. Aug 11, 2005
### Subhasish Mandal
What is the relation between energy & time period of a simple pendulam while not considering the small angle approximation? Please show the graph also them.
7. Aug 11, 2005
### Subhasish Mandal
problem on SHM
undefined :rofl:
8. Aug 12, 2005
### asdf1
?
i am referring to the Stirling approximation(sorry, i forgot to add that at the end of my question)...
i saw that equation in the "advanced engineering mathematics" book by kreyszig as part of the solution to a problem...
but what i wonder is how did the stirling approximation come from?
9. Aug 12, 2005
### lurflurf
$$\log(x!)=\sum_{n=1}^x \log(n) \sim \int_0^x \log(t) dt=x\log(x)-x$$
where ~ here means goes to asymptotically for large x
that is the integral becomes a good approximation of the sum as x becomes large.
Last edited: Aug 12, 2005
10. Aug 13, 2005
### asdf1
Why? That's the part that I don't understand...
11. Aug 13, 2005
### Galileo
You can approximate the sum by an integral. If you draw a graph, the sum $\sum_{n=1}^{x}\ln x$ is equal to the area of x rectangles, each of width 1. The heights are ln(1),ln(2),...,ln(x).
So you can approximate this area by the integral $\int_1^x \ln t dt$. Drawing a picture may help.
12. Aug 13, 2005
### lurflurf
It is a Riemann sum we partition (0,x) into (we assume here x is a natural number)
[0,1],[1,2],[2,3],...,[n-2,n-1],[n-1,n]
and chose as the point of evaluation for each interval the right boundry
we can consider one term in the Reimann sum as an approximation to the integral over the region of the term.
$$\log(n) \sim \int_{n-1}^n \log(x)dx=\log(e^{-1}(1+\frac{1}{n-1})^{n-1}n)$$
clearly this will be a good approximation if x is large and not so good is n is not large. Thus the approximation over (0,x) cannot make up for its poor start, but the relative error gets better and better. So we have asymptodic convergence. The absolute error will never be small, but the relative error will. Often since x! grows rapidly we do not mind the absolute error being high (or moderate) so long as the relative error is low.
Last edited: Aug 13, 2005
13. Aug 14, 2005
### asdf1
thanks! It makes a lot more sense now...
but there's still one I don't get:
What's the difference between the absolute and relative error?
14. Aug 14, 2005
### HallsofIvy
Staff Emeritus
?? That's a completely different question!!
Suppose in measuring a distance of 100 meters, I make an error of 10 cm.
The absolute error is 10 cm. The relative error is that "relative" to the entire measurement: 10 cm/100m = 0.1 m/100m= 0.001 (and, of course, has no units).
There is an Engineering rule of thumb: when you add measurements, the absolute errors add. When you multiply measurements, the relative errors add.
That is, if I measure distance y with absolute error at most Δy and distance x with absolute error at most Δx, then the true values of x and y might be as low as x- Δx and y- Δy. The true value of x+ y might be as low as (x-Δx)+(y-Δy)= (x+y)- (Δx+Δy) The true values of x and y might be as large as x+&Deltax and y+Δy. The true value of x+ y might be as large as (x+Δx)+ (y+Δy)= (x+y)+(Δx+Δy). That is, the error in x+y might be as large as Δx+ Δy.
On the other hand, if I multiply instead of adding, the true value of xy might be as low as (x- Δx)(y- Δy)= xy- (xΔy+ yΔx)+ (ΔxΔy) which, ignoring the "second order" term ΔxΔy (that's why this is a "rule of thumb" rather than an exact formula), is xy- (xΔy+ yΔy). The true value of xy might be as large as xy+ (xΔy+ xΔx). The absolute error might be as large as xΔy+ yΔx which depend on x and y as well as the absolute errors Δx and Δy. However, the "relative" error in xy is (xΔy+ yΔx)/xy= Δy/y+ Δx/x, the sum of the two relative errors in x and y.
15. Aug 14, 2005
### lurflurf
like HallsofIvy said
absolute error=|approximate-exact|
relative error=|approximate-exact|/exact
think about approximating (x+1)^2 with x^2 for large
the relative error becomes small
the absolute error grows
the approximation
log(x!)~log(x)-x
does the same
16. Aug 15, 2005
thanks! :) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9702963829040527, "perplexity": 2039.9601951089096}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698544678.42/warc/CC-MAIN-20161202170904-00464-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://ec.gateoverflow.in/635/gate-ece-2015-set-1-question-34 | 31 views
A MOSFET in saturation has a drain current of $1$ mA for $V_{DS} =0.5 \: V$. If the channel length modulation coefficient is $0.05 \: V^{-1}$, the output resistance (in $k \Omega$) of the MOSFET is ___________. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.919950008392334, "perplexity": 2858.8652484504023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337524.47/warc/CC-MAIN-20221004184523-20221004214523-00766.warc.gz"} |
http://math.stackexchange.com/questions/243367/checking-that-rho-is-a-representation | # Checking that $\rho$ is a representation
Let $z$ be a generator of the cyclic group $\mathbb{Z}_3 = \{ 1,z,z^2 \}$. Prove that a representation $\rho$ of $\mathbb{Z}_3$ in the $2$-dimensional complex vector space $\mathbb{C}^2$ can be defined by $$\rho(z) = \begin{bmatrix} -1 & 1 \\ -1 & 0 \end{bmatrix}.$$
I know I need to check that $\rho : \mathbb{Z}_3 \to GL(\mathbb{C}^2)$ is a homomorphism, but how do I check that $\rho(g \circ h) = \rho(g) \circ \rho(h)$ for any $g,h \in \mathbb{Z}_3$?
-
Since any $g,h$ can be written as $g=z^a.h=z^b$ we have $\rho(gh)=\rho(z^{a+b}).$ Implicit in the question is that $\rho(z^c):=A^c$ for any $c=0,1,2,$ where $A$ is the given matrix. Now you know that $\rho(gh)=A^{a+b}=A^aA^b=\rho(g)\rho(h).$
The important property that you need to check is that $A^3=I$ is the identity matrix.
-
First thing to do is make sure that $\rho(z)=\left(\begin{array}{cc}-1&1\\-1&0\end{array}\right)$ is really an element of $GL_2(\mathbb{C})$. It's obvious that $\rho(z)\in M_{2\times 2}(\mathbb{C})$, but you should prove it's invertible.
Then, just write it out. $\mathbb{Z}_3$ is cyclic, so the entire representation is determined by $\rho(z)$. If you can prove that $\rho(z)$ has order $3$ in $GL_2(\mathbb{C})$ you're done. So, calculate $\rho(z)^2$ and $\rho(z)^3$ and make sure the latter is the identity matrix and the former is not. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9522873163223267, "perplexity": 42.804369907520716}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163988740/warc/CC-MAIN-20131204133308-00041-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/infinite-solutions-to-homogeneous-system.633803/ | # Infinite solutions to homogeneous system?
• Start date
• #1
hivesaeed4
217
0
Could someone explain the following theorem to me:
Given a homogeneous system of n linear equations in m unknowns if m>n (i.e. there are more unknowns than equations) there will be infinitely many solutions to the system.
• #2
4,815
134
Hey hivesaeed4.
Think about when you have a unique solution (and the system is not inconsistent): for a unique solution you have as many elements in the vector as you have variables.
If m > n then as long as the system is not inconsistent, then many of the rows are going to dissappear and become 0 vectors when you do row reduction (like Gaussian elimination) since some of those vectors will be linearly dependent on the other vectors.
But if m < n it means that as long as you have one non-zero vector it means that at most you will never be able to get a unique solution (since you need minimum m = n for this to happen).
So you will need to introduce a parameter and then relate those free parameters to the final solution.
Geometrically if you want to think of a solution of n n-dimensional plane equations it means that for a unique solution, all points have to intersect at the same point.
Think about the 3D case for the moment: if you have two planes intersect then it means that you either have the two planes being equal (which means the intersection gives the whole plane) or you get an intersection which gives a line as your solution.
If you add one more plane that is not linearly dependent on the other two then your line will reduce to a point and a point is a unique solution.
So until we get m = n independent vectors, we have situations where we can have many solutions instead of 1 and this is the visual explanation of why the statement you are wondering about actually holds.
• #3
Homework Helper
43,021
971
We can, of course, always write a system of equations as a matrix problem of the form Ax= b. In particular, if A is a square matrix, mapping, say, Rn to Rn, there are three possibilities for the equation Ax= b.
1) the equation has a unique solution.
2) the equation has NO solution.
3) the equation has an infinite number of solutions.
If A is invertible, we can multiply both sides of Ax= b by its inverse to get $A^{-1}Ax= x= A^{-1}b$. That clearly is a solution to the equation and, because x must be equal to that, the solution is unique.
If A is not invertible, then it has rank, k, less than n. That means that A maps all of Rn into a subspace of Rn of dimension k< n. It also means that A has null space (space of vectors, v, such that Av= 0) of dimension n- k.
In that case, if b does not happen to lie in that subspace, the equation, Ax= b, has no solution. For any x, Ax is in that subspace and b is not. If b does happen to lie in that subspace, there is a solution, a vector x such that Ax= b, but it is also true that for any v in the null space of A, A(x+ v)= Ax+ Av= b+ 0= b so there are an infinite number of solutions.
Since that is true for square matrices, what does it have to do with this problem? Well, if we have m equations, in n unknowns, with m< n, we can represent it has the matrix equation Ax= b where A is Not square. It has m rows and n columns. But we can make it square by adding n- m rows of all zeros and adding n- m 0s to the vector b. Since that matrix obviously has determinant 0, it is not invertible, which means it has either no solution or an infinite number of solutions. Because the system is homogeneous, Ax= 0, it has the obvious solution, x= 0, which means "no solution" cannot apply. We are left with an infinite number of solutions.
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971 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8887413740158081, "perplexity": 316.5112684545685}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500058.1/warc/CC-MAIN-20230203154140-20230203184140-00623.warc.gz"} |
https://www.vedantu.com/question-answer/a-letter-is-taken-out-at-random-from-the-word-class-12-maths-cbse-5f4f0d61d1fa823ca5b596f3 | Question
# A letter is taken out at random from the word ASSISTANT and another from STATISTICS. The probability that they are the same letters is :A. $\dfrac{13}{90}$ B. $\dfrac{17}{90}$ C. $\dfrac{19}{90}$ D. $\dfrac{15}{90}$
Hint: For finding the probability, we will first the letters that represent in both of the words, ASSISTANT and STATISTICS. Then we will find out the probability that a letter is taken out from the first word and will find out the same letter from the other word. As we have to find a letter that is present in both the words, we will use the formula, $P\left( {{A}_{1}}.{{B}_{1}} \right)=P\left( {{A}_{1}} \right).P\left( {{B}_{1}} \right)$, where P (A) and P (B) are the probabilities of any two events A and B. But we have to find the probability for all the words that are common in both the words, so we have to add the probability of each word that are taken out from both the words. The formula for the probability of any event A is given as, $P\left( A \right)=\dfrac{\text{ number of favourable outcomes of A}}{\text{total number of outcomes}}$.
Complete step by step solution:
We have been given two words ASSISTANT and STATISTICS and have been asked to find the probability of taking out the same letters. We can see that the letters that are common in both the words, ASSISTANT and STATISTICS are A, S, I and T. So, we will find the probability of taking out each of these common letters in both the words one by one, using the formula for the probability of any event A as, $P\left( A \right)=\dfrac{\text{ number of favourable outcomes of A}}{\text{total number of outcomes}}$.
We will find the probability of taking letter A from the first word, ASSISTANT. There are 2 As in this word and the total number of letters is 9. So, the probability of taking letter A from the first word is $\dfrac{2}{9}$.
Now, for finding the probability of taking letter A from the second word, STATISTICS, we know that is 1 A and the total number of letters here are 10. So, the probability of taking letter A from the second word is $\dfrac{1}{10}$.
So, now, we will find the probability that letter A is taken out from both the words, using the formula, $P\left( {{A}_{1}}.{{B}_{1}} \right)=P\left( {{A}_{1}} \right).P\left( {{B}_{1}} \right)$. So, we get,
$P\left( A \right)=\dfrac{2}{9}\times \dfrac{1}{10}=\dfrac{2}{90}\ldots \ldots \ldots \left( i \right)$
Now, we will find the probability of taking letter S from the first word, ASSISTANT. There are 3 S in this word and the total number of letters is 9. So, the probability of taking letter S from the first word is $\dfrac{3}{9}$.
And, for finding the probability of taking letter S from the second word, STATISTICS, we have 3 S here and the total number of letters as 10. So, the probability of taking letter S from the second word is $\dfrac{3}{10}$.
So, we will get the probability that letter S is taken out from both the words, using the formula as,
$P\left( S \right)=\dfrac{3}{9}\times \dfrac{3}{10}=\dfrac{9}{90}\ldots \ldots \ldots \left( ii \right)$
Now we have to find the probability of taking a letter I from the first word, ASSISTANT. There is 1 I in this word and the total number of letters is 9. So, the probability of taking a letter I from the first word is $\dfrac{1}{9}$.
Now, for the probability of taking letter I from the second word, STATISTICS, we know that is 2 I and the total number of letters here are 10. So, the probability of taking a letter I from the second word is $\dfrac{2}{10}$.
So, we can find the probability that letter I am taken out from both the words, using the formula as,
$P\left( I \right)=\dfrac{1}{9}\times \dfrac{2}{10}=\dfrac{2}{90}\ldots \ldots \ldots \left( iii \right)$
And finally, we will find the probability of taking letter T from the first word, ASSISTANT. There are 2 T in this word and the total number of letters is 9. So, the probability of taking a letter T from the first word is $\dfrac{2}{9}$.
Now, for finding the probability of taking letter T from the second word, STATISTICS, we know that there are 3 T and the total number of letters are 10. So, the probability of taking a letter T from the second word is $\dfrac{3}{10}$.
So, we will find the probability that letter T is taken out from both the words, using the formula as,
$P\left( T \right)=\dfrac{2}{9}\times \dfrac{3}{10}=\dfrac{6}{90}\ldots \ldots \ldots \left( iv \right)$
Now, that we have found the probabilities of each of the common letters being taken from both the words, we can find the probability of the same letters being taken as,
$P\left( A \right)+P\left( S \right)+P\left( I \right)+P\left( T \right)$
We have these values from equations (i), (ii), (iii) and (iv). So, we will get the probability as,
$\dfrac{2}{90}+\dfrac{9}{90}+\dfrac{2}{90}+\dfrac{6}{90}=\dfrac{19}{90}$
So, the probability of the same letters being taken is $\dfrac{19}{90}$.
Therefore, the correct answer is option C.
Note: There are chances that the students might get confused with the formula and use $P\left( A.B \right)=P\left( A \right)+P\left( B \right)$ for finding the probability of any letter being taken out from both the words instead of $P\left( A.B \right)=P\left( A \right).P\left( B \right)$. But students should know when two events are independent, the probability of both occurring is the product of the probabilities of the individual events, that is, $P\left( A\text{ }and\text{ }B \right)P\left( A.B \right)=P\left( A \right).P\left( B \right)$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.968262255191803, "perplexity": 164.89045382167657}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00643.warc.gz"} |
https://quizlet.com/33255436/the-atmosphere-flash-cards/ | 15 terms
# The atmosphere
Earth Science
###### PLAY
atmosphere
The mixture of gases which surrounds a planet.
troposphere
The closest layer of gasses to the Earth's surface; 0-16 km above Earth's surface; this is the weather occurs, it contains most atmospheric water vapor; temperature and pressure decreases with increasing altitude.
stratosphere
The second closest layer of atmosphere; extends from about 16 to 50 km up; location of ozone layer; absorbs ~97% of Ultraviolet radiation; temperature increases with increasing altitude.
mesosphere
The third closest later of the atmosphere to Earth; extends from 50 to 80 km above the Earth, most meteorites burn up in this layer; temperature decreases with increasing altitude.
thermosphere
The fourth layer of the atmosphere; extends from 80 to 500 km; this is where the auroras form; temperature increases with increasing altitude.
exosphere
The outermost atmospheric layer; 500km+ above the Earth's surface; composed of the lightest gasses; satellites orbit here; temperature increases with altitude; lowest air pressure
ionosphere
Layer of electrically charged particles located in the thermosphere and mesosphere; absorbs harmful solar energy including X-rays and gamma rays.
ozone
Is a gassed formed by three atoms of oxygen (O3). The ozone layer is found in the stratosphere. It is the ozone layer which filters the UV radiation from the Sun. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8489369750022888, "perplexity": 3077.7060365746343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084893629.85/warc/CC-MAIN-20180124090112-20180124110112-00468.warc.gz"} |
https://en.m.wikipedia.org/wiki/Uniform_distribution_(discrete) | Discrete uniform distribution
(Redirected from Uniform distribution (discrete))
In probability theory and statistics, the discrete uniform distribution is a symmetric probability distribution whereby a finite number of values are equally likely to be observed; every one of n values has equal probability 1/n. Another way of saying "discrete uniform distribution" would be "a known, finite number of outcomes equally likely to happen".
Notation Probability mass functionn = 5 where n = b − a + 1 Cumulative distribution function ${\displaystyle {\mathcal {U}}\{a,b\}}$ or ${\displaystyle \mathrm {unif} \{a,b\}}$ ${\displaystyle a\in \{\dots ,-2,-1,0,1,2,\dots \}\,}$${\displaystyle b\in \{\dots ,-2,-1,0,1,2,\dots \},b\geq a}$${\displaystyle n=b-a+1\,}$ ${\displaystyle k\in \{a,a+1,\dots ,b-1,b\}\,}$ ${\displaystyle {\frac {1}{n}}}$ ${\displaystyle {\frac {\lfloor k\rfloor -a+1}{n}}}$ ${\displaystyle {\frac {a+b}{2}}\,}$ ${\displaystyle {\frac {a+b}{2}}\,}$ N/A ${\displaystyle {\frac {(b-a+1)^{2}-1}{12}}}$ ${\displaystyle 0\,}$ ${\displaystyle -{\frac {6(n^{2}+1)}{5(n^{2}-1)}}\,}$ ${\displaystyle \ln(n)\,}$ ${\displaystyle {\frac {e^{at}-e^{(b+1)t}}{n(1-e^{t})}}\,}$ ${\displaystyle {\frac {e^{iat}-e^{i(b+1)t}}{n(1-e^{it})}}}$
A simple example of the discrete uniform distribution is throwing a fair die. The possible values are 1, 2, 3, 4, 5, 6, and each time the dice is thrown the probability of a given score is 1/6. If two dice are thrown and their values added, the resulting distribution is no longer uniform since not all sums have equal probability.
The discrete uniform distribution itself is inherently non-parametric. It is convenient, however, to represent its values generally by all integers in an interval [a,b], so that a and b become the main parameters of the distribution (often one simply considers the interval [1,n] with the single parameter n). With these conventions, the cumulative distribution function (CDF) of the discrete uniform distribution can be expressed, for any k ∈ [a,b], as
${\displaystyle F(k;a,b)={\frac {\lfloor k\rfloor -a+1}{b-a+1}}}$
Estimation of maximumEdit
This example is described by saying that a sample of k observations is obtained from a uniform distribution on the integers ${\displaystyle 1,2,\dotsc ,N}$ , with the problem being to estimate the unknown maximum N. This problem is commonly known as the German tank problem, following the application of maximum estimation to estimates of German tank production during World War II.
The uniformly minimum variance unbiased (UMVU) estimator for the maximum is given by
${\displaystyle {\hat {N}}={\frac {k+1}{k}}m-1=m+{\frac {m}{k}}-1}$
where m is the sample maximum and k is the sample size, sampling without replacement.[1] This can be seen as a very simple case of maximum spacing estimation.
This has a variance of[1]
${\displaystyle {\frac {1}{k}}{\frac {(N-k)(N+1)}{(k+2)}}\approx {\frac {N^{2}}{k^{2}}}{\text{ for small samples }}k\ll N}$
so a standard deviation of approximately ${\displaystyle {\tfrac {N}{k}}}$ , the (population) average size of a gap between samples; compare ${\displaystyle {\tfrac {m}{k}}}$ above.
The sample maximum is the maximum likelihood estimator for the population maximum, but, as discussed above, it is biased.
If samples are not numbered but are recognizable or markable, one can instead estimate population size via the capture-recapture method.
DerivationEdit
For any integer m such that k ≤ m ≤ N, the probability that the sample maximum will be equal to m can be computed as follows. The number of different groups of k tanks that can be made from a total of N tanks is given by the binomial coefficient ${\displaystyle {\tbinom {N}{k}}}$ . Since in this way of counting the permutations of tanks are counted only once, we can order the serial numbers and take note of the maximum of each sample. To compute the probability we have to count the number of ordered samples that can be formed with the last element equal to m and all the other k-1 tanks less or equal to m-1. The number of samples of k-1 tanks that can be made from a total m-1 tanks is given by the binomial coefficient ${\displaystyle {\tbinom {m-1}{k-1}}}$ , so the probability of having a maximum m is ${\displaystyle P(m)={\tbinom {m-1}{k-1}}{\big /}{\tbinom {N}{k}}}$ .
Given the total number N and the sample size k, the expected value of the sample maximum is
{\displaystyle {\begin{aligned}\mu =\mathrm {E} [m]&=\sum _{m=k}^{N}m{\frac {\tbinom {m-1}{k-1}}{\tbinom {N}{k}}}\\&={\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\frac {m!}{(m-k)!}}\\&={\frac {k!}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\tbinom {m}{k}}\\&=k{\frac {\tbinom {N+1}{k+1}}{\tbinom {N}{k}}}\\&={\frac {k(N+1)}{k+1}},\end{aligned}}}
where the hockey-stick identity ${\displaystyle \sum _{m=k}^{N}{\tbinom {m}{k}}={\tbinom {N+1}{k+1}}}$ was used.
From this equation, the unknown quantity N can be expressed in terms of expectation and sample size as
{\displaystyle {\begin{aligned}N&=\mu \left(1+k^{-1}\right)-1.\end{aligned}}}
By linearity of the expectation, it is obtained that
{\displaystyle {\begin{aligned}\mu \left(1+k^{-1}\right)-1&=\mathrm {E} \left[m\left(1+k^{-1}\right)-1\right],\end{aligned}}}
and so an unbiased estimator of N is obtained by replacing the expectation with the observation,
{\displaystyle {\begin{aligned}{\hat {N}}&=m\left(1+k^{-1}\right)-1.\end{aligned}}}
Besides being unbiased this estimator also attains minimum variance. To show this, first note that the sample maximum is a sufficient statistic for the population maximum since the probability P(m;N) is expressed as a function of m alone. Next it must be shown that the statistics m is also a complete statistic, a special kind of sufficient statistics (demonstration pending). Then the Lehmann–Scheffé theorem implies that ${\displaystyle {\hat {N}}}$ is the minimum-variance unbiased estimator of N.[2]
The variance of the estimator is calculated from the variance of the sample maximum
{\displaystyle {\begin{aligned}\mathrm {Var} [{\hat {N}}]&={\frac {(k+1)^{2}}{k^{2}}}\mathrm {Var} [m].\end{aligned}}}
The variance of the maximum is in turn calculated from the expected values of ${\displaystyle m}$ and ${\displaystyle m^{2}}$ . The calculation of the expected value of ${\displaystyle m^{2}}$ is,
{\displaystyle {\begin{aligned}\mathrm {E} [m^{2}]&=\sum _{m=k}^{N}m^{2}{\frac {\tbinom {m-1}{k-1}}{\tbinom {N}{k}}}\\&={\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}m{\frac {m!}{(m-k)!}}\\&={\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}(m+1-1){\frac {m!}{(m-k)!}}\\&={\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\frac {(m+1)!}{(m-k)!}}-{\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\frac {m!}{(m-k)!}}\end{aligned}}}
where the second term is the expected value of ${\displaystyle m}$ . The first term can be expressed in terms of k and N,
{\displaystyle {\begin{aligned}{\frac {1}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\frac {(m+1)!}{(m-k)!}}&={\frac {(k+1)!}{(k-1)!{\tbinom {N}{k}}}}\sum _{m=k}^{N}{\tbinom {m+1}{k+1}}\\&={\frac {k(k+1)}{\tbinom {N}{k}}}\sum _{n=k+1}^{N+1}{\tbinom {n}{k+1}}\\&={\frac {k(k+1)}{\tbinom {N}{k}}}{\tbinom {N+2}{k+2}}\\&={\frac {k(N+2)(N+1)}{(k+2)}}\end{aligned}}}
where the replacement ${\displaystyle n=m+1}$ was made and the hockey-stick identity used. Replacing this result and the expectation of ${\displaystyle m}$ in the equation of ${\displaystyle E[m^{2}]}$ ,
{\displaystyle {\begin{aligned}\mathrm {E} [m^{2}]&={\frac {k(N+2)(N+1)}{(k+2)}}-{\frac {k(N+1)}{k+1}}\\&=k(N+1){\Big (}{\frac {N+2}{k+2}}-{\frac {1}{k+1}}{\Big )}\\&={\frac {k(N+1)(kN+k+N)}{(k+1)(k+2)}}\end{aligned}}}
The variance of ${\displaystyle m}$ is then obtained,
{\displaystyle {\begin{aligned}\mathrm {Var} [m]&=\mathrm {E} [m^{2}]-\mathrm {E} [m]^{2}\\&={\frac {k(N+1)}{(k+1)}}{\Big (}{\frac {kN+k+N}{k+2}}-{\frac {k(N+1)}{k+1}}{\Big )}\\&={\frac {k(N+1)}{(k+1)}}{\frac {(N-k)}{(k+2)(k+1)}}\\&={\frac {k(N+1)(N-k)}{(k+1)^{2}(k+2)}}\end{aligned}}}
Finally the variance of the estimator ${\displaystyle {\hat {N}}}$ can be calculated,
{\displaystyle {\begin{aligned}\mathrm {Var} [{\hat {N}}]&={\frac {(k+1)^{2}}{k^{2}}}\mathrm {Var} [m]\\&={\frac {(k+1)^{2}}{k^{2}}}{\frac {k(N+1)(N-k)}{(k+1)^{2}(k+2)}}\\&={\frac {(N+1)(N-k)}{k(k+2)}}.\end{aligned}}}
Random permutationEdit
See rencontres numbers for an account of the probability distribution of the number of fixed points of a uniformly distributed random permutation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9749999642372131, "perplexity": 1244.4561988389098}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583510853.25/warc/CC-MAIN-20181016155643-20181016181143-00289.warc.gz"} |
http://mathhelpforum.com/calculus/81933-control-systems-ode-confusion.html | # Math Help - control systems - ODE confusion
1. ## control systems - ODE confusion
Suppose that is the solution to the initial value problem
where is the state transition matrix.
Find for the following
I've done:
So
But since the left side is with respect to x_1 and not x_2 I'm not sure I can do this, and not sure where to go from here?
2. You have x1'= -x1- 4x2 and x2'= -x1- x2. From the first equation, 4x2= -x1'- x1. If you differentiate the first equation again, you get x1"= -x1'- 4x2'. From the second equation, x2'= -x1- x2 so that is x1"= -x1'- 4(-x1- x2)= -x1'- 4x1+ 4x2= -x1'- 4x1+ (-x1'- x1)= -2x1'- 5x1.
Can you solve the single equation x1"= -2x1'- 5x1 which is the same as x1"+ 2x1'+ 5x1= 0? If so then you can find x2 from 4x2= -x1'- 1.
Another way to do this problem, more in the "matrix" spirit, would be to find the eigenvalues and eigenvectors of $\begin{bmatrix}-1 & -4 \\ -1 & -1\end{bmatrix}$
3. Thank you so much that post was really helpful.
I'm having trouble with the same question but for matrix:
I've done:
So
which gives rise to:
However I can't see anyway of finding out what x_3 is, am I missing something?!
Thanks again. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9845672845840454, "perplexity": 3160.5820546704726}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397873.63/warc/CC-MAIN-20160624154957-00086-ip-10-164-35-72.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/integral-with-legendre-generating-function.273207/ | # Integral with legendre generating function
1. Nov 19, 2008
### buffordboy23
1. The problem statement, all variables and given/known data
Use the Legendre generating function to show that for A > 1,
$$\int^{\pi}_{0} \frac{\left(Acos\theta + 1\right)sin\thetad\theta}{\left(A^{2}+2Acos\theta+1\right)^{1/2}} = \frac{4}{3A}$$
2. Relevant equations
The Legendre generating function
$$\phi\left(-cos\theta,A\right) = \left(A^{2}+2Acos\theta+1\right)^{-1/2}} = \sum^{\infty}_{n=0}P_{n}\left(-cos\theta\right)A^{n}$$
3. The attempt at a solution
Pretty much clueless, and the book makes no mention of anything else. I know that with these parameters the series should be convergent for |A| < 1, and that the interval now runs from [-pi, pi]. The only thing on the internet that I could find that deals with with integrals of Legendre functions is for orthogonality, which doesn't seem to apply here.
2. Nov 19, 2008
### Pere Callahan
Using the trick the integral becomes
$$/int_0^\pi{\left(A\cos\theta+1\right)\sin\theta\sum_{n=0}^\infty{P_n(-\cos\theta)A^n}}$$
I suggest you are courageous and interchange the integration and summation (or maybe you can argue that this is allowed in our case?) then you expand $A\cos\theta +1$ in terms of Legendre functions. You should then be able to use some orthoganality property.
3. Dec 4, 2008
### buffordboy23
I'm still stuck with this problem. Here is my work.
Let $$x = -cos\theta \rightarrow d\theta = dx/sin\theta$$. Substitution gives
$$\int^{1}_{-1} \frac{\left(1-Ax\right)dx}{\left(A^{2}-2Ax+1\right)^{1/2}} = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{k}\right)\left[1-Ax\right]dx = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{k}\right)\left[P_{0}\left(x\right)-P_{1}\left(x\right)A\right]dx = \int^{1}_{-1} \left(P_{0}\left(x\right)^{2}-P_{1}\left(x\right)^{2}A^{2}\right)dx$$
where the last equation exploited orthogonality. Yet,
$$\int^{1}_{-1} \left(P_{0}\left(x\right)^{2}-P_{1}\left(x\right)^{2}A^{2}\right)dx = 2 - \frac{2A^{2}}{3}$$
and is not the answer. What am I missing?
I know that the legendre generating function requires |A|<1, but this problem says A>1. This implies that the series is divergent, but does that necessarily matter in this context?
4. Dec 4, 2008
### gabbagabbahey
For A>1, this is not the generating function.
Hint: if A>1, then 1/A<1
5. Dec 4, 2008
### buffordboy23
Here we go.
$$\int^{1}_{-1} \frac{\left(1-Ax\right)dx}{\left(A^{2}-2Ax+1\right)^{1/2}} = \int^{1}_{-1} \frac{\left(1/A-x\right)dx}{\left(1-2x/A+1/A^{2}\right)^{1/2}} =\int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{-k}\right)\left[1/A-x\right]dx = \int^{1}_{-1} \left(\sum^{\infty}_{k=0}P_{k}\left(x\right)A^{-k}\right)\left[P_{0}\left(x\right)/A-P_{1}\left(x\right)\right]dx$$
$$= \int^{1}_{-1} \left(\frac{P_{0}\left(x\right)^{2}}{A}-\frac{P_{1}\left(x\right)^{2}}{A}\right)dx = \frac{1}{A}\int^{1}_{-1} \left(1-x^{2}\right)dx = \frac{4}{3A}$$
Thanks.
Last edited: Dec 4, 2008 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.944153368473053, "perplexity": 1121.9573942088748}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171070.80/warc/CC-MAIN-20170219104611-00446-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://undergroundmathematics.org/vector-geometry/lots-of-vector-lines | Fluency exercise
Problem
Here are the equations of 12 straight lines.
$\mathbf{r}= \begin{pmatrix} 0 \\ -6 \\ \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 4 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ -3 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 8 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 2 \\ 9.5 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -8 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 1 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ 4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 4 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 6 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ 6 \\ \end{pmatrix} + \mu \begin{pmatrix} -2 \\ 10 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} -4 \\ 2 \\ \end{pmatrix} + \mu \begin{pmatrix} 5 \\ -1 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ -4 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 6 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 0 \\ 11 \\ \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -6 \end{pmatrix}$
$\mathbf{r}= \begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \end{pmatrix}$
These 12 straight lines can be divided up into six pairs, each pair matching one of the following descriptions:
• These lines are parallel.
• These lines are perpendicular.
• These lines have the same $y$-intercept.
• These lines have the same $x$-intercept.
• These lines both go through the point $(1, 5)$.
• These lines …
Can you sort them into the correct pairs and complete the final description? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9704426527023315, "perplexity": 1455.1254422045622}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00613.warc.gz"} |
https://kar.kent.ac.uk/view/subjects/QA241.type.html | # Items where Subject is "Q Science > QA Mathematics (inc Computing science) > QA150 Algebra > QA241 Number theory"
Group by: Creator's name | Item Type | Date | No Grouping
Number of items at this level: 21.
## Article
Fedorov, Yuri N., Hone, Andrew N.W. (2016) Sigma-function solution to the general Somos-6 recurrence via hyperelliptic Prym varieties. Journal of Integrable Systems, 1 (1). ISSN 2058-5985. (doi:10.1093/integr/xyw012) (KAR id:53527) +1 more...
Hone, Andrew N.W. (2015) Continued fractions for some transcendental numbers. Monatshefte fur Mathematik, . ISSN 0026-9255. E-ISSN 1436-5081. (doi:10.1007/s00605-015-0844-2) (KAR id:50921)
Hone, Andrew N.W. (2015) Curious Continued Fractions, Nonlinear Recurrences and Transcendental Numbers. Journal of Integer Sequences, 18 (15.8.4). pp. 1-10. ISSN 1530-7638. (KAR id:50523)
Hone, Andrew N.W. (2006) Diophantine non-integrability of a third order recurrence with the Laurent property. Journal of Physics A: Mathematical and General, 39 (12). L171-L177. ISSN 0305-4470. (doi:10.1088/0305-4470/39/12/L01) (KAR id:41492)
Hone, Andrew N.W. (2007) Laurent Polynomials and Superintegrable Maps. Symmetry, Integrability and Geometry: Methods and Applications, 3 (022). pp. 1-18. ISSN 1815-0659. (KAR id:41491)
Hone, Andrew N.W. (2016) On the continued fraction expansion of certain Engel series. Journal of Number Theory, 164 . pp. 269-281. ISSN 0022-314X. (doi:10.1016/j.jnt.2015.12.024) (KAR id:53528)
Hone, Andrew N.W. (2007) Sigma function solution of the initial value problem for Somos 5 sequences. Transactions of the American Mathematical Society, 359 (10). pp. 5019-5034. ISSN 0002-9947. (doi:10.1090/S0002-9947-07-04215-8) (KAR id:771)
Hone, Andrew N.W., Varona, Juan Luis (2021) Continued fractions for strong Engel series and Lüroth series with signs. Acta Arithmetica, 199 . ISSN 0065-1036. (doi:10.4064/aa200529-26-11) (KAR id:89769)
Lampe, Philipp (2016) Diophantine equations via cluster transformations. Journal of Algebra, 462 . pp. 320-337. ISSN 0021-8693. (doi:10.1016/j.jalgebra.2016.04.033) (KAR id:67640)
Woodcock, Chris F. (1987) A 2 variable riemann-zeta function. Journal of Number Theory, 27 (2). pp. 212-221. ISSN 0022-314X. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23540)
Woodcock, Chris F. (1979) Convolutions on the ring of p-adic integers. Journal of the London Mathematical Society, 20 . pp. 101-108. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23550)
Woodcock, Chris F. (1974) Fourier analysis for p-adic Lipschitz functions. Journal of the London Mathematical Society, 7 . pp. 681-693. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23554)
Woodcock, Chris F. (1984) Spectral properties of tame functions on the ring of p-adic integers. Journal of the London Mathematical Society, 30 . pp. 407-418. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23547)
Woodcock, Chris F. (1974) An invariant p-adic integral on Zp. Journal of the London Mathematical Society, 8 . pp. 731-734. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23553)
Woodcock, Chris F. (1975) A note on some congruences for the bernoulli numbers bm. Journal of the London Mathematical Society, 11 . p. 256. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23552)
Woodcock, Chris F. (1996) A p-adic analog of Wirtinger's inequality. Michigan Mathematical Journal, 43 (2). pp. 349-354. ISSN 0026-2285. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23524)
Woodcock, Chris F. (1982) A spectral valuation on the ring of p-adic integers. Journal of the London Mathematical Society, 25 . pp. 223-234. ISSN 0024-6107. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23549)
## Book section
Hone, Andrew N.W. (2021) ECM Factorization with QRT Maps. In: Advances in Software Engineering, Education, and e-Learning. Transactions on Computational Science and Computational Intelligence. Springer. ISBN 978-3-030-70872-6. (doi:10.1007/978-3-030-70873-3_28) (KAR id:81861)
Woodcock, Chris F. (1992) p-Adic Fourier Series. In: Baker, Andrew and Plymen, Roger J., eds. P-adic methods and their applications. Oxford University Press, pp. 167-193. ISBN 0198535945:9780198535942. (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23528)
## Conference or workshop item
Hone, Andrew N.W. (2008) Algebraic curves, integer sequences and a discrete Painleve transcendent. In: SIDE 6, 19-24 June 2004, Helsinki, Finland. (Unpublished) (KAR id:41489)
Woodcock, Chris F. (1978) Parseval's formula and p-adic interpolation. In: Proceedings of the conference on p-adic analysis. Report. pp. 205-217. , Katholieke Univ. Nijmegen, Nijmegen (The full text of this publication is not currently available from this repository. You may be able to access a copy if URLs are provided) (KAR id:23551)
This list was generated on Mon Oct 3 22:17:09 2022 BST. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8915332555770874, "perplexity": 4749.630342389328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337473.26/warc/CC-MAIN-20221004023206-20221004053206-00217.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/38891-homomorphisms.html | # Math Help - Homomorphisms
1. ## Homomorphisms
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James
$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image
Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here
Thanks
2. Originally Posted by alittletouched
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James
$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image
Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here
Thanks
How is $\Sigma_n$ defined?
3. $\Sigma_n$ is the symmetric group of n letters
As examples
$\Sigma_2=\left\{1_{\Sigma_2},(1 2)\right\}$
and
$\Sigma_3=\left\{1_{\Sigma_3},(1 2),(2 3),(1 3),(1 2 3),(1 3 2)\right\}$
Thank! James
4. Originally Posted by alittletouched
Hello all. I am stuck trying to figure out if these are homomorphisms and what there images/kernels are if that is the case. Any help would be great! James
$G=\Sigma_5, X=\Sigma_5$. Then the homomorphism takes $\sigma\rightarrow (3 5 2 4 1)*\sigma*(3 1 4 2 5)$. I know that this is a homomorphism, I am just stuck on the kernel/image
Oh! i always used the notation $S_n$. anyway, the Kernel of the homomorphism is only contains the identity.
note that $(3~5~2~4~1)^{-1} = (3~1~4~2~5)$
to get an output of $(1)$ we need $\sigma * (3~1~4~2~5) = (3~5~2~4~1)^{-1}$
clearly $\sigma$ has to be the identity permutation
5. Originally Posted by alittletouched
Working in $\Sigma_8$. Then $G=<(1 2 3)(8 4 6 7 5)>$. If $\sigma=(1 2 3)(8 4 6 7 5)$, the homomorphism maps to $\mathbb{Z}_5$ by $\sigma^i\rightarrow i+5\mathbb{Z}$. I'm not sure what to do from here
Thanks
as for this one. if we are mapping to $\mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?
so you mean $\sigma^i \mapsto [i + 5 \mathbb{Z}]$
the order of $\sigma$ is 8, so that $0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$
6. Originally Posted by Jhevon
as for this one. if we are mapping to $\mathbb{Z}_5$ then we should be mapping to equivalence classes mod 5, right?
so you mean $\sigma^i \mapsto [i + 5 \mathbb{Z}]$
the order of $\sigma$ is 8, so that $0 \le i \le 8$, and then everything repeats for higher values. thus, to find the kernel (which i suppose is what you want), it would be $\{ \sigma^i \mid i + 5 \mathbb{Z} \text{ is a multiple of 5}\}$
just a couple of points:
1) the order of $\sigma$ is 15 not 8.
2) it is important to note that $\sigma^i \rightarrow [i + 5\mathbb{Z}]$ is basically well-defined. that's because 15 (the order of $\sigma$) is divisible by 5.
3) then the map is obviously homomorphism and onto and its kernel is $\{(1), (1 \ 2 \ 3), (1 \ 3 \ 2) \}.$
7. Originally Posted by NonCommAlg
just a couple of points:
1) the order of $\sigma$ is 15 not 8.
yes, you're right. i added when i should have multiplied
8. That's great thanks
How would I find the images for the homomorphisms? I expect that the image of the second homomorphism is merely the integers but Im not sure about the first | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9451363682746887, "perplexity": 212.9891787086147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398444228.5/warc/CC-MAIN-20151124205404-00284-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://arxiv.org/abs/1703.03535 | math.LO
(what is this?)
# Title: Continuous Varieties of Metric and Quantitative Algebras
Authors: Wataru Hino
Abstract: A metric algebra is a metric variant of the notion of $\Sigma$-algebra, first introduced in universal algebra to deal with algebras equipped with metric structures such as normed vector spaces. In this paper, we showed metric versions of the variety theorem, which characterizes strict varieties (classes of metric algebras defined by metric equations) and continuous varieties (classes defined by a continuous family of basic quantitative inferences) by means of closure properties. To this aim, we introduce the notion of congruential pseudometric on a metric algebra, which corresponds to congruence in classical universal algebra, and we investigate its structure.
Subjects: Logic (math.LO) Cite as: arXiv:1703.03535 [math.LO] (or arXiv:1703.03535v1 [math.LO] for this version)
## Submission history
From: Wataru Hino [view email]
[v1] Fri, 10 Mar 2017 03:48:15 GMT (26kb) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9299566745758057, "perplexity": 2717.444382001069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688103.0/warc/CC-MAIN-20170922003402-20170922023402-00705.warc.gz"} |
http://mathhelpforum.com/calculus/185276-differential.html | 1. ## differential
the eccentricity of the ellipse x^2/4+y^2/3 is changed at the rate of 0.1/sec.the time at which it will touch the auxillary circle is
i know auxilary circle equation that is
x^2+y^2=a^2+b^2
and eccentricity also
2. ## Re: differential
Originally Posted by prasum
the eccentricity of the ellipse x^2/4+y^2/3 is changed at the rate of 0.1/sec.the time at which it will touch the auxillary circle is
i know auxilary circle equation that is
x^2+y^2=a^2+b^2
and eccentricity also
Could you transcribe the exact formulation of the problem? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.854034960269928, "perplexity": 4535.915545421802}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891105.83/warc/CC-MAIN-20180122054202-20180122074202-00740.warc.gz"} |
https://www.cs.tau.ac.il/~turkel/notes/wiener_theory.html | Next: Program Up: Project Home Previous: Introduction
# Theory of Wiener Filtering
The Wiener Filter is a noise filter based on Fourier iteration. its main advantage is the short computational time it takes to find a solution.
Consider a situation such that there is some underlying, uncorrupted singal u(t) that is required to measure. Error occur in the measurement due to imperfection in equipments, thus the output signal is corrupted. There are two ways the signal can be corrupted: First, the equipment can convolve, or 'smear' the signal. This occurs when the equipment doesn't have a perfect, delta function response to the signal. Let s(t) be the smear signal and r(t) be the known response that cause the convoltion. Then s(t) is related to u(t) by:
or S(f) = R(f)U(f)
(1)
where S, R, U are Fourier Transform of s, r, and u.
The second source of signal corruption is the unknown background noise n(t). Therefor the measured signal c(t) is a sum of s(t) and n(t):
c(t) = s(t) + n(t)
(2)
To deconvolve s to find u, simply divide S(f) by R(f), i.e. in the absense of noise n. To deconvolve c where n is present then one need to find an optimum filter function , or , which filters out the noise and gives a signal by:
(3)
where is as close to the original signal as possible.
For to be similar to u, their differences squared is as close to zero as possible, i.e.
or is minimised
(4)
Substituting equation (1), (2) and (3), the Fourier version becomes:
(5)
after rearranging. The best filter is one where the above integral is a minimum at every value of f. This is when,
(6)
Now, where , and are the power spectrum of C, S, and N. Therefore,
Figure 1 shows a plot of . It can be seen that and can be extrapolated from the graph.
Figure 1. Plot of power spectrum of signal plut noise
From the above theory, it can be seen that a program can be written to Wiener Filter signal from noise using Fourier Transform. This is what Jean is investigating with her part. There is another way to Wiener Filtering a signal but this time without Fourier Transform the data. The latter is what I set out to do with my part of programming, and this is the Mean-Squared Method.
## The Mean-Squared Method
The Mean-squared Methods uses the fact that the Wiener Filter is one that is based on the least-squared principle, i.e. the filter minimises the error between the actual output and the desired output. To eliminate noise, this filter works by narrow the width of the distribution curve of figure 1.
To do this, first the varience of the data matrix is to be found. Then, a box of certain size is passed around the matrix, moving one pixel at a time. For every box, the local mean and variance is found. And the filtered value of each pixel is found by the following formula:
where Ai,j is the filtered signal, is the local mean, is the local variance, s2 is the noise variance of the entire data matrix, and Ni,j is the original signal value.
From the above formula, it can be see that if the original signal is similar to the local varience then the filtered value will be that of the local mean, if the original signal is very much different from the local mean, then it will be filtered to give a higher/lower intensity signal depends on the differences. Also, if the local variance is similar to the matrix variance, which is around 1 (i.e. only noise exist in the box) then the filtered signal will be that of the local mean, which should be close to zero. But if the local variance is much bigger than the matrix variance (i.e. when the box is at the actual signal), then the signal will be amplified.
As the box moves through the entire matrix, it will calculate the solution to each pixel using the above formula, and thus filters the data. The program for the routine is shown in the next section: Program.
Next: Program Up: Project Home Previous: Introduction
Rosalind Wang
2000-05-27 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9049092531204224, "perplexity": 747.5593360088325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487636559.57/warc/CC-MAIN-20210618104405-20210618134405-00056.warc.gz"} |
https://www.physicsforums.com/threads/the-relationship-between-time-taken-per-oscillation-and-mass.485009/ | # The relationship between time taken per oscillation and mass
• #1
3
0
In an experiment, a ruler is connected to the table and some weights are bounded to one end of the ruler. The ruler is then flicked and the time taken per oscillation is measured.
I have plotted a graph with the data I have collected, with the mass on the y-axis and time on x-axis. The graph produced appears to be a curve. I have tried altering the values on the x-axis; I have squared it, 1 over the square of it, square rooted it, and I found that the graph becomes linear when the values are squared. So the mass should be proportional to 1 over the square of the time taken.
I have tried finding out a mathematical relationship for this, but I am not sure if this is correct or not.
Well, if we make w=angular velocity, then w=θ/t, with θ being angular displacement and t being the time period. Since θ belongs in a circle, then it is safe to say that w=2π/t (?)
Also, if the force of an oscillation is proportional to -displacement (x), then it is true to say that F=-kx, with k being a constant.
Since F also = ma, then ma=-kx.
According the the simple harmonic wave equation for acceleration is a=-xw2sinwt. Since the formula for displacement(x) = x sinwt and a=-w2(x sinwt), then a=-xw2
So ma=-kx will become m(-xw2)=-kx, then using some algebra, m=k/w2. Since w=2π/t, then m=kT2/4π2.
Since k/4π2 is a constant, I can ignore that and say m is proportional to t2.
Is my reasoning true? I feel like I am wrong in quite a few spots.
Thanks for helping
Last edited:
## Answers and Replies
• #2
Gokul43201
Staff Emeritus
Gold Member
7,083
18
Yes, this is essentially correct, but you take a few confusing steps because of a poor choice of notation. For instance, you use x to represent displacement as well as the amplitude (or maximum displacement). You should use different symbols for these - it will help keep things clear for you as well.
You write: x = x sin(wt), which looks nonsensical.
Better would be something like: x = A sin (wt), or x = x0 sin (wt).
Also, next time, a question like this is better suited for the Intro Physics section.
• #3
3
0
Thank you.
• #4
4
0
Yes thank you very much! It is so helpful!
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3K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8995233774185181, "perplexity": 1149.201944613358}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00461.warc.gz"} |
https://math.stackexchange.com/questions/3303947/closed-form-for-an-infinite-series-involving-lower-incomplete-gamma-functions/3304097#3304097 | # Closed form for an infinite series involving lower incomplete gamma functions
I need to evaluate the inverse Laplace transform $$Q(t) = \mathcal{L}^{-1}\big\{\frac{e^{b/s}}{s(s-a)}\big\}(t).$$ Using the identity $$\mathcal{L}^{-1}\{\frac{f(s)}{s-a}\}(t)= e^{at}\int_0^tdu e^{-au}\mathcal{L}^{-1}\{f(s)\}(u)$$ with knowledge of the inverse transform $$\mathcal{L}^{-1}\{\frac{e^{b/s}}{s}\}(u) = I_0(2\sqrt{bu})$$, the series representation of the modified Bessel function $$I_0(z) = \sum_{k=0}^\infty \frac{1}{k!k!}\big(\frac{z}{2}\big)^{2k}$$, and the definition of the lower incomplete gamma function $$\gamma(k,x) = \int_0^x t^{k-1}e^{-t}dt$$ provides $$Q(t)$$ in the form $$Q(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!k!}\gamma(k+1,at).$$
Is this as good as it gets? Is there an approach I could use to evaluate this sum? So far I have tried expressing the incomplete gamma function in terms of hypergeometric functions, but this does not seem to provide any traction.
One option is to introduce the identity $$\gamma(k+1,at) = k!(1-e^{-at} \sum_{l=0}^k \frac{(at)^k}{k!})$$ obtaining $$Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].$$ The second term of this resembles a Humbert series $$\Phi_3(\beta,\gamma,x,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(\beta)_m}{(\gamma)_{m+n}m!n!}x^my^n$$ with the wrong summation limits. Does anyone see a path here? I suppose taking negative values in the Pockhammer symbols might produce a correspondence.
In any case I expect some hypergeometric function representation of this sum. Can anyone offer guidance? I have found several related problems Closed-form Solution for series involving incomplete Gamma Function and Any way to simplify integral of Confluent Hypergeometric Function of the First Kind?
$$Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].$$
I'll blindly try to reverse the order of summation and see what happens.
$$\begin{array}\\ S(u, v) &=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\sum_{k=l}^\infty \frac{u^lv^k}{k!l!}\\ &=\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=l}^\infty \frac{v^k}{k!}\\ &=\sum_{l=0}^\infty\frac{u^l}{l!}(e^v-\sum_{k=0}^{l-1} \frac{v^k}{k!})\\ &=\sum_{l=0}^\infty\frac{u^l}{l!}e^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\ &=e^ue^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1} \frac{v^k}{k!}\\ &=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}(\sum_{k=0}^{l} \frac{v^k}{k!}-\frac{v^l}{l!})\\ &=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{u^l}{l!}\frac{v^l}{l!}\\ &=e^{u+v}-\sum_{l=0}^\infty\sum_{k=0}^{l}\frac{u^l}{l!} \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{(uv)^l}{l!^2}\\ &=e^{u+v}-S(v, u)+I_0(2\sqrt{uv}) \\ \end{array}$$
where $$I_0$$ is the modified Bessel function of the first kind.
So this isn't a evaluation but we get the relation
$$S(u, v)+S(v, u) =e^{u+v}+I_0(2\sqrt{uv})$$.
Then
$$\begin{array}\\ Q(t) &= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big]\\ &= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}S(at, b/a)\Big]\\ &= \frac{1}{a}\Big[e^{at+b/a}-S(at, b/a)\Big]\\ &= \frac{1}{a}\Big[e^{at+b/a}-(e^{at+b/a}-S(b/a, at)+I_0(2\sqrt{(at)(b/a)}))\Big]\\ &= \frac{1}{a}\Big[S(b/a, at)-I_0(2\sqrt{tb})\Big]\\ \end{array}$$
Again, not an evaluation, but a possibly useful alternative expression.
This reminds me very much of some work I did over forty years ago on the Marcum Q-function. You might look that up and follow the references. You can start here:
https://en.wikipedia.org/wiki/Marcum_Q-function
• Thanks @marty cohen! This is excellent. I have to study a few steps more carefully to fully understand. I searched for your work on Q-functions but couldn't find it: can you offer a bit more info? I'd love to have a look at it Jul 25, 2019 at 23:41
• It was done when I worked for a private company, and I don't have my work any more. Sorry. Jul 26, 2019 at 2:00
To recap my findings from @martycohen's guidance, I got to this result for the inverse Laplace transform I need: $$\mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!}\frac{\gamma(k+1,at)}{\Gamma(k+1)}.$$ The book "An Introduction to the Classical Functions of Mathematical Physics" by Temme (1996) provides the definition $$Q_\mu(u,v) = 1- e^{-u}\sum_{k=0}^\infty\frac{u^k}{k!}\frac{\gamma(\mu+k,v)}{\Gamma(\mu+k)}$$ for the non-central $$\chi^2$$ distribution, also known as the "generalized Marcum $$Q$$-function", or the just the "Marcum $$Q$$-function" when $$\mu=1$$. Marty's suggestion provides $$\mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{1}{a}e^{at+b/a}[1-Q_1(b/a,at)].$$ There is a representation of this function as an infinite superposition of modified Bessel functions of the first kind, zeroth order: $$Q_\mu(u,v) = 1-\int_0^v \Big(\frac{z}{u}\Big)^{\frac{1}{2}(\mu-1)}e^{-z-x}I_{\mu-1}(2\sqrt{xz}).$$ This makes perfect sense in context of the problem which led to the need for this inverse Laplace transform. Thanks Marty! This helps my research. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9478708505630493, "perplexity": 160.2590088293965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00249.warc.gz"} |
https://neurips.cc/Conferences/2021/ScheduleMultitrack?event=27454 | `
Timezone: »
Poster
NEO: Non Equilibrium Sampling on the Orbits of a Deterministic Transform
Achille Thin · Yazid Janati El Idrissi · Sylvain Le Corff · Charles Ollion · Eric Moulines · Arnaud Doucet · Alain Durmus · Christian X Robert
Thu Dec 09 12:30 AM -- 02:00 AM (PST) @ Virtual
Sampling from a complex distribution $\pi$ and approximating its intractable normalizing constant $\mathrm{Z}$ are challenging problems. In this paper, a novel family of importance samplers (IS) and Markov chain Monte Carlo (MCMC) samplers is derived. Given an invertible map $\mathrm{T}$, these schemes combine (with weights) elements from the forward and backward Orbits through points sampled from a proposal distribution $\rho$. The map $\mathrm{T}$ does not leave the target $\pi$ invariant, hence the name NEO, standing for Non-Equilibrium Orbits. NEO-IS provides unbiased estimators of the normalizing constant and self-normalized IS estimators of expectations under $\pi$ while NEO-MCMC combines multiple NEO-IS estimates of the normalizing constant and an iterated sampling-importance resampling mechanism to sample from $\pi$. For $\mathrm{T}$ chosen as a discrete-time integrator of a conformal Hamiltonian system, NEO-IS achieves state-of-the art performance on difficult benchmarks and NEO-MCMC is able to explore highly multimodal targets. Additionally, we provide detailed theoretical results for both methods. In particular, we show that NEO-MCMC is uniformly geometrically ergodic and establish explicit mixing time estimates under mild conditions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9319068193435669, "perplexity": 2239.503223835344}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00550.warc.gz"} |
https://www.lmfdb.org/L/2/2016/7.4 | ## Results (1-50 of at least 1000)
Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim arith $\mathbb{Q}$ self-dual $\operatorname{Arg}(\epsilon)$ $r$ First zero Origin
2-2016-28.11-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 28.11 $$0.0 0 -0.198 0 0.977336 Modular form 2016.1.cd.a.991.2 2-2016-252.79-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 252.79$$ $0.0$ $0$ $0.392$ $0$ $2.00194$ Modular form 2016.1.dd.a.1087.1
2-2016-252.79-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 252.79 $$0.0 0 -0.357 0 0.682370 Modular form 2016.1.dd.a.1087.2 2-2016-252.67-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 252.67$$ $0.0$ $0$ $0.357$ $0$ $1.85783$ Modular form 2016.1.dd.a.319.1
2-2016-252.67-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 252.67 $$0.0 0 -0.392 0 0.616480 Modular form 2016.1.dd.a.319.2 2-2016-252.247-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 252.247$$ $0.0$ $0$ $-0.162$ $0$ $1.18272$ Modular form 2016.1.bw.a.1759.2
2-2016-252.247-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 252.247 $$0.0 0 0.0871 0 0.984191 Modular form 2016.1.bw.a.1759.1 2-2016-252.151-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 252.151$$ $0.0$ $0$ $0.162$ $0$ $1.94791$ Modular form 2016.1.bw.a.1663.2
2-2016-252.151-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 252.151 $$0.0 0 -0.0871 0 0.888242 Modular form 2016.1.bw.a.1663.1 2-2016-224.69-c0-0-2 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.69$$ $0.0$ $0$ $0.218$ $0$ $1.63003$ Modular form 2016.1.dp.c.1189.1
2-2016-224.69-c0-0-1 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.69 $$0.0 0 -0.281 0 0.775588 Modular form 2016.1.dp.a.1189.1 2-2016-224.69-c0-0-0 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.69$$ $0.0$ $0$ $-0.0312$ $0$ $0.692095$ Modular form 2016.1.dp.b.1189.1
2-2016-224.181-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.181 $$0.0 0 0.468 0 2.41015 Modular form 2016.1.dp.a.181.1 2-2016-224.181-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.181$$ $0.0$ $0$ $-0.0312$ $0$ $1.16715$ Modular form 2016.1.dp.c.181.1
2-2016-224.181-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.181 $$0.0 0 -0.281 0 1.15600 Modular form 2016.1.dp.b.181.1 2-2016-224.13-c0-0-2 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.13$$ $0.0$ $0$ $0.281$ $0$ $1.50608$ Modular form 2016.1.dp.a.685.1
2-2016-224.13-c0-0-1 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.13 $$0.0 0 -0.218 0 1.21396 Modular form 2016.1.dp.c.685.1 2-2016-224.13-c0-0-0 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.13$$ $0.0$ $0$ $0.0312$ $0$ $1.20749$ Modular form 2016.1.dp.b.685.1
2-2016-224.125-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.125 $$0.0 0 0.281 0 1.86729 Modular form 2016.1.dp.b.1693.1 2-2016-224.125-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 224.125$$ $0.0$ $0$ $0.0312$ $0$ $0.935488$ Modular form 2016.1.dp.c.1693.1
2-2016-224.125-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 224.125 $$0.0 0 -0.468 0 0.553472 Modular form 2016.1.dp.a.1693.1 2-2016-168.83-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 168.83$$ $0.0$ $0$ $0.222$ $0$ $1.75719$ Modular form 2016.1.e.a.1007.1
2-2016-168.83-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 168.83 $$0.0 0 0.0270 0 1.20435 Modular form 2016.1.e.a.1007.2 2-2016-168.83-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 168.83$$ $0.0$ $0$ $-0.0270$ $0$ $1.14333$ Modular form 2016.1.e.a.1007.3
2-2016-168.83-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 168.83 $$0.0 0 -0.222 0 0.744749 Modular form 2016.1.e.a.1007.4 2-2016-7.6-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 7.6$$ $0.0$ $0$ $0.250$ $0$ $1.83053$ Modular form 2016.1.f.a.1441.1
2-2016-7.6-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 7.6 $$0.0 0 0.250 0 1.66575 Modular form 2016.1.f.a.1441.2 2-2016-7.6-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 7.6$$ $0.0$ $0$ $-0.250$ $0$ $0.911336$ Modular form 2016.1.f.a.1441.4
2-2016-7.6-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 7.6 $$0.0 0 -0.250 0 0.683158 Modular form 2016.1.f.a.1441.3 2-2016-672.83-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.83$$ $0.0$ $0$ $-0.441$ $0$ $1.93035$ Modular form 2016.1.dn.b.755.1
2-2016-672.83-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.83 $$0.0 0 0.00420 0 1.66567 Modular form 2016.1.dn.d.755.1 2-2016-672.83-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.83$$ $0.0$ $0$ $-0.245$ $0$ $0.872598$ Modular form 2016.1.dn.c.755.1
2-2016-672.83-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.83 $$0.0 0 -0.191 0 0.813611 Modular form 2016.1.dn.a.755.1 2-2016-672.587-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.587$$ $0.0$ $0$ $-0.00420$ $0$ $1.45534$ Modular form 2016.1.dn.b.1259.1
2-2016-672.587-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.587 $$0.0 0 -0.0582 0 1.34491 Modular form 2016.1.dn.d.1259.1 2-2016-672.587-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.587$$ $0.0$ $0$ $0.191$ $0$ $1.22051$ Modular form 2016.1.dn.c.1259.1
2-2016-672.587-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.587 $$0.0 0 -0.254 0 0.510009 Modular form 2016.1.dn.a.1259.1 2-2016-672.419-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.419$$ $0.0$ $0$ $0.0582$ $0$ $1.68764$ Modular form 2016.1.dn.d.1763.1
2-2016-672.419-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.419 $$0.0 0 0.254 0 1.40997 Modular form 2016.1.dn.a.1763.1 2-2016-672.419-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.419$$ $0.0$ $0$ $-0.191$ $0$ $1.27611$ Modular form 2016.1.dn.c.1763.1
2-2016-672.419-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.419 $$0.0 0 0.00420 0 0.752721 Modular form 2016.1.dn.b.1763.1 2-2016-672.251-c0-0-3 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.251$$ $0.0$ $0$ $0.191$ $0$ $1.63816$ Modular form 2016.1.dn.a.251.1
2-2016-672.251-c0-0-2 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.251 $$0.0 0 -0.00420 0 1.56054 Modular form 2016.1.dn.d.251.1 2-2016-672.251-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 672.251$$ $0.0$ $0$ $0.245$ $0$ $1.18996$ Modular form 2016.1.dn.c.251.1
2-2016-672.251-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 672.251 $$0.0 0 0.441 0 0.343911 Modular form 2016.1.dn.b.251.1 2-2016-56.5-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 56.5$$ $0.0$ $0$ $0.103$ $0$ $1.59517$ Modular form 2016.1.ce.a.145.2
2-2016-56.5-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 56.5 $$0.0 0 -0.396 0 0.525284 Modular form 2016.1.ce.a.145.1 2-2016-56.45-c0-0-1 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 56.45$$ $0.0$ $0$ $0.396$ $0$ $2.10298$ Modular form 2016.1.ce.a.1585.1
2-2016-56.45-c0-0-0 $1.00$ $1.00$ $2$ $2^{5} \cdot 3^{2} \cdot 7$ 56.45 $$0.0 0 -0.103 0 1.28797 Modular form 2016.1.ce.a.1585.2 2-2016-56.13-c0-0-2 1.00 1.00 2 2^{5} \cdot 3^{2} \cdot 7 56.13$$ $0.0$ $0$ $0.250$ $0$ $1.66556$ Artin representation 2.2016.8t11.c.b Modular form 2016.1.l.b.433.1
Next | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9607576727867126, "perplexity": 445.22427568611005}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303709.2/warc/CC-MAIN-20220121192415-20220121222415-00686.warc.gz"} |
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Conclusions: The X-ray properties of HD 45314 indicate that this star is a new member of the class of γ Cas analogs, the first one among the original category of Oe stars. Based on observations collected with XMM-Newton, an ESA Science Mission with instruments and contributions directly funded by ESA Member States and the USA (NASA), and observations collected at the European Southern Observatory (La Silla, Chile) and the Observatoire de Haute Provence (France). [less ▲]Detailed reference viewed: 8 (2 ULg) Massive Stars in the Gaia-ESO SurveyBlomme, Ronny; Fremat, Yves; Lobel, Alex et alin Massive Stars: From alpha to Omega (2013, June 01)The Gaia-ESO Survey (GES) is an ambitious project to study the formation and evolution of the Milky Way and its stellar populations. It is led by Gerry Gilmore and Sofia Randich and includes about 350 Co ... [more ▼]The Gaia-ESO Survey (GES) is an ambitious project to study the formation and evolution of the Milky Way and its stellar populations. It is led by Gerry Gilmore and Sofia Randich and includes about 350 Co-Investigators. During 300 nights (spread over 5 years) of order 10^5 Giraffe spectra and 10^4 UVES spectra will be taken. As part of the survey, about 13 clusters will be observed that were chosen specifically for their massive-star content. We report on the preliminary analysis of GES data from two such clusters: NGC 3293 and NGC 6705. We determine stellar parameters for the B-type stars in NGC 3293 and compare the A-type stars population between the two clusters. We also use a repeat observation to study binarity and use the radial velocity information to study cluster membership in NGC 3293. We also list our plans for future observations, which include the Carina nebula region. [less ▲]Detailed reference viewed: 16 (3 ULg) An Observational Study of Mixing in Fast-Rotating Massive Stars: Description of the Method and Very First ResultsMorel, Thierry ; Palate, Matthieu ; Rauw, Grégor in Massive Stars: From alpha to Omega (2013, June 01)Collecting He and CNO abundances for a large sample of fast-rotating massive stars is crucial for addressing the efficiency of rotational mixing in these objects. This is becoming a pressing issue ... [more ▼]Collecting He and CNO abundances for a large sample of fast-rotating massive stars is crucial for addressing the efficiency of rotational mixing in these objects. This is becoming a pressing issue following the recent recognition based on observations by the VLT-FLAMES Survey of massive stars that fast rotators can show no signs of deep mixing contrary to the predictions of models. We have embarked in a project aiming at determining the abundances of the key elements indicators of mixing in a sample of bright, Galactic OB dwarfs. We present here the method and the very first results for two of the fastest OB stars known vsin i = 400 km/s : HD 93521 and zeta Oph. [less ▲]Detailed reference viewed: 5 (1 ULg) Solar-like oscillations in distant stars as seen by CoRoT : the special case of HD 42618, a solar sisterBarban, C.; Deheuvels, S.; Goupil, M. J. et alin Journal of Physics: Conference Series (2013), 440We report the observations of a main-sequence star, HD 42618 (T[SUB]eff[/SUB] = 5765 K, G3V) by the space telescope CoRoT. This is the closest star to the Sun ever observed by CoRoT in term of its ... [more ▼]We report the observations of a main-sequence star, HD 42618 (T[SUB]eff[/SUB] = 5765 K, G3V) by the space telescope CoRoT. This is the closest star to the Sun ever observed by CoRoT in term of its fundamental parameters. Using a preliminary version of CoRoT light curves of HD 42618, p modes are detected around 3.2 mHz associated to l = 0, 1 and 2 modes with a large spacing of 142 μHz. Various methods are then used to derive the mass and radius of this star (scaling relations from solar values as well as comparison between theoretical and observationnal frequencies) giving values in the range of (0.80 - 1.02)M[SUB]solar[/SUB] and (0.91 - 1.01)R[SUB]solar[/SUB]. A preliminary analysis of l = 0 and 1 modes allows us also to study the amount of penetrative convection at the base of the convective envelope. [less ▲]Detailed reference viewed: 14 (1 ULg) Abundance study of the two solar-analogue CoRoT targets HD 42618 and HD 43587 from HARPS spectroscopyMorel, Thierry ; Rainer, M.; Poretti, E. et alin Astronomy and Astrophysics (2013), 552We present a detailed abundance study based on spectroscopic data obtained with HARPS of two solar-analogue main targets for the asteroseismology programme of the CoRoT satellite: HD 42618 and HD 43587 ... [more ▼]We present a detailed abundance study based on spectroscopic data obtained with HARPS of two solar-analogue main targets for the asteroseismology programme of the CoRoT satellite: HD 42618 and HD 43587. The atmospheric parameters and chemical composition are accurately determined through a fully differential analysis with respect to the Sun observed with the same instrumental set-up. Several sources of systematic errors largely cancel out with this approach, which allows us to narrow down the 1-σ error bars to typically 20 K in effective temperature, 0.04 dex in surface gravity, and less than 0.05 dex in the elemental abundances. Although HD 42618 fulfils many requirements for being classified as a solar twin, its slight deficiency in metals and its possibly younger age indicate that, strictly speaking, it does not belong to this class of objects. On the other hand, HD 43587 is slightly more massive and evolved. In addition, marked differences are found in the amount of lithium present in the photospheres of these two stars, which might reveal different mixing properties in their interiors. These results will put tight constraints on the forthcoming theoretical modelling of their solar-like oscillations and contribute to increase our knowledge of the fundamental parameters and internal structure of stars similar to our Sun. Based on observations collected at the La Silla Observatory, ESO (Chile) with the HARPS spectrograph at the 3.6-m telescope, under programme LP185.D-0056.Tables 1 and 2 are available in electronic form at http://www.aanda.org [less ▲]Detailed reference viewed: 14 (0 ULg) Differential population studies using asteroseismology: Solar-like oscillating giants in CoRoT fields LRc01 and LRa01Miglio, A.; Chiappini, C.; Morel, Thierry et alin European Physical Journal Web of Conferences (2013, March 01)Solar-like oscillating giants observed by the space-borne satellites CoRoT and Kepler can be used as key tracers of stellar populations in the Milky Way. When combined with additional photometric ... [more ▼]Solar-like oscillating giants observed by the space-borne satellites CoRoT and Kepler can be used as key tracers of stellar populations in the Milky Way. When combined with additional photometric/spectroscopic constraints, the pulsation spectra of solar-like oscillating giant stars not only reveal their radii, and hence distances, but also provide well-constrained estimates of their masses, which can be used as proxies for the ages of these evolved stars. In this contribution we provide supplementary material to the comparison we presented in Miglio et al. (2013) between populations of giants observed by CoRoT in the fields designated LRc01 and LRa01. [less ▲]Detailed reference viewed: 13 (3 ULg) An abundance study of the red giants in the seismology fields of the CoRoT satelliteMorel, Thierry ; Miglio, A.; Lagarde, N. et alin European Physical Journal Web of Conferences (2013, March 01)A precise characterisation of the red giants in the seismology fields of the CoRoT satellite is a prerequisite for further in-depth seismic modelling. The optical spectra obtained for 19 targets have been ... [more ▼]A precise characterisation of the red giants in the seismology fields of the CoRoT satellite is a prerequisite for further in-depth seismic modelling. The optical spectra obtained for 19 targets have been used to accurately estimate their fundamental parameters and chemical composition. The extent of internal mixing is also investigated through the abundances of Li, CNO and Na (as well as [SUP]12[/SUP]C/[SUP]13[/SUP]C in a few cases). [less ▲]Detailed reference viewed: 11 (1 ULg) High-precision CoRoT space photometry and fundamental parameter determination of the B2.5V star HD 48977Thoul, Anne ; Degroote, Pieter; Catala, Claude et alin Astronomy and Astrophysics (2013), 551We present the CoRoT light curve of the bright B2.5V star HD 48977 observed during a short run of the mission in 2008, as well as a high-resolution spectrum gathered with the HERMES spectrograph at the ... [more ▼]We present the CoRoT light curve of the bright B2.5V star HD 48977 observed during a short run of the mission in 2008, as well as a high-resolution spectrum gathered with the HERMES spectrograph at the Mercator telescope. We use several time series analysis tools to explore the nature of the variations present in the light curve. We perform a detailed analysis of the spectrum of the star to determine its fundamental parameters and its element abundances. We find a large number of high-order g-modes, and one rotationally induced frequency. We find stable low-amplitude frequencies in the p-mode regime as well. We conclude that HD 48977 is a new Slowly Pulsating B star with fundamental parameters found to be Teff = 20000 $\pm$ 1000 K and log(g)=4.2 $/pm$ 0.1. The element abundances are similar to those found for other B stars in the solar neighbourhood. HD 48977 was observed during a short run of the CoRoT satellite implying that the frequency precision is insufficient to perform asteroseismic modelling of the star. Nevertheless, we show that a longer time series of this star would be promising for such modelling. Our present study contributes to a detailed mapping of the instability strips of B stars in view of the dominance of g-mode pulsations in the star, several of which occur in the gravito-inertial regime. [less ▲]Detailed reference viewed: 21 (13 ULg) Galactic archaeology: mapping and dating stellar populations with asteroseismology of red-giant starsMiglio, A.; Chiappini, C.; Morel, Thierry et alin Monthly Notices of the Royal Astronomical Society (2013), 429Our understanding of how the Galaxy was formed and evolves is severely hampered by the lack of precise constraints on basic stellar properties such as distances, masses and ages. Here, we show that solar ... [more ▼]Our understanding of how the Galaxy was formed and evolves is severely hampered by the lack of precise constraints on basic stellar properties such as distances, masses and ages. Here, we show that solar-like pulsating red giants represent a well-populated class of accurate distance indicators, spanning a large age range, which can be used to map and date the Galactic disc in the regions probed by observations made by the CoRoT and Kepler space telescopes. When combined with photometric constraints, the pulsation spectra of such evolved stars not only reveal their radii, and hence distances, but also provide well-constrained estimates of their masses, which are reliable proxies for the ages of the stars. As a first application, we consider red giants observed by CoRoT in two different parts of the Milky Way, and determine precise distances for ˜2000 stars spread across nearly 15 000 pc of the Galactic disc, exploring regions which are a long way from the solar neighbourhood. We find significant differences in the mass distributions of these two samples which, by comparison with predictions of synthetic models of the Milky Way, we interpret as mainly due to the vertical gradient in the distribution of stellar masses (hence ages) in the disc. In the future, the availability of spectroscopic constraints for this sample of stars will not only improve the age determination, but also provide crucial constraints on age-velocity and age-metallicity relations at different Galactocentric radii and heights from the plane. [less ▲]Detailed reference viewed: 24 (2 ULg) Are Magnetic OB Stars More Prone to Mixing? Still an Unsettled IssueMorel, Thierry in Astronomical Society of the Pacific Conference Series (2013, January 01)We review our knowledge of the mixing properties of magnetic OB stars and discuss whether the observational data presently available support, as predicted by some theoretical models, the idea that ... [more ▼]We review our knowledge of the mixing properties of magnetic OB stars and discuss whether the observational data presently available support, as predicted by some theoretical models, the idea that magnetic phenomena favour the transport of chemical elements. A (likely statistical) relationship between enhanced mixing and the existence of a field has been emerging over the last few years. As discussed in this contribution, however, a clear answer to this question is presently hampered by the lack of large and well-defined samples of magnetic and non-magnetic stars. [less ▲]Detailed reference viewed: 10 (2 ULg) GAUFRE: a tool for an automated determination of atmospheric parameters from spectroscopyValentini, Marica ; Morel, Thierry ; Miglio, Andrea et alin 40th Liege International Astrophysical Colloquium 'Ageing low-mass stars: from red giants to white dwarfs' (2013, January 01)We present an automated tool for measuring atmospheric parameters (T_eff, log(g), [Fe/H]) for F-G-K dwarf and giant stars. The tool, called GAUFRE, is written in C++ and composed of several routines ... [more ▼]We present an automated tool for measuring atmospheric parameters (T_eff, log(g), [Fe/H]) for F-G-K dwarf and giant stars. The tool, called GAUFRE, is written in C++ and composed of several routines: GAUFRE-RV measures radial velocity from spectra via cross-correlation against a synthetic template, GAUFRE-EW measures atmospheric parameters through the classic line-by-line technique and GAUFRE-CHI2 performs a chi^2 fitting to a library of synthetic spectra. A set of F-G-K stars extensively studied in the literature were used as a benchmark for the program: their high signal-to-noise and high resolution spectra were analysed by using GAUFRE and results were compared with those present in literature. The tool is also implemented in order to perform the spectral analysis after fixing the surface gravity (log(g)) to the accurate value provided by asteroseismology. A set of CoRoT stars, belonging to LRc01 and LRa01 fields was used for first testing the performances and the behaviour of the program when using the seismic log(g). [less ▲]Detailed reference viewed: 12 (2 ULg) CoRoT Observations of O Stars: Diverse Origins of VariabilityBlomme, R.; Briquet, Maryline ; Degroote, P. et alin Astronomical Society of the Pacific Conference Series (2013, January 01)Six O-type stars were observed continuously by the CoRoT satellite during a 34.3-day run. The unprecedented quality of the data allows us to detect even low-amplitude stellar pulsations in some of these ... [more ▼]Six O-type stars were observed continuously by the CoRoT satellite during a 34.3-day run. The unprecedented quality of the data allows us to detect even low-amplitude stellar pulsations in some of these stars (HD 46202 and the binaries HD 46149 and Plaskett's star). These cover both opacity-driven modes and solar-like stochastic oscillations, both of importance to the asteroseismological modeling of O stars. Additional effects can be seen in the CoRoT light curves, such as binarity and rotational modulation. Some of the hottest O-type stars (HD 46223, HD 46150 and HD 46966) are dominated by the presence of red-noise: we speculate that this is related to a sub-surface convection zone. [less ▲]Detailed reference viewed: 38 (18 ULg) 2009: A Colliding-Wind OdysseyFahed, R.; Moffat, A. F. J.; Zorec, J. et alin Astronomical Society of the Pacific Conference Series (2013, January 01)We present the results from two optical spectroscopic campaigns on colliding-wind binaries (CWB) which both occurred in 2009. The first one was on WR 140 (WC7pd + O5.5fc), the archetype of CWB, which ... [more ▼]We present the results from two optical spectroscopic campaigns on colliding-wind binaries (CWB) which both occurred in 2009. The first one was on WR 140 (WC7pd + O5.5fc), the archetype of CWB, which experienced periastron passage of its highly elliptical 8-year orbit in January. The WR 140 campaign consisted of a unique and constructive collaboration between amateur and professional astronomers and took place at half a dozen locations, including Teide Observatory, Observatoire de Haute Provence, Dominion Astrophysical Observatory, Observatoire du Mont-Mégantic and at several small private observatories. The second campaign was on a selection of 5 short-period WR + O binaries not yet studied for colliding-wind effects: WR 12 (WN8h), WR 21 (WN5o + O7 V), WR 30 (WC6 + O7.5 V), WR 31 (WN4o + O8), and WR 47 (WN6o + O5). The campaign took place at Leoncito Observatory, Argentina, during 1 month. We provide updated values of most of these systems for the orbital parameters, new estimates for the WR and O star masses and new constraints on the mass-loss rates and colliding wind geometry. [less ▲]Detailed reference viewed: 8 (0 ULg) Multisite spectroscopic seismic study of the β Cep star V2052 Ophiuchi: inhibition of mixing by its magnetic fieldBriquet, Maryline ; Neiner, C.; Aerts, C. et alin Monthly Notices of the Royal Astronomical Society (2012), 427We used extensive ground-based multisite and archival spectroscopy to derive observational constraints for a seismic modelling of the magnetic β Cep star V2052 Ophiuchi. The line-profile variability is ... [more ▼]We used extensive ground-based multisite and archival spectroscopy to derive observational constraints for a seismic modelling of the magnetic β Cep star V2052 Ophiuchi. The line-profile variability is dominated by a radial mode (f1 = 7.148 46 d-1) and by rotational modulation (P_rot = 3.638 833 d). Two non-radial low-amplitude modes (f2 = 7.756 03 d-1 and f3 = 6.823 08 d-1) are also detected. The four periodicities that we found are the same as the ones discovered from a companion multisite photometric campaign and known in the literature. Using the photometric constraints on the degrees ℓ of the pulsation modes, we show that both f_2 and f_3 are prograde modes with (ℓ, m) = (4, 2) or (4, 3). These results allowed us to deduce ranges for the mass (M ∈ [8.2, 9.6] M_sun) and central hydrogen abundance (X_c ∈ [0.25, 0.32]) of V2052 Oph, to identify the radial orders n1 = 1, n2 = -3 and n3 = -2, and to derive an equatorial rotation velocity v_eq ∈ [71, 75] km s-1. The model parameters are in full agreement with the effective temperature and surface gravity deduced from spectroscopy. Only models with no or mild core overshooting (α_ov ∈ [0, 0.15] local pressure scale heights) can account for the observed properties. Such a low overshooting is opposite to our previous modelling results for the non-magnetic β Cep star θ Oph having very similar parameters, except for a slower surface rotation rate. We discuss whether this result can be explained by the presence of a magnetic field in V2052 Oph that inhibits mixing in its interior. [less ▲]Detailed reference viewed: 16 (6 ULg) The nature of the high Galactic latitude O-star HD 93521: new results from X-ray and optical spectroscopyRauw, Grégor ; Morel, Thierry ; Palate, Matthieu in Astronomy and Astrophysics (2012), 546Context. Owing to its unusual location and its isolation, the nature of the high Galactic latitude O9.5 Vp object HD 93521 is still uncertain. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8414002060890198, "perplexity": 3533.5570359018425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500832052.6/warc/CC-MAIN-20140820021352-00375-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://talkstats.com/threads/typing-in-formulas.2596/ | # Typing in Formulas
#### Humphrey
##### New Member
Hi all,
I'm typing a paper using Microsoft WORD and need to type in some statistical formulas. However, I don't know how to do it in word. I need radicals, exponents,division lines, etc.
I'd be thankful if you tell me how to write formulas in Word.
Humphrey
#### PeterVincent
##### New Member
Use equation editor which is included in Word, Google for a tutorial.
#### Davellll
##### New Member
Insert-->Object-->Microsoft equation editor
#### Humphrey
##### New Member
thanks i found it.
however, it doesn't let writing exponents and subscripts.
how can i write say, x raised to the power of 2?
or put a five on top of a radical.
when the equation under the radical is long how can i extend the line of the radical to cover the whole length of the formula?
how can i draw long horizental division lines? underlining doesn't work.
thanks
humphrey | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9543157815933228, "perplexity": 2602.389327638036}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488567696.99/warc/CC-MAIN-20210625023840-20210625053840-00607.warc.gz"} |
http://math.stackexchange.com/questions/253075/confused-about-lipschitz-functions | Given a function $f(x,y)$, I want to show whether the function satisfies a uniform Lipschitz condition with respect to $y$, and determine the Lipschitz constant $L$.
Questions:
1.) I know that $f(x,y)$ is Lipschitz with respect to $y$ if $$|f(x,y_1) - f(x,y_2)|\le L|(x,y_1)-(x,y_2)|$$ But what does uniform Lipschitz condition mean?
2.) What is a good strategy to determine whether a function $f(x,y)$ is uniformly Lipschitz with respect to $y$? (Maybe somebody can explain it by using a few examples)....
Thank you!
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"Uniform with respect to y" means only that it is Lipschitz with the same constant L for any x – bonext Dec 7 '12 at 14:13
If $f$ is continuous and $\frac{\partial f}{\partial y}$ exists and is uniformly bounded on a convex region $\Omega$, then $f$ is uniformly Lipschitz on $\Omega$. To see this, apply the mean value theorem at points $(x,y_1), (x,y_2)\in\Omega$, with $y_1<y_2$. The mvt tells us that there exists $y_3\in(y_1,y_2)$ such that $$\frac{\partial f}{\partial y}(x,y_3)=\frac{f(x,y_2)-f(x,y_1)}{y_2-y_1}.$$ If the partial derivative is uniformly bounded on $\Omega$ by $M$, we can then say that $$\frac{|f(x,y_2)-f(x,y_1)|}{|y_2-y_1|}\leq M\quad\hbox{for all}\quad (x,y_1), (x,y_2)\in\Omega.$$
Consider your example $$f(x,y)=x\sin y,$$ and lets look at this on the rectangle $$\Omega=\{(x,y): -a\leq x\leq a, -b\leq y\leq b\}$$ where $a$ and $b$ are positive. We have $$\frac{\partial f}{\partial y} = x\cos y$$ and so $$\left|\frac{\partial f}{\partial y}\right| = |x\cos y|\leq |x| \leq a \quad \hbox{for all}\quad (x,y)\in\Omega.$$ Thus $\frac{\partial f}{\partial y}$ is uniformly bounded by $M=a$ on $\Omega$.
(Definition: The function $g$ is uniformly bounded on the set $A$ if there is a number $K$ such that $|g(p)|\leq K$ for all points $p\in A$.)
A set $\Omega$ is convex if for every pair of points $p,q$ in the set, the line joining $p$ and $q$ is also contained in the set. In $\mathbb{R}^2$, examples include rectangles, parallelograms, discs, ellipses and $\mathbb{R}^2$ itself - but not stars or a disc with a bite taken out of it.
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Tnx. So for example $f(x, y) = x sin(y)$, which is continuous, df/dy exists. But how can I say if it's uniformly bounded on a convex region? What does that mean exactly? – MSKfdaswplwq Dec 7 '12 at 15:05
Good questions - I've edited my answer to try to address them. – user12477 Dec 7 '12 at 15:23
Is the following set convex? $$\text{The rectangle of }[x_0,x_0+a], (-\infty, +\infty)$$ – MSKfdaswplwq Dec 7 '12 at 21:05
@Hempo I wouldn't trust that book! $\frac{y}{(1+x)^2}$ is undefined on the line $x=-1$, and so its domain cannot be $\mathbb{R}^2$. The rectangle you mention is indeed convex: if you think about what it looks like - it's an infinite vertical strip between the vertical lines $x=x_0$ and $x=x_0+a$ - it should be clear that any pair of points in the set are connected by a segment that stays inside the set. It is also fairly straightforward to prove this formally. – user12477 Dec 8 '12 at 12:22
To find \underline{a} domain on which a function $f(x,y)$ satisfies the Lipschitz condition is to choose a number $M>0$ and then find the largest convex domain on which the inequality $|\frac{\partial f}{\partial y}|<M$ is satisfied. If you are interested in the largest domain on which the function is Lipschitz, then ask: what is the largest value that $M$ may have? This just describes one possible and non-exhaustive approach: there isn't a general prescription for doing what you ask. – user12477 Dec 8 '12 at 12:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.930658221244812, "perplexity": 109.10884263476717}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678702919/warc/CC-MAIN-20140313024502-00049-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/two-problems-which-have-been-giving-me-problems.218857/ | # Two problems which have been giving me problems
1. Feb 29, 2008
### end3r7
1. The problem statement, all variables and given/known data
1) Test the following series for Uniform Convergence
$$\sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}}$$
2) Let f(n,x) = $$\sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}$$
a) Test for absolutely convergence on [0,1]
b) Test for uniformly convergence on [0,1]
c) Is $$\sum\limits_{n = 1}^{\inf } {|f(n,x)|}$$ absolutely convergent on [0,1]?
2. Relevant equations
3. The attempt at a solution
For the first, I'm utterly lost. Is there an easy way to deal with such series?
For the second, could I just argue that for all 0<=x<1, there exists a, s.t. x < a <1
and thus
$$|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}$$
and for x = 1 and any a > 0
$$|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}$$
This would prove all 3 right? But can I argue taht way? Can I fix my 'x' ahead of time, or does my argument have to work for all x simultaneously? Cuz if it does, then all I would have to do is choose x between a and 1 and the argument would break down.
2. Feb 29, 2008
### end3r7
I think I figured out two (will post my work later).
But for 1, could I argue that the sequences of partial sums cannot be uniformly cauchy since they are unbounded near x = 1.
3. Feb 29, 2008
### end3r7
I confess I might have jumped the gun in saying that I figure out number two. Any help on either would still be greatly appreciated (also, if I could get a mod to give this thread a more decriptive title.... I don't think I can change the title).
4. Feb 29, 2008
### e(ho0n3
For the first one, how about splitting the sum into two: one for n even, one for n odd. Then you can use the p-series test on each.
The argument you use when 0 <= x < 1 works. When x = 1, what's inside the sum is 0 so the whole sum is 0. I find it unnecessary to consider the absolute value of the sum.
5. Mar 1, 2008
### end3r7
Call me stupid (I'd rather you don't though =P), but I'm not sure I understand your approach for the first.
What I did was show that I can get x arbitrarily close to 1, and since log(x) is continuous, I can get arbitrarily close 1. Basically, I showed that for any 'n', I can make $$log(x) < \frac{1}{ne}$$
so for x sufficiently close to 1
$$|\frac{1}{{n^{x}\ln (x)}}| > \frac{1}{|{n||\ln (x)|}} > |\frac{ne}{n}| = e$$
where first inequality holds because x is between 0 and 1.
Therefore the terms of the series do not go to zero uniformly on its domain, so it can't converge.
Is that a valid argument?
6. Mar 1, 2008
### e(ho0n3
What I wrote is for testing convergence, not uniform convergence. Sorry about that. I'm not familiar with uniform convergence.
7. Mar 1, 2008
### end3r7
No problemo. =)
(Btw, uniform convergence is essentially the same, but it has to work for arbitrary x in the interval).
8. Mar 1, 2008
### e(ho0n3
If x is a fixed constant greater than 0, by the alternating series test, the sum in 1) converges. Does this mean that the sum converges uniformly in the interval $(0, \infty)$?
9. Mar 1, 2008
### end3r7
Oops, I forgot to say that x belongs to [0,1].
Note that log(1) = 0, so it can't converge.
To converge uniformly, it has to converge for every x in the interval.
10. Mar 1, 2008
### end3r7
Please, if a mod could, edit the first one so x belongs to the closed interval [0,1].
Thanks =)
Similar Discussions: Two problems which have been giving me problems | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9227167367935181, "perplexity": 694.2946845637179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105922.73/warc/CC-MAIN-20170819201404-20170819221404-00206.warc.gz"} |
http://wpd.ugr.es/~iemath/events/event/geometry-seminar-8/ | # Geometry Seminar
#### Event Details
Date: April 9, 2021
Time: 12h-13h
Author:Marilena Moruz (Al.I. Cuza University of Iasi)
Title: Ruled real hypersurfaces in $$\mathbb CP^n_p$$
Summary: H. Anciaux and K. Panagiotidou [1] initiated the study of non-degenerate real hypersurfaces in non-flat indefinite complex space forms in 2015. Next, in 2019 M. Kimura and M. Ortega [2] further developed their ideas, with a focus on Hopf real hypersurfaces in the indefinite complex projective space $$\mathbb CP^n_p$$. In this work we are interested in the study of non-degenerate ruled real hypersurfaces in $$\mathbb CP^n_p$$. We first define such hypersurfaces, then give basic characterizations. We also construct their parameterization. They are described as follows. Given a regular curve $$\alpha$$ in $$\mathbb CP^n_p$$, then the family of the complete, connected, complex $$(n - 1)$$-dimensional totally geodesic submanifolds orthogonal to $$\alpha'$$ and $$J\alpha'$$, where $$J$$ is the complex structure, generates a ruled real hypersurface. This representation agrees with the one given by M. Lohnherr and H. Reckziegel in the Riemannian case [3]. Further insights are given into the cases when the ruled real hypersurfaces are minimal or have constant sectional curvatures. The present results are part of a joint work together with prof. M. Ortega and prof. J.D. Pérez.
[1] H. Anciaux, K. Panagiotidou, Hopf Hypersurfaces in pseudo-Riemannian complex and para-complex space forms, Diff. Geom. Appl. 42 (2015) 1-14.
[2] M. Kimura, M. Ortega, Hopf Real Hypersurfaces in Indefinite Complex Projective, Mediterr. J. Math. (2019) 16:27.
[3] M. Lohnherr, H. Reckziegel, On ruled real hypersurfaces in complex space forms. Geom. Dedicata 74 (1999), no. 3, 267–286.
Where: https://oficinavirtual.ugr.es/redes/SOR/SALVEUGR/accesosala.jsp?IDSALA=22960561 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8659082055091858, "perplexity": 1334.0233051944454}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038464045.54/warc/CC-MAIN-20210417192821-20210417222821-00399.warc.gz"} |
https://www.risk.net/journal-of-computational-finance/2275487/tracking-value-at-risk-through-derivative-prices | # Tracking value-at-risk through derivative prices
## Simon I. Hill
#### Abstract
ABSTRACT
Using observations of an underlying instrument's price series and of derivative prices, we consider the filtering problem of jointly tracking real-world measure parameters and stochastic discount factor parameters. A state-space model of the evolution of the price processes is used, and the filtering is performed through sequential Monte Carlo. Variance gamma and normal inverse Gaussian models of the price process are used as examples. The filter output is used to find diagnostic values such as value-at-risk and expected price change. Both models track these realistically; implementations are presented illustrating the gain in information obtained over standard methods. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9127696752548218, "perplexity": 1432.8862297972357}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806760.43/warc/CC-MAIN-20171123070158-20171123090158-00306.warc.gz"} |
https://asmedigitalcollection.asme.org/ICEM/proceedings-abstract/ICEM2007/43390/885/329724 | Waste tributyl phosphate (TBP) and normal dodecane generated from R&D activities on recycle of nuclear fuel has been stored in Japan Atomic Energy Agency (JAEA). If it is incinerated, a large quantity of contaminated phosphorous compounds will be generated as radioactive secondary wastes. The objective of this study is to reduce the generation of the radioactive secondary wastes by the treatment of the waste TBP/dodecane using steam reforming system. We constructed the demonstration scale steam reforming system which consists of a gasification chamber for vaporization of wastes, a metal mesh filter for removal of radioactive nuclides from gasified wastes, a combustion chamber, and scrubbers for removal of phosphorous oxides. We conducted process demonstration tests using waste TBP/dodecane with 0.07 g/L of uranium. We studied the temperature dependence of the gasification ratio of inorganic phosphorus compounds formed by pyrolysis of TBP in the gasification chamber and removal of uranium by the filter. As the results, more than 90% of phosphorus compounds were gasified from the gasification chamber at temperature of 600°C or more, and the uranium concentration in the waste water generated from the off-gas treatment system is under the detection limits. The waste water containing the separated phosphorus compounds can be discharged into the river or the sea as the liquid wastes in which uranium concentration is under the regulatory level. These results show the steam reforming system is effective in the reduction of radioactive secondary waste in the treatment of TBP/dodecane.
This content is only available via PDF. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8192059993743896, "perplexity": 3732.99718534814}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00406.warc.gz"} |
http://math.stackexchange.com/users/36205/jisang-yoo | # Jisang Yoo
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bio website yoo2080.wordpress.com location age member for 1 year, 9 months seen Apr 11 at 14:19 profile views 18
2 Mutually singular measures with the same support 1 Simpler proof - Non atomic measures 1 Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? 0 Construction of a Borel set with positive but not full measure in each interval
# 141 Reputation
+10 Simpler proof - Non atomic measures +10 Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions? +20 Mutually singular measures with the same support
# 0 Questions
This user has not asked any questions
# 5 Tags
4 measure-theory × 4 1 examples-counterexamples 2 singular-measures 1 proof-verification 1 real-analysis × 2
# 9 Accounts
Stack Overflow 1,137 rep 29 Super User 237 rep 127 Mathematics 141 rep 4 TeX - LaTeX 133 rep 4 English Language & Usage 128 rep 3 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8432998657226562, "perplexity": 2188.505815903174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00490-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://q4interview.com/imp_formulas.php?t=42 | Arithmetic Aptitude :: Simplification
Home > Arithmetic Aptitude > Simplification > Important Formulas
Simplification Important Formulas
1. 'BODMAS' Rule:
This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of given expression.
B - Bracket,
O - of,
D - Division,
M - Multiplication,
S - Subtraction
Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order ( ), { } and [ ].
After removing the brackets, we must use the following operations strictly in the order:
(i) of
(ii) Division
(iii) Multiplication
(v) Subtraction.
2. Modulus of a Real Number:
Modulus of a real number a is defined as
$$\left| x \right| = \begin{cases} \ \ \ a, \ if \ a > 0 \\ -a, \ if \ a < 0 \\ \end{cases}$$
Ex. |5| = 5 and |-5| = -(-5) = 5.
3. Virnaculum (or Bar):
When an expression contains Virnaculum, before applying the 'BODMAS' rule, we simplify the expression under the Virnaculum. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9595155715942383, "perplexity": 4977.703183336274}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250589861.0/warc/CC-MAIN-20200117152059-20200117180059-00270.warc.gz"} |
https://www.computer.org/csdl/trans/td/2001/03/l0272-abs.html | The Community for Technology Leaders
ABSTRACT
<p><b>Abstract</b>—Torus/mesh-based machines have received increasing attention. It is natural to identify the maximum healthy submeshes in a faulty torus/mesh so as to lower potential performance degradation, because the time for executing a parallel algorithm tends to depend on the size of the assigned submesh. This paper proposes an efficient approach for identifying all the maximum healthy submeshes present in a faulty torus/mesh. The proposed approach is based on manipulating set expressions, with the search space reduced considerably by taking advantage of the interesting properties of a faulty torus/mesh. This procedure is a distributed one, because every healthy node performs the same procedure independently and concurrently. We show that the proposed scheme may outperform previous methods.</p>
INDEX TERMS
Distributed algorithms, faulty meshes, faulty tori, prime submeshes, reconfiguration, set expressions.
CITATION
Hsing-Lung Chen, Shu-Hua Hu, "Submesh Determination in Faulty Tori and Meshes", IEEE Transactions on Parallel & Distributed Systems, vol. 12, no. , pp. 272-282, March 2001, doi:10.1109/71.914767 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8965178728103638, "perplexity": 1939.6061200934812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560282110.46/warc/CC-MAIN-20170116095122-00141-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://gyros.biz/index.asp?p=85 | Previous Page Index(For best results view this page as a PDF) Next Page
172 NAVIGATIONAL COMPASSES
face inclined to the horizontal at the proper angle and the clock with the attached disk turned with respect to the lubber line through an angle equal to that of the desired course from the meridian. Then, the proper course is maintained by steering
the airplane or ship so that the shadow of the shadowpin is maintained along the axis of the clock hand.
When the course is not exactly north or south, the apparent time changes. In this case either the clock may be reset after each change of a few degrees longitude, or it may be set permanently to the apparent time of the middle meridian.
105. The Apparent Motion of the Spin-Axle of an Un
constrained Gyroscope Due to the Rotation of the Earth. - According to the First Law of Gyrodynamics (Art. 36), the spin-axle of an unconstrained spinning gyro, unacted upon by any torque, remains fixed in space. This law is also called the law of rigidity
of plane of the gyro, or the law
of the fixity in space of the spin-axle. If the gyro-axle of a spinning gyro at the equator is parallel to the geographic axis of the earth, then the axle will remain horizontal and in the meridian plane. If the gyroaxle of a spinning gyro at the equator is horizontal in the east-west position, as at 0, Fig. 134, then, although the gyroaxle preserves its direction in
space as the earth rotates, it appears to an observer on the earth to make one complete turn each twenty-four hours about the horizontal axis in the meridian plane.
If an unconstrained spinning gyro be situated between the equator and either pole, then the spin-axle will appear to an ob
THE VARIOUS TYPES 173
server on the earth to move about both a horizontal and a vertical axis, Fig. 135. At two times during twenty-four hours, the gyroaxle is horizontal and at two times it is in the meridian plane. The motion of the spin-axle relative to the earth is about a cone having the center of the gyro as apex.
The spin-axle of an unconstrained gyroscope continues to be directed toward the same fixed star but it does not continue to be directed toward the same fixed point on the earth. The fixity of the spin-axle in space of an unconstrained gyroscope is inadequate for the production of an instrument that will indicate directions on the earth.
106. The Meridian-Seeking Tendency of a Pendulous Gyroscope. - It has been shown that if a gyroscope be rotated about an axis about which the turning of the gyro-axle is prevented (Art. 49), the axle will set itself parallel to the axis of rotation with the spin of the gyro-wheel in the same direction as the rotation of the gyroscope. It might be imagined that this effect would be adequate for causing the axle of a spinning gyro-wheel, with one degree of angular freedom suppressed, to set itself parallel to the earth's axis. For example, if a spinning gyro-wheel on the earth is mounted so that angular motion about every horizontal axis is suppressed, the spin-axle will tend to set itself parallel to the earth's axis and will turn toward the meridian plane. If, however, the instrument is on a moving ship, the spin-axle will tend to set itself parallel to the axis of the resultant of the angular velocity of the earth and that of the ship with respect to the earth. Since the angular velocity of the ship is often much greater than that of the earth, a gyro-wheel mounted in this manner would not give correct indications. To be of value for use as a compass, the gyroaxle (a) should be urged toward the meridian by a torque sufficient to bring it into the meridian within a reasonable length of time after being set into spinning motion, (b) should quickly return to the meridian when displaced therefrom, (c) should be nearly horizontal when in the meridian.
If, at some instant, the spin-axle of a gyroscope north of the equator is nearly horizontal and in the meridian plane, then at succeeding instants the north-seeking end will have an apparent motion away from the meridian toward the east. In order that the gyro-axle may maintain its position in the meridian, the northseeking end must be given a westerly precessional velocity equal to the vertical component of the earth's angular velocity. This
Previous Page Index(For best results view this page as a PDF) Next Page | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8079842329025269, "perplexity": 675.4205024626202}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145742.20/warc/CC-MAIN-20200223001555-20200223031555-00110.warc.gz"} |
https://gateoverflow.in/335209/ugc-net-cse-january-2017-part-3-question-29 | 588 views
In RSA public key cryptosystem suppose $n=p*q$ where $p$ and $q$ are primes. $(e,n)$ and $(d,n)$ are public and private keys respectively. Let $M$ be an integer such that $o< M< n$ and $\phi(n)=(p-1)(q-1)$.
Which of the following equations represent RSA public key cryptosystem?
1. $\begin{array}{} C \equiv M^{e} \text{(mod n)} \\ M \equiv C^{d} \text{(mod n)} \end{array} \\$
2. $ed\equiv1 \text{(mod n)} \\$
3. $ed\equiv1 ( \text{mod } \phi( n)) \\$
4. $\begin{array}{} C\equiv M^{e} ( \text{mod } \phi(n)) \\ M \equiv C^{d}( \text{mod } \phi (n)) \end{array}$
1. I and II
2. I and III
3. II and III
4. I and IV
Comparing it with RSA algorithm , I and III is correct ahere...see the example
• Choose p = 3 and q = 11
• Compute n = p * q = 3 * 11 = 33
• Compute φ(n) = (p - 1) * (q - 1) = 2 * 10 = 20
• Choose e such that 1 < e < φ(n) and e and n are coprime. Let e = 7
• Compute a value for d such that (d * e) % φ(n) = 1. One solution is d = 3 [(3 * 7) % 20 = 1]
• Public key is (e, n) => (7, 33)
• Private key is (d, n) => (3, 33)
• The encryption of m = 2 is c = 27 % 33 = 29
• The decryption of c = 29 is m = 293 % 33 = 2
so B is correct answer here
by
Comparing it with RSA algorithm , I and III is correct ahere...see the example
• Choose p = 3 and q = 11
• Compute n = p * q = 3 * 11 = 33
• Compute φ(n) = (p - 1) * (q - 1) = 2 * 10 = 20
• Choose e such that 1 < e < φ(n) and e and n are coprime. Let e = 7
• Compute a value for d such that (d * e) % φ(n) = 1. One solution is d = 3 [(3 * 7) % 20 = 1]
• Public key is (e, n) => (7, 33)
• Private key is (d, n) => (3, 33)
• The encryption of m = 2 is c = 27 % 33 = 29
• The decryption of c = 29 is m = 293 % 33 = 2
so B is correct answer here
In RSA algorithm
1) Cipher text is obtained as $Cipher Text = (Plain Text)^{public\ key} \ mod n$
2) Plain Text is obtained as $Plain Text = (Cipher Text)^{private\ key} \ mod n$
3) e should be chosen in such a way that it satisfy $e*d = 1(mod)\phi (n)$
Hence option (2) is correct
1 vote | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8722947835922241, "perplexity": 968.0507519873212}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00796.warc.gz"} |
http://gradestack.com/MCAT-Complete-Tutor/Body-Momentum/Conservation-of-Momentum/18207-3322-28418-study-wtw | # Conservation of Momentum
Conservation is one of those words in physics which has a special meaning. When a physicist says that momentum is conserved, she means that momentum ("movage") has a kind of permanence, that it cannot be created from nothing nor destroyed. The formal statement is in the box:
If a system of objects is isolated (external forces are balanced), then the total momentum of the system stays constant over time, that is, the total momentum is constant. In particular,
………….(3)
An external force is a force on one of the objects in the system by an outside agent, while an internal force is a force between two objects in the system.
Example-1
A Mack truck (9000 kg) going north at 10 m/s encounters a Porsche (1000 kg) going south at 20 m/s. What is the velocity (speed and direction) of the resulting fused mass of metal?
Solution
We DRAW A DIAGRAM (Figure 8-3). Let's discuss the collision itself, shown in the middle part of Figure 8-3. There are two normal forces and two gravitational forces which are all external. These, however, are balanced. The force of the truck on the car and that of the car on the truck are internal forces.
Figure 8-3
Momentum is conserved, and we write (in one dimension)
Pbefore = Pafter
mMvM + mpvp = (mM + mp)vf
(9000 kg)(10 m/s) + (1000 kg)(–20 m/s) = (10,000 kg)vf
vf = 7 m/s
Notice we used vp = –20 m/s, since it was going south, and we chose north to be positive. We have to pay attention to signs because momentum is a vector quantity.
Example-2
During the collision in Section A, the external forces are the tensions in the threads and gravity, and these are balanced. The internal forces are all the complicated forces among the balls. So momentum is conserved. We do not yet know enough to show why exactly one ball jumps off the right end. Momentum would be conserved, for example, if all five balls headed to the right at one fifth the impact velocity of the left ball.
Example-3
Is momentum conserved while the left ball is swinging from its initial height on its way to collision? (See Figure 8-4.)
Figure 8-4
Solution
It certainly does not seem so, since the ball starts with zero momentum and achieves a maximum momentum just before impact. Figure 8-5 shows why momentum is not conserved. The external forces on the ball are unbalanced.
Figure 8-5
Example-4
Is momentum conserved for a crocodile dropped from a ladder? If not, what is the external force? Try doing this one yourself. (See Figure 8-6.)
Figure 8-6
Here is a major hint:
Whenever a problem involves a collision, especially one with crunching, crashing, or sticking, you will probably need to use conservation of momentum.
Example-5
A car (1000 kg) going north (10 m/s) collides with a truck (1500 kg) going east (20 m/s). Assume there is negligible friction at the time of the collision. What is the final speed and direction of the combined car/truck?
Solution
We DRAW A DIAGRAM (Figure 8-7).
Figure 8-7
The total momentum before the collision can be read from Figure 8-8.
From the Pythagorean theorem we find the magnitude of the total momentum ptot = 3.2 x 104 kg m/s. The magnitude of the velocity is v = (3.2 x 104 kg m/s)/(2500 kg) = 13 m/s. The direction we obtain from tanφ = (1 x 104 kg m/s)/(3 x 104 kg m/s). Thus φ = 18˚ north of east.
Figure 8-8 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9250341057777405, "perplexity": 778.8590209861268}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00581-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://archive.org/stream/TreatiseOnAnalysisVolII/TXT/00000158.txt | Full text of "Treatise On Analysis Vol-Ii"
See other formats
```9 MEASURABLE FUNCTIONS 139
A function/: X -> K is a step function if it takes only & finite number of
n
values ak (1 ^ k g n). In that case we have/= ]£ ak cpAk, where Ak =f~1(ak)9
with the convention that if ak is ± oo, the value of ak(pAk is aA in Ak and 0 in
X — Ak (13.11). It follows immediately from this and from (13.9.9) that/is
measurable if and only if each of the Ak is measurable.
(13.9.12) Le£/:X-»R be measurable. Then there exists a sequence (gn)
of universally measurable step functions with compact support, such that
\gn(x)\ <£ \f(x)\ for all x e X and such that the sequence (gn(x)} converges al-
most everywhere to f(x) (which implies that f is equivalent to a universally
measurable function).
Consider a partition of X consisting of a sequence (Kn) of compact sets
and a negligible set N. For each integer i £ n, there exists a finite covering
(Uij\£j£qtn of K£ by sets which are open in Ki9 such that the oscillation of
/in each U\$ is g l/n (with respect to a distance on K, cf. (3.16.5)). Now we
have the following lemma:
(13.9.12.1) Let (Fh\^h^p be a finite family of integrable sets. Then there
exists a finite family (Gk)igkgr of pairwise disjoint integrable sets, such that
each Fh is the union of some of the Gk.
p
Consider the 2P - 1 sets of the form (°| Zh, where each Zh is either F,, or
X — Fh, and Zh = F,, for at least one index h. Then the family of distinct non-
empty sets of this form satisfies the conditions of the lemma, because these sets
p
are integrable by (13.7.6) and each F/, is the union of the sets p) Zz for which
By applying this lemma to each of the finite families (^Jij\^j^qin9 we
obtain a. partition (A^X^^,.^ of Kf into universally measurable, integrable
sets. Let | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9799660444259644, "perplexity": 2581.6286394387917}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463609061.61/warc/CC-MAIN-20170527210417-20170527230417-00132.warc.gz"} |
https://indico.cern.ch/event/773241/contributions/3535505/ | # The 6th Conference of the Polish Society on Relativity
Sep 23 – 26, 2019
Other Institutes
Europe/Warsaw timezone
## Quantum gravity on a torus - an update
Sep 23, 2019, 5:30 PM
20m
Other Institutes
#### Other Institutes
Maritime Academy, Wały Chrobrego 1-2, Szczecin, Poland
Talk/Seminar Parallel Sessions
### Speaker
Jakub Gizbert-Studnicki (Jagiellonian University)
### Description
During last POTOR's conference I discussed a quantum gravity model defined by Causal Dynamical Triangulations (CDT) where spatial slices of equal proper time have fixed topology of a 3-torus. Identification of phase structure and order of the phase transitions constitute first steps in the quest for a continuum limit of CDT where, following the asymptotic safety conjecture, the resulting theory of quantum gravity becomes nonperturbatively renormalizable. Initial study of the toroidal model, presented last year, showed that one can identify the same phase structure as earlier observed for the spherical spatial topology, including phase 'C', where quantum fluctuations of spatial volume are well described by mini-superspace like models. Now I will also discuss the recent study of the order of phase transitions in the toroidal case, which is crucial in defining the continuum limit of CDT. If time permits I will comment on the possibility of reintroducing spatial coordinates in the CDT model and on the impact of scalar fields with non-trivial boundary conditions on the toroidal CDT results.
### Primary author
Jakub Gizbert-Studnicki (Jagiellonian University)
### Presentation materials
There are no materials yet. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8846254944801331, "perplexity": 2467.476181356187}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585916.29/warc/CC-MAIN-20211024081003-20211024111003-00081.warc.gz"} |
http://lt-jds.jinr.ru/record/61717?ln=en | / nucl-ex CERN-PH-EP-2013-135
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
Published in: Phys.Lett.
Year: 2014
Vol.: B728
Page No: 25-38
Pages: 14
Year: 2014
Abstract: In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured by the ALICE detector at the LHC, are reported. The transverse momentum distributions exhibit a hardening as a function of event multiplicity, which is stronger for heavier particles. This behavior is similar to what has been observed in pp and Pb--Pb collisions at the LHC. The measured $p_T$ distributions are compared to results at lower energy and with predictions based on QCD-inspired and hydrodynamic models.
Note: 23 pages, 7 captioned figures, 5 tables, author from page 19 to page 23, new version
Total numbers of views: 2371
Numbers of unique views: 823
DOI: 10.1016/j.physletb.2013.11.020 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.983792245388031, "perplexity": 3519.2497230842555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578626296.62/warc/CC-MAIN-20190424034609-20190424060609-00211.warc.gz"} |
http://math.stackexchange.com/questions/251354/find-the-direct-expression-of-ft | Find the direct expression of $f(t)$
I am trying to find the direct expression of the function $f(t)$ given by $$f(t)= \int_{1}^\infty \frac{\arctan (tx)}{x^2\sqrt{x^2-1}}dx$$ It's hard for me to calculate the integration directly.Should I try the method of interchanging $\frac{d}{dx}$ with$\int$ or$\int$ with $\int$ ? Thanks for help.
-
What does it mean to exchange $\int$ with $\int$? – Andrew Uzzell Dec 5 '12 at 9:29
Here is an answer obtained by maple
$$-\frac{\pi}{2} \, \left( -1-t+\sqrt {1+{t}^{2}} \right) .$$
-
Exams become easier if the maple was allowed.Thank you all the same. – C Weid Dec 5 '12 at 9:19
@CWeid: You are welcome. – Mhenni Benghorbal Dec 5 '12 at 10:01
@CWeid: By the way, where did this problem come from? – Mhenni Benghorbal Dec 5 '12 at 10:03
Entrance exam in Chern Institute of Mathematics,2009 – C Weid Dec 5 '12 at 14:24
Presumably this was intended to be done with pen and paper, so here is what they might have had in mind. First note that convergence is excellent, so we may differentiate with respect to $t$ to obtain $$f'(t) = \int_1^\infty \frac{x}{1+t^2x^2} \frac{1}{x^2\sqrt{x^2-1}} dx.$$ Now let $x^2-1 = u^2$ so that $x\, dx = u\, du$ and $$f'(t) = \int_0^\infty \frac{u}{1+t^2(u^2+1)} \frac{1}{(u^2+1)\sqrt{u^2}} du = \int_0^\infty \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1} du.$$ Putting $$g(u) = \frac{1}{2} \frac{1}{1+t^2(u^2+1)} \frac{1}{u^2+1} \quad \text{we seek to evaluate} \quad \int_{-\infty}^\infty g(u) du.$$ This may be done with the Cauchy Residue theorem applied to a half-circle contour in the upper half plane of radius $R$ (the contribution of the circle vanishes in the limit), giving $$2 \pi i \left(\operatorname{Res}_{u=i} g(u) + \operatorname{Res}_{u=\sqrt{-\frac{1}{t^2}-1}} g(u) \right).$$ These poles are both simple and we obtain $$2\pi i \left(- \frac{1}{4} i - \frac{1}{4} \frac{1}{\sqrt{-\frac{1}{t^2}-1}}\right) = \frac{\pi}{2} - i \frac{\pi}{2}\frac{1}{\sqrt{-\frac{1}{t^2}-1}} = \frac{\pi}{2} - \frac{\pi}{2}\frac{t}{\sqrt{t^2+1}} \quad( t\geqslant 0)$$ Integrating we have $$f(t) = C + \frac{\pi}{2} t - \frac{\pi}{2} \sqrt{t^2+1}.$$ Finally, to determine $C$, note that $f(0) = 0$ or $$0 = C - \frac{\pi}{2}$$ and hence $C = \frac{\pi}{2}$ giving the result $$f(t) = \frac{\pi}{2} ( 1 + t - \sqrt{t^2+1} ).$$ When $t<0,$same calculation shows $$f(t)= \frac{\pi}{2}(-1 + t +\sqrt{t^2+1})$$
-
To differentiate the function under the integral,we need to check the convergence of $f'(t)$,whether it's uniform,rather than $f(x)$ itself.Thank you for this excellent answer,I am trying to find a simpler one. – C Weid Dec 6 '12 at 6:45
I've found another method.The idea is to find the differential equation which $f(x)$ and $f'(x)$ satisfy.It's a first-order linear ODE,solve the Cauchy problem then we get $f(x)$.Still it costs quite some pen and paper. – C Weid Dec 6 '12 at 10:42
There is a little mistake in your solution.When $t>0$ the answer is correct,the $t<0$ case need to be more careful about the residue at $u=\sqrt{-\frac{1}{t^2}-1}$. – C Weid Dec 6 '12 at 14:04
I've corrected it. – C Weid Dec 6 '12 at 14:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9398435354232788, "perplexity": 503.2319799119676}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701163512.72/warc/CC-MAIN-20160205193923-00296-ip-10-236-182-209.ec2.internal.warc.gz"} |
http://en.wikibooks.org/wiki/High_School_Trigonometry/Circular_Functions_of_Real_Numbers | # High School Trigonometry/Circular Functions of Real Numbers
In this lesson you will view the trigonometric ratios of angles of rotation around the coordinate grid as a continuous, circular function. The connection will be made between how the ratios change as the angle of rotation increases or decreases, and how the graph of the function depicts that change.
## Learning Objectives
• Identify the 6 basic trigonometric ratios as continuous functions of the angle of rotation around the origin.
• Identify the domain and range of the six basic trigonometric functions.
• Identify the radian and degree measure, as well as the coordinates of points on the unit circle for the quadrant angles, and those with reference angles of 30°, 45°, and 60°.
## y = sin(x), The Sine Graph
By now, you have become very familiar with the specific values of sine, cosine, and tangents for certain angles of rotation around the coordinate grid. In mathematics, we can often learn a lot by looking at how one quantity changes as we consistently vary another. In this case, what will happen to the value of, let's say, the sine of the angle as we gradually rotate around the coordinate grid. We would be looking at the sine value as a function of the angle of rotation around the coordinate grid. We refer to any such function as a circular function, because they can be defined using the unit circle. First of all, you may recall from earlier sections that the sine of an angle in standard position in the coordinate grid is the ratio of $\tfrac{y}{r}$, where y is the y-coordinate of any point on the angle and r is the distance from the origin to that point.
Because the ratios are the same for a given angle, regardless of the length of the radius r, we can use the unit circle to make things a little more convenient.
The denominator is now 1, so we have the simpler expression sin(θ) = y. The advantage to this is that we can use the y-coordinate of the point on the unit circle to trace the value of sin(θ) through a complete rotation. Imagine if we start at 0 and then rotate counter-clockwise through gradually increasing angles. Since the y-coordinate is the sine value, watch the height of the point as you rotate.
Through Quadrant I that height gets larger, starting at 0, increasing quickly at first, then slower until the angle reaches 90°, at which point, the height is at its maximum value, 1.
As you rotate into the third quadrant, the change in the height now reverses itself and starts to decrease towards 0.
When you start to rotate into the third and fourth quadrants, the length of the segment increases, but this time in a negative direction, growing to −1 at 270° and heading back toward 0 at 360°.
After one complete rotation, even though the angle continues to increase, the sine values will simply repeat themselves. The same would have been true if we chose to rotate clockwise to investigate negative angles, and this explains why the sine function is periodic. The period is 2π radians or 360°, because that is the angle measure required before the sine of the angle will simply repeat the previous sequence of values.
Let's translate this circular motion into a graph of the sine value vs. the angle of rotation. The following sequence of pictures demonstrates the connection. As the angle of rotation increases, watch the y-coordinate of the point on the angle as it traces horizontally. Ignore the values along the horizontal axis at this point as they just relative. What is important is that you make the connection between the circular rotation and the change in the height of the point.
Notice that once we rotate around once, the point traces back over the same values again. The red curve that you see is one period of a sine "wave". The below animation shows this happen in "real time".
Let's look at some specific values so we can graph the sine function more precisely. Since we already know what happens in between, you can draw a fairly accurate sketch by plotting the points for the quadrant angles (0, $\tfrac{\pi}{2}$, π, $\tfrac{3\pi}{2}$, 2π).
The value of sin(θ) goes from 0 to 1 to 0 to −1 and back to 0. Graphed along a horizontal axis showing θ, it would look like this:
Filling in the gaps in between and allowing for multiple rotations as well as negative angles results in the graph of y = sin(x) where x is any angle of rotation (usually expressed in radians):
As we have already mentioned, sin(x) has a period of 2π. You should also note that the y-values never go above 1 or below −1, so the range of a sine wave is {−1 ≤ y ≤ 1}. Because we can continue to spin around the circle forever, there is no restriction on the angle x, so the domain of sin(x) is all reals.
## y = cos(x), The Cosine Graph
In chapter 1, you learned that sine and cosine are very closely related. The cosine of an angle is the same as the sine of the complementary angle. So, it should not surprise you that sine and cosine waves are very similar in that they are both periodic with a period of 2π, a range from −1 to 1, and a domain of all real angles.
The cosine of an angle is the ratio of $\tfrac{x}{r}$, so in the unit circle, the cosine is the x-coordinate of the point of rotation. If we trace the x−coordinate through a rotation, you will notice the change in the distance is similar to sin(x), but cos(x) starts in a different place. The x−coordinate of a 0° angle is 1 and the x-coordinate for 90° is 0, so the cosine value is decreasing from 1 to 0 through the 1st quadrant.
Here is a similar sequence of rotations to the one we used for sine. This time compare the x coordinate of the point of rotation with the height of the point as it traces along the horizontal.
The graph of y = cos(x) has a period of 2π. Just like sin(x), the x-values never "escape" from the unit circle, so they stay between −1 and 1. The range of a cosine wave is also {−1 ≤ y ≤ 1}. And also just like the sine function, there is no restriction on the angle of rotation, so the domain of cos(x) is all reals.
## y = tan(x), The Tangent Graph
The graph of the tangent ratio as a function of the angle of rotation presents a few complications. First of all, the domain is no longer all real angles. As you may remember there are some angles (90° and 270°, for example) for which the tangent is not defined. As we will see in this section, the range of tan(x) is actually all real numbers.
The measurement of each of the six trig functions can be found by using a single segment from the unit circle, however, the remaining functions are not as obvious as sine and cosine. The name of the tangent function comes from the tangent line, which is a line that is perpendicular to the radius of a circle at a point on the circle so that the line touches the circle at exactly one, and only one, point. So, to create the tangent segment, first we draw a tangent line perpendicular to the x axis.
If we extend angle θ through the unit circle so that it intersects with the tangent line, the tangent function is defined as the length of the red segment.
The dashed segment is 1 because it is the radius of the unit circle. Recall that the tangent of θ is $\tfrac{y}{x}$, so we can verify that this segment is indeed the tangent by using similar triangles.
$\tan(\theta) = \frac{y}{x} = \frac{t}{1} = t$
$\tan(\theta) = t\,\!$
So, as we increase the angle of rotation, think about how this segment changes. When the angle is 0, the segment has no length. As we begin to rotate through the first quadrant, it will increase, very slowly at first.
But, you can see very soon that the value increases past one. As the angle gets closer to 90°, the segment will need to stretch quite high in order to intersect the extension of the angle and it will grow at a faster and faster rate.
As we get very close to the yaxis that the segment gets infinitely large, until when the angle really hits 90°, at which point the extension of the angle and the tangent line will actually be parallel and therefore never meet!
This means there is no definition for the length of the tangent segment, or as it may be helpful to think of it, the tangent segment is infinitely large.
Before continuing, let's take a look at this portion of the graph through the first quadrant. The tangent starts at 0, for a 0° angle, then increases slowly at first. That increase gets much steeper and as we approach a 90° rotation.
Again, just a small break in the x axis on these graphs will make it more clear that these two concepts to note lie side-by-side on the same coordinate grid.
In fact as we get infinitely close to 90°, the tangent value increases without bound, until when we actually reach 90°, at which point the tangent is undefined. A line that a graph gets infinitely close to without touching is called an asymptote. So the tangent function has an asymptote at 90°.
As we rotate past 90°, now the intersection of the extension of the angle and the tangent line is actually below the x axis. This fits nicely with what we know about the tangent for a 2nd quadrant angle being negative. It will be first be very, very negative, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the positive numbers as the angle enters the 3rd quadrant.
The segment will again get infinitely large as it approaches 270°. After being undefined at 270°, the angle crosses into the 4th quadrant and once again changes from being infinitely negative, to approaching zero as we complete a full rotation.
So, this motion graphed over several rotations would look like this:
Notice that the x axis is measured in radians (not in terms of π). Our asymptotes occur every π radians, starting at $\tfrac{\pi}{2}$. The period of the graph is therefore π radians. The domain is all reals except for the "holes" at $\tfrac{\pi}{2}$, $\tfrac{3\pi}{2}$, −$\tfrac{\pi}{2}$, etc. and the range is all real numbers.
## The Three Reciprocal Functions: cot(x), csc(x), and sec(x)
### Cotangent
Cotangent is the reciprocal of tangent, so it makes sense to generate the circular function for cotangent by drawing the tangent line at a point on the y axis and extending the angle, instead of the x axis.
We can verify that this is the case again by using similar triangles. Because the purported cotangent segment is parallel to the base of the yellow triangle, then angle θ is in the opposite corner and the triangles are indeed similar, even though their positions are reversed.
So,
$\cot(\theta) = \frac{x}{y} = \frac{t}{1} = t$
$\cot(\theta) = t\,\!$
Now that we have established the cotangent segment, think about how this segment changes as we rotate around the coordinate grid starting at 0°. First of all, at 0° itself, the cotangent is undefined because the segment is parallel to the ray of the angle θ. As we begin to increase the angle of rotation, the segment will be extremely large and begin to get smaller as we approach 90°, very quickly at first, but then slowing down as it gets closer to 0 length at 90°.
After passing 90°, the segment will again start to lengthen, but this time it will be in the negative direction, increasing slowly at first, then getting infinitely large in the negative direction until 180°, at which point it is again undefined.
After passing this point, the periodic behavior kicks in and the function now repeats the same sequence of values as we rotate from 180°, back to 360°.
Tracing this motion on the graph over several rotations gives:
Remember that cotangent and tangent are reciprocals of each other, so any point at which the tangent was equal to 0, the cotangent will be undefined and any point at which the tangent was undefined, the cotangent is equal to 0.
You might also notice that the graphs consistently intersect at 1 and −1. These are the angles that have 45° reference angles, which always have tangents and cotangents equal to 1 or −1. It makes sense that 1 and −1 are the only values for which a function and its reciprocal are the same. Keep this in mind as we look at cosecant and secant compared to their reciprocals of sine and cosine.
The cotangent function has a domain of all real angles except multiples of π {… − 2π. − π.0, π, 2π …} The range is all real numbers.
### Cosecant
There are many ways possible to find the cosecant segment. One approach is to look at the right triangle formed by the cotangent segment and use the Pythagorean Theorem to generate the cosecant.
$a^2 + b^2 = c^2\,\!$ $1^2 + \left (\frac{x}{y}\right )^2 = c^2$ $\left (\frac{y}{y}\right)^2 + \left (\frac{x}{y}\right )^2 = c^2$ $\frac{y^2}{y^2} + \frac{x^2}{y^2} = c^2$ $\frac{y^2 + x^2}{y^2} = c^2$
From the original triangle in the unit circle, y2 + x2 = r2
$\frac{r^2}{y^2} = c^2$ $\left (\frac{r}{y} \right )^2 = c^2$
Since $\tfrac{r}{y}$ is cosecant, then the cosecant must be the same as side c.
Tracing the length of this segment, it is undefined at 0°, infinitely large for very small angles, decreasing to 1 at 90° and then increasing infinitely until it is undefined at 180°. The process repeats from 180° to 360°, however, the segment starts infinitely negative, increases to −1 at 270° before approaching an infinitely negative length.
The period of the function is therefore 2π with a domain of all real angles except multiples of π {… −2π, −π, 0, π, 2π …}. The range is all real numbers greater than 1 or less than −1.
The graph then would look as follows:
Here is the graph of y = sin(x) as well:
Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs at 1 and −1. Many students are reminded of parabolas when they look at the half period of the cosecant graph. While they are similar in that they each have a local minimum or maximum and they begin and end in the same direction the comparisons end there and they shouldn't be referred to as parabolic. The mathematics that defines the values, and therefore shape, of the graph is completely different from the quadratic function of a parabola.
### Secant
Much like the relationship between sine and cosine, secant and cosecant share many similarities. The segment used to generate y = sec(x) is shown below:
You will be asked to demonstrate this in the exercises section. This segment is 1 unit for 0°, then grows through the first quadrant, and is undefined at 90°. It is infinitely negative shrinking down to −1 through the 2nd quadrant, before lengthening back towards infinite negativity and is undefined at 270°. Translating this motion to a graph of y = sec(x) gives us:
Comparing it with the cosine graph:
The period is 2π, the range is the same as y = csc(x) {y : y ≥ 1 or y ≤ −1}, and the domain is all real angles except multiples of $\tfrac{\pi}{2}$ {… −$\tfrac{3\pi}{2}$, −$\tfrac{\pi}{2}$, $\tfrac{\pi}{2}$, $\tfrac{3\pi}{2}$ …}.
## Lesson Summary
The six trigonometric functions defined by the ratios in a right triangle can be placed in the context of the coordinate grid by thinking of them in terms of a point (x, y) rotating around a circle centered at the origin with a radius of one. This circle is called the unit circle. The sine of the angle of rotation is the y−coordinate of the point, the cosine of the angle is the x coordinate, and the tangent is $\tfrac{y}{x}$. The values of the other three ratios; cotangent, cosecant, and secant can also be found in terms of their reciprocal relationships, but all of these values can be constructed geometrically as various segments around the angle of rotation on the unit circle. Instead of finding isolated values, we can look at each ratio as a function of the angle of rotation. These are called circular functions. Here are the domains and ranges of the six circular trigonometric functions.
Table 2.3
Function Domain Range
sin(x) all reals {y : −1 ≤ y ≤ 1}
cos(x) all reals {y : −1 ≤ y ≤ 1}
tan(x) {x : xn · $\tfrac{\pi}{2}$, where n is any odd integer} all reals
csc(x) {x : xnπ, where n is any integer} {y : y > 1 or y < −1}
sec(x) {x : xn · $\tfrac{\pi}{2}$, where n is any odd integer} {y : y > 1 or y < −1}
cot(x) {x : xn · $\tfrac{\pi}{2}$, where n is any odd integer} all reals
## Review Questions
1. Show that side A in this drawing is equal to sec(θ).
2. In Chapter 1, you learned that tan2(θ) + 1 = sec2(θ). Use the drawing and results from question 1 to demonstrate this identity.
3. This diagram shows a unit circle with all the angles that have reference angles of 30°, 45°, and 60°, as well as the quadrant angles. Label the coordinates of all points on the unit circle. On the smallest circle, label the angles in degrees, and on the middle circle, label the angles in radians.
4. Draw and label the line segments in the following drawing that represent the six trigonometric functions (sine, cosine, tangent, cosecant, secant, cotangent).
5. Which of the following shows functions that are both increasing as x increases from 0 to $\tfrac{\pi}{2}$?
(a) sin(x) and cos(x)
(b) tan(x) and csc(x)
(c) sec(x) and cot(x)
(d) csc(x) and sec(x)
6. Which of the following statements are true as x increases from $\tfrac{3\pi}{2}$ to 2π?
(a) cos(x) approaches 0
(b) tan(x) gets infinitely large
(c) cos(x) < sin(x)
(d) cot(x) gets infinitely small
$\frac{x}{1} =$ $\frac{1}{A}$ $Ax =\,\!$ $1\,\!$ $A =\,\!$ $\frac{1}{x}$ $\cos(\theta) =\,\!$ $x\,\!$ $\frac{1}{\cos(\theta)} =$ $\frac{1}{x}$ $\frac{1}{\cos(\theta)} =$ $\sec(\theta) = \frac{1}{x}$ $\therefore \sec(\theta) =\,\!$ $A\,\!$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 46, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8915484547615051, "perplexity": 327.45257590201464}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802765698.11/warc/CC-MAIN-20141217075245-00115-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://notes.reasoning.page/html/helpful-things-know-about-exponential-form | # A calculus of the absurd
#### 14.6 Helpful things to know about the exponential form
Complex numbers sometimes form geometric series (particularly when trigonometric functions are translated into exponential form). For example,
$$2 + 2e^{\frac {\pi }{10}i} + 2e^{\frac {2\pi }{10}i} + 2e^{\frac {3\pi }{10}i} + 2e^{\frac {4\pi }{10}i}$$
forms a geometric series with $$a=2$$ and $$r=2e^{\frac {\pi }{10}i}$$. This means that the sum (as shown in the "sequences and series" section of these notes) is equal to
$$\frac {2\left ( 1 - \left (e^{\frac {\pi }{10}}\right )^5 \right )} {1-e^{\frac {\pi }{10}}}$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9580609798431396, "perplexity": 725.7053711478849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00280.warc.gz"} |
http://www.aimath.org/pastworkshops/complexitycrmap.html | # Complexity of mappings in CR geometry
September 11 to September 15, 2006
at the
American Institute of Mathematics, Palo Alto, California
organized by
John P. D'Angelo and Peter Ebenfelt
## Original Announcement
This workshop will study mappings in CR geometry. Participants will focus on a quite specific part of this general subject, namely the relationship between the complexity of a CR mapping and the complexity of the CR structures on the domain and target manifolds.
The workshop has three principal goals. The first goal is to determine the fundamental notions of CR complexity and to prove sharp results about these notions. The second goal is to organize CR complexity theory into a broad framework that will be useful in CR geometry and also apply to other parts of mathematics. The third goal is to bring active senior researchers and young mathematicians together to work in a focused manner that will forge interactions and guide future research.
We illustrate the idea using the unit sphere, which is in some precise sense the simplest CR manifold. Consider a proper holomorphic mapping between balls in complex Euclidean spaces of possibly different dimensions. When the domain dimension is at least two, and the mapping is smooth at the boundary, then it must be a rational mapping. Its restriction to the boundary sphere defines a non-constant CR mapping between spheres. One may ask the following natural questions: How complicated can a CR mapping between spheres be, given the domain and target dimensions? What are the appropriate measures of complexity?
Let us first consider the equidimensional case. In one dimension we are considering finite Blaschke products, whose complexity can be measured by the number of factors, or equivalently, the degree of the divisor. In higher dimensions, it is a well-known result (Pinchuk and Alexander) that a proper holomorphic mapping between equidimensional balls is an automorphism, and hence a rational mapping of degree one. Moreover, the mapping is spherically equivalent to the identity mapping, and hence it is not complicated. When the target space has lower dimension than the domain space, the only CR mappings between spheres are constant, and thus even less complicated. On the other hand, as the dimension of the target space becomes large relative to that of the domain, the dimension of the moduli space of CR mappings between spheres is unbounded. One can create rational CR mappings between spheres of arbitrarily large degree (and arbitrary complexity in other senses) by allowing the target dimension to be large enough (D'Angelo).
On the other hand, in low codimension there are restrictions on the degree (Faran, Webster, Huang-Ji). Thus there is a relationship between the complexity of a rational CR mapping (as measured by the degree in this case) between spheres and its domain and target dimensions. Making this relationship quantitative and precise is a difficult and fundamental problem that, along with other related problems, will guide the workshop.
## Material from the workshop
A list of participants.
The workshop schedule.
A report on the workshop activities.
There is a short glossary of terms related to the workshop topic. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222188591957092, "perplexity": 443.21357150405976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982295192.28/warc/CC-MAIN-20160823195815-00002-ip-10-153-172-175.ec2.internal.warc.gz"} |
https://forum.freecodecamp.org/t/not-able-to-figure-out-the-question-escape-sequences-in-strings/8884 | # Not able to figure out the question...Escape Sequences in Strings
Encode the following sequence, separated by spaces:
backslash tab tab carriage-return new-line and assign it to myStr
should i use the code from previous challenge or
var myStr;
how to assign backslash to the var ?
There is a diagram with everything explained on the challenge page :
Common escaped characters :
’ single quote
" double quote
\ backslash
\n new line
\r carriage return
\t tab
\b backspace
\f form feed
You need to assign a string to myStr using theses characters, delimited by a single space between each.
var myStr;
myStr = “\ \ \t \t \t \t \r \r \n \n”;
this is not working
got it working after some tweeks… thank you
1 Like
var myStr;
myStr = " \\ \t \t \r \n ";
but it didn’t work for me.
what’s wrong with this?
got it work after googling this https://github.com/FreeCodeCamp/FreeCodeCamp/issues/5828
What tweaks did you make? I tried the same code and it wouldn’t work. Thanks!
what is your code? so we can understand where you got stuck…
var myStr = “Here is a backslash: \.\n\t\t\r Here is a new line with two tabs.”;
you need to give spaces between each escape sentence
Thanks, that worked!
1 Like
var myStr = “Here is a backslash: \. \n \t \t Here is a new line with two tabs.”; // Change this line
Hello! Final code must look like this? I spend whole day trying to solve it, but it didnt work(
I can’t find a solution, can you help me guys?
this is my code:
myStr = ‘Here is a backslash: \ \t \t \r \n Here is a new line with two tabs.’;
Hey guys I figure it out like this
text: Here is a backslash: .
Here is a new line with two tabs.
answer: var myStr = “Here is a backslash: \.\n\t\tHere is a new line with two tabs.”;
Hope it works to someone else
What am I doing wrong?!
var myStr = “Here is a backslash: \ . \n \t \t Here is a new line with two tabs.”; // Change this line
When I c&p this code there were two backslashes behind: “backslash: -> \\ <-” Like so. (Im having to put three to get two to show up.
Hi…Not sure what you mean.
The problem requires one backslash to appear in the text followed by the period.
To do this then one additional backslash is needed immediately before the one you wish to appear.
\.
var myStr = “Here is a backslash: \.\n\t\tHere is a new line with two tabs.”;
hopefully that makes sense
Thanks.
I was trying to explain the correct answer to the original poster not say my code wasn’t working lol
have no idea how i added you, and your code worked perfectly sorry for the mix up, i took it down
No need for any sorry!
I’ll be the one saying sorry on the 10th time I ask for help in th future!!! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8734925389289856, "perplexity": 4937.211305393923}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104672585.89/warc/CC-MAIN-20220706121103-20220706151103-00259.warc.gz"} |
http://cms.math.ca/cmb/kw/Riemannian%20manifolds | location: Publications → journals
Search results
Search: All articles in the CMB digital archive with keyword Riemannian manifolds
Expand all Collapse all Results 1 - 2 of 2
1. CMB 2005 (vol 48 pp. 340)
Andruchow, Esteban
Short Geodesics of Unitaries in the $L^2$ Metric Let $\M$ be a type II$_1$ von Neumann algebra, $\tau$ a trace in $\M$, and $\l2$ the GNS Hilbert space of $\tau$. We regard the unitary group $U_\M$ as a subset of $\l2$ and characterize the shortest smooth curves joining two fixed unitaries in the $L^2$ metric. As a consequence of this we obtain that $U_\M$, though a complete (metric) topological group, is not an embedded riemannian submanifold of $\l2$ Keywords:unitary group, short geodesics, infinite dimensional riemannian manifolds.Categories:46L51, 58B10, 58B25
2. CMB 2002 (vol 45 pp. 378)
Fernández-López, Manuel; García-Río, Eduardo; Kupeli, Demir N.
The Local Möbius Equation and Decomposition Theorems in Riemannian Geometry A partial differential equation, the local M\"obius equation, is introduced in Riemannian geometry which completely characterizes the local twisted product structure of a Riemannian manifold. Also the characterizations of warped product and product structures of Riemannian manifolds are made by the local M\"obius equation and an additional partial differential equation. Keywords:submersion, Möbius equation, twisted product, warped product, product Riemannian manifoldsCategories:53C12, 58J99
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https://www.physicsforums.com/threads/reduction-formulae-question-in-x-cos-n-x.911117/ | # Reduction Formulae Question : In= ∫x(cos^n(x))
Tags:
1. Apr 12, 2017
### k1902
1. The problem statement, all variables and given/known data
Let In = ∫x(cos^n(x)) with limits between x=π/2, x=0 for n≥0
i) Show that nIn=(n-1)In-2 -n^-1 for n≥2
ii) Find the exact value of I3
2. Relevant equations
∫u'v = uv-∫uv' is what I use for these questions
3. The attempt at a solution
Rewritten as ∫ xcos^n-1(x) cosx
u'=cosx v= xcos^n-1x
u= sinx v'= cos^n-1 -x(n-1)sinxcos^n-2x
But I can't seem to write it in the form it asks for.
2. Apr 12, 2017
### LCKurtz
Your questions are indecipherable. Strange notation and no parentheses.
3. Apr 12, 2017
### Staff: Mentor
Part of the problem is that if you type i in brackets, the browser things you mean the change the font type to italics. Besides that, it's hard to tell exactly what the problem is you're trying to solve.
Here's what I think you meant.
Let $I_n = \int_0^{\pi/2} x \cos^n(x) dx$, with $n \ge 0$
i) Show that $n I_n = (n - 1)I_{n-2} - n^{n - 1}$, for $n \ge 2$.
ii) Find the exact value of $I_3$.
Is this anywhere close to what you're asking?
PS - I used LaTeX to format what I wrote. We have a tutorial here: https://www.physicsforums.com/help/latexhelp/. This is under the INFO menu, under Help/How-to.
4. Apr 13, 2017
### k1902
Sorry, I'm new here and have no idea how to format the text.
But yes that's what I was trying to write , except part i) is i) Show that $n I_n = (n - 1)I_{n-2} - n^{- 1}$, for $n \ge 2$.
Thanks for the help with formatting :)
5. Apr 13, 2017
### k1902
Apologies for that, here's the cleaned up, comprehensible version of the question (thanks to Mark44):
Let $I_n = \int_0^{\pi/2} x \cos^n(x) dx$, with $n \ge 0$
i) Show that $n I_n = (n - 1)I_{n-2} - n^{- 1}$, for $n \ge 2$.
ii) Find the exact value of $I_3$.
6. Apr 13, 2017
### Staff: Mentor
I would try integration by parts twice, starting with $u = \cos^n(x), dv = xdx$. After the second integration by parts, you should have an equation that you can solve algebraically for $I_n$ in terms of $I_{n - 2}$ and other terms.
Draft saved Draft deleted
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http://math.stackexchange.com/questions/299104/showing-that-fx3-gx3-hx3-3fxgxhx-1-for-functions-f | # Showing that $f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1$ for functions $f$, $g$, and $h$ defined by certain power series [duplicate]
I'm having trouble with this question, I have found the interval of convergence of $h(x)$ to be $(-\infty, \infty)$, but I don't know how to use that for the question as well as the hint. Any help would be appreciated. Thanks!
If $$f(x) =\sum_{n = 0}^\infty \frac{x^{3n}}{(3n)!},\qquad g(x) =\sum_{n = 0}^\infty \frac{x^{3n+1}}{(3n+1)!},\qquad h(x) = \sum_{n = 0}^\infty \frac{x^{3n+2}}{(3n+2)!}$$ show that $$f(x)^3 + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1.$$ Hint: show that $h'(x) = g(x)$.
-
## marked as duplicate by Qiaochu YuanFeb 12 '13 at 4:11
Are you saying you don't know how to prove the hint? Once you have the hint, you can prove something similar for $g'$ and for $f'$, and then you can differentiate the left side of the equation you're asked to prove, and see what happens. – Gerry Myerson Feb 10 '13 at 1:48
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Feb 10 '13 at 1:49
No I can prove the hint, I just need help with the "show that" part and thanks for the edit! – user61653 Feb 10 '13 at 1:54
Thank you! I got it now! Just didn't know what to do with the derivatives – user61653 Feb 10 '13 at 2:05
Note that $f'(x) = h(x)$, $g'(x)=f(x)$, and $h'(x)=g(x)$. Then consider
$$\frac{d}{dx} [f^3 + g^3 + h^3 - 3 f g h]$$
and show that it is zero using the above derivatives. Viz.,
\begin{align}\frac{d}{dx} [f^3 + g^3 + h^3 - 3 f g h] &= 3 f^2 f'+3 g^2 g'+3 h^2 h' - 3 f'gh-3 f g'h - 3 f g h'\\ &= 3 f^2 h+3 g^2 f + 3 h^2 g - 3 f'gh-3 f g'h - 3 f g h'\\ &= 0 \\\end{align}
Also note that $f(0)=1$, $g(0)=0$, and $h(0)=0$. Thus the integration constant of the above equation is 1, and
$$f^3(x) + g^3(x) + h^3(x) - 3 f(x) g(x) h(x)=1$$
-
Aren't we trying to show that it is equal to 1 though? – user61653 Feb 10 '13 at 1:56
Yep. See the rest of the post (sorry it was fragmented). – Ron Gordon Feb 10 '13 at 2:00
No worries, and thank you! It makes sense! I just didn't know what to do when I had the derivatives! – user61653 Feb 10 '13 at 2:04
Notice that $h'=g$ and $g'=f$ and $f'=h$, now consider $$(fgh)'=f(g'h+gh')+f'gh$$ $$=fg'h+fgh'+f'gh$$ $$=f^2h+fg^2+gh^2$$ whcih implies that $$fgh=\int f^2h+fg^2+gh^2$$$$=\frac{f^3}{3}+\frac{g^3}{3}+\frac{h^3}{3}+c$$ which implies that $$f^3(x)+g^3(x)+h^3(x)-3f(x)g(x)h(x)=c$$ for all $x\in \mathbb R$, substitute x=0 to get that c=1 and hence you get the result.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.956993043422699, "perplexity": 529.246298612727}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736682773.27/warc/CC-MAIN-20151001215802-00116-ip-10-137-6-227.ec2.internal.warc.gz"} |
https://emba.gnu.org/emacs/emacs/-/commit/ebc6903b2989b2de3154afd59fece90ba3882306 | Commit ebc6903b by Richard M. Stallman
### *** empty log message ***
parent 40f40667
... ... @@ -335,7 +335,7 @@ overlaps the overlay on exit from the search. During the search, such overlays are made temporarily visible by temporarily modifying their invisible and intangible properties. If you want this to be done differently for a certain overlay, give it a want this to be done differently for a certain overlay, give it an @code{isearch-open-invisible-temporary} property which is a function. The function is called with two arguments: the first is the overlay, and the second is @code{t} to make the overlay visible, or @code{nil} to ... ...
... ... @@ -867,9 +867,10 @@ correctly; Edebug will tell you when you have tried enough different conditions that each form has returned two different values. Coverage testing makes execution slower, so it is only done if @code{edebug-test-coverage} is non-@code{nil}. Whether or not coverage testing is enabled, frequency counting is performed for all execution of an instrumented function, even if the execution mode is Go-nonstop. @code{edebug-test-coverage} is non-@code{nil}. Frequency counting is performed for all execution of an instrumented function, even if the execution mode is Go-nonstop, and regardless of whether coverage testing is enabled. Use @kbd{M-x edebug-display-freq-count} to display both the coverage information and the frequency counts for a definition. ... ... @@ -1185,8 +1186,9 @@ elements must all match or none, use @code{&optional @item &rest @kindex &rest @r{(Edebug)} All following elements in the specification list are repeated zero or more times. In the last repetition, however, it is ok if the expression runs out before matching all of the elements of the specification list. more times. In the last repetition, however, it is not a problem if the expression runs out before matching all of the elements of the specification list. To repeat only a few elements, use @code{[&rest @var{specs}@dots{}]}. To specify several elements that must all match on every repetition, use ... ...
... ... @@ -698,7 +698,7 @@ Frames * Pop-Up Menus:: Displaying a menu for the user to select from. * Dialog Boxes:: Displaying a box to ask yes or no. * Pointer Shapes:: Specifying the shape of the mouse pointer. * Window System Selections::Transferring text to and from other window. * Window System Selections::Transferring text to and from other windows. * Color Names:: Getting the definitions of color names. * Resources:: Getting resource values from the server. * Server Data:: Getting info about the X server. ... ...
... ... @@ -55,7 +55,7 @@ See @code{/} and @code{%} in @ref{Numbers}. @xref{Read Only Buffers}. @item cyclic-function-indirection @code{"Symbol's chain of function indirections@* contains a loop"}@* @code{"Symbol's chain of function indirections\@* contains a loop"}@* @xref{Function Indirection}. @item end-of-buffer ... ...
... ... @@ -576,10 +576,10 @@ The argument @var{pretend} has the same meaning as in @findex set-screen-height @findex set-screen-width The old-fashioned functions @code{set-screen-height} and @code{set-screen-width}, which were used to specify the height and width of the screen in Emacs versions that did not support multiple frames, are still usable. They apply to the selected frame. The older functions @code{set-screen-height} and @code{set-screen-width} were used to specify the height and width of the screen, in Emacs versions that did not support multiple frames. They are semi-obsolete, but still work; they apply to the selected frame. @defun x-parse-geometry geom @cindex geometry specification ... ...
... ... @@ -1182,8 +1182,8 @@ frame at a time. @defvar mode-line-buffer-identification This variable identifies the buffer being displayed in the window. Its default value is @code{("%12b")}, which means that it usually displays twelve characters of the buffer name. default value is @code{("%12b")}, which displays the buffer name, padded with spaces to at least 12 columns. @end defvar @defvar global-mode-string ... ... @@ -1484,7 +1484,8 @@ For example, Fortran mode uses it this way: The @code{imenu-generic-expression} patterns can then use @samp{\\sw+} instead of @samp{\\(\\sw\\|\\s_\\)+}. Note that this technique may be inconvenient to use when the mode needs to limit the initial character of a name to a smaller set of characters of a name to a smaller set of characters than are allowed in the rest of a name. Setting this variable makes it buffer-local in the current buffer. @end defvar ... ...
... ... @@ -704,13 +704,13 @@ systems used for I/O to a subprocess. @tindex select-safe-coding-system @defun select-safe-coding-system from to &optional preferred-coding-system This function selects a coding system for encoding the between This function selects a coding system for encoding the text between @var{from} and @var{to}, asking the user to choose if necessary. The optional argument @var{preferred-coding-system} specifies a coding system try first. If it can handle the text in the specified region, then it is used. If this argument is omitted, the current buffer's value of @code{buffer-file-coding-system} is tried first. system to try first. If that one can handle the text in the specified region, then it is used. If this argument is omitted, the current buffer's value of @code{buffer-file-coding-system} is tried first. If the region contains some multibyte characters that the preferred coding system cannot encode, this function asks the user to choose from ... ... | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9563875198364258, "perplexity": 3510.9743556818776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00596.warc.gz"} |
https://www.physicsforums.com/threads/finding-the-angle-of-3-dimensional-vectors.186421/ | # Finding the angle of 3-dimensional vectors.
1. Sep 23, 2007
### niyati
How would the angle between two vectors be found, if, for each vector, three components (i, j, k) were given?
Ex. Given that vector A = 2.0 i + 4.0 j - 7.0 k and vector B = 5.0 i - 3.0 j + 1.0 k, what is the angle between A and B?
2. Sep 23, 2007
### neutrino
Use the definition of the scalar (dot) product.
Last edited: Sep 23, 2007
3. Sep 23, 2007
### FedEx
Ok.
We can find it by dot product.We know that for two vectors A and B
$$\vec{A} \cdot \vec{B} = AB\cos\theta$$
Hence find A dot B and divide it by AB. And take its arccosine and you will get your angle.
4. Sep 23, 2007
### niyati
...dot...product?
All right. I'm pulling out some of my old Pre-Cal stuff, when I learned that. D
Thank you.
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https://piping-designer.com/index.php/properties/classical-mechanics/2868-radius-of-gyration | Written by Jerry Ratzlaff on . Posted in Classical Mechanics
Radius of gyration, abbreviated as k or r, also called gyradius, is the distance from the axis of rotation to a point where the total mass of the body is supposed to be concentrated.
$$\large{ k = \sqrt{ \frac{ I }{ A } } }$$
### Where:
$$\large{ k }$$ = radius of gyration
$$\large{ A }$$ = area cross-section of material
$$\large{ I }$$ = secont moment of area | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9748700261116028, "perplexity": 1441.1193211661146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00363.warc.gz"} |
https://www.physicsforums.com/threads/four-vectors-and-lorentz-transformations.714554/ | # Four vectors and lorentz transformations
1. Oct 5, 2013
### smallgirl
1. Consider a four vector $$x^{\mu}$$, that is timelike (i.e $$x^{2}>0$$. show that it is always possible to find a frame where the coordinates of x are of the form $$(x^{0'},0)$$. Determine the lorentz transformation relating the initial frame to this particular frame
3. I figured that assuming that the starting 4 vector could be of the form $$(x^{0},0,0,x^{3})$$ then the resulting answer would be $$x^{'\mu}=\Lambda_{\nu}^{\mu}x^{\mu}$$
2. Oct 5, 2013
### Staff: Mentor
It would probably be easier to work backwards starting with $(x^{0'},0)$.
3. Oct 5, 2013
### smallgirl
And then find the Lorentz transformation that would give me x?
What I'm also curious about is, how do I deal with x=y=0 in the Lorentz transformation matrix? Would I just put 0 for each component? If so, I dont know how to write a transformation that only acts on t but leaves z' the same as z...
4. Oct 5, 2013
### Staff: Mentor
Just use the inverse lorentz transformation matrix on it in terms of the lambdas. See what that gives.
5. Oct 6, 2013
### vanhees71
I'd think more intuitively. Suppose your time-like vector is the time-like momentum of a (massive!) particle. Then, how do you have to change the inertial reference frame to make the spatial part of the four-momentum (i.e., its usual three-momentum vector) vanish? Then it's easy to write down the prove for the corresponding Lorentz-transformation matrix.
PS: I cannot stress it often enough in this forum: The concise notation of the Lorentz matrices in the index notation is crucial, i.e., you should type
Code (Text):
{\Lambda^{\mu}}_{\nu}
resulting in ${\Lambda^{\mu}}_{\nu}$ with properly ordered upper and lower indices!
6. Oct 6, 2013
### smallgirl
$$x'_{\mu}=(\Lambda^{-1})_{\mu}^{\alpha}x_{\alpha}$$
I'm trying to figure out in latex how to write it properly, as I'm aware of what it's meant to look like anyways.
7. Oct 6, 2013
### BruceW
vanhees' post has the way to write the ordered indices (in latex). Anyway, you are getting to the answer. I'm guessing your plan is that since you know the Lorentz transform, you can then just write down the inverse? Also, I'm guessing that for this problem, you are only allowed to use proper orthochronous linear transformations. The problem doesn't specify this, but it would be a weird question if it allowed any kind of Poincare transform.
8. Oct 6, 2013
### smallgirl
Yeah I tried adding an underscore but it came out funny.. Oh well....
Yes I am assuming I can only use orthochronous lorentz transforms.
I don't understand the latter part of what you said though. When you say the inverse do you mean the inverse of $$(\Lambda^{-1})_{\mu}^{\alpha}$$ ?
9. Oct 6, 2013
### vanhees71
Sure, the intuitive idea to think about I suggested in my previous posting, automatically shows that a proper orthochronous Lorentz transform is sufficient to achieve the goal. It's really obvious!
10. Oct 6, 2013
### smallgirl
$$\Lambda_{\alpha}^{\mu}$$
11. Oct 6, 2013
### BruceW
What I meant was: if you know the form of the proper orthochronous Lorentz transform going from the vector $(x_0,0)$ to another vector, then the answer for the problem is the inverse of this transform. But as vanhees is saying, it might actually be easier to just think of how to transform to the $(x_0,0)$ vector, from another vector. (using proper orthochronous transform).
edit: p.s. 'proper orthochronous Lorentz transform' is probably what most people simply call 'Lorentz transform'. Even though they are technically not the same thing.
12. Oct 6, 2013
### vela
Staff Emeritus
You can't assume this. The problem tells you only that $x^\mu$ is timelike.
13. Oct 6, 2013
### smallgirl
I've been told in the question to assume this....
14. Oct 6, 2013
### BruceW
it will be easier to write it like $(x^0,x^1)$ if you are allowed, since the matrix will be smaller ;) But you can write it like $(x^0,0,0,x^3)$ if that makes more sense to you.
15. Oct 7, 2013
### smallgirl
I don't know how to make the spatial part of the transformation vanish.
I'm wondering whether the matrix would be made of cosh and sinh because then I guess I could have the spatial part vanish provided the angle was right...?
Last edited: Oct 7, 2013
16. Oct 7, 2013
### BruceW
it is more simple than you are thinking. write down the usual Lorentz transform matrix between two reference frames with some velocity $\beta$. And now just choose $\beta$ to get the specific transform you want.
17. Oct 7, 2013
### Staff: Mentor
That's why I'm suggesting using the inverse transformation. Have you learned the form of the Lorentz Transformation when the relative velocity vector for the two frames of reference is not in the same direction as one of their spatial coordinate axes? If not, just use the standard configuration inverse Lorentz Transformation to start with.
Ax=γ(Ax'+βAt')
At=γ(At'+βAx')
In your system, Ax' is equal to zero.
Chet
18. Oct 7, 2013
### vanhees71
Again my suggestion: Suppose your time-like vector is the four-momentum of a particle. Setting $c=1$ it's given by $p=(p^0,\vec{p})$. The three-velocity of the particle is given by $\vec{v}=\vec{p}/p^0$. Now, with which Lorentz transform can you go over to a reference frame, where the particle is at rest? It's really very simple!
Then you write down the appropriate proper orthochronous Lorentz-trafo matrix as an ansatz and prove that it really fulfills the task to bring the particle to rest in the new reference frame, i.e.,
$$p'=(p'{}^0,0,0,0).$$
19. Oct 9, 2013
### Staff: Mentor
Continuing from where I left off in #17, if Ax'=0, you get
Ax=βγAt'
At=γAt'
20. Oct 14, 2013
### smallgirl
Hey,
Well I managed to solve it a few days ago :-)
Was dead simple... I was massively overthinking it...
Draft saved Draft deleted
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https://www.physicsforums.com/threads/evaluate-the-definite-integral-for-the-area-of-the-surface.407480/ | # Evaluate the definite integral for the area of the surface.
1. Jun 2, 2010
### lude1
1. The problem statement, all variables and given/known data
Evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.
y=(x3/6) + (1/2x), [1,2]
2. Relevant equations
2π∫[r(x)](1+[f'(x)2])
3. The attempt at a solution
First I found the derivative.
f'(x)= (x2/2) + (1/2x2)dx
And since y is a function of x, r(x) is
r(x)= (x3/6) + (1/2x)
Then I plug everything in and get
2π∫ [(x3/6) + (1/2x)] * {1 + [(x2/2) + (1/2x2)]2}1/2}dx
And then I'm stuck. The book tells me that I am suppose to get
2π∫ [(x3/6) + (1/2x)] * [(x2/2) + (1/2x2)]dx
But I have no idea how they got that. Specifically, I don't know how they got rid of the radical...
2. Jun 2, 2010
### mmmboh
For starters you derived it wrong, the derivative of f(x) is f'(x)=x2/2-1/2x2. Maybe this is the problem. Unless you just wrote your f(x) wrong.
Edit: Yes that is the problem, use the correct f'(x), and then expand f'(x)2 and put everything over a common denominator, and then you can make an equation which is being squared that is equal to that, and then the root cancels out the squares.
Last edited: Jun 2, 2010
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https://www.physicsforums.com/threads/monochromator-and-changing-grating-question.575303/ | # Homework Help: Monochromator and changing grating question.
1. Feb 7, 2012
### hlee
1. The problem statement, all variables and given/known data
A simple grating monochromator requires the incident light to arrive from the entrance slit at angle "A incident" relative to the "neutral" position, and only light reflected at angle "Aout" will travel through the exit slit. For this monochromator, Ain = Aout = 50 degrees.
If the grating has a blaze density of 1200 lines/mm, at what angle would the grating need to be rotated to to allow the first order diffraction peak for light of wavelength 615 nm to pass through the exit slit?
Hint: You will need to use the sum-angle trigonometric identity to solve this question
2. Relevant equations
nλ=d (sini + sinj)
sin(a+b)=sin(a)cos(b)+sin(b)cos(a)
3. The attempt at a solution
using using nλ=d (sini + sinj) I found that when grating angle is not changed, the 1st order reflection would be at -1.6 degrees. I am not quite sure how to use sum angle to find angle of rotation of grating so that light will exit thru the exit slit.
I would assume grating normal changes thus incident light changes and thus output wont produce -1.6 degree?
any help would be greatful. Thank you. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8077629804611206, "perplexity": 1618.4304928683184}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794866326.60/warc/CC-MAIN-20180524131721-20180524151721-00472.warc.gz"} |
http://math.stackexchange.com/questions/138541/left-inverse-implies-right-inverse-in-a-finite-ring/138543 | Left inverse implies right inverse in a finite ring
Let $R$ be a finite ring, and assume $\exists x,y\in R$ such that $xy=1$. How can I show it implies $yx=1$?
-
$ya=yb\Rightarrow y(a-b)=0\Rightarrow xy(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$ – anon Jul 21 '13 at 10:54
Hint: $xy=1$ implies that left multiplication by $y$ is one-to-one. Can you draw a conclusion whether or not there is a $z$ such that $yz=1$?
If so, you can complete the argument by showing that $z=x$.
-
Easy exercise: modify the proof for the case where $R$ is a finite $k$-algebra, where $k$ is field. Hard exercise: find a proof that works for both at once. (Hint: invent $\mathbb{F}_1$.) – Zhen Lin Apr 29 '12 at 19:37
@Zhen You may find of interest work by Vasconcelos, Armendariz et. al on analogous questions for finitely generated modules. For references see my 2007/12/12 sci.math post – Bill Dubuque Apr 29 '12 at 20:33
Hint $\$ As often occurs, this result on numbers is a special case of a result on functions. namely, consider $\rm\:x,y\:$ as left-multiplication maps $\rm\:f(r) = xr,\ g(r) = yr,\:$ then apply the following
Lemma $\rm\ fg = 1\ \Rightarrow\ gf = 1\$ for maps $\rm\:f,g\:$ on a finite set $\rm\:R.$
$\rm(1)\ \ \ fg = 1\ \Rightarrow\ g\ is\ 1\!-\!1\:$ by $\rm\:f\:$ of $\rm\:g(a) = g(b)\ \Rightarrow\ a = b$
$\rm(2)\ \ \ g\ is\ 1\!-\!1\ \Rightarrow\ g\:$ is onto, since $\rm\:R\:$ is finite
$\rm(3)\ \ \ g\ is\ onto\ \Rightarrow\ gf = 1\:$ by $\rm\ a = g(b) = g(fg(b)) = gf(a)$
Remark $\$ In fact we may view the ring as the set of such maps (left-regular representation), where the elements of $\rm\:R\:$ are essentially viewed as $1$-dimensional matrices. Then the above is analogous to a well-known result about matrices, e.g. see my post here where I prove $\rm\ AB = I\:\Rightarrow\; BA = 1,\:$ or, equivalently, $\rm\:B\:$ injective $\rm \Rightarrow$ $\rm\: B\:$ surjective, by exploiting the pigeonhole principle. See also other posts in that thread which clarify the fundamental role played by the pigeonhole principle. See also this question on Dedekind-finite rings, i.e. rings where $\rm\:xy = 1\:\Rightarrow\: yx = 1.$
-
Let $f_y\colon:R\rightarrow R,\ z\mapsto yz$ then: $$f_y(z)=f_y(t)\iff yz=yt\Rightarrow x(yz)=x(yt)\Rightarrow (xy)z=(xy)t\Rightarrow z=t$$ hence $f_y$ is one to one. Now since $R$ is finite then the map $f_y$ is bijective hence there's a unique $z\in R$ s.t. $f_y(z)=yz=1$ so $x(yz)=(xy)z=z=x$ and conclude.
-
Another way to understand this argument is functionally: if left-multiplication by $y$ were non-injective, then post-composing it with left-multiplication by $x$ would necessarily yield a non-injective map, meaning multiplication by $xy$, or $1$, absurd. – anon Jul 21 '13 at 19:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9657203555107117, "perplexity": 251.75806223378555}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398446218.95/warc/CC-MAIN-20151124205406-00296-ip-10-71-132-137.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/149521-integrating-r-2-exp-2r.html | 1. ## Integrating r^2 exp(-2r/a)
Integrating in the limits: infinity -> 0
I have tried integration by substitution and get -ar^2/2 exp(-2r/a) + k, which then gives me zero when I apply the limits. Apparently, the final answer is a^3/4.
Please could I have some help? I am new to integration by substitution btw.
Thanks.
2. Originally Posted by Lex
Integrating in the limits: infinity -> 0
I have tried integration by substitution and get -ar^2/2 exp(-2r/a) + k, which then gives me zero when I apply the limits. Apparently, the final answer is a^3/4.
Please could I have some help? I am new to integration by substitution btw.
Thanks.
When you write "exp(-2r/a)" do you mean:
$[A]\;\;\;r^2e^{\frac{-2r}{a}}$
or
$[B]\;\;\;(r^2)^{\frac{-2r}{a}}$
?
3. It's the first one - [A].
4. Try integration by parts-- twice.
Let $u = r^2$, so $du = 2r\,dr$, $dv = \exp(-2r/a)$, and $v = -\frac{a}{2}\exp(-2r/a)$.
$\lim_{b \to \infty}\left[-\frac{ar^2}{2}\exp(-2r/a)\right]_0^b + \lim_{b \to \infty} \int_0^b ar\exp(-2r/a)\,dr$
Now we have to assume a > 0 for the limit in the left term to exist. If we do this, we can compute it to 0.
Use integration by parts on the remaining integral. I computed it to $a^3/4$ as you said the answer should be.
5. Originally Posted by gosualite
Try integration by parts-- twice.
Let $u = r^2$, so $du = 2r\,dr$, $dv = \exp(-2r/a)$, and $v = -\frac{a}{2}\exp(-2r/a)$.
$\lim_{b \to \infty}\left[-\frac{ar^2}{2}\exp(-2r/a)\right]_0^b + \lim_{b \to \infty} \int_0^b ar\exp(-2r/a)\,dr$
Now we have to assume a > 0 for the limit in the left term to exist. If we do this, we can compute it to 0.
Use integration by parts on the remaining integral. I computed it to $a^3/4$ as you said the answer should be.
Hey Lex, just incase you haven't been introduced to "Integration by Parts" yet, just know that this is what it entails (if you have been introduced and understand it, just skip this post):
Given an integral of the form:
$\int f(x)g'(x)dx$
We can go back to the definition of the power rule to minipulate the equation, and add integration symbols to both sides, and do some more minipulation (all of this is "mathematically legal" of course. If you want me to show you how we can arrive at the following formula, just post here and I'll show you why integration by parts works.) We can arrive at the following formula for an integral of the above form:
$\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$
The above is more commonly written, with alittle less obvious notation, as the following:
$\int\;u\;dv = uv\;-\;\int\;v\;du$
One thing to notice in both equations is the $g'(x)$ in the first and the $dv$ in the second. Realizing this, it just means that to find $g(x)$ (i.e. $v$ in the second equation) on the right side of the equation, we just need to integrate it (i.e. $g(x) = \int\;g'(x)$. That may seem really obvious but allot of people miss that and get confused. And as for finding the $f'(x)$ (i.e. $du$ in the second equation) it should be obvious from the notation that $f'(x) = \frac{d}{dx}[f(x)]$, but it might be slightly less obvious in the notation of the second equation (du = $\frac{d}{dx}[u]$). Again, this may seem obvious, but I've seen many people get confused with this part of the notation as well. I perfer writing the equation in the first form, because then people do not so easily forget that $u$ and $v$ are both functions of $x$, and they don't so easily mix which "part" of the "parts" to integrate and which to differentiate.
Another note, you may ask "Which part of the equation do I define as $u$ and which as $dv$?" Its simple to decide. Follow the "LIATE" rule (proposed by Herbert Kasube of Bradley University). Out of the two functions of x you have in your integrand, the first function's type to appear on the following list should be $u$. Obviously, the other function is then, by default, $dv$. Heres the LIATE list:
[1] - L - Logarithmic Functions: [i.e. $log(x)\;,\;ln(x)\;,\;etc.$]
[2] - I - Inverse Trigonomic Functions: [i.e. $arctan(x)\;,\;sin^{-1}(x)\;,\;etc.$]
[3] - A - Algebraic Functions: [i.e. $x^2+6x\;,\;4x^16-x^2+12\;,\;etc.$]
[4] - T - Trigonomic Functions: [i.e. $cos(x)\;,\;cot(x)\;,\;etc.$]
[5] - E - Exponential Functions: [i.e. $e^x\;,\;3^{x}\;,\;etc.$]
So, as an example, say you have:
$\int cos(x)e^xdx$
and you want to know if $cos(x)$ or $e^x$ should be $u$. Follow your finger down the list: neither function is logarithmic, so move down. Niether function is inverse trigonomic so move down. Niether function is algebraic (remeber, not just a function involving algebra, but "algebraic" as defined in the list) so we move down. Bingo! We see that $cos(x)$ is a trigonomic function! So then $u=cos(x)$ and by default $e^x = dv$. Also note that $e^x$ is exponential, and last on the list, so it follows that if you went backwards (i.e. up the list) the first function type you ran into would be your $dv$.
I hope this serves as an ample explaination of integration by parts. Now try to finish the integral problem you originally posted. Use the first few steps that gosualite posted and then go from there. Good luck. Any more questions, post here and I'll be happy to help.
6. Thanks to both of you. I managed to get to the solution. mfetch22, please could you explain how to use the LIATE rule in gosualite's method? I got to the answer by choosing u and dv by trial and error, but it seems that dv must be taken as the exponential function in each integration by parts step, right?
7. Originally Posted by Lex
Thanks to both of you. I managed to get to the solution. mfetch22, please could you explain how to use the LIATE rule in gosualite's method? I got to the answer by choosing u and dv by trial and error, but it seems that dv must be taken as the exponential function in each integration by parts step, right?
I'd be glad to. I am assuming you are refering to the first part of his solution, where he makes the definitions of u and dv? This part right
Originally Posted by gosualite
Try integration by parts-- twice.
Let $u = r^2$, so $du = 2r\,dr$, $dv = \exp(-2r/a)$, and $v = -\frac{a}{2}\exp(-2r/a)$.
if you meant some other part of his work, post here and let me know and I'll show the correct part for which you were reffering to. Assuming you meant the above part in the quoation, this is how we apply the LIATE rule. First, make sure to be checking the post where I detailed each part of LIATE as I go through the explaination. So, we have the following integral for which we need to evaluate which "part" to define as $u$ and which as $dv$ :
$\int\;r^2e^{\frac{-2r}{a}}$
Going through the LIATE process, we first look for any logarithmic functions (hence the L), and neither $r^2$ or $e^{\frac{-2r}{a}}$ are logarithimc functions. Moving on. Then we check for inverse trigonomic functions. Same story, niether function fits "I" for inverse trigonomic. Then we check for "A", algebraic functions, and we get a match. $r^2$ falls under the "algebraic" definitions in LIATE. Therefore, as the rules of LIATE dictate, the first type of function that we come across as we move down the LIATE letters is the "algebraic" type, flagged by the expression $r^2$ in the integrand. Thus, $u=r^2$, and as a consequence the other function is $dv$, giving $dv = e^{\frac{-2r}{a}}$. And, as I explained in my other post, finding $v$ and $du$ is simply a matter of integrating $dv$ and differentiating $u$, as gosualite showed a couple posts above.
Lex, if you have any further questions, post back here and I'll be happy to assist
8. Thanks again. Makes sense. For some reason I didn't recognise r^2 as algebraic, but now I do!
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# integrate e^2/r^2 infinity to r
Click on a term to search for related topics. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 65, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9777206778526306, "perplexity": 454.2128574823668}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560279657.18/warc/CC-MAIN-20170116095119-00135-ip-10-171-10-70.ec2.internal.warc.gz"} |
http://compgroups.net/comp.text.tex/setting-tex-text-variable-depending-on-keyval-g/1920147 | f
#### setting TeX text variable depending on keyval-generated param? (low-level TeX question)
I am using keyval to set a "scaled" parameter and am having trouble
getting the right behavior, which is:
If I \usepackage{package} with no [scaled] option, I want to set a
variable to a compile-time-specified default, say "s*[a.b]".
If I \usepackage[scaled]{package}, same thing.
And if I \usepackage[scaled=x.y], I want to set the variable to
"s*[x.y]".
(This is for font scaling, and I've used as my model the Helvetica
scaled parameter.)
The variable name (for the purposes of this conversation) is
pnm@scaled. What I have that's not working correctly is:
\expandafter\ifx\csname pmn@scaled\endcsname\relax
\let\pmn@scaled\@empty
\else
\edef\pmn@scaled{s*[1.2]}%
\fi
I think what this says is "if pnm@scaled is not defined (case 1), leave
it empty, otherwise set it to "s[1.2]". But the else clause is not
working correctly, and I don't know how to differentiate between
"defined by keyval but with no value" and "defined by keyval but with a
value". Also I'm pretty sure I don't really know the difference between
\let and \edef.
Any help gratefully appreciated. Thanks.
JDO
0
john_owens (39)
4/10/2006 10:02:08 PM
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[email protected] schrieb:
> I am using keyval to set a "scaled" parameter and am having trouble
> getting the right behavior, which is:
>
> If I \usepackage{package} with no [scaled] option, I want to set a
> variable to a compile-time-specified default, say "s*[a.b]".
>
> If I \usepackage[scaled]{package}, same thing.
>
> And if I \usepackage[scaled=x.y], I want to set the variable to
> "s*[x.y]".
>
> (This is for font scaling, and I've used as my model the Helvetica
> scaled parameter.)
>
> The variable name (for the purposes of this conversation) is
> pnm@scaled. What I have that's not working correctly is:
>
> \expandafter\ifx\csname pmn@scaled\endcsname\relax
> \let\pmn@scaled\@empty
> \else
> \edef\pmn@scaled{s*[1.2]}%
> \fi
>
> I think what this says is "if pnm@scaled is not defined (case 1), leave
> it empty, otherwise set it to "s[1.2]". But the else clause is not
> working correctly, and I don't know how to differentiate between
> "defined by keyval but with no value" and "defined by keyval but with a
> value". Also I'm pretty sure I don't really know the difference between
> \let and \edef.
I'm not sure I have understand the problem, but to get the option
[scaled] be equivalent to the option [scaled=1.2] you can use the
optional argument that sets a default
\define@key{prefix}{scaled}[1.2]{...}
To get the \if-\else-\fi correct, I guess it would be enough if you
define \pnm@scaled to be s*[1.2] at the start of your package, and then
evaluate the package options. Either they will overwrite the predefined
value or not.
I would also suggest that you take a look at the xkeyval package. With
it you can give keys preset values.
--
Ulrike Fischer
e-mail: zus�tzlich meinen Vornamen vor dem @ einf�gen.
e-mail: add my first name between the news and the @.
0
news9686 (1970)
4/11/2006 7:44:48 AM
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http://tex.stackexchange.com/questions/115432/best-practice-for-typesetting-quantifiers | # Best practice for typesetting quantifiers?
When I work with quantifiers I noted that are very close to the other symbols and the result does not look good, for example
$\exists a\in\mathbb{R}\exists b\in\mathbb{R}\forall c\in\mathbb{R}\forall d\in\mathbb{R}$
Which is the proper form to write quantifiers?
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There exist real scalars a,b for all real scalars c,d – percusse May 21 '13 at 18:56
I would recommend using $\exists a\in\mathbb{R}$, $\exists b\in\mathbb{R}$, $\forall c\in\mathbb{R}$, and $\forall b\in\mathbb{R}$, or perhaps $\exists a, b \in\mathbb{R}$, $\forall c, d \in\mathbb{R}$. – Peter Grill May 21 '13 at 19:02
@PeterGrill Breaking down (the beginning of) such a mathematical statement into multiple math-mode parts seems odd to me... – Jubobs May 21 '13 at 19:13
Sometimes even a space $\exists a\in\mathbb{R}\ \exists b\in\mathbb{R}$ can help. I agree with @percusse though. – marczellm May 21 '13 at 19:19
@percusse the problem is I can't always use the metalanguage working in logic. – Gastón Burrull May 21 '13 at 19:47
It depends on the context.
If this is part of a piece of text, then you might consider Peter Grill's suggestion:
$\exists a\in\mathbb{R}$, $\exists b\in\mathbb{R}$,
$\forall c\in\mathbb{R}$, and $\forall b\in\mathbb{R}$
On the other hand, if the quantifiers are part of a logical formula, you might consider a dot between the quantifiers, like this:
$\exists a\in\mathbb{R}\ldotp\exists b\in\mathbb{R}\ldotp \forall c\in\mathbb{R}\ldotp\forall b\in\mathbb{R}\ldotp P$
This dot notation is inherited, I think, from Russell and Whitehead's Principia Mathematica, and is quite widely used, particularly in computer science. A comma between quantifiers is quite unusual, though it does appear in the syntax of the Coq theorem prover.
$\exists a\in\mathbb{R}, \exists b\in\mathbb{R}, \forall c\in\mathbb{R}, \forall d\in\mathbb{R}, P$
The comma notation becomes awkward when you want to quantify several variables at the same time, because then you have two different types of comma in the same formula:
$\exists a,b\in\mathbb{R}, \forall c,d\in\mathbb{R}, P$
In such cases, you might consider putting just a space between the variables, like this:
$\exists a\;b\in\mathbb{R}, \forall c\;d\in\mathbb{R}, P$
The idea of putting spaces between variables, rather than commas, is taken from the syntax of the Isabelle theorem prover.
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I strongly disagree about using dots between quantifiers. Commas are fine, though. – Jubobs May 21 '13 at 19:35
I liked the second one, I prefer commas but is there a code for commas instead of using \ldotp? What about simple spaces "\ "? – Gastón Burrull May 21 '13 at 19:37
this answer is the closest to what I want, because what I want is a unique formula, not a separation into two parts what do you think about use of "\ " or "," instead of "\ldotp"? – Gastón Burrull May 21 '13 at 20:08
\ and , are fine alternatives. I incorporated , into my answer. – John Wickerson May 22 '13 at 7:42
Simply make these characters what they should be: Operators. They aren't arithmetic operators but logical ones, but that doesn't make any difference here:
\documentclass{article}
\usepackage{amsmath,amssymb}
\DeclareMathOperator{\Exists}{\exists}
\DeclareMathOperator{\Forall}{\forall}
\begin{document}
$\Exists a\in\mathbb{R}\Exists b\in\mathbb{R}\Forall c\in\mathbb{R}\Forall d\in\mathbb{R}$
$\Exists a\in\mathbb{R}:\Exists b\in\mathbb{R}:\Forall c\in\mathbb{R}:\Forall d\in\mathbb{R}$
$\Exists a,b\in\mathbb{R}:\Forall c,d\in\mathbb{R}$
\end{document}
Last but not least, it's equivalent but easier to grasp, if the both "exists" and "foralls" are grouped. R^2 would be wrong in this case, because a and b should each be in R. (a,b) would be in R^2, but that's not written.
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Logical conjunction ∧ is an operator because if P and Q are formulae, then so is (P)∧(Q). ∃x is an operator because if P is a formula then so is ∃x(P). ∃x∈R is an operator for the same reason. But ∃, by itself, is not an operator in this sense, so I don't think it should be declared as one. – John Wickerson May 22 '13 at 7:28
\colon is better than : when writing for example "For every x there exists y such that...". – Jori Mäntysalo May 22 '13 at 9:42
@JohnWickerson: You are right. But ∃x is not a symbol by itself and so cannot be an operator in a typographical sense. The same is true for the integral: if f(x) is an formula, then \int f(x) is not a formula, but \int f(x)dx is. Yet, \int is a typographical operator. So \exists alone is not a logical operator, but \exists x\in M:P(x) is. Yet, \exists should be a typhographical operator. – Toscho May 22 '13 at 12:26
In my opinion, the real issue with quantifiers is that it's hard to obtain consistent spacing, as I explained in this answer. The most striking example I found: $\forall W\forall A$ gives
Of course there should be more space before the second quantifier; a single space \ will usually be OK. The problem is the spacing after the quantifiers. There is no simple solution to this, other than using manual kerning where needed. In this case, $\forall\mkern2mu W\ \forall\mkern-1mu A$ looks quite alright:
Let me point out that I'd use quantifiers only in displayed formulas, never in inline math.
-
Another possibility is:
$\exists\ a,b \in \mathbb{R},\ \forall\ c, b \in\mathbb{R}$
-
I liked use of comma. I probably will use this in the future $\exists a\in\mathbb{R}, \exists b\in\mathbb {R}, \forall c\in \mathbb{R}, \forall d\in\mathbb{R}$. Since I don't like the space "\ " after the quantifier. – Gastón Burrull May 21 '13 at 19:52
The disadvantage of using commas, at least in the example above, is that you now have two different types of comma in your formula, with two different meanings, and this could make the formula a bit hard to understand. – John Wickerson May 22 '13 at 7:31
I don't know if this is what you are asking, but it's related.
In my opinion it's horrible the space after the quantifiers (they look very close to the next letter). I always edit them and add an small space
\let\existstemp\exists
\let\foralltemp\forall
\renewcommand*{\exists}{\existstemp\mkern2mu}
\renewcommand*{\forall}{\foralltemp\mkern2mu}
By the way, as others are saying, it depends on the situation. If it's inline I would go for There exist real scalars a,b for all real scalars c,d (Percusse's comment). But if it's inside a \displaymath I would go for the symbols.
First of all, I usually space my math with \quads (this is personal taste, and you have to choose what you use). And, in second place, I don't know how your example should be read:
• If it's read There exist real scalars a,b for all real scalars c,d I would change the order and write For all real scalars c,d there exist real scalars a,b… and write \forall c,d \in \mathbb{R} \quad \exists a,b \in \mathbb{R}.
• And if it's read as There exist real scalars a,b such that for all real scalars c,d… then I would write \exists a,b \in \mathbb{R}, \quad \forall c,d \in \mathbb{R}
Here it is a full example.
\documentclass{article}
\usepackage{amssymb}
\let\existstemp\exists
\let\foralltemp\forall
\begin{document}
$\exists a,b \in \mathbb{R}, \quad \forall c,d \in \mathbb{R}$
\renewcommand*{\exists}{\existstemp\mkern2mu}
\renewcommand*{\forall}{\foralltemp\mkern2mu}
$\exists a,b \in \mathbb{R}, \quad \forall c,d \in \mathbb{R}$
$\forall c,d \in \mathbb{R} \quad \exists a,b \in \mathbb{R}$
\end{document}
In order to justify the \quads instead of the \s, here is another example which, in my opinion, shows my idea (and why in displaymaths \quads are useful):
I think that the first line is far more readable than the second one.
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I'm interested in space between \mathbb{R} and \exists. Writting "\mathbb{R} \exists" is horrible and "\mathbb{R}\quad \exists" is exaggerated, I prefer "\mathbb{R}\ \exists" or "\mathbb{R}\ \exists". About your suggestion, what about $\forall\, c$? "\," is a little space after quantifier also. – Gastón Burrull May 21 '13 at 20:00
@GastónBurrull About the \,, yes, it works (I used \mkern2mu to show how to adjust it). By the way the \quad if it's in a \displaymath I think it's much better than \ because it clearly separates the sentence. – Manuel May 21 '13 at 20:03
In your first item the meaning changes drastically if you swap the order. – percusse May 22 '13 at 7:55
@percusse My answer to that is: Of course. But then I think, may be I misunderstood part of the question. Shouldn't it change if I swap the order? May be in logic (which I don't know) it shouldn't. My point was only to add the space after the quantifiers and show the \quads as useful mathematical spaces. If I'm wrong, please correct me, it's true I know nothing about logic. – Manuel May 22 '13 at 10:02
@Manuel Sure. I learned it the hard way so I have an eye for that structure from my PhD :) One says there are fixed a,b for all c,d if you swap the order. The other one says for each a and b you can find some c and d. And that caused me a lot of trouble in thepast because they don't teach that in engineering heh. – percusse May 22 '13 at 10:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9121256470680237, "perplexity": 1071.104156096304}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997883466.67/warc/CC-MAIN-20140722025803-00067-ip-10-33-131-23.ec2.internal.warc.gz"} |
http://xrpp.iucr.org/Da/ch2o4v0001/ | International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier
International Tables for Crystallography (2006). Vol. D, ch. 2.4, pp. 329-335
https://doi.org/10.1107/97809553602060000641
## Chapter 2.4. Brillouin scattering
R. Vachera* and E. Courtensa
aLaboratoire des Verres, Université Montpellier 2, Case 069, Place Eugène Bataillon, 34095 Montpellier CEDEX, France
Correspondence e-mail: [email protected]
Brillouin scattering of light originates from the interaction of an incident radiation with thermal acoustic vibrations in matter and probes the long-wavelength acoustic phonons. In this chapter, calculations of the sound velocities and scattered intensities for the most commonly investigated vibrational modes in bulk condensed matter are presented. Tables list the geometries for longitudinal and transverse acoustic modes in the eleven Laue classes. The current state of the art for Brillouin spectroscopy is also briefly summarized.
### 2.4.1. Introduction
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Brillouin scattering originates from the interaction of an incident radiation with thermal acoustic vibrations in matter. The phenomenon was predicted by Brillouin in 1922 (Brillouin, 1922) and first observed in light scattering by Gross (Gross, 1930a,b). However, owing to specific spectrometric difficulties, precise experimental studies of Brillouin lines in crystals were not performed until the 1960s (Cecchi, 1964; Benedek & Fritsch, 1966; Gornall & Stoicheff, 1970) and Brillouin scattering became commonly used for the investigation of elastic properties of condensed matter with the advent of laser sources and multipass Fabry–Perot interferometers (Hariharan & Sen, 1961; Sandercock, 1971). More recently, Brillouin scattering of neutrons (Egelstaff et al., 1989) and X-rays (Sette et al., 1998) has been observed.
Brillouin scattering of light probes long-wavelength acoustic phonons. Thus, the detailed atomic structure is irrelevant and the vibrations of the scattering medium are determined by macroscopic parameters, in particular the density ρ and the elastic coefficients . For this reason, Brillouin scattering is observed in gases, in liquids and in crystals as well as in disordered solids.
Vacher & Boyer (1972) and Cummins & Schoen (1972) have performed a detailed investigation of the selection rules for Brillouin scattering in materials of various symmetries. In this chapter, calculations of the sound velocities and scattered intensities for the most commonly investigated vibrational modes in bulk condensed matter are presented. Brillouin scattering from surfaces will not be discussed. The current state of the art for Brillouin spectroscopy is also briefly summarized.
### 2.4.2. Elastic waves
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#### 2.4.2.1. Non-piezoelectric media
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The fundamental equation of dynamics (see Section 1.3.4.2 ), applied to the displacement u of an elementary volume at r in a homogeneous material is Summation over repeated indices will always be implied, and T is the stress tensor. In non-piezoelectric media, the constitutive equation for small strains S is simply The strain being the symmetrized spatial derivative of u, and c being symmetric upon interchange of k and , the introduction of (2.4.2.2) in (2.4.2.1) gives (see also Section 1.3.4.2 ) One considers harmonic plane-wave solutions of wavevector Q and frequency ω,For small compared with the wavelength , the total derivative can be replaced by the partial in (2.4.2.3). Introducing (2.4.2.4) into (2.4.2.3), one obtains where is the unit vector in the propagation direction, is the unit tensor and , where is the phase velocity of the wave. This shows that is an eigenvector of the tensor . For a given propagation direction , the three eigenvalues are obtained by solving To each there is an eigenvector given by (2.4.2.5) and an associated phase velocity
The tensor is symmetric upon interchange of the indices () because . Its eigenvalues are real positive, and the three directions of vibration are mutually perpendicular. The notation indicates a unit vector. The tensor is also invariant upon a change of sign of the propagation direction. This implies that the solution of (2.4.2.5) is the same for all symmetry classes belonging to the same Laue class.
For a general direction , and for a symmetry lower than isotropic, is neither parallel nor perpendicular to , so that the modes are neither purely longitudinal nor purely transverse. In this case (2.4.2.6) is also difficult to solve. The situation is much simpler when is parallel to a symmetry axis of the Laue class. Then, one of the vibrations is purely longitudinal (LA), while the other two are purely transverse (TA). A pure mode also exists when belongs to a symmetry plane of the Laue class, in which case there is a transverse vibration with perpendicular to the symmetry plane. For all these pure mode directions, (2.4.2.6) can be factorized to obtain simple analytical solutions. In this chapter, only pure mode directions are considered.
#### 2.4.2.2. Piezoelectric media
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In piezoelectric crystals, a stress component is also produced by the internal electric field E, so that the constitutive equation (2.4.2.2) has an additional term (see Section 1.1.5.2 ), where e is the piezoelectric tensor at constant strain.
The electrical displacement vector D, related to E by the dielectric tensor , also contains a contribution from the strain, where is at the frequency of the elastic wave.
In the absence of free charges, , and (2.4.2.9) provides a relation between E and S, For long waves, it can be shown that E and Q are parallel. (2.4.2.10) can then be solved for E, and this value is replaced in (2.4.2.8) to give Comparing (2.4.2.11) and (2.4.2.2), one sees that the effective elastic tensor now depends on the propagation direction . Otherwise, all considerations of the previous section, starting from (2.4.2.6), remain, with c simply replaced by .
### 2.4.3. Coupling of light with elastic waves
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#### 2.4.3.1. Direct coupling to displacements
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The change in the relative optical dielectric tensor produced by an elastic wave is usually expressed in terms of the strain, using the Pockels piezo-optic tensor p, as The elastic wave should, however, be characterized by both strain S and rotation A (Nelson & Lax, 1971; see also Section 1.3.1.3 ):The square brackets on the left-hand side are there to emphasize that the component is antisymmetric upon interchange of the indices, . For birefringent crystals, the rotations induce a change of the local in the laboratory frame. In this case, (2.4.3.1) must be replaced by where is the new piezo-optic tensor given by One finds for the rotational partIf the principal axes of the dielectric tensor coincide with the crystallographic axes, this gives This is the expression used in this chapter, as monoclinic and triclinic groups are not listed in the tables below.
For the calculation of the Brillouin scattering, it is more convenient to use which is valid for small .
#### 2.4.3.2. Coupling via the electro-optic effect
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Piezoelectric media also exhibit an electro-optic effect linear in the applied electric field or in the field-induced crystal polarization. This effect is described in terms of the third-rank electro-optic tensor r defined by Using the same approach as in (2.4.2.10), for long waves can be expressed in terms of , and (2.4.3.8) leads to an effective Pockels tensor accounting for both the piezo-optic and the electro-optic effects: The total change in the inverse dielectric tensor is then The same equation (2.4.3.7) applies.
### 2.4.4. Brillouin scattering in crystals
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#### 2.4.4.1. Kinematics
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Brillouin scattering occurs when an incident photon at frequency interacts with the crystal to either produce or absorb an acoustic phonon at , while a scattered photon at is simultaneously emitted. Conservation of energy gives where positive corresponds to the anti-Stokes process. Conservation of momentum can be written where Q is the wavevector of the emitted phonon, and , are those of the scattered and incident photons, respectively. One can define unit vectors q in the direction of the wavevectors k by where n and are the appropriate refractive indices, and is the vacuum wavelength of the radiation. Equation (2.4.4.3b) assumes that so that is not appreciably changed in the scattering. The incident and scattered waves have unit polarization vectors and , respectively, and corresponding indices n and . The polarization vectors are the principal directions of vibration derived from the sections of the ellipsoid of indices by planes perpendicular to and , respectively. We assume that the electric vector of the light field Eopt is parallel to the displacement Dopt. This is exactly true for many cases listed in the tables below. In the other cases (such as skew directions in the orthorhombic group) this assumes that the birefringence is sufficiently small for the effect of the angle between and to be negligible. A full treatment, including this effect, has been given by Nelson et al. (1972).
After substituting (2.4.4.3) in (2.4.4.2), the unit vector in the direction of the phonon wavevector is given by The Brillouin shift is related to the phonon velocity V by Since , from (2.4.4.5) and (2.4.4.3), (2.4.4.4) one finds where is the angle between and .
#### 2.4.4.2. Scattering cross section
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The power , scattered from the illuminated volume V in a solid angle , where and are measured inside the sample, is given by where is the incident light intensity inside the material, is the appropriate elastic constant for the observed phonon, and the factor results from taking the fluctuation–dissipation theorem in the classical limit for (Hayes & Loudon, 1978). The coupling coefficient M is given by In practice, the incident intensity is defined outside the scattering volume, , and for normal incidence one can write Similarly, the scattered power is observed outside as , and again for normal incidence. Finally, the approximative relation between the scattering solid angle , outside the sample, and the solid angle , in the sample, is Substituting (2.4.4.9a,b,c) in (2.4.4.7), one obtains (Vacher & Boyer, 1972) where the coupling coefficient is In the cases of interest here, the tensor is diagonal, without summation on i, and (2.4.4.11) can be written in the simpler form
### 2.4.5. Use of the tables
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The tables in this chapter give information on modes and scattering geometries that are in most common use in the study of hypersound in single crystals. Just as in the case of X-rays, Brillouin scattering is not sensitive to the presence or absence of a centre of symmetry (Friedel, 1913). Hence, the results are the same for all crystalline classes belonging to the same centric group, also called Laue class. The correspondence between the point groups and the Laue classes analysed here is shown in Table 2.4.5.1. The monoclinic and triclinic cases, being too cumbersome, will not be treated here.
Table 2.4.5.1| top | pdf | Definition of Laue classes
Crystal systemLaue classPoint groups
Cubic
Hexagonal
Tetragonal
Trigonal
Orthorhombic O
For tensor components and , the tables make use of the usual contracted notation for index pairs running from 1 to 6. However, as the tensor is not symmetric upon interchange of , it is necessary to distinguish the order and . This is accomplished with the following correspondence:
Geometries for longitudinal modes (LA) are listed in Tables 2.4.5.2 to 2.4.5.8. The first column gives the direction of the scattering vector that is parallel to the displacement . The second column gives the elastic coefficient according to (2.4.2.6). In piezoelectric materials, effective elastic coefficients defined in (2.4.2.11) must be used in this column. The third column gives the direction of the light polarizations and , and the last column gives the corresponding coupling coefficient [equation (2.5.5.11)]. In general, the strongest scattering intensity is obtained for polarized scattering (), which is the only situation listed in the tables. In this case, the coupling to light () is independent of the scattering angle , and thus the tables apply to any value.
Table 2.4.5.2| top | pdf | Cubic Laue classes and : longitudinal modes
This table, written for the class , is also valid for the class with the additional relation . It can also be used for the spherical system where , .
C
Table 2.4.5.3| top | pdf | Tetragonal and hexagonal Laue classes: longitudinal modes
This table, written for the class , is also valid for the class with the additional relations ; .
C
Table 2.4.5.4| top | pdf | Hexagonal Laue class : longitudinal modes
C
Table 2.4.5.5| top | pdf | Tetragonal Laue class : longitudinal modes
C
Table 2.4.5.6| top | pdf | Orthorhombic Laue class O: longitudinal modes
C
Table 2.4.5.7| top | pdf | Trigonal Laue class : longitudinal modes
C
Table 2.4.5.8| top | pdf | Trigonal Laue class : longitudinal modes
C
Tables 2.4.5.9 to 2.4.5.15 list the geometries usually used for the observation of TA modes in backscattering (). In this case, is always perpendicular to (pure transverse modes), and is not necessarily parallel to . Cases where pure TA modes with in the plane perpendicular to are degenerate are indicated by the symbol D in the column for . For the Pockels tensor components, the notation is if the rotational term vanishes by symmetry, and it is otherwise.
Table 2.4.5.9| top | pdf | Cubic Laue classes and : transverse modes, backscattering
This table, written for the class , is also valid for the class with the additional relation . It can also be used for the spherical system where , .
C
D
Table 2.4.5.10| top | pdf | Tetragonal and hexagonal Laue classes: transverse modes, backscattering
This table, written for the class , is also valid for the class with the additional relations ; .
C
Table 2.4.5.11| top | pdf | Hexagonal Laue class : transverse modes, backscattering
; .
C
Table 2.4.5.12| top | pdf | Tetragonal Laue class : transverse modes, backscattering
C
Table 2.4.5.13| top | pdf | Orthorhombic Laue class O: transverse modes, backscattering
C
Table 2.4.5.14| top | pdf | Trigonal Laue class : transverse modes, backscattering
; .
C
D
D
Table 2.4.5.15| top | pdf | Trigonal Laue class : transverse modes, backscattering
C
D
D
Tables 2.4.5.16 to 2.4.5.22 list the common geometries used for the observation of TA modes in 90° scattering. In these tables, the polarization vector is always perpendicular to the scattering plane and is always parallel to the incident wavevector of light q. Owing to birefringence, the scattering vector does not exactly bisect and [equation (2.4.4.4)]. The tables are written for strict 90° scattering, , and in the case of birefringence the values of to be used are listed separately in Table 2.4.5.23. The latter assumes that the birefringences are not large, so that the values of are given only to first order in the birefringence.
Table 2.4.5.16| top | pdf | Cubic Laue classes and : transverse modes, right-angle scattering
This table, written for the class , is also valid for the class with the additional relation . It can also be used for the spherical system where , .
CScattering plane
D
Table 2.4.5.17| top | pdf | Tetragonal and hexagonal Laue classes: transverse modes, right-angle scattering
This table, written for the class , is also valid for the class with the additional relations ; .
CScattering plane
D
Table 2.4.5.18| top | pdf | Hexagonal Laue class: transverse modes, right-angle scattering
; .
CScattering plane
D
Table 2.4.5.19| top | pdf | Tetragonal Laue class: transverse modes, right-angle scattering
CScattering plane
D
Table 2.4.5.20| top | pdf | Orthorhombic Laue class O: transverse modes, right-angle scattering
CScattering plane
Table 2.4.5.21| top | pdf | Trigonal Laue class : transverse modes, right-angle scattering
; .
CScattering plane
D
Table 2.4.5.22| top | pdf | Trigonal Laue class : transverse modes, right-angle scattering
; .
CScattering plane
D
D
Table 2.4.5.23| top | pdf | Particular directions of incident light used in Tables 2.4.5.17 to 2.4.5.22
, , .
Notation
### 2.4.6. Techniques of Brillouin spectroscopy
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Brillouin spectroscopy with visible laser light requires observing frequency shifts falling typically in the range ∼1 to ∼100 GHz, or ∼0.03 to ∼3 cm−1. To achieve this with good resolution one mostly employs interferometry. For experiments at very small angles (near forward scattering), photocorrelation spectroscopy can also be used. If the observed frequency shifts are cm−1, rough measurements of spectra can sometimes be obtained with modern grating instruments. Recently, it has also become possible to perform Brillouin scattering using other excitations, in particular neutrons or X-rays. In these cases, the coupling does not occur via the Pockels effect, and the frequency shifts that are observed are much larger. The following discussion is restricted to optical interferometry.
The most common interferometer that has been used for this purpose is the single-pass planar Fabry–Perot (Born & Wolf, 1993). Upon illumination with monochromatic light, the frequency response of this instrument is given by the Airy function, which consists of a regular comb of maxima obtained as the optical path separating the mirrors is increased. Successive maxima are separated by . The ratio of the maxima separation to the width of a single peak is called the finesse F, which increases as the mirror reflectivity increases. The finesse is also limited by the planarity of the mirrors. A practical limit is . The resolving power of such an instrument is , where is the optical thickness. Values of R around to can be achieved. It is impractical to increase above ∼5 cm because the luminosity of the instrument is proportional to . If higher resolutions are required, one uses a spherical interferometer as described below.
A major limitation of the Fabry–Perot interferometer is its poor contrast, namely the ratio between the maximum and the minimum of the Airy function, which is typically ∼1000. This limits the use of this instrument to samples of very high optical quality, as otherwise the generally weak Brillouin signals are masked by the elastically scattered light. To avert this effect, several passes are made through the same instrument, thus elevating the Airy function to the corresponding power (Hariharan & Sen, 1961; Sandercock, 1971). Multiple-pass instruments with three, four or five passes are common. Another limitation of the standard Fabry–Perot interferometer is that the interference pattern is repeated at each order. Hence, if the spectrum has a broad spectral spread, the overlap of adjacent orders can greatly complicate the interpretation of measurements. In this case, tandem instruments can be of considerable help. They consist of two Fabry–Perot interferometers with combs of different periods placed in series (Chantrel, 1959; Mach et al., 1963). These are operated around a position where the peak transmission of the first interferometer coincides with that of the second one. The two Fabry–Perot interferometers are scanned simultaneously. With this setup, the successive orders are reduced to small ghosts and overlap is not a problem. A convenient commercial instrument has been designed by Sandercock (1982).
To achieve higher resolutions, one uses the spherical Fabry–Perot interferometer (Connes, 1958; Hercher, 1968). This consists of two spherical mirrors placed in a near-confocal configuration. Their spacing is scanned over a distance of the order of . The peculiarity of this instrument is that its luminosity increases with its resolution. One obvious drawback is that a change of resolving power, i.e. of , requires other mirrors. Of course, the single spherical Fabry–Perot interferometer suffers the same limitations regarding contrast and order overlap that were discussed above for the planar case. Multipassing the spherical Fabry–Perot interferometer is possible but not very convenient. It is preferable to use tandem instruments that combine a multipass planar instrument of low resolution followed by a spherical instrument of high resolution (Pine, 1972; Vacher, 1972). To analyse the linewidth of narrow phonon lines, the planar standard is adjusted dynamically to transmit the Brillouin line and the spherical interferometer is scanned across the line. With such a device, resolving powers of ∼ have been achieved. For the dynamical adjustment of this instrument one can use a reference signal near the frequency of the phonon line, which is derived by electro-optic modulation of the exciting laser (Sussner & Vacher, 1979). In this case, not only the width of the phonon, but also its absolute frequency shift, can be determined with an accuracy of ∼1 MHz. It is obvious that to achieve this kind of resolution, the laser source itself must be appropriately stabilized.
In closing, it should be stressed that the practice of interferometry is still an art that requires suitable skills and training in spite of the availability of commercial instruments. The experimenter must take care of a large number of aspects relating to the optical setup, the collection and acceptance angles of the instruments, spurious reflections and spurious interferences, etc. A full list is too long to be given here. However, when properly executed, interferometry is a fine tool, the performance of which is unequalled in its frequency range.
### References
Benedek, G. & Fritsch, K. (1966). Brillouin scattering in cubic crystals. Phys. Rev. 149, 647–662.
Born, M. & Wolf, E. (1993). Principles of optics. Sixth corrected edition. Oxford: Pergamon Press. Reissued (1999) by Cambridge University Press.
Brillouin, L. (1922). Diffusion de la lumière et des rayons X par un corps transparent homogène. Influence de l'agitation thermique. Ann. Phys. Paris, 17, 88–122.
Cecchi, L. (1964). Etude interférométrique de la diffusion Rayleigh dans les cristaux – diffusion Brillouin. Doctoral Thesis, University of Montpellier.
Chantrel, H. (1959). Spectromètres interférentiels à un et deux étalons de Fabry–Perot. J. Rech. CNRS, 46, 17–33.
Connes, P. (1958). L'étalon de Fabry–Perot sphérique. J. Phys. Radium, 19, 262–269.
Cummins, H. Z. & Schoen, P. E. (1972). Linear scattering from thermal fluctuations. In Laser handbook, Vol. 2, edited by F. T. Arecchi & E. O. Schulz-Dubois, pp. 1029–1075. Amsterdam: North-Holland.
Egelstaff, P. A., Kearley, G., Suck, J.-B. & Youden, J. P. A. (1989). Neutron Brillouin scattering in dense nitrogen gas. Europhys. Lett. 10, 37–42.
Friedel, G. (1913). Sur les symétries cristallines que peut révéler la diffraction des rayons Röntgen. C. R. Acad. Sci. Paris, 157, 1533–1536.
Gornall, W. S. & Stoicheff, B. P. (1970). The Brillouin spectrum and elastic constants of xenon single crystals. Solid State Commun. 8, 1529–1533.
Gross, E. (1930a). Change of wave-length of light due to elastic heat waves at scattering in liquids. Nature (London), 126, 201–202.
Gross, E. (1930b). The splitting of spectral lines at scattering of light by liquids. Nature (London), 126, 400.
Hariharan, P. & Sen, J. (1961). Double-passed Fabry–Perot interferometer. J. Opt. Soc. Am. 51, 398–399.
Hayes, W. & Loudon, R. (1978). Scattering of light by crystals. New York: Wiley.
Hercher, M. (1968). The spherical mirror Fabry–Perot interferometer. Appl. Opt. 7, 951–966.
Mach, J. E., McNutt, D. P., Roessler, F. L. & Chabbal, R. (1963). The PEPSIOS purely interferometric high-resolution scanning spectrometer. I. The pilot model. Appl. Opt. 2, 873–885.
Nelson, D. F. & Lax, M. (1971). Theory of photoelastic interaction. Phys. Rev. B, 3, 2778–2794.
Nelson, D. F., Lazay, P. D. & Lax, M. (1972). Brillouin scattering in anisotropic media: calcite. Phys. Rev. B, 6, 3109–3120.
Pine, A. S. (1972). Thermal Brillouin scattering in cadmium sulfide: velocity and attenuation of sound; acoustoelectric effects. Phys. Rev. B, 5, 2997–3003.
Sandercock, J. R. (1971). The design and use of a stabilised multipassed interferometer of high contrast ratio. In Light scattering in solids, edited by M. Balkanski, pp. 9–12. Paris: Flammarion.
Sandercock, J. R. (1982). Trends in Brillouin scattering: studies of opaque materials, supported films, and central modes. In Light scattering in solids III. Topics in appied physics, Vol. 51, edited by M. Cardona & G. Güntherodt, pp. 173–206. Berlin: Springer.
Sette, F., Krisch, M. H., Masciovecchio, C., Ruocco, G. & Monaco, G. (1998). Dynamics of glasses and glass-forming liquids studied by inelastic X-ray scattering. Science, 280, 1550–1555.
Sussner, H. & Vacher, R. (1979). High-precision measurements of Brillouin scattering frequencies. Appl. Opt. 18, 3815–3818.
Vacher, R. (1972). Contribution à l'étude de la dynamique du réseau cristallin par analyse du spectre de diffusion Brillouin. Doctoral Thesis, University of Montpellier II.
Vacher, R. & Boyer, L. (1972). Brillouin scattering: a tool for the measurement of elastic and photoelastic constants. Phys. Rev. B, 6, 639–673. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8908558487892151, "perplexity": 2028.725777436007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823588.0/warc/CC-MAIN-20181211061718-20181211083218-00435.warc.gz"} |
http://math.stackexchange.com/questions/18668/mapping-from-x-to-s4/18676 | # Mapping from X to $S^4$
I found this question in a book (Topology II: homotopy and homology: classical manifolds)
Show that the quotient space $X = S^2 \times S^2 / [(x_1,x_2) \sim (Rx_1,Rx_2)]$ where R is the reflection in the equatorial plane, is homeomorphic to $S^4$.
I am still in the process of learning topology and I really don't think I can prove this result (or even understand a proof if someone were generous enough to provide me with one). I apologize in advance if this is trivial, but it will be of great help if someone could give a homeomorphism. I would really like to use the mapping of the space $X$ to $S^4$ in my work. If you are aware of the proof and would provide it in your answer, I will definitely make an effort to understand it.
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I think it might be helpful to restrict your $x_1$ coordinate to be on or above the equator. As long as $x_1$ isn't on the equator, this gives a unique representative for every point, so you get $\{x_1\}\times S^2$. When $x_1$ is on the equator, you can apply $R$ freely to the second coordinate, so you get $\{x_1\}\times D^2$. The puzzle is how to fit these together...
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I would add that thinking about why $S^1\times S^1$ modulo simultaneous reflection in each coordinate is equal to $S^2$ is helpful for developing intuition about the question. – Grumpy Parsnip Jan 24 '11 at 13:16
Thanks for the suggestions and comments. I did think about the $S^1 \times S^1$ case and you are right, it does seem very intuitive that $S^1 \times S^1 / (x_1,x_2) \sim (Rx_1,Rx_2)$ be homeomorphic to $S^2$. It seemed obvious enough that I guessed the $S^2 \times S^2$ case would be straightforward. But I could never figure it out. – Srikanth Jan 24 '11 at 17:03
I don't have enough rep points to respond directly to Eric, but that should not be the case, since $S^4$ doesn't have a boundary either. I think you're thinking of $S^2 / (x \tilde{} Rx) \times S^2 / (x \tilde{} Rx),$ which is not exactly the same space as $X$.
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Thanks. I was thinking exactly as you said and just realized it and deleted my comment. – Eric O. Korman Jan 24 '11 at 1:17
Trying to follow up on Aaron's hints.
So one way to look at $X$ is as the product space $D^2 \times S^2$ with an equivalence relation $E$ on the boundary $\partial D^2 \times S^2 = S^1 \times S^2$.
The Equivalence relation $E$ on the boundary is : $({ b_1 \in S^1 }, { b_2 \in S^2 }) \sim ({ b_1 \in S^1 }, { Rb_2 \in S^2 })$
SO $X = (D^2 \times S^1) /E$.
Now consider the following subspace $Y$ of $X$:
$Y = { [-1,1] \times S^2 }$ with $E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and $E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$ as equivalence relations on $Y$
$I=[-1,1]$ is a line passing through the origin of $D^2$. I think I could prove $Y$ to be homeomorphic to $S^3$.
Please check: This is probably a very crude way to show the homeomorphism, but this is the only way I know how to do this. May be there are some algebraic methods to prove this result ?
Visualize $Y$ as a combination two solid cylinders of $I \times D^2$. So I have two solid cylinders, $C_1$ and $C_2$, when joined along the cylindrical surfaces form $Y$. So each point on the cylindrical surface of $C_1$ has an equivalent point on the cylindrical surface of $C_2$, since they belong to the same sphere $S^2$.
The disks that belong to the top and bottom surfaces of cylinder $C_1$ are identified with the disks of cylinder $C_2$ due to equivalence relations $E_1,E_2$ on the boundaries of subspace $I \times S^2$ that it inherits from $E$. The relations are:
$E_1 = (-1;y \in S^2) \sim (-1;Ry \in S^2)$ and $E_2 = (1;y \in S^2) \sim (1;Ry \in S^2)$. \
Hence I have two cylinders $C_1$ and $C_2$ where any point on the surface of $C_1$ has one and only one equivalent point on the surface of $C_2$. A solid cylinders is homeomorphic to a solid ball, hence the subspace $Y$ is equivalent to two solid spheres where the surface points of one solid sphere are identified with surface points of the other solid sphere. This space is equivalent to $S^3$.
May be there is some way to get $S^4$ by extending the subspace $(I \times S^2)/[E_1,E_2]$ to the entire $(D^2 \times S^2)/E$ ??
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https://au.mathematicstip.com/5773-147-maximum-and-minimum-values-mathematics.html | # 14.7: Maximum and Minimum Values - Mathematics
Learning Objectives
• Use partial derivatives to locate critical points for a function of two variables.
• Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
• Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.
One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.
## Critical Points
For functions of a single variable, we defined critical points as the values of the variable at which the function's derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.
Definition: Critical Points
Let (z=f(x,y)) be a function of two variables that is differentiable on an open set containing the point ((x_0,y_0)). The point ((x_0,y_0)) is called a critical point of a function of two variables (f) if one of the two following conditions holds:
1. (f_x(x_0,y_0)=f_y(x_0,y_0)=0)
2. Either (f_x(x_0,y_0) ; ext{or} ; f_y(x_0,y_0)) does not exist.
Example (PageIndex{1}): Finding Critical Points
Find the critical points of each of the following functions:
1. (f(x,y)=sqrt{4y^2−9x^2+24y+36x+36})
2. (g(x,y)=x^2+2xy−4y^2+4x−6y+4)
Solution
a. First, we calculate (f_x(x,y) ; ext{and} ; f_y(x,y):)
[egin{align*} f_x(x,y) &=dfrac{1}{2}(−18x+36)(4y^2−9x^2+24y+36x+36)^{−1/2} [4pt] &=dfrac{−9x+18}{sqrt{4y^2−9x^2+24y+36x+36}} end{align*}]
[egin{align*} f_y(x,y) &=dfrac{1}{2}(8y+24)(4y^2−9x^2+24y+36x+36)^{−1/2} [4pt] &=dfrac{4y+12}{sqrt{4y^2−9x^2+24y+36x+36}} end{align*}.]
Next, we set each of these expressions equal to zero:
[egin{align*} dfrac{−9x+18}{sqrt{4y^2−9x^2+24y+36x+36}} &=0 [4pt] dfrac{4y+12}{sqrt{4y^2−9x^2+24y+36x+36}} &=0. end{align*}]
Then, multiply each equation by its common denominator:
[egin{align*} −9x+18 &=0 [4pt] 4y+12 &=0. end{align*}]
Therefore, (x=2) and (y=−3,) so ((2,−3)) is a critical point of (f).
We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:
[4y^2−9x^2+24y+36x+36=0. label{critical1}]
Equation ef{critical1} represents a hyperbola. We should also note that the domain of (f) consists of points satisfying the inequality
[4y^2−9x^2+24y+36x+36≥0.]
Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:
[egin{align*} 4y^2−9x^2+24y+36x+36 &=0 [4pt] 4y^2−9x^2+24y+36x &=−36 [4pt] 4y^2+24y−9x^2+36x &=−36 [4pt] 4(y^2+6y)−9(x^2−4x) &=−36 [4pt] 4(y^2+6y+9)−9(x^2−4x+4) &=−36−36+36 [4pt] 4(y+3)^2−9(x−2)^2 &=−36.end{align*}]
Dividing both sides by (−36) puts the equation in standard form:
[egin{align*} dfrac{4(y+3)^2}{−36}−dfrac{9(x−2)^2}{−36} &=1 [4pt] dfrac{(x−2)^2}{4}−dfrac{(y+3)^2}{9} &=1. end{align*}]
Notice that point ((2,−3)) is the center of the hyperbola.
Thus, the critical points of the function (f) are ( (2, -3) ) and all points on the hyperbola, (dfrac{(x−2)^2}{4}−dfrac{(y+3)^2}{9}=1).
b. First, we calculate (g_x(x,y)) and (g_y(x,y)):
[egin{align*} g_x(x,y) &=2x+2y+4 [4pt] g_y(x,y) &=2x−8y−6. end{align*}]
Next, we set each of these expressions equal to zero, which gives a system of equations in (x) and (y):
[egin{align*} 2x+2y+4 &=0 [4pt] 2x−8y−6 &=0. end{align*}]
Subtracting the second equation from the first gives (10y+10=0), so (y=−1). Substituting this into the first equation gives (2x+2(−1)+4=0), so (x=−1).
Therefore ((−1,−1)) is a critical point of (g). There are no points in (mathbb{R}^2) that make either partial derivative not exist.
Figure (PageIndex{1}) shows the behavior of the surface at the critical point.
Exercise (PageIndex{1})
Find the critical point of the function (f(x,y)=x^3+2xy−2x−4y.)
Hint
Calculate (f_x(x,y)) and (f_y(x,y)), then set them equal to zero.
The only critical point of (f) is ((2,−5)).
The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.
Definition: Global and Local Extrema
Let (z=f(x,y)) be a function of two variables that is defined and continuous on an open set containing the point ((x_0,y_0).) Then (f) has a local maximum at ((x_0,y_0)) if
[f(x_0,y_0)≥f(x,y)]
for all points ((x,y)) within some disk centered at ((x_0,y_0)). The number (f(x_0,y_0)) is called a local maximum value. If the preceding inequality holds for every point ((x,y)) in the domain of (f), then (f) has a global maximum (also called an absolute maximum) at ((x_0,y_0).)
The function (f) has a local minimum at ((x_0,y_0)) if
[f(x_0,y_0)≤f(x,y)]
for all points ((x,y)) within some disk centered at ((x_0,y_0)). The number (f(x_0,y_0)) is called a local minimum value. If the preceding inequality holds for every point ((x,y)) in the domain of (f), then (f) has a global minimum (also called an absolute minimum) at ((x_0,y_0)).
If (f(x_0,y_0)) is either a local maximum or local minimum value, then it is called a local extremum (see the following figure).
In Calculus 1, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.
Fermat’s Theorem for Functions of Two Variables
Let (z=f(x,y)) be a function of two variables that is defined and continuous on an open set containing the point ((x_0,y_0)). Suppose (f_x) and (f_y) each exist at ((x_0,y_0)). If f has a local extremum at ((x_0,y_0)), then ((x_0,y_0)) is a critical point of (f).
## Second Derivative Test
Consider the function (f(x)=x^3.) This function has a critical point at (x=0), since (f'(0)=3(0)^2=0). However, (f) does not have an extreme value at (x=0). Therefore, the existence of a critical value at (x=x_0) does not guarantee a local extremum at (x=x_0). The same is true for a function of two or more variables. One way this can happen is at a saddle point. An example of a saddle point appears in the following figure.
Figure (PageIndex{3}): Graph of the function (z=x^2−y^2). This graph has a saddle point at the origin.
In this graph, the origin is a saddle point. This is because the first partial derivatives of f((x,y)=x^2−y^2) are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to (y=0) is (z=x^2) (a parabola opening upward), but the vertical trace corresponding to (x=0) is (z=−y^2) (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.
Given the function (z=f(x,y),) the point (ig(x_0,y_0,f(x_0,y_0)ig)) is a saddle point if both (f_x(x_0,y_0)=0) and (f_y(x_0,y_0)=0), but (f) does not have a local extremum at ((x_0,y_0).)
The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant (D) that replaces (f''(x_0)) in the second derivative test for a function of one variable.
Second Derivative Test
Let (z=f(x,y)) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ((x_0,y_0)). Suppose (f_x(x_0,y_0)=0) and (f_y(x_0,y_0)=0.) Define the quantity
[D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−ig(f_{xy}(x_0,y_0)ig)^2.]
Then:
1. If (D>0) and (f_{xx}(x_0,y_0)>0), then f has a local minimum at ((x_0,y_0)).
2. If (D>0) and (f_{xx}(x_0,y_0)<0), then f has a local maximum at ((x_0,y_0)).
3. If (D<0), then (f) has a saddle point at ((x_0,y_0)).
4. If (D=0), then the test is inconclusive.
See Figure (PageIndex{4}).
To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.
Problem-Solving Strategy: Using the Second Derivative Test for Functions of Two Variables
Let (z=f(x,y)) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ((x_0,y_0).) To apply the second derivative test to find local extrema, use the following steps:
1. Determine the critical points ((x_0,y_0)) of the function (f) where (f_x(x_0,y_0)=f_y(x_0,y_0)=0.) Discard any points where at least one of the partial derivatives does not exist.
2. Calculate the discriminant (D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)−ig(f_{xy}(x_0,y_0)ig)^2) for each critical point of (f).
3. Apply the four cases of the test to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.
Example (PageIndex{2}): Using the Second Derivative Test
Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:
1. (f(x,y)=4x^2+9y^2+8x−36y+24)
2. (g(x,y)=dfrac{1}{3}x^3+y^2+2xy−6x−3y+4)
Solution
a. Step 1 of the problem-solving strategy involves finding the critical points of (f). To do this, we first calculate (f_x(x,y)) and (f_y(x,y)), then set each of them equal to zero:
[egin{align*} f_x(x,y) &=8x+8 [4pt] f_y(x,y) &=18y−36. end{align*}]
Setting them equal to zero yields the system of equations
[egin{align*} 8x+8 &=0 [4pt] 18y−36 &=0. end{align*}]
The solution to this system is (x=−1) and (y=2). Therefore ((−1,2)) is a critical point of (f).
Step 2 of the problem-solving strategy involves calculating (D.) To do this, we first calculate the second partial derivatives of (f:)
[egin{align*} f_{xx}(x,y) &=8 [4pt] f_{xy}(x,y) &=0 [4pt] f_{yy}(x,y) &=18. end{align*}]
Therefore, (D=f_{xx}(−1,2)f_{yy}(−1,2)−ig(f_{xy}(−1,2)ig)^2=(8)(18)−(0)^2=144.)
Step 3 states to apply the four cases of the test to classify the function's behavior at this critical point.
Since (D>0) and (f_{xx}(−1,2)>0,) this corresponds to case 1. Therefore, (f) has a local minimum at ((−1,2)) as shown in the following figure.
Figure (PageIndex{5}): The function (f(x,y)) has a local minimum at ((−1,2,−16).) Note the scale on the (y)-axis in this plot is in thousands.
b. For step 1, we first calculate (g_x(x,y)) and (g_y(x,y)), then set each of them equal to zero:
[egin{align*} g_x(x,y) &=x^2+2y−6 [4pt] g_y(x,y) &=2y+2x−3. end{align*}]
Setting them equal to zero yields the system of equations
[egin{align*} x^2+2y−6 &=0 [4pt] 2y+2x−3 &=0. end{align*}]
To solve this system, first solve the second equation for (y). This gives (y=dfrac{3−2x}{2}). Substituting this into the first equation gives
[egin{align*} x^2+3−2x−6 &=0 [4pt] x^2−2x−3 &=0 [4pt] (x−3)(x+1) &=0. end{align*}]
Therefore, (x=−1) or (x=3). Substituting these values into the equation (y=dfrac{3−2x}{2}) yields the critical points (left(−1,frac{5}{2} ight)) and (left(3,−frac{3}{2} ight)).
Step 2 involves calculating the second partial derivatives of (g):
[egin{align*} g_{xx}(x,y) &=2x [4pt] g_{xy}(x,y) &=2[4pt] g_{yy}(x,y) &=2. end{align*}]
Then, we find a general formula for (D):
[egin{align*} D(x_0, y_0) &=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)−ig(g_{xy}(x_0,y_0)ig)^2 [4pt] &=(2x_0)(2)−2^2[4pt] &=4x_0−4.end{align*}]
Next, we substitute each critical point into this formula:
[egin{align*} Dleft(−1, frac{5}{2} ight) &=(2(−1))(2)−(2)^2=−4−4=−8 [4pt] Dleft(3,− frac{3}{2} ight) &=(2(3))(2)−(2)^2=12−4=8. end{align*}]
In step 3, we note that, applying Note to point (left(−1,frac{5}{2} ight)) leads to case (3), which means that (left(−1,frac{5}{2} ight)) is a saddle point. Applying the theorem to point (left(3,−frac{3}{2} ight)) leads to case (1), which means that (left(3,−frac{3}{2} ight)) corresponds to a local minimum as shown in the following figure.
Exercise (PageIndex{2})
Use the second derivative test to find the local extrema of the function
[ f(x,y)=x^3+2xy−6x−4y^2. onumber]
Hint
Follow the problem-solving strategy for applying the second derivative test.
(left(frac{4}{3},frac{1}{3} ight)) is a saddle point, (left(−frac{3}{2},−frac{3}{8} ight)) is a local maximum.
## Absolute Maxima and Minima
When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.
Extreme Value Theorem
A continuous function (f(x,y)) on a closed and bounded set (D) in the plane attains an absolute maximum value at some point of (D) and an absolute minimum value at some point of (D).
Now that we know any continuous function (f) defined on a closed, bounded set attains its extreme values, we need to know how to find them.
Finding Extreme Values of a Function of Two Variables
Assume (z=f(x,y)) is a differentiable function of two variables defined on a closed, bounded set (D). Then (f) will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:
1. The values of (f) at the critical points of (f) in (D).
2. The values of (f) on the boundary of (D).
The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of (D), then it is located at an interior point of (D). But an interior point ((x_0,y_0)) of (D) that’s an absolute extremum is also a local extremum; hence, ((x_0,y_0)) is a critical point of (f) by Fermat’s theorem. Therefore the only possible values for the global extrema of (f) on (D) are the extreme values of (f) on the interior or boundary of (D).
Problem-Solving Strategy: Finding Absolute Maximum and Minimum Values
Let (z=f(x,y)) be a continuous function of two variables defined on a closed, bounded set (D), and assume (f) is differentiable on (D). To find the absolute maximum and minimum values of (f) on (D), do the following:
1. Determine the critical points of (f) in (D).
2. Calculate (f) at each of these critical points.
3. Determine the maximum and minimum values of (f) on the boundary of its domain.
4. The maximum and minimum values of (f) will occur at one of the values obtained in steps (2) and (3).
Finding the maximum and minimum values of (f) on the boundary of (D) can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in Example (PageIndex{3}). The same approach can be used for other shapes such as circles and ellipses.
If the boundary of the set (D) is a more complicated curve defined by a function (g(x,y)=c) for some constant (c), and the first-order partial derivatives of (g) exist, then the method of Lagrange multipliers can prove useful for determining the extrema of (f) on the boundary which is introduced in Lagrange Multipliers.
Example (PageIndex{3}): Finding Absolute Extrema
Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:
1. (f(x,y)=x^2−2xy+4y^2−4x−2y+24) on the domain defined by (0≤x≤4) and (0≤y≤2)
2. (g(x,y)=x^2+y^2+4x−6y) on the domain defined by (x^2+y^2≤16)
Solution
a. Using the problem-solving strategy, step (1) involves finding the critical points of (f) on its domain. Therefore, we first calculate (f_x(x,y)) and (f_y(x,y)), then set them each equal to zero:
[egin{align*} f_x(x,y) &=2x−2y−4 [4pt] f_y(x,y) &=−2x+8y−2. end{align*}]
Setting them equal to zero yields the system of equations
[egin{align*} 2x−2y−4 &=0[4pt] −2x+8y−2 &=0. end{align*}]
The solution to this system is (x=3) and (y=1). Therefore ((3,1)) is a critical point of (f). Calculating (f(3,1)) gives (f(3,1)=17.)
The next step involves finding the extrema of (f) on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:
(L_1) is the line segment connecting ((0,0)) and ((4,0)), and it can be parameterized by the equations (x(t)=t,y(t)=0) for (0≤t≤4). Define (g(t)=fig(x(t),y(t)ig)). This gives (g(t)=t^2−4t+24). Differentiating (g) leads to (g′(t)=2t−4.) Therefore, (g) has a critical value at (t=2), which corresponds to the point ((2,0)). Calculating (f(2,0)) gives the (z)-value (20).
(L_2) is the line segment connecting ((4,0)) and ((4,2)), and it can be parameterized by the equations (x(t)=4,y(t)=t) for (0≤t≤2.) Again, define (g(t)=fig(x(t),y(t)ig).) This gives (g(t)=4t^2−10t+24.) Then, (g′(t)=8t−10). g has a critical value at (t=frac{5}{4}), which corresponds to the point (left(0,frac{5}{4} ight).) Calculating (fleft(0,frac{5}{4} ight)) gives the (z)-value (27.75).
(L_3) is the line segment connecting ((0,2)) and ((4,2)), and it can be parameterized by the equations (x(t)=t,y(t)=2) for (0≤t≤4.) Again, define (g(t)=fig(x(t),y(t)ig).) This gives (g(t)=t^2−8t+36.) The critical value corresponds to the point ((4,2).) So, calculating (f(4,2)) gives the (z)-value (20).
(L_4) is the line segment connecting ((0,0)) and ((0,2)), and it can be parameterized by the equations (x(t)=0,y(t)=t) for (0≤t≤2.) This time, (g(t)=4t^2−2t+24) and the critical value (t=frac{1}{4}) correspond to the point (left(0,frac{1}{4} ight)). Calculating (fleft(0,frac{1}{4} ight)) gives the (z)-value (23.75.)
We also need to find the values of (f(x,y)) at the corners of its domain. These corners are located at ((0,0),(4,0),(4,2)) and ((0,2)):
[egin{align*} f(0,0) &=(0)^2−2(0)(0)+4(0)^2−4(0)−2(0)+24=24 [4pt] f(4,0) &=(4)^2−2(4)(0)+4(0)^2−4(4)−2(0)+24=24 [4pt] f(4,2) &=(4)^2−2(4)(2)+4(2)^2−4(4)−2(2)+24=20[4pt] f(0,2) &=(0)^2−2(0)(2)+4(2)^2−4(0)−2(2)+24=36. end{align*}]
The absolute maximum value is (36), which occurs at ((0,2)), and the global minimum value is (20), which occurs at both ((4,2)) and ((2,0)) as shown in the following figure.
b. Using the problem-solving strategy, step (1) involves finding the critical points of (g) on its domain. Therefore, we first calculate (g_x(x,y)) and (g_y(x,y)), then set them each equal to zero:
[egin{align*} g_x(x,y) &=2x+4 [4pt] g_y(x,y) &=2y−6. end{align*}]
Setting them equal to zero yields the system of equations
[egin{align*} 2x+4 &=0 [4pt] 2y−6 &=0. end{align*}]
The solution to this system is (x=−2) and (y=3). Therefore, ((−2,3)) is a critical point of (g). Calculating (g(−2,3),) we get
[g(−2,3)=(−2)^2+3^2+4(−2)−6(3)=4+9−8−18=−13.]
The next step involves finding the extrema of g on the boundary of its domain. The boundary of its domain consists of a circle of radius (4) centered at the origin as shown in the following graph.
The boundary of the domain of (g) can be parameterized using the functions (x(t)=4cos t,, y(t)=4sin t) for (0≤t≤2π). Define (h(t)=gig(x(t),y(t)ig):)
[egin{align*} h(t) &=gig(x(t),y(t)ig) [4pt] &=(4cos t)^2+(4sin t)^2+4(4cos t)−6(4sin t) [4pt] &=16cos^2t+16sin^2t+16cos t−24sin t[4pt] &=16+16cos t−24sin t. end{align*}]
Setting (h′(t)=0) leads to
[egin{align*} −16sin t−24cos t &=0 [4pt] −16sin t &=24cos t[4pt] dfrac{−16sin t}{−16cos t} &=dfrac{24cos t}{−16cos t} [4pt] an t &=−dfrac{3}{2}. end{align*}]
This equation has two solutions over the interval (0≤t≤2π). One is (t=π−arctan (frac{3}{2})) and the other is (t=2π−arctan (frac{3}{2})). For the first angle,
[egin{align*} sin t &=sin(π−arctan( frac{3}{2}))=sin (arctan ( frac{3}{2}))=dfrac{3sqrt{13}}{13} [4pt] cos t &=cos (π−arctan ( frac{3}{2}))=−cos (arctan ( frac{3}{2}))=−dfrac{2sqrt{13}}{13}. end{align*}]
Therefore, (x(t)=4cos t =−frac{8sqrt{13}}{13}) and (y(t)=4sin t=frac{12sqrt{13}}{13}), so (left(−frac{8sqrt{13}}{13},frac{12sqrt{13}}{13} ight)) is a critical point on the boundary and
[egin{align*} gleft(− frac{8sqrt{13}}{13}, frac{12sqrt{13}}{13} ight) &=left(− frac{8sqrt{13}}{13} ight)^2+left( frac{12sqrt{13}}{13} ight)^2+4left(− frac{8sqrt{13}}{13} ight)−6left( frac{12sqrt{13}}{13} ight) [4pt] &=frac{144}{13}+frac{64}{13}−frac{32sqrt{13}}{13}−frac{72sqrt{13}}{13} [4pt] &=frac{208−104sqrt{13}}{13}≈−12.844. end{align*}]
For the second angle,
[egin{align*} sin t &=sin (2π−arctan ( frac{3}{2}))=−sin (arctan ( frac{3}{2}))=−dfrac{3sqrt{13}}{13} [4pt] cos t &=cos (2π−arctan ( frac{3}{2}))=cos (arctan ( frac{3}{2}))=dfrac{2sqrt{13}}{13}. end{align*}]
Therefore, (x(t)=4cos t=frac{8sqrt{13}}{13}) and (y(t)=4sin t=−frac{12sqrt{13}}{13}), so (left(frac{8sqrt{13}}{13},−frac{12sqrt{13}}{13} ight)) is a critical point on the boundary and
[egin{align*} gleft( frac{8sqrt{13}}{13},− frac{12sqrt{13}}{13} ight) &=left( frac{8sqrt{13}}{13} ight)^2+left(− frac{12sqrt{13}}{13} ight)^2+4left( frac{8sqrt{13}}{13} ight)−6left(− frac{12sqrt{13}}{13} ight) [4pt] &=dfrac{144}{13}+dfrac{64}{13}+dfrac{32sqrt{13}}{13}+dfrac{72sqrt{13}}{13}[4pt] &=dfrac{208+104sqrt{13}}{13}≈44.844. end{align*}]
The absolute minimum of (g) is (−13,) which is attained at the point ((−2,3)), which is an interior point of (D). The absolute maximum of (g) is approximately equal to 44.844, which is attained at the boundary point (left(frac{8sqrt{13}}{13},−frac{12sqrt{13}}{13} ight)). These are the absolute extrema of (g) on (D) as shown in the following figure.
Exercise (PageIndex{3}):
Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function
[f(x,y)=4x^2−2xy+6y^2−8x+2y+3 onumber]
on the domain defined by (0≤x≤2) and (−1≤y≤3.)
Hint
Calculate (f_x(x,y)) and (f_y(x,y)), and set them equal to zero. Then, calculate (f) for each critical point and find the extrema of (f) on the boundary of (D).
The absolute minimum occurs at ((1,0): f(1,0)=−1.)
The absolute maximum occurs at ((0,3): f(0,3)=63.)
Example (PageIndex{4}): Profitable Golf Balls
Pro-(T) company has developed a profit model that depends on the number (x) of golf balls sold per month (measured in thousands), and the number of hours per month of advertising (y), according to the function
[ z=f(x,y)=48x+96y−x^2−2xy−9y^2,]
where (z) is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is (50,000), and the maximum number of hours of advertising that can be purchased is (25). Find the values of (x) and (y) that maximize profit, and find the maximum profit.
Solution
Using the problem-solving strategy, step (1) involves finding the critical points of (f) on its domain. Therefore, we first calculate (f_x(x,y)) and (f_y(x,y),) then set them each equal to zero:
[egin{align*} f_x(x,y) &=48−2x−2y [4pt] f_y(x,y) &=96−2x−18y. end{align*}]
Setting them equal to zero yields the system of equations
[egin{align*} 48−2x−2y &=0 [4pt] 96−2x−18y &=0. end{align*}]
The solution to this system is (x=21) and (y=3). Therefore ((21,3)) is a critical point of (f). Calculating (f(21,3)) gives (f(21,3)=48(21)+96(3)−21^2−2(21)(3)−9(3)^2=648.)
The domain of this function is (0≤x≤50) and (0≤y≤25) as shown in the following graph.
(L_1) is the line segment connecting ((0,0)) and ((50,0),) and it can be parameterized by the equations (x(t)=t,y(t)=0) for (0≤t≤50.) We then define (g(t)=f(x(t),y(t)):)
[egin{align*} g(t) &=f(x(t),y(t)) [4pt] &=f(t,0)[4pt] &=48t+96(0)−y^2−2(t)(0)−9(0)^2[4pt] &=48t−t^2. end{align*}]
Setting (g′(t)=0) yields the critical point (t=24,) which corresponds to the point ((24,0)) in the domain of (f). Calculating (f(24,0)) gives (576.)
(L_2) is the line segment connecting ((50,0)) and ((50,25)), and it can be parameterized by the equations (x(t)=50,y(t)=t) for (0≤t≤25). Once again, we define (g(t)=fig(x(t),y(t)ig):)
[egin{align*} g(t) &=fig(x(t),y(t)ig)[4pt] &=f(50,t)[4pt] &=48(50)+96t−50^2−2(50)t−9t^2 [4pt] &=−9t^2−4t−100. end{align*}]
This function has a critical point at (t =−frac{2}{9}), which corresponds to the point ((50,−29)). This point is not in the domain of (f).
(L_3) is the line segment connecting ((0,25)) and ((50,25)), and it can be parameterized by the equations (x(t)=t,y(t)=25) for (0≤t≤50). We define (g(t)=fig(x(t),y(t)ig)):
[egin{align*} g(t) &=fig(x(t),y(t)ig)[4pt] &=f(t,25) [4pt] &=48t+96(25)−t^2−2t(25)−9(25^2) [4pt] &=−t^2−2t−3225. end{align*}]
This function has a critical point at (t=−1), which corresponds to the point ((−1,25),) which is not in the domain.
(L_4) is the line segment connecting ((0,0)) to ((0,25)), and it can be parameterized by the equations (x(t)=0,y(t)=t) for (0≤t≤25). We define (g(t)=fig(x(t),y(t)ig)):
[egin{align*} g(t) &=fig(x(t),y(t)ig) [4pt] &=f(0,t) [4pt] &=48(0)+96t−(0)^2−2(0)t−9t^2 [4pt] &=96t−9t^2. end{align*}]
This function has a critical point at (t=frac{16}{3}), which corresponds to the point (left(0,frac{16}{3} ight)), which is on the boundary of the domain. Calculating (fleft(0,frac{16}{3} ight)) gives (256).
We also need to find the values of (f(x,y)) at the corners of its domain. These corners are located at ((0,0),(50,0),(50,25)) and ((0,25)):
[egin{align*} f(0,0) &=48(0)+96(0)−(0)^2−2(0)(0)−9(0)^2=0[4pt] f(50,0) &=48(50)+96(0)−(50)^2−2(50)(0)−9(0)^2=−100 [4pt] f(50,25) &=48(50)+96(25)−(50)^2−2(50)(25)−9(25)^2=−5825 [4pt] f(0,25) &=48(0)+96(25)−(0)^2−2(0)(25)−9(25)^2=−3225. end{align*}]
## MAXIMUM AND MINIMUMVALUES
F INDING a maximum or a minimum (Lesson 10) has its application in pure mathematics, where for example we could find the largest rectangle that has a given perimeter. It also has its application to commercial problems, such as finding the least dimensions of a carton that is to contain a given volume.
Example 1. Find the dimensions of the rectangle that, for a given perimeter, will have the largest area.
Solution . Let the base of the rectangle be x , let its height be y , let A be its area, and let P be the given perimeter. Then
Since we are going to maximize A , we would like to have A as a function only of x . And we can do that because in the expression for P we can solve for y :
A = x (½ P &minus x ) = ½ x P &minus x 2 . On taking the derivative of A and setting it equal to 0, dA dx = ½ P &minus 2 x = 0, x = ¼ P .
The base is one quarter of the perimeter. We can now find the value of y :
The height is also one quarter of the perimeter. That figure is a square The rectangle that has the largest area for a given perimeter
is a square.
(Note: The value we found is a maximum because the second derivative is negative.)
All maximum-minimum problems follow this same procedure:
&bull Write the function whose maximum or minimum value is to be determined . (In the Example, we wrote A = xy .) &bull The resulting expression will typically contain more than one variable. Use the information given in the problem to express every variable in terms of a single variable. (In the Example, we expressed y in terms of x .) &bull Find the critical value of that single variable by taking the derivative and setting it equal to 0. (In the Example, we took the derivative of A with respect to x .) &bull If necessary, determine the values of the other variables. (In the Example, we evaluated y by substituting the critical value of x .)
In the following, notice how we follow these steps.
Example 2. A box having a square base and an open top is to contain 108 cubic feet. What should its dimensions be so that the material to make it will be a minimum? That is, what dimensions will cost the least?
Solution . Let x be the side of the square base, and let y be its height. Then
Area of base = x 2 . Area of four sides = 4 xy . Let M be the total amount of material. Then M = x 2 + 4 xy .
Now, how shall we express y in terms of x ?
We have not yet used the fact that the volume must be 108 cubic feet. The volume is equal to
y = 108 x 2 and therefore in the expression for M , 4 xy = 4 x · 108 x 2 = M = = This implies, on multiplying through by the denominator x 2 : 2 x 3 &minus 432 = 0 x 3 = 216 x = 6 feet. We can now evaluate y : y =
These are the dimensions that will cost the least.
Example 3. Find the dimensions of the rectangle with the most area that can be inscribed in a semi-circle of radius r . Show, in fact, that the area of that rectangle is r 2 .
Solution . First, it should be clear that there is a rectangle with the
greatest area, as the figures above show.
Let x be the base of the rectangle, and let y be its height. Then, since r is the radius:
= r 2 = r 2 x 2 + 4 y 2 = 4 r 2 . Therefore, y =
Let A be the area we want to maximize. A = xy . That is,
A = According to the product rule and the chain rule (Example 1): dA dx =
&minus x 2 + (4 r 2 &minus x 2 ) = 0. This implies: x 2 = 2 r 2 x = This is the base of the largest rectangle. As for the height y : y = y = y =
The area of this largest rectangle, then, is
Problem 1. Find two numbers whose sum is 42 and whose product will be the largest.
(Hint: Call the two numbers x and y . For convenience, call the product something. You will then have two equations in two unknowns. Express the product as function of a single variable, and find its maximum.)
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
The two numbers are 21 and 21.
Problem 2. You have a given length of fence. Using the wall of a house as one side of a rectangular fence, how would you place the fence around the other three sides in order to enclose the largest possible area?
Place half the fence parallel to the house.
Problem 3. Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r . Show, in fact, that that area will be 2 r 2 .
That figure is a square. Each side of the square is r .
Problem 4. A can is to be constructed in the form of a right circular cylinder. If it is to contain a given volume V , what dimensions will require the least amount of material?
Show, in fact, that the height h of the can must equal its width, which is twice the radius r .
First, the top and bottom of the can are each a circle. And the area of the lateral surface is equivalent to a rectangle whose dimensions are 2 &pi r × h . Therefore,
The area A of the material = 2 &pi r 2 + 2 &pi rh .
From the formula for V , express h in terms of r.
A therefore has a minimum at
Compare Lesson 29 of Algebra, Problem 2.
The height of the can must equal its width, which is 2 r .
Problem 5. Find the volume V of the largest right circular cone that can be inscribed in a sphere of radius r .
V = 1 3 (area of the base) · (height).
Let P be the center of the sphere of radius r . Let APB be the height of the cone, and call that height h . Then PB = h &minus r .
Let s be the radius of the cone. Then
We want to maximize V as a function of h alone. Therefore we must express s in terms of h and the constant r .
Now, s and h &minus r are the sides of a right triangle. Therefore,
s 2 + ( h &minus r ) 2 = r 2 s 2 + h 2 &minus 2 hr + r 2 = r 2 s 2 = 2 hr &minus h 2 .
V = 1 3 &pi (2 hr &minus h 2 ) h . = &pi 3 (2 h 2 r &minus h 3 ). dV dh = &pi 3 (4 h r &minus 3 h 2 ).
V therefore has a maximum at
Upon substituting that value for h in the expression for V above, that maximum volume is:
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Even $1 will help. ## IB Mathematics HL SL-Maxima and Minima In the previous post, IB Maths Tutors discussed how to find the equation of tangents and normal to a curve. There are a few more Applications of Derivatives in IB Mathematics HL SL, ‘Maxima and Minima’ is one of them. Maxima and Minima:- 1. A function f(x) is said to have a maximum at x = a if f(a) is greater than every other value assumed by f(x) in the immediate neighbourhood of x = a. Symbolically gives maxima for a sufficiently small positive h. Similarly, a function f(x) is said to have a minimum value at x = b if f(b) is least than every other value assumed by f(x) in the immediate neighbourhood at x = b. Symbolically If x = b gives minima for a sufficiently small positive h. The maximum & minimum values of a function are also known as local/relative maxima or local/relative minima as these are the greatest & least values of the function relative to some neighborhood of the point in question. The term ‘extremum’ or (extremal) or ‘turning value’ is used both for maximum or a minimum value. A maximum (minimum) value of a function may not be the greatest (least) value in a finite interval. A function can have several maximum & minimum values & a minimum value may even be greater than a maximum value. Maximum & minimum values of a continuous function occur alternately & between two consecutive maximum values, there is a minimum value & vice versa. 2. A Necessary Condition For Maximum & Minimum:- If f(x) is a maximum or minimum at x = c & if f'(c) exists then f'(c) = 0. The set of values of x for which f'(x) = 0 are often called stationary points or critical points . The rate of change of function is zero at a stationary point. In IB Mathematics HL SL questions are asked on these points In case f'(c) does not exist f(c) may be a maximum or a minimum & in this case left hand and right-hand derivatives are of opposite signs. The greatest (global maxima) and the least (global minima) values of a function f in an interval [a, b] are f(a) or f(b) or are given by the values of x for which f'(x) = 0. Critical points are those where f'(x)= 0 if it exists, or it fails to exist either by virtue of a vertical tangent or by virtue of a geometrical sharp corner but not because of discontinuity of function. 3. Sufficient Condition For Extreme Values:- x=c is a point of local maxima where f'(c)=0 similarly, x=c is a point of local maxima where f'(c)=0. Here ‘h’ is a sufficiently small positive quantity. If f'(x) does not change sign i.e. has the same sign in a certain complete neighbourhood of c, then f(x) is either strictly increasing or decreasing throughout this neighborhood implying that f(c) is not an extreme value of the given function. 4.Use Of Second Order Derivative In Ascertaining The Maxima Or Minima:- (a) f(c) is a minimum value of the function f, if f'(c) = 0 & f”(c) > 0. (b) f(c) is a maximum value of the function f, if f'(c) = 0 & f”(c) < 0. If f'(c) = 0 then the test fails. Revert back to the first order derivative check for ascertaining the maxima or minima. If y = f make the function (x) is a quantity to be maximum or minimum, find those values of x for which f'(x)=0 Test each value of x for which f'(x) = 0 to determine whether it provides a maximum or minimum or neither. The usual tests are : (a) If is positive when y is minimum. If is negative when y is maximum. If when the test fails. (b) If is positive for then a maximum occurs at (c)If is zero for then a maximum occurs at (d)If is negative for then a maximum occurs at (e) But if dy/dx changes sign from negative to zero to positive as x advances through xo there is a minimum. If dy/dx does not change sign, neither a maximum nor a minimum. Such points are called Points of Inflection. Point of inflexion is a point where the shape of f(x) changes from concave to convex or convex to concave. This concept is usually asked in IB Mathematics HL SL If the derivative fails to exist at some point, examine this point as possible maximum or minimum. If the function y = f (x) is defined for only a limited range of values then we should examine x = a & x=b for possible extreme values. If the derivative fails to exist at some point, we should examine this point as possible maximum or minimum. Important Notes -: Given a fixed point A(x1, y1) and a moving point P(x, f (x)) on the curve y = f(x). Then AP will be maximum or minimum if it is normal to the curve at P If the sum of two positive numbers x and y is constant than their product is maximum if they are equal, i.e. x + y = c , x > 0 , y > 0 , then xy = [(x + y) 2 – (x – y) 2 ] If the product of two positive numbers is constant then their sum is least if they are equal.i.e. (x+y) 2 =(x-y) 2 + 4xy 6. Useful Formulae Of Mensuration To Remember a. The volume of a cuboid = lbh b. Surface area of a cuboid = 2 (lb + bh + hl) c. The volume of a prism = area of the base x height. d. The lateral surface of a prism = perimeter of the base x height. e. The total surface of a prism = lateral surface + 2 area of the base (Note that lateral surfaces of a prism are all rectangles) f. ## 14.7: Maximum and Minimum Values - Mathematics Statistics: double gsl_stats_max (const double data [], size_t stride , size_t n ) This function returns the maximum value in data , a dataset of length n with stride stride . The maximum value is defined as the value of the element @math which satisfies @c <$x_i ge x_j$>@math for all @math. If you want instead to find the element with the largest absolute magnitude you will need to apply fabs or abs to your data before calling this function. Statistics: double gsl_stats_min (const double data [], size_t stride , size_t n ) This function returns the minimum value in data , a dataset of length n with stride stride . The minimum value is defined as the value of the element @math which satisfies @c <$x_i le x_j\$>@math for all @math.
If you want instead to find the element with the smallest absolute magnitude you will need to apply fabs or abs to your data before calling this function.
Statistics: void gsl_stats_minmax (double * min , double * max , const double data [], size_t stride , size_t n ) This function finds both the minimum and maximum values min , max in data in a single pass.
Statistics: size_t gsl_stats_max_index (const double data [], size_t stride , size_t n ) This function returns the index of the maximum value in data , a dataset of length n with stride stride . The maximum value is defined as the value of the element @math which satisfies @math for all @math. When there are several equal maximum elements then the first one is chosen.
Statistics: size_t gsl_stats_min_index (const double data [], size_t stride , size_t n ) This function returns the index of the minimum value in data , a dataset of length n with stride stride . The minimum value is defined as the value of the element @math which satisfies @math for all @math. When there are several equal minimum elements then the first one is chosen.
Statistics: void gsl_stats_minmax_index (size_t * min_index , size_t * max_index , const double data [], size_t stride , size_t n ) This function returns the indexes min_index , max_index of the minimum and maximum values in data in a single pass.
## Maxima and Minima in Calculus
The local maxima are the largest values (maximum) that a function takes in a point within a given neighborhood.
The local minima are the smallest values (minimum), that a function takes in a point within a given neighborhood.
Definition of local maximum and local minimum
A function f has a local maximum (or relative maximum) at c, if f(c) &ge f(x) where x is near c.
Definition of global maximum or global minimum
A function f has a global maximum (or absolute maximum) at c if f(c) &ge f(x) for all x in D, where D is the domain of f. The number f(c) is called the maximum value of f on D.
Similarly, f has a global minimum (or absolute minimum) at c if f(c) &le f(x) for all x in D and the number f(c) is called the minimum value of f on D.
The maximum and minimum values of f are called the extreme values of f
The following diagram illustrates local minimum, global minimum, local maximum, global maximum. Scroll down the page for examples and solutions.
If f has a local maximum or minimum at c, and if f &lsquo(c) exists then
f &lsquo
(c) = 0
Definition of critical number
A critical number of a function f is a number c in the domain of f such that either f &lsquo(c) = 0 of f &lsquo(c) does not exists.
Example:
Find the critical numbers of
So, the critical numbers are and 0.
### The Extreme Value Theorem
If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some number c and d in [a, b]
### The Closed Interval Method
These are the steps to find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:
Step 1: Find the values of f at the critical numbers of f in (a, b).
Step 2: Find the values of f at the endpoints of the interval.
Step 3: The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.
Find the absolute maximum and minimum value of the function
Since f is continuous on , we can use the Closed Interval Method.
So, the critical numbers are x = 0 and x = 2
The values of f at these critical numbers are
f(0) = 1 and f(2) = &ndash3
Step 2: Find the values of f at the endpoints of the interval.
The values of f at the endpoints of the interval are
Step 3: The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.
Comparing the four numbers, we see that the absolute maximum value is f(4) = 17 and the absolute minimum is f(2) = &ndash3.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9072805643081665, "perplexity": 945.9082475756235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00305.warc.gz"} |
http://www.ebroadcast.com.au/lookup/encyclopedia/eu/Euler_characteristic.html | Contents
Euler characteristic
The Euler characteristic of a polyhedron is V - E + F where V, E, and F are respectively the numbers of vertices, edges, and faces. (The first word is pronounced "oiler"; see Leonhard Euler).
The Euler characteristic of any polyhedron homeomorphic to a sphere is 2. For instance, for a cube we have 8 - 12 + 6 = 2 and for a tetrahedron we have 4 - 6 + 4 = 2.
In general, the Euler characteristic is a topological invariant, i.e., any two polyhedra that are homeomorphic to each other have the same Euler characteristic. One can therefore extend the definition to more general surfaces than polyhedra, and speak of the Euler characteristic of, for example, a torus, which would be the Euler characteristic of any polyhedron homeomorphic to a torus. In this sense, a torus has Euler characteristic 0.
One can also define the concept of Euler characteristic of manifolds of dimension other than 2. One approach is to define the Euler characteristic of any simplicial complex as the alternating sum
number of points - number of 1-simplices + number of 2-simplices - number of 3-simplices +...
and then define the Euler characteristic of a manifold as the Euler characteristic of any simplicial complex homeomorphic to it. With this definition, circles and squares have Euler characteristic 0 and solid balls have Euler characteristic 1.
The Euler characteristic χ of a manifold is closely related to its genus g: if the manifold is orientable, we have
$\chi=2-2g$
and if it is not orientable, we have
$\chi=2-g.$
For two-dimensional orientable Riemannian manifolds, the Euler characteristic can also be found by integrating the curvature, see the Gauss-Bonnet theorem.
More generally still, for any topological space, we can define the n-th Betti number[?] bn as the rank of the n-th simplicial homology group. The Euler characteristic can then be defined as the alternating sum
b0 - b1 + b2 - b3 + ...
if these Betti numbers are all finite and zero beyond a certain index n0. Two topological spaces which are homotopy equivalent have isomorphic homology groups and hence the same Euler characteristic. This notion generalizes all the definitions given above and can be extended to other homology theories outside of topology.
The concept of Euler characteristic of a bounded finite partially ordered set is another generalization, important in combinatorics. A poset is "bounded" if it has smallest and largest elements, which let us call 0 and 1. The Euler characteristic of such a poset is μ(0,1), where μ is the Möbius function in that poset's incidence algebra. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9452295899391174, "perplexity": 198.60662591386253}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368702452567/warc/CC-MAIN-20130516110732-00087-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/184052-question-surjective-map.html | # Math Help - Question on a surjective map
1. ## Question on a surjective map
If $f:A \rightarrow B$ is surjective and A is finite, then B is finite. True or false?
I think this is false but my book says it is true.
I think its false by counter-example:
Let A= $(-\frac{\pi}{2},\frac{\pi}{2})$ and let f(x) = tan(x). Then B is infinite?
2. ## Re: Question on a surjective map
But your A is infinite as well despite the assumption.
3. ## Re: Question on a surjective map
Originally Posted by worc3247
If $f:A \rightarrow B$ is surjective and A is finite, then B is finite. True or false?
Easy to prove:
Since A is finite, let h be a finite enumeration of A.
Let f be a surjection from A onto B.
Define a function g from B as follows:
g(y) = the least n such that f(h(n)) = y.
So g is an injection from B into a finite set. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9444687366485596, "perplexity": 1036.3665490601782}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394010856682/warc/CC-MAIN-20140305091416-00052-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://geodesick.com/tag/path-integral-qft/ | # Feynman Rules for Ordinary Integrals
Path integral quantum field theory, perhaps single handedly constructed by Richard Feynman, remains both elusive to undergraduates that wish to study the subject and immensely useful for performing the calculations found in quantum field theory due to the path integral formulation being manifestly symmetric between space and time. Here, we will enter the world of path integral quantum field theory by meticulously performing the calculations so that it is accessible to an advanced undergraduate.
# Tensorial Propagators (Part III)
In Part I, we discovered that tensorial propagators in quantum field theory are often non-invertible which poses an issue when one is trying to determine the form of the propagator. Then, in Part II, we discussed possible techniques involving both symmetries and invariants of the Lagrangian. Here, we aim to reveal how invariant quantities may be useful for propagators, and gauge fields specifically, and we will ultimately derive the form of the photon propagator.
# Tensorial Propagators (Part I)
Here we study some issues with tensorial propagators that are often encountered in the study of quantum field theory. We will use the photon propagator from electromagnetism as an example to guide us through the troublesome calculations.
# Path Integral Free Propagator
In quantum field theory, the propagator gives a probability amplitude for a particle traveling from some point $(t_i,x_i)$ to $(t_f,x_f)$ with a certain energy and momentum. These propagators are the first steps into quantum field theory, I aim to bring these to the masses. The one that is most interesting and easy to grasp is the free propagator for a free field theory. The following derivation is inspired by Anthony Zee’s Quantum Field Theory in a Nutshell of UC Santa Barbara. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9653410315513611, "perplexity": 389.5450080781014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00522.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=100&t=20855&p=59979 | ## Quiz 2
Sue_Park_2D
Posts: 13
Joined: Wed Sep 21, 2016 2:57 pm
### Quiz 2
On the last question for quiz #2, I stated that the rate limiting step was step 2, but got it wrong.
I thought step 1 could not be the rate limiting step because it was at equilibrium, shown by the double arrows between the reactants and products. I don't understand why step 1 would have to be the rate limiting step instead of step 2.
Thank you!
Milan Hirpara 3K
Posts: 20
Joined: Sat Jul 09, 2016 3:00 am
### Re: Quiz 2
Just because the step shows the reaction is at equilibrium does not mean that step is the rate limiting step. Rate limiting step is determined by the slow step, so whichever step had the lower k, rate constant, is going to be the rate determining step.
SoJeong Lee
Posts: 10
Joined: Wed Sep 21, 2016 2:55 pm
### Re: Quiz 2
Does anyone know the answer for quiz 2 #4-c which asks for the rate law?
Amber Duong 1B
Posts: 10
Joined: Wed Sep 21, 2016 2:55 pm
### Re: Quiz 2
The rate limiting step was the first step because its reaction constant was less than the second step, meaning that it was slower. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8587834239006042, "perplexity": 2491.7733390690682}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578528058.3/warc/CC-MAIN-20190419201105-20190419223105-00368.warc.gz"} |
https://www.physicsforums.com/threads/help-with-satellites-and-planets-orbital-motion.554605/ | # Help with satellites and planets orbital motion
1. Nov 27, 2011
### jolynnnicole
A Martian civilization is attempting to develp the capabilities for space travel. They have built a prototype satellite with a mass of 578kg that they plan to put in orbit Mars a distance of 602km above the surface of mars.
A) Using the data provided, calculate the orbital velocity of the satellite.
B)Find the period of the satellite.
C)How would you answer to (a) change if the satellite was twice as massive? Explain how you know.
D) How would you answer to (a) change if the satellite was to be twice as far from the surface of Mars?
Please explain thoroughly. I really don't understand any of this stuff and need help studying.
The attempt at a solution
I could only do part A and I used the square root of Gm/r which equals
the square root of (6.67E-11)*(6.42E23)/3398km which equals 3398000m and my answer was 3549.921453 m/s
Everything else I'm soooo lost and going to cry bc I don't understand
2. Nov 27, 2011
### Zula110100100
Knowing the tangential velocity you should find the angular velocity and that should give you the period, for C and D just redo the equations with the new data and describe the change
I guess you don't need to use angular velocity, you could use v = 2∏r/T, where T is the period. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8765888810157776, "perplexity": 1044.2162053079296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583680452.20/warc/CC-MAIN-20190119180834-20190119202834-00570.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/197949-system-equations-using-mod.html | # Math Help - System of equations using mod
1. ## System of equations using mod
If i have two equations ,
11a+b(mod26)=17
&
22a+b(mod26)=22
How do i begin to solve for a and b? Ive been trying to make vectors out of them and i get for a=11/5, but this doesn't sound right, i think i should be getting whole numbers shouldn't i?
2. ## Re: System of equations using mod
$22a+b \equiv 22$ (mod 26)
$11a+b \equiv 17$ (mod 26)
$\Rightarrow 11a \equiv 5$(mod 26)
$\Rightarrow a \equiv 5.11^{-1}$(mod 26)
$11^{-1}=19$ Since $11.19 \equiv 1$ (mod 26)
$\Rightarrow a \equiv 5.19$(mod 26)
$\Rightarrow a \equiv 17$ (mod 26) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8983082175254822, "perplexity": 1114.2774802270224}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042987155.85/warc/CC-MAIN-20150728002307-00246-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://minnesota-staging.pure.elsevier.com/en/publications/a-reciprocal-branching-problem-for-automorphic-representations-an | # A reciprocal branching problem for automorphic representations and global Vogan packets
Dihua Jiang, Baiying Liu, Bin Xu
Research output: Contribution to journalArticlepeer-review
1 Scopus citations
## Abstract
Let G be a group and let H be a subgroup of G. The classical branching rule (or symmetry breaking) asks: For an irreducible representation πof G, determine the occurrence of an irreducible representation σ of H in the restriction of πto H. The reciprocal branching problem of this classical branching problem is to ask: For an irreducible representation σ of H, find an irreducible representation πof G such that σ occurs in the restriction of πto H. For automorphic representations of classical groups, the branching problem has been addressed by the well-known global Gan-Gross-Prasad conjecture. In this paper, we investigate the reciprocal branching problem for automorphic representations of special orthogonal groups using the twisted automorphic descent method as developed in [13]. The method may be applied to other classical groups as well.
Original language English (US) 249-277 29 Journal fur die Reine und Angewandte Mathematik 2020 765 https://doi.org/10.1515/crelle-2019-0016 Published - Aug 1 2020
### Bibliographical note
Funding Information:
The first named author is partially supported by NSF grant DMS-1600685 and DMS-1901802. The second named author is partially supported by NSF grants DMS-1702218, DMS-1848058, and by start-up funds from the Department of Mathematics at Purdue University. The third named author is partially supported by NSFC grant No.11501382 and by the Fundamental Research Funds for the Central Universities. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9412620067596436, "perplexity": 548.0014234220764}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00023.warc.gz"} |
http://link.springer.com/article/10.1007%2Fs11467-009-0080-0 | , Volume 5, Issue 1, pp 107-122
Date: 18 Dec 2009
# Aquantitative assessment of stochastic electrodynamics with spin (SEDS): Physical principles and novel applications
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## Abstract
Stochastic electrodynamics (SED) without spin, denoted as pure SED, has been discussed and seriously considered in the literature for several decades because it accounts for important aspects of quantum mechanics (QM). SED is based on the introduction of the nonrenormalized, electromagnetic stochastic zero-point field (ZPF), but neglects the Lorentz force due to the radiation random magnetic field Br. In addition to that rather basic limitation, other drawbacks remain, as well: i) SED fails when there are nonlinear forces; ii) it is not possible to derive the Schrödinger equation in general; iii) it predicts broad spectra for rarefied gases instead of the observed narrow spectral lines; iv) it does not explain double-slit electron diffraction patterns. We show in this short review that all of those drawbacks, and mainly the first most basic one, can be overcome in principle by introducing spin into stochastic electrodynamics (SEDS). Moreover, this modification of the theory also explains four observed effects that are otherwise so far unexplainable by QED, i.e., 1) the physical origin of the ZPF, and its natural upper cutoff; 2) an anomaly in experimental studies of the neutrino rest mass; 3) the origin and quantitative treatment of 1/f noise; and 4) the high-energy tail (∼ 1021 eV) of cosmic rays. We review the theoretical and experimental situation regarding these things and go on to propose a double-slit electron diffraction experiment that is aimed at discriminating between QM and SEDS. We show that, in the context of this experiment, for the case of an electron beam focused on just one of the slits, no interference pattern due to the other slit is predicted by QM, while this is not the case for SEDS. A second experiment that could discriminate between QED and SEDS regards a transversely large electron beam including both slits obtained in an insulating wall, where the ZPF is reduced but not vanished. The interference pattern according to SEDS should be somewhat modified with respect to QED’s. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8294692039489746, "perplexity": 1125.4512792908308}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447549109.94/warc/CC-MAIN-20141224185909-00014-ip-10-231-17-201.ec2.internal.warc.gz"} |
https://math.cornell.edu/alex-townsend | # Alex Townsend
Goenka Family Assistant Professor
Malott Hall, Room 589
[email protected]
### Educational Background
• DPhil (2014) University of Oxford
• Mathematics
• Mathematics
## Research
Numerical analysis, scientific computing, and deep learning
I am interested in the study and development of numerical algorithms in applied mathematics. I mainly work in the following three areas: (1) Novel spectral methods for the solution of differential equations, (2) Low-rank techniques, and (3) Theoretical aspects of deep learning.
## Publications
• Dense networks that do not synchronize and sparse ones that do (with M. Stillman and S. H. Strogatz), Chaos, 30 (2020), 083142.
• Fast algorithms using orthogonal polynomials (with S. Olver and R. M. Slevinsky), Acta Numerica, 29 (2020), pp. 573-699.
• Stable extrapolation of analytic functions (with L. Demanet), Foundations of Computational Mathematics, 19 (2019), pp. 297-331.
• Bounds on the singular values of matrices with displacement structure (with B. Beckermann), SIAM Review, 61 (2019), pp. 319-344.
• Why are big data matrices approximately low rank? (with M. Udell), SIAM Journal on Mathematics of Data Science, 1 (2019), pp. 144-160. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8997763395309448, "perplexity": 3277.303786949945}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585997.77/warc/CC-MAIN-20211024111905-20211024141905-00681.warc.gz"} |
https://math.boisestate.edu/m287/author/kencoiteux/ | # Author Archives: Ken Coiteux
We have already looked at iterations of linear functions, so a natural extension would be to look at quadratic functions in the form $f(x)=ax(x-1)$. There are some familiar questions, such as what happens when $a$ is varied with a fixed $x_0$ and vice versa.
There is a little twist, though. Iteration around a fixed may cause nearby points to converge towards that fixed point. That point is then called an attractor. Nearby points could also diverge away from the fixed point. If so, the point is then called a repeller. Sometimes there is no pattern at all and can look quite chaotic……..
Chaos will be looked at and that is the major draw for me to consider this chapter. Let’s have some random fun!!!
# Absolute values on Q (Part II)
Continuing on from absolute values on Q part I/ , I am going to show that the definition of p-adic absolute value satisfies the three properties of absolute values.
Property 1
Let $x=0$ and p be a fixed prime. By definition then, $|x|_p = 0$. This satisfies the first property ($|x|_p=0$ if and only if $x=0$).
Suppose now $x \ne 0$. Then $|x|_p = p^{-ord_p(x)}$. Using Conjecture 1, $|x|_p=|p^n \cdot a|_p =p^{-n}$. Since $p^{-n}$ cannot be 0, $p^{-n} > 0$.
Property 2
Let $x=0$ and $y \ne 0$ and p be a fixed prime. Then $|x \cdot y|_p = |0|_p = 0 = |x|_p \cdot |y|_p$.
Suppose $|x|_p = |p^n \cdot a|$ and $|y|_p = |p^n \cdot b|$. In other words, x and y have the same exponent on p. Then, by definition,
$|x \cdot y|_p = |p^n \cdot a \cdot p^n \cdot b|_p = |p^{2n} \cdot ab|_p = p^{-2n}$.
By the second property,
$|x \cdot y|_p = |x|_p \cdot |y|_p = |p^n c\dot a|_p \cdot |p^n \cdot b|_p = p^{-n}p{-n}=p{-2n}$
Suppose x and y have different exponents on the p. $|x|_p=|p^n \cdot a|_p and |y|_p=|p^m cdot b|$. Multiply x by y….
$|xy|_p=|p^n \cdot a \cdot p^m \cdot b|_p=|p^{n+m} \cdot ab|_p=p^{-n-m}$
By the second property,
$|xy|_p=|x|_p \cdot |y|_p = p^{-n} \cdot p^{-m}=p^{-n-m}$.
Property 3
Let $x=y=0$ and p be a fixed prime. Then by definition $|x+y|_p = 0$. This partly satisfies the property for $0 \le 0+0$ Now consider $x \ne 0$. Then $|x+y|_p = |x|_p = |p^n \cdot a|_p = p^{-n}$. This contributes to satisfying the property for $p^{-n} \le p^{-n} + 0$.
Now let x and y have the same exponent on p.
$|x+y|_p = |p^n \cdot a + p^n \cdot b|_p = |p^n(a+b)|_p=p^{-n}$
If the definition satisfies the property, then $p^{-n}$ should be less than $|x|_p+|y|_p$. In fact this is true because $|x|_p+|y|_p = p^{-n}+p^{-n}$ which is larger than $p^{-n}$.
Last case is when x and y have different exponents on p.
$|x+y|_p=|p^n \cdot a + p^m \cdot b|_p=p^{-n}$ where $n<m$
Now consider property 3.
$|x|_p+|y|_p=p^{-n}+p^{-m}$
Now that p-adic absolute values do satisfy the properties of absolute values, there is one more thing to look at: Archimedean absolute values. A little warning for you, the following definition is counter-intuitive.
Definition 2: An absolute value on Q is said to be non-Archimedean if the properties of absolute values are satisfied along with an additional property:
$|x+y| \le max(|x|,|y|)$ for all x,y \in Q.
Absolute values that satisfy the three properties but not the fourth are said to be Archimedean.
Theorem 1: p-adic absolute values are non-Archimedean.
It has already been shown that p-adic absolute values satisfy the first three conditions. Now all that needs to be shown is that p-adic absolute values satisfy the fourth property. Assume p is a fixed prime
Case 1: $x=y=0$
$|x+y|_p = |0|_p = 0 \le max(|x|_p,|y|_p) = 0$
Case 2: $x \ne 0, y=0$
$|x+y|_p = |x|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, 0) = p^{-n}$
Case 3: x and y have the same exponent on p
$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-n}) = p^{-n}$
Case 4: x and y have different exponents on p (n<m)
$|x+y|_p = p^{-n} \le max(|x|_p,|y|_p) = max(p^{-n}, p^{-m}) = p^{-m}$
QED
# Absolute Values on Q (Part I)
Real numbers, Q, can be constructed from the rational numbers Q={a/b where a,b ϵ the integers Z}. Construction of p-adic numbers, Qp, is done exactly the same way.
Let’s now consider absolute values. An absolute value is a map from Q to [0,∞) that has the following properties for any x,y ϵ Q:
1. |x| ≥ 0, and |x| = 0 if and only if x = 0,
2. |x∙y| = |x|∙|y|,
3. |x+y| ≤ |x|+|y| (The triangle inequality).
Now that we have some properties, we can look at types of absolute values that can be put on the rationals. Alexander Ostrowski showed in 1935 that there are only three types.
Definition 1: Let x be a rational number.
The trivial absolute value of x, denoted |x|0, is defined by
$|x|_0 = \left\{ \begin{array}{l l} 1 & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$
The usual absolute value of x, denoted |x|, is defined by
$|x|_{\infty} = \left\{ \begin{array}{l l} x & \quad \text{if x\ge0 }\\ -x & \quad \text{if x\le0} \end{array} \right.$
The p-adic absolute value of x, denoted |x|p, is defined for a given prime p by
$|x|_p = \left\{ \begin{array}{l l} 1/p^{ord_p(x)} & \quad \text{if x\ne0 }\\ 0 & \quad \text{if x=0} \end{array} \right.$
The quantity ordp(x) is called the order of x. It is the highest power of p that divides x. For example, let p=5 and let x=75. Then ord5(75)=2. Since 25=52 and 25 is the highest power of 5 that divides 75, then the order of x is 2. Confused yet? Here are some examples of p-adic absolute values:
|75|5 = |52 ∙ 3|5 = 5-2
|10|5 = |51 ∙ 2|5 = 5-1
|13|5 = |50 ∙ 13|5 = 1
$|\frac{2}{75}|5 = |5-2 ∙ \frac{2}{3} |5 = 52$
|-375|5 = |53 ∙ -3|5 = 5-3
In these examples, we were looking at 5-adic absolute values. We broke the number up into the form 5n ∙ a. The 5-adic absolute value of 5n ∙ a is 5-n.
Conjecture 1: Given a rational number x, the p-adic absolute value of x is p-n, where pn divides x.
|x|p = |pn ∙ a|p = p-n
In the introduction post (http://math.boisestate.edu/m287/quick-intro-to-p-adics/), it was mentioned that large p-adic numbers are closer to 0 than smaller p-adic numbers. If one was to think of absolute value as a measure of distance, notice in the above examples that the larger the exponent on p, the resulting absolute value (or distance from 0) gets smaller.
What are p-adic numbers? They are a different set of numbers first introduced by Kurt Hensel in 1897. The motivation at that time was an attempt to bring the ideas and methods of power series methods into number theory. They have been used in proving Fermat’s Last Theorem and have other applications in number theory. See http://mathworld.wolfram.com/p-adicNumber.html for more information.
A little terminology needs to be introduced. The p in p-adic represents any prime number. For each prime, there is a new and different set of p-adic numbers. Q2 identifies the 2-adics, Q5 represents the 5-adics, Q17 represents the 17-adics. To keep the same notation, Q will represent the real numbers.
Another term to consider is “close.” The basic idea is that given a number n, it is close to 0 if it is highly divisible by a prime p. Consider the numbers 25 and 625. Relatively speaking, 625 has more factors of 5 than does 25, or in other words, 625 has a higher divisibility by 5. Therefore 625 is closer to 0 than 25. This idea will be made a little bit clearer in future postings.
Iteration of linear functions was the first lab we covered. Towards the end of the lab, I felt like this was a precursor to fractal geometry. After all, the Mandelbrot set is based on an iteration of a quadratic function with the use of complex numbers. Chapter 14 deals with iteration of quadratic functions. This is one step closer to fractal geometry…..and a precursor to chaos theory, both of which are very interesting to me.
Early definitions in the chapter are:
• fixed point: given a function $f(x)$, a point $u$ is a fixed point of $f$ if $f(u)=u$.
• attractor: a fixed point is an attractor when all nearby points move towards it under iteration.
• repeller: a fixed point is a repeller when all nearby points move away from it under iteration
Here’s a type of question from the chapter:
Given a function $f(x)=ax(x-1)$, how do various values of $a$ affect fixed points, attractors, repellers, and zeroes. What about changing initial values?
# Tous!!!
Tous. Ken nee’eesih’inoo. Neeyeiheihinoo heesou’sii’ii.
Hi! My name is Ken. I am studying math.
A couple things that interest me are fractals and cultural math.
http://en.wikipedia.org/wiki/Fractal
The equation Mandelbrot used to create his famous picture is $z_{n+1}=(z_n)^2+c$
http://www.amazon.com/Native-American-Mathematics-MIchael-Closs/dp/0292711859
This book mostly covers various number systems found among Native American tribes……one group had a base-8 number system!!!! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.93451327085495, "perplexity": 1036.6793944304204}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00387.warc.gz"} |
http://tex.stackexchange.com/questions/15334/getting-numbers-to-appear-in-circles | Getting numbers to appear in circles
I have a linked question that I previously found on here, but I cannot make any comment yet to post. The link being: Good way to make \textcircled numbers?
My question was how could we use this command more then once in our TeX file in different parts of our file without generating an error.
Example Code:
\documentclass{article}
\usepackage{tikz}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
\node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
\begin{document}
Numbers aligned with the text: \circled{1} \circled{2} \circled{3} end.
\end{document}
The error message is this:
! LaTeX Error: Command \circled already defined.
Or name \end... illegal, see p.192 of the manual.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
1.222 ...circle,draw,inner sep=2pt] (char) {#1};}}
When removing the \newcommand* it shows this next error message:
! You can't use 'macro parameter character #' in restricted horizontal mode.
<argument> ...ircle,draw,inner sep=2pt] (char {##
1}
Can someone help out please. Thank You so much.
-
Are you getting those error messages with the exact code you posted? I ask you because that code compiles OK in my system. – Gonzalo Medina Apr 8 '11 at 17:49
Like @Gonzalo already said: The above example code compiles fine. You seem to define \circled multiple times, instead of using it. Make sure that you only have one \newcommand and that it isn't inside a macro you use. – Martin Scharrer Apr 8 '11 at 17:53
Also: welcome to TeX.SX! A tip: you can use backticks to format inline code or package names, like I did in my edit. We also prefer to not have any opening or closing lines. Thanks. – Martin Scharrer Apr 8 '11 at 17:59
Yea, i used that entire code in two separate locations with the new command in each of them in the TeX file. I just want to put the circle somewhere different in the file without saying it is used multiple times. Thanks, for the warm welcome. – night owl Apr 8 '11 at 18:01
@night owl. I quote: "I used that entire code in two separate locations with the new command in each of them in the teX file." That is definitely not what you should do (unless I misunderstand what you mean). You just need one single definition in the preamble. You also just need one preamble. Does the code snipplet you quoted by itself compile fine for you? If so, search your TeX file for duplicate definitions of \circled. – Willie Wong Apr 8 '11 at 18:15
As is mentioned in the comments, your example code compiles fine, but you mustn't use the \newcommand*\circled more than once. If you want more circled numbers, just use \circled{n} for each instance of a circled n.
My guess is that you did the following: You tried some code like
\documentclass{article}
\usepackage{tikz}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
\node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
\begin{document}
Numbers aligned with the text: \circled{1} \circled{2} \circled{3} end.
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
\node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
Numbers aligned with the text: \circled{1} \circled{2} \circled{3} end.
\end{document}
and got ! LaTeX Error: Command \circled already defined. This is expected behaviour; you should define \circled only once in the preamble. Then you tried removing the second \newcommand*, but you only removed that. Then indeed you get the error message
! You can't use macro parameter character #' in restricted horizontal mode.
What you have to do is remove the full two lines of the definition, i.e., the lines
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
\node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9643519520759583, "perplexity": 3530.8994964397257}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657135080.9/warc/CC-MAIN-20140914011215-00261-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an%3A0672.05055 | ×
# zbMATH — the first resource for mathematics
Fast parallel algorithms for chordal graphs. (English) Zbl 0672.05055
This paper presents fast parallel algorithms concerning problems on chordal graphs, where efficient sequential algorithms are known.
One of the problems is the recognition of chordal graphs. The time bound is polylogarithmic and the processor bound is polynomial but not linear. The best known sequential algorithm needs only linear time. But the processor bound of this parallel algorithm is better than that of an earlier parallel algorithm of A. Edenbrandt. Here also fast parallel algorithms are presented, which construct a corresponding collection of subtrees of a tree, an elimination ordering, and minimum colouring, and a maximum weight independent set for any chordal graph.
P. Klein presented later a polylog time parallel algorithm for the construction of an elimination ordering for each chordal graph, which needs only a linear number of processors, and therefore also a recognition algorithm for chordal graphs with a linear processor bound. Only the time bound is a little bit worse than that of the algorithm of J. Naor, M. Naor, and A. Schäffer.
Reviewer: E.Dahlhaus
##### MSC:
05C38 Paths and cycles 68R10 Graph theory (including graph drawing) in computer science
##### Keywords:
fast parallel algorithms; chordal graphs; recognition
Full Text: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9402334690093994, "perplexity": 1114.511861761922}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00690.warc.gz"} |
https://bio.libretexts.org/Learning_Objects/Laboratory_Experiments/General_Biology_Labs/Unfolding_the_Mystery_of_Life_-_Biology_Lab_Manual_for_Non-Science_Majors_(Genovesi_Blinderman_and_Natale)/06%3A_Enzymes | # 6: Enzymes
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## Learning Objectives
• To examine the function and properties of enzymes in living cells
• To distinguish between substrate, enzyme, and product in a reaction
• To collect and analyze data on tissue-specific catalase activity
• To collect and analyze data on the effect of temperature on catalase action
Thumbnail: Dihydrofolate reductase is inhibited by methotrexate which prevents binding of its substrate, folic acid. (CC BY 4.0 International; Thomas Shafee (modified) via Wikipedia)
This page titled 6: Enzymes is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ellen Genovesi, Laura Blinderman, & Patrick Natale via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8203243017196655, "perplexity": 4631.352460475092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00761.warc.gz"} |
http://www.jiskha.com/search/index.cgi?query=Under+certain+conditions%2C+neon+(Ne)+gas+diffuses+at+a+rate+of+7.0+centimeters+per+second.+Under+the+same+conditions%2C+an+unknown+gas+diffuses+at+a+rate+of+4.9+centimeters+per+second.+What+is+the+approximate+molar+mass+of+the+unknown+gas%3F | Friday
January 30, 2015
# Search: Under certain conditions, neon (Ne) gas diffuses at a rate of 7.0 centimeters per second. Under the same conditions, an unknown gas diffuses at a rate of 4.9 centimeters per second. What is the approximate molar mass of the unknown gas?
Number of results: 70,833
chemistry
Under certain conditions, neon (Ne) gas diffuses at a rate of 7.0 centimeters per second. Under the same conditions, an unknown gas diffuses at a rate of 4.9 centimeters per second. What is the approximate molar mass of the unknown gas?
October 31, 2010 by smith
chemistry
Under certain conditions, neon (Ne) gas diffuses at a rate of 4.5 centimeters per second. Under the same conditions, an unknown gas diffuses at a rate of 10.1 centimeters per second. What is the approximate molar mass of the unknown gas?
May 26, 2011 by Carly
chemistry
Under certain conditions, neon (Ne) gas diffuses at a rate of 7.0 centimeters per second. Under the same conditions, an unknown gas diffuses at a rate of 4.9 centimeters per second. What is the approximate molar mass of the unknown gas? 6.5 g/mol 11 g/mol 20 g/mol 41 g/mol
October 31, 2010 by smith
chemistry
A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 52 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 24 s for 1.0 L of O2 gas to effuse. Calculate the molar mass ...
October 10, 2010 by Jennifer
chemistry
Ima Chemist found the density of Freon-11 (CFCl3) to be 5.58 g/L under her experimental conditions. Her measurements showed that the density of an unknown gas was 4.38 g/L under the same conditions. What is the molar mass of the unknown?
chemistry
If the ratio of the rate of effusion of an unknown gas to that of CO under the same conditions is 0.7978, calculate the molar mass (g/mol) of the gas
January 15, 2010 by eng
chemistry
If 0.45 mol of neon gas occupies a volume of 789 mL at a certain temperature and pressure, what volume would 0.376 mols of neon occupy under the same conditions? what Law is used to solve?
April 18, 2012 by Anonymous
Chemistry
"The analyzer said that at 20degreesC the unknown gas was 82.8% carbon by weight and 17.2% hydrogen. The analyzer also diffused the gas through a special apparatus and measured the rate of its diffusion compared to how fast oxygen gas diffused under the same conditions. The ...
January 17, 2011 by Anonymous
Chemistry
"The analyzer said that at 20degreesC the unknown gas was 82.8% carbon by weight and 17.2% hydrogen. The analyzer also diffused the gas through a special apparatus and measured the rate of its diffusion compared to how fast oxygen gas diffused under the same conditions. The ...
January 17, 2011 by Anonymous
chemistry
The rate of effusion of a particular gas was measured and found to be 25.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH4) gas is 46.4 mL/min. What is the molar mass of the unknown gas?
October 31, 2010 by Mattie
chemistry
The rate of effusion of a particular gas was measured and found to be 23.0 mL/min. Under the same conditions, the rate of effusion of pure methane (CH4) gas is 46.2 mL/min. What is the molar mass of the unknown gas?
November 7, 2010 by Anonymous
chemistry
during an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles in another gas under the same conditions. What is the molar mass of the unknown gas?
July 5, 2014 by Brooklynne
chem
An unknown gas Q requires 1.80 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q?
October 12, 2009 by Sheena
chemistry
An unknown gas Q requires 2.17 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q?
March 4, 2011 by ria
chemistry
what is the molar mass of the unknown gas that diffuses 0.250 times as fast as hydrogen gas under the same circumstances?
August 9, 2011 by Aamna Qamar
Chemistry
Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 4.92 mL of O2 had passed through the membrane, but only 3.04 mL of of the unknown gas had ...
April 15, 2014 by Peter
chemistry
If 4.83 mL of an unknown gas effuses through a hole in a plate in the same time it takes 9.23 mL of argon, Ar, to effuse through the same hole under the same conditions, what is the molecular weight of the unknown gas?
October 31, 2012 by Zoe
Chemistry
A gas, Y, diffuses 1.05 times faster than SO2 gas at the same conditions. What is the molecular weight of gas Y?
December 3, 2013 by Krystal
chem 101
it takes a certain gas 1.98 times as long to diffuse as an equal volume of helium under the same conditions. what is the calculated molecular weight of the gas?
November 5, 2011 by Anonymous
Chemistry
A sample of unknown gas effuses in 10.6 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas?
October 7, 2010 by Missy*
Chemistry
It takes 200. seconds for a sample of carbon dioxide to effuse through a porous plug and 280. seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas? Enter your answer to 3 significant figures.
February 18, 2013 by Olivia
chem
takes 190. seconds for a sample of carbon dioxide to effuse through a porous plug and 295 seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas? Enter your answer to 3 significant figures.
April 29, 2013 by sarah
chemistry
Effusion/Diffusion Chemistry Help!!! If 4.83mL of an unknown gas effuses through a hole in a plate in the same time it takes 9.23mL of Argon, Ar, to effuse through the same hole under the same conditions, what is the molecular mass of the unknown gas? (146amu)
June 17, 2009 by Capacino
chemistry
Density of phosgene gas, COCl2(MM=99) ander a certain condition is 4.5g/L. What is the density if H2S gas(MM=34) under the same conditions?
October 20, 2008 by annemarie
Chemistry #27
If He(g) has an average kinetic energy of 4410 J/mol under certain conditions, what is the root mean square speed of O2(g) molecules under the same conditions?
October 13, 2013 by Chelsea
chemistry
If He(g) has an average kinetic energy of 7590 J/mol under certain conditions, what is the root mean square speed of F2(g) molecules under the same conditions?
September 28, 2014 by A
chemistry
If He(g) has an average kinetic energy of 7590 J/mol under certain conditions, what is the root mean square speed of F2(g) molecules under the same conditions?
September 28, 2014 by B
Chemistry
Metals lose electrons under certain conditions to attain a noble-gas electron configuration. How many electrons must be lost by the element Ca? Is it 2 e^-? Which noble-gas electron configuration is attained in this process? argon radon krypton xenon helium neon
October 28, 2012 by Bill
Chemistry
a sample of gas of unknown pressure occupies 0.766 L at the temperture of 298k the same sample of gas is than tested under known conditions and has the pressure of 32.6kPa and occupies 0.664 L at 303 k. What is the original pressure of gas
July 13, 2011 by Rocky
AP Chemistry
The rate of effusion of unknown gas X is found to be about 1.5 times that of SF6 gas (MW=146 g/mol) at the same conditions of temperature and pressure. What is the molecular weight of gas X? Answer in units of g/mol
February 16, 2014 by Gabriella
Chemistry
It takes 32 seconds for a certain number of moles of H2 to effuse through a leak in a tank. An unknown gas was later places under the same conditions; it took 1:39 for the same number of moles to effuse. Find the molecular mass of the gas. I set it up like this: rate of ...
January 7, 2007 by John
grahams law
If a molecule of O3 diffuses a certain distance in 3500 s, how long (s) would it take a molecule of Cl2 to diffuse the same distance at the same temperature and under the same conditions?
April 12, 2011 by Matty
chemistry
Effusion/Diffusion Chemistry Help!!! If 4.83mL of an unknown gas effuses through a hole in a plate in the same time it takes 9.23mL of Argon, Ar, to effuse through the same hole under the same conditions, what is the molecular mass of the unknown gas? (146amu) Responses: * ...
June 18, 2009 by Capacino
Chemistry
If 3.21 mol of a gas occupies 56.2L at 44C and a pressure of 793 torr, what volume will 5.29 mol of this gas occupy under the same conditions? (A) 61.7 L (B) 30.9 L (C) 14.7 L (D) 92.6 L (E) 478 L
January 1, 2010 by Anonymous
Chemistry
If 3.21 mol of a gas occupies 56.2L at 44C and a pressure of 793 torr, what volume will 5.29 mol of this gas occupy under the same conditions? (A) 61.7 L (B) 30.9 L (C) 14.7 L (D) 92.6 L (E) 478 L is it d, 92.6 L?
January 3, 2010 by Anonymous
Chem hw!
1. What pressure would be needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume? 2. If the pressure on a 1.04-L sample of gas is doubled at constant temperature, what will be the new volume of the gas? 3. A 1.04-L sample of gas at 759 mm Hg pressure ...
March 11, 2010 by Teisha
Chem hw!
1. What pressure would be needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume? 2. If the pressure on a 1.04-L sample of gas is doubled at constant temperature, what will be the new volume of the gas? 3. A 1.04-L sample of gas at 759 mm Hg pressure ...
March 11, 2010 by Teisha
Chemistry
At pressures greater than 60,000 kPa, how does the volume of a real gas compare with the volume of an ideal gas under the same conditions? A.)It is much greater. B.)It is much less. C.)There is no difference. D.)It depends on the type of gas. I think it's A. Thank You.
January 17, 2013 by Deborah
Chemistry
a real gas would behave most like an ideal gas under conditions of
November 18, 2011 by tom
At pressures greater than 60,000 kPa, how does the volume of a real gas compare with the volume of an ideal gas under the same conditions? A.)It is much greater. B.)It is much less. C.)There is no difference. D.)It depends on the type of gas. I think it is A. Thank You.
January 20, 2013 by Deborah
sciencee
1. What pressure would be needed to compress 25.1 mL of hydrogen at 1.01 atm to 25% of its original volume? 2. If the pressure on a 1.04-L sample of gas is doubled at constant temperature, what will be the new volume of the gas? 3. A 1.04-L sample of gas at 759 mm Hg pressure ...
March 9, 2010 by Teisha
Chem
If 4.0 grams of hydroven, 6.02 x 10^22 atoms of helium and 4.48 liters of argon, measured at STP are placed in a 2.0 liter flask, at 27 degrees C, what is the final pressure? Under controlled conditions the velocity of hydrogen molecules was found to be 250 meters/sec, and the...
November 23, 2008 by Anonymous
chem
compund containing only C, H, and N yeilds the following data- 1. complete combustion of 34.3 mg of the compund producedmg of CO2 and 40.3 mg of H2O 2. a 65.2mg sample was analyzed by the dumas mthod, giving 35.6 mL of N2 at 740 torr and 25C 3. effusion rate as a gas was 24....
November 16, 2007 by mel
Chemistry
Nitrogen and hydrogen, with volumes of 12.0 L each, are injected into a sealed rigid reaction chamber containing a catalyst. The pressure of the chamber is 8.00 atmospheres and at a known temperature. The reaction is allowed to proceed to completion while being maintained at a...
August 10, 2011 by Neha
Chemistry
Consider the following chemical equation: 2NO2-->N2O4 If 25.0mL of NO2 gas is completely converted to N2O4 gas under the same conditions, what volume will the N2O4 occupy?
July 2, 2010 by Victoria
A gas, while expanding under isobaric conditions, does 660 J of work. The pressure of the gas is 1.40 x 105 Pa, and its initial volume is 1.70 x 10-3 m3. What is the final volume of the gas? Would I multiply the two values?
June 21, 2012 by Hannah
chemistry
A 0.12-mole sample of nitrogen gas occupies a volume of 2.55 L. What is the volume of 0.32 mol of nitrogen gas under the same conditions?
December 9, 2011 by BB
Chem 130
If 2.03 g of helium gas occupies a volume of 12.0 L at 22°C, what volume will 6.19 g of helium gas occupy under the same conditions?
May 5, 2014 by michelle
Chemistry
A sample of gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of to effuse under the same conditions.
December 31, 2014 by Claire
physics/chemistry(kinetic ..gasses)
How many degreed of freedom have the gas molecules,if under standard conditions the gas dendity is 1.3kg/m^3 and velocity of sound propagation on it is v=330m/s?
May 13, 2013 by duke
chemistry
Nitric acid can be produced by the reaction of gaseous nitrogen dioxide with water. 3NO2(g) + H2O() −! 2HNO3() + NO(g) If 745 L of NO2 gas react with water, what volume of NO gas will be produced? As- sume the gases are measured under the same conditions.
February 7, 2012 by Aaron
Chemistry
I am totally confused. Does my answer make sense or do I need to start over? The more I read it the more I doubt my answer. I am also confused when it states Please be specific. Is it looking for a specific number for the temperature and pressure or just the conditions? Under ...
January 31, 2008 by Jamie
Biochemistry
In yeast, ethanol is produced from glucose under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions? A cell-free yeast extract is placed in a solution that contains 3.50 × 102 mmol ...
April 8, 2013 by Josh
Biochem
In yeast, ethanol is produced from glucose under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions? A cell-free yeast extract is placed in a solution that contains 2.50 × 102 mmol ...
October 30, 2013 by B
BioChem
In yeast, ethanol is produced from glucose under anaerobic conditions. What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions? A cell-free yeast extract is placed in a solution that contains 2.50 × 102 mmol ...
October 30, 2013 by B
Chemistry
A toy balloon originally held 1.00 g of helium gas and had a radius of 10.0 cm. During the night, 0.25 g of the gas effused from the balloon. Assuming ideal gas behavior under these constant pressure and temperature conditions, what was the radius of the balloon the next morning?
September 12, 2012 by Dia B
chemistry
Solid white P melts then vaporizes at high temperatures. Gaseous white P effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions. How many atoms are in a molecule of gaseous white phosphorus?
February 20, 2013 by sally
chemistry
What is the volume of 1.0g of hydrogen gas at 0 degrees celsisus and 1 atmosphere? The density of hydrogen under these conditions is 0.090g per L.
September 17, 2010 by Tammie
Statistics
Suppose in a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate...
June 14, 2010 by John
Chemistry
The equilibrium constant for the reaction of bromine and chlorine to form bromine chloride under certain conditions is 4.7x10-2. Cl2(g) + Br2(g) --> 2BrCl(g) K= 4.7 x 10-2 What is the equilibrium constant for the following reaction under the same conditions? 1/2 Cl2(g) + 1/...
November 3, 2007 by Dan
Chemistry
1.Calculate the density of oxygen, O2, under each of the following conditions: STP 1.00 atm and 20.0 ∘C 2.To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.3-L bulb, then filled it with the gas at 1.90atm and 20.0...
November 7, 2014 by LO
math
A car traveled 180 miles and use 6 gallons of gas if the same car traveled 1260 miles under similar conditions how many gallons of gas would be used
January 7, 2015 by thaylorah
Biochem
What is the maximum amount of ethanol (in millimoles) that could theoretically be produced under the following conditions? Condition: A cell-free yeast extract is placed in a solution that contains 3.50 × 102 mmol glucose, 0.30 mmol ADP, 0.30 mmol Pi, 0.60 mmol ATP, 0.20 mmol...
April 3, 2013 by Mark
Chemistry
A compound containing only C, H, and N yields the following data. (i) Complete combustion of 34.8 mg of the compound produced 33.3 mg of CO2 and 40.9 mg of H2O. (ii) A 66.3-mg sample of the compound was analyzed for nitrogen by the Dumas method, giving 36.2 mL of N2 at 740. ...
November 20, 2012 by Darrel
Chemistry
what volume of oxygen gas measured at 25C and 760 torr is required to react with 1.0L of methane measured under the same conditions of temperature and pressure?
March 24, 2012 by Mitch
Chemistry
A sample of methane (CH4) gas contains a small amount of helium. Calculate the volume percentage of helium if the density of the sample is 0.70902 g/L t 273K and 1.00 atm. I know that the sample is at STP and that the 1 mole of gas under these conditions occupies 22.4L, but ...
October 29, 2014 by Chris
chemistry
Under standard conditions, the Gibbs free energy of the reactants G,std(reactants) in a reaction in the gas phase is 232.94 kJ and the Gibbs free energy of the products G,std(products) is 211.56 kJ. Calculate the value of the equilibrium constant for this reaction under ...
April 28, 2014 by bekah
chemistry
a sample containing 1.50 mol neon gas has a volume of 8.00L.what is the new volume of the gas in liters when the following changes occur in the quantity of the gas at constant pressure and temperature? a.a leak allows on-half of the neon atoms to escape. b. a sample of 25.0g ...
June 27, 2010 by Jetta
Chemistry
12. Solutions cans be supersaturated if they contain more dissolved solute than saturated solutions under the same conditions.- T? 16. Heating a solution decreases the solubility of the gas in it.- T? Thanks -MC
October 27, 2010 by mysterychicken
ap chemistry
Under certain conditions, the reaction 3A + 2B -> 4C was observed to proceed at a rate of 0.00348 M· s−1. What was the corresponding rate of change in reactant A?
November 6, 2013 by gabriella
chemistry
If 1.00 {\rm mol} of {\rm{N}}_2 has a volume of 47.0 {\rm L} under the reaction conditions, how many liters of gas can be formed by heating 38.5 g of {\rm{NaN}}_3
September 30, 2010 by Anonymous
Math - Variation
under certain conditions, the thrust T of a propeller varies jointly as the fourth power of it's diameter, d, and the square of the number, n, of revolutions per second. What happens to the thrust if n is doubled and d is halved?
May 23, 2014 by Kiirsty
Statistics
Suppose in a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate...
June 16, 2010 by Jacob
Statistics
Suppose in a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate...
June 17, 2010 by Jacob
chemistry
A fixed amount of gas initially at 100 K and pressure P1 is expanded from 1 L to 100 L in a piston. The temperature, T2, of the gas at this point is 50 K. The piston is then locked at constant volume V2=100 L and half the gas is pumped out at a constant temperature of T2 . The...
December 5, 2011 by laura
chemistry
A 9.96 mol sample of methane gas is maintained in a 0.836 L container at 296 K. What is the pressure in atm calculated using the van der Waals' equation for CH4 gas under these conditions? For CH4, a = 2.25 L2atm/mol2 and b = 4.28E-2 L/mol.
January 29, 2012 by jim
Chem 1000
A Cylinder is filled with 1.20 mol of He(g) at 25 C under ambient pressure of 0.966 atm. The gas in the cylinder is then heated with 1.430 KJ of heat, and the piston is raised by the expanding gas under the constant ambient pressure. Calculate the following after expansion. a...
October 16, 2010 by J Velji
College Chemistry
A Cylinder is filled with 1.20 mol of He(g) at 25 C under ambient pressure of 0.966 atm. The gas in the cylinder is then heated with 1.430 KJ of heat, and the piston is raised by the expanding gas under the constant ambient pressure. Calculate the following after expansion. a...
October 16, 2010 by J Velji
chemistry
Under what conditions of temperature and pressure do you think that the gas laws would work best? List specifics.
January 29, 2008 by Cindy
chemisrty
Under what conditions of temperature and pressure do you think that the gas laws would work best? Please be specific
September 29, 2009 by Anonymous
chemistry
O2 effuses at a rate that is _____ times that of Xe under the same conditions. rate 02/rate x2 = ?
June 21, 2013 by Annie
Chem
In an experiment, 25.0 ml of a gas with a pressure of 1. atm is contained in a balloon at 25.0 c. The Balloon is then cooled to 5.0 c, and the pressure is found to be .750 atm. What is the volume of the gas under the new conditions?
June 29, 2011 by Chris
Chemistry
In an experiment, 25.0 mL of a gas with a pressure of 1.00 atm is contained in a balloon at 25.0°C. The balloon is then cooled to 5.0°C, and the pressure is found to be 0.750 atm. What is the volume of the gas under the new conditions?
November 6, 2012 by Brittany
Neon
I need to find out the cost of the inert gas, Neon and it's Normal Phase to finish off a project. I have searched and searched but can't come up with the info needed. Neon is a rare gas, obtained by liquifing air. Its cost for very pure form is about thirty dollars per ten ...
October 16, 2006 by Holly
ethics
If you are an employee of a business, and learn that it is violating the law, what would you do in this situation? What are your ethical obligations? Under what conditions would you report the illegal activities? Under what conditions would you keep quiet?
November 30, 2010 by Anonymous
Chemistry
A given volume of nitrogen, N2, rewuired 68.3 s to effuse from a hole in a chamber. Under the same conditions, another gas required 85.6 s for the samw volume to effuse. What is the molecular mass ?
March 3, 2010 by Anonymous
physics conduction
a one brick thick wall of 10cm thickness has temperatures 40deg C and 5deg C at the surfaces. if the thermal conductivity of the brick is 0.6W per m per K, what is the rate of heat flow per unit area W per m square through the bricks if steady state conditions apply? explain ...
June 1, 2012 by ridhi
ap chemistry
A 6.9 gram sample of a gaseous substance occupies 10 L at 63C and 537 torr. What is the density of the gas under these conditions? Answer in units of g/L
February 9, 2013 by cheri
chemistry
A sample of neon effuses from a container in 73 seconds. The same amount of an unknown noble gas requires 149 seconds. Identify the gas.
October 16, 2008 by nikki
Chemistry
A sample of neon effuses from a container in 80 seconds. The same amount of an unknown noble gas requires 163 seconds Identify the gas.
August 10, 2013 by Anonymous
chemistry
i reallly need help with combining the gas laws. here are a few questions i am struggling with if anyone could please help! 1. A steel container holds 750.L of O2 gas at STP. a) how many moles of O2 does the container hold? b)how many grams of O2 does the container hold? c) if...
May 22, 2011 by me
chemistry
if 3.21 mol of a gas occupies 56.2 liters at 44 degrees celcius and 793 torr, 5.29 mol of this gas occupies how many liters under these conditions?
January 20, 2010 by Anonymous
Krypton gas is four times denser than neon gas at the same temperature and pressure. Which gas is predicted to effuse faster? i got neon which is correct but How much faster? how many times? i dont know how to get htis
January 8, 2013 by Samantha
AP Chemistry
A 2.8 gram sample of a gaseous substance occupies 12 L at 47 degrees C and 683 torr. What is the density of the gas under these conditions? Answer in units of g/L
February 16, 2014 by Gabriella
managerial economics
"Deciding on which price to charge is more difficult under less than perfectly competitive conditions (i.e., monopoly) than it is under perfectly competitive conditions." Discuss
December 8, 2008 by ryan
Statistics
Suppose in a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate...
June 21, 2010 by Jacob
Statistics
Suppose in a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate...
June 21, 2010 by Jacob
Chemistry
Consider a sample of helium and a sample of neon, both at 30ºC and 1.5atm. Both samples have a volume of 5.0 liters. Which statement concerning these samples is not true? a- each weighs the same amount b- each has the same # of atoms of gas c- the density of the neon is ...
November 6, 2009 by John S. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.811957061290741, "perplexity": 1555.198135900518}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-06/segments/1422115855561.4/warc/CC-MAIN-20150124161055-00077-ip-10-180-212-252.ec2.internal.warc.gz"} |
http://math.stackexchange.com/questions/248676/in-svd-why-u-i-av-i-sigma-i/248680 | # In SVD, why $u_i=Av_i/\sigma_i$?
In SVD, $$u_i=Av_i/\sigma_i$$
so we conclude that $A=U\Sigma V^{T}$.
I read this part several times but still hard to understand what they are trying to explain. Can anyone help me to understand this topic?
-
The SVD of the matrix $A$ is given by $A = U\Sigma V^T$, which can be rewritten as $$A = \sum_j \sigma_j u_jv_j^T\ \ (1)$$ where $U$ and $V$ are Hermetian. Multiplying (1) from the right by some $v_i$, we obtain $$Av_i = {(\sum_j \sigma_j u_jv_j^T)}v_i$$ but because all the $(v_j)_i$ are orthonormal ($v_i^Tv_j = \delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta) we obtain $$Av_i = \sigma_i u_i$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9860666990280151, "perplexity": 64.74403068107425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645208021.65/warc/CC-MAIN-20150827031328-00279-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/two-black-holes-falling-head-on.126939/ | # Two black holes falling head-on
1. ### Jorrie
816
This was posted before, way down in another thread and went unanswered. So I thought I will start a new thread of the interesting question.
Quote: Originally Posted by Farsight
I release a test particle of mass mp a given distance away from a black hole of mass M and measure their initial closing acceleration as a.
If I now substitute the test particle with a second black whole of mass M, is the initial closing acceleration of the two black holes a or 2a or something else?.
I think the locally Lorentz acceleration of a test particle starting from rest outside a black hole with mass M at circumferential distance 2 Pi r is given by: (with c=G=1)
a = -(1-2M/r)^(-0.5) M/r^2
What is the equivalent for two black holes, momentarily at rest relative to each other, like Farsight asked above? I am pretty sure we can't just add the calculated accelerations like in the Newton case.
Isn't there a "commandment" in relativity that says: "Thou shalt not add thine own acceleration directly to that of thine fellow traveler"?
2. ### pervect
7,948
Staff Emeritus
I'd suggest using the equivalent Newtonian situation as a rough guide.
As far as departures from Newtonian behavior - as far as I know colliding black holes have only been handled numerically, not analytically.
Once you've got the numerical simulation down :-), you still have to pin down how, exactly, you are going to measure the acceleration to answer the question. You'd have to specify all the gory details - the idea of an accelration presupposes some sort of coordinate system, and you'll have to specify either the coordinate system or the measurement technique to get an answer.
3. ### Jorrie
816
Thanks pervect, but yea, gory problem, especially the coordinate choice! The nice "locally Lorentz" spacetime surrounding a test particle is down the (BH) drain.
Perhaps, since the stated problem is completely static, 'pseudo Schwarzschild' coordinates with the origin in-between the two BHs can be used in a numerical attempt. There will of course be two event horizons...
4. ### pervect
7,948
Staff Emeritus
If the BH are actually free-falling, the problem isn't static :-(.
If you have two charged BH maintaining a constant distance, you can probably find the force between them. The static metric allows you to think in terms of forces. Of course the electric fields are contributing to the metric in this case, so you can't really think of the solution as being just that of the two BH - it's the combined solution of the two black holes and their electromagnetic fields.
I thought there was possibly an analytic solution known for this case, but I'm not sure.
"The double-Kerr solution of Kramer and Neugebauer"
but so far I haven't dug up an actual metric.
5. ### Jorrie
816
OK, bad word choice. If the two BHs are momentarily stationary, we do not have to worry about relative velocity or movement in curved space. Then the problem should be simpler, but probably not easy! I’m also searching for a metric, but have not found anything yet.
BTW, for simplicity, let's leave charge and rotation out for the moment...
6. ### Jorrie
816
Two black holes, momentarily stationary...
The Newtonian equivalent suggests that if a single black hole with mass M causes a momentarily stationary particle to fall with a starting acceleration a at a radial distance r, then two momentarily stationary black holes with that same mass and at that same distance apart, will fall towards each other with a starting relative acceleration of 2a. In the Newtonian case, a = -M/r^2, geometrically.
In the relativistic (Schwarzschild) case:
a = -(M/r^2)/(1-2M/r)^(0.5), i.e., the Newtonian case divided by the gravitational redshift factor.
Since we are not working with colliding black holes, but rather with momentarily stationary ones, will it be a reasonable assumption to just double this Schwarzschild acceleration, like in the Newtonian case?
As in a previous threat https://www.physicsforums.com/showthread.php?t=125715&page=3), we have to ignore the effect of tidal deformation of the objects for simplicity. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8793780207633972, "perplexity": 818.050471555887}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207929899.62/warc/CC-MAIN-20150521113209-00025-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://labs.tib.eu/arxiv/?author=Delphine%20Perrodin | • ### Radio pulsars: testing gravity and detecting gravitational waves(1709.02816)
Sept. 8, 2017 astro-ph.HE
Pulsars are the most stable macroscopic clocks found in nature. Spinning with periods as short as a few milliseconds, their stability can supersede that of the best atomic clocks on Earth over timescales of a few years. Stable clocks are synonymous with precise measurements, which is why pulsars play a role of paramount importance in testing fundamental physics. As a pulsar rotates, the radio beam emitted along its magnetic axis appears to us as pulses because of the lighthouse effect. Thanks to the extreme regularity of the emitted pulses, minuscule disturbances leave particular fingerprints in the times-of-arrival (TOAs) measured on Earth with the technique of pulsar timing. Tiny deviations from the expected TOAs, predicted according to a theoretical timing model based on known physics, can therefore reveal a plethora of interesting new physical effects. Pulsar timing can be used to measure the dynamics of pulsars in compact binaries, thus probing the post-Newtonian expansion of general relativity beyond the weak field regime, while offering unique possibilities of constraining alternative theories of gravity. Additionally, the correlation of TOAs from an ensemble of millisecond pulsars can be exploited to detect low-frequency gravitational waves of astrophysical and cosmological origins. We present a comprehensive review of the many applications of pulsar timing as a probe of gravity, describing in detail the general principles, current applications and results, as well as future prospects.
• ### The Scaling of the RMS with Dwell Time in NANOGrav Pulsars(1510.09084)
Oct. 30, 2015 astro-ph.IM
Pulsar Timing Arrays (PTAs) are collections of well-timed millisecond pulsars that are being used as detectors of gravitational waves (GWs). Given current sensitivity, projected improvements in PTAs and the predicted strength of the GW signals, the detection of GWs with PTAs could occur within the next decade. One way we can improve a PTA is to reduce the measurement noise present in the pulsar timing residuals. If the pulsars included in the array display uncorrelated noise, the root mean square (RMS) of the timing residuals is predicted to scale as $\mathrm{T}^{-1/2}$, where T is the dwell time per observation. In this case, the sensitivity of the array can be increased by increasing T. We studied the 17 pulsars in the five year North American Nanohertz Observatory for Gravitational Waves (NANOGrav) data set to determine if the noise in the timing residuals of the pulsars observed was consistent with this property. For comparison, we performed the same analysis on PSR B1937+21, a pulsar that is known to display red noise. With this method, we find that 15 of the 17 NANOGrav pulsars have timing residuals consistent with the inverse square law. The data also suggest that these 15 pulsars can be observed for up to eight times as long while still exhibiting an RMS that scales as root T.
• ### European Pulsar Timing Array Limits on Continuous Gravitational Waves from Individual Supermassive Black Hole Binaries(1509.02165)
We have searched for continuous gravitational wave (CGW) signals produced by individually resolvable, circular supermassive black hole binaries (SMBHBs) in the latest EPTA dataset, which consists of ultra-precise timing data on 41 millisecond pulsars. We develop frequentist and Bayesian detection algorithms to search both for monochromatic and frequency-evolving systems. None of the adopted algorithms show evidence for the presence of such a CGW signal, indicating that the data are best described by pulsar and radiometer noise only. Depending on the adopted detection algorithm, the 95\% upper limit on the sky-averaged strain amplitude lies in the range $6\times 10^{-15}<A<1.5\times10^{-14}$ at $5{\rm nHz}<f<7{\rm nHz}$. This limit varies by a factor of five, depending on the assumed source position, and the most constraining limit is achieved towards the positions of the most sensitive pulsars in the timing array. The most robust upper limit -- obtained via a full Bayesian analysis searching simultaneously over the signal and pulsar noise on the subset of ours six best pulsars -- is $A\approx10^{-14}$. These limits, the most stringent to date at $f<10{\rm nHz}$, exclude the presence of sub-centiparsec binaries with chirp mass $\cal{M}_c>10^9$M$_\odot$ out to a distance of about 25Mpc, and with $\cal{M}_c>10^{10}$M$_\odot$ out to a distance of about 1Gpc ($z\approx0.2$). We show that state-of-the-art SMBHB population models predict $<1\%$ probability of detecting a CGW with the current EPTA dataset, consistent with the reported non-detection. We stress, however, that PTA limits on individual CGW have improved by almost an order of magnitude in the last five years. The continuing advances in pulsar timing data acquisition and analysis techniques will allow for strong astrophysical constraints on the population of nearby SMBHBs in the coming years.
• ### European Pulsar Timing Array Limits On An Isotropic Stochastic Gravitational-Wave Background(1504.03692)
Sept. 9, 2015 astro-ph.CO, astro-ph.IM
We present new limits on an isotropic stochastic gravitational-wave background (GWB) using a six pulsar dataset spanning 18 yr of observations from the 2015 European Pulsar Timing Array data release. Performing a Bayesian analysis, we fit simultaneously for the intrinsic noise parameters for each pulsar, along with common correlated signals including clock, and Solar System ephemeris errors, obtaining a robust 95$\%$ upper limit on the dimensionless strain amplitude $A$ of the background of $A<3.0\times 10^{-15}$ at a reference frequency of $1\mathrm{yr^{-1}}$ and a spectral index of $13/3$, corresponding to a background from inspiralling super-massive black hole binaries, constraining the GW energy density to $\Omega_\mathrm{gw}(f)h^2 < 1.1\times10^{-9}$ at 2.8 nHz. We also present limits on the correlated power spectrum at a series of discrete frequencies, and show that our sensitivity to a fiducial isotropic GWB is highest at a frequency of $\sim 5\times10^{-9}$~Hz. Finally we discuss the implications of our analysis for the astrophysics of supermassive black hole binaries, and present 95$\%$ upper limits on the string tension, $G\mu/c^2$, characterising a background produced by a cosmic string network for a set of possible scenarios, and for a stochastic relic GWB. For a Nambu-Goto field theory cosmic string network, we set a limit $G\mu/c^2<1.3\times10^{-7}$, identical to that set by the {\it Planck} Collaboration, when combining {\it Planck} and high-$\ell$ Cosmic Microwave Background data from other experiments. For a stochastic relic background we set a limit of $\Omega^\mathrm{relic}_\mathrm{gw}(f)h^2<1.2 \times10^{-9}$, a factor of 9 improvement over the most stringent limits previously set by a pulsar timing array.
• ### Timing Noise Analysis of NANOGrav Pulsars(1311.3693)
We analyze timing noise from five years of Arecibo and Green Bank observations of the seventeen millisecond pulsars of the North-American Nanohertz Observatory for Gravitational Waves (NANOGrav) pulsar timing array. The weighted autocovariance of the timing residuals was computed for each pulsar and compared against two possible models for the underlying noise process. The first model includes red noise and predicts the autocovariance to be a decaying exponential as a function of time lag. The second model is Gaussian white noise whose autocovariance would be a delta function. We also perform a nearest-neighbor" correlation analysis. We find that the exponential process does not accurately describe the data. Two pulsars, J1643-1224 and J1910+1256, exhibit weak red noise, but the rest are well described as white noise. The overall lack of evidence for red noise implies that sensitivity to a (red) gravitational wave background signal is limited by statistical rather than systematic uncertainty. In all pulsars, the ratio of non-white noise to white noise is low, so that we can increase the cadence or integration times of our observations and still expect the root-mean-square of timing residual averages to decrease by the square-root of observation time, which is key to improving the sensitivity of the pulsar timing array. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.813408374786377, "perplexity": 1327.6750250340929}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347390755.1/warc/CC-MAIN-20200526081547-20200526111547-00129.warc.gz"} |
http://mathonline.wikidot.com/vector-dot-product-euclidean-inner-product | Vector Dot Product Euclidean Inner Product
# Dot Product (Euclidean Inner Product)
Definition: The Dot Product (for vectors in 2-space and 3-space), also known as the Euclidean Inner Product (for vectors in n-space) of two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ denoted $\vec{u} \cdot \vec{v}$ is a method of multiplying two vectors resulting in a numerical quantity defined by $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta$ or $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$ (component form of the dot product).
Note that we have two forms of the dot product. We will now go on to prove the component form of the dot product, that is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$ in Euclidean 3-space is equivalent to our original definition so $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta = u_1v_1 + u_2v_2 +... + u_nv_n$.
Theorem 1: If $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\mid \vec{u} \mid \mid \vec{v} \mid \cos \theta = u_1v_1 + u_2v_2 + ... + u_nv_n$.
• Proof in 3-Space: Let $\vec{u} = (u_1, u_2, u_3), \vec{v} = (v_1, v_2, v_3) \in \mathbb{R}^3$, and position $\vec{u}$ and $\vec{v}$ so that their initial points are at the origin. Now construct a vector that connects these vectors together. This vector will be $\vec{v} - \vec{u}$ as illustrated:
• Recall the formula for the law of cosines ($a^2 = b^2 + c^2 - 2bc \cos(\theta_a)$) that relates the angle of two connected sides to the lengths of the three sides of the triangle. In this case, the length of the three sides of our triangle will be $\| \vec{u} \|$, $\| \vec{v} \|$ and $\| \vec{v} - \vec{u} \|$. Substituting what we know into the law of cosines we get that:
(1)
\begin{align} a^2 = b^2 + c^2 - 2bc \cos(\theta_a) \\ \| \vec{v} - \vec{u} \|^2 = \| \vec{u} \|^2 + \| \vec{v} \|^2 - 2 \| \vec{u} \| \| \vec{v} \| \cos (\theta_{\| \vec{u} \|}) \\ 2\| \vec{u} \| \| \vec{v} \| \cos(\theta) = \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 \\ \| \vec{u} \| \| \vec{v} \| \cos(\theta) = \frac{1}{2} ( \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 ) \end{align}
• We defined the dot product of two vectors to be $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta$, so we can make that substitution from our equation.
(2)
\begin{align} \vec{u} \cdot \vec{v} = \frac{1}{2} ( \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 ) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} \left (\left( \sqrt{u_{1}^2 + u_{2}^2 + u_{3}^2} \right)^2 + \left( \sqrt{v_{1}^2 + v_{2}^2 + v_{3}^2} \right)^2 - \left(\sqrt{(v_{1} - u_{1})^2 + (v_{2} - u_{2})^2 + (v_{3} - u_{3})^2}\right)^2 \right) \\ \quad \vec{u} \cdot \vec{v} = \frac{1}{2} ( [u_{1}^2 + u_{2}^2 + u_{3}^2] + [v_{1}^2 + v_{2}^2 + v_{3}^2] - [(v_{1}^2 - 2v_{1}u_{1} + u_{1}^2) + (v_{2}^2 - 2v_{2}u_{2} + u_{2}^2 ) + (v_{3}^2 - 2v_{3}u_{3} + u_{3}^2)]) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} ( u_{1}^2 + u_{2}^2 + u_{3}^2 + v_{1}^2 + v_{2}^2 + v_{3}^2 - v_{1}^2 + 2v_{1}u_{1} - u_{1}^2 - v_{2}^2 + 2v_{2}u_{2} - u_{2}^2 - v_{3}^2 + 2v_{3}u_{3} - u_{3}^2) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} (2v_{1}u_{1} + 2v_{2}u_{2}+ 2v_{3}u_{3}) \\ \vec{u} \cdot \vec{v} = v_{1}u_{1} + v_{2}u_{2} + v_{3}u_{3} \\ \vec{u} \cdot \vec{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} \\ \blacksquare \end{align}
# Angles Between Vectors
The original formula for the dot product is helpful in finding the angle between two vectors $\vec{u}$ and $\vec{v}$. For example, if we take the formula for the dot product and rearrange it to isolate cosine, we get that:
(3)
\begin{align} cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \mid \mid \vec{u} \mid \mid \mid \mid \vec{v} \mid \mid } \end{align}
## Acute, Obtuse, and Perpendicular Vector Angles
Recall the following three classifications of angles:
Name of Angle Range of Angle Sign of Cosine
Acute Angles $0 < \theta < \frac{pi}{2}$ $\cos \theta > 0$
Perpendicular (Right) Angles $\theta = \frac{pi}{2}$ $\cos \theta = 0$
Obtuse Angles $\frac{\pi}{2} < \theta < \pi$ $\cos \theta < 0$
We also note that the norm of any vector is always positive, hence, we have now found an important property of the vector dot product. If we know the sign of the dot product between two vectors, then we also know if the angle between them is acute or obtuse since the value of cosine affects the entire sign of the dot product. Furthermore, if $\vec{u} \cdot \vec{v} = 0$, then these two vectors are perpendicular to each other.
(4)
\begin{align} \mathrm{if} \quad \vec{u} \cdot \vec{v} > 0, \quad 0 < \theta < \frac{\pi}{2} \\ \mathrm{if} \quad \vec{u} \cdot \vec{v} < 0, \quad \frac{\pi}{2} < \theta < \pi \\ \mathrm{if} \quad \vec{u} \cdot \vec{v} = 0, \quad \theta = \frac{\pi}{2} \end{align}
# Properties of the Dot Product
Theorem 2 (Commutativity of the Dot Product): For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$, their dot product is commutative, that is $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$.
• Proof: We will use the component form of the dot product to prove this. Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. The proof will follow straight forward:
(5)
\begin{align} \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n \\ \vec{u} \cdot \vec{v} = v_1u_1 + v_2u_2 + ... + v_nu_n \\ \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \\ \blacksquare \end{align}
Theorem 3 (Distributivity of the Dot Product): For any three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$, their dot product is distributive, that is $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$.
• Proof: Let $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$.
(6)
\begin{align} \vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}(v_{1} + w_{1}) + u_{2}(v_{2} + w_{2}) + ... + u_{n}(v_{n} + w_{n}) \\ \vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}v_{1} + u_{1}w_{1} + u_{2}v_{2} + u_{2}w_{2} + ... + u_{n}v_{n} + u_{n}w_{n}\\ \vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}v_{1} + u_{2}v_{2} + ... + u_{n}v_{n} + u_{1}w_{1} + u_{2}w_{2} + ... + u_{n}w_{n} \\ \vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \\ \blacksquare \end{align}
Theorem 4: For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ and some scalar $k$, it follows that $k(\vec{u} \cdot \vec{v}) = (k\vec{u}) \cdot \vec{v}) = \vec{u} \cdot (k \vec{v})$.
• Proof: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ and let $k$ be a scalar.
(7)
\begin{align} k(\vec{u} \cdot \vec{v}) = k(u_{1}v_{1} + u_{2}v_{2} + ... + u_{n}v_{n} \\ k(\vec{u} \cdot \vec{v}) = (ku_{1}v_{1} + ku_{2}v_{2} + ... + ku_{n}v_{n}) \\ k(\vec{u} \cdot \vec{v}) = u_{1}(kv_{1}) + u_{2}(kv_{2}) + ... + u_{n}(kv_{n}) \\ k(\vec{u} \cdot \vec{v}) = \vec{u} \cdot (k\vec{v}) \\ \blacksquare \end{align}
We will omit the rest of the proof as it is practically identical to the above.
Theorem 5: Given any vector $\vec{u} \in \mathbb{R}^n$, it follows that $\vec{u} \cdot \vec{u} ≥ 0$.
• Proof: We first note that $\vec{u} \cdot \vec{u} = u_1^2 + u_2^2 + ... + u_n^2$. All of the elements in this sum are nonnegative so the dot product is not negative either. We note that if $\vec{u} = \vec{0}$, then $\vec{u} \cdot \vec{u} = 0$. $\blacksquare$
# Alternate Form of the Euclidean Inner Product
We will now look at yet another form of the dot product which can be useful in certain circumstances.
Theorem 6: For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\vec{u} \cdot \vec{v} = \frac{1}{4} \mid \mid \vec{u} + \vec{v} \mid \mid^2 - \frac{1}{4} \mid \mid \vec{u} - \vec{v} \mid \mid^2$.
• Proof: We will prove our claim by showing that the righthand side is equal to the lefthand side.
(8)
\begin{align} \quad \frac{1}{4} \mid \mid \vec{u} + \vec{v} \mid \mid ^2 - \frac{1}{4} \mid \mid \vec{u} - \vec{v} \mid \mid ^2 \\ \quad = \frac{1}{4} (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) - \frac{1}{4}(\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) \\ \quad = \frac{1}{4} (\mid \mid \vec{u} \mid \mid ^2 + 2(\vec{u} \cdot \vec{v}) + \mid \mid \vec{v}\mid \mid ^2) - \frac{1}{4}(\mid \mid \vec{u} \mid \mid ^2 - 2(\vec{u} \cdot \vec{v}) + \mid \mid \vec{v}\mid \mid ^2) \\ \quad = \frac{1}{4} \mid \mid \vec{u} \mid \mid ^2 + \frac{1}{2}\vec{u} \cdot \vec{v} + \frac{1}{4}\mid \mid \vec{v}\mid \mid ^2 - \frac{1}{4}\mid \mid \vec{u} \mid \mid ^2 + \frac{1}{2}\vec{u} \cdot \vec{v} - \frac{1}{4}\mid \mid \vec{v}\mid \mid ^2 \\ \quad = \frac{1}{2} \vec{u} \cdot \vec{v} + \frac{1}{2}\vec{u} \cdot \vec{v} \\ \quad = \vec{u} \cdot \vec{v} \\ \blacksquare \end{align} | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 8, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000098943710327, "perplexity": 1630.339901809693}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00322.warc.gz"} |
https://www.physicsforums.com/threads/block-moving-on-an-inclined-plane.753170/ | # Block moving on an inclined plane
1. May 11, 2014
### utkarshakash
1. The problem statement, all variables and given/known data
A block is placed on a plane inclined at an angle θ. The coefficient of friction between the block and the plane is µ = tan θ. The block is given a kick so that it initially moves with speed V horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?
3. The attempt at a solution
The force of friction exactly balances the gravitational force along the incline. So there will be no acceleration along the inclined plane and the trajectory of the block must be a straight line and it continues to move in its original direction of motion. The frictional force will act opposite to its velocity and after a long time the block should stop. So the final velocity should be 0. But this is not the correct answer. I can't figure out what's wrong with my reasoning.
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2. May 11, 2014
### voko
You are ignoring the fact that the force of kinetic friction is always aligned with the velocity. So initially it is purely horizontal and does not oppose the force of gravity that pulls the block sideways, so the velocity vector gets a sideways component, so does the force of friction, etc.
3. May 12, 2014
### utkarshakash
So how should I proceed?
4. May 12, 2014
### voko
The first step here would be to write down the components of the net force acting on the particle in the plane of the incline.
5. May 12, 2014
### utkarshakash
The net force is mgsinα down the incline and frictional force = mgsinα opposite the direction of initial velocity
6. May 12, 2014
### voko
It would be useful to have that written as differential equations.
7. May 12, 2014
### ehild
The force of friction is opposite to the instantaneous velocity.
ehild
8. May 12, 2014
### haruspex
9. May 12, 2014
### voko
There is a very simple integral that can be reached via a mathematical device, or physical intuition.
10. May 12, 2014
### ehild
You do not need to solve the differential equation. The problem asks the speed after very long time.
If you write the differential equations for the velocity components along the slope(vx) and in the horizontal direction (vy), and also the derivative of the speed v, you get a very simple relation between dv/dt and dvx/dt. Also, it is easy to figure out the long-term behaviour of vy.
But I would like to see the solution for v(t)...
ehild
11. May 12, 2014
### dauto
Agreed. I would like to see the complete solution as well.
12. May 13, 2014
### haruspex
Ok.
The forces are mg sin(θ) down the plane, and likewise in the direction opposite to the velocity.
In (s, ψ) coordinates, with ψ as the angle of the velocity to the horizontal in the plane,
$\ddot s = (\sin(\psi)-1)g \sin(\theta)$
$\dot s \dot \psi = g \sin(\theta) \cos(\psi)$
Writing $\dot s = g \sin(\theta) u$, this simplifies to
$\dot u = (\sin(\psi)-1)$
$u \dot \psi = \cos(\psi)$
Eliminating u we get:
$\dot\psi^2(1-2 \sin(\psi)) = \ddot \psi \cos(\psi)$, with initial conditions $\psi(0) = 0, \dot \psi(0) = \frac 1{u(0)} = \frac v{g \sin(\theta)}$
Rewriting that as $\dot\psi(\sec(\psi)-2 \tan(\psi)) = \frac{\ddot \psi }{\dot \psi }$ we can integrate to obtain
$\dot \psi = A \cos(\psi)(1+\sin(\psi))$, $u = \frac 1{A(1+\sin(\psi))}$
The initial conditions give $A = \frac{g \sin(\theta)}{v}$, so $\dot s = \frac v {(1+\sin(\psi))}$
We want $\mathop{\lim}_{t\rightarrow \infty}\dot s = \mathop{\lim}_{\psi\rightarrow \frac{\pi}{2}}\dot s = \frac v2$
(What's a better way to write limits in LaTex?)
You could also solve to find t as a function of ψ.
13. May 13, 2014
### ehild
$$\mathop{\lim}_{t\rightarrow \infty}\dot s = \mathop{\lim}_{\psi\rightarrow \frac{\pi}{2}}\dot s = \frac v2$$
Use $$instead of $, but then it becomes a new line. ehild 14. May 13, 2014 ### voko Nicely done, haruspex. For limits, there is a "limits" modifier that forces indices above and below the current symbol:$\lim\limits^{up}_{down} \int\limits_{from}^{to}\sum\limits_{begin}^{end}$. 15. May 13, 2014 ### utkarshakash Can you please explain what have you done here? It's difficult for me to follow your approach. What does (s,ψ) coordinates mean here? 16. May 13, 2014 ### voko You do not need to follow that. Use the approach mentioned by ehild. 17. May 13, 2014 ### TSny 18. May 13, 2014 ### utkarshakash 19. May 13, 2014 ### haruspex As ehild and voko point out, there's a much easier way to obtain$\dot s = \frac{v}{1+\sin(\psi)}$. If not, I would have given hints, not posted my calculus solution in detail. In (s,ψ) coordinates, s is the distance along the path from some fixed origin, and ψ is the angle the path makes to some fixed direction at that point. (I chose that to be the initial direction of motion, with positive ψ being down the plane.) The link I posted shows how to convert to and from other systems. It gives much simpler differential equations in some cases, like this one. The acceleration along the path,$\ddot s $, comes from the frictional force plus the sin(ψ) component of the downplane gravitational force:$\ddot s = (\sin(\psi)-1)g \sin(\theta)$The acceleration normal to the path,$\dot s \dot \psi$, comes from the cos(ψ) component of the downplane gravitational force:$\dot s \dot \psi = g \sin(\theta) \cos(\psi)$Since g sin(θ) occurs as a common factor, it can be eliminated by incorporating it into s. 20. May 13, 2014 ### ehild Cartesian coordinates might be used, too, and that approach might be more familiar to the OP. The x axis is horizontal, directed along the initial velocity. The y axis is parallel with the slope, pointing downward. See picture. The downward force is Fg=mgsinθ. The magnitude of the force of friction is also mgsinθ. The force of friction is opposite to the velocity v. It can be written as the negative of the magnitude multiplied by the unit vector in the direction of the velocity \vec F_f=-mgsinθ \frac{\vec v}{v} If vx, vy are the components of the velocity and v is the speed$ v= \sqrt{v_x^2+v_y^2}$the horizontal component of the friction is$-mg\sinθ \frac{v_x}{v}$and the vertical component is$-mg\sinθ \frac{v_y}{v}##. Write out the x and y components of acceleration, and also the time derivative of the speed v. Use the notation A=gsinθ.$$\dot v_x=-A\frac{v_x}{v}\dot v_y=A(1-\frac{v_y}{v})\dot v=\frac{v_x \dot v_x +v_y \dot v_y}{v}
Substitute the first two equations for vx, vy into the third one. You will see that the time derivative of vy is equal and opposite to that of v. Integrate.
It can be proved mathematically, but it is clear that the horizontal component of the velocity tends to zero with time, so the speed becomes equal to the vertical velocity component v→vy.
ehild
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https://brilliant.org/practice/uniform-acceleration/?subtopic=applications-of-integration&chapter=displacement-2 | Calculus
# Uniform Acceleration
A ball is currently $1000m$ above the ground and held at rest. It is dropped at $t = 0$, and experiences gravitational acceleration of $9.8 \, m / s^2$.
When $3$ seconds is passed, what is the height of the ball from the ground?
Object P with an initial velocity of $9\text{ m/s}$ starts to move eastward along a straight line under a constant acceleration of $8\text{ m/s}^2$ from point A. At the same time, another object Q starts to move eastward under a constant acceleration of $9\text{ m/s}^2$ from rest at point A. How far does object Q travel in meters when the two objects meet again?
An object with an initial velocity of $9\text{ m/s}$ moves along the $x$-axis with constant acceleration. After $8$ seconds, its velocity is $49\text{ m/s}.$ How far did it travel during the $8$ seconds in meters?
The speed of a bullet is measured to be $640 \text{ m/s}$ as the bullet emerges from its $1.20 \text{ m}$ long barrel. Assuming a constant acceleration, find the time that the bullet spends in the barrel after it is fired.
A car that is initially traveling at $20\text{ m/s}$ accelerates uniformly in a straight line for $5$ seconds at a rate of $6\text{ m/s}^2.$ How far in meters does the car travel during the $5$ second period?
× | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8092201948165894, "perplexity": 342.72490406019983}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00109.warc.gz"} |
http://mathhelpforum.com/calculus/209966-related-rate.html | Math Help - Related Rate
1. Related Rate
A lighthouse is .8 miles away from shore. The light rotates 6 times a minute. How fast in MPH is the spot of light moving along the shore when it is .6 miles away?
So I did the problem and I'm not sure but I think I got 1/.6. is that right?
2. Re: Related Rate
Let's let $D$ be the perpendicular distance from the lighthouse to the shore (a straight line) and $x$ be the distance of the spot of light from the point on the shore intersected by the distance line segment. Let $\theta$ be the angle subtended by the distance line segment and the beam of light. Let $\omega=\frac{d\theta}{dt}$. We may then state:
$\tan(\theta)=\frac{x}{D}$
Differentiating with respect to time $t$, we find:
$\omega\sec^2(\theta)=\frac{1}{D}\cdot\frac{dx}{dt}$
Solving for $\frac{dx}{dt}$ and using $\sec(\theta)=\frac{\sqrt{x^2+D^2}}{D}$ we have:
$\frac{dx}{dt}=\frac{\omega(x^2+D^2)}{D}$
For this problem, we are given:
$\omega=6\cdot2\pi\frac{\text{rad}}{\text{min}} \cdot\frac{60\text{ min}}{1\text{ hr}}=720\pi\,\frac{1}{\text{hr}}$
$x=\frac{3}{5}\,\text{mi}$
$D=\frac{4}{5}\,\text{mi}$
Now, plug 'n' chug. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870100617408752, "perplexity": 358.2510988503963}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500826025.8/warc/CC-MAIN-20140820021346-00368-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://par.nsf.gov/biblio/10304428-erratum-azimuthal-anisotropy-relativistic-heavy-ion-collider-first-fourth-harmonics-phys-rev-lett | Erratum: Azimuthal Anisotropy at the Relativistic Heavy Ion Collider: The First and Fourth Harmonics [Phys. Rev. Lett. 92 , 062301 (2004)]
Authors:
; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; more »
Award ID(s):
Publication Date:
NSF-PAR ID:
10304428
Journal Name:
Physical Review Letters
Volume:
127
Issue:
6
ISSN:
0031-9007
1. Abstract We show that for some even $k\leqslant 3570$ and all $k$ with $442720643463713815200|k$, the equation $\phi (n)=\phi (n+k)$ has infinitely many solutions $n$, where $\phi$ is Euler’s totient function. We also show that for a positive proportion of all $k$, the equation $\sigma (n)=\sigma (n+k)$ has infinitely many solutions $n$. The proofs rely on recent progress on the prime $k$-tuples conjecture by Zhang, Maynard, Tao, and PolyMath.
2. A bstract We report the first measurement of the exclusive cross sections e + e − → $$B\overline{B}$$ B B ¯ , e + e − → $$B{\overline{B}}^{\ast }$$ B B ¯ ∗ , and e + e − → $${B}^{\ast }{\overline{B}}^{\ast }$$ B ∗ B ¯ ∗ in the energy range from 10 . 63 GeV to 11 . 02 GeV. The B mesons are fully reconstructed in a large number of hadronic final states and the three channels are identified using a beam-constrained-mass variable. The shapes of the exclusive cross sections showmore » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9835489988327026, "perplexity": 971.7013675591182}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335303.67/warc/CC-MAIN-20220929003121-20220929033121-00775.warc.gz"} |
https://www.physicsforums.com/threads/find-the-limit.331310/ | # Find the limit
1. Aug 18, 2009
### yanjt
Hi,
I am wondering whether i answer the question correctly.
The question is,either find the limit of explain why there is no limit exists as k goes to infinity for
k sin(k$$\pi$$/3).
My answer for this question is that there is no limit exists since infinity multiply by any number will become infinity.Hence,regardless what value we get from sin(k$$\pi$$/3),it will become infinity when it is multiplied by infinity.Can I explain in this way?Thanks!
2. Aug 18, 2009
### JG89
Assuming by number you mean a constant value, then that is a good way to think about it intuitively. You know that for some values of k, sin(kpi/3) will be positive, and for other values of k it will be negative. k is always positive, so the expression k*sin(kpi/3) will switch between large positive values and large negative values, depending on the value of k.
If you need to make this into a rigorous argument, it shouldn't be too hard.
3. Aug 18, 2009
### yanjt
So there is no limit for this question?I can just write the statement out without showing any working?
4. Aug 18, 2009
### Дьявол
sin(kп/3) $\in$ [-1,1]
So k multiplied by any number in the interval [-1,1] (if k goes to infinity) is again infinity.
5. Aug 18, 2009
### Cyosis
This is simply not true, $-1*\infty=-\infty$. Also the OPs original argument is not true, because a limit resulting in infinity is defined just fine. The problem here as explained by JG80 is that the function oscillates and this oscillation goes on forever.
They ask for an explanation so yes explaining in words as to why there is no limit would be just fine.
6. Aug 18, 2009
### lanedance
one way you could look a litttle more rigourously is to pick some arbitrary montonic sequences {kn}, that tend to infinity. Then consider the sequences {f(kn)},
If you can find sequences with different limits, or a sequence that does not converge, you have shown the limit does not exist.
Or in similar fashion, u could find positivek,k' with N<k<k', such that f(k)>N, f(k')<-N, then as N is arbitrary, you have shown f(k) is unbounded from both above & below as k tends to infinity and hence has no limit, which is effectivey what cyosis & JG89 have said
Last edited: Aug 18, 2009
7. Aug 18, 2009
### Дьявол
I do not know what is wrong with my statement.
$-1*\infty$ equals infinity just negative. It is again infinity.
My statement was meant to express $x\in [-1,1] * \infty = \pm \infty$
By saying infinity, I mean $\pm \infty$.
Regards.
8. Aug 18, 2009
### Elucidus
There are errors in saying "infinity times a number is some infinity." The problem comes from the whether the other factor can equal zero or not or changes sign.
In the limit $$\lim_{x \rightarrow \infty}xe^{-x}$$ one could argue that since x approaches infinity and that e-x is a positve number then this limit diverges to infinity. Unfortunately this argument would be grossly wrong. The limit is in fact 0.
The issue with the limit being discussed is that the sine function is oscillating between -1 and 1 (and sometimes equalling 0). The explanation as to why the limit is undefined is that there is no finite number to which it converges and it does not diverge to either infinity. Both of these can be demonstrated by examing sequence of values that behave differently.
One needs to show why it does not converge and why it does not diverge to some infinity.
(BTW: a limit that diverges to infinity is still defined.)
--Elucidus
9. Aug 18, 2009
### HallsofIvy
The difficulty with saying $\pm \infty= \infty$ is that it mixes concepts of "infinity". The "one point compactification" of the real numbers adds the single point "$\infty$" and is topologically equivalent to a circle while the "Stone-Cech" compactification add $+\infty$ and $-\infty$ and is topologically equivalent to a closed line segment.
Also, some texts would say that {n} "diverges to infinity" while {-n} "diverges to negative infinity" and {$(-1)^{n}n$} simply "diverges". | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9444091320037842, "perplexity": 674.779863411043}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645538.8/warc/CC-MAIN-20180318052202-20180318072202-00795.warc.gz"} |
https://au.answers.yahoo.com/question/index?qid=20190615233712AA2j49h | Britney asked in Science & MathematicsPhysics · 5 months ago
# A 3-kg rock swings in a circle of radius 5 m. making one complete trip every 2 seconds. What is the centripetal acceleration of the rock?
Relevance
• oubaas
Lv 7
5 months ago
ω = 2PI/T = 2PI/2 = PI rad/sec
ac = ω^2*r = 3.1416^2*5 = 49.35 m/sec^2
• 5 months ago
Centripetal acceleration = v^2 ÷ r
During the two seconds, the rock moves a distance of the circumference of a circle that has a radius.5 meters..
d = 2 * π * 5 = 10 * π meters
v = d ÷ t = 10 * π ÷ 2 = 5 * π
The speed of the rock is approximately 15.7 m/s.
Centripetal acceleration = 25 * π^2 ÷ 5 = 5 * π^2
This is approximately 49.3 m/s^2. I hope this is helpful for you.
• 5 months ago
Centripetal acceleration is v²/r where v is velocity given by time taken to traverse the circumference in one rotation which is 2π*5 / 2 m/s = 5π² m/s.
So we substitute (5π)² /5 to get Centripetal acceleration is 5π² m/s²
Given that the mass of the rock is 3 kg, we may determine the force due to centripetal acceleration using the formula Force = mass x acceleration
F = ma, = 3*5π² = 148.0 m/s²
• 5 months ago
The radius is 5 m, so the circumference is:
C = 2πr
C = 10π
The rock completes one revolution in 2 seconds, so its velocity is:
v = C / t
v = 10π / 2
v = 5π
We can now find the centripetal acceleration:
a = v² / r
a = (5π)² / 5
a = 5π²
a ≈ 49.3 m/s²
Still have questions? Get answers by asking now. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9170839190483093, "perplexity": 2586.0086247013433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668716.69/warc/CC-MAIN-20191116005339-20191116033339-00493.warc.gz"} |
https://groupprops.subwiki.org/wiki/Index | # Index of a subgroup
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This article is about a basic definition in group theory. The article text may, however, contain advanced material.
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## Definition
### Symbol-free definition
The index of a subgroup in a group is the following equivalent things:
1. The number of left cosets of the subgroup
2. The number of right cosets of the subgroup
The collection of left cosets is sometimes termed the coset space, so in this language, the index of a subgroup is the cardinality of its coset space.
### Definition with symbols
Given a subgroup $H$ of a group $G$, the index of $H$ in $G$, denoted $[G:H]$, is defined in the following ways:
1. It is the number of left cosets of $H$ in $G$, i.e. the number of sets of the form $xH$.
2. It is the number of right cosets of $H$ in $G$, i.e. the number of sets of the form $Hx$.
The collection of left cosets of $H$ in $G$ is sometimes termed the coset space, and is denoted $G/H$. With this notation, the index of $H$ in $G$, is the cardinality $\left|G/H\right|$.
### Equivalence of definitions
The equivalence of definitions follows from the fact that there is a natural bijection between the collection of left cosets of a subgroup, and the collection of its right cosets, given by the map $g \mapsto g^{-1}$
Further information: Left and right coset spaces are naturally isomorphic
### Further note for finite groups
When the group is finite, then by Lagrange's theorem, the index of a subgroup is the ratio of the order of the group to the order of the subgroup.
## Facts
### Multiplicativity of the index
Further information: Index is multiplicative If $H \le K \le G$, then we have:
$[G:K][K:H] = [G:H]$
In other words, the number of cosets of $H$ in $G$ equals the number of cosets of $H$ in $K$, times the number of cosets of $K$ in $G$.
In fact, more is true. We can set up a bijection as follows:
$G/K \times K/H \to G/H$
However, this bijection is not a natural one, and, in order to define it, we first need to choose a system of coset representatives of $H$.
### Effect of intersection on the index
Further information: Conjugate-intersection index theorem If $H_1$ and $H_2$ are two subgroups of $G$, then the index of $H_1 \cap H_2$ is bounded above by the product of the indices of $H_1$ and of $H_2$.
This follows as a consequence of the product formula. Note that equality holds if and only if $H_1H_2 = G$.
Note that in case $H_1$ and $H_2$ are conjugate subgroups of index $r$, the index of $H_1 \cap H_2$ is bounded above by $r(r-1)$.
## References
### Textbook references
• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Page 57, Point (6.8) (definition in paragraph, defined as number of left cosets)
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 90 (formal definition, defined as number of left cosets) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 40, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9740868806838989, "perplexity": 242.01935415252777}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00553.warc.gz"} |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-7-functions-and-graphs-7-5-formulas-applications-and-variation-7-5-exercise-set-page-488/75 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\color{blue}{y=\dfrac{4wx^2}{z}}$
RECALL: (1) When $y$ varies directly as $x$, the equation of the variation is $y=kx$ . (2) When $y$ varies inversely as $x$, the equation of the variation is $y=\frac{k}{x}$. (3) When $y$ varies jointly as $x$ and $z$, the equation of the variation is $y=kxz$. $y$ varies jointly as $w$ and the square of $x$ and inversely as $z$. Thus, the equation of the variation is $y=\dfrac{kwx^2}{z}$. To find the value of $k$, substitute the given values to obtain: $$y=\dfrac{kwx^2}{z} \\49=\dfrac{k\cdot3\cdot 7^2}{12} \\49=\dfrac{3k\cdot 49}{12} \\49=\dfrac{147k}{12} \\\frac{12}{147} \cdot 49=\frac{147k}{12} \cdot \frac{12}{147} \\4=k$$ Thus, the equation of the variation is: $\color{blue}{y=\dfrac{4wx^2}{z}}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9794525504112244, "perplexity": 163.48688351924906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583514314.87/warc/CC-MAIN-20181021181851-20181021203351-00394.warc.gz"} |
http://science.sciencemag.org/content/189/4201/457 | Reports
# Solar Proton Events: Stratospheric Sources of Nitric Oxide
See allHide authors and affiliations
Science 08 Aug 1975:
Vol. 189, Issue 4201, pp. 457-459
DOI: 10.1126/science.189.4201.457
## Abstract
The production of nitric oxide (NO) in the stratosphere during each of the solar proton events of November 1960, September 1966, and August 1972 is calculated to have been comparable to or larger than the total average annual production of NO by the action of galactic cosmic rays. It is therefore very important to consider the effect of solar proton events on the temporal and spatial distribution of ozone in the stratosphere. A study of ozone distribution after such events may be particularly important for validating photochemical-diffusion models. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.978796124458313, "perplexity": 1787.2878016972259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00536.warc.gz"} |
https://www.mathcounterexamples.net/2015/05/ | # Pointwise convergence and properties of the limit (part 1)
We look here at the continuity of a sequence of functions that converges pointwise and give some counterexamples of what happens versus uniform convergence.
### Recalling the definition of pointwise convergence
We consider here real functions defined on a closed interval $$[a,b]$$. A sequence of functions $$(f_n)$$ defined on $$[a,b]$$ converges pointwise to the function $$f$$ if and only if for all $$x \in [a,b]$$ $$\displaystyle \lim\limits_{n \to +\infty} f_n(x) = f(x)$$. Pointwise convergence is weaker than uniform convergence.
### Pointwise convergence does not, in general, preserve continuity
Suppose that $$f_n \ : \ [0,1] \to \mathbb{R}$$ is defined by $$f_n(x)=x^n$$. For $$0 \le x <1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 0$$, while if $$x = 1$$ then $$\displaystyle \lim\limits_{n \to +\infty} x^n = 1$$. Hence the sequence $$f_n$$ converges to the function equal to $$0$$ for $$0 \le x < 1$$ and to $$1$$ for $$x=1$$. Although each $$f_n$$ is a continuous function of $$[0,1]$$, their pointwise limit is not. $$f$$ is discontinuous at $$1$$. We notice that $$(f_n)$$ doesn't converge uniformly to $$f$$ as for all $$n \in \mathbb{N}$$, $$\displaystyle \sup\limits_{x \in [0,1]} \vert f_n(x) - f(x) \vert = 1$$. That's reassuring as uniform convergence of a sequence of continuous functions implies that the limit is continuous! Continue reading Pointwise convergence and properties of the limit (part 1)
# The set of all commutators in a group need not be a subgroup
I here provide a simple example of a group whose set of commutators is not a subgroup. The counterexample is due to P.J. Cassidy (1979).
### Description of the group $$G$$
Let $$k[x,y]$$ denote the ring of all polynomials in two variables over a field $$k$$, and let $$k[x]$$ and $$k[y]$$ denote the subrings of all polynomials in $$x$$ and in $$y$$ respectively. $$G$$ is the set of all upper unitriangular matrices of the form
$A=\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)$ where $$f(x) \in k[x]$$, $$g(y) \in k[y]$$, and $$h(x,y) \in k[x,y]$$. The matrix $$A$$ will also be denoted $$(f,g,h)$$.
Let’s verify that $$G$$ is a group. The products of two elements $$(f,g,h)$$ and $$(f^\prime,g^\prime,h^\prime)$$ is
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & f^\prime(x) & h^\prime(x,y) \\ 0 & 1 & g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$
$=\left(\begin{array}{ccc} 1 & f(x)+f^\prime(x) & h(x,y)+h^\prime(x,y)+f(x)g^\prime(y) \\ 0 & 1 & g(y)+g^\prime(y) \\ 0 & 0 & 1 \end{array}\right)$ which is an element of $$G$$. We also have:
$\left(\begin{array}{ccc} 1 & f(x) & h(x,y) \\ 0 & 1 & g(y) \\ 0 & 0 & 1 \end{array}\right)^{-1} = \left(\begin{array}{ccc} 1 & -f(x) & f(x)g(y) – h(x,y) \\ 0 & 1 & -g(y) \\ 0 & 0 & 1 \end{array}\right)$ proving that the inverse of an element of $$G$$ is also an element of $$G$$. Continue reading The set of all commutators in a group need not be a subgroup
# A topological vector space with no non trivial continuous linear form
We consider here the $$L^p$$- spaces of real functions defined on $$[0,1]$$ for which the $$p$$-th power of the absolute value is Lebesgue integrable. We focus on the case $$0 < p < 1$$. We'll prove that those $$L^p$$-spaces are topological vector spaces on which there exists no continuous non-trivial linear forms (i.e. not vanishing identically). Continue reading A topological vector space with no non trivial continuous linear form
# A nowhere locally bounded function
In that article, I described some properties of Thomae’s function$$f$$. Namely:
• The function is discontinuous on $$\mathbb{Q}$$.
• Continuous on $$\mathbb{R} \setminus \mathbb{Q}$$.
• Its right-sided and left-sided limits vanish at all points.
Let’s modify $$f$$ to get function $$g$$ defined as follow:
$g: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & q \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.$ $$f$$ and $$g$$ both vanish on the set of irrational numbers, while on the set of rational numbers, $$g$$ is equal to the reciprocal of $$f$$. We now consider an open subset $$O \subset \mathbb{R}$$ and $$x \in O$$. As $$f$$ right-sided and left-sided limits vanish at all points, we have $$\lim\limits_{n \to +\infty} f(x_n) = 0$$ for all sequence $$(x_n)$$ of rational numbers converging to $$x$$ (and such a sequence exists as the rational numbers are everywhere dense in the reals). Hence $$\lim\limits_{n \to +\infty} g(x_n) = + \infty$$ as $$f$$ is positive.
We can conclude that $$g$$ is nowhere locally bounded. The picture of the article is a plot of function $$g$$ on the rational numbers $$r = \frac{p}{q}$$ in lowest terms for $$0 < r < 1$$ and $$q \le 50$$.
# A function continuous at all irrationals and discontinuous at all rationals
Let’s discover the beauties of Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.
Thomae’s function is a real-valued function defined as:
$f: \left|\begin{array}{lrl} \mathbb{R} & \longrightarrow & \mathbb{R} \\ x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\ \frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0 \end{array}\right.$
### $$f$$ is periodic with period $$1$$
This is easy to prove as for $$x \in \mathbb{R} \setminus \mathbb{Q}$$ we also have $$x+1 \in \mathbb{R} \setminus \mathbb{Q}$$ and therefore $$f(x+1)=f(x)=0$$. While for $$y=\frac{p}{q} \in \mathbb{Q}$$ in lowest terms, $$y+1=\frac{p+q}{q}$$ is also in lowest terms, hence $$f(y+1)=f(y)=\frac{1}{q}$$. Continue reading A function continuous at all irrationals and discontinuous at all rationals
# Generating the symmetric group with a transposition and a maximal length cycle
Can the symmetric group $$\mathcal{S}_n$$ be generated by any transposition and any $$n$$-cycle for $$n \ge 2$$ integer? is the question we deal with.
We first recall some terminology:
Symmetric group
The symmetric group $$\mathcal{S}_n$$ on a finite set of $$n$$ symbols is the group whose elements are all the permutations of the $$n$$ symbols. We’ll denote by $$\{1,\dots,n\}$$ those $$n$$ symbols.
Cycle
A cycle of length $$k$$ (with $$k \ge 2$$) is a cyclic permutation $$\sigma$$ for which there exists an element $$i \in \{1,\dots,n\}$$ such that $$i, \sigma(i), \sigma^2(i), \dots, \sigma^k(i)=i$$ are the only elements moved by $$\sigma$$. We’ll denote the cycle $$\sigma$$ by $$(s_0 \ s_1 \dots \ s_{k-1})$$ where $$s_0=i, s_1=\sigma(i),\dots,s_{k-1}=\sigma^{k-1}(i)$$.
Transposition
A transposition is a cycle of length $$2$$. We denote below the transposition of elements $$a \neq b$$ by $$(a \ b)$$ or $$\tau_{a,b}$$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9829190969467163, "perplexity": 116.80256023770487}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250606226.29/warc/CC-MAIN-20200121222429-20200122011429-00455.warc.gz"} |
http://mathhelpforum.com/advanced-applied-math/191328-derive-differential-equation-boundary-conditions-heat-transfer.html | ## Derive differential equation and boundary conditions for heat transfer.
I am trying to derive the differential equation along with boundary conditions for steady state operation of the following system. This is an extended surface (see picture) attached to a cooling tower to enhance heat transfer to the ambient. The tower to which it is attached is 200 C and the ambient temperature is 5 C.
I am very lost and am unable to begin this problem. Any tips or suggestions for these problems in general as well as a path to follow for solving this particular problem would be greatly appreciated. I am capable of doing the math, just not the equally important task of deriving the equation. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.865546464920044, "perplexity": 133.17994022930137}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123635.74/warc/CC-MAIN-20170423031203-00298-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://arxiv.org/abs/1408.5067 | physics.plasm-ph
(what is this?)
# Title: Simulations of an ultracold, neutral plasma with equal masses
Abstract: The results of a theoretical investigation of an ultracold, neutral plasma composed of equal mass positive and negative charges are reported. In our simulations, the plasma is created by the fast dissociation of a neutral particle. The temperature of the plasma is controlled by the relative energy of the dissociation. We studied the early time evolution of this system where the initial energy was tuned so that the plasma is formed in the strongly coupled regime. In particular, we present results on the temperature evolution and three body recombination. In the weakly coupled regime, we studied how an expanding plasma thermalizes and how the scattering between ions affects the expansion. Because the expansion causes the density to drop, the velocity distribution only evolves for a finite time with the final distribution depending on the number of particles and initial temperature of the plasma.
Subjects: Plasma Physics (physics.plasm-ph); Atomic Physics (physics.atom-ph) DOI: 10.1088/0953-4075/47/24/245701 Cite as: arXiv:1408.5067 [physics.plasm-ph] (or arXiv:1408.5067v1 [physics.plasm-ph] for this version)
## Submission history
From: Francis Robicheaux [view email]
[v1] Thu, 21 Aug 2014 17:00:33 GMT (28kb) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9248225092887878, "perplexity": 1536.3069165799611}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647530.92/warc/CC-MAIN-20180320185657-20180320205657-00429.warc.gz"} |
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