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https://www.chromforum.org/viewtopic.php?f=5&t=40121&p=192751 | 1,606,698,327,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00432.warc.gz | 596,966,999 | 12,190 | unknown volume hplc - Chromatography Forum
## unknown volume hplc
Basic questions from students; resources for projects and reports.
### unknown volume hplc
I have a known concentration and peak area is it possible to work out the volume?
### Re: unknown volume hplc
Perhaps. It would be;
Concentration (let's say ug/mL) x Injection Volume (in mL) = ug (Mass)
where 1 mL = 1000 uL
Peak area has nothing to do with the calculation. Remember, HPLC is a 'mass' device and not volume. Why do you want to know?
### Re: unknown volume hplc
To calculate the actual volume of the peak you will need to know the detector response versus the concentration of the target molecule. It should be a linear curve.
### Re: unknown volume hplc
There's more than a bit of ambiguity here:
1. Where/how is your "known concentration" determined? Is it the concentration of the sample that was injected? Is it the concentration at the peak maximum (and if so, how did you determine it)?
2. What units are you using for peak area? Integrator counts? mV?
3. What volume are you trying to determine (the volume injected? the volume of the peak itself?)
4. If you're after the volume of the peak, what fraction of the analyte do you wish to account for? Chromatographic peaks are, to a good first approximation, Gaussian. The "baseline" width of a peak is approximately 4 x sigma, and so includes roughly 95% of the area. Is that what you want, or are you looking at 99%, or . . . ?
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
### Re: unknown volume hplc
i have a calibration curve of conc vs peak area
I am trying to determine how much volume of a drink you have to consume to be toxic (500mM)
the peak area units is (iss)
### Re: unknown volume hplc
Ok. So you inject a sample of your drink and measure the peak area of the compound of interest. Use your calibration curve to convert that to concentration.
Concentration is mass (or moles) divided by volume. You should be able to take it from there.
-- Tom Jupille
LC Resources / Separation Science Associates
tjupille@lcresources.com
+ 1 (925) 297-5374
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Subscribe to our eNewsletter with daily, weekly or monthly updates: Food, Environmental, (Bio)Pharmaceutical, Bioclinical, Liquid Chromatography, Gas Chromatography and Mass Spectrometry. | 715 | 2,919 | {"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} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.910288 |
https://www.klipinterest.com/members/casinobtc16100485/activity/22643/ | 1,675,550,056,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00627.warc.gz | 856,383,292 | 32,490 | • Dice games of chance, dice games english posted an update 2 years ago
Dice games of chance
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The Secret of the Opera – 110.5 ltc | 1,977 | 8,570 | {"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} | 2.65625 | 3 | CC-MAIN-2023-06 | latest | en | 0.96325 |
http://de.metamath.org/mpegif/nn2ge.html | 1,597,309,816,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738964.20/warc/CC-MAIN-20200813073451-20200813103451-00278.warc.gz | 27,492,070 | 6,464 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > nn2ge Structured version Unicode version
Theorem nn2ge 10601
Description: There exists a positive integer greater than or equal to any two others. (Contributed by NM, 18-Aug-1999.)
Assertion
Ref Expression
nn2ge
Distinct variable groups: , ,
Proof of Theorem nn2ge
StepHypRef Expression
1 nnre 10583 . . 3
3 nnre 10583 . . 3
5 leid 9711 . . . . . . 7
65biantrud 505 . . . . . 6
76biimpa 482 . . . . 5
83, 7sylan 469 . . . 4
9 breq2 4399 . . . . . 6
10 breq2 4399 . . . . . 6
119, 10anbi12d 709 . . . . 5
1211rspcev 3160 . . . 4
138, 12syldan 468 . . 3 | 273 | 667 | {"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} | 2.625 | 3 | CC-MAIN-2020-34 | latest | en | 0.185878 |
https://www.physicsforums.com/threads/magnetism-in-infinite-conducting-slab.38818/ | 1,526,944,519,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00143.warc.gz | 820,437,117 | 14,800 | # Homework Help: Magnetism in infinite conducting slab
1. Aug 9, 2004
### daveed
"a conducting slab has infinite extent in the x and y directions and thickness L in the z direction. The slab is centered at z=0 and carries a uniform current density J=Ji where i, j, and k are unit vectors in the x, y, and z directiosn."
-Find the magnetic field B at all points.
-A square loop of side a is placed distance b above the slab. The loop has unit normal vector n=sin(q)i+cos(q)j and applied current J. what is the net force and net torque to the loop as a function of q?
-the applied current I is now removed from the loop and the current density in the slab J=Ji is reduced to zero over time T. The wire used to construct the loop has resistance/unit S. How much charge flows through each cross section of the loop wire due to the reduction in current density.
2. Aug 10, 2004
### Locrian
What have you tried so far?
It seems to me ampere's law would work well in determining the magnetic field; use a square loop such that you know exactly how much current passing through it and the only field is at the top and bottom. Integrating a line of wires across the surface and then again through the depth is another option, but sounds horrible to me.
Once you have the magnetic field, the second part should come together easily enough.
3. Dec 9, 2004
### daveed
i really dont know how to go about doing this when its not a single wire... sorry =(
4. Dec 9, 2004
### Tide
Have you considered the integral form of Ampere's Law? | 375 | 1,535 | {"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} | 2.96875 | 3 | CC-MAIN-2018-22 | longest | en | 0.941107 |
http://www.quantstart.com/articles/Bayesian-Inference-of-a-Binomial-Proportion-The-Analytical-Approach/ | 1,660,786,957,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00463.warc.gz | 83,058,537 | 12,132 | Bayesian Inference of a Binomial Proportion - The Analytical Approach
Bayesian Inference of a Binomial Proportion - The Analytical Approach
Updated for Python 3.8, April 2022
In the previous article on Bayesian statistics we examined Bayes' rule and considered how it allowed us to rationally update beliefs about uncertainty as new evidence came to light. We mentioned briefly that such techniques are becoming extremely important in the fields of data science and quantitative finance.
In this article we are going to expand on the coin-flip example that we studied in the previous article by discussing the notion of Bernoulli trials, the beta distribution and conjugate priors.
Our goal in this article is to allow us to carry out what is known as "inference on a binomial proportion". That is, we will be studying probabilistic situations with two outcomes (e.g. a coin-flip) and trying to estimate the proportion of a repeated set of events that come up heads or tails.
Our goal is to estimate how fair a coin is. We will use that estimate to make predictions about how many times it will come up heads when we flip it in the future.
While this may sound like a rather academic example, it is actually substantially more applicable to real-world applications than may first appear. Consider the following scenarios:
• Engineering: Estimating the proportion of aircraft turbine blades that possess a structural defect after fabrication
• Social Science: Estimating the proportion of individuals who would respond "yes" on a census question
• Medical Science: Estimating the proportion of patients who make a full recovery after taking an experimental drug to cure a disease
• Corporate Finance: Estimating the proportion of transactions in error when carrying out financial audits
• Data Science: Estimating the proportion of individuals who click on an ad when visiting a website
As can be seen, inference on a binomial proportion is an extremely important statistical technique and will form the basis of many of the articles on Bayesian statistics that follow.
## The Bayesian Approach
While we motivated the concept of Bayesian statistics in the previous article, I want to outline first how our analysis will proceed. This will motivate the following (rather mathematically heavy) sections and give you a "bird's eye view" of what a Bayesian approach is all about.
As we stated above, our goal is estimate the fairness of a coin. Once we have an estimate for the fairness, we can use this to predict the number of future coin flips that will come up heads.
We will learn about the specific techniques as we go while we cover the following steps:
1. Assumptions - We will assume that the coin has two outcomes (i.e. it won't land on its side), the flips will appear randomly and will be completely independent of each other. The fairness of the coin will also be stationary, that is it won't alter over time. We will denote the fairness by the parameter $\theta$.
2. Prior Beliefs - To carry out a Bayesian analysis, we must quantify our prior beliefs about the fairness of the coin. This comes down to specifying a probability distribution on our beliefs of this fairness. We will use a relatively flexible probability distribution called the beta distribution to model our beliefs.
3. Experimental Data - We will carry out some (virtual) coin-flips in order to give us some hard data. We will count the number of heads $z$ that appear in $N$ flips of the coin. We will also need a way of determining the probability of such results appearing, given a particular fairness, $\theta$, of the coin. For this we will need to discuss likelihood functions, and in particular the Bernoulli likelihood function.
4. Posterior Beliefs - Once we have a prior belief and a likelihood function, we can use Bayes' rule in order to calculate a posterior belief about the fairness of the coin. We couple our prior beliefs with the data we have observed and update our beliefs accordingly. Luckily for us, if we use a beta distribution as our prior and a Bernoulli likelihood we also get a beta distribution as a posterior. These are known as conjugate priors.
5. Inference - Once we have a posterior belief we can estimate the coin's fariness $\theta$, predict the probability of heads on the next flip or even see how the results depend upon different choices of prior beliefs. The latter is known as model comparison.
At each step of the way we will be making visualisations of each of these functions and distributions using the Seaborn plotting package for Python. Seaborn sits "on top" of Matplotlib, but has far better defaults for statistical plotting. For some interesting examples of what Seaborn can do, take a look at the gallery.
## Assumptions of the Approach
As with all models we need to make some assumptions about our situation.
• We are going to assume that our coin can only have two outcomes, that is it can only land on its head or tail and never on its side
• Each flip of the coin is completely independent of the others, i.e. we have independent and identically distributed (i.i.d.) coin flips
• The fairness of the coin does not change in time, that is it is stationary
With these assumptions in mind, we can now begin discussing the Bayesian procedure.
## Recalling Bayes' Rule
In the previous article we outlined Bayes' rule. I've repeated the box callout here for completeness:
#### Bayes' Rule for Bayesian Inference
\begin{eqnarray} P(\theta|D) = P(D|\theta) \; P(\theta) \; / \; P(D) \end{eqnarray}
Where:
• $P(\theta)$ is the prior. This is the strength in our belief of $\theta$ without considering the evidence $D$. Our prior view on the probability of how fair the coin is.
• $P(\theta|D)$ is the posterior. This is the (refined) strength of our belief of $\theta$ once the evidence $D$ has been taken into account. After seeing 4 heads out of 8 flips, say, this is our updated view on the fairness of the coin.
• $P(D|\theta)$ is the likelihood. This is the probability of seeing the data $D$ as generated by a model with parameter $\theta$. If we knew the coin was fair, this tells us the probability of seeing a number of heads in a particular number of flips.
• $P(D)$ is the evidence. This is the probability of the data as determined by summing (or integrating) across all possible values of $\theta$, weighted by how strongly we believe in those particular values of $\theta$. If we had multiple views of what the fairness of the coin is (but didn't know for sure), then this tells us the probability of seeing a certain sequence of flips for all possibilities of our belief in the coin's fairness.
Note that we have three separate components to specify, in order to calcute the posterior. They are the likelihood, the prior and the evidence. In the following sections we are going to discuss exactly how to specify each of these components for our particular case of inference on a binomial proportion.
## The Likelihood Function
We have just outlined Bayes' rule and have seen that we must specify a likelihood function, a prior belief and the evidence (i.e. a normalising constant). In this section we are going to consider the first of these components, namely the likelihood.
### Bernoulli Distribution
Our example is that of a sequence of coin flips. We are interested in the probability of the coin coming up heads. In particular, we are interested in the probability of the coin coming up heads as a function of the underlying fairness parameter $\theta$.
This will take a functional form, $f$. If we denote by $k$ the random variable that describes the result of the coin toss, which is drawn from the set $\{1,0\}$, where $k=1$ represents a head and $k=0$ represents a tail, then the probability of seeing a head, with a particular fairness of the coin, is given by:
\begin{eqnarray} P(k = 1 | \theta) = f(\theta) \end{eqnarray}
We can choose a particularly succint form for $f(\theta)$ by simply stating the probability is given by $\theta$ itself, i.e. $f(\theta) = \theta$. This leads to the probability of a coin coming up heads to be given by:
\begin{eqnarray} P(k = 1 | \theta) = \theta \end{eqnarray}
And the probability of coming up tails as:
\begin{eqnarray} P(k = 0 | \theta) = 1-\theta \end{eqnarray}
This can also be written as:
\begin{eqnarray} P(k | \theta) = \theta^k (1 - \theta)^{1-k} \end{eqnarray}
Where $k \in \{1, 0\}$ and $\theta \in [0,1]$.
This is known as the Bernoulli distribution. It gives the probability over two separate, discrete values of $k$ for a fixed fairness parameter $\theta$.
In essence it tells us the probability of a coin coming up heads or tails depending on how fair the coin is.
### Bernoulli Likelihood Function
We can also consider another way of looking at the above function. If we consider a fixed observation, i.e. a known coin flip outcome, $k$, and the fairness parameter $\theta$ as a continuous variable then:
\begin{eqnarray} P(k | \theta) = \theta^k (1 - \theta)^{1-k} \end{eqnarray}
tells us the probability of a fixed outcome $k$ given some particular value of $\theta$. As we adjust $\theta$ (e.g. change the fairness of the coin), we will start to see different probabilities for $k$.
This is known as the likelihood function of $\theta$. It is a function of a continuous $\theta$ and differs from the Bernoulli distribution because the latter is actually a discrete probability distribution over two potential outcomes of the coin-flip $k$.
Note that the likelihood function is not actually a probability distribution in the true sense since integrating it across all values of the fairness parameter $\theta$ does not actually equal 1, as is required for a probability distribution.
We say that $P(k | \theta) = \theta^k (1 - \theta)^{1-k}$ is the Bernoulli likelihood function for $\theta$.
### Multiple Flips of the Coin
Now that we have the Bernoulli likelihood function we can use it to determine the probability of seeing a particular sequence of $N$ flips, given by the set $\{k_1, ..., k_N\}$.
Since each of these flips is independent of any other, the probability of the sequence occuring is simply the product of the probability of each flip occuring.
If we have a particular fairness parameter $\theta$, then the probability of seeing this particular stream of flips, given $\theta$, is given by:
\begin{eqnarray} P(\{k_1, ..., k_N\} | \theta) &=& \prod_{i} P(k_i | \theta) \\ &=& \prod_{i} \theta^{k_i} (1 - \theta)^{1-{k_i}} \end{eqnarray}
What about if we are interested in the number of heads, say, in $N$ flips? If we denote by $z$ the number of heads appearing, then the formula above becomes: \begin{eqnarray} P(z, N | \theta) = {N \choose z} \theta^z (1 - \theta)^{N-z} \end{eqnarray}
That is, the probability of seeing $z$ heads, in $N$ flips, assuming a fairness parameter $\theta$. We will use this formula when we come to determine our posterior belief distribution later in the article.
## Quantifying our Prior Beliefs
An extremely important step in the Bayesian approach is to determine our prior beliefs and then find a means of quantifying them.
In the Bayesian approach we need to determine our prior beliefs on parameters and then find a probability distribution that quantifies these beliefs.
In this instance we are interested in our prior beliefs on the fairness of the coin. That is, we wish to quantify our uncertainty in how biased the coin is.
To do this we need to understand the range of values that $\theta$ can take and how likely we think each of those values are to occur.
$\theta=0$ indicates a coin that always comes up tails, while $\theta = 1$ implies a coin that always comes up heads. A fair coin is denoted by $\theta=0.5$. Hence $\theta \in [0,1]$. This implies that our probability distribution must also exist on the interval $[0,1]$.
The question then becomes - which probability distribution do we use to quantify our beliefs about the coin?
### Beta Distribution
In this instance we are going to choose the beta distribution. The probability density function of the beta distribution is given by the following:
\begin{eqnarray} P(\theta | \alpha, \beta) = \theta^{\alpha - 1} (1 - \theta)^{\beta - 1} / B(\alpha, \beta) \end{eqnarray}
Where the term in the denominator, $B(\alpha, \beta)$ is present to act as a normalising constant so that the area under the PDF actually sums to 1.
I've plotted a few separate realisations of the beta distribution for various parameters $\alpha$ and $\beta$ below:
To plot the image yourself, you will need to install seaborn:
pip install seaborn
The Python code to produce the plot is given below:
import numpy as np
from scipy.stats import beta
import matplotlib.pyplot as plt
import seaborn as sns
if __name__ == "__main__":
sns.set_palette("deep", desat=.6)
sns.set_context(rc={"figure.figsize": (8, 4)})
x = np.linspace(0, 1, 100)
params = [
(0.5, 0.5),
(1, 1),
(4, 3),
(2, 5),
(6, 6)
]
for p in params:
y = beta.pdf(x, p[0], p[1])
plt.plot(x, y, label="$\\alpha=%s$, $\\beta=%s$" % p)
plt.xlabel("$\\theta$, Fairness")
plt.ylabel("Density")
plt.legend(title="Parameters")
plt.show()
Essentially, as $\alpha$ becomes larger the bulk of the probability distribution moves towards the right (a coin biased to come up heads more often), whereas an increase in $\beta$ moves the distribution towards the left (a coin biased to come up tails more often).
However, if both $\alpha$ and $\beta$ increase then the distribution begins to narrow. If $\alpha$ and $\beta$ increase equally, then the distribution will peak over $\theta=0.5$, i.e. when the coin is far.
Why have we chosen the beta function as our prior? There are a couple of reasons:
• Support - It's defined on the interval $[0,1]$, which is the same interval that $\theta$ exists over.
• Flexibility - It possesses two shape parameters known as $\alpha$ and $\beta$, which give it significant flexibility. This flexibility provides us with a lot of choice in how we model our beliefs.
However, perhaps the most important reason for choosing a beta distribution is because it is a conjugate prior for the Bernoulli distribution.
#### Conjugate Priors
In Bayes' rule above we can see that the posterior distribution is proportional to the product of the prior distribution and the likelihood function:
\begin{eqnarray} P(\theta | D) \propto P(D|\theta) P(\theta) \end{eqnarray}
A conjugate prior is a choice of prior distribution, that when coupled with a specific type of likelihood function, provides a posterior distribution that is of the same family as the prior distribution.
The prior and posterior both have the same probability distribution family, but with differing parameters.
Conjugate priors are extremely convenient from a calculation point of view as they provide closed-form expressions for the posterior, thus negating any complex numerical integration.
In our case, if we use a Bernoulli likelihood function AND a beta distribution as the choice of our prior, we immediately know that the posterior will also be a beta distribution.
Using a beta distribution for the prior in this manner means that we can carry out more experimental coin flips and straightforwardly refine our beliefs. The posterior will become the new prior and we can use Bayes' rule successively as new coin flips are generated.
If our prior belief is specified by a beta distribution and we have a Bernoulli likelihood function, then our posterior will also be a beta distribution.
Note however that a prior is only conjugate with respect to a particular likelihood function.
### Why Is A Beta Prior Conjugate to the Bernoulli Likelihood?
We can actually use a simple calculation to prove why the choice of the beta distribution for the prior, with a Bernoulli likelihood, gives a beta distribution for the posterior.
As mentioned above, the probability density function of a beta distribution, for our particular parameter $\theta$, is given by:
\begin{eqnarray} P(\theta | \alpha, \beta) = \theta^{\alpha - 1} (1 - \theta)^{\beta - 1} / B(\alpha, \beta) \end{eqnarray}
You can see that the form of the beta distribution is similar to the form of a Bernoulli likelihood. In fact, if you multiply the two together (as in Bayes' rule), you get:
\begin{eqnarray} \theta^{\alpha - 1} (1 - \theta)^{\beta - 1} / B(\alpha, \beta) \times \theta^{k} (1 - \theta)^{1-k} \propto \theta^{\alpha + k - 1} (1 - \theta)^{\beta + k} \end{eqnarray}
Notice that the term on the right hand side of the proportionality sign has the same form as our prior (up to a normalising constant).
### Multiple Ways to Specify a Beta Prior
At this stage we've discussed the fact that we want to use a beta distribution in order to specify our prior beliefs about the fairness of the coin. However, we only have two parameters to play with, namely $\alpha$ and $\beta$.
How do these two parameters correspond to our more intuitive sense of "likely fairness" and "uncertainty in fairness"?
Well, these two concepts neatly correspond to the mean and the variance of the beta distribution. Hence, if we can find a relationship between these two values and the $\alpha$ and $\beta$ parameters, we can more easily specify our beliefs.
It turns out (see this Cross Validated post and the Wikipedia article on the beta distribution) that the mean $\mu$ is given by:
\begin{eqnarray} \mu = \frac{\alpha}{\alpha + \beta} \end{eqnarray}
While the standard deviation $\sigma$ is given by:
\begin{eqnarray} \sigma = \sqrt{\frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}} \end{eqnarray}
Hence, all we need to do is re-arrange these formulae to provide $\alpha$ and $\beta$ in terms of $\mu$ and $\sigma$. $\alpha$ is given by:
\begin{eqnarray} \alpha = \left( \frac{1-\mu}{\sigma^2} - \frac{1}{\mu} \right) \mu^2 \end{eqnarray}
While $\beta$ is given by:
\begin{eqnarray} \beta = \alpha \left( \frac{1}{\mu} - 1 \right) \end{eqnarray}
Note that we have to be careful here, as we should not specify a $\sigma > 0.289$, since this is the standard deviation of a uniform density (which itself implies no prior belief on any particular fairness of the coin).
Let's carry out an example now. Suppose I think the fairness of the coin is around 0.5, but I'm not particularly certain. I may specify a standard deviation of around 0.1. What beta distribution is produced as a result?
Plugging in the numbers into the above formulae gives us $\alpha = 12$ and $\beta = 12$ and the beta distribution in this instance looks like the following:
Notice how the peak is centred around 0.5 but that there is significant uncertainty in this belief, represented by the width of the curve.
## Using Bayes' Rule to Calculate a Posterior
We are now finally in a position to be able to calculate our posterior beliefs using Bayes' rule.
Bayes' rule in this instance is given by:
\begin{eqnarray} P(\theta | z, N) = P(z, N | \theta) P(\theta) / P(z, N) \end{eqnarray}
This says that the posterior belief in $\theta$, given $z$ heads in $N$ flips, is equal to the likelihood of seeing $z$ heads in $N$ flips, given a fairness $\theta$, multiplied by our prior belief in $\theta$, normalised by the evidence.
If we substitute in the values for the likelihood function calculated above, as well as our prior belief beta distribution, we get:
\begin{eqnarray} P(\theta | z, N) &\propto& P(z, N | \theta) P(\theta) \\ &\propto& \theta^z (1- \theta)^{N-z} \theta^{\alpha - 1} (1-\theta)^{\beta - 1} \\ &=& \theta^{z + \alpha -1} (1- \theta)^{N-z + \beta - 1} \\ &=& \text{beta}(\theta | z + \alpha, N - z + \beta) \end{eqnarray}
If our prior is given by $\text{beta}(\theta|\alpha,\beta)$ and we observe $z$ heads in $N$ flips subsequently, then the posterior is given by $\text{beta}(\theta|z+\alpha, N-z+\beta)$.
This is an incredibly straightforward (and useful!) updating rule. All we need do is specify the mean $\mu$ and standard deviation $\sigma$ of our prior beliefs, carry out $N$ flips and observe the number of heads $z$ and we automatically have a rule for how our beliefs should be updated.
As an example, suppose we consider the same prior beliefs as above for $\theta$ with $\mu=0.5$ and $\sigma=0.1$. This gave us the prior belief distribution of $\text{beta}(\theta|12,12)$.
Now suppose we observe $N=50$ flips and $z=10$ of them come up heads. How does this change our belief on the fairness of the coin?
We can plug these numbers into our posterior beta distribution to get:
\begin{eqnarray} \text{beta}(\theta |z+\alpha, N-z+\beta) &=& \text{beta}(\theta |10+12, 50-10+12) \\ &=& \text{beta}(\theta |22, 52) \end{eqnarray}
We can plot the prior and posterior belief distributions. I have used a blue dotted line for the prior belief and a green solid line for the posterior:
Notice how the peak shifts dramatically to the left since we have only observed 10 heads in 50 flips. In addition, notice how the width of the peak has shrunk, which is indicative of the fact that our belief in the certainty of the particular fairness value has also increased.
At this stage we can compute the mean and standard deviation of the posterior in order to produce estimates for the fairness of the coin. In particular, the value of $\mu_{\text{post}}$ is given by: \begin{eqnarray} \mu_{\text{post}} &=& \frac{\alpha}{\alpha + \beta} \\ &=& \frac{22}{22+52} \\ &=& 0.297 \end{eqnarray}
While the standard deviation $\sigma_{\text{post}}$ is given by:
\begin{eqnarray} \sigma_{\text{post}} &=& \sqrt{\frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}} \\ &=& \sqrt{\frac{22 \times 52}{(22 + 52)^2 (22 + 52 + 1)}} \\ &=& 0.053 \end{eqnarray}
In particular the mean has sifted to approximately 0.3, while the standard deviation (s.d.) has halved to approximately 0.05. A mean of $\theta=0.3$ states that approximately 30% of the time, the coin will come up heads, while 70% of the time it will come up tails. The s.d. of 0.05 means that while we are more certain in this estimate than before, we are still somewhat uncertain about this 30% value.
If we were to carry out more coin flips, the s.d. would reduce even further as $\alpha$ and $\beta$ continued to increase, representing our continued increase in certainty as more trials are carried out.
Note in particular that we can use a posterior beta distribution as a prior distribution in a new Bayesian updating procedure. This is another extremely useful benefit of using conjugate priors to model our beliefs.
In the next article we are going to discuss what happens in the case where a beta distribution is insufficient to model our prior beliefs about uncertainty, in which case we will need to use a numerical method, known as Markov Chain Monte Carlo sampling, in order to construct our posterior distribution. | 5,536 | 22,812 | {"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} | 3.4375 | 3 | CC-MAIN-2022-33 | latest | en | 0.923829 |
http://gregmankiw.blogspot.com/2015_02_01_archive.html | 1,529,445,954,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00398.warc.gz | 136,666,499 | 10,716 | Thursday, February 26, 2015
Sentence of the Day
"genetic differences explained roughly 33% of the variations in individual savings rates."
Thursday, February 19, 2015
What matters more--the productivity slowdown or the inequality increase?
The Economic Report of the President was released today. A friend draws my attention to Table 1-3 on page 34, which presents several historical counterfactuals. It finds:
1. If productivity growth had not slowed after 1973, the median household would have \$30,000 of additional income today.
2. If income inequality had not increased after 1973, the median household would have \$9,000 of additional income today.
So, which is the bigger problem? (Of course, neither has an easy solution.)
Tuesday, February 17, 2015
Nobel Prize for Sale
I don't know the story behind this, but apparently a Kuznets heir is selling his Nobel Prize.
Update 2/24: With less than 2 days to go, no one has offered the minimum bid of \$150,000.
Update 2/26: Someone offers the minimum bid.
Update 2/27: It goes for \$390,848.
Friday, February 06, 2015
Good News
The Crimson reports on the good judgment of Harvard students:
Thursday, February 05, 2015
Defending Pete Carroll
Justin Wolfers says that, by the logic of game theory, the losing Superbowl coach does not deserve all the opprobrium he has been getting. I have been thinking the same thing.
Wednesday, February 04, 2015
If you are an undergraduate, this conference may be of interest.
Monday, February 02, 2015
Competition and Cooperation
A nice essay by Tim Taylor, very appropriate for introductory students.
The Rise of the Inequality Debate
Professor Lars Syll thinks I made of fool of myself in a previous post when I wondered why we have only recently started discussing income inequality so extensively, even though the increase in inequality occurred mainly between 1980 and 2000. He writes, "Wonder on which planet Greg has been living the last twenty years."
Of course, we economists have been discussing the topic for a long time. Indeed, I had a whole chapter on income inequality in the first edition of my favorite textbook, which came out about 20 years ago. But the public has been discussing the topic widely only recently.
To document this fact for Professor Syll, I used the NY Times's very cool chronicle website to generate the chart below. As you can see, the percentage of NYT articles that uses the word "inequality" has increased more than ten-fold in the past few years. So has the percentage that uses the phrase "income inequality."
By the way, the earlier blip in the use of "inequality" was in 1866, the year of the Civil Rights Act of 1866. The inequality being discussed then was political, not economic. The wide discussion of "income inequality" is unprecedented and very recent. | 662 | 2,830 | {"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} | 2.515625 | 3 | CC-MAIN-2018-26 | latest | en | 0.96128 |
https://www.era.lib.ed.ac.uk/handle/1842/27900?show=full | 1,561,580,072,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00095.warc.gz | 725,267,525 | 5,837 | dc.contributor.author Bhatt, [unknown] en dc.date.accessioned 2018-01-31T11:42:29Z dc.date.available 2018-01-31T11:42:29Z dc.date.issued 1948 en dc.identifier.uri http://hdl.handle.net/1842/27900 dc.description.abstract en dc.description.abstract CHAPTER I: en Transformation of Statistical Variate: 1.1 The two approaches to the problem of skewness; 1.2 Transformation of statistical variate; 1.3 Macalister's curve for geometric mean; 1.4 Edgeworth's method of translation; 1.5 Kapteyn's theory and curves; 16 A basis for Kapteyn's general curve in symbolic generating functions for non-additive variates; 1.7 Conclusion. dc.description.abstract CHAPTER II: Polynomial transformation of normal en variate: 2.1 Polynomial transformation of normal variate; 2.2 Fitting by moments; 2.3 Fitting by quantiles; 2.4 Prof. Aitken's solution of a set of inconsistent linear equations; 2.5 Prof. Aitken's method of fitting polynomials to correlated data; 2.6 The Reciprocal of the variance matrix for quantiles; 2.7 Standard matrices for W and solutions; 2.8 Variances of solutions; 2.9 Changes in Standard Results for negative skewness; 2.10 Fitting the transformation to the data; 2.11 Method of finding the quartiles; 2.12 An alternative basis for standardizing W matrices. dc.description.abstract CHAPTER III: Exponential transformation of normal en variate: 3.1 Exponential transformation of normal variate; 3.2 Macalister's curve: Estimation of the parameters and properties: 3.3 The general case. dc.description.abstract CHAPTER IV Numerical Examples: 4.1 Distribution of weights of 7749 en Adults born in U.K.; 4.2 Distribution of weights of 11382 Glasgow school children. dc.description.abstract APPENDIX I Normal Sextiles and Octiles with their en powers; APPENDIX II Abscissae and Ordinates at Sextiles and Octiles of Pearson's Type III curve; APPENDIX III Standard Results for Sextiles and Octiles from level of skewness 0.1 to 1.1; APPENDIX IV Variances of Solutions; APPENDIX V Details of computation of Standard Results for Octiles, level of skew .=0.5; APPENDIX VI A note on the distribution of n-iles in large samples; References. dc.publisher The University of Edinburgh en dc.relation.isreferencedby en dc.subject Annexe Thesis Digitisation Project 2017 Block 16 en dc.title Transformation of a normal statistical variate, with applications to curve-fitting en dc.type Thesis or Dissertation en dc.type.qualificationlevel en dc.type.qualificationname PhD Doctor of Philosophy en
| 628 | 2,491 | {"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} | 2.53125 | 3 | CC-MAIN-2019-26 | latest | en | 0.700157 |
https://studylib.net/doc/25356144/work-power-mechanic-energy | 1,653,018,484,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00369.warc.gz | 634,199,988 | 11,011 | # Work,Power,Mechanic energy
```scalar
reference
GPE=mgh
Gravitational potential energy
system
Nonnegative value
scalar
No nonconservative force do
work
unit
Potential energy
EPE=1/2kx^2
change of Kinetic energy
= -change of potential
energy
No nonconservative force work
KE=1/2mv^2
Mechanic energy
Elastic potential
MEi=MEf
joule
Kinetic energy
KE
Work
Joule
scalar
No negative value
Work-Energy principle
W=ΔKE
Wnet=Wc+Wnc
P=W/t
Power
Watt
scalar
W=Fdcosθ
scalar
``` | 162 | 469 | {"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} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.565297 |
http://www.mathworks.com/matlabcentral/fileexchange/40804-contemporary-communications-systems-matlab-files/content/MatlabFiles%20for%20Contemporary%20Communicatios%20Books/ip_07_10.m | 1,438,577,209,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042989507.42/warc/CC-MAIN-20150728002309-00044-ip-10-236-191-2.ec2.internal.warc.gz | 576,448,154 | 10,278 | Code covered by the BSD License
# Contemporary Communications Systems Matlab Files
### Omar Ruiz (view profile)
Matlab Files in this book
ip_07_10.m
```% MATLAB script for Illustrative Problem 10, Chapter 7.
clear
echo on
K=10;N=2*K;T=100;variance=1;
noise=sqrt(variance)*randn(1,N);
a=rand(1,36);
a=sign(a-0.5);
b=reshape(a,9,4);
% Generate the 16QAM points
XXX=2*b(:,1)+b(:,2)+j*(2*b(:,3)+b(:,4));
XX=XXX';
X=[0 XX 0 conj(XX(9:-1:1))];
x=zeros(1,N);
for n=0:N-1
for k=0:N-1
x(n+1)=x(n+1)+1/sqrt(N)*X(k+1)*exp(j*2*pi*n*k/N);
echo off
end
end
echo on
r=x+noise;
Y=zeros(1,10);
for k=1:9
for n=0:N-1
Y(1,k+1)=Y(1,k+1)+1/sqrt(N)*r(n+1)*exp(-j*2*pi*k*n/N);
echo off
end
end
echo on
% Detect the nearest neighbor in the 16QAM constellation
for k=1:9
if real(Y(1,k+1))>0
if real(Y(1,k+1))>2
Z(1,k+1)=3;
else
Z(1,k+1)=1;
end
else
if real(Y(1,k+1))<-2
Z(1,k+1)=-3;
else
Z(1,k+1)=-1;
end
end
if imag(Y(1,k+1))>0
if imag(Y(1,k+1))>2
Z(1,k+1)=Z(1,k+1)+3*j;
else
Z(1,k+1)=Z(1,k+1)+j;
end
else
if imag(Y(1,k+1))<-2
Z(1,k+1)=Z(1,k+1)-3*j;
else
Z(1,k+1)=Z(1,k+1)-j;
end
end
echo off
end
echo on
error=max(size(find(Z(1,2:10)-X(1,2:10))));
``` | 543 | 1,132 | {"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} | 3.234375 | 3 | CC-MAIN-2015-32 | longest | en | 0.426502 |
https://db0nus869y26v.cloudfront.net/en/Gram | 1,723,688,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641141870.93/warc/CC-MAIN-20240815012836-20240815042836-00369.warc.gz | 144,361,108 | 14,946 | gram
The mass of this pen cap is about 1 gram. A weight scale such as this can give an accurate reading of mass for many objects (see Weight vs. mass).
General information
Unit systemSI
Unit ofMass
Symbolg
Conversions
1 g in ...... is equal to ...
SI base units 10−3 kilograms
CGS units 1 gram
Imperial units
U.S. customary
0.0352740 ounces
Atomic mass units 6.02214076×1023 Da
The gram (originally gramme;[1] SI unit symbol g) is a unit of mass in the International System of Units (SI) equal to one thousandth of a kilogram.
Originally defined as of 1795 as "the absolute weight of a volume of pure water equal to the cube of the hundredth part of a metre [1 cm3], and at the temperature of melting ice",[2] the defining temperature (≈0 °C) was later changed to 4 °C, the temperature of maximum density of water.
By the late 19th century, there was an effort to make the base unit the kilogram and the gram a derived unit. In 1960, the new International System of Units defined a gram as one one-thousandth of a kilogram (i.e., one gram is 1×10−3 kg). The kilogram, as of 2019, is defined by the International Bureau of Weights and Measures from the fixed numerical value of the Planck constant (h).[3][4]
## Official SI symbol
The only unit symbol for gram that is recognised by the International System of Units (SI) is "g" following the numeric value with a space, as in "640 g" to stand for "640 grams" in the English language. The SI disallows use of abbreviations such as "gr" (which is the symbol for grains),[5]: C-19 "gm" ("g⋅m" is the SI symbol for gram-metre) or "Gm" (the SI symbol for gigametre).
## History
The word gramme was adopted by the French National Convention in its 1795 decree revising the metric system as replacing the gravet (introduced in 1793 simultaneously with a base measure called grave, of which gravet was a subdivision). Its definition remained that of the weight of a cubic centimetre of water.[6][7]
French gramme was taken from the Late Latin term gramma. This word—ultimately from Greek γράμμα (grámma), "letter"—had adopted a specialised meaning in Late Antiquity of "one twenty-fourth part of an ounce" (two oboli),[8] corresponding to about 1.14 modern grams. This use of the term is found in the carmen de ponderibus et mensuris ("poem about weights and measures") composed around 400 AD.[a] There is also evidence that the Greek γράμμα was used in the same sense at around the same time, in the 4th century, and survived in this sense into Medieval Greek,[10] while the Latin term died out in Medieval Latin and was recovered in Renaissance scholarship.[b]
The gram was the base unit of mass in the 19th-century centimetre–gram–second system of units (CGS). The CGS system coexisted with the metre–kilogram–second system of units (MKS), first proposed in 1901, during much of the 20th century, but the gram was displaced by the kilogram as the base unit for mass when the MKS system was chosen for the SI base units in 1960.
## Uses
The gram is the most widely used unit of measurement for non-liquid ingredients in cooking and grocery shopping worldwide.[11][12] Liquid ingredients are often measured by volume rather than mass.
Many standards and legal requirements for nutrition labels on food products require relative contents to be stated per 100 g of the product, such that the resulting figure can also be read as a percentage.
## Conversion factors
• 1 gram (g) ≈ 15.43236 grains (gr)
• 1 grain (gr) ≈ 0.0647989 grams
• 1 avoirdupois ounce (oz) ≈ 28.3495 grams
• 1 troy ounce (ozt) = 31.1034768 g (exact, by definition)
• 100 grams (g) ≈ 3.52740 ounces (oz)
• 1 carat (ct) = 0.2 grams
• 1 gamma (γ) = 10−6 grams[13][14]
• 1 undecimogramme = 1 "eleventh-gram" = 10−11 grams in the historical quadrant–eleventh-gram–second system (QES system) a.k.a. hebdometre–undecimogramme–second system (HUS system) [15]
• 500 grams (g) = 1 jin in the Chinese units of measurement.
## Notes
1. ^ The date and authorship of this Late Latin didactic poem are both uncertain; it was attributed to Priscian but is now attributed to Rem(m)ius Favinus/Flav(in)us.[9] The poem's title is reflected in the French phrase poids et mesures ("weights and mesures") in the title of the 1795 National Convention decree, Décret relatif aux poids et aux mesures that introduced the gram, and indirectly in the name of the General Conference on Weights and Measures responsible for the modern definition of the metric units.
2. ^ In the Renaissance, the carmen de ponderibus et mensuris was received as a work of the 1st-century grammarian Remmius Palaemon edited in 1528 by Johann Setzer of Hagenau, together with works by Celsius, Priscian and Johannes Caesarius; Aurelij Cornelij Celsi, De re medica, libri octo eruditissimi. Q. Sereni Samonici Praecepta medica, uersibus hexametris. Q. Rhemnij Fannij Palaemonis, De ponderibus [et] mensuris, liber rarus [et] utilissimus
## References
1. ^ "Weights and Measures Act 1985 (c. 72)". The UK Statute Law Database. Office of Public Sector Information. Archived from the original on 12 September 2008. Retrieved 26 January 2011. §92.
2. ^ "Décret relatif aux poids et aux mesures" (in French). 1795. Archived from the original on 25 February 2013.
3. ^ Draft Resolution A "On the revision of the International System of units (SI)" to be submitted to the CGPM at its 26th meeting (2018) (PDF), archived from the original (PDF) on 29 April 2018, retrieved 17 May 2020
4. ^ Decision CIPM/105-13 (October 2016) Archived 24 August 2017 at the Wayback Machine. The day is the 144th anniversary of the Metre Convention.
5. ^ National Institute of Standards and Technology (October 2011). Butcher, Tina; Cook, Steve; Crown, Linda et al. eds. "Appendix C – General Tables of Units of Measurement" Archived 2016-06-17 at the Wayback Machine (PDF). Specifications, Tolerances, and Other Technical Requirements for Weighing and Measuring Devices Archived 2016-08-23 at the Wayback Machine. NIST Handbook. 44 (2012 ed.). Washington, D.C.: U.S. Department of Commerce, Technology Administration, National Institute of Standards and Technology. ISSN 0271-4027 Archived 25 December 2022 at the Wayback Machine. OCLC OCLC 58927093. Retrieved 30 June 2012.
6. ^ "Décret relatif aux poids et aux mesures du 18 germinal an 3 (7 avril 1795)" [Decree of 18 Germinal, year III (April 7, 1795) regarding weights and measures]. Grandes lois de la République (in French). Digithèque de matériaux juridiques et politiques, Université de Perpignan. Archived from the original on 10 May 2013. Retrieved 3 November 2011.
7. ^ Convention nationale, décret du 1er août 1793, ed. Duvergier, Collection complète des lois, décrets, ordonnances, règlemens avis du Conseil d'état, publiée sur les éditions officielles du Louvre, vol. 6 (2nd ed. 1834), p. 70 Archived 2015-04-02 at the Wayback Machine. The metre (mètre) on which this definition depends was itself defined as the ten-millionth part of a quarter of Earth's meridian, given in traditional units as 3 pieds, 11.44 lignes (a ligne being the 12th part of an pouce (inch), or the 144th part of a pied.
8. ^ Charlton T. Lewis, Charles Short, A Latin Dictionary s.v. "gramma" Archived 2015-07-17 at the Wayback Machine, 1879
9. ^ Knorr, Wilbur R. (1996). "Carmen de ponderibus et mensuris". In Hornblower, Simon; Spawforth, Antony (eds.). The Oxford Classical Dictionary (3rd ed.). Oxford: Oxford University Press. p. 292. ISBN 019866172X.
10. ^ Henry George Liddell. Robert Scott. A Greek-English Lexicon (revised and augmented edition, Oxford, 1940) s.v. γράμμα Archived 2015-07-17 at the Wayback Machine, citing the 10th-century work Geoponica and a 4th-century papyrus edited in L. Mitteis, Griechische Urkunden der Papyrussammlung zu Leipzig, vol. i (1906), 62 ii 27.
11. ^ Pat Chapman (2007). India Food and Cooking: The Ultimate Book on Indian Cuisine. London: New Holland Publishers (UK) Ltd. p. 64. ISBN 978-1845376192. Retrieved 20 November 2014. Most of the world uses the metric system to weigh and measure. This book puts metric first, followed by imperial because the US uses it (with slight modifications which need not concern us).
12. ^ Gisslen, Wayne (2010). Professional Cooking, College Version. New York: Wiley. p. 107. ISBN 978-0-470-19752-3. Retrieved 20 April 2011. The system of measurement used in the United States is complicated. Even when people have used the system all their lives, they still sometimes have trouble remembering things like how many fluid ounces are in a quart or how many feet are in a mile. ... The United States is the only major country that uses almost exclusively the complex system of measurement we have just described.
13. ^ 5th SI Brochure (1985), p. 78
14. ^ "NIST Special Publication 811 – NIST Guide to the SI, Chapter 5: Units Outside the SI". NIST. 28 January 2016. Archived from the original on 12 August 2016. Retrieved 10 December 2022.
15. ^ "System of Measurement Units – Engineering and Technology History Wiki". ethw.org. 24 April 2012. Archived from the original on 29 April 2018. Retrieved 29 April 2018.
16. ^ "Circulating Coin Designs". Japan Mint. Archived from the original on 18 September 2009. Retrieved 7 March 2010. | 2,548 | 9,215 | {"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} | 2.875 | 3 | CC-MAIN-2024-33 | latest | en | 0.919757 |
http://mathhelpforum.com/number-theory/88338-help-problem-do-logic-please-print.html | 1,506,012,124,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687833.62/warc/CC-MAIN-20170921153438-20170921173438-00648.warc.gz | 226,419,220 | 3,155 | Help on a problem to do with logic please
• May 9th 2009, 06:54 PM
the undertaker
Help on a problem to do with logic please
Problem:
3 positive integers are written on a board:
x y z
Person A calculated that the HCF of 2 of the integers and got 1 000 004
Person B did the same and got 1 000 006
Person C did the same and got 1 000 008.
Person D is sure that at least person A, B or C made a mistake, despite the fact they calculated the HCF of different numbers.
Is she right?
Thx.
• May 10th 2009, 12:10 AM
Opalg
Quote:
Originally Posted by the undertaker
Problem:
3 positive integers are written on a board:
x y z
Person A calculated that the HCF of 2 of the integers and got 1 000 004
Person B did the same and got 1 000 006
Person C did the same and got 1 000 008.
Person D is sure that at least person A, B or C made a mistake, despite the fact they calculated the HCF of different numbers.
Is she right? | 266 | 920 | {"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} | 2.640625 | 3 | CC-MAIN-2017-39 | longest | en | 0.949093 |
https://www.physicsforums.com/threads/ideal-gas-law-problem-help-needed.614422/ | 1,606,594,766,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195745.90/warc/CC-MAIN-20201128184858-20201128214858-00482.warc.gz | 801,647,287 | 15,014 | # Ideal gas law problem, help needed!
## Homework Statement
A rigid container holds 2.80 mol of gas at a pressure of 1.20 atm and a temperature of 60.0 degrees Celsius. What is the container's volume?
PV=nRT
## The Attempt at a Solution
(2.80*8.31*333)/1.20 = 6457m^3
This answer seems way too big, what am I doing wrong?
Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
Watch units.
Should use N/m^2 rather than atm?
Should I**
TSny
Homework Helper
Gold Member
When you write 8.31 for R, what are the units that go with that number?
J/mol*K
TSny
Homework Helper
Gold Member
What system of units uses Joules, moles, and Kelvins? Then decide what the units of pressure are for that system of units.
Ahhhh okay, I got it now. Thanks! | 221 | 788 | {"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} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.905084 |
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# Magnetic field at point P due to given current distribution:A. $\dfrac{{{\mu _0}I}}{{4\pi a}} \odot$B. $\dfrac{{{\mu _0}I}}{{2\pi a}} \odot$C. $\dfrac{{{\mu _0}I}}{{\pi a}} \otimes$D. Zero
Last updated date: 23rd Jul 2024
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Answer
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384.9k+ views
Hint:The magnetic field at point P due to lower wire will be zero since the point lies on the wire. Use Biot-Savart’s law to determine the magnetic field due to a finite wire at a point P which is at a distance r from the wire. Use the right-hand thumb rule to determine the direction of the magnetic field.
Formula used:
$B = \dfrac{{{\mu _0}I}}{{4\pi d}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)$
Here, ${\mu _0}$ is the permeability of free space, I is the current, d is the distance of point from the wire, ${\theta _1}$ and ${\theta _2}$ are the angles made by the wire with point P.
Complete step by step answer:
We can see in the given current distribution, the current forms a rectangle. Let the current through wire 1 is such that the point P is on the wire 1 and the wire 2 makes an angle $45^\circ$ to the point P as shown in the figure below.
The magnetic field at point P due to wire 1 will be zero since the point lies on the wire 1.
Let us calculate the magnetic field due to wire 2 whose one end is infinite and another end makes an angle $45^\circ$ to the point P. If we consider the horizontal and vertical section of wire 2, then both sections are making the angle $45^\circ$ to the point P. The magnetic field at point P will be the addition of magnetic fields due to current in both sections of wire
2. Therefore, the current will be twice of I that is $2I$.
Using Biot-Savart’s law, the magnetic field at a point P due to current carrying wire, we have,
${B_2} = \dfrac{{{\mu _0}I}}{{4\pi r}}\left( {\sin {\theta _1} + \sin {\theta _2}} \right)$
Here, ${\mu _0}$ is the permeability of the free space.
Substituting $I = 2I$, $r = \sqrt 2 a$, ${\theta _1} = 45^\circ$ and ${\theta _2} = 90^\circ$ in the above equation, we get,
${B_2} = \dfrac{{{\mu _0}2I}}{{4\pi \left( {\sqrt 2 a} \right)}}\left( {\sin \left( {45} \right) + \sin \left( {45} \right)} \right)$
$\Rightarrow {B_2} = \dfrac{{{\mu _0}2I}}{{4\pi \left( {\sqrt 2 a} \right)}}\left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}} \right)$
$\Rightarrow {B_2} = \dfrac{{{\mu _0}2I}}{{4\pi \left( {\sqrt 2 a} \right)}}\left( {\dfrac{2}{{\sqrt 2 }}} \right)$
$\Rightarrow {B_2} = \dfrac{{{\mu _0}I}}{{2\pi a}}$
Now, let us determine the direction of the magnetic field at point P due to wire 2. We have from Right-hand thumb rule, if we hold a current carrying conductor in our right hand such that the thumb points in the direction of current; then the curled fingers around the conductor denotes the direction of the magnetic field. Thus, we can see that the curled fingers at point P points outward from the page. Therefore, the direction of the magnetic field is out of the page.
So, the correct answer is option B.
Note: We have determined the magnetic field due to whole wire 2 which is bent perpendicular along its length rather than determining the magnetic field due to each horizontal and vertical section of the wire. The magnetic field at a point lying on the length of wire is zero because both the angles in Biot-Savart’s law becomes $90^\circ$ and $180^\circ$. The sine value for both angles is zero. | 1,036 | 3,480 | {"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} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.844056 |
https://nag.com/numeric/MB/manual64_24_1/html/F04/f04intro.html | 1,516,438,760,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00558.warc.gz | 783,120,777 | 18,815 | Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
# NAG Toolbox Chapter IntroductionF04 — Simultaneous Linear Equations
## Scope of the Chapter
This chapter is concerned with the solution of the matrix equation AX = B$AX=B$, where B$B$ may be a single vector or a matrix of multiple right-hand sides. The matrix A$A$ may be real, complex, symmetric, Hermitian, positive definite, positive definite Toeplitz or banded. It may also be rectangular, in which case a least squares solution is obtained.
Much of the functionality of this chapter has been superseded by functions from Chapters F07 and F08 (Lapack routines) as those chapters have grown and have included driver and expert driver functions.
For a general introduction to sparse systems of equations, see the F11 Chapter Introduction, which currently provides functions for large sparse systems. Some functions for sparse problems are also included in this chapter; they are described in Section [Sparse Matrix s].
## Background to the Problems
A set of linear equations may be written in the form
Ax = b $Ax=b$
where the known matrix A$A$, with real or complex coefficients, is of size m$m$ by n$n$ (m$m$ rows and n$n$ columns), the known right-hand vector b$b$ has m$m$ components (m$m$ rows and one column), and the required solution vector x$x$ has n$n$ components (n$n$ rows and one column). There may also be p$p$ vectors bi${b}_{\mathit{i}}$, for i = 1,2,,p$\mathit{i}=1,2,\dots ,p$, on the right-hand side and the equations may then be written as
AX = B, $AX=B,$
the required matrix X$X$ having as its p$p$ columns the solutions of Axi = bi$A{x}_{\mathit{i}}={b}_{\mathit{i}}$, for i = 1,2,,p$\mathit{i}=1,2,\dots ,p$. Some functions deal with the latter case, but for clarity only the case p = 1$p=1$ is discussed here.
The most common problem, the determination of the unique solution of Ax = b$Ax=b$, occurs when m = n$m=n$ and A$A$ is not singular, that is rank(A) = n$\mathrm{rank}\left(A\right)=n$. This is discussed in Section [Unique Solution of ] below. The next most common problem, discussed in Section [The Least Squares Solution of , , ] below, is the determination of the least squares solution of Axb$Ax\simeq b$ required when m > n$m>n$ and rank(A) = n$\mathrm{rank}\left(A\right)=n$, i.e., the determination of the vector x$x$ which minimizes the norm of the residual vector r = bAx$r=b-Ax$. All other cases are rank deficient, and they are treated in Section [Rank-deficient Cases].
### Unique Solution of Ax=b
Most functions in this chapter solve this particular problem. The computation starts with the triangular decomposition A = PLU$A=PLU$, where L$L$ and U$U$ are respectively lower and upper triangular matrices and P$P$ is a permutation matrix, chosen so as to ensure that the decomposition is numerically stable. The solution is then obtained by solving in succession the simpler equations
Ly = PTb Ux = y
$Ly = PTb Ux = y$
the first by forward-substitution and the second by back-substitution.
If A$A$ is real symmetric and positive definite, U = LT$U={L}^{\mathrm{T}}$, while if A$A$ is complex Hermitian and positive definite, U = LH$U={L}^{\mathrm{H}}$; in both these cases P$P$ is the identity matrix (i.e., no permutations are necessary). In all other cases either U$U$ or L$L$ has unit diagonal elements.
Due to rounding errors the computed ‘solution’ x0${x}_{0}$, say, is only an approximation to the true solution x$x$. This approximation will sometimes be satisfactory, agreeing with x$x$ to several figures, but if the problem is ill-conditioned then x$x$ and x0${x}_{0}$ may have few or even no figures in common, and at this stage there is no means of estimating the ‘accuracy’ of x0${x}_{0}$.
There are three possible approaches to estimating the accuracy of a computed solution.
One way to do so, and to ‘correct’ x0${x}_{0}$ when this is meaningful (see next paragraph), involves computing the residual vector r = bAx0$r=b-A{x}_{0}$ in extended precision arithmetic, and obtaining a correction vector d$d$ by solving PLUd = r$PLUd=r$. The new approximate solution x0 + d${x}_{0}+d$ is usually more accurate and the correcting process is repeated until (a) further corrections are negligible or (b) they show no further decrease.
It must be emphasized that the ‘true’ solution x$x$ may not be meaningful, that is correct to all figures quoted, if the elements of A$A$ and b$b$ are known with certainty only to say p$p$ figures, where p$p$ is less than full precision. The first correction vector d$d$ will then give some useful information about the number of figures in the ‘solution’ which probably remain unchanged with respect to maximum possible uncertainties in the coefficients.
An alternative approach to assessing the accuracy of the solution is to compute or estimate the condition number of A$A$, defined as
κ(A) = ‖A‖ . ‖A − 1‖ . $κ(A) = ‖A‖ . ‖A-1‖ .$
Roughly speaking, errors or uncertainties in A$A$ or b$b$ may be amplified in the solution by a factor κ(A)$\kappa \left(A\right)$. Thus, for example, if the data in A$A$ and b$b$ are only accurate to 5$5$ digits and κ(A)103$\kappa \left(A\right)\approx {10}^{3}$, then the solution cannot be guaranteed to have more than 2$2$ correct digits. If κ(A)105$\kappa \left(A\right)\ge {10}^{5}$, the solution may have no meaningful digits.
To be more precise, suppose that
Ax = b and (A + δA)(x + δx) = b + δb. $Ax=b and (A+δA)(x+δx)=b+δb.$
Here δA$\delta A$ and δb$\delta b$ represent perturbations to the matrices A$A$ and b$b$ which cause a perturbation δx$\delta x$ in the solution. We can define measures of the relative sizes of the perturbations in A$A$, b$b$ and x$x$ as
ρA = (‖δA‖)/(‖A‖), ρb = (‖δb‖)/(‖b‖) and ρx = (‖δx‖)/(‖x‖) respectively. $ρA=‖δA‖ ‖A‖ , ρb=‖δb‖ ‖b‖ and ρx=‖δx‖ ‖x‖ respectively.$
Then
ρx ≤ (κ (A))/(1 − κ (A)ρA)(ρA + ρb) $ρx≤κ (A) 1-κ (A)ρA (ρA+ρb)$
provided that κ(A)ρA < 1$\kappa \left(A\right){\rho }_{A}<1$. Often κ(A)ρA1$\kappa \left(A\right){\rho }_{A}\ll 1$ and then the bound effectively simplifies to
ρx ≤ κ(A)(ρA + ρb). $ρx≤κ(A)(ρA+ρb).$
Hence, if we know κ(A)$\kappa \left(A\right)$, ρA${\rho }_{A}$ and ρb${\rho }_{b}$, we can compute a bound on the relative errors in the solution. Note that ρA${\rho }_{A}$, ρb${\rho }_{b}$ and ρx${\rho }_{x}$ are defined in terms of the norms of A$A$, b$b$ and x$x$. If A$A$, b$b$ or x$x$ contains elements of widely differing magnitude, then ρA${\rho }_{A}$, ρb${\rho }_{b}$ and ρx${\rho }_{x}$ will be dominated by the errors in the larger elements, and ρx${\rho }_{x}$ will give no information about the relative accuracy of smaller elements of x$x$.
A third way to obtain useful information about the accuracy of a solution is to solve two sets of equations, one with the given coefficients, which are assumed to be known with certainty to p$p$ figures, and one with the coefficients rounded to (p1$p-1$) figures, and to count the number of figures to which the two solutions agree. In ill-conditioned problems this can be surprisingly small and even zero.
### The Least Squares Solution of Ax≃b, m>n, rankA=n
The least squares solution is the vector $\stackrel{^}{x}$ which minimizes the sum of the squares of the residuals,
S = (b − Ax̂)T(b − Ax̂) = ‖b − Ax̂‖22. $S=(b-Ax^)T(b-Ax^)=‖b-Ax^‖22.$
The solution is obtained in two steps.
(a) Householder transformations are used to reduce A$A$ to ‘simpler form’ via the equation QA = R$QA=R$, where R$R$ has the appearance
((R̂)/0) $(R^0)$
with $\stackrel{^}{R}$ a nonsingular upper triangular n$n$ by n$n$ matrix and 0$0$ a zero matrix of shape (mn)$\left(m-n\right)$ by n$n$. Similar operations convert b$b$ to Qb = c$Qb=c$, where
c = ((c1)/(c2)) $c=(c1c2)$
with c1${c}_{1}$ having n$n$ rows and c2${c}_{2}$ having (mn$m-n$) rows.
(b) The required least squares solution is obtained by back-substitution in the equation
R̂x̂ = c1. $R^x^=c1.$
Again due to rounding errors the computed 0${\stackrel{^}{x}}_{0}$ is only an approximation to the required $\stackrel{^}{x}$, but as in Section [Unique Solution of ], this can be improved by ‘iterative refinement’. The first correction d$d$ is the solution of the least squares problem
Ad = b − Ax̂0 = r $Ad=b-Ax^0=r$
and since the matrix A$A$ is unchanged, this computation takes less time than that of the original 0${\stackrel{^}{x}}_{0}$. The process can be repeated until further corrections are (a) negligible or (b) show no further decrease.
### Rank-deficient Cases
If, in the least squares problem just discussed, rank(A) < n$\mathrm{rank}\left(A\right), then a least squares solution exists but it is not unique. In this situation it is usual to ask for the least squares solution ‘of minimal length’, i.e., the vector x$x$ which minimizes x2${‖x‖}_{2}$, among all those x$x$ for which bAx2${‖b-Ax‖}_{2}$ is a minimum.
This can be computed from the Singular Value Decomposition (SVD) of A$A$, in which A$A$ is factorized as
A = QDPT $A=QDPT$
where Q$Q$ is an m$m$ by n$n$ matrix with orthonormal columns, P$P$ is an n$n$ by n$n$ orthogonal matrix and D$D$ is an n$n$ by n$n$ diagonal matrix. The diagonal elements of D$D$ are called the ‘singular values’ of A$A$; they are non-negative and can be arranged in decreasing order of magnitude:
d1 ≥ d2 ≥ ⋯ ≥ dn ≥ 0. $d1≥d2≥⋯≥dn≥0.$
The columns of Q$Q$ and P$P$ are called respectively the left and right singular vectors of A$A$. If the singular values dr + 1,,dn${d}_{r+1},\dots ,{d}_{n}$ are zero or negligible, but dr${d}_{r}$ is not negligible, then the rank of A$A$ is taken to be r$r$ (see also Section [The Rank of a Matrix]) and the minimal length least squares solution of Axb$Ax\simeq b$ is given by
x̂ = D†QTb $x^=D†QTb$
where D${D}^{†}$ is the diagonal matrix with diagonal elements d11,d21,,dr1,0,,0${d}_{1}^{-1},{d}_{2}^{-1},\dots ,{d}_{r}^{-1},0,\dots ,0$.
The SVD may also be used to find solutions to the homogeneous system of equations Ax = 0$Ax=0$, where A$A$ is m$m$ by n$n$. Such solutions exist if and only if rank(A) < n$\mathrm{rank}\left(A\right), and are given by
n x = ∑ αipi i = r + 1
$x=∑i=r+1nαipi$
where the αi${\alpha }_{i}$ are arbitrary numbers and the pi${p}_{i}$ are the columns of P$P$ which correspond to negligible elements of D$D$.
The general solution to the rank-deficient least squares problem is given by + x$\stackrel{^}{x}+x$, where $\stackrel{^}{x}$ is the minimal length least squares solution and x$x$ is any solution of the homogeneous system of equations Ax = 0$Ax=0$.
### The Rank of a Matrix
In theory the rank is r$r$ if nr$n-r$ elements of the diagonal matrix D$D$ of the singular value decomposition are exactly zero. In practice, due to rounding and/or experimental errors, some of these elements have very small values which usually can and should be treated as zero.
For example, the following 5$5$ by 8$8$ matrix has rank 3$3$ in exact arithmetic:
22 14 − 1 − 3 9 9 2 4 10 7 13 − 2 8 1 − 6 5 2 10 − 1 13 1 − 7 6 0 3 0 − 11 − 2 − 2 5 5 − 2 7 8 3 4 4 − 1 1 2
.
$22 14 -1 -3 9 9 2 4 10 7 13 -2 8 1 -6 5 2 10 -1 13 1 -7 6 0 3 0 -11 -2 -2 5 5 -2 7 8 3 4 4 -1 1 2 .$
On a computer with 7$7$ decimal digits of precision the computed singular values were
3.5 × 101, 2.0 × 101, 2.0 × 101, 1.3 × 10 − 6, 5.5 × 10 − 7 $3.5×101, 2.0×101, 2.0×101, 1.3×10-6, 5.5×10-7$
and the rank would be correctly taken to be 3$3$.
It is not, however, always certain that small computed singular values are really zero. With the 7$7$ by 7$7$ Hilbert matrix, for example, where aij = 1 / (i + j1)${a}_{ij}=1/\left(i+j-1\right)$, the singular values are
1.7, 2.7 × 10 − 1, 2.1 × 10 − 2, 1.0 × 10 − 3, 2.9 × 10 − 5, 4.9 × 10 − 7, 3.5 × 10 − 9. $1.7, 2.7×10-1, 2.1×10-2, 1.0×10-3, 2.9×10-5, 4.9×10-7, 3.5×10-9.$
Here there is no clear cut-off between small (i.e., negligible) singular values and larger ones. In fact, in exact arithmetic, the matrix is known to have full rank and none of its singular values is zero. On a computer with 7$7$ decimal digits of precision, the matrix is effectively singular, but should its rank be taken to be 6$6$, or 5$5$, or 4$4$?
It is therefore impossible to give an infallible rule, but generally the rank can be taken to be the number of singular values which are neither zero nor very small compared with other singular values. For example, if there is a sharp decrease in singular values from numbers of order unity to numbers of order 107${10}^{-7}$, then the latter will almost certainly be zero in a machine in which 7$7$ significant decimal figures is the maximum accuracy. Similarly for a least squares problem in which the data is known to about four significant figures and the largest singular value is of order unity then a singular value of order 104${10}^{-4}$ or less should almost certainly be regarded as zero.
It should be emphasized that rank determination and least squares solutions can be sensitive to the scaling of the matrix. If at all possible the units of measurement should be chosen so that the elements of the matrix have data errors of approximately equal magnitude.
### Generalized Linear Least Squares Problems
The simple type of linear least squares problem described in Section [The Least Squares Solution of , , ] can be generalized in various ways.
1. Linear least squares problems with equality constraints:
find x to minimize S = ‖c − Ax‖22 subject to Bx = d, $find x to minimize S=‖c-Ax‖22 subject to Bx=d,$
where A$A$ is m$m$ by n$n$ and B$B$ is p$p$ by n$n$, with pnm + p$p\le n\le m+p$. The equations Bx = d$Bx=d$ may be regarded as a set of equality constraints on the problem of minimizing S$S$. Alternatively the problem may be regarded as solving an overdetermined system of equations
( A ) B
x =
( c ) d
,
$A B x= c d ,$
where some of the equations (those involving B$B$) are to be solved exactly, and the others (those involving A$A$) are to be solved in a least squares sense. The problem has a unique solution on the assumptions that B$B$ has full row rank p$p$ and the matrix
( A ) B
$\left(\begin{array}{c}A\\ B\end{array}\right)$ has full column rank n$n$. (For linear least squares problems with inequality constraints, refer to Chapter E04.)
2. General Gauss–Markov linear model problems:
minimize ‖y‖2 subject to d = Ax + By, $minimize ‖y‖2 subject to d=Ax+By,$
where A$A$ is m$m$ by n$n$ and B$B$ is m$m$ by p$p$, with nmn + p$n\le m\le n+p$. When B = I$B=I$, the problem reduces to an ordinary linear least squares problem. When B$B$ is square and nonsingular, it is equivalent to a weighted linear least squares problem:
find x to minimize ‖B − 1(d − Ax)‖2. $find x to minimize ‖B-1(d-Ax)‖2.$
The problem has a unique solution on the assumptions that A$A$ has full column rank n$n$, and the matrix (A,B)$\left(A,B\right)$ has full row rank m$m$.
### Calculating the Inverse of a Matrix
The functions in this chapter can also be used to calculate the inverse of a square matrix A$A$ by solving the equation
AX = I $AX=I$
where I$I$ is the identity matrix. However, solving the equations AX = B$AX=B$ by calculation of the inverse of the coefficient matrix A$A$, i.e., by X = A1B$X={A}^{-1}B$, is definitely not recommended.
Similar remarks apply to the calculation of the pseudo-inverse of a singular or rectangular matrix.
### Estimating the 1-norm of a Matrix
The 1-norm of a matrix A$A$ is defined to be:
m ‖A‖1 = max ∑ |aij| 1 ≤ j ≤ n i = 1
$‖A‖1 = max 1≤j≤n ∑ i=1 m |aij|$
(1)
Typically it is useful to calculate the condition number of a matrix with respect to the solution of linear equations, or inversion. The higher the condition number the less accuracy might be expected from a numerical computation. A condition number for the solution of linear equations is A . A1$‖A‖.‖{A}^{-1}‖$. Since this might be a relatively expensive computation it often suffices to estimate the norm of each matrix.
## Recommendations on Choice and Use of Available Functions
See also Section [Recommendations on Choice and Use of Available Functions] in the F07 Chapter Introduction for recommendations on the choice of available functions from that chapter.
### Black Box and General Purpose Functions
Most of the functions in this chapter are categorised either as Black Box functions or general purpose functions.
Black Box functions solve the equations Axi = bi$A{x}_{\mathit{i}}={b}_{\mathit{i}}$, for i = 1,2,,p$\mathit{i}=1,2,\dots ,p$, in a single call with the matrix A$A$ and the right-hand sides, bi${b}_{i}$, being supplied as data. These are the simplest functions to use and are suitable when all the right-hand sides are known in advance and do not occupy too much storage.
General purpose functions, in general, require a previous call to a function in Chapters F01, F03 or F07 to factorize the matrix A$A$. This factorization can then be used repeatedly to solve the equations for one or more right-hand sides which may be generated in the course of the computation. The Black Box functions simply call a factorization function and then a general purpose function to solve the equations.
The function nag_linsys_real_gen_sparse_lsqsol (f04qa) which uses an iterative method for sparse systems of equations does not fit easily into this categorization, but is classified as a general purpose function in the decision trees and indexes.
### Systems of Linear Equations
Most of the functions in this chapter solve linear equations Ax = b$Ax=b$ when A$A$ is n$n$ by n$n$ and a unique solution is expected (see Section [Unique Solution of ]). The matrix A$A$ may be ‘general’ real or complex, or may have special structure or properties, e.g., it may be banded, tridiagonal, almost block-diagonal, sparse, symmetric, Hermitian, positive definite (or various combinations of these).
It must be emphasized that it is a waste of computer time and space to use an inappropriate function, for example one for the complex case when the equations are real. It is also unsatisfactory to use the special functions for a positive definite matrix if this property is not known in advance.
Functions are given for calculating the approximate solution, that is solving the linear equations just once, and also for obtaining the accurate solution by successive iterative corrections of this first approximation using additional precision arithmetic, as described in Section [Unique Solution of ]. The latter, of course, are more costly in terms of time and storage, since each correction involves the solution of n$n$ sets of linear equations and since the original A$A$ and its LU$LU$ decomposition must be stored together with the first and successively corrected approximations to the solution. In practice the storage requirements for the ‘corrected’ functions are about double those of the ‘approximate’ functions, though the extra computer time may not be prohibitive since the same matrix and the same LU$LU$ decomposition is used in every linear equation solution.
A number of the Black Box functions in this chapter return estimates of the condition number and the forward error, along with the solution of the equations. But for those functions that do not return a condition estimate two functions are provided – nag_linsys_real_gen_norm_rcomm (f04yd) for real matrices, nag_linsys_complex_gen_norm_rcomm (f04zd) for complex matrices – which can return a cheap but reliable estimate of A1$‖{A}^{-1}‖$, and hence an estimate of the condition number κ(A)$\kappa \left(A\right)$ (see Section [Unique Solution of ]). These functions can also be used in conjunction with most of the linear equation solving functions in Chapter F11: further advice is given in the function documents.
Other functions for solving linear equation systems, computing inverse matrices, and estimating condition numbers can be found in Chapter F07, which contains LAPACK software.
### Linear Least Squares Problems
The majority of the functions for solving linear least squares problems are to be found in Chapter F08.
For the case described in Section [The Least Squares Solution of , , ], when mn$m\ge n$ and a unique least squares solution is expected, there are two functions for a general real A$A$, one of which (nag_linsys_real_gen_solve (f04jg)) computes a first approximation and the other (nag_linsys_real_gen_lsqsol (f04am)) computes iterative corrections. If it transpires that rank(A) < n$\mathrm{rank}\left(A\right), so that the least squares solution is not unique, then nag_linsys_real_gen_lsqsol (f04am) takes a failure exit, but nag_linsys_real_gen_solve (f04jg) proceeds to compute the minimal length solution by using the SVD (see below).
If A$A$ is expected to be of less than full rank then one of the functions for calculating the minimal length solution may be used.
For mn$m\gg n$ the use of the SVD is not significantly more expensive than the use of functions based upon the QR$QR$ factorization.
Problems with linear equality constraints can be solved by nag_lapack_dgglse (f08za) (for real data) or by nag_lapack_zgglse (f08zn) (for complex data), provided that the problems are of full rank. Problems with linear inequality constraints can be solved by nag_opt_lsq_lincon_solve (e04nc) in Chapter E04.
General Gauss–Markov linear model problems, as formulated in Section [Generalized Linear Least Squares Problems], can be solved by nag_lapack_dggglm (f08zb) (for real data) or by nag_lapack_zggglm (f08zp) (for complex data).
### Sparse Matrix Functions
Functions specifically for sparse matrices are appropriate only when the number of nonzero elements is very small, less than, say, 10% of the n2${n}^{2}$ elements of A$A$, and the matrix does not have a relatively small band width.
Chapter F11 contains functions for both the direct and iterative solution of real sparse linear systems. There are two functions in Chapter F04 for solving sparse linear equations (nag_linsys_real_sparse_fac_solve (f04ax) and nag_linsys_real_gen_sparse_lsqsol (f04qa)). nag_linsys_real_sparse_fac_solve (f04ax) utilizes a factorization of the matrix A$A$ obtained from nag_matop_real_gen_sparse_lu (f01br) or nag_matop_real_gen_sparse_lu_reuse (f01bs), while nag_linsys_real_gen_sparse_lsqsol (f04qa) uses an iterative technique and requires a user-supplied function to compute matrix-vector products Ac$Ac$ and ATc${A}^{\mathrm{T}}c$ for any given vector c$c$.
nag_linsys_real_gen_sparse_lsqsol (f04qa) solves sparse least squares problems by an iterative technique, and also allows the solution of damped (regularized) least squares problems (see the function document for details).
## Decision Trees
The name of the function (if any) that should be used to factorize the matrix A$A$ is given in brackets after the name of the function for solving the equations.
### Tree 1: Black Box functions for unique solution of Ax = b$\mathbit{A}\mathbit{x}=\mathbit{b}$ (Real matrix)
Is A$A$ symmetric? _yes Is A$A$ positive definite? _yes Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_linsys_real_posdef_tridiag_solve (f04bg) (see Note 1) or nag_lapack_dptsv (f07ja) or nag_lapack_dptsvx (f07jb) (see Note 2) | | | no| | | | nag_linsys_real_posdef_band_solve (f04bf) (see Note 1) or nag_lapack_dpbsv (f07ha) or nag_lapack_dpbsvx (f07hb) (see Note 2) | | no| | | Is A$A$ a Toeplitz matrix? _yes Are the equations the Yule–Walker equations? _yes nag_linsys_real_toeplitz_yule (f04fe) | | | no| | | | nag_linsys_real_toeplitz_solve (f04ff) | | no| | | Do you require an accurate solution using iterative refinement? _yes nag_linsys_real_posdef_solve_ref (f04ab) or nag_linsys_real_posdef_solve_1rhs (f04as) (see Note 3) | | no| | | Is one triangle of A$A$ stored as a linear array? _yes nag_linsys_real_posdef_packed_solve (f04be) (see Note 1) or nag_lapack_dppsv (f07ga) or nag_lapack_dppsvx (f07gb) (see Note 2) | | no| | | nag_linsys_real_posdef_solve (f04bd) (see Note 1) or nag_lapack_dposv (f07fa) or nag_lapack_dposvx (f07fb) (see Note 2) | no| | Is one triangle of A$A$ stored as a linear array? _yes nag_linsys_real_symm_packed_solve (f04bj) (see Note 1) or nag_lapack_dspsv (f07pa) or nag_lapack_dspsvx (f07pb) (see Note 2) | no| | nag_linsys_real_symm_solve (f04bh) (see Note 1) or nag_lapack_dsysv (f07ma) or nag_lapack_dsysvx (f07mb) (see Note 2) no| Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_linsys_real_tridiag_solve (f04bc) (see Note 1) or nag_lapack_dgtsv (f07ca) or nag_lapack_dgtsvx (f07cb) (see Note 2) | no| | nag_linsys_real_band_solve (f04bb) (see Note 1) or nag_lapack_dgbsv (f07ba) or nag_lapack_dgbsvx (f07bb) (see Note 2) no| Do you require an accurate solution using iterative refinement? _yes nag_linsys_real_square_solve_ref (f04ae) or nag_linsys_real_square_solve_1rhs (f04at) (see Note 3) no| nag_linsys_real_square_solve (f04ba) (see Note 1) or nag_lapack_dgesv (f07aa) or nag_lapack_dgesvx (f07ab) (see Note 2)
### Tree 2: Black Box functions for unique solution of Ax = b$\mathbit{A}\mathbit{x}=\mathbit{b}$ (Complex matrix)
Is A$A$ Hermitian? _yes Is A$A$ positive definite? _yes Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_linsys_complex_posdef_tridiag_solve (f04cg) (see Note 1) or nag_lapack_zptsv (f07jn) or nag_lapack_zptsvx (f07jp) (see Note 2) | | | no| | | | nag_linsys_complex_posdef_band_solve (f04cf) (see Note 1) or nag_lapack_zpbsv (f07hn) or nag_lapack_zpbsvx (f07hp) (see Note 2) | | no| | | Is one triangle of A$A$ stored as a linear array? _yes nag_linsys_complex_posdef_packed_solve (f04ce) (see Note 1) or nag_lapack_zppsv (f07gn) or nag_lapack_zppsvx (f07gp) (see Note 2) | | no| | | nag_linsys_complex_posdef_solve (f04cd) (see Note 1) or nag_lapack_zposv (f07fn) or nag_lapack_zposvx (f07fp) (see Note 2) | no| | Is one triangle of A$A$ stored as a linear array? _yes nag_linsys_complex_herm_packed_solve (f04cj) (see Note 1) or nag_lapack_zhpsv (f07pn) or nag_lapack_zhpsvx (f07pp) (see Note 2) | no| | nag_linsys_complex_herm_solve (f04ch) (see Note 1) or nag_lapack_zhesv (f07mn) or nag_lapack_zhesvx (f07mp) (see Note 2) no| Is A$A$ symmetric? _yes Is one triangle of A$A$ stored as a linear array? _yes nag_linsys_complex_symm_packed_solve (f04dj) (see Note 1) or nag_lapack_zspsv (f07qn) or nag_lapack_zspsvx (f07qp) (see Note 2) | no| | nag_linsys_complex_symm_solve (f04dh) (see Note 1) or nag_lapack_zsysv (f07nn) or nag_lapack_zsysvx (f07np) (see Note 2) no| Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_linsys_complex_tridiag_solve (f04cc) (see Note 1) or nag_lapack_zgtsv (f07cn) or nag_lapack_zgtsvx (f07cp) (see Note 2) | no| | nag_linsys_complex_band_solve (f04cb) (see Note 1) or nag_lapack_zgbsv (f07bn) or nag_lapack_zgbsvx (f07bp) (see Note 2) no| nag_linsys_complex_square_solve (f04ca) (see Note 1) or nag_lapack_zgesv (f07an) or nag_lapack_zgesvx (f07ap) (see Note 2)
### Tree 3: General purpose functions for unique solution of Ax = b$\mathbit{A}\mathbit{x}=\mathbit{b}$ (Real matrix)
Is A$A$ a sparse matrix and not banded? _yes Chapter F11 or nag_linsys_real_sparse_fac_solve (f04ax) (nag_matop_real_gen_sparse_lu (f01br) or nag_matop_real_gen_sparse_lu_reuse (f01bs)) or nag_linsys_real_gen_sparse_lsqsol (f04qa) no| Is A$A$ symmetric? _yes Is A$A$ positive definite? _yes Is A$A$ band matrix? _yes Is A$A$ tridiagonal? _yes nag_lapack_dpttrs (f07je) (nag_lapack_dpttrf (f07jd)) | | | no| | | | Is A$A$ variable band width? _yes nag_linsys_real_posdef_vband_solve (f04mc) (nag_matop_real_vband_posdef_fac (f01mc)) | | | no| | | | nag_lapack_dpbtrs (f07he) (nag_lapack_dpbtrf (f07hd)) | | no| | | Is A$A$ a Toeplitz matrix? _yes Are the equations the Yule–Walker equations? _yes nag_linsys_real_toeplitz_yule_update (f04me) | | | no| | | | nag_linsys_real_toeplitz_update (f04mf) | | no| | | Is one triangle of A$A$ stored as a linear array? _yes nag_lapack_dpptrs (f07ge) (nag_lapack_dpptrf (f07gd)) | | no| | | nag_lapack_dpotrs (f07fe) (nag_lapack_dpotrf (f07fd)) | no| | Is one triangle of A$A$ stored as a linear array? _yes nag_lapack_dsptrs (f07pe) (nag_lapack_dsptrf (f07pd)) | no| | nag_lapack_dsytrs (f07me) (nag_lapack_dsytrf (f07md)) no| Is A$A$ triangular? _yes Is A$A$ a band matrix? _yes nag_lapack_dtbtrs (f07ve) | no| | Is A$A$ stored as a linear array? _yes nag_lapack_dtptrs (f07ue) | no| | nag_lapack_dtrtrs (f07te) no| Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_linsys_real_tridiag_fac_solve (f04le) (nag_matop_real_gen_tridiag_lu (f01le)) or nag_lapack_dgttrs (f07ce) (nag_lapack_dgttrf (f07cd)) | no| | Is A$A$ almost block diagonal? _yes nag_linsys_real_blkdiag_fac_solve (f04lh) (nag_matop_real_gen_blkdiag_lu (f01lh)) | no| | nag_lapack_dgbtrs (f07be) (nag_lapack_dgbtrf (f07bd)) no| nag_lapack_dgetrs (f07ae) (nag_lapack_dgetrf (f07ad))
### Tree 4: General purpose functions for unique solution of Ax = b$\mathbit{A}\mathbit{x}=\mathbit{b}$ (Complex matrix)
Is A$A$ a sparse matrix and not banded? _yes Chapter F11 no| Is A$A$ Hermitian? _yes Is A$A$ positive definite? _yes Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_lapack_zpttrs (f07js) (nag_lapack_zpttrf (f07jr)) | | | no| | | | nag_lapack_zpbtrs (f07hs) (nag_lapack_zpbtrf (f07hr)) | | no| | | Is one triangle of A$A$ stored as a linear array? _yes nag_lapack_zpptrs (f07gs) (nag_lapack_zpptrf (f07gr)) | | no| | | nag_lapack_zpotrs (f07fs) (nag_lapack_zpotrf (f07fr)) | no| | Is one triangle of A$A$ stored as a linear array? _yes nag_lapack_zhptrs (f07ps) (nag_lapack_zhptrf (f07pr)) | no| | nag_lapack_zhetrs (f07ms) (nag_lapack_zhetrf (f07mr)) no| Is A$A$ symmetric? _yes Is one triangle of A$A$ stored as a linear array? _yes nag_lapack_zsptrs (f07qs) (nag_lapack_zsptrf (f07qr)) | no| | nag_lapack_zsytrs (f07ns) (nag_lapack_zsytrf (f07nr)) no| Is A$A$ triangular? _yes Is A$A$ a band matrix? _yes nag_lapack_ztbtrs (f07vs) | no| | Is A$A$ stored as a linear array? _yes nag_lapack_ztptrs (f07us) | no| | nag_lapack_ztrtrs (f07ts) no| Is A$A$ a band matrix? _yes Is A$A$ tridiagonal? _yes nag_lapack_zgttrs (f07cs) (nag_lapack_zgttrf (f07cr)) | no| | nag_lapack_zgbtrs (f07bs) (nag_lapack_zgbtrf (f07br)) no| nag_lapack_zgetrs (f07as) (nag_lapack_zgetrf (f07ar))
### Tree 5: General purpose functions for least squares and homogeneous equations (without constraints)
Is the problem Ax = 0$Ax=0$? _yes nag_lapack_dgelss (f08ka) no| Is A$A$ sparse? _yes nag_linsys_real_gen_sparse_lsqsol (f04qa) no| Is rank(A) = n$\mathrm{rank}\left(A\right)=n$? _yes Are storage and time more important than accuracy? _yes nag_linsys_real_gen_solve (f04jg) | no| | nag_linsys_real_gen_lsqsol (f04am) no| Is m > n$m>n$? _yes nag_linsys_real_gen_solve (f04jg) or nag_lapack_dgelss (f08ka) no| nag_lapack_dgelss (f08ka)
Note: there are also functions in Chapter F08 for solving least squares problems.
Note 1: also returns an estimate of the condition number and the forward error.
Note 2: also returns an estimate of the condition number, the forward error and the backward error. Requires additional workspace.
Note 3: for a single right-hand side only.
## Functionality Index
Black Box functions, Ax = b,
complex general band matrix nag_linsys_complex_band_solve (f04cb)
complex general matrix nag_linsys_complex_square_solve (f04ca)
complex general tridiagonal matrix nag_linsys_complex_tridiag_solve (f04cc)
complex Hermitian matrix,
packed matrix format nag_linsys_complex_herm_packed_solve (f04cj)
standard matrix format nag_linsys_complex_herm_solve (f04ch)
complex Hermitian positive definite band matrix nag_linsys_complex_posdef_band_solve (f04cf)
complex Hermitian positive definite matrix,
packed matrix format nag_linsys_complex_posdef_packed_solve (f04ce)
standard matrix format nag_linsys_complex_posdef_solve (f04cd)
complex Hermitian positive definite tridiagonal matrix nag_linsys_complex_posdef_tridiag_solve (f04cg)
complex symmetric matrix,
packed matrix format nag_linsys_complex_symm_packed_solve (f04dj)
standard matrix format nag_linsys_complex_symm_solve (f04dh)
real general band matrix nag_linsys_real_band_solve (f04bb)
real general matrix,
multiple right-hand sides,
iterative refinement using additional precision nag_linsys_real_square_solve_ref (f04ae)
multiple right-hand sides, standard precision nag_linsys_real_square_solve (f04ba)
single right-hand side,
iterative refinement using additional precision nag_linsys_real_square_solve_1rhs (f04at)
real general tridiagonal matrix nag_linsys_real_tridiag_solve (f04bc)
real symmetric matrix,
packed matrix format nag_linsys_real_symm_packed_solve (f04bj)
standard matrix format nag_linsys_real_symm_solve (f04bh)
real symmetric positive definite band matrix nag_linsys_real_posdef_band_solve (f04bf)
real symmetric positive definite matrix,
multiple right-hand sides,
iterative refinement using additional precision nag_linsys_real_posdef_solve_ref (f04ab)
multiple right-hand sides, standard precision nag_linsys_real_posdef_solve (f04bd)
packed matrix format nag_linsys_real_posdef_packed_solve (f04be)
single right-hand side,
iterative refinement using additional precision nag_linsys_real_posdef_solve_1rhs (f04as)
real symmetric positive definite Toeplitz matrix,
general right-hand side nag_linsys_real_toeplitz_solve (f04ff)
Yule–Walker equations nag_linsys_real_toeplitz_yule (f04fe)
real symmetric positive definite tridiagonal matrix nag_linsys_real_posdef_tridiag_solve (f04bg)
General Purpose functions, Ax = b,
real almost block-diagonal matrix nag_linsys_real_blkdiag_fac_solve (f04lh)
real band symmetric positive definite matrix, variable bandwidth nag_linsys_real_posdef_vband_solve (f04mc)
real sparse matrix,
direct method nag_linsys_real_sparse_fac_solve (f04ax)
iterative method nag_linsys_real_gen_sparse_lsqsol (f04qa)
real symmetric positive definite Toeplitz matrix,
general right-hand side, update solution nag_linsys_real_toeplitz_update (f04mf)
Yule–Walker equations, update solution nag_linsys_real_toeplitz_yule_update (f04me)
real tridiagonal matrix nag_linsys_real_tridiag_fac_solve (f04le)
Least squares and Homogeneous Equations,
real m by n matrix,
m ≥ n, rank = n or minimal solution nag_linsys_real_gen_solve (f04jg)
rank = n, iterative refinement nag_linsys_real_gen_lsqsol (f04am)
real sparse matrix nag_linsys_real_gen_sparse_lsqsol (f04qa)
Service Functions,
complex rectangular matrix,
norm and condition number estimation nag_linsys_complex_gen_norm_rcomm (f04zd)
real matrix,
covariance matrix for linear least squares problems nag_linsys_real_gen_lsq_covmat (f04ya)
real rectangular matrix,
norm and condition number estimation nag_linsys_real_gen_norm_rcomm (f04yd)
## References
Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore
Lawson C L and Hanson R J (1974) Solving Least Squares Problems Prentice–Hall
Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag | 11,063 | 34,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 324, "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} | 3.203125 | 3 | CC-MAIN-2018-05 | latest | en | 0.854056 |
https://stats.stackexchange.com/questions/115716/why-is-my-confidence-interval-wrong/115717 | 1,576,095,300,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540532624.3/warc/CC-MAIN-20191211184309-20191211212309-00504.warc.gz | 477,678,144 | 29,270 | # Why is my confidence interval wrong?
I'm working through the questions in this book. One question asks to construct a 90% confidence interval when the variable has a mean of 11 and a standard deviation of 1.8, and n = 4.
Here's my workings out:
sd/sqrt(n) = 1.8/sqrt(4) = 1.8/2 = 0.9
Upper = 11 + 1.645 x 0.9 = 12.48
Lower = 11 + 1.645 x 0.9 = 9.52
So my 90% confidence interval is 9.5-12.5.
But the book says the 90% confidence interval is 8.9-13.1.
Where am I going wrong? | 172 | 484 | {"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} | 3.875 | 4 | CC-MAIN-2019-51 | latest | en | 0.917006 |
https://brainly.ph/question/6984 | 1,487,621,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170609.0/warc/CC-MAIN-20170219104610-00052-ip-10-171-10-108.ec2.internal.warc.gz | 708,327,472 | 10,467 | # 1. how to add fractions with different denominators?2. how to add improper fractions?3. how to add mixed numbers?
1
by jhansen
2014-04-04T11:57:19+08:00
1. In adding fractions with different denominators, you have to take the LCM (Least Common Multiple) of the denominators and then divide that LCM with each denominator and multiply them by the numerator.
Example:
The LCM of 3 and 4 is 12, so we will have:
Then we divide 12 by the denominator 3, which gives us 4 (12/3=4), then multiply that answer to 2, which makes it 8 (4*2).
But we still have to add the other fraction. Do the same with it. Try it!
If you did the correct procedure, you will be able to get the answer
2. Adding improper fractions is just the same with adding our usual fractions, whether they have the same denominator or not.
3. If you are given with a mixed number and another fraction, you have to make the mixed number an improper fraction first.
In order to do that, you just multiply the denominator by the whole number and add the numerator, and the value that you get will be your new numerator.
Example:
The second case is that if you have two mixed numbers. This is easier.
You just have to add the whole numbers together and then add the fractions and combine them.
Example:
But, you can also do it by making both of them improper fractions then doing the addition. The answer is the same. :)
Though the problem with that process is that you will get larger values to add and multiply which will take a bit of time. But you have the freedom to choose.
Hope that helps. Good luck! | 380 | 1,582 | {"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} | 4.3125 | 4 | CC-MAIN-2017-09 | latest | en | 0.926643 |
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Show 40 post(s) from this thread on one page
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• Jul 28th 2010, 03:10 PM
SpiffyEh
linear algebra - concept proofs
I attached the 3 problems i'm having trouble with. They're easier to see in the attachment rather than if i typed them.
Can someone please explain these to me step by step or guide me through them? I'm not even sure where to begin.
Thank you
A_4 is the matrix
[.1 .6
.9 .4]
• Jul 28th 2010, 06:43 PM
tonio
Quote:
Originally Posted by SpiffyEh
I attached the 3 problems i'm having trouble with. They're easier to see in the attachment rather than if i typed them.
Can someone please explain these to me step by step or guide me through them? I'm not even sure where to begin.
Thank you
I think questions 1,2 are practically impossible to understand: what´s $q\,,\,A_4^n...$ ? I suppose $x_1\,,\,x_2$ are the components of x...??
And what's $A^H\,,\,\sigma(A)$ ? Is the last one the signature of a matrix A?
Tonio
• Jul 28th 2010, 08:08 PM
SpiffyEh
They're basically proofs, I'm supposed to show how those things lead to one another butr I don't know how to go about doing that
• Jul 29th 2010, 01:46 AM
Ackbeet
For the spectrum of self-adjoint matrices, you can proceed as follows:
Assume $A=A^{\dagger}.$ (This is the more common notation for Hilbert adjoint.) Further assume that $Ax=\lambda x$ for some nonzero vector $x$. Then $\lambda\in\sigma(A).$
Now examine the inner product $\langle x|Ax\rangle=\langle x|\lambda x\rangle=\lambda\langle x|x\rangle.$ However, it's also true that $\langle x|Ax\rangle=\langle Ax|x\rangle,$ by adjointness. Hence,
$\langle x|Ax\rangle=\langle \lambda x|x\rangle=\overline{\lambda}\langle x|x\rangle.$
Can you see where to go from here?
• Jul 29th 2010, 02:43 AM
tonio
Quote:
Originally Posted by SpiffyEh
They're basically proofs, I'm supposed to show how those things lead to one another butr I don't know how to go about doing that
This doesn't answer my question: what are the symbols you're using, anyway? Without that it's impossible to answer questions.
Tonio
• Jul 29th 2010, 08:16 AM
SpiffyEh
I really don't see where it's going from here. I understand the relationships but I don't see where they're going
• Jul 29th 2010, 08:24 AM
SpiffyEh
Quote:
Originally Posted by tonio
This doesn't answer my question: what are the symbols you're using, anyway? Without that it's impossible to answer questions.
Tonio
I'm pretty sure q is a steady state vector.
Sorry, I completely forgot to include A_4
A_4 is the matrix
[.1 .6
.9 .4]
A^H is the hermitian of A
And i'm guessing X_1 and X_2 are the components of x, i'm not sure about that one
Lastly, I honestly don't know what sigma(A) is
• Jul 29th 2010, 08:28 AM
Ackbeet
Ok. I've just shown that $\lambda\langle x|x\rangle=\overline{\lambda}\langle x|x\rangle,$ and we know that $x\not=0.$ So what can you do with this equation?
By the way, $\sigma(A)$ is a standard notation in functional analysis for the spectrum of the operator $A$. If $A$ is a finite dimensional matrix, then $\sigma(A)=\sigma_{p}(A)=\{\lambda|Ax=\lambda x\land x\not=0\}.$ The middle term there is called the point spectrum, and the last set there is just the eigenvalues. The eigenvalues, by definition, are equal to the point spectrum, and that's true of any operator.
• Jul 29th 2010, 08:32 AM
Ackbeet
For the first problem, I would diagonalize $A_{4}$. That should allow you to take the limit quite easily.
• Jul 29th 2010, 08:39 AM
SpiffyEh
I hoestly still don't understand the spectrum one. I just don't see where i'm supposed to be going with it. We were never told what sigma(A) means or what its called so I couldn't find it in the book, thank you for explaing that.
As for the first problem, I diagonalized it for another problem and got
[1 0
0 -.5]
The problem before the first one told me that q = [2/5 3/5]^T for A_4. I don't see the limit going to that.
• Jul 29th 2010, 09:14 AM
Ackbeet
You can divide out by the $\langle x|x\rangle,$ and you're left with the equation $\lambda=\overline{\lambda}.$ What does that tell you?
About the Limits of Time Series problem: when you diagonalized $A_{4}$, you found an invertible matrix $P$ such that $A_{4}=PDP^{-1},$ where $D$ is the diagonal matrix.
So, proving that
$\displaystyle{\lim_{n\to\infty}A_{4}^{n}x=q}$ is the same as proving that
$\displaystyle{\lim_{n\to\infty}PD^{n}P^{-1}x=q}.$
About the Connection to Transposition problem: are there any assumptions about the size of $A$?
• Jul 29th 2010, 09:21 AM
SpiffyEh
Quote:
Originally Posted by Ackbeet
You can divide out by the $\langle x|x\rangle,$ and you're left with the equation $\lambda=\overline{\lambda}.$ What does that tell you?
Doesn't that mean that lambda is a constant, a real constant since its conjugate is the same?
Quote:
Originally Posted by Ackbeet
About the Limits of Time Series problem: when you diagonalized $A_{4}$, you found an invertible matrix $P$ such that $A_{4}=PDP^{-1},$ where $D$ is the diagonal matrix.
So, proving that
$\displaystyle{\lim_{n\to\infty}A_{4}^{n}x=q}$ is the same as proving that
$\displaystyle{\lim_{n\to\infty}PD^{n}P^{-1}x=q}.$
I'll try that. So can I just take diagonal entries of the matrix to the nth power? Also, I don't see that going to the q from the problem before this one.
Quote:
Originally Posted by Ackbeet
About the Connection to Transposition problem: are there any assumptions about the size of $A$?
I think its a square matrix. I thought about this one, since det(A) = det(A^T) you can relate that to finding eigenvalues, which means the eigenvalues would be the same. So there would be n-many eigenvectors correct? Does that logic make sense and prove it?
• Jul 29th 2010, 10:08 AM
Ackbeet
Quote:
Doesn't that mean that lambda is a constant, a real constant since its conjugate is the same?
Well, eigenvalues are assumed to be constants already. But yes, if a number is equal to its complex conjugate, then it's real. This means your'e done with that problem: you can now show that if a number is an eigenvalue, then it's real. That proves the set inclusion property you were asked to show.
Quote:
I'll try that. So can I just take diagonal entries of the matrix to the nth power? Also, I don't see that going to the q from the problem before this one.
You tell me whether the nth power of a diagonal matrix can be computed by taking the nth power of the numbers on the diagonal. Hint: try squaring a diagonal matrix. You'll see what happens. Once you square it, try cubing it. Etc. Incidentally, I wouldn't recommend taking the limit of the matrix and then computing the LHS. I think the diagonalization will allow you to compute everything on the LHS of the equation that is to the right of the limit sign, and then you'd take the limit. The result should be a column vector. Show me what you have, and we'll see where that goes.
Quote:
I think its a square matrix. I thought about this one, since det(A) = det(A^T) you can relate that to finding eigenvalues, which means the eigenvalues would be the same.
The eigenvalues would be the same, I agree.
Quote:
So there would be n-many eigenvectors correct? Does that logic make sense and prove it?
I think it likely. But it needs a proof. The eigenvectors of a matrix would not be, I think, the same as the eigenvectors of its transpose. Therefore, it's not inherently obvious, at least to me, that there would be the same number of linearly independent eigenvectors.
I asked if you knew what the size of $A$ was. Of course it's square, or the whole eigenvalue process would be undefined. I'm wondering if it's n x n or not. Because if it is, there might be a very nice way of relating the eigenvectors of $A$ to those of $A^{T}.$
• Jul 29th 2010, 11:06 AM
SpiffyEh
Quote:
Originally Posted by Ackbeet
Well, eigenvalues are assumed to be constants already. But yes, if a number is equal to its complex conjugate, then it's real. This means your'e done with that problem: you can now show that if a number is an eigenvalue, then it's real. That proves the set inclusion property you were asked to show.
Oh ok, that makes alittle more sense. I'm still a little confused about when you said $\langle x|Ax\rangle=\langle x|\lambda x\rangle=\lambda\langle x|x\rangle.$ How do you go from the 2nd part to the 3rd? Is it because lambda is just a constant?
Quote:
Originally Posted by Ackbeet
You tell me whether the nth power of a diagonal matrix can be computed by taking the nth power of the numbers on the diagonal. Hint: try squaring a diagonal matrix. You'll see what happens. Once you square it, try cubing it. Etc. Incidentally, I wouldn't recommend taking the limit of the matrix and then computing the LHS. I think the diagonalization will allow you to compute everything on the LHS of the equation that is to the right of the limit sign, and then you'd take the limit. The result should be a column vector. Show me what you have, and we'll see where that goes.
I tried multiplying diagonal matricies and it is te nth power on the diagonal. So I got
P = P^(-1) =
[ 2 -1 [ 1 1
3 1] -3 2]
When I compute P * D^n * P^(-1)
I get
[.5^n 3^n
4.5^n 2^n]
Am I on the right track?
Quote:
Originally Posted by Ackbeet
I think it likely. But it needs a proof. The eigenvectors of a matrix would not be, I think, the same as the eigenvectors of its transpose. Therefore, it's not inherently obvious, at least to me, that there would be the same number of linearly independent eigenvectors.
I asked if you knew what the size of $A$ was. Of course it's square, or the whole eigenvalue process would be undefined. I'm wondering if it's n x n or not. Because if it is, there might be a very nice way of relating the eigenvectors of $A$ to those of $A^{T}.$
I think it is n x n. I'm not sure how to relate the eigenvectors, but I understand the eigenvalue part.
• Jul 29th 2010, 12:16 PM
Ackbeet
Quote:
How do you go from the 2nd part to the 3rd?
This is one of the axioms of inner products. In physics, at least, we assume that the inner product is linear in the second term. That is, $\langle x|ay\rangle=a\langle x|y\rangle$ and $\langle x|(y+z)\rangle=\langle x|y\rangle+\langle x|z\rangle$ for all scalars (yes, constants) $a$ and vectors $x$, $y$, and $z.$ With the inner product having conjugate symmetry, you can show that the inner product is conjugate linear in the first term. That's where the complex conjugate of $\lambda$ came from.
Moving on to the second problem, I agree with your $P$, but not with your $P^{-1}$. You need to multiply the matrix you have by 1/5 to get the correct inverse. I also think your $PD^{n}P^{-1}$ needs a little more careful work. Each entry in the matrix there is going to be the sum of two different elements to the nth power. Incidentally, the q you mentioned in post # 10 is correct.
About the third problem. See here for a very interesting discussion of left and right eigenvectors. They have much to do with the transpose matrix. You might find either what you need, or an idea. You might try playing around with determinants, perhaps of Equation (18).
Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last | 3,105 | 11,178 | {"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": 52, "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} | 3.515625 | 4 | CC-MAIN-2017-26 | longest | en | 0.921294 |
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Contents
## Big M method
When the simplex has artificial variables, it is possible not to find an obvious starting solution (test if the origin is in the definition field). In this case, a starting solution must be found using the big M method.
This method computes an auxiliary problem before solving the simplex of the linear program, which is why it is called a two-phase simplex.
## Why an artificial variable?
Before explaining how to obtain a solution to start the simplex, it is important to understand why it is necessary to add an artificial variable in addition to the slack variable.
Take as an example the following linear program:
The simplex starts by taking the base variables at zero, so with the following vector (0, 0, -19, 32). This is impossible because X3 cannot take a negative value. This is why a new variable must be added, which will therefore be artificial.
Here the vector (0, 0, 0, 19, 32) is a feasible solution. But since it is an artificial variable, it must be possible to remove it from the Simplex in order to be able to find a global solution. It is therefore necessary that X5 = 0. To check if this is possible, we are looking to minimize X5.
## Construction of Phase 1
We prepare in a similar way to the initial table of the Simplex method, but with some differences. By taking the previous example, we try to solve the following problem:
We know that only X4 and X5 are in the base, so the first step is to express the objective function with the non-base variables (here X1, X2, X3). In the example, this is possible thanks to the first constraint.
The first step is to solve the following linear program:
## Phase 2
The second phase of the Two-Phase method is developed exactly like the Simplex method, except that before starting the iterations it is necessary to remove the columns corresponding to the artificial variables, and to reconstruct the original table.
• Delete the columns of artificial variables: If we have come to the conclusion that the initial problem has a solution, we need to prepare our table for the second phase. This step is very simple, you just have to delete the columns corresponding to the artificial variables.
• Reconstruction of the initial table: The initial array, in this case, remains roughly equal to the last array of the first phase. The line of the objective function should be modified only by that of the initial problem and recalculate the line Z (in the same way as in the first table of phase 1).
The stopping condition is the same as in the Simplex method. In other words, when in the indicator line none of the values of the reduced costs is negative (because we are in the case of a maximization).
The previous simplex gives for optimal solution Z = 0 with the vector (0, 19/3, 0, 20/3, 0). That means that it is possible to do without the artificial variable and that the solution found is in the space of definition of the initial problem (since X5 does not intervene and that it is out of base). The final solution gives the following simplex:
The last step of phase 1 consists of taking the initial objective function (here -2X1 - X2) and replacing all the base variables with non-base variables (here X2 and X4 are in the base, X1 and X3 are outside the bases ).
We have for the first constraint X2 = 19/3 - 2 / 3X1 + 1 / 3X3, we can therefore rewrite the objective function by: Z = - 4 / 3X1 - 1 / 3X2 -19/3.
From this point, all the iterations, until reaching the optimal solution of the problem, show no difference with the Simplex method.
## Simplex algorithm
Here is the revisited simplex algorithm:
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Hello, how's it going? I have a car physics system in place which works really well. All I need some help with is that at low velocities, say 1-5mph, the car jitters all over the place (the suspension rocks all over and the velocity is unstable). I know this is a common problem but I'm not sure what controls I need in place to fix it. Any help is greatly appreciated.
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There are two things that immediately come to mind.
First, if you are using simple Euler integration, the suspension springs will make your simulation unstable. Simple Euler is almost never a good idea. Try Verlet instead, or even 2nd or 4th order Runge-Kutta.
You can make Euler semi-work by adding damping into your system. Everywhere you have a spring, you should also have a damper that applies a force in the opposite direction of the car velocity along the direction of the spring. The damper force is:
F_damper = -c * V * (V/fabsf(V));
where c is a damping coefficient, and V is the relative motion of the car to the other end of the spring. For example, if we're talking about a spring at the front right wheel, assuming the wheel stays planted on the ground, the velocity of the car at the front-right corner will be equal to the upward or downward velocity of the front-right corner of the car. The damping force will be in the opposite direction, e.g., if the car front-right is moving down, the damping force will be acting upward on the car at that point.
Damping can help stabilize Euler (and other) integration, but time step will still need to be fairly small.
Second, if you are using a Coulomb friction model, be careful how you deal with the transition between dynamic and static friction.
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How do you handle tire forces?
If you use a slipRatio type model for longitudinal forces, the equation will start to become worthless at low speeds. If you solve that with a differential system then you may still need a damper constant to avoid oscillations of wheel velocity. Maybe that is jerking the car around?
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also damping forces may make things unstable if you use Euler. The best way is, as Graham said, to swich to RK4 or Verlet.
If you have high damping force proportional to velocity (not velocity squared), that is, damping_force=-velocity*k;
it may make things be unstable if mass is small.
In that case, use
next_frame_velocity=velocity*exp(-dt*k/m)+acceleration_due_to_other_forces*dt;
- it's more precise result of damping over certain timestep.
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Quote:
Original post by Dmytryalso damping forces may make things unstable if you use Euler. The best way is, as Graham said, to swich to RK4 or Verlet.
Damping is actually required to stabilize simple Euler if there are spring forces (which there are for a suspension system). If you have damping, then Euler becomes conditionally stable, meaning it will be stable if your time step is small enough. But, for any given time step---no matter how small, if you increase damping sufficiently, simpler Euler will become unstable----requiring that you further reduce time step to stabilize it again.
Quote:
Original post by DmytryAnd as about dampipng. If you have high damping force proportional to velocity (not velocity squared), that is, damping_force=-velocity*k; it may make things be unstable if mass is smallIn that case, usenext_frame_velocity=velocity*exp(-dt*k/m)+acceleration_due_to_other_forces*dt;- it's more precise result of damping over certain timestep.
Good point! That accounts to the deterioration of velocity of the duration of the time step, thus minimizing the chance of velocity reversal, which is physically wrong.
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Quote:
Original post by grhodes_at_work
Quote:
Original post by Dmytryalso damping forces may make things unstable if you use Euler. The best way is, as Graham said, to swich to RK4 or Verlet.
Damping is actually required to stabilize simple Euler if there are spring forces (which there are for a suspension system). If you have damping, then Euler becomes conditionally stable, meaning it will be stable if your time step is small enough. But, for any given time step---no matter how small, if you increase damping sufficiently, simpler Euler will become unstable----requiring that you further reduce time step to stabilize it again.
The even more practically important thing, for sufficiently small mass Euler will be unstable.
Quote:
Quote:
Original post by DmytryAnd as about dampipng. If you have high damping force proportional to velocity (not velocity squared), that is, damping_force=-velocity*k; it may make things be unstable if mass is smallIn that case, usenext_frame_velocity=velocity*exp(-dt*k/m)+acceleration_due_to_other_forces*dt;- it's more precise result of damping over certain timestep.
Good point! That accounts to the deterioration of velocity of the duration of the time step, thus minimizing the chance of velocity reversal, which is physically wrong.
Yes, and velocity reversal due to this exp alone is inpossible, exp(-dt*k/m) is always <1 , so it always works as damper.
It's especially important for -v2 drag. In that case, for any timestep, if object have sufficiently high velocity, it will do bad things.
In that case,
next_frame_velocity=velocity*exp(-dt*Length(velocity)*k/m)+acceleration_due_to_other_forces*dt;
gives alot better results, i.e. no matter how big velocity is, it will damp Euler well, for small dt it gives exactly same result.
If it's possible to analitically solve
dv/dt=-a*v2
that would be even better.
In summary, using analitical solution, or even anything that works *like* analitical solution for given timestep without assuming it's very small gives alot better results. By "works *like*" i mean produces same results with very small dt and produces results that makes some sense with big dt. For instance, clamping damping results so velocity can't be negated by damping is bit better than straightforward approach. And analitical solution, formule with exp, is even better.
Recently there was thread about buoyancy, i found semi-analitical solution to be _very_ stable (instead of calculating impact forces i just did pulse conservation, taking into account added mass of water, like it's inelastic collision).
It's good not only for Euler, but for other integration methods as well. (it's not very simple to do with RK4, but definitely possible with modified RK4 that allows to have analitical/semi-analitical solutions for some variables)
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Problem Lang User Status Time Memory
Partition Function cpp pekempey AC 2869 ms 131.75 MiB
ケース詳細
Name Status Time Memory
0_00 AC 92 ms 32.67 MiB
100000_00 AC 688 ms 59.89 MiB
10000_00 AC 163 ms 35.77 MiB
1000_00 AC 96 ms 32.92 MiB
100_00 AC 93 ms 32.71 MiB
1_00 AC 94 ms 32.75 MiB
200000_00 AC 1344 ms 80.66 MiB
300000_00 AC 2841 ms 129.48 MiB
400000_00 AC 2842 ms 130.60 MiB
500000_00 AC 2869 ms 131.75 MiB
example_00 AC 93 ms 32.67 MiB
#include <bits/stdc++.h> #define rep(i, n) for (int i = 0; i < (n); i++) #define repr(i, n) for (int i = (n) - 1; i >= 0; i--) #define range(a) a.begin(), a.end() using namespace std; using ll = long long; constexpr int MOD = 998244353; class mint { int n; public: mint(int n_ = 0) : n(n_) {} explicit operator int() { return n; } friend mint operator-(mint a) { return -a.n + MOD * (a.n != 0); } friend mint operator+(mint a, mint b) { int x = a.n + b.n; return x - (x >= MOD) * MOD; } friend mint operator-(mint a, mint b) { int x = a.n - b.n; return x + (x < 0) * MOD; } friend mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; } friend mint &operator+=(mint &a, mint b) { return a = a + b; } friend mint &operator-=(mint &a, mint b) { return a = a - b; } friend mint &operator*=(mint &a, mint b) { return a = a * b; } friend bool operator==(mint a, mint b) { return a.n == b.n; } friend bool operator!=(mint a, mint b) { return a.n != b.n; } friend istream &operator>>(istream &i, mint &a) { return i >> a.n; } friend ostream &operator<<(ostream &o, mint a) { return o << a.n; } }; mint operator "" _m(unsigned long long n) { return n; } mint modpow(mint a, long long b) { mint res = 1; while (b > 0) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } mint modinv(mint n) { int a = (int)n, b = MOD; int s = 1, t = 0; while (b != 0) { int q = a / b; a -= q * b; s -= q * t; swap(a, b); swap(s, t); } return s >= 0 ? s : s + MOD; } template<int N> struct FFT { complex<double> rots[N]; FFT() { const double pi = acos(-1); for (int i = 0; i < N / 2; i++) { rots[i + N / 2].real(cos(2 * pi / N * i)); rots[i + N / 2].imag(sin(2 * pi / N * i)); } for (int i = N / 2 - 1; i >= 1; i--) { rots[i] = rots[i * 2]; } } inline complex<double> mul(complex<double> a, complex<double> b) { return complex<double>( a.real() * b.real() - a.imag() * b.imag(), a.real() * b.imag() + a.imag() * b.real() ); } void fft(vector<complex<double>> &a, bool rev) { const int n = a.size(); int i = 0; for (int j = 1; j < n - 1; j++) { for (int k = n >> 1; k > (i ^= k); k >>= 1); if (j < i) { swap(a[i], a[j]); } } for (int i = 1; i < n; i *= 2) { for (int j = 0; j < n; j += i * 2) { for (int k = 0; k < i; k++) { auto s = a[j + k + 0]; auto t = mul(a[j + k + i], rots[i + k]); a[j + k + 0] = s + t; a[j + k + i] = s - t; } } } if (rev) { reverse(a.begin() + 1, a.end()); for (int i = 0; i < n; i++) { a[i] *= 1.0 / n; } } } vector<long long> convolution(vector<long long> a, vector<long long> b) { int t = 1; while (t < a.size() + b.size() - 1) t *= 2; vector<complex<double>> z(t); for (int i = 0; i < a.size(); i++) z[i].real(a[i]); for (int i = 0; i < b.size(); i++) z[i].imag(b[i]); fft(z, false); vector<complex<double>> w(t); for (int i = 0; i < t; i++) { auto p = (z[i] + conj(z[(t - i) % t])) * complex<double>(0.5, 0); auto q = (z[i] - conj(z[(t - i) % t])) * complex<double>(0, -0.5); w[i] = p * q; } fft(w, true); vector<long long> ans(a.size() + b.size() - 1); for (int i = 0; i < ans.size(); i++) { ans[i] = round(w[i].real()); } return ans; } vector<mint> convolution(vector<mint> a, vector<mint> b) { int t = 1; while (t < a.size() + b.size() - 1) t *= 2; vector<complex<double>> A(t), B(t); for (int i = 0; i < a.size(); i++) A[i] = complex<double>((int)a[i] & 0x7fff, (int)a[i] >> 15); for (int i = 0; i < b.size(); i++) B[i] = complex<double>((int)b[i] & 0x7fff, (int)b[i] >> 15); fft(A, false); fft(B, false); vector<complex<double>> C(t), D(t); for (int i = 0; i < t; i++) { int j = (t - i) % t; auto AL = (A[i] + conj(A[j])) * complex<double>(0.5, 0); auto AH = (A[i] - conj(A[j])) * complex<double>(0, -0.5); auto BL = (B[i] + conj(B[j])) * complex<double>(0.5, 0); auto BH = (B[i] - conj(B[j])) * complex<double>(0, -0.5); C[i] = AL * BL + AH * BL * complex<double>(0, 1); D[i] = AL * BH + AH * BH * complex<double>(0, 1); } fft(C, true); fft(D, true); vector<mint> ans(a.size() + b.size() - 1); for (int i = 0; i < ans.size(); i++) { long long l = (long long)round(C[i].real()) % MOD; long long m = ((long long)round(C[i].imag()) + (long long)round(D[i].real())) % MOD; long long h = (long long)round(D[i].imag()) % MOD; ans[i] = (l + (m << 15) + (h << 30)) % MOD; } return ans; } }; FFT<1 << 21> fft; typedef vector<mint> poly; poly operator-(poly a) { for (int i = 0; i < a.size(); i++) { a[i] = -a[i]; } return a; } poly operator+(poly a, poly b) { for (int i = 0; i < a.size(); i++) { a[i] += b[i]; } return a; } poly operator-(poly a, poly b) { for (int i = 0; i < a.size(); i++) { a[i] -= b[i]; } return a; } poly &operator+=(poly &a, poly b) { return a = a + b; } poly &operator-=(poly &a, poly b) { return a = a - b; } poly pinv(poly a) { const int n = a.size(); poly x = {modinv(a[0])}; for (int i = 1; i < n; i *= 2) { vector<mint> tmp(min(i * 2, n)); for (int j = 0; j < tmp.size(); j++) { tmp[j] = a[j]; } auto e = -fft.convolution(tmp, x); e[0] += 2; x = fft.convolution(x, e); x.resize(i * 2); } x.resize(n); return x; } poly plog(poly a) { const int n = a.size(); vector<mint> b(n); for (int i = 1; i < n; i++) { b[i - 1] = i * a[i]; } a = fft.convolution(pinv(a), b); for (int i = n - 1; i >= 1; i--) { a[i] = modinv(i) * a[i - 1]; } a[0] = 0; a.resize(n); return a; } // g = exp(f(x)) // log g - f(x) = 0 // g - g * (log g - f(x))) // g * (1 - log g + f(x)) poly pexp(poly a) { const int n = a.size(); poly x = {1}; for (int i = 1; i < n; i *= 2) { x.resize(i * 2); auto e = -plog(x); e[0] += 1; for (int j = 0; j < min<int>(n, e.size()); j++) { e[j] += a[j]; } x = fft.convolution(x, e); x.resize(i * 2); } x.resize(n); return x; } int main() { cin.tie(nullptr); ios::sync_with_stdio(false); int N; cin >> N; N++; // (1+x+x^2+...)(1+x^2+x^4+...) // 1/(1-x) 1/(1-x^2) // -log(1-x) -log(1-x^2) - ... vector<mint> I(N); I[1] = 1; for (int i = 2; i < N; i++) { I[i] = I[MOD % i] * (MOD - MOD / i); } vector<mint> f(N); for (int i = 1; i < N; i++) { for (int j = 1; i * j < N; j++) { f[i * j] += I[j]; } } f = pexp(f); rep(i, N) cout << f[i] << " \n"[i == N - 1]; } | 2,592 | 6,463 | {"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} | 3 | 3 | CC-MAIN-2020-40 | latest | en | 0.19132 |
https://academicpapers.net/6-dr-barrett-has-run-a-test-to-predict-verbal-abilities-0-8-from-number-of-siblings-0-6-he-th/ | 1,669,694,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00103.warc.gz | 109,215,364 | 12,569 | # 6. Dr. Barrett has run a test to predict verbal abilities (0-8) from number of siblings (0-6). He th
6. Dr. Barrett has run a test to predict verbal abilities (0-8) from number of siblings (0-6). He thinks that the more people you have to talk to when you are young (number of siblings) the more verbal you will be. He has already run the test in SPSS and below is the output. (Total 21 pts)MeanSDNVerbalAbility4.251.96360Siblings3.451.81760ModelSum of SquaresdfMean SquareFSig.1Regression181.8741181.874232.470.000aResidual45.37658.782Total227.25059ModelUnstandardized CoefficientsStandardized CoefficientstSig.BStd. ErrorBeta1(Constant).917.2473.717.000Siblings.966.063.89515.247.000A. What are the independent and dependent variables and how are each scaled (1 pt each; Total 4 pts)? Independent:________________________ Scaling:_____________________________ Dependent:__________________________ Scaling:_____________________________B. What test should Dr. Barrett use (4 pts)?__________________________________________________C. Why that test (2 pts)?D. For the Independent Variable (2 pts): Mean___________ Standard Deviation_________________E. For the Dependent Variable (2 pts): Mean___________ Standard Deviation_________________F. Should Dr. Barrett (2 pts)? Retain H0 Reject H0H. Interpret the findings (5 pts): | 341 | 1,324 | {"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} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.722117 |
https://baahkast.com/what-are-the-7-base-quantities-and-their-base-units/ | 1,718,550,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00130.warc.gz | 99,546,734 | 8,787 | # What are the 7 base quantities and their base units?
## What are the 7 base quantities and their base units?
Seven Base SI Units
Unit Physical Quantity Symbol
ampere electric current A
kelvin thermodynamic temperature K
mole amount of a substance mol
candela luminous intensity cd
### How many base units are there in the SI?
seven SI base units
The seven SI base units, which are comprised of: Length – meter (m) Time – second (s) Amount of substance – mole (mole)
What are the 4 base units of the metric system?
The metric system is a system of measurement that uses the meter, liter, and gram as base units of length (distance), capacity (volume), and weight (mass) respectively.
What are the 7 base dimensions?
In total, there are seven primary dimensions. Primary (sometimes called basic) dimensions are defined as independent or fundamental dimensions, from which other dimensions can be obtained. The primary dimensions are: mass, length, time, temperature, electric current, amount of light, and amount of matter.
## What are base measurements?
A base quantity is one of a conventionally chosen subset of physical quantities, where no quantity in the subset can be expressed in terms of the others. The SI units, or Systeme International d’unites which consists of the metre, kilogram, second, ampere, Kelvin, mole and candela are base units.
### How many base units are in the International System of units?
There are seven base units that form the basis of the Système International d’Unités or International System of Units. SI, International System of Units Includes: SI base units SI units & symbols SI / metric prefixes Unit definitions SI (metric) / Imperial conversion
What are the 7 base units of the metric system?
The Base Units of the Metric System. Each unit is considered to be dimensionally independent of each other. These dimensions are described as the measurements of length, mass, time, electric current, temperature, amount of a substance, and luminous intensity. This list has the current definitions of each of the seven base units.
What are the base units used in the SI system?
The base used in SI system is 10, which makes the conversion easier. Latin and Greek prefixes are used in SI system and these refer to the numbers. Without the use of conversion factors, the SI units can be derived from one another. What are the common System of Units used in measurements?
## What are the base units of measurement in science?
The base units include: mass, length, time, temperature, amount of substance, electric current, and luminous intensity. In order that each of the SI units and quantities can be standardised across the globe, it is necessary to have exact definitions of each of them. | 578 | 2,743 | {"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} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.909143 |
https://www.888poker.com/magazine/strategy/poker-winrate | 1,601,293,940,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401600771.78/warc/CC-MAIN-20200928104328-20200928134328-00499.warc.gz | 620,167,590 | 28,367 | Poker players like to measure their success at the tables through a metric known as winrate.
• So, what exactly is winrate in poker?
• How is it calculated?
• How essential is it?
In this guide, we will discuss answers to the most common questions regarding winrate.
## What is Winrate in Poker?
Winrate in poker measures how fast a player wins. In cash games, it is measured in terms of the number of big blinds made per 100 hands (on average).
In tournaments, the calculation of winrate is slightly different. It’s based on the average return a player makes on each buy-in invested. It’s known as return on investment or ROI
Imagine a player makes (on average) \$0.10 for every \$1 tournament he enters. His ROI (return on investment) is 10%.
## Calculating Winrate in Poker
In the modern era, many players use software to calculate their winrate. This situation is especially the case for online players. Their poker tracker will calculate winrate or ROI automatically based on the hand histories it receives.
However, what if we play live or don’t make use of tracking software. How do we manually calculate our winrate?
For cash game players we will need access to the following –
1. Total number of hands we have played
2. Total profit in big blinds
Unfortunately, without these two values, it won’t be possible to calculate winrate. However, we can use estimated figures. For example, we might know that we have played live poker for around 1,000 hours and that we usually get dealt about 30 hands per hour.
Here is the formula:
Calculating Cash Game Winrate
Cash game winrate in bb/100 = (Profit in bb / Number of hands) * 100
Let’s test our knowledge –
We have played 100,000 hands of poker and won 3500 big blinds. What is our winrate in bb/100 hands?
(3500 / 100,000) * 100 = 3.5bb / 100 hands
1. Our tournament winnings.
2. Amount spent on tournament buy-ins.
Let’s see the formula in action.
We have \$10000 in total earnings but have spent around \$7500 on tournament buy-ins. What is our ROI?
(\$10000 - \$7500) / (\$7500 * 100) = ROI
\$2500 / \$750000 = 0.0333 or 3.33% ROI.
Note the following:
1. (Tournament Winnings – Tournament Buy-ins) is identical to our overall tournament profits. If we know our total profits, we can simply use that value for the first part of the formula rather than subtracting one value from another.
2. The formula outputs the answer as a decimal, 0.333 in this case. We multiply that by 100 to see our ROI in its percentage form.
## What’s a Good Winrate in Poker?
One of the most commonly asked questions is, “what’s a good winrate?”. The truth is, it depends on a wide variety of factors such as stakes, toughness of games, and rake.
If we were forced to give specific values as an answer to the question, it would generally be as follows.
• Online Cash Games - 0-10bb/100 hands with 10+ being considered exceptional in most games.
• Live Cash Games - 0-30bb/100 hands with 30+ being considered exceptional in most games.
• Online Tournaments ROI - 0-30% with 30%+ being considered exceptional in most games.
• Live Tournaments ROI- 0-100% with 100%+ being considered exceptional in most games.
Note that in each case we have specified a winrate range starting from zero because any positive winrate is a good winrate. With so many inflated brags in the online poker forums, it can be easy to forget that over 90% of poker players are losing players.
Even a 1bb/100 winrate in online cash games should not be looked down upon. Depending on the stakes being played such a winrate might be skilled enough to generate a healthy living.
## Winrate by Position
Cash game players like to breakdown their bb/100 winrate by position as a form of analysis. This type of analysis would rarely be done manually. However, it’s something that players use tracking software to monitor. It can help establish if play from one specific position at the table is sub-optimal.
Here is a rough breakdown of expected cash game winrate by position on a full ring table.
• Earlier positions - Marginally positive or break-even.
• Lojack - 5
• Hijack - 10
• Cutoff - 25
• Button - 40
• Small Blind - (-15)
• Big Blind - (-30)
Note that the average winrate in the small blind and big blind are negative. This rate is perfectly normal. If we were to fold every time we are in the big blind automatically, our winrate for that position would be -100bb/100 hands.
Hence, -30bb/100 hands is a significant improvement on that winrate, despite being negative overall.
The winrate from the earlier positions is more or less negligible. This statistic is one reason why many players prefer shorthanded tables.
## Variance in Poker
It’s crucial to account for variance when determining our winrate. In the context of poker, the term variance refers to the way our results can differ from our actual winrate over a period of time.
A losing poker player can go on a winning streak that lasts a lot longer than the average person might expect.
For example, imagine we know the precise winrate of a losing player, and it’s -2bb/100 hands. Over 100,000 hands, what are the chances that such a player makes money?
Well, it does depend on a value known as “standard deviation” which can be obtained using tracking software, but most would assume that the -2bb/100 loser has virtually zero chance of winning over such a large sample.
Calculations indicate that the percentage chance of such a player being in profit is a massive 21.5%!
Running 1,000 trial simulations, we occasionally see our -2bb/100 loser winning as much as 60 buy-ins (60 * 100bb stacks) over the sample. On the flip side,this player can also lose as many as 80 stacks despite his expected value being around 20 stacks lost over the 100,000-hand sample.
So, when a live player plays 1,000 hands then hits the forums to brag about his “sick winrate”, we should be able to see that such a sample size is non-representative.
Ideally, we’d have at least 300,000 hands (preferably more) before being able to know our actual winrate. (Even then, our -2bb/100 loser has around 8.5% chance of making a profit over the sample).
As we might imagine, this presents live players with a bit of a puzzle. Assuming 30 hands per hour as a standard rate of dealing at a brick and mortar casino it would take around 10,000 hours of grinding to guess our winrate.
Even if we played for 12 hours a day, seven days a week, we would still have to grind for over two years to reach such a sample. Assuming a more regular 40-hour working week, it would take closer to five years to achieve such a sample size.
To put this into context, a high-volume online professional could reach such a sample in around three months assuming 100,000 hands played per month (which is not unreasonable when mass multi-tabling).
## Winrate vs Other Metrics
With a heavy emphasis on improving winrate, it can be very easy to lose sight of what is essential. Although a high winrate is great for bragging rights, the goal of most players is to make as much money as possible not to generate the highest winrate.
As such, our hourly rate is a more critical metric than our winrate.
Scenario 1 – We play one table online and agonize over every decision, generating our highest possible winrate of 10bb/100 hands. We play around 100 hands per hour.
Scenario 2 – We play six tables online and accept that we may make some mistakes here and there. Our winrate drops to around 7bb/100 hands, but we are now playing 500 hands per hour.
There is no question that scenario 1 produces the highest winrate and the coolest looking graph (if we use tracking software). However, it should also be clear that scenario 2 is a more profitable moneymaking venture.
Scenario 1 generates around 10bb per hour. Scenario 2 generates 35bb per hour due to the higher volume.
When seeing the result on a forum, players might be quick to declare the player in scenario 1 a “better player” due to the higher winrate. But who really is the better player? The guy making 10bb/hour or the guy making 35bb/hour?
There isn’t a set answer here since it depends on what our goals are, but the majority of players would take scenario 2 over scenario 1 any day.
The same can be said for ROI in tournaments. A higher ROI doesn’t necessarily mean a more substantial amount of profit. Online players maximize their earnings by striking a delicate balance between winrate and volume.
Timothy "Ch0r0r0" Allin is a professional player, coach, and author. Since the beginning in 2006 he has built his roll from the lowest limits online without depositing a single dollar. After competing in some of world's toughest lineups (and winning) he now shares his insights and strategies with the 888poker magazine.
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Can You Win More with an Exploitative Poker Strategy Style of Play? | 2,314 | 9,889 | {"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} | 3.421875 | 3 | CC-MAIN-2020-40 | longest | en | 0.957484 |
https://www.unitconverters.net/flow/ounce-second-to-kilogram-day-gasoline-at-15-5-b0c.htm | 1,721,417,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00049.warc.gz | 888,197,224 | 3,388 | Home / Flow Conversion / Convert Ounce/second to Kilogram/day (Gasoline At 15.5%b0C)
# Convert Ounce/second to Kilogram/day (Gasoline At 15.5%b0C)
Please provide values below to convert ounce/second [oz/s] to kilogram/day (Gasoline at 15.5%b0C), or vice versa.
From: ounce/second To: kilogram/day (Gasoline at 15.5%b0C)
### Ounce/second to Kilogram/day (Gasoline At 15.5%b0C) Conversion Table
Ounce/second [oz/s]Kilogram/day (Gasoline At 15.5%b0C)
0.01 oz/s18.8909882292 kilogram/day (Gasoline at 15.5%b0C)
0.1 oz/s188.9098822919 kilogram/day (Gasoline at 15.5%b0C)
1 oz/s1889.0988229189 kilogram/day (Gasoline at 15.5%b0C)
2 oz/s3778.1976458377 kilogram/day (Gasoline at 15.5%b0C)
3 oz/s5667.2964687566 kilogram/day (Gasoline at 15.5%b0C)
5 oz/s9445.4941145944 kilogram/day (Gasoline at 15.5%b0C)
10 oz/s18890.988229189 kilogram/day (Gasoline at 15.5%b0C)
20 oz/s37781.976458377 kilogram/day (Gasoline at 15.5%b0C)
50 oz/s94454.941145944 kilogram/day (Gasoline at 15.5%b0C)
100 oz/s188909.88229189 kilogram/day (Gasoline at 15.5%b0C)
1000 oz/s1889098.8229189 kilogram/day (Gasoline at 15.5%b0C)
### How to Convert Ounce/second to Kilogram/day (Gasoline At 15.5%b0C)
1 oz/s = 1889.0988229189 kilogram/day (Gasoline at 15.5%b0C)
1 kilogram/day (Gasoline at 15.5%b0C) = 0.0005293529 oz/s
Example: convert 15 oz/s to kilogram/day (Gasoline at 15.5%b0C):
15 oz/s = 15 × 1889.0988229189 kilogram/day (Gasoline at 15.5%b0C) = 28336.482343783 kilogram/day (Gasoline at 15.5%b0C) | 603 | 1,480 | {"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} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.754084 |
https://www.thestudentroom.co.uk/showthread.php?page=22&t=4120683 | 1,513,533,958,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597295.74/warc/CC-MAIN-20171217171653-20171217193653-00032.warc.gz | 796,734,128 | 42,471 | You are Here: Home >< Maths
# Edexcel IGCSE Mathematics A - Paper 3H - 2016 - Unofficial Mark Scheme Watch
1. (Original post by cornflakeexpert)
pls!
OK I'll let you know once I have it.
Remind me on weekend.
2. (Original post by IGCSETANK)
Worst school? Ok then big **** search charter house and then tell me I got to the worst in the country ****** hahaha. Funny how you are self consciously trying to reassure yourself that you got it right by correcting other people that probably got it right, thick piece of ****. Really reflects your intellectual aptitude doesn't it?
I don't care what you are trying to say or if you're angry who the **** do you think you are using the word ****** like that. That's disgusting.
3. OK, sorry for misleading you. But we worked it out in our maths lesson (with our teacher), and the correct answer was 102.4cm^3.
4. (Original post by WhiteX)
for the set notaion what did you say for the last part because there is no commen letter bettwen the two right ?
Yes. Set A shares no members with Set B and vice versa.
5. (Original post by sivthasan)
OK, sorry for misleading you. But we worked it out in our maths lesson (with our teacher), and the correct answer was 102.4cm^3.
yh bro thats what i got
6. (Original post by sivthasan)
OK, sorry for misleading you. But we worked it out in our maths lesson (with our teacher), and the correct answer was 102.4cm^3.
Just saying, but it asked for that question to be to 3sf i believe, so it was just 102 cm ^3
7. Anyone remember the lengths of the sides of the pentagon - if so i can post some proof of 188.
8. It didn't ask for 3 significant figures, because it was an exact answer.
9. (Original post by sivthasan)
OK, sorry for misleading you. But we worked it out in our maths lesson (with our teacher), and the correct answer was 102.4cm^3.
I also got 102.4 the
volume factor was 15.625
and thus the working was 1600 / 15.625
10. Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
11. (Original post by HKHASSAN)
Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
I think it was there. ^
12. (Original post by HKHASSAN)
Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
it was 2100 + 1000 = 3100
then 3100 / 6000 = 0.516
X100 =52%
13. (Original post by Martins1)
Anyone remember the lengths of the sides of the pentagon - if so i can post some proof of 188.
8, 12, 12, 8, 13 angles 105
14. (Original post by BOBQ)
it was 2100 + 1000 = 3100
then 3100 / 6000 = 0.516
X100 =52%
Why did you add 1000 currency? Did it say somewhere in the question that he already had 1000?
15. (Original post by HKHASSAN)
Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
Yeah, it was 2100+1000=3100.
You will probably still get two or three out of five.
16. (Original post by HKHASSAN)
Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
I think you need to add his original share onto 2100 rupees. He originally had 1000 rupees (t00x2). That's what I did anyway and I got 52
17. (Original post by HKHASSAN)
Why did you add 1000 currency? Did it say somewhere in the question that he already had 1000?
He had 2 of the 12 parts in the ratio.
18. (Original post by HKHASSAN)
Why did you add 1000 currency? Did it say somewhere in the question that he already had 1000?
It said " Imran gives Bhavin 3/5 of his money
19. (Original post by Martins1)
He had 2 of the 12 parts in the ratio.
Ohh I wish I read the question properly sometimes and not just dive into it :/
I thought Bhavin was some extra person :[
I don't think this affects my A* seeing as, overall, I lost about 7-8 marks max.
20. (Original post by HKHASSAN)
Can someone tell me if I got the ratio question right
The question was A, B and C share 6000 (insert currency here) into the ratio 2 : 3 : 7
C then gives 3/5 of his share to D.
What percentage of the original total currency does D now have.
People are saying it is 38% but I got 35% here is how:
2 + 3 + 7 = 12
Therefore 1 share = 6000/12 = 500
C got 7 shares so he got 3500 (currency)
C then gave 3/5 of his share to D so I did 3500 x 3/5 = 2100
Then I did 2100/6000 = 0.35 which is 35%.
Where did I go wrong?? I'm a bit concerned. :/
Didnt he already have like 1000 or something
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 2,371 | 7,940 | {"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} | 2.9375 | 3 | CC-MAIN-2017-51 | latest | en | 0.967608 |
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# locally compact space
a topological space in which each point has a neighborhood that is compact.
* * *
Universalium. 2010.
### Look at other dictionaries:
• Locally compact space — In topology and related branches of mathematics, a topological space is called locally compact if, roughly speaking, each small portion of the space looks like a small portion of a compact space.Formal definitionLet X be a topological space. The… … Wikipedia
• locally compact space — Math. a topological space in which each point has a neighborhood that is compact … Useful english dictionary
• Locally regular space — In mathematics, particularly topology, a topological space X is locally regular if intuitively it looks locally like a regular space. More precisely, a locally regular space satisfies the property that each point of the space belongs to a subset… … Wikipedia
• Locally normal space — In mathematics, particularly topology, a topological space X is locally normal if intuitively it looks locally like a normal space. More precisely, a locally normal space satisfies the property that each point of the space belongs to a… … Wikipedia
• Locally connected space — In this topological space, V is a neighbourhood of p and it contains a connected neighbourhood (the dark green disk) that contains p. In topology and other branches of mathematics, a topological space X is locally connected if every point admits… … Wikipedia
• Compact space — Compactness redirects here. For the concept in first order logic, see compactness theorem. In mathematics, specifically general topology and metric topology, a compact space is an abstract mathematical space whose topology has the compactness… … Wikipedia
• Locally compact group — In mathematics, a locally compact group is a topological group G which is locally compact as a topological space. Locally compact groups are important because they have a natural measure called the Haar measure. This allows one to define… … Wikipedia
• locally compact — adjective (of a topological space) That for every point of the given topological space, there is a neighborhood of that point whose closure is compact … Wiktionary
• Σ-compact space — In mathematics, a topological space is said to be sigma; compact if it is the union of countably many compact subspaces. A space is said to be sigma; locally compact if it is both sigma; compact and locally compact.Properties and Examples* Every… … Wikipedia
• Feebly compact space — In mathematics, in the realm of topology, a topological space is said to be feebly compact if every locally finite cover by nonempty open sets is finite.Some facts:* Every compact space is feebly compact. * Every feebly compact paracompact space… … Wikipedia | 572 | 2,758 | {"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} | 2.765625 | 3 | CC-MAIN-2020-50 | latest | en | 0.883471 |
http://everything.explained.today/Bridge_(graph_theory)/ | 1,679,991,016,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00350.warc.gz | 16,229,509 | 5,783 | Bridge (graph theory) explained
In graph theory, a bridge, isthmus, cut-edge, or cut arc is an edge of a graph whose deletion increases the graph's number of connected components.[1] Equivalently, an edge is a bridge if and only if it is not contained in any cycle. For a connected graph, a bridge can uniquely determine a cut. A graph is said to be bridgeless or isthmus-free if it contains no bridges.
This type of bridge should be distinguished from an unrelated meaning of "bridge" in graph theory, a subgraph separated from the rest of the graph by a specified subset of vertices; see .
Trees and forests
A graph with
n
nodes can contain at most
n-1
bridges, since adding additional edges must create a cycle. The graphs with exactly
n-1
bridges are exactly the trees, and the graphs in which every edge is a bridge are exactly the forests.
In every undirected graph, there is an equivalence relation on the vertices according to which two vertices are related to each other whenever there are two edge-disjoint paths connecting them. (Every vertex is related to itself via two length-zero paths, which are identical but nevertheless edge-disjoint.) The equivalence classes of this relation are called 2-edge-connected components, and the bridges of the graph are exactly the edges whose endpoints belong to different components. The bridge-block tree of the graph has a vertex for every nontrivial component and an edge for every bridge.[2]
Relation to vertex connectivity
Bridges are closely related to the concept of articulation vertices, vertices that belong to every path between some pair of other vertices. The two endpoints of a bridge are articulation vertices unless they have a degree of 1, although it may also be possible for a non-bridge edge to have two articulation vertices as endpoints. Analogously to bridgeless graphs being 2-edge-connected, graphs without articulation vertices are 2-vertex-connected.
In a cubic graph, every cut vertex is an endpoint of at least one bridge.
Bridgeless graphs
A bridgeless graph is a graph that does not have any bridges. Equivalent conditions are that each connected component of the graph has an open ear decomposition,[3] that each connected component is 2-edge-connected, or (by Robbins' theorem) that every connected component has a strong orientation.[3]
An important open problem involving bridges is the cycle double cover conjecture, due to Seymour and Szekeres (1978 and 1979, independently), which states that every bridgeless graph admits a multi-set of simple cycles which contains each edge exactly twice.[4]
Tarjan's bridge-finding algorithm
The first linear time algorithm for finding the bridges in a graph was described by Robert Tarjan in 1974.[5] It performs the following steps:
• Find a spanning forest of
G
• Create a rooted forest
F
from the spanning forest
• Traverse the forest
F
in preorder and number the nodes. Parent nodes in the forest now have lower numbers than child nodes.
• For each node
v
in preorder (denoting each node using its preorder number), do:
• Compute the number of forest descendants
ND(v)
for this node, by adding one to the sum of its children's descendants.
• Compute
L(v)
, the lowest preorder label reachable from
v
by a path for which all but the last edge stays within the subtree rooted at
v
. This is the minimum of the set consisting of the preorder label of
v
, of the values of
L(w)
at child nodes of
v
and of the preorder labels of nodes reachable from
v
by edges that do not belong to
F
.
• Similarly, compute
H(v)
, the highest preorder label reachable by a path for which all but the last edge stays within the subtree rooted at
v
. This is the maximum of the set consisting of the preorder label of
v
, of the values of
H(w)
at child nodes of
v
and of the preorder labels of nodes reachable from
v
by edges that do not belong to
F
.
• For each node
w
with parent node
v
, if
L(w)=w
and
H(w)<w+ND(w)
then the edge from
v
to
w
is a bridge.
Bridge-finding with chain decompositions
A very simple bridge-finding algorithm[6] uses chain decompositions.Chain decompositions do not only allow to compute all bridges of a graph, they also allow to read off every cut vertex of G (and the block-cut tree of G), giving a general framework for testing 2-edge- and 2-vertex-connectivity (which extends to linear-time 3-edge- and 3-vertex-connectivity tests).
Chain decompositions are special ear decompositions depending on a DFS-tree T of G and can be computed very simply: Let every vertex be marked as unvisited. For each vertex v in ascending DFS-numbers 1...n, traverse every backedge (i.e. every edge not in the DFS tree) that is incident to v and follow the path of tree-edges back to the root of T, stopping at the first vertex that is marked as visited. During such a traversal, every traversed vertex is marked as visited. Thus, a traversal stops at the latest at v and forms either a directed path or cycle, beginning with v; we call this pathor cycle a chain. The ith chain found by this procedure is referred to as Ci. C=C1,C2,... is then a chain decomposition of G.
The following characterizations then allow to read off several properties of G from C efficiently, including all bridges of G.[6] Let C be a chain decomposition of a simple connected graph G=(V,E).
1. G is 2-edge-connected if and only if the chains in C partition E.
2. An edge e in G is a bridge if and only if e is not contained in any chain in C.
3. If G is 2-edge-connected, C is an ear decomposition.
4. G is 2-vertex-connected if and only if G has minimum degree 2 and C1 is the only cycle in C.
5. A vertex v in a 2-edge-connected graph G is a cut vertex if and only if v is the first vertex of a cycle in C - C1.
6. If G is 2-vertex-connected, C is an open ear decomposition. | 1,345 | 5,855 | {"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} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.943972 |
https://deceptivelyeducational.blogspot.com/2014/03/pot-of-gold-multiplication-game.html | 1,638,461,910,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00252.warc.gz | 267,383,306 | 19,732 | Friday, March 14, 2014
Pot of Gold Multiplication Game
It's almost St. Patrick's Day and in honor of that amazing holiday, I created a fun BINGO-like multiplication game to play with my 3rd grader.
What You Need
Multiplication Fact Flash Cards 1-10 (print some here if you don't already have a deck)
game pieces (use poker chips, glass baubles, buttons, etc.)
Download 6 pages of free game cards from Google Drive here. Page 1 of this 7-page PDF is blank so you can adapt the game to practice other skills.
How to Play
Simply shuffle the flash cards and place them on the table face down. Turn the cards over one and by and solve the problem. Each player places a game piece on their Leprechaun pot/rainbow if the answer is noted there. The first player to fill their pot wins!
Variations
If your child is still struggling with multiplication, print a multiplication table available for them to check their answers. Also have players take turns drawing a card and answering the problem.
If your child is proficient in math and is playing with adults, have them solve all the problems and call them out.
This is a good game for siblings to play. The older sibling can solve the problem and the younger sibling can practice identifying large and small numbers.
My son had a fun with this, even using an auctioneer-like voice to call out the answers. We were neck-and-neck, but he filled his pot of gold before I did and was declared one very lucky winner!
Want another St. Patty's Day-inspired multiplication game? Check out Leprechaun's Luck! | 348 | 1,552 | {"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} | 2.640625 | 3 | CC-MAIN-2021-49 | latest | en | 0.963311 |
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Topic: true; the secondpower is not germaine toe the tetragon, but the tetrahedron
Replies: 10 Last Post: Apr 9, 2017 8:44 PM
Messages: [ Previous | Next ]
thugstyle103@gmail.com Posts: 61 Registered: 1/13/17
Re: true; the secondpower is not germaine to the tetragon, but the tetrahedron
Posted: Apr 9, 2017 11:57 AM
it is 23 tetravols, or 23/3 hexavols;
23 is the first nontwinprime, other than t00
> viz, what is the volume of the truncated tetrrahedron?
> > so, let's get the extrema on those spheres
> > > the general tetrahedron is tangent to three spheres
> > > w.r.t the three component skew tetragona | 246 | 822 | {"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} | 2.65625 | 3 | CC-MAIN-2018-13 | longest | en | 0.837478 |
http://www.regiononehealth.org/2019/05/14/how-to-get-interest-rate/ | 1,566,653,267,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00016.warc.gz | 309,537,909 | 49,624 | # How To Get Interest Rate
Lowest 20 Year Mortgage Rates 20-year mortgage rates – Compare. | Bankrate.com – The 20-year fixed rate mortgage usually has a lower interest rate than a 30-year fixed-rate mortgage. Additionally, since you’ll be paying less interest over a shorter amount of time, your total interest payment will be significantly lower. This is attractive to folks who want to pay less interest over the life.
How to Convert APY to Interest Rate | Pocketsense – Multiply the result from step 5 by 100 to convert to a percentage to find the interest rate. For example, you would multiply 0.053660387 by 100 to find the interest rate equals about 5.366 percent if the APY is 5.5 percent and interest is compounded monthly.
How to Calculate Interest Rates on Bank Loans – There are many methods banks use to calculate interest rates, and each method will change the amount of interest you pay. If you know how to calculate interest rates, you will better understand your loan contract with your bank. You also will be in a better position to negotiate your interest rate with your bank.
Excel formula: Calculate interest rate for loan | Exceljet – One use of the RATE function is to calculate the periodic interest rate when the amount, number of payment periods, and payment amount are known. For this example, we want to calculate the interest rate for \$5000 loan, and with 60 payments of \$93.22 each.
Is A High Interest Rate Good What is a Good Interest Rate on a Credit Card? (with pictures) – These rates are not expected when people’s credit is less than perfect. Even with imperfect credit, many people can get cards that loan at about a 20% rate, and it probably isn’t a good idea to accept a card that charges more than that.
Do your research and learn what kind of interest rates a person with your credit score can qualify for. Tell the issuer what you want and what you can get from other issuers. Need to transfer a.
Private Mortgage Lending Rates Loan Options That Work for You Mortgage – Mortgages | Fifth Third Bank – For mortgages, home loans, mortgage rates & information on loan types, contact a. and adjustable rate mortgage (arm) loans available; No Private Mortgage.
4 Ways You Can Get A Lower Credit Card Interest Rate – One credit card can go through a lot of APRs. If you got a low-interest rate in a new credit card, you need to know that this will not last. The creditor usually changes the rate and makes it higher after 6 to 12 months. Even the 0% interest that is common in balance transfer cards will not last forever.
How to Find Simple Interest Rate: Definition, Formula. – Find the maturity value for a simple interest loan of \$4,000 at an annual interest rate of 10.5% to be repaid in 105 days. It is common practice for banks to assume there are 360 days in a year.
3 Ways to Get a Lower student loan interest rate – If you have high-interest federal or private student loans, refinancing can be a useful tool to get a lower student loan interest rate and save money. With refinancing, you work with a private lender to take out a new loan to repay some or all of your current debt with low-interest student loans. | 689 | 3,172 | {"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} | 2.71875 | 3 | CC-MAIN-2019-35 | latest | en | 0.950906 |
https://www.globalpostalcodesystem.info/mail/1-package-of-yeast-is-how-many-teaspoons.html | 1,669,697,891,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00027.warc.gz | 831,959,168 | 31,018 | # 1 Package Of Yeast Is How Many Teaspoons?
Yeast Variety Conversions
1 Packet (Envelope) of Active Dry Yeast Equals:
Unit Amount
Weight 1/4 oz.
Volume 2 1/4 tsp.
Instant Yeast 1 envelope or 1/4 oz. or 2 1/4 tsp.
How many teaspoons in one package of yeast? On average, one package of yeast contains 2.25 teaspoons of yeast. When you purchase yeast from a market, they generally sell three packages of yeast together.
## How many teaspoons are in a packet of dry yeast?
One packet of yeast contains 1/4 ounce or approximately 2 1/4 teaspoons of yeast. Subsequently, question is, how many tablespoons are in a packet of dry yeast?
## How many teaspoons of yeast are in a bread machine yeast?
Rapid Rise Yeast, Fast Rising Yeast, or Bread Machine Yeast contains 2 1/4 teaspoons of yeast that equals 7 grams, or 1/4 ounce. Yeast is a fungus that consists of a single cell.
## How long does yeast last in packages?
Yeast sold in packets, such as active dry or instant, can be stored at room temperature in a dry spot for 12 to 18 months. Dry yeast lasts about two to four months past the best by date; after that, it will likely expire. In this brief guide, we answered the question ‘How many teaspoons are in a packet of yeast?’.
## How much does a packet of cake yeast weigh?
If we talk of cake yeast or fresh compressed, it also comes in packets that weigh 7 grams, roughly equal to 2 1/4 teaspoons. Cake Yeast is also called wet yeast or fresh.
## What is 1 package of active dry yeast equivalent to?
1 envelope (or packet) of Active Dry Yeast, Instant Yeast, Rapid Rise Yeast, Fast Rising Yeast or Bread Machine Yeast weighs 7 grams, or 1/4 ounce and equals 2 1/4 teaspoons (11 mL).
## How many teaspoons is 2 yeast packets?
While older recipes call for 1 tablespoon or 1 packet of active dry yeast, the new yeast packets contain 2-1/4 teaspoons yeast.
## How many teaspoons are in a package of Fleischmann’s yeast?
“How much dry yeast is in a ¼-ounce packet?” About 2-¼ teaspoons.
## How much is a pkg of dry yeast?
When you buy yeast, it is generally sold in a packet. It is common for three packets to be sold together. Each packet of yeast contains ¼ ounce, which is the equivalent to seven grams or 2 ¼ teaspoons.
## How many tbsp are in a package of yeast?
The standard package of yeast contains 2 1/4 teaspoons of yeast, therefore 2.25/3=. 75, so there is 3/4 tablespoon in a package of yeast.
## How much is a package of yeast?
We all know that recipes typically call for a “packet” of yeast. But how much yeast is actually hiding inside? One yeast packet will typically contain 2 ¼ teaspoons of yeast, which can also be measured as ¼ ounce or seven grams.
## How much is in an envelope of Fleischmann yeast?
1 envelope dry yeast (1/4 oz) = 2-1/4 teaspoons = 1 cake fresh yeast (0.6 oz).
## How many teaspoons are in a .25 ounce package of yeast?
ANSWER: One (0.25-ounce) package of active dry yeast is equal to 2 1/4 teaspoons of yeast.
## Can I use active yeast in a bread machine?
You can use active dry yeast in your bread machine, but it should be dissolved in water before being used. In contrast, bread machine yeast can be mixed in with other dry ingredients.
## How many teaspoons is 7g of yeast?
(1) A packet of yeast is typically 7g exactly. So if you’re buying yeast by the packet, use one packet. But assuming you will be measuring from bulk yeast, the correct measure by volume would be 2 1/4 tsp instead of 2 1/2 tsp.
## How much yeast is in a sachet?
When converting recipes, one sachet of fast action dried yeast is equivalent to 15g of fresh yeast, or 7g of ordinary dried yeast. Each sachet of fast action dried yeast is sufficient for making bread using 500g of bread flour, so for a complete 1kg bag flour, use two sachets.
## Is active yeast the same as instant yeast?
A Quick Primer. Dry yeast comes in two forms: active and instant. ‘Active’ describes any dry yeast that needs to be activated prior to use, while ‘instant dry yeast’ describes any dry yeast that’s ready for use the instant you open the package.
## How many tablespoons are in a package of dry yeast?
Typically, a standard packet of yeast contains 2-2tsp+14% tsp yeast. A packet of bread yeast normally weighs 7-8 grams and includes 2.25 teaspoons of yeast. 2 1/4 teaspoons yeast equals 7 grams, or 1/4 ounce, in Rapid Rise Yeast, Fast Rising Yeast, or Bread Machine Yeast.
## How many spoons of yeast are there in 1 package?
One packet of ¼ oz yeast is equivalent to 2.25 teaspoons. When you purchase yeast from the market, they generally sell three packets of yeast together. Each of these three packages weighs 7 grams which is equivalent to 2.25 teaspoons. If you are fond of baking bread, you can buy bags or jars of yeast.
## How many teaspoon does 1 package of active dry?
It is important to add active dry yeast into the water before adding it to the dry ingredients. 1 package of active dry yeast contains ¼ oz. of it, which is equivalent to 2.25 teaspoons. Instant yeast is made of finer granules. It has a slightly different taste than that of active dry yeast so you can use them interchangeably in recipes.
## How many cups of yeast are in one packet?
What is a Typical Yeast to Flour Ratio? One packet of dry yeast (2 and 1/4 teaspoons) will raise up to 4 cups of flour. How do you measure yeast? Envelopes of yeast generally weigh 1/4 ounce each and measure approximately 2-1/4 teaspoons.
## What is one package of dry yeast equal to?
1. 1 envelope (or packet) of Active Dry Yeast, Instant Yeast, Rapid Rise Yeast, Fast Rising Yeast, or Bread Machine Yeast (depending on your preference).
2. A quarter ounce of yeast weighs 7 grams, or 1/4 ounce, and is equal to 2 1/4 teaspoons (11 mL).
3. Similarly, how much yeast is contained in a packet?
4. One package of yeast includes 1/4 ounce of yeast, which is equal to roughly 2 1/4 teaspoons.
5. The following question is: how many teaspoons of dry yeast are included in a package of dry yeast?
Conversions of Yeast Varieties
1 Packet (Envelope) of Active Dry Yeast Equals:
Volume 2 1/4 tsp.
Instant Yeast 1 envelope or 1/4 oz. or 2 1/4 tsp.
Bread machine Yeast 1 envelope or 1/4 oz. or 2 1/4 tsp.
Rapid Rise Yeast 1 envelope or 1/4 oz. or 2 1/4 tsp.
1. What is the amount of yeast contained in a Fleischmann packet?
2. (1/4 cup dry yeast) = 2-1/4 teaspoons fresh yeast = 1 cake fresh yeast (1/4 cup dry yeast) (0.6 oz.) Which dry yeast is the most effective?
3. When it comes to ordinary bread baking, instant yeast is a popular alternative.
4. Instant yeast, sometimes known as rapid-rise yeast, quick-rise yeast, or even bread machine yeast, is available in tiny packets or jars and is suitable for use in everyday bread manufacturing (you can easily substitute instant for active dry yeast in most recipes).
### What happens if you use too much yeast?
When there is too much yeast present, the dough may become flat because the yeast releases gas before the flour is ready to expand. If you leave the dough to rise for an excessive amount of time, it will develop a yeasty or beery fragrance and taste, and it will eventually deflate or rise poorly in the oven, resulting in a light crust.
### What is the best yeast to use?
SAF instant yeast is the yeast of choice in our test kitchen, and it is well regarded for its strength and adaptability. SAF Red is an all-purpose yeast that works well in recipes that call for ″normal″ yeast, such as our Classic Sandwich Bread, Beautiful Burger Buns, and No-Knead Crusty White Bread, among others.
### How do I convert dry yeast to instant yeast?
1. To swap active dry yeast with instant (or fast rise) yeast, follow these steps: Use 25 percent more active dry than you would normally.
2. For example, if a recipe asks for 1 teaspoon of quick yeast, 1 1/4 teaspoons of active dry yeast should be used in its place.
3. And don’t forget to ″prove″ the yeast by dissolving it in a quantity of the recipe’s water that has been heated to 105 degrees, as described above.
### How do I convert instant yeast to active dry yeast?
Instant Yeast Can Be Used to Replace Active Dry Yeast in Baking
1. To determine how much instant yeast to use, multiply the amount of active dry yeast in the recipe by 0.75 to get the amount of instant yeast you need.
2. 1 package active dry yeast (2 1/4 teaspoons) Equals 1 2/3 teaspoons instant yeast
3. 1 teaspoon active dry yeast = 3/4 teaspoon instant yeast
4. 1 teaspoon active dry yeast = 3/4 teaspoon instant yeast
### What is the difference between active dry yeast and instant yeast?
The most common type of yeast used in baking is active-dry yeast, which may be found in most recipes. Instant dry yeast, on the other hand, does not require proofing in warm water and may be added immediately to dry components like as flour and salt without any further steps. Instant yeast particles are smaller than regular yeast particles, allowing them to dissolve more quickly.
### How do you activate dry yeast?
It is not necessary to use boiling water to activate the yeast. It is recommended to use a modest amount of water that is either room temperature or slightly warm. Toss the mixture with a spoon or fork for a minute or two, until the yeast has completely dissolved.
### How do I convert fresh yeast to dry?
To convert from fresh yeast to active dry yeast, double the fresh yeast quantity by 0.4, and vice versa. Before being added into a dough, active dried yeast must be soaked in warm water for at least 15 minutes. The fresh yeast amount must be multiplied by 0.33 in order to convert to instant dry yeast quantity.
### How do you know yeast is good?
You may check your yeast’s activity by adding 1 teaspoon of sugar and 2 1/4 teaspoons of yeast (from one packet) to 1/4 cup of warm water and letting it sit for 10 minutes. After that, you must wait 10 minutes. Even if the liquid begins to boil and have a yeasty fragrance, the yeast is still viable.
### Does yeast expire?
Compressed or Cake Yeast has a shelf life of around a year. It’s important to remember that yeast, like many other baking items, generally has a best before date rather than a use by date or an expiration date on the package. This distinction allows you to safely utilize yeast for baking purposes for a period of time after the best before date has passed.
### Why is yeast bad for you?
A small amount of yeast in your system is beneficial. Excessive consumption might result in infections and other health issues. If you take antibiotics too frequently or use oral contraceptives, your body may begin to produce an excessive amount of yeast. Gas, bloating, mouth sores, foul breath, a film on your tongue, and itching rashes are all common side effects of this.
### Can you make yeast?
Step 1: In a small mixing basin, whisk together equal parts flour and water. Step 2: Cover the bowl loosely with a cover or a cloth and let it out on the counter to cool to room temperature for 30 minutes. The process of the yeast and bacteria colonizing your batter will be accelerated if you keep it in a somewhat warm, but not too hot, environment.
## How Many Teaspoons In 1 Package Of Yeast? (+3 Step Test)
In this post, we will address the topic ″How many teaspoons are in a packet of yeast?″ as well as the question ″How should yeast be stored?″
## How many teaspoons in 1 package of yeast?
For example, a conventional box of yeast will typically include 2-2tsp+1/4 tsp of yeast per package. In most cases, a packet of bread yeast weighs roughly 7-8 grams and includes 2.25 teaspoons of yeast per packet. It includes 2 1/4 teaspoons of yeast, which is equal to 7 grams of yeast, or 1/4 ounce. Other names for this product are Rapid Rise Yeast and Fast Rising Yeast.
## The function of Yeast in baking
1. Yeast is a kind of fungus that is made up of a single cell.
2. In baking, it serves as a leavening agent in the same way as baking powder does.
3. It is, however, a biological process that results in the production of carbon dioxide.
4. The biological mechanism, known as fermentation, takes advantage of the starch in bread and converts it into gas, alcohol, and water, among other things.
5. Unlike baking powder, which evaporates during baking, alcohol aids in the development of gluten and the addition of a great deal of flavor to the bread.
The gas produced aids in the raising of the bread.
## Different types of Yeast used in baking
### Instant yeast
When compared to active dry yeast, it has smaller grain size and a higher quantity of live yeast cells per unit volume of the product. As the name implies, this yeast does not require rehydration and may be used immediately to the dough mixture without any additional steps.
### Osmotolerant yeast
Yeast development is hampered by high concentrations of salt and sugar. As a result, osmotolerant yeast comes to the rescue when baking bread with a high sugar concentration.
### Cream yeast
Cream yeast is just yeast that has been compressed and placed in a liquid condition. It is commonly utilized in the bread industry because of its compacted nature.
### Active dry yeast
Compared to other kinds of yeast, it is more stable, and the grain size is larger than that of quick yeast. It is sold in a latent state and must be proofed before being used in the dough.
### Brewer’s yeast
It is used in the production of alcohol by brewers, as the name indicates. Brewer’s yeast is also utilized for its nutritional properties, which are well documented. It contains significant levels of chromium, which helps to maintain the digestive tract healthy.
### Rapid-rise yeast
It is a form of instant yeast that has a smaller particle size and a faster dissolve rate than regular instant yeast. It generates a significant amount of carbon dioxide (CO2).
### Compressed yeast
It is essentially dry cream yeast that has been formed into little or big blocks and covered with aluminum foil. Despite the fact that it is extremely perishable, it is still employed in bakeries.
## Other FAQs about Yeast which you may be interested in.
What is the shelf life of yeast? Does nutritional yeast have a shelf life? Is it possible to utilize yeast that has expired?
## How to perform a freshness test for yeast?
It is impossible to tell if yeast is alive or dead simply by looking at it. Check to see if the yeast is alive and well by doing the test listed below.
1. 14 cup lukewarm water should be placed in a cup.
2. It is preferable to use a thermometer to check the temperature of the water and ensure that it is between 110 and 115 degrees Fahrenheit.
3. A temperature of more than 120°F can kill your yeast, so keep it cool.
4. Yeast will not be able to reach its full potential at temperatures lower than 110° Fahrenheit.
5. 1 14 teaspoon of yeast and 1 teaspoon of granulated sugar should be added to the water.
Give it a good swirl until the yeast is completely dissolved. Wait for 10-15 minutes before continuing.
In order for the yeast to be active, it must produce bubbles or foam on the surface of the liquid and emit a strong yeasty odor. If this is not the case, the yeast is dead.
See also: How Long Does The Post Office Hold Packages?
Alternatively, a little portion of dough (the size of a walnut) can be used for this proving test.
## How to store yeast?
• Dry yeast and fresh yeast are the two varieties of yeast that are most commonly used by home bakers and commercial bakeries, respectively:Dry yeast and Fresh yeast
1. A package of dried yeast that has not been opened will remain excellent for several months in the pantry if it is kept away from the heat of the stove, direct sunshine, and moisture.
2. Immediately upon opening the packet, the dormant yeast granules become susceptible to deterioration.
3. As a result, it must be stored in a refrigerator.
4. It is not possible to refrigerate dried yeast that has been opened.
5. The moist climate of the refrigerator will degrade the quality of the yeast that is stored inside.
Alternatively, you may move the yeast to a zip-lock freezer bag and preserve it there.Fresh yeast, often known as cake yeast or compressed yeast, is more prone to deterioration than dried yeast.It is only available in chilled form, and it must be kept in the refrigerator in order to remain alive.
## Conclusion
How many teaspoons of yeast are there in a packet of yeast? And how should yeast be stored are covered in detail in this article.
## References
Hello, my name is Charlotte, and I enjoy cooking. In a prior life, I worked as a chef. I add some of my culinary expertise to the dishes on this hub and am available to answer any food-related queries.
## How Many Teaspoons Are In A Packet Of Yeast? (3 Brand Kinds)
• Throughout this brief tutorial, we will provide an answer to the query ‘How many teaspoons are there in a packet of yeast?’ We will look at the many types of yeast that are available for purchase, as well as the number of teaspoons that can be found in various brands of yeast. Yeast may be divided into four types: Cake yeast, Active Dry Yeast, Quick Rise Yeast, Instant (or fast-rising) Yeast are all terms used to describe yeast.
### How many teaspoons are in a packet of yeast?
1. An average package includes 7 grams of sugar, which is used nearly entirely in recipes.
2. If a recipe asks for one packet of yeast, you may anticipate that you will need to use two and a quarter teaspoonfuls of yeast.
3. A package of yeast may not always include 2.25 teaspoons; it might alternatively contain either 2.5 teaspoons or 3.25 teaspoons.
4. Yeast is a vital ingredient in many culinary endeavors, including baking, brewing, and fermentation.
5. Yeast is a vital element in the production of various varieties of bread.
Yeast permits the bread to rise, which in turn contributes to the bread’s light and fluffy texture.When yeast is introduced to bread dough, it breaks down huge starch molecules into simple sugars, which is what makes the bread taste so good.It is at this point that carbon dioxide and ethyl alcohol are released, resulting in the formation of air bubbles in the dough, which causes it to rise.
### How many teaspoons are generally present in a packet of yeast?
For every 1/4 oz.of yeast in a package, there are roughly 2 1/4 teaspoons of sugar in it.Yeast is packed according to its weight.
• This measurement falls between 2 and 2 1/4 teaspoons due to the possibility that the particle size of the yeast may vary.
• Instant yeast, bread machine yeast, and rapid rise yeast are all types of yeast that are commonly used.
• Two and a quarter teaspoons of yeast are contained in a package of each of these kinds.
• If we’re talking about Yeast starter, Sponge, and Biga, each package contains 1 cup of active yeast.
One cup of yeast is equal to around 48 teaspoons.One gram of yeast is equal to 0.35 teaspoons, and you can figure out how many grams of yeast are in a package by looking at the label on the yeast envelope itself.Calculating the number of total teaspoons that you estimate to be included within your packet is as simple as multiplying 0.35 by the number of grams printed on your packaging.On average, yeast packages contain 7 grams of yeast, which indicates that this packet of instant yeast will include 1 3/4 teaspoons of the active ingredient in it.If your yeast package has 6 grams of yeast, it will include 1 1/2 teaspoons of yeast, and if it contains 5 grams of yeast, it will contain 1 1/4 teaspoons of yeast.Each packet of Fleischmanns’ RR yeast, which is a type of rapid rise yeast, contains 2 1/2 teaspoons of active yeast per packet.
• The packet of Red Star instant yeast contains 1 1/2 teaspoons of yeast.
• The active dry yeast, instant yeast, rapid rise yeast, fast rising yeast, and bread machine yeast, all made in the United States, each include 2 1/4 teaspoons of active dry yeast.
• The yeast packages in question weigh 7 grams or 11 milliliters.
• When it comes to cake yeast or fresh compressed yeast, it is also available in packets weighing 7 grams, which is approximately equal to 2 1/4 teaspoons.
### Cake Yeast:
Cake yeast is often referred to as wet yeast or fresh yeast. Because of the high moisture content of the yeast, it is referred to as wet yeast. It is highly perishable and must be kept refrigerated in order to avoid spoilage.
## Other FAQs about Yeast which you may be interested in.
What is the cost of a packet of yeast?
## Active Dry Yeast:
Active dry yeast aids in the leavening of bread and the creation of an airy, light texture.Before using active dry yeast, it is necessary to dissolve it in water.Individual packets of active dry yeast or tiny glass jars of active dry yeast are available for purchase.
• It’s important to remember to refrigerate the latter after opening to ensure that the yeast remains fresh and active.
• Follow the steps outlined below to create active dry yeast, such as Fleischmann’s Yeast: Yeast and Sorbitan Monostearate are the only two substances that may be found in the package.
• 1/4 cup warm water should be added to the contents of the envelope.
• The heated water should be between 100 and 110 degrees Fahrenheit.
A kitchen thermometer may be used to get more precise results.Stir add 1 teaspoon of sugar and let the mixture aside for 10 minutes to allow the yeast to froth.The sugar provides nourishment for the yeast, and the warm temperature encourages the yeast to grow and reproduce.When the volume of the mixture doubles, it signals that the yeast is active and ready to be used as a baking aid in the recipe.
### Quick Rise/ Instant Yeast:
Instant yeast dissolves and activates in a short period of time. You may also mix it directly into the dry ingredients and get the same outcome. It is not necessary to prove it before using it, as is the case with active dry yeast.
### How to store yeast?
Despite the fact that yeast has an expiration date, it has a lengthy shelf life.In a dry place at room temperature for 12 to 18 months, yeast marketed in packets such as active dry or instant can be kept fresh for up to two years.Dry yeast has a shelf life of two to four months after it has passed its best before date; after that, it will most certainly expire.
• The short answer to the question ‘How many teaspoons are there in a package of yeast?’ is provided in this tutorial.
• We looked at the many types of yeast that were available for purchase, as well as the number of teaspoons that were present in various brands of yeast.
## Citations
My name is Charlotte, and I enjoy cooking. In a prior life, I worked as a chef. I add some of my culinary expertise to the dishes on this hub and am available to answer any food-related queries.
## Question: 1 Package Of Yeast Equals How Many Teaspoons
1/4 cup (or 2 1/4 teaspoons) of yeast is contained in each container. This is around 7 grams (11 mL) of liquid.
## What is 1 package of active dry yeast equivalent to?
Conversions of Yeast Varieties An active dry yeast packet (or an envelope) contains the following ingredients: 1/4 oz. in weight 2 1/4 teaspoons of volume Instant 1 envelope (or 1/4 oz. or 2 1/4 tsp.) active dry yeast Machine that makes bread 1 envelope (or 1/4 oz. or 2 1/4 tsp.) active dry yeast
## How many teaspoons are in a packet of yeast?
Each packet of yeast includes 14 ounces, which is the equivalent of seven grams or 2 14 teaspoons of active dry yeast per packet. Though the amount of yeast required varies from recipe to recipe, the majority of breads ask for one packet or somewhat less than that amount. You may also get yeast in jars or sacks, which is very convenient if you bake a lot of bread.
## How many tablespoons are in a packet of active yeast?
One fourteenth of an ounce of yeast is the equal to seven grams or two and a quarter teaspoons. In most bread recipes, one packet of yeast or somewhat less than that is called for, however the exact amount varies from one to the next. You may also get yeast in jars or sacks, which is very convenient if you bake a lot.
## How much yeast is in a package of dry yeast?
Thanks. One (0.25-ounce) packet of active dry yeast is equal to about 2 1/4 teaspoons of active dry yeast.
## How many teaspoons is 2 yeast packets?
The new yeast packets contain 2-1/4 teaspoons of active dry yeast, whereas prior recipes called for 1 tablespoon or one packet of active dry yeast.
## Is 7g of yeast a teaspoon?
(1) A package of yeast normally contains exactly 7g of yeast. Consequently, if you’re purchasing yeast in a packet, use one packet. The proper volume measurement, assuming that you will be measuring from bulk yeast, is 2 1/4 teaspoons instead of 2 1/2 teaspoon.
## How much yeast is in a Fleischmann packet?
″Can you tell me how much dry yeast is in a 14.5-ounce packet?″ Approximately 2-1/4 teaspoons
## What if I put too much yeast in my bread?
When there is too much yeast present, the dough may become flat because the yeast releases gas before the flour is ready to expand. If you leave the dough to rise for an excessive amount of time, it will develop a yeasty or beery fragrance and taste, and it will eventually deflate or rise poorly in the oven, resulting in a light crust.
## How much Fleischmann yeast is in an envelope?
(1/4 cup dried yeast) equals 2-1/4 teaspoons fresh yeast, which equals 1 cake fresh yeast (0.6 oz).
## Is active dry yeast the same as instant yeast?
Dry yeast is available in two varieties: active and quick. ″Active dry yeast″ refers to any dry yeast that must be activated prior to use, whereas ″instant dry yeast″ refers to any dry yeast that is ready to use as soon as the container is opened.
## How much is half a packet of yeast?
Active yeast and quick yeast are the two types of dry yeast available. If you have a packet of active dry yeast, it means that it must be activated before use, whereas ″instant dry yeast″ means that it may be used immediately after it has been opened.
## How many tablespoons is a cube of yeast?
When one cube of fresh yeast is converted to a tablespoon, the result is 4.49 tablespoons. How many teaspoons of fresh yeast are contained within a single cube? The solution is as follows: 1 cube euro (cube) unit in a fresh yeast measure equals 4.49 tbsp (tablespoon) in the equivalent measure for the same fresh yeast type and 1 cube euro (cube) unit in the comparable measure.
## Can you use half a packet of yeast?
Make a 50- to 80-percent reduction in the amount of yeast you use in your recipe. Consider the possibility that you might create the same loaf of bread with half the amount of yeast — or even less? You may undoubtedly do so – but not by following the same approach as you would normally do.
## How do I convert dry yeast to instant yeast?
To swap active dry yeast with instant (or fast rise) yeast, follow these steps: Use 25 percent more active dry than you would normally. For example, if a recipe asks for 1 teaspoon of quick yeast, 1 1/4 teaspoons of active dry yeast should be used in its place.
## How much is in a sachet of yeast?
The amount of quick action dry yeast needed to convert a recipe is equal to 15g of fresh yeast or 7g of conventional dried yeast when converting the recipe. Each sachet of quick action dry yeast is adequate for baking one loaf of bread using 750g of bread flour; thus, two sachets are required for a 1.5kg bag of bread flour.
## What temp kills yeast?
Thermocouples are used to measure temperature. While there are some drawbacks to using water that is a bit too chilly for the yeast, using water that is too warm—between 130 and 140°F—is lethal to the yeast and should be avoided.
## How long does yeast last in fridge?
Recently acquired yeast (with a valid purchase date) can be stored in a cool area (pantry or cabinet), refrigerated, frozen, or kept at room temperature for up to 2 years. Once the yeast has been opened, it is best kept in the refrigerator to be used within four months, or in the freezer to be used within six months if kept frozen.
## How many teaspoons is 4 grams of yeast?
One gram of active dry yeast is equal to 0.35 teaspoon when translated to teaspoon measurements. How many tablespoons of active dry yeast are there in one gram of active dry yeast? As a result, the change of 1 g (gram) unit in an active dry yeast measure equals 0.35 tsp (teaspoon) when measured in the equivalent measure and when measured in the same active dry yeast type.
## How do you measure dry yeast?
YEAST THAT HAS BEEN DRYED.Packets of yeast typically contain 2 and 1/4 teaspoons, which is equal to 1/4 ounce.If your recipe asks for more or less than one standard packet of yeast (or if you are measuring out of a jar or container), measure the yeast in the same way you would measure baking powder or baking soda in the same container.
• Dry yeast is available in two forms: active-dry and immediate.
## How much is 3 grams of yeast in teaspoons?
One teaspoon of quick yeast equals 3.15 grams when converted to gram form.
## How Many Teaspoons Are In A Packet Of Dry Yeast
1/4 cup (or 2 1/4 teaspoons) of yeast is contained in each container. This is around 7 grams (11 mL) of liquid.
## What is 1 package of active dry yeast equivalent to?
Conversions of Yeast Varieties An active dry yeast packet (or an envelope) contains the following ingredients: 1/4 oz. in weight 2 1/4 teaspoons of volume Instant 1 envelope (or 1/4 oz. or 2 1/4 tsp.) active dry yeast Machine that makes bread 1 envelope (or 1/4 oz. or 2 1/4 tsp.) active dry yeast
## How many teaspoons are in a packet of dried yeast?
Each packet of yeast includes 14 ounces, which is the equivalent of seven grams or 2 14 teaspoons of active dry yeast per packet. Though the amount of yeast required varies from recipe to recipe, the majority of breads ask for one packet or somewhat less than that amount. You may also get yeast in jars or sacks, which is very convenient if you bake a lot of bread.
## How many teaspoons are in a packet of Fleischmann’s yeast?
″Can you tell me how much dry yeast is in a 14.5-ounce packet?″ Approximately 2-1/4 teaspoons
## How many tablespoons are in a packet of dry yeast?
, I’ve been preparing Southern cuisine for 45 years. One package of active dry yeast contains 3/4 of a tablespoon of active dry yeast. A packet contains 2.25 teaspoons of powder, and a tablespoon has 3 teaspoons of powder.
See also: What Is On The Post Office Exam?
## How many teaspoons is 2 yeast packets?
The new yeast packets contain 2-1/4 teaspoons of active dry yeast, whereas prior recipes called for 1 tablespoon or one packet of active dry yeast.
## What happens if you put too much yeast in bread?
When there is too much yeast present, the dough may become flat because the yeast releases gas before the flour is ready to expand. If you leave the dough to rise for an excessive amount of time, it will develop a yeasty or beery fragrance and taste, and it will eventually deflate or rise poorly in the oven, resulting in a light crust.
## How much yeast is in a sachet?
The amount of quick action dry yeast needed to convert a recipe is equal to 15g of fresh yeast or 7g of conventional dried yeast when converting the recipe. Each sachet of quick action dry yeast is adequate for baking one loaf of bread using 750g of bread flour; thus, two sachets are required for a 1.5kg bag of bread flour.
## Is 7g of yeast a teaspoon?
(1) A package of yeast normally contains exactly 7g of yeast. Consequently, if you’re purchasing yeast in a packet, use one packet. The proper volume measurement, assuming that you will be measuring from bulk yeast, is 2 1/4 teaspoons instead of 2 1/2 teaspoon.
## Is active dry yeast the same as instant yeast?
Dry yeast is available in two varieties: active and quick. ″Active dry yeast″ refers to any dry yeast that must be activated prior to use, whereas ″instant dry yeast″ refers to any dry yeast that is ready to use as soon as the container is opened.
## How much is in a packet of dry yeast?
2 1/4 teaspoons of active dry yeast is equal to 0.25 ounces (0.25-ounce) packet of active dry yeast
## Can you use instant yeast in a bread machine?
Despite the fact that there are many different varieties of yeast available at the grocery store, and despite the fact that many different bread recipes call for different types of yeast, there are actually only two forms of yeast. As a result, bread machine yeast, fast rise yeast, and instant yeast can all be used interchangeably in the same recipe without alteration.
## How do you use bread machine yeast instead of active dry?
If you want to use active dry yeast in a recipe instead of instant (bread machine) yeast, double the amount of yeast by 1.25 times. 1 teaspoon quick yeast (for use in a bread machine) equals 1 1/4 teaspoons dry active yeast.
## What temp kills yeast?
Thermocouples are used to measure temperature. While there are some drawbacks to using water that is a bit too chilly for the yeast, using water that is too warm—between 130 and 140°F—is lethal to the yeast and should be avoided.
## How much is half a packet of yeast?
1/4 cup (or 2 1/4 teaspoons) of yeast is contained in each container. This is around 7 grams (11 mL) of liquid. This is indicative of all of the major brands in the United States.
## How many tablespoons is a cube of yeast?
When one cube of fresh yeast is converted to a tablespoon, the result is 4.49 tablespoons. How many teaspoons of fresh yeast are contained within a single cube? The solution is as follows: 1 cube euro (cube) unit in a fresh yeast measure equals 4.49 tbsp (tablespoon) in the equivalent measure for the same fresh yeast type and 1 cube euro (cube) unit in the comparable measure.
## Can instant yeast be used for active dry yeast?
If you want to use instant yeast instead of active dry yeast, you may skip the process of dissolving the yeast in liquid and just add it to the dough right away. You should mix in the water or other liquid that was intended for activating with your liquid components so that you keep the same total amount of liquid as you did before.
## Can you use half a packet of yeast?
Make a 50- to 80-percent reduction in the amount of yeast you use in your recipe. Consider the possibility that you might create the same loaf of bread with half the amount of yeast — or even less? You may undoubtedly do so – but not by following the same approach as you would normally do.
## How much water do you add to yeast?
Two and a quarter teaspoons of yeast are included in a little foil package. A quarter cup of warm water is usually sufficient to activate that amount of yeast, which results in around 1/2 cup of fully active yeast.
## How much yeast should I use for bread?
Depending on the recipe and rising time, you can use as little as 1 teaspoon of instant yeast per pound (about 4 cups) of flour, or as much as 2 1/4 teaspoons (occasionally more) of instant yeast per pound (about 4 cups).
## How much yeast do I use for 3 cups of flour?
Use half a teaspoon of yeast per cup of flour in a typical cycle machine for best results. The sum charged for one-hour or express machines may be 2-3 times more. Only 3/4 teaspoon of active dry yeast per cup of flour can be substituted for the standard cycle in the baking process. Some brands of yeast allow you to use both instant and bread machine yeast in the same recipe.
## What makes a bread soft and fluffy?
Amazingly delicate texture is achieved by making a soft bread with a moist, close-knit crumb. In order to do this, the moisture in the crumb is retained while cooling, which would otherwise escape. Increased wetness can also be achieved by using additional water in the recipe or by including a tenderizing agent to make the gluten soft and fluffy. The date is May 16, 2021.
## Equivalents of Different Yeast Varieties
When baking bread or certain cakes, recipes will typically ask for a specific amount of yeast—which may or may not be the variety you have on hand in the kitchen.In addition to fresh yeast, which is available as crushed cakes or blocks, dry yeast is available in the form of dried granules.In addition to the immediate variety, there are other bread machine and quick rise types of the dry yeast available for purchase.
• There are also yeast starters (which aid in the growth of yeast or the reactivation of dormant yeast), sponges (a combination of flour, water, and yeast that is commonly employed in the production of sourdough bread), and biga.
• Biga is a form of bread that is commonly seen in Italian baking, particularly for loaves with several holes, such as ciabatta.
• In the baking industry, fresh yeast is more commonly utilized by professional bakers, but dried yeast is more commonly used by home cooks.
• But which type of dried yeast is the most convenient to keep on hand?
Because no one maintains six different types of yeast on hand in their kitchen, this handy conversion chart will assist you in substituting the type of yeast you have on hand with the type of yeast asked for in the recipe you’re making.It doesn’t matter if the yeast is specified by volume, weight, or the number of envelopes in the recipe; this chart has you covered!
## Tips for Using Yeast
• Many individuals find the process of utilizing yeast intimidating: determining whether or not the yeast is still active, obtaining the proper water temperature, and deciding whether or not to add sugar. However, yeast does not have to be a frightening substance. By following a few simple guidelines, you may learn to feel comfortable using yeast in your baking projects. When it comes to storing yeast, you never have to be concerned if your envelope of yeast has gone dormant. By keeping the yeast in the freezer, you may ensure that it remains active for several months.
• Temperature of the water—in order to activate the yeast, you must use ″lukewarm water.″ If it is not necessary to have a certain temperature, it is sufficient if it feels somewhat warm, similar to room temperature Allow it to settle for a minute or two after adding the water, then whisk with a fork until smooth.
• Adding sugar—adding a pinch of the sweet stuff to the dough is something of a wives’ tale, since it does not aid in the rising of the dough. However, if the yeast is still alive, it may generate bubbles, thus if you are unsure whether the yeast is expired or not, this is an useful test to perform
• Obtaining the proper temperature for rising—Yeast needs a temperature between 70 and 80 degrees Fahrenheit to reproduce, so if your kitchen is cooler than that, you may place the dough in an oven that has been previously preheated and then turned off. If your home is excessively warm, a cold oven may be the ideal solution. Just be careful not to expose the dough to temperatures below 50 degrees Fahrenheit, since this will cause the yeast to fall dormant.
• Certain additives cause the dough to rise more slowly
• for example, if your dough contains eggs, dairy, oil, or salt, the rising time will be longer. A basic dough made from flour and water will rise more quickly
## Yeast Conversions
Yeast Conversions I was wondering if you could tell me what the current day equivalency would be for an old-fashioned yeast cake? Thanks, Mary Sanders Hi Mary, Here’s all you need to know about yeast. This excerpt is taken from the January 2003 Home Cook’n Newsletter issue on bread making. Hope it helps, Desi @ DVO Yeast Conversions. When recipes call for a specific type of yeast, you can substitute the type you prefer easily. If you wish to use active-dry yeast in recipes calling for instant yeast, increase the amount of yeast called for by 25 percent. Vice versa, if you wish to use instant yeast in recipes calling for active-dry, decrease the amount of yeast called for by 25 percent. Instant yeast is much more potent than active-dry. Increasing or reducing the yeast types will encourage a better-flavored product than if you just used the yeast types interchangeably.** (4) 1 packet (2-1/4 teaspoons) active dry yeast* =.6-ounce cake of compressed yeast = 1 packet (2-1/4 teaspoons) or scant 1-3/4 teaspoons instant yeast** * Because of improvements in yeast properties and potency by manufacturers, the amount of yeast in packets has decreased. While older recipes call for 1 tablespoon or 1 packet of active dry yeast, the new yeast packets contain 2-1/4 teaspoons yeast. You can reduce the amount of yeast used in any of these older recipes by substituting 2-1/4 teaspoons per tablespoon of yeast. **Packets of instant yeast contain 2-1/4 teaspoons of yeast, but using only 1-3/4 teaspoons when substituting for active dry yeast will give better results in your bread; although you may use the entire packet if you wish. Alternatively, add 2-3/4 teaspoons of active dry yeast in recipes calling for 1 packet instant yeast. Do you have something to share with other Cook’n readers? Email your thoughts to us. Tell us about you and your family, and send us a picture.We’d love to hear from you.and who knows.perhaps you will be the star of the next newsletter!
## frequently asked questions
The following are some of the most frequently asked questions we receive, along with extensive answers and explanations to assist bakers of all skill levels. In addition, if you are unable to locate the information you want, please check our Baking Basics area for more resources.
### “How should I store yeast?”
Unopened yeast should be stored in a cold, dry location, such as a pantry (or refrigerator). The activity of the yeast is reduced when it is exposed to oxygen, heat, or dampness. Refrigerate in an airtight container at the back of the refrigerator, away from drafts, once it’s been opened. Use within three to four months of purchase; freezing is not advised.
### ″I’m not sure if my yeast is expired, what should I do?″
″Can I use yeast that has expired in my recipe?″ ″Can you tell me how to prove yeast in order to test for activity?″ ″Is it possible to save dough that has failed to rise?″ Active Dry is a trademark of Active Dry, Inc.
1. Combine in a large, warm mixing basin the following ingredients for each package of yeast called for in the recipe: 14 cup lukewarm water (100°–110°F), 1 teaspoon sugar, and 1 packet (2-1/4 teaspoons) of yeast. Stir until the mixture is completely dissolved.
2. Slowly incorporate small (walnut-sized) pieces of dough into the yeast mixture using an electric mixer until approximately one-half of the dough has been included.
3. Stir in the rest of the dough with a wooden spoon. Mix in just enough flour to keep the dough from becoming sticky.
4. Allow the dough to rise, shape, and bake according to the directions in the recipe.
• RapidRise® Use fresh yeast to make a small amount of loose dough that will be added to your bigger batch of dough. To do this, combine the following elements:
• (1) Fleischmann’s® Instant Yeast, Fleischmann’s® RapidRise® Instant Yeast
• 12 to 1 cup all-purpose flour
• 1 teaspoon sugar
• 14 cup lukewarm water (120°–130°F) should be added to the dry ingredients
• the remaining dough should be stirred in with a spoon. Mix in just enough flour to keep the dough from becoming sticky.
• Pour in the heated slurry and mix until it creates a slurry, then add it to your dough. You may need to add a little more flour until the dough reaches the consistency you prefer. Make sure not to overwork your dough. Allow your dough to rise for 30 to 45 minutes in a warm, draft-free environment.
### “What’s the difference between different types of yeast and can I make substitutions between them?”
In Active Dry recipes, may RapidRise® and Bread Machine Yeast be used instead of active dry yeast?What is the best way to utilize Active Dry Yeast in RapidRise® recipes?The difference between Instant Yeast, Bread Machine Yeast, and RapidRise® Instant Yeast is as follows: The difference between rapid rising yeast (RapidRise®/Bread Machine Yeast) and active dry yeast is explained in detail here.
• If you have any questions about using Fresh Active Yeast, please let us know.
• If you have any questions about substituting dried yeast, please let us know.
• ″Can Active Dry Yeast be used in bread machines?″
### “Can dough be stored in the refrigerator or freezer?”
″Can any dough be refrigerated?″ says the cook. ″Can I put my dough in the freezer?″ How does frozen dough come to be made?
### ″What are no-knead breads and how do the recipes work?″
What is the mechanism by which loaves of bread may knead themselves? ″Can you tell me where the concept of ‘kneadless’ breads came from?″
### “Should recipes be adjusted for high altitudes?”
Yes. However, there are no specific guidelines for altering yeast breads while baking at high elevations. The materials, as well as the entire bread-making process, are affected by altitude. When baking at elevations more than 3,000 feet, we recommend the following basic guidelines:
### “I would rather use my heavy-duty stand mixer than mix the dough by hand. Is this okay?”
Yes!Simply combine the dry ingredients in a large mixing bowl, then add the liquid while mixing on low speed with the paddle attachment.The second stirring is normally extremely quick (just long enough to deflate the dough), but you may alternatively do this with a heavy-duty mixer if you want to save time.
• But don’t forget to switch to the dough hook after the first rise since the dough will be rubbery after that (from the gluten development).
See also: How Do I Know If My Tacoma Has A Tow Package?
### “My family loves these breads, but my aunt insists they all come out browner than her loaves used to. Why would that be?”
Your aunt’s breads were most likely produced using the traditional, more known ″direct″ approach, which you are familiar with.Breads baked in this manner just do not brown as deeply or as soon as breads prepared in the cold, slow-rise approach, for whatever reason.The lengthy, steady rise provides more time for specific chemical processes to occur, which in turn enhance browning and a rich, pleasant flavor as a result.
• Deep browning is one of the distinguishing characteristics of today’s artisan-style breads; artisan bakers believe that greater browning results in fuller taste.
• Nonetheless, this is a question of personal preference.
• If you like a somewhat lighter hue, simply reduce the oven temperature by 25 degrees Fahrenheit from what the recipe calls for and bake for a few minutes longer.
• Also, at the moment where you want to slow down the browning of the loaf top, cover it with aluminum foil (shiny side out).
### “I am reducing the salt in my diet. Can I omit the salt from these recipes?”
You can limit the amount of salt you use by one-third, but you should not eliminate it entirely.Salt not only enhances the texture and flavor of the dish, but it also prevents the yeast from getting excessively bubbly and hyperactive during fermentation.This holds true for all yeasted doughs, including pizza dough.
• When lowering the amount of salt used, keep in mind that the dough may rise more quickly than usual.
### “What happened to my Recipe Box?”
While our website has a wealth of fresh and useful information that is both entertaining and simple to use, it does not offer the Recipe Box function.If you are unable to locate a certain recipe from your Recipe Box despite knowing the title, please call us at 1-800-777-4959 or send us an email with the recipe’s title and we will gladly search our recipe archives for you.This commitment to Fleischmann’s® Yeast motivates us to continually enhance your online experience, and we sincerely regret for any trouble this may give you.
• We hope you will continue to enjoy all of our delectable recipes in the future.
## How Much Yeast Is In A Packet? Accurate Measurement for Yeast
If you’re a bread enthusiast, you’re probably aware that yeast is a crucial element in many different varieties of bread.Yeast permits the bread to rise, which in turn contributes to the bread’s light and fluffy texture.If you’re thinking about creating your own bread, you might be wondering how much yeast is included in a single packet.
• When baking bread, bakers have used yeast for thousands of years to get the desired result.
• For the best results, make sure you use the correct amount of yeast for the bread recipe you’re working on.
• Bread’s texture is dependent on the use of yeast, which is essential for this.
## How Much Yeast Is In A Packet: How Yeast Works
While it may come as a surprise to some, the yeast used to produce bread is truly alive.Single-celled organism that not only permits the bread to rise but also contributes to the flavor and texture of the finished product.Additionally, it contributes to the strengthening of the proteins in bread.
• When you look at bread recipes, you’ll find that many of them will instruct you to add warm water to the yeast before mixing it together.
• It is the addition of warm water to yeast that causes it to reactivate, allowing it to work its magic throughout the bread-making process.
• To allow bread to rise properly, the component yeast is utilized.
• When yeast is introduced to bread dough, it breaks down huge starch molecules into simple sugars, which is what makes the bread taste so good.
It is at this point that carbon dioxide and ethyl alcohol are produced, which causes air bubbles to form in the dough, causing it to rise.Kneading and rising are critical phases when working with yeast in a dough recipe.Gluten develops during the kneading process, while the rising process enables the yeast to work its magic.Both of these procedures are necessary for a chewy, airy loaf of bread.
## How Much Yeast is in a Packet?
It is common for yeast to be offered in a packet when it is purchased.It is normal for three packets to be sold together in a single transaction.Each packet of yeast includes 14 ounces, which is the equivalent of seven grams or 2 14 teaspoons of active dry yeast per packet.
• Though the amount of yeast required varies from recipe to recipe, the majority of breads ask for one packet or somewhat less than that amount.
• You may also get yeast in jars or sacks, which is very convenient if you bake a lot of bread.
• Packets of yeast, on the other hand, are frequently the best option because there is no need to worry about moisture getting into the yeast and affecting it.
## How Much Yeast Is In A Packet: Different Types
It is commonly accepted that there are four main forms of yeast that may be employed in the process of creating bread.Active dry yeast, quick yeast, fast rise yeast, and fresh yeast are the varieties of yeast that you will find for baking bread with.Certain recipes may call for a specific type of yeast, but others may allow for the use of any variety of yeast.
• Active dry yeast and quick yeast are the two most prevalent varieties of yeast that you may utilize in your baking projects.
• In grocery shops, they are easily accessible in packets and may be purchased in bulk.
• The process of making active dry yeast involves removing the water from the yeast and granulating it.
• In most cases, active dry yeast must be dissolved in water before it can be added to the dry ingredients in the recipe.
Instant yeast is similar to active dry yeast in appearance, but it is composed of smaller granules.It is usually not necessary to add it to water before using it; instead, it may be put immediately to the dry components.Due to the fact that they come from distinct strands of yeast, there is a little difference in flavor between active dry yeast and quick yeast, but they may be used interchangeably in bread recipes in most cases.When it comes to preparing bread, the quick rise and fresh yeast are less usually utilized.Rapid rise yeast is comparable to instant yeast; however, it is derived from a distinct strain of yeast that produces a more vigorous rise in the dough.Cinnamon buns and other baked goods requiring only a single short rise are perfect candidates for rapid rise yeast.
• Fresh yeast might be difficult to come by, but it can be an excellent alternative if you do a lot of baking and need a quick fix.
• It is also known as compressed yeast or cake yeast, and it may only be kept in the refrigerator for approximately a week at a time.
• It differs from the other yeasts in that it comes in a solid block form that must be broken and dissolved in water before use.
## Storing Yeast
Despite the fact that yeast has an expiration date, it has a lengthy shelf life.It is possible to store active dry or instant yeast at room temperature in a dry area for 12 to 18 months, while it may be possible to store it for longer depending on the expiration date indicated on the container.Dry yeast is normally good for two to four months after it has passed its best before date; after that, it is unlikely to be suitable for usage.
• Instant or active dry yeast can be stored in an airtight bag or container for up to three to four months in the refrigerator and up to six months in the freezer once it has been opened.
• Store unopened packets of yeast in the freezer for several years if you want to increase the shelf life of your supplies.
• Test your yeast by mixing one teaspoon of sugar with 14 cup of water in a measuring cup with one packet of yeast.
• If your yeast is still good, you may use this method to make bread.
After 10 minutes, the yeast should have foamed up to the half-cup mark, but it is still safe to use.
## Using Yeast
Despite the fact that yeast might be frightening to work with, there are many delicious bread recipes that are simple to create with it.When it comes to yeast, it might be difficult to work with, but as long as you follow the instructions to the letter, you will end up with a loaf of wonderful bread.Because most recipes just call for the amount of yeast found in a packet, it is quite convenient to employ this method.
• When it comes to baking bread, active dry yeast and quick yeast will be your go-to yeasts since they are readily available, simple to use, and often interchangeable.
• Active dry yeast and quick yeast are both available in packages of 14 ounces, allowing you to utilize either one in the majority of recipes.
• Most grocery stores will offer the yeast in three-packs, which is the standard size.
• These packets are handy to use and keep because of their small size.
## How Much Yeast is in a Packet (and How Does Yeast Work?)
It is necessary to use yeast when preparing bread.Yeast is crucial for making your bread rise and taste fantastic, as well as having a nice fluffy texture and a lovely fluffy texture.It goes without saying that you can’t make bread without yeast – at least not successfully – without it.
• But how much yeast is included in a single packet?
• Hello there, pastry chefs!
• My name is Michelle, and I like baking bread in my spare time.
• I prepare a variety of desserts, whether they are sweet or savory.
Having said that, yeast is something I work with on a regular basis.The amount of yeast included in a package has been determined, and I would want to share my findings with you!Have you ever been interested as to how much yeast is contained within a package of yeast?Then you’ll want to continue reading.You’ll learn how much yeast is contained in a package, as well as some other interesting facts about yeast that you won’t want to miss reading about below.Everyone, let’s speak about yeast for a minute.
## What is Yeast, and How Does it Work?
If you’re new to the world of bread baking, you might be scratching your brain about what yeast is in the first place.Don’t worry, you’re not alone.While the majority of people have heard of yeast and understand its critical role in bread production, many bakers are unfamiliar with the component.
• Let’s have a little discussion about it.
• Yeast is a single-celled fungus that can only be seen under a microscope because it is alive.
• When utilized to make bread, yeast performs two functions: it ferments glucose and releases carbon dioxide into the atmosphere.
• This allows gasses in the dough to expand, causing the dough to rise as a result.
Not only is yeast necessary for allowing your bread to rise in the manner in which it should, but it also aids in the production of the proper flavor and texture in your bread.Make sure you use enough yeast in your bread recipe if you want a light and fluffy loaf of bread!
## How Much Yeast is in a Packet?
Okay, let’s get down to the business of why we’re all here today: to discuss how much yeast is included in a package. We’re all familiar with the fact that most recipes ask for a ″packet″ of yeast. Is it possible to tell how much yeast is truly hidden inside? The amount of yeast in a yeast package is normally 2 14 teaspoons, which is equal to 1 14 ounce or seven grams of yeast.
### Are All Packets of Yeast Created Equal?
There are several different forms of yeast available on the market, including active dry yeast (the most common), instant yeast, quick rise yeast, and fast-rising yeast.Active dry yeast is the most common type of yeast.Because there are so many alternatives, you could assume that the amount varies depending on the type of yeast.
• However, this is not the case in this instance.
• Whatever sort of yeast you use, the results will be the same.
• Regardless of whether you’re using active dry yeast, quick yeast, or another sort of yeast, the amount will stay the same — 2 14 teaspoons or 1 1/4 ounce of yeast.
• Additionally, the brand has no effect on the amount of yeast contained within a packet.
So, when you’re in the grocery store and realize you’ve forgotten to buy yeast, don’t freak out.All packets are made equal, and they will all perform admirably when used in your recipe.
## FAQs
No matter what sort of yeast is used, a packet of yeast includes 2 12 teaspoons of yeast per package. Depending on what you’re cooking, your recipe will normally call for one or two packages of instant pudding mix. If you still have questions regarding how much yeast is contained in a packet, have a look at the list of frequently asked questions below.
### How many teaspoons are in a packet of yeast?
Every package of yeast contains 2 14 tablespoons of active yeast (which can also be referred to as envelopes in some recipes). The amount of yeast contained within will not alter depending on the type of yeast used. Active dry yeast, instant rise yeast, and all other types of yeast will all have the same number of teaspoons of yeast in their respective packages.
### What is 1 gram of yeast in teaspoons?
If you need to know how much 1 gram of yeast equals in teaspoons for whatever reason, the answer is 0.35 teaspoons. This can be a useful piece of information to have, especially if your recipe calls for it. Keep this information in your baking toolbox!
### How many yeast cells are in a packet?
Although a yeast packet includes 2 14 teaspoons of yeast, the number of yeast cells contained in a packet is not the same as the amount of yeast in a packet. A average yeast packet will contain an amazing 69 billion yeast cells, which is rather remarkable and quite intriguing to think about.
### How many mL is a packet of yeast?
When we check at the amount of yeast contained within a package of yeast, we find that it contains 2 14 teaspoons. This measurement corresponds to around 11 mL.
## Final Thoughts
When it comes to preparing bread, yeast is a must.Each package of yeast contains the same amount of yeast – 2 14 teaspoons – regardless of the kind.Make sure to follow the instructions to the letter, and don’t omit the yeast.
• Otherwise, you will wind up with a loaf of bread that is flat, thick, and crumbly, which is not desired.
• Yuck!
• Did you realize that a packet of yeast included a certain amount of yeast?
• Which sort of yeast do you find yourself using the most frequently?
Please share your thoughts with us in the comments section below! | 13,624 | 60,056 | {"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} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.940967 |
https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/176/2/91857/the-number-of-unimodular-zeros-of-self-reciprocal-polynomials-with-coefficients-in-a-finite-set | 1,670,194,600,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00415.warc.gz | 878,937,358 | 7,718 | A+ CATEGORY SCIENTIFIC UNIT
# Publishing house / Journals and Serials / Acta Arithmetica / All issues
## Acta Arithmetica
PDF files of articles are only available for institutions which have paid for the online version upon signing an Institutional User License.
## The number of unimodular zeros of self-reciprocal polynomials with coefficients in a finite set
### Volume 176 / 2016
Acta Arithmetica 176 (2016), 177-200 MSC: 11C08, 41A17, 26C10, 30C15. DOI: 10.4064/aa8442-7-2016 Published online: 17 October 2016
#### Abstract
Let ${\rm NZ}(T_n)$ denote the number of real zeros of a trigonometric polynomial $$T_n(t) = \sum_{j=0}^n{a_{j,n} \cos(jt)}, \ \quad a_{j,n} \in {\mathbb C},$$ in a period $[a,a+2\pi)$, $a \in {\mathbb R}$. Let ${\rm NZ}(P_n)$ denote the number of zeros of an algebraic polynomial $$P_n(z) = \sum_{j=0}^n{p_{j,n} z^j}, \ \quad p_{j,n} \in {\mathbb C},$$ that lie on the unit circle of ${\mathbb C}$. Let $${\rm NC}_k(P_n) := \Big|\Big\{u: 0 \leq u \leq n-k+1, \, \sum_{j=u}^{u+k-1}{p_{j,n}} \neq 0 \Big\}\Big|.$$ One of the highlights of this paper states that $\lim_{n \rightarrow \infty}{ {\rm NZ}(T_n)} = \infty$ whenever the set $\{a_{j,n}: j \in \{0,1,\ldots,n\}, \, n \in {\mathbb N}\} \subset [0,\infty)$ is finite and $$\lim_{n \rightarrow \infty}{|\{j \in \{0,1,\ldots,n\}:a_{j,n} \neq 0\}|} = \infty.$$ This follows from a more general result stating that $$\lim_{n \rightarrow \infty}{{\rm NZ}(P_{2n})} = \infty$$ whenever $P_{2n}$ is self-reciprocal, the set $\{p_{j,2n}: j \in \{0,1,\ldots,2n\}, \, n \in {\mathbb N}\} \subset {\mathbb R}$ is finite, and $\lim_{n \rightarrow \infty}{{\rm NC}_k(P_{2n})} = \infty$ for every $k \in {\mathbb N}$.
#### Authors
• Tamás ErdélyiDepartment of Mathematics
Texas A&M University
College Station, TX 77843, U.S.A.
e-mail
## Search for IMPAN publications
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http://braungardt.trialectics.com/mathematics/topology/knot-theory/ | 1,508,758,651,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825900.44/warc/CC-MAIN-20171023111450-20171023131450-00030.warc.gz | 58,097,826 | 47,658 | Knot theory
Knot theory is a very fast growing field of mathematics. Knots are not natural phenomena, and there exists only a finite number of distinct knots in three-dimensional space. Knots define spaces, because we can think of a knot as a way in which different dimensions can be connected. Mathematicians are working on notation systems for knots, which leads to a form of arithmetic for knots. This has fascinating consequences for other disciplines, and for our understanding of reality in general (see the heading “Virtual knots”.)
What is a knot?
Complex knots can oftentimes be simplified with a few moves, which the German mathematician Reidemeister organized into three categories. (Reidemeister moves.) Once the knot is simplified and no further crossing can be removed, the knot is classified by the number of crossings that remain. This is called a “prime knot.” For example, the trefoil knot is classified by its fewest number of crossings – three (see the diagram below).
It is possible to have more than one prime knot with the same number of crossings. In this case, we usually use subscripts to denote different knots with the same number of crossings, such as the 51 and 52 knots in the diagram below:
There is no known formula for giving the number of distinct prime knots as a function of the number of crossings. The numbers of distinct prime knots having n= 1, 2, 3,… crossings are 0, 0, 1, 1, 2, 3, 7, 21, 49, 165, 552, 2176, 9988.
Virtual Knots?
The following quote gives first a short introduction to knots, and then it discusses the discovery of virtual knots. From: A revolution in knot theory (physorg.com). Provided by American Mathematical Society.
“As sailors have long known, many different kinds of knots are possible; in fact, the variety is infinite. A *mathematical* knot can be imagined as a knotted circle: Think of a pretzel, which is a knotted circle of dough, or a , which is the “un-knot” because it is not knotted. Mathematicians study the patterns, symmetries, and asymmetries in knots and develop methods for distinguishing when two knots are truly different.
Mathematically, one thinks of the string out of which a knot is formed as being a one-dimensional object, and the knot itself lives in three-dimensional space. Drawings of knots, like the ones done by Tait, are projections of the knot onto a two-dimensional plane. In such drawings, it is customary to draw over-and-under crossings of the string as broken and unbroken lines. If three or more strands of the knot are on top of each other at single point, we can move the strands slightly without changing the knot so that every point on the plane sits below at most two strands of the knot. A planar knot diagram is a picture of a knot, drawn in a two-dimensional plane, in which every point of the diagram represents at most two points in the knot. Planar knot diagrams have long been used in mathematics as a way to represent and study knots.
As Nelson reports in his article, mathematicians have devised various ways to represent the information contained in knot diagrams. One example is the Gauss code, which is a sequence of letters and numbers wherein each crossing in the knot is assigned a number and the letter O or U, depending on whether the crossing goes over or under. The Gauss code for a simple knot might look like this: O1U2O3U1O2U3.
In the mid-1990s, mathematicians discovered something strange. There are Gauss codes for which it is impossible to draw planar knot diagrams but which nevertheless behave like knots in certain ways. In particular, those codes, which Nelson calls *nonplanar Gauss codes*, work perfectly well in certain formulas that are used to investigate properties of knots. Nelson writes: “A planar Gauss code always describes a [knot] in three-space; what kind of thing could a nonplanar Gauss code be describing?” As it turns out, there are “virtual knots” that have legitimate Gauss codes but do not correspond to knots in three-dimensional space. These virtual knots can be investigated by applying combinatorial techniques to knot diagrams.
Just as new horizons opened when people dared to consider what would happen if -1 had a square root—and thereby discovered complex numbers, which have since been thoroughly explored by mathematicians and have become ubiquitous in physics and engineering—mathematicians are finding that the equations they used to investigate regular knots give rise to a whole universe of “generalized knots” that have their own peculiar qualities. Although they seem esoteric at first, these generalized knots turn out to have interpretations as familiar objects in mathematics. “Moreover,” Nelson writes, “classical knot theory emerges as a special case of the new generalized knot theory.”
Knot Tables
When we consider all the available knots up to 9 crossings, we get the following table with this system of notation:
KNOT TABLE 01-939
01 31 41 51 52 61 62 63 71 72 73 74 75 76 77 81 82 83 84 85 86 87 88 89 810 811 812 813 814 815 816 817 818 819 820 821 91 92 93 94 95 96 97 98 99 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 | 1,207 | 5,206 | {"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} | 4.375 | 4 | CC-MAIN-2017-43 | longest | en | 0.937928 |
https://math.stackexchange.com/questions/104260/rotations-and-the-parallel-postulate?noredirect=1 | 1,582,142,458,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144167.31/warc/CC-MAIN-20200219184416-20200219214416-00514.warc.gz | 462,018,744 | 33,543 | # Rotations and the parallel postulate.
If we take a full rotation to be $360^\circ$, then it seems that we can prove the following Starting from the red point, we walk clockwise along the triangle. At each vertex, we must turn through the green angles marked to proceed down the adjacent sides of the triangle. When we return to the red point, we will have turned through one full rotation. This means that the sum of the exterior angles is given as $360^\circ$, implying the interior angles of the triangle sums of $180^\circ$.
The fact that the angles of a triangle sum to $180^\circ$ is well known to be equivalent to the parallel postulate and this made me wonder whether if the fact that a full rotation being $360^\circ$ is also equivalent to the parallel postulate?
I avoided stating the question using "exterior angles of a triangle sums to $360^\circ$" and instead used the more ambiguous term "rotations" to emphasize the fact that rotations seem to be more general. We can for example show that the interior angles of a heptagram sum to $180^\circ$ by noting that three full rotations are made while "walking" thge heptagram. This should generalize to arbitrary closed polygons and seems stronger than the fact that the exterior angles sum to $180^\circ$.
In summary, I would be interested in knowing the connections that this technique has to the parallel postulate as well as if this technique is a "rigorous" way of finding the internal angles of more complex shapes such as the heptagram.
• The parallel postulate is only false in non-Euclidean geometries. I'm not sure if it even can fail to the extent that these external angles sum to a full $360^\circ$. Angles are always defined in terms of (roughly speaking) "local rotations" so the statement is fine. – anon Jan 31 '12 at 15:27
• en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem#Triangles – Will Jagy Jan 31 '12 at 19:49
Notice that, in this case, the sum of the exterior angles is $270^\circ$, not $360^\circ$.
However, in answer to your question about the sum of the interior angles of a polygon, since an external and the corresponding interior angle sum to $180^\circ$, the sum of the exterior angles and the interior angles is $180^\circ\times$ the number of sides. Since, as you have noted, in the Euclidean plane, the sum of the exterior angles is $360^\circ$, we get that the sum of the interior angles of a polygon with $n$ sides is $(n-2)180^\circ$.
• No, locally, a complete rotation has $360^\circ$. However, the parallel postulate is not a local postulate. A triangle whose external angles sum to $270^\circ$ cannot be local. Very small spherical triangles have exterior angles that are just a bit below $360^\circ$. In fact, the sum of the exterior angles of a spherical triangle falls short of $2\pi$ radians by the area of the triangle in steradians. – robjohn Jan 31 '12 at 18:21 | 681 | 2,878 | {"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} | 3.5 | 4 | CC-MAIN-2020-10 | latest | en | 0.930296 |
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Algebra 1-Chapter 1
Algebra 1-Chapter 1 Vocabulary
QuestionAnswer
A letter or symbol that represents one or more numbers. variable
A group of numbers, symbols, and variables that express an operation or a series of operations. algebraic expression
An expression that represents repeated multiplication of the same factor. power
The number or expression that is used as a factor in a repeated multiplication. base
The number or variable that represents the number of times the base of a power is used as a factor. exponent
Rules for evaluating an expression involving more than one operation. order of operations
Describes a real-world situation using words as labels and using math symbols to relate the words. verbal model
A fraction that compares two quantities measured in different units. rate
A fraction that compares two quantities measured in different units, in which the denominator of the fraction is 1. unit rate
A mathematical sentence formed by placing the symbol = between two expressions. equation
A mathematical sentence formed by placing one of the symbols: less than, less than or equal to, greater than, or greater than or equal to, between two expressions. inequality
An equation or inequality that contains an algebraic expression. open sentence
A number that produces a true statement when substituted for the variable in an equation. solution of an equation
A number that produces a true statement when substituted for the variable in an inequality. solution of an inequality
An equation that relates two or more quantities. formula
A rule that establishes a relationship between two quantities, called input and output, such that each input is paired with exactly one output. function
The set of all inputs of a function. domain
The set of all outputs of a function. range
The input variable of a function. independent variable
The output variable of a function. dependent variable
To find the value of an algebraic expression by substituting numbers for variables, performing the operation(s) and simplifying the result. evaluate an algebraic expression
To combine like terms and complete all operations. simplify
A compact way of writing an expression involving repeated multiplication of the same number. exponential form
Created by: aelohaus | 619 | 3,013 | {"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} | 2.984375 | 3 | CC-MAIN-2019-51 | latest | en | 0.912974 |
https://math.stackexchange.com/questions/1757189/are-generic-smooth-functions-analytic | 1,716,468,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00401.warc.gz | 323,799,102 | 38,744 | # Are generic smooth functions analytic?
It is well-known that generic continuous functions are differentiable almost nowhere. I was somewhat surprised to learn in my functional analysis course that the same is true of $\alpha$-Holder functions for $0<\alpha<1$. I am aware that there exist functions that are smooth but analytic almost nowhere, so I've been wondering if this too is the generic situation. My intuition is thoroughly conflicted so I'm wondering if anyone has any insight.
By generic I mean a set whose complement is of first category. Any similar definition would be fine.
Please note: I am fully aware that there exist smooth functions that are nowhere analytic. My question--which as of August 2, 2016 has not been resolved--is whether a generic smooth function in this topological sense is analytic everywhere/nowhere/somewhere.
• The domain can be the reals, a compact interval, or anything else that's convenient. Apr 24, 2016 at 21:17
• Analytic a.e. on its whole domain Apr 24, 2016 at 21:17
• where can I find an intuitive description of "generic continuous or smooth function" ? Apr 24, 2016 at 21:41
• @user1952009 "generic" is a topological property of a set that says it is in some way the typical case. en.m.wikipedia.org/wiki/Generic_property Apr 25, 2016 at 0:08
• This answer (especially the first link provided) gives an answer for genericity in the Baire sense. The strongest result is that (I paraphrase) the "every point of a Baire-typical $C^\infty$ function is a point at which the Taylor series has zero radius of convergence." Aug 3, 2016 at 2:47
Generically, no. For every sequence of real numbers (a set which has obviously a very large cardinality), there exists a smooth function such that the Taylor series of the function at the origin is that sequence. Since there are many more sequences of real numbers than sequences corresponding to convergent Taylor series, it follows that generically smooth functions are not analytic (for example just consider the fact that a necessary condition for the convergence of the Taylor series is that the coefficients approach 0, whereas most sequences of real numbers which have limits don't approach 0, and in fact most sequences don't even have limits).
(Beginning of Rigorous Argument)
https://en.wikipedia.org/wiki/Non-analytic_smooth_function
Look at Theorem 2 of the attached link: http://math.bard.edu/belk/math351/FunctionSpaces.pdf
Since the pointwise limit (as functions from $\mathbb{N} \to \mathbb{R}$) of any sequence of real-valued sequences with limit zero is necessarily again a sequence with limit zero, it follows as a corollary of Theorem 2 that $c_0$ is a closed subset of $\mathbb{R}^{\infty}$ under the standard product topology on that space, i.e. $c_0$ is its own closure. The interior of $c_0$ in the product topology, however, is necessarily empty, since any open subset of $\mathbb{R}^{\infty}$ must contain a basis set; however $c_0$ cannot contain any basis set of the product topology, since any sequence converging to 0 has unrestricted values for any finite number of coordinates (since the only restriction is on its behavior in the limit).
Therefore $c_0$ is nowhere dense in the product topology of $\mathbb{R}^{\infty}$, and thus meager.
By uniqueness of Taylor series for analytic functions, the space of all functions analytic at the origin corresponds to a proper subspace of $c_0$. In contrast, since uniqueness of Taylor series does not apply for non-analytic smooth functions, $\mathbb{R}^{\infty}$ corresponds to a strict subspace of all smooth functions. Hence the space of all analytic functions is a strict subspace of a space meager in a strict subspace of the space of all smooth functions, hence is itself necessarily meager in the space of all smooth functions.
(End of Rigorous Argument)
For example, if we were to define a measure on the space $c$ such that the measure of the set of all sequences with limit in the Lebesgue-measurable set $A \subset \mathbb{R}$ had measure equal to the Lebesgue measure of $A$, then the set of all analytic functions would correspond to a strict subset of the null-set $c_0$, whereas the space of all smooth functions would not even be contained inside $c$.
In fact, we could construct a measure on the space of all real sequences such that the measure of all sequences with $\liminf z_n =x$ and $\limsup z_n =y$ with $(x,y)\in B \subset \mathbb{R}^2$ has the measure equal to the Lebesgue measure of $B$ (assuming B is Lebesgue-measurable). Then the space $c$, in which the analytic functions are a null set, corresponds to the line $y=x$, and hence has measure 0 (since any line in the plane has Lebesgue measure 0). Yet the space of all smooth functions would still not be a subspace of the entire space $\mathbb{R}^{\infty}$, a set of infinite measure.
EDIT: From the above, it follows that:
The set of functions analytic at the origin is strictly smaller than the set of sequences of real numbers with limit 0 (the space of null sequences $c_0$), which is strictly smaller (by a lot) than the set of all real sequences which have a limit (the space of all convergent sequences $c$), which is again strictly smaller than the set of all real sequences $\mathbb{R}^{\infty}$ (by a lot), which corresponds to a subspace of smooth functions.
As a corollary, to each individual function analytic at the origin $f$ (which is a much larger set than the set of functions analytic on the entire real line), there exists a unique set of non-analytic smooth functions corresponding to $\mathbb{R}^2$.
(First, take all sequences of the form $(a_n +x)$ where x is a real number, and $(a_n)$ is the sequence going to 0 corresponding to the Taylor coefficients of the expansion of $f$ at 0. Then each $(a_n)$ corresponds to a smooth function with Taylor coefficients $(a_n +x)$ at the origin. All but one such sequence has limit $x\not=0$, which means it must correspond to a non-analytic smooth function, since its Taylor series clearly does not converge.
Then for each $(a_n +x)$ we can define for every real number $y$ sequences $(a_n,\frac{y}{x}[a_n+x])$ which do not converge but which have limit superior x and limit inferior y if $x<y$, or $\limsup y$ and $\liminf x$ if $y<x$.
Each such sequence again corresponds to a unique non-convergent series and hence a unique non-analytic smooth function. So there are a LOT LOT more non-analytic smooth functions than analytic smooth functions.
• I don't quite see this cardinality argument. Is this just to show that such a smooth non-analytic function exists? Apr 25, 2016 at 0:05
• I don't study set theory, but I believe these spaces all have the cardinality of the continuum (see previous stackexchange posts). Also, it's very easy to see that a generic set can have a smaller cardinality than the whole space (eg. the empty set is always generic in an at most countable space with the discrete topology). Apr 25, 2016 at 1:48
• Ok the basis for my claim is that the cardinality of continuous functions on R is the continuum (since a continuous function is determined by its values on Q) so the cardinality of the smooth functions is not greater. Apr 25, 2016 at 2:14
• It is not true that $c_0$ is closed in $\Bbb {R}^\Bbb {N}$. Apr 25, 2016 at 8:15
• By Taylor coefficient, do you mean the $a_k$ in $$\sum a_k x^k$$ or in $$\sum \frac{1}{k!} a_k x^k?$$ I think William meant the first case, while @zhw meant the second case. This would explain the apparent contradiction. Aug 3, 2016 at 2:26 | 1,859 | 7,530 | {"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} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.933355 |
https://math.answers.com/Q/How_many_faces_of_square_-_based_pyramid | 1,656,477,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00313.warc.gz | 429,897,532 | 40,605 | 0
# How many faces of square - based pyramid?
Wiki User
2014-05-15 12:31:51
In a square- based pyramid there are 5 faces
There are five faces: The square base, and the four triangles from each side of the square to the top (or bottom) vertex.
Wiki User
2014-05-15 12:31:51
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February 8, 2023, 13:01 2D Finite Volume code in MATLAB #1 New Member Chris Join Date: Nov 2021 Posts: 12 Rep Power: 4 Hey everyone, I'm trying to implement a finite volume code in MATLAB for a ellitical dgl -u'' = f in 2d. The 1d case worked out for me: Code: ```function unum = solvFVM(mesh, prob) % This function probves an elliptical problem -u''(x) = f(x) % We need a matriz of size: nloc = mesh.N-1; A = sparse(nloc,nloc); % One cell, one line hri = mesh.hri; % Here we need a loop. ii = 1; A(ii,ii ) = hri(ii) + 1./(mesh.xr(1)-mesh.xg(1)); A(ii,ii+1) = -hri(ii); for ii = 2:nloc-1 A(ii,ii-1) = -hri(ii-1); A(ii,ii ) = hri(ii) + hri(ii-1); A(ii,ii+1) = -hri(ii); end ii = nloc; A(ii,ii-1) = -hri(ii-1); A(ii,ii ) = 1./(mesh.xg(end)-mesh.xr(end)) + hri(ii-1); % spy(A) % RHS fi = mesh.f(mesh.xr) .* mesh.hg; % Valor de la funcion tamaņo intervalo fi(1 ) = fi(1 ) - 1./(mesh.xr(1)-mesh.xg(1)) * prob.u0; fi(end) = fi(end) - 1./(mesh.xg(end)-mesh.xr(end)) * prob.u1; %unum = sparse(A)\fi'; unum = [prob.u0; -sparse(A)\fi'; prob.u1];``` For the 2d case my code is as follows: Code: ```.rtcContent { padding: 30px; }.lineNode {font-size: 10pt; font-family: Menlo, Monaco, Consolas, "Courier New", monospace; font-style: normal; font-weight: normal; }function unum = solvFVM2D(mesh, prob) % This function probves an elliptical problem -u''(x) = f(x) % We need a matriz of size: nxr = mesh.Nx-1; nyr = mesh.Ny-1; nloc = (nxr)*(nyr); A = sparse(nloc,nloc); % One cell, one line fi = zeros(1, nloc); % Here we need a loop. for jj = 1:nyr for ii = 1:nxr fi(ii + (jj-1)*nxr) = mesh.f(mesh.xr(ii), mesh.yr(jj)) * mesh.hxg(ii) * mesh.hyg(jj); if ii > 1 W = mesh.hyg(jj)/mesh.hxr(ii-1); else W = mesh.hyg(jj)/(mesh.xr(ii)-mesh.xg(ii)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hyg(jj)./(mesh.xr(ii)-mesh.xg(ii)) * mesh.usol(mesh.xp(ii), mesh.yp(jj)); end if ii <= nxr-1 E = mesh.hyg(jj)/mesh.hxr(ii); else if ii == 1 E = mesh.hyg(jj)/(mesh.xg(ii)-mesh.xr(ii)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hyg(jj)./(mesh.xg(ii)-mesh.xr(ii)) * mesh.usol(mesh.xp(ii), mesh.yp(jj)); elseif ii < nxr E = mesh.hyg(jj)/(mesh.xg(ii)-mesh.xr(ii)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hyg(jj)./(mesh.xg(ii)-mesh.xr(ii)) * mesh.usol(mesh.xr(ii), mesh.yr(jj)); else E = mesh.hyg(jj)/(mesh.xg(ii+1)-mesh.xr(ii)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hyg(jj)./(mesh.xg(ii+1)-mesh.xr(ii)) * mesh.usol(mesh.xg(ii+1), mesh.yg(jj+1)); end end if jj > 1 S = mesh.hxg(ii)/mesh.hyr(jj-1); else S = mesh.hxg(ii)/(mesh.yr(jj)-mesh.yg(jj)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hxg(ii)./(mesh.yr(jj)-mesh.yg(jj)) * mesh.usol(mesh.xp(ii), mesh.yp(jj)); end if jj <= nyr-1 N = mesh.hxg(ii)/mesh.hyr(jj); else if jj == 1 N = mesh.hxg(ii)/(mesh.yg(jj)-mesh.yr(jj)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hxg(ii)./(mesh.yg(jj)-mesh.yr(jj)) * mesh.usol(mesh.xp(ii), mesh.yp(jj)); elseif jj < nyr N = mesh.hxg(ii)/(mesh.yg(jj)-mesh.yr(jj)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hxg(ii)./(mesh.yg(jj)-mesh.yr(jj)) * mesh.usol(mesh.xg(ii), mesh.yr(jj)); else N = mesh.hxg(ii)/(mesh.yg(jj+1)-mesh.yr(jj)); fi(ii + (jj-1)*nxr) = fi(ii + (jj-1)*nxr) - mesh.hxg(ii)./(mesh.yg(jj+1)-mesh.yr(jj)) * mesh.usol(mesh.xg(ii+1), mesh.yg(jj+1)); end end P = N + S + E + W; indx = ii + (jj-1) * nxr; A(indx, indx) = P; if indx < nloc A(indx, indx + 1) = -E; end if indx > 1 A(indx, indx - 1) = -W; end if indx <= nloc-nxr A(indx, indx + nxr) = -N; end if indx > nxr A(indx, indx - nxr) = -S; end end end %% spy(A) %unum = [prob.u0; sparse(A)\fi'; prob.u1]; sol = -sparse(A)\fi'; sol_mat = reshape(sol, [nxr, nyr]); unum = [mesh.usol(mesh.xg(1), mesh.yp(2:end-1))+ zeros(1, nyr); sol_mat; mesh.usol(mesh.xg(end), mesh.yp(2:end-1)) + zeros(1, nyr)]; unum = [mesh.usol(mesh.xp(:), mesh.yp(1))+zeros(nxr+2, 1), unum, mesh.usol(mesh.xp(:), mesh.yp(end))+zeros(nxr+2, 1)];``` The code only works for a quasi 1D case (only one cell in x or y direction), but fails when I'm trying to increase the cells (see picture below). Can anyone give me a hint what the error could be? Thanks in advance and regards! Chris
February 8, 2023, 13:06 #2 New Member Chris Join Date: Nov 2021 Posts: 12 Rep Power: 4 The files can be found here: https://drive.google.com/drive/folde...6R?usp=sharing
Tags elliptic pde, finite volume method, matlab
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Contact Us - CFD Online - Privacy Statement - Top | 1,968 | 5,305 | {"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} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.447074 |
https://en.wikipedia.org/wiki/Newform | 1,505,920,082,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687324.6/warc/CC-MAIN-20170920142244-20170920162244-00092.warc.gz | 665,480,360 | 11,737 | # Atkin–Lehner theory
(Redirected from Newform)
In mathematics, Atkin–Lehner theory is part of the theory of modular forms, in which the concept of newform is defined in such a way that the theory of Hecke operators can be extended to higher level. A newform is a cusp form 'new' at a given level N, where the levels are the nested congruence subgroups:
${\displaystyle \Gamma _{0}(N)=\left\{{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\in {\text{SL}}(2,\mathbf {Z} ):c\equiv 0{\pmod {N}}\right\}}$
of the modular group, with N ordered by divisibility. That is, if M divides N, Γ0(N) is a subgroup of Γ0(M). The oldforms for Γ0(N) are those modular forms f(τ) of level N of the form g(d τ) for modular forms g of level M with M a proper divisor of N, where d divides N/M. The newforms are defined as a vector subspace of the modular forms of level N, complementary to the space spanned by the oldforms, i.e. the orthogonal space with respect to the Petersson inner product.
The Hecke operators, which act on the space of all cusp forms, preserve the subspace of newforms and are self-adjoint and commuting operators (with respect to the Petersson inner product) when restricted to this subspace. Therefore, the algebra of operators on newforms they generate is a finite-dimensional C*-algebra that is commutative; and by the spectral theory of such operators, there exists a basis for the space of newforms consisting of eigenforms for the full Hecke algebra.
## Atkin–Lehner involutions
Consider a Hall divisor e of N, which means that not only does e divide N, but also e and N/e are relatively prime (often denoted e||N). If N has s distinct prime divisors, there are 2s Hall divisors of N; for example, if N = 360 = 23⋅32⋅51, the 8 Hall divisors of N are 1, 23, 32, 51, 23⋅32, 23⋅51, 32⋅51, and 23⋅32⋅51.
For each Hall divisor e of N, choose an integral matrix We of the form
${\displaystyle W_{e}={\begin{pmatrix}ae&b\\cN&de\end{pmatrix}}}$
with det We = e. These matrices have the following properties:
• The elements We normalize Γ0(N): that is, if A is in Γ0(N), then WeAW−1
e
is in Γ0(N).
• The matrix W2
e
, which has determinant e2, can be written as eA where A is in Γ0(N). We will be interested in operators on cusp forms coming from the action of We on Γ0(N) by conjugation, under which both the scalar e and the matrix A act trivially. Therefore the equality W2
e
= eA implies that the action of We squares to the identity; for this reason, the resulting operator is called an Atkin-Lehner involution.
• If e and f are both Hall divisors of N, then We and Wf commute modulo Γ0(N). Moreover, if we define g to be the Hall divisor g = ef/(e,f)2, their product is equal to Wg modulo Γ0(N).
• If we had chosen a different matrix We instead of We, it turns out that WeWe modulo Γ0(N), so We and We would determine the same Atkin-Lehner involution.
We can summarize these properties as follows. Consider the subgroup of GL(2,Q) generated by Γ0(N) together with the matrices We; let Γ0(N)+ denote its quotient by positive scalar matrices. Then Γ0(N) is a normal subgroup of Γ0(N)+ of index 2s (where s is the number of distinct prime factors of N); the quotient group is isomorphic to (Z/2Z)s and acts on the cusp forms via the Atkin-Lehner involutions. | 927 | 3,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "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} | 2.765625 | 3 | CC-MAIN-2017-39 | latest | en | 0.903747 |
https://www.physicsforums.com/threads/changing-earth-orbit-july-06-is-this-for-real.81503/ | 1,576,074,630,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540531917.10/warc/CC-MAIN-20191211131640-20191211155640-00488.warc.gz | 831,824,721 | 17,944 | # Changing Earth Orbit July 06! Is this for real?
## Main Question or Discussion Point
This site has a whole theory about changing the orbit of the earth by having 600 million people jumping at the same time.
I am very skeptical about this. I mean it sounds like the violation of the conservation of energy/momentum if you can change the earth's orbit without any mass ejection or a collision of some asteroid with the earth.
Can someone try to explain how this can be possible?
Related Other Physics Topics News on Phys.org
Well, there WOULD be a mass ejection: the 600,000,000 people.
The problem is that this ejection returns.
Consider this scenario... 300 passengers on a 747(which is going about smoothly) suddenly decide to stand up and jump up in concert. If done properly, a collective force will be exerted on the 747 pressing it down. Then, just afterwards, the 747 experiences a lack of weight equal to the 300 passengers, so the 747 rises. Then the 300 passengers hit the floor and the 747 experiences a sudden weight gain and thus moves down.
On average, there is no net effect.
However, perturbations of the craft during this time most certainly occur.
Ok so basically if they gather enough people with no brains out there and they actually jump, it wouldn't help the earth's orbit at all but just result in some stupid earthquake! :surprised
russ_watters
Mentor
Crumbles said:
Ok so basically if they gather enough people with no brains out there and they actually jump, it wouldn't help the earth's orbit at all but just result in some stupid earthquake! :surprised
Well, lets calculate the energy involved real quick: Lets say the average of these jumpers is a svelte high-jumper who has a mass of 50kg and jumps 1m (yeah, I'm lazy). 600 million times 50 times 1 is 30 billion (giga) joules or 0.03 TJ. According to http://www.geop.itu.edu.tr/~onur/seis/eq_energy/ [Broken] site, thats half the energy of a 4.0 magnitude earthquake and an insignificant fraction of the energy of a nuclear bomb. So it'd be detectable, but you probably wouldn't feel it.
But there's a catch: earthquakes are localized. They happen in a pretty small area and the energy travels outward as a wave. So the energy of those 600 million jumpers would need to be concentrated either by having them all stand on the same football-field sized area or jump in a precise sequence to cause the waves to constructively interfere.
Last edited by a moderator:
russ_watters said:
But there's a catch: earthquakes are localized. They happen in a pretty small area and the energy travels outward as a wave. So the energy of those 600 million jumpers would need to be concentrated either by having them all stand on the same football-field sized area or jump in a precise sequence to cause the waves to constructively interfere.
Ha ha! I like it! Can't wait for the 20th July of next year to watch a bunch of blisfully ignorant fools jumping thinking it's gonna do something!! :rofl:
even if that small number of peole could cause such a force by jumping, it wouldn't do any good....the earth has gravity, so they will come back and counteract the force they made by jumping.
What is the purpose of altering the Earth's orbit?
Deek said:
What is the purpose of altering the Earth's orbit?
If we can alter the earth's orbit to be slightly further from the sun, it could solve climate issues such as global warming etc..
Pengwuino
Gold Member
Crumbles said:
Ha ha! I like it! Can't wait for the 20th July of next year to watch a bunch of blisfully ignorant fools jumping thinking it's gonna do something!! :rofl:
I thought we already got this every november 2nd since the 60's :rofl: :rofl:
Crumbles said:
If we can alter the earth's orbit to be slightly further from the sun, it could solve climate issues such as global warming etc..
Or we could investigate alleviations to pollution. Which one seems more sensible to you? | 900 | 3,917 | {"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} | 3.0625 | 3 | CC-MAIN-2019-51 | longest | en | 0.924671 |
https://brainmass.com/physics/conduction/energy-you-are-considering-to-replace-the-single-glass-in-your-house-148993 | 1,481,186,291,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542455.45/warc/CC-MAIN-20161202170902-00027-ip-10-31-129-80.ec2.internal.warc.gz | 830,474,350 | 21,951 | Share
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# Energy - You are considering to replace the single glass in your house
CFL - IS AN ENERGY-EFFICIENT LIGHT
You are considering to replace the single glass in your house with double insulated glass.
The single glass has an R value of 0.91 m2-K/W and the double glass has a R value of 2.04 m2-K/W.
Calculate the heat loss for each type of glass from a room through a 2 m by 3 m window if the indoor temperature is 20 °C while it is 5 °C outdoors. [3]
How many 60 W incandescent bulbs must be replaced by 17 W CFLs before you get a MW saving? Assume all is switched on at the same time. [2]
Assuming 80 % of these lights are utilised during peak times and that Eskom is willing to pay R 3 million per peak MW saved. Further assume that Eskom is distributing these CFLs for free. What is the cost of each CFL to make such a rollout worth the effort? [3]
If each CLF cost R11 (bulk price) and the total project cost for distribution is R100,000. Is a CFL rollout worth the effort? [2]
#### Solution Preview
(a) - Glass --
As nothing is said about convective heat conductivity, we assume that it is to be disregarded.
The heat loss for a single glass is
2m * 3m * (20 °C - 5 °C) / (0.91 m2-K/W) = 98.9 W
And for the double glass it is
2m * 3m * (20 °C - 5 °C) / (2.04 m2-K/W) = 44.1 W
(b) CFL bulbs
Each bulb ...
#### Solution Summary
You are considering to replace the single glass in your house with double insulated glass.
The single glass has an R value of 0.91 m2-K/W and the double glass has a R value of 2.04 m2-K/W.
\$2.19 | 442 | 1,566 | {"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} | 3.828125 | 4 | CC-MAIN-2016-50 | latest | en | 0.93267 |
https://zwaltman.wordpress.com/2017/01/04/84-some-properties-of-transition-matrices-probability/ | 1,521,747,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648000.93/warc/CC-MAIN-20180322190333-20180322210333-00206.warc.gz | 951,174,199 | 16,594 | Today I Learned
Some of the things I've learned every day since Oct 10, 2016
84: Some Properties of Transition Matrices (Probability)
The following are some miscellaneous related properties of transition matrices describing stochastic processes such as Markov chains.
A ‘transition matrix’ is a square matrix $A$ with real entries in $[0, 1]$ such that the sum of the entries of any given of $A$ is equal to 1, the interpretation being that $A_{i, j}$ is the probability of a stochastic process transitioning from state $j$ to state $i$. A ‘regular’ matrix is a square matrix $A$ such that $A^n$ contains strictly positive entries for some $n > 1$.
For any transition matrix:
1. Every transition matrix has $1$ as an eigenvalue.
For regular transition matrices:
1. The algebraic multiplicity of $1$ as an eigenvalue of $A$ is $1$.
2. The columns of $\lim _{n \rightarrow \infty} A^n$ are identical, all being the unique $A$-invariant probability vector $v$ such that the sum of the entries of $v$ is $1$ (because it’s a probability vector) and $A v = v$.
3. With $v$ defined as above, for any probability vector $w$$\lim _{n \rightarrow \infty} A^n w = v$. That is, the system or process described by $A$ eventually converges to a distribution described by $v$, regardless of it’s initial state. | 335 | 1,303 | {"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": 24, "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} | 3.375 | 3 | CC-MAIN-2018-13 | latest | en | 0.887931 |
https://philosophy.stackexchange.com/questions/32614/is-%E2%88%83xax-bx-the-same-as-%E2%88%80x-ax-bx/32615 | 1,618,481,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00547.warc.gz | 499,709,474 | 34,841 | # Is ∃x(Ax -> Bx) the same as ∀x Ax -> Bx [closed]
Because according to this book it does:
https://textbookequity.org/Textbooks/Magnus_forallx.pdf
Page 64 - > Quantifiers and scope - paragraph 4.
I personally think that the former translates into, 'there exists an x, such that if Ax, then Bx', and the latter saying 'if an x is Ax then Bx'.
In this specific example, the symbolization key is:
``````UD: people
Gx: x can play guitar
Rx: x is a rock star
l: Lenny
``````
So here, it's:
``````∃x(Gx -> Gl)
∀xGx -> Gl
``````
So for me the former is stating 'There exists someone such that if that someone plays guitar, then so can Lenny'. Where the latter is stating 'if everyone can play the guitar, then so can Lenny'.
So, the former seems to be picking out a specific member of a set, not just any member, but some particular person. The latter seems to be referring to every member of a set.
• The equivalence is proved in predicate logic. See this post. – Mauro ALLEGRANZA Mar 2 '16 at 19:36
• Obviously, the for-all case implies the there-exists case. The other way around is not true. Can you make the philosophical problem you're trying to solve more clear? – James Kingsbery Mar 2 '16 at 19:45
• @JamesKingsbery - as per Jo's answer below, the title of the question is WRONG. – Mauro ALLEGRANZA Mar 2 '16 at 20:30
• This is related to what Raymond Smullyan calls the Drinker's Principle (sometimes called the Drinker's Paradox). "There is somebody in the pub such that if that person is drinking then everyone is drinking." It is true if and only if the pub is non-empty. It holds in standard predicate logic, but not in free logics. – Bumble Mar 2 '16 at 20:57
• Where in the paragraph you are linking to is there a variable x on both the sides of the implication? – Marco Disce Mar 3 '16 at 9:03
You're right, this is extremely counterintuitive. However, remember, that statements are equivalent in the case that they have the same truth conditions.
``````∃x(Gx -> Gl)
``````
is true if there exists someone who is not a guitarist or if Lemmy is a guitarist. It is false just in the case that Lemmy is not a guitarist and all other people are.
``````∀xGx -> Gl
``````
is true in the same cases and false in the same cases.
I have to admit, it took me a long time to think through this one. It might be easier to remember that A -> B is equivalent to !A V B (with ! for NOT)
So one statement is
`````` ∃x(!Gx V Gl)
``````
One statement is
``````!∀xGx V Gl
``````
In a world where Lemmy is a guitarist, both statements are true. In a world where Lemmy is not a guitarist, the truth value of the first reduces to ∃x!Gx (there exists someone who is not a guitarist) and the second to !∀xGx (not all people are guitarists). Those are equivalent.
IMPORTANT As Jo mentioned, ∃x(!Gx -> Gl) and ∃x(!Gx -> Bx) are quite different. In one, the consequent is a constant, in the other it is a variable. That DOES make a difference.
We can prove it intuitively considering the possible cases (of course, I'll refer to the correct example of Magnus' text: page 64, changing l with a for readibility).
(A) ∀xGx → Ga is true.
We ave two possible sub-cases:
(Ai) ∀xGx is true; thus (truth-table for conditional) also Ga is true.
If Ga is true, then also the conditional Gx → Ga is, and thus also ∃x(Gx → Ga) is true.
(Aii) ∀xGx is false; thus, for some value c for x, Gc is false.
Thus Gc → Ga is true (truth-table for conditional) and so ∃x(Gx → Ga) is true.
(B) ∀xGx → Ga is false.
This can happens only when ∀xGx is true and Ga is false. But if ∀xGx is true, then Gc is true for any value c for x; thus, Gc → Ga is false (truth-table for conditional) for any value c, and so ∃x(Gx → Ga) is false.
Conclusion: the two formuale are equivalent.
Of course, the equivalence holds also with any other predicate letter in place of the G in the consequent (i.e.Ga).
The key fact is:
∀xGx → R is equivalent to ∃x(Gx → R) provided that x is not free in R.
Note that the quoted passage of the textbook does not deal with the formulas from your quote.
It does not speak about "Bx" but about "Gl" with l being a fixed individual (Lemmy) not a variable. | 1,180 | 4,174 | {"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} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.940791 |
http://stackoverflow.com/questions/11188560/decimal-positions-to-decimal-via-hex-python/11188624 | 1,432,324,342,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207926620.50/warc/CC-MAIN-20150521113206-00287-ip-10-180-206-219.ec2.internal.warc.gz | 226,876,829 | 17,974 | # decimal positions to decimal via hex? (python)
Sorry for confusing title, but i did not know what to call this...
So in python I'm reading a binary file via an addon (intelhex).
This gets me the values for each byte loopin through it
``````for x in range(start_addr,end_addr):
print ih[x]
``````
gives:
``````1
79
60
246
``````
which is the same as:
``````01
4F
3C
F6
``````
I want the decimal value of `014F3CF6 = 21970166`
Is the best approach to just convert the decimals to hex and then concatenate the hex values and the convert to decimal again? Best being most understandable(pythonic) and/or most efficient
EDIT: To clarify what i want:
I want to convert [1,79,60,246] to 21970166
(since the list is the same as `[01,4F,3C,F6] which is 014F3CF6 which is 21970166)`
-
realised that as well, fixed now i hope – Vixen Jun 25 '12 at 12:11
Looking at `intelhex`, I presume you are doing something like
``````import intelhex
ih = intelhex.IntelHex('myfile.hex')
``````
``````>>> ih[0x01c200]
224
>>> ih[0x01c201]
165
>>> ih[0x01c202]
230
>>> ih[0x01c203]
246
``````
you can do
``````s = ih.gets(0x01c200, 4) # "\xe0\xa5\xe6\xf6"
``````
then convert to int like
``````import struct
i = struct.unpack('>I', s) # (3768968950L,)
``````
This can then be packaged up as
``````def getInt(ih, addr):
getInt(ih, 0x01c200) # -> 3768968950
``````
Edit:
Be aware that `ih[undefined_addr]` returns 255, while `ih.gets(undefined_addr, 1)` throws a NotEnoughDataError instead. In fact, `ih.gets(addr, n)` throws NotEnoughDataError if any byte in [addr:addr+n] is undefined.
If you do ih.dump('dumpfile.txt'), any bytes not defined in the .hex file show up as '--'; this may make it significantly easier to debug. I suggest you do a file dump, pull it up in a text editor, and take a look at bytes 0x20, 0x21, 0x22, 0x23.
-
Thanks a lot! This is what i really wanted, but i did not manage to find it myself! – Vixen Jun 25 '12 at 14:15
Also made use of this, try: mystring = ih.gets(0x20,4) except intelhex.NotEnoughDataError: print "There is not enough data at that location" – Vixen Jun 25 '12 at 14:30
If you do `ih.dump()`, what does it show at memory location 0x20? Are you sure you are looking for a 4-byte int, not a 2-byte shortint? – Hugh Bothwell Jun 25 '12 at 16:56
I'm certain of the value I'm getting, since the script is comparing it against an excel sheet to find differences and similarities. – Vixen Jun 26 '12 at 8:03
In python integers are not "decimals" or "hex". They are string representations of numbers. To convert a decimal string to int, use `int('12345', 10)`; to convert a hex string, use `int('1234ABC', 16)`. To convert integer to decimal string representation, you use `str(12345)`, to convert to hex string, use `hex(12345)`.
Furthermore, you should see the module `struct` and consider using it to convert binary data to integers.
-
I would do something like
``````s = 0
for i in bytes:
s = s * 256 + i
print s
``````
Besides, 79 is `0x4F`, not `0xF4`. `0xF4` is 244.
-
You can use `int`:
``````>>> int('014F3CEC', 16)
21970156
``````
This means you should concatenate the strings to one string which contains the hexadecimal number. You can do that as follows:
``````>>> ih = [1, 79, 60, 236]
>>> ih_s = [hex(i)[2:].zfill(2) for i in ih]
>>> ih_s
['01', '4f', '3c', 'ec']
>>> hex_string = ''.join(ih_s)
>>> hex_string
'014f3cec'
``````
-
Great, took some time to understand hex(i)[2:].zfill(2) though – Vixen Jun 25 '12 at 12:23
An alternative to the `zfill` would be to use string formatting: `'{0:02x}'.format(i)` – Mark Dickinson Jun 25 '12 at 13:48 | 1,173 | 3,630 | {"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} | 3.09375 | 3 | CC-MAIN-2015-22 | latest | en | 0.794121 |
https://www.coursehero.com/file/6366705/File-Assignment3pdf/ | 1,485,303,485,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285315.77/warc/CC-MAIN-20170116095125-00277-ip-10-171-10-70.ec2.internal.warc.gz | 909,106,871 | 92,179 | File Assignment3.pdf
# File Assignment3.pdf - ASSIGNMENT NO.3 Assignment problem...
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ASSIGNMENT NO.3 Assignment problem #1 (35 out of 100) Two cables each having a length L of approximately 40m, support a loaded container of weight W . The cables, whch have effective cross-sectional area A = 48mm 2 and effective modulus of Elasticity E = 160GPa, are identical except one cable islonger than the other when they are hanging separetely and unloaded. The difference in lengths is d = 100mm. The cables are made of steel having an elastoplastic stress-strain diagram with 500 Y σ = MPa. Assume that the weight W is initially zero and is slowly increased by the addition of material to the container. (a) Determine the weight Y W that first produces yielding of the shorter cable. Also, determine the corresponding elongation Y δ of the shorter cable. (b) Determine the weight P W that produces yielding of both cables. Also, determine the elongation P of the shorter cable when the weight W just reaches the value P W . Assignment problem #2 (35 out of 100)
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Ask a homework question - tutors are online | 286 | 1,238 | {"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} | 2.859375 | 3 | CC-MAIN-2017-04 | longest | en | 0.905891 |
https://drostlab.github.io/philentropy/reference/KL.html | 1,603,610,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888402.81/warc/CC-MAIN-20201025070924-20201025100924-00538.warc.gz | 293,950,466 | 4,409 | This function computes the Kullback-Leibler divergence of two probability distributions P and Q.
KL(x, test.na = TRUE, unit = "log2", est.prob = NULL)
## Arguments
x a numeric data.frame or matrix (storing probability vectors) or a numeric data.frame or matrix storing counts (if est.prob = TRUE). See distance for details. a boolean value indicating whether input vectors should be tested for NA values. a character string specifying the logarithm unit that shall be used to compute distances that depend on log computations. method to estimate probabilities from a count vector. Default: est.prob = NULL.
## Value
The Kullback-Leibler divergence of probability vectors.
## Details
$$KL(P||Q) = \sum P(P) * log2(P(P) / P(Q)) = H(P,Q) - H(P)$$
where H(P,Q) denotes the joint entropy of the probability distributions P and Q and H(P) denotes the entropy of probability distribution P. In case P = Q then KL(P,Q) = 0 and in case P != Q then KL(P,Q) > 0.
The KL divergence is a non-symmetric measure of the directed divergence between two probability distributions P and Q. It only fulfills the positivity property of a distance metric.
Because of the relation KL(P||Q) = H(P,Q) - H(P), the Kullback-Leibler divergence of two probability distributions P and Q is also named Cross Entropy of two probability distributions P and Q.
## References
Cover Thomas M. and Thomas Joy A. 2006. Elements of Information Theory. John Wiley & Sons.
H, CE, JSD, gJSD, distance
Hajk-Georg Drost
## Examples
# Kulback-Leibler Divergence between P and Q
P <- 1:10/sum(1:10)
Q <- 20:29/sum(20:29)
x <- rbind(P,Q)
KL(x)
#> Metric: 'kullback-leibler' using unit: 'log2'; comparing: 2 vectors.#> kullback-leibler
#> 0.1392629
# Kulback-Leibler Divergence between P and Q using different log bases
KL(x, unit = "log2") # Default
#> Metric: 'kullback-leibler' using unit: 'log2'; comparing: 2 vectors.#> kullback-leibler
#> 0.1392629 KL(x, unit = "log")
#> Metric: 'kullback-leibler' using unit: 'log'; comparing: 2 vectors.#> kullback-leibler
#> 0.09652967 KL(x, unit = "log10")
#> Metric: 'kullback-leibler' using unit: 'log10'; comparing: 2 vectors.#> kullback-leibler
#> 0.0419223
# Kulback-Leibler Divergence between count vectors P.count and Q.count
P.count <- 1:10
Q.count <- 20:29
x.count <- rbind(P.count,Q.count)
KL(x, est.prob = "empirical")
#> Metric: 'kullback-leibler' using unit: 'log2'; comparing: 2 vectors.#> kullback-leibler
#> 0.1392629
# Example: Distance Matrix using KL-Distance
Prob <- rbind(1:10/sum(1:10), 20:29/sum(20:29), 30:39/sum(30:39))
# compute the KL matrix of a given probability matrix
KLMatrix <- KL(Prob)
#> Metric: 'kullback-leibler' using unit: 'log2'; comparing: 3 vectors.
# plot a heatmap of the corresponding KL matrix
heatmap(KLMatrix) | 834 | 2,806 | {"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} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.692537 |
https://answersdrive.com/how-much-does-1-million-in-100-bills-weigh-4680398 | 1,606,284,100,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00599.warc.gz | 191,839,508 | 8,095 | 28th November 2019
18
# How much does \$1 million in \$100 bills weigh?
Watching "Dumb & Dumber" (for some unknown reason) and since they are carrying around the case with \$1,000,000, I wondered how much it weighs. I know everyone on the OT carries \$1,000,000 in their front pocket, so here is what you are carrying around. \$1M in \$100 bills = 22.05 pounds.
\$1M in \$20 bills = 110.5 pounds.
People also ask, how much is 1 million dollars in 100 dollar bills?
A packet of one hundred \$100 bills is less than 1/2" thick and contains \$10,000. Fits in your pocket easily and is more than enough for week or two of shamefully decadent fun. Believe it or not, this next little pile is \$1 million dollars (100 packets of \$10,000).
How heavy is a 100 dollar bill?
A friend of mine with a scientific scale tells me that \$100 bills weigh about one gram each. It takes just a few steps to calculate how many tons of \$100 bills there are in \$20 billion. There are 200,000,000 one hundred dollar bills in \$20 billion. For simplicity sake, let's say that 500 grams equals one pound.
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1 G R E Which of these is the number 5,005,0? five million, five hundred, fourteen five million, five thousand, fourteen five thousand, five hundred, fourteen five billion, five million, fourteen LIFORNI STNRS TEST SM00 The estimated cost to build a new baseball stadium is ninety-four million dollars. What is this number in standard form? Which decimal should be placed in the box to have the numbers in order from least to greatest? ? SM00 \$90,00 \$9,000 \$90,00,000 \$9,000,000 Which of the following has the greatest value? SM What is 67,8,59 rounded to the nearest hundred thousand? 67,000,000 67,800,000 67,80,000 67,900,000 SM SM06 6 What is 58,607 rounded to the nearest hundred? 58,000 58,600 58,700 8,000 SM06 8 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
2 LIFORNI STNRS TEST G R E 7 Which fraction represents the largest part of a whole? 9 Which square is shaded? 6 8 Which fraction means the same as 0.7? SM SM0057 SM067 0 The numbers in the pattern decrease by the same amount each time. What are the next three numbers in this pattern? 0, 8, 6,,,,, 0,, 0,, 0,, 0,, SM096 9 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
3 G R E Which symbol is located at on the number line below? LIFORNI STNRS TEST Kira owes Mark \$5, and Mark owes Kira \$7. Which statement means the same thing? Kira owes Mark \$. Kira owes Mark \$. Mark owes Kira \$. Mark owes Kira \$. SM088 What fraction is best represented by point P on this number line? 0 P SM057 Marisol is counting by s. If she starts counting at 0, what two numbers are missing below? 0, 7,,,,, 8, 5 9, 7 0,, SM05 SM059 0 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
4 LIFORNI STNRS TEST G R E 5 On the number line below, what number does point M represent? M 7 Which letter represents 0.80 on the number line below? W X Y Z W X Y Z SM SM07 8 On Thursday hris drove 67 miles, on Friday he drove 68 miles, and on Saturday he drove 7 miles. pproximately how many miles did hris drive in the three days? 6 Look at the number line. 00 miles 00 miles 00 miles 0 etween which two shapes is? between and between and 00 miles = SM068 between and between and SM SM08 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
5 G R E 0 The total length of a vehicle is 05.8 inches. What is the length of the vehicle rounded to the nearest whole number? 00 inches 05 inches 06 inches 0 inches LIFORNI STNRS TEST SM06 Jonathan read 5 pages during his summer reading program. In order to reach his goal of 650 pages, how many more pages does he need to read? SM08 The sales tax for an item is \$0.7. What is the amount of tax rounded to the nearest dime? \$0.0 \$0.5 \$0.7 \$ = R R SM069 SM = SM05 6 Justin solved the problem below. Which expression could be used to check his answer? (5 )+ (5 )+ (5 + ) (5 + ) 5 r )6 SM079 SM77 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
6 LIFORNI STNRS TEST G R E 7 There are 58 cases of soda in a warehouse. If there are cans of soda in each case, how many cans of soda are in the warehouse? SM06 8 There are 0 teachers at a school. Each teacher is provided with 500 sheets of paper. How many sheets of paper is this in all? 0,000 00,000,000,000 0,000,000 SM09 9 year has 65 days, and a day has hours. How many hours are in 65 days? SM080 0 Valley Transport has been hired to deliver new seats to the Oak Valley Sports Stadium. The company will use trucks to move the seats. If each truck holds 05 seats, how many seats will be delivered to the stadium?,85,95,05,5 SM067 There are 9 rows of seats in a theater. Each row has the same number of seats. If there is a total of 6 seats, how many seats are in each row? SM0 Maria read a 0-page book in 7 days. She read the same number of pages each day. How many pages did she read each day? 0 6 SM070 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
7 G R E ) Jeb paid \$7 for a magazine subscription. If he is paying \$ for each issue of the magazine, how many issues of the magazine will he receive? R 7 9 R 5 85 = = LIFORNI STNRS TEST SM066 SM Which statement is true? The only factors of 8 are and 8. The only factors of 9 are and 9. The only factors of 0 are and 0. The only factors of are and. 8 Which is a prime number? Which number is represented by n? n =8 SM00 SM07 SM9 SM98 6 Which of these is another way to write the product 7? SM00 0 What is the value of the expression below? ( + ) (7 ) 0 0 SM0 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
8 LIFORNI STNRS TEST G R E What is the value of the expression below if a =? 5 (a + 8) What is the value of 6 ( )? SM ( + ) ( ) = What is the value of x? ( 6 ) ( ) = x SM05 SM75 SM ( 8 ) = SM nna bought bags of red gumballs and 5 bags of white gumballs. Each bag of gumballs had 7 pieces in it. Which expression could nna use to find the total number of gumballs she bought? ( 7 ) +5= ( 7 5) += 7 (5 + )= 7 + (5 ) = SM087 ( + ) = ( 9 + ) 6 = SM06 SM This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
9 G R E 9 Which equation below represents the area () of the rectangle in square centimeters? 9 cm LIFORNI STNRS TEST 5 Look at the problem below. If x =, what is y? x + y = 0 5 = 9 = cm = ( 5 ) + ( 9) 5 = ( ) + ( 9) 8 6 SM Look at the problem below. If + = 7, what is,? 7, = + + SM09 SM The sum of x plus y equals 6. If x = 7, which equation can be used to find the value of y? y 7 = y = 6 x y=6 x + 7= 6 5 The letters S and T stand for numbers. If S 00 = T 00, which statement is true? S= T S> T S= T +00 S> T = 5 + SM097 SM098 SM This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
10 LIFORNI STNRS TEST G R E 55 Tina and erek collect baseball cards. Each has the same number of cards. If Roberto gives Tina and erek 5 more baseball cards each, who will have the greater number of baseball cards, Tina or erek? Tina erek Tina and erek will have the same number of baseball cards. There is not enough information to answer the question. SM Which number should be put in the box to make this equation true? =, + 58 If 7 = 7 a, what is the value of a? = = SM059 SM SM 57 What number goes in the box to make this number sentence true? (7 ) 5 =, 5 7 SM This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
11 G R E LIFORNI STNRS TEST 60 Gabrielle wants to cover her square garden with mulch to protect her plants. Which bag of mulch will Gabrielle need to buy to exactly cover the entire garden area? feet MULH MULH 80 square feet 0 square feet MULH MULH 0 square feet square feet SM06 8 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
12 LIFORNI STNRS TEST G R E 6 Which statement about the figures is true? 6 Which statement about the figures is true? 5 Rectangle Figure Figure Rectangle They both have the same area. They both have the same width. They both have the same length. They both have the same perimeter. They have different areas. They have the same area. They have the same length. They have different perimeters. SM0700 SM070 9 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
13 G R E 6 This figure is made of three squares joined together. LIFORNI STNRS TEST 6 hu plotted points on a grid. The points were all on the same straight line. in. in. in. in What is the area of the figure in square inches? 9 square inches 8 square inches 7 square inches 8 square inches SM If she plots another point on the line, what could be its coordinates? (, 5) (, ) (6, ) ( 7, ) SM This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
14 LIFORNI STNRS TEST G R E 65 Look at the line segment shown below. y 66 Look at the graph. Point S is at (5, 8). Point F is at (5, ). y (, ) (6, ) x 0 What is the length of the line segment? unit units units 6 units SM S F How can you find the number of units from point S to point F? dd: 5 +8 dd: + 8 Subtract: 8 5 Subtract: 8 x SM089 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
15 G R E LIFORNI STNRS TEST 67 What is the length of the line segment shown on the grid? 68 Which figures below show pairs of lines that appear to be parallel? y Figure (,) x Figure (,, 5) 9 units 7 units 5 units units Figure SM0059 Figure only Figure only Figure and Figure Figure and Figure SM0976 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
16 LIFORNI STNRS TEST G R E 69 Look at the circle with center O. 70 Which pair of shapes is congruent? O The line segment appears to be an arc. a perimeter. a radius. a diameter. SM068 SM95 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
17 G R E LIFORNI STNRS TEST 7 Which figure has a line of symmetry and rotational symmetry? 7 When it is 0:0, what kind of angle is formed by the hands of the clock? acute obtuse right straight SM0707 SM0709 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
18 LIFORNI STNRS TEST G R E 7 Which figure can form a pyramid when folded on the dotted lines without overlapping? 75 Which shape must have four equal sides and four right angles? square rectangle rhombus parallelogram SM066 SM0 7 What kind of a triangle always has acute angles and sides the same length? isosceles right equilateral scalene SM This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
19 G R E LIFORNI STNRS TEST 76 Pietro surveyed 5 students about their favorite colors and made this bar graph. Number of Students Favorite olors lue Green Red Yellow Purple Which of the following tally charts did he use to make this graph? Favorite olors lue Green Red Yellow Purple Favorite olors lue Green Red Yellow Purple Favorite olors lue Green Red Yellow Purple Favorite olors lue Green Red Yellow Purple SM97 6 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
20 LIFORNI STNRS TEST G R E 77 What is the mode of this set of numbers? 6 {,,,,,, 6} SN t a local school, the fourth, fifth, and sixth graders sold flowers as a fundraiser. The bar graph below shows how many flowers were sold by each grade. 79 Royce has a bag with 8 red marbles, blue marbles, 5 green marbles, and 9 yellow marbles all the same size. If he pulls out marble without looking, which color is he most likely to choose? red blue green yellow SN005 Flowers Number of Flowers Grade Grade 5 Grade 6 How many flowers did the students sell in all? SM090 7 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
21 G R E 80 Phan placed these animal cards into a bag. LIFORNI STNRS TEST What is the probability that she will draw a card with a tiger? SM069 8 This is a sample of alifornia Standards Test questions. This is NOT an operational test form. Test scores cannot be projected based on performance on released test questions. opyright 008 alifornia epartment of Education.
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MATH BOOKMAKING IDEAS TO FLIP, FLAP, AND FOLD CONTRIBUTING WRITERS Karen Bauer, Jan Brennan, Rosa Drew, Ronda Howley, Heidi Meyer, Tiffani Mugurassa, and Brenda Wyma EDITOR Alaska Hults ILLUSTRATOR Jane | 10,574 | 40,393 | {"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} | 3.0625 | 3 | CC-MAIN-2020-45 | latest | en | 0.892241 |
http://www.internetdict.com/answers/how-many-liters-in-one-gallon.html | 1,516,639,559,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00693.warc.gz | 456,542,667 | 7,612 | # How Many Liters in One Gallon?
Because a gallon is a much larger volume than one liter, it takes a few liters to make on gallon. To be exact, there are 3.7854 liters in one gallon. | 49 | 183 | {"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} | 2.6875 | 3 | CC-MAIN-2018-05 | latest | en | 0.91452 |
https://astarmathsandphysics.com/gcse-physics-notes/712-capacitors-and-capacitance.html?tmpl=component&print=1&page= | 1,531,890,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590051.20/warc/CC-MAIN-20180718041450-20180718061450-00414.warc.gz | 582,559,466 | 2,409 | ## Capacitors and Capacitance
Suppose we take a battery and connect one terminal to a piece of metal. Charge will pass from the terminal to the metal until the metal is charged to the potential of the terminal. If we connect the other terminal to another piece of metal then that terminal will be charged to the potential of that other terminal. No charge can pass from on piece of metal to another is the two pieces are not connected in some way. The situataion is illustrated below.
Obviously if the voltage of the cell increases the potential difference between the two pieces of metal will also increase. In fact up to certain limits the charge on the pieces of metal is proportional to the voltage across it. The constant of proportionality is called the capacitance, written
The name for a device which stores charge in this way is capacitor The symbol is shown below.
The field between the plate reflects the force on a positive charge and is shown from left to right. | 195 | 979 | {"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} | 3.28125 | 3 | CC-MAIN-2018-30 | latest | en | 0.948829 |
https://www.doubtnut.com/question-answer-physics/a-geostationary-satellite-orbits-the-earth-at-a-height-of-nearly-36000-km-from-the-surface-of-earth--643743456 | 1,653,124,716,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539049.32/warc/CC-MAIN-20220521080921-20220521110921-00033.warc.gz | 868,386,957 | 79,570 | HomeEnglishClass 11PhysicsChapterGravitation
A geostationary satellite orbi...
# A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of earth. What is the potential due to earth's gravity at the site of this satellite? ( Take the potential energy at infinity to be zero). Mass of the earth 6.0 xx 10^(24) kg, radius = 6400km.
Updated On: 17-04-2022
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Text Solution
-9.338 xx 10^(7) J kg^(-1)
Solution :
V = -G M // R + h = - 6.67 xx 10^(-11) xx 6 xx 10^(24) // 6400 xx 10^(3) + 36000 xx 10^(3) = - 9.338 xx 10^(7) J kg^(-1). | 220 | 672 | {"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} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.673266 |
https://math.stackexchange.com/questions/4485857/probability-of-rolling-exactly-1-number-exactly-3-times-in-6-rolls-of-a-fair-die | 1,660,928,872,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00385.warc.gz | 349,967,875 | 68,141 | # Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die
I'm trying to calculate the probability based on the size of the event space divided by the size of the sample space $$P=\frac{|E|}{|S|}$$
I know that $$|S|=6^6$$, but am not sure what exactly the event space consists. Currently my thoughts are that we have 6 choices for our favorable event(the triples) and for the remaining 3 numbers we have $$5\times5\times4=100$$, $$4$$ because we do not want to include the possibility of having 3 same numbers two times, and to consider all possible arrangements, there are then $$\frac{6!}{3!1!1!1!}$$ possibilities.
This leads to our final equation of : $$P=\frac{|E|}{|S|}=\frac{6!}{3!1!1!1!} \times \frac{6\times100}{6^6}$$
But the problem is that this exceeds 1, which is clearly wrong but I couldn't really figure out what is the fix for my equation.
Thanks:)
• Try and take cases: 3+2+1, 3+3(unless stated that it shouldn’t be counted), 3+1+1+1. Jul 4 at 8:25
• I have simplified and abbreviated my approach. Jul 6 at 20:37
First, select the number that appears exactly $$3$$ times: $$6$$ ways. We have $$\binom{6}{3}$$ ways to place them. Now, in the rest of the $$3$$ positions, you have $$5$$ options. So, count = $$6 \cdot \binom{6}{3} \cdot 5^3$$
But, in this we are also counting the number arrangements of $$3 + 3$$ (so two numbers appear $$3$$ times), so subtract $$\binom{6}{2} \cdot 2 \cdot \binom{6}{3}$$ This is subtracting the number of ways to pick $$2$$ numbers and place them in $$3 + 3$$.
EDIT: My explanation was wrong and thanks to @N.F.Taussig for pointing it out. I will just put their comment that explains where I am wrong:
The factor of $$2\binom{6}{2}\binom{6}{3}$$ is twice the number of arrangements in which two numbers each appear three times. That is what we want to subtract since those patterns are counted twice among the $$\binom{6}{1}\binom{6}{3}5^3$$ arrangements in which a number appears three times, and we don't want to count such arrangements at all. However, your wording suggests that the term you are subtracting is the number of arrangements in which two numbers each appear three times, which is not the case.
So, answer = $$\frac{120 \cdot 125 - 40 \cdot 15}{6^6} = 0.31$$
Currently my thoughts are that we have $$6$$ choices for our favorable event (the triples)
That is correct.
and for the remaining $$3$$ numbers we have $$5×5×4=100$$, $$4$$ because we do not want to include the possibility of having $$3$$ same numbers two times
You are ordering the combinations later on, but when multiplying $$5 × 5 × 4$$ like that, you are not just picking the elements but also ordering them, something which you are doing later on. So, this over-counts. Also, it's not $$5 × 5 × 4$$, What if the first element (when you multiply by $$5$$) is the count-$$3$$-element itself?
and to consider all possible arrangements, there are then $$\frac{6!}{3!1!1!1!}$$ possibilities.
This is true only if the counts look like $$6 = 3 + 1 + 1 + 1$$ but what if $$6 = 3 + 2 + 1$$?
Let me know if you have any questions)))
• The factor of $2\binom{6}{2}\binom{6}{3}$ is twice the number of arrangements in which two numbers each appear three times. That is what we want to subtract since those patterns are counted twice among the $\binom{6}{1}\binom{6}{3}5^3$ arrangements in which a number appears three times, and we don't want to count such arrangements at all. However, your wording suggests that the term you are subtracting is the number of arrangements in which two numbers each appear three times, which is not the case. Jul 4 at 10:03
• @N.F.Taussig Sorry, I messed up the explanation. Thanks for pointing it out. Jul 4 at 10:59
Since there are six possible values for each of the six rolls, there are indeed $$6^6$$ elements in the sample space.
For the favorable cases, since we want to calculate the number of cases in which exactly one number appears three times, there are two possibilities:
• One number appears three times and three other numbers each appear once.
• One number appears three times, a second number appears twice, and a third number appears once.
One number appears three times and three other numbers each appear once: There are six ways to select the number which appears three times, $$\binom{6}{3}$$ ways to select which three of the six positions in the sequence of rolls that number occupies, $$\binom{5}{3}$$ ways to select which three of the other five numbers each appear once, and $$3!$$ ways to arrange those three distinct numbers in the remaining positions of the sequence. Hence, there are $$\binom{6}{1}\binom{6}{3}\binom{5}{3}3!$$ such cases.
One number appears three times, a second number appears twice, and a third number appears once: There are six ways to select the number which appears three times, $$\binom{6}{3}$$ ways to select which three of the six positions in the sequence of rolls that number occupies, five ways to select which of the remaining numbers appears twice, $$\binom{3}{2}$$ ways to select which two of the three positions in the sequence of rolls that number occupies, and four ways to select which of the remaining numbers fills the remaining position in the sequence. Hence, there are $$\binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$ such cases.
Since the above cases are mutually exclusive and exhaustive, the number of favorable cases is $$\binom{6}{1}\binom{6}{3}\binom{5}{3}3! + \binom{6}{1}\binom{6}{3}\binom{5}{1}\binom{3}{2}\binom{4}{1}$$
Hence, the probability that one number appears exactly three times in six rolls of a fair six-sided die is $$\frac{\dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{3}3! + \dbinom{6}{1}\dbinom{6}{3}\dbinom{5}{1}\dbinom{3}{2}\dbinom{4}{1}}{6^6}$$
Number of cases when there are exactly three 1's is $$N_1={6\choose 3}5^3$$. Number of cases when there are exactly three 1's and exactly three 2's is $$N_2={6\choose 3}$$. Number of cases when there are exactly three 1's and no more threes is $$N_3=N_1-5N_2={6\choose 3}(5^3-5)$$. Number of cases when exactly 1 number is rolled exactly 3 times is $$N_4=6N_3={6\choose 3}(5^3-5)6=14400$$. Probability is $$\frac{N_4}{6^6}=\frac{25}{81}\approx 0.31$$
For patterns matching stipulations, we shall use the format
[Choose numbers for pattern]$$\times$$[Permute]
$$3-2-1\; pattern:$$
$$\left[\binom61\binom51\binom41\right]\times\left[\frac{6!}{3!2!}\right] = 7200$$
$$3-1-1-1\; pattern:$$
$$\left[\binom61\binom53\right]\times\left[\frac{6!}{3!}\right]= 7200$$
Thus $$Pr = \dfrac{7200+7200}{6^6} \approx 0.31$$ | 1,943 | 6,569 | {"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": 53, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2022-33 | latest | en | 0.94276 |
https://www.scribd.com/document/99934459/Filter-kalman-vs-complementary | 1,569,116,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00559.warc.gz | 1,023,220,605 | 68,683 | You are on page 1of 20
# The Balance Filter
A Simple Solution for Integrating Accelerometer and Gyroscope Measurements for a Balancing Platform
## Rev.1: Submitted as a Chief Delphi white paper - June 25, 2007.
Sensors
Y X
2-Axis Accelerometer: Measures acceleration, but really force per unit mass. (F = ma, so a = F/m) Can be used to measure the force of gravity. Above, X-axis reads 0g, Y-axis reads -1g. Can be used to measure tilt:
Gyroscope: Measures angular rate (speed of rotation). Reads zero when stationary. Reads positive or negative when rotating:
Y X X
Y sees slightly less gravity. Is Y useful information? Probably not: a) It is far less sensitive to small changes in angle than X. b) It does not depend on direction of tilt.
## x_acc = (float)(x_acc_ADC x_acc_offset) * x_acc_scale; gyro = (float)(gyro_ADC gyro_offset) * gyro_scale;
x_acc float-type
gyro float-type
*Even though neither the ADC result nor the offset can be negative, they will be subtracted, so it couldnt hurt to make them signed variables now. Units could be degrees or radians [per second for the gyro]. They just have to be consistent.
## A bit more about the accelerometer
If it was necessary to have an estimate of angle for 360 of rotation, having the Y-axis measurement would be useful, but not necessary. With it, we could use trigonometry to find the inverse tangent of the two axis readings and calculate the angle. Without it, we can still use sine or cosine and the X-axis alone to figure out angle, since we know the magnitude of gravity. But trig kills processor time and is non-linear, so if it can be avoided, it should. For a balancing platform, the most important angles to measure are near vertical. If the platform tilts more than 30 in either direction, theres probably not much the controller can do other than drive full speed to try to catch it. Within this window, we can use small angle approximation and the X-axis to save processor time and coding complexity: Platform is tilted forward by and angle , but stationary (not accelerating horizontally).
X
X-axis reads: (1g) sin() small angle approximation: sin() , in radians This works well (within 5%) up to = /6 = 30. So in the following bit of code, x_acc = (float)(x_acc_ADC x_acc_offset) * x_acc_scale; x_acc will be the angle in radians if x_acc_scale is set to scale the output to 1[g] when the X-axis is pointed straight downward. To get the angle in degrees, x_acc_scale should be multiplied by 180/.
Desired Measurements
In order to control the platform, it would be nice to know both the angle and the angular velocity of the base platform. This could be the basis for an angle PD (proportional/derivative) control algorithm, which has been proven to work well for this type of system. Something like this:
## Motor Output = Kp (Angle) + Kd (Angular Velocity)
What exactly Motor Output does is another story. But the general idea is that this control setup can be tuned with Kp and Kd to give stability and smooth performance. It is less likely to overshoot the horizontal point than a proportional-only controller. (If angle is positive but angular velocity is negative, i.e. it is heading back toward being horizontal, the motors are slowed in advance.)
Kp
In effect, the PD control scheme is like adding an adjustable spring and damper to the Segway.
Kd
Mapping Sensors
Y X
Angle
Angular Velocity
Best approach?
Mapping Sensors
Y X
Most Obvious
Angle
Angular Velocity
Pros: Intuitive. Easy to code. Gyro gives fast and accurate angular velocity measurement.
Cons: Noisy. X-axis will read any horizontal acceleration as a change in angle. (Imagine the platform is horizontal, but the motors are causing it to accelerate forward. The accelerometer cannot distinguish this from gravity.)
Mapping Sensors
Y X
## Quick and Dirty Fix
Low-Pass Filter*
Angle
Angular Velocity
*Could be as simple as averaging samples: angle = (0.75)*(angle) + (0.25)*(x_acc); 0.75 and 0.25 are example values. These could be tuned to change the time constant of the filter as desired. Pros: Still Intuitive. Still easy to code. Filters out short-duration horizontal accelerations. Only long-term acceleration (gravity) passes through. Cons: Angle measurement will lag due to the averaging. The more you filter, the more it will lag. Lag is generally bad for stability.
Mapping Sensors
Y X
Single-Sensor Method
Angle
Numeric Integration*
Angular Velocity
*Simple physics, dist. = vel. time. Accomplished in code like this: angle = angle + gyro * dt; Requires that you know the time interval between updates, dt. Pros: Only one sensor to read. Fast, lag is not a problem. Not subject to horizontal accelerations. Still easy to code. Cons: The dreaded gyroscopic drift. If the gyro does not read perfectly zero when stationary (and it wont), the small rate will keep adding to the angle until it is far away from the actual angle.
Mapping Sensors
Y X
Kalman Filter
Angle
Physical Model
Magic?
Angular Velocity
Pros: Supposedly the theoretically-ideal filter for combining noisy sensors to get clean, accurate estimates. Takes into account known physical properties of the system (mass, inertia, etc.).
Cons: I have no idea how it works. Its mathematically complex, requiring some knowledge of linear algebra. There are different forms for different situations, too. Probably difficult to code. Would kill processor time.
Mapping Sensors
Y X
Complementary Filter
Low-Pass Filter
High-Pass Filter
Angle
Numeric Integration
Angular Velocity
*Luckily, its more easily-said in code: angle = (0.98)*(angle + gyro * dt) + (0.02)*(x_acc); More explanation to come
Pros: Can help fix noise, drift, and horizontal acceleration dependency. Fast estimates of angle, much less lag than low-pass filter alone. Not very processor-intensive.
Cons: A bit more theory to understand than the simple filters, but nothing like the Kalman filter.
## More on Digital Filters
There is a lot of theory behind digital filters, most of which I dont understand, but the basic concepts are fairly easy to grasp without the theoretical notation (z-domain transfer functions, if you care to go into it). Here are some definitions: Integration: This is easy. Think of a car traveling with a known speed and your program is a clock that ticks once every few milliseconds. To get the new position at each tick, you take the old position and add the change in position. The change in position is just the speed of the car multiplied by the time since the last tick, which you can get from the timers on the microcontroller or some other known timer. In code: position += speed*dt;, or for a balancing platform, angle += gyro*dt;. Low-Pass Filter: The goal of the low-pass filter is to only let through long-term changes, filtering out short-term fluctuations. One way to do this is to force the changes to build up little by little in subsequent times through the program loop. In code: angle = (0.98)*angle + (0.02)*x_acc; If, for example, the angle starts at zero and the accelerometer reading suddenly jumps to 10, the angle estimate changes like this in subsequent iterations:
Iter. 1 0.20 2 0.40 3 0.59 4 0.78 5 0.96 6 1.14 7 1.32 8 1.49 9 1.66 10 1.83
If the sensor stays at 10, the angle estimate will rise until it levels out at that value. The time it takes to reach the full value depends on both the filter constants (0.98 and 0.02 in the example) and the sample rate of the loop (dt).
## More on Digital Filters
High-Pass Filter: The theory on this is a bit harder to explain than the low-pass filter, but conceptually it does the exact opposite: It allows short-duration signals to pass through while filtering out signals that are steady over time. This can be used to cancel out drift. Sample Period: The amount of time that passes between each program loop. If the sample rate is 100 Hz, the sample period is 0.01 sec. Time Constant: The time constant of a filter is the relative duration of signal it will act on. For a low-pass filter, signals much longer than the time constant pass through unaltered while signals shorter than the time constant are filtered out. The opposite is true for a highpass filter. The time constant, , of a digital low-pass filter, y = (a)*(y) + (1-a)*(x);, running in a loop with sample period, dt, can be found like this*:
a dt a 1 a dt
So if you know the desired time constant and the sample rate, you can pick the filter coefficient a. Complementary: This just means the two parts of the filter always add to one, so that the output is an accurate, linear estimate in units that make sense. After reading a bit more, I think the filter presented here is not exactly complementary, but is a very good approximation when the time constant is much longer than the sample rate (a necessary condition of digital control anyway).
*http://en.wikipedia.org/wiki/Low-pass_filter#Passive_digital_realization
A Closer Look at the Angle Complementary Filter angle = (0.98)*(angle + gyro*dt) + (0.02)*(x_acc);
Integration. Low-pass portion acting on the accelerometer.
Something resembling a high-pass filter on the integrated gyro angle estimate. It will have approximately the same time constant as the low-pass filter.
If this filter were running in a loop that executes 100 times per second, the time constant for both the low-pass and the high-pass filter would be:
## a dt 0.98 0.01sec 0.49 sec 1 a 0.02
This defines where the boundary between trusting the gyroscope and trusting the accelerometer is. For time periods shorter than half a second, the gyroscope integration takes precedence and the noisy horizontal accelerations are filtered out. For time periods longer than half a second, the accelerometer average is given more weighting than the gyroscope, which may have drifted by this point.
## A Closer Look at the Angle Complementary Filter
For the most part, designing the filter usually goes the other way. First, you pick a time constant and then use that to calculate filter coefficients. Picking the time constant is the place where you can tweak the response. If your gyroscope drifts on average 2 per second (probably a worst-case estimate), you probably want a time constant less than one second so that you can be guaranteed never to have drifted more than a couple degrees in either direction. But the lower the time constant, the more horizontal acceleration noise will be allowed to pass through. Like many other control situations, there is a tradeoff and the only way to really tweak it is to experiment. Remember that the sample rate is very important to choosing the right coefficients. If you change your program, adding a lot more floating point calculations, and your sample rate goes down by a factor of two, your time constant will go up by a factor of two unless you recalculate your filter terms. As an example, consider using the 26.2 msec radio update as your control loop (generally a slow idea, but it does work). If you want a time constant of 0.75 sec, the filter term would be:
## 0.75 sec 0.966 dt 0.75 sec 0.0262 sec
So, angle = (0.966)*(angle + gyro*0.0262) + (0.034)*(x_acc);. The second filter coefficient, 0.034, is just (1 - 0.966).
## A Closer Look at the Angle Complementary Filter
Its also worthwhile to think about what happens to the gyroscope bias in this filter. It definitely doesnt cause the drifting problem, but it can still effect the angle calculation. Say, for example, we mistakenly chose the wrong offset and our gyroscope reports a rate of 5 /sec rotation when it is stationary. It can be proven mathematically (I wont here) that the effect of this on the angle estimate is just the offset rate multiplied by the time constant. So if we have a 0.75 sec time constant, this will give a constant angle offset of 3.75. Besides the fact that this is probably a worst-case scenario (the gyro should never be that far offset), a constant angle offset is much easier to deal with than a drifting angle offset. You could, for example, just rotate the accelerometer 3.75 in the opposite direction to accommodate for it.
## Enough theory. Time for some experimental results.
Control Platform: Data Acquisition: Gyroscope: Accelerometer: Custom PIC-based wireless controller, 10-bit ADCs. Based on the Machine Science XBoard*. Over a serial USB radio, done in Visual Basic. ADXRS401, Analog Devices iMEMS 75 /sec angular rate sensor ADXL203, Analog Devices iMEMS 2-axis accelerometer
*http://www.machinescience.org
## Time Constant: 0.62 sec
Notice how the filter handles both problems: horizontal acceleration disturbances while not rotating (highlighted blue) and gyroscope drift (highlighted red).
## Time Constant: 0.58 sec
Two things to notice here: First, the unanticipated startup problem (blue highlight). This is what can happen if you dont initialize your variables properly. The long time constant means the first few seconds can be uncertain. This is easily fixed by making sure all important variables are initialized to zero, or whatever a safe value would be. Second, notice the severe gyro offset (red highlight), about 6 /sec, and how it creates a constant angle offset in the angle estimate. (The angle offset is about equal to the gyro offset multiplied by the time constant.) This is a good worst-case scenario example.
Conclusions
I think this filter is well-suited to D.I.Y. balancing solutions for the following reasons: 1. It seems to work. The angle estimate is responsive and accurate, not sensitive to horizontal accelerations or to gyroscope drift. 2. It is microprocessor-friendly. It requires a small number of floating-point operations, but chances are you are using these in your control code anyway. It can easily be run in control loops at or above 100 Hz. 3. It is intuitive and much easier to explain the theory than alternatives like the Kalman filter. This might not have anything to do with how well it works, but in educational programs like FIRST, it can be an advantage.
Before I say with 100% certainty that this is a perfect solution for balancing platforms, Id like to see it tested on some hardwareperhaps a D.I.Y. Segway? Also, Im not sure how much of this applies to horizontal positioning. I suspect not much: without gravity, there is little an accelerometer can do to give an absolute reference. Sure, you can integrate it twice to estimate position, but this will drift a lot. The filtering technique, though, could be implemented with a different set of sensors maybe an accelerometer and an encoder set but the scenario is not exactly analogous. (Encoders are not absolute positioning devicesthey can drift too if wheels lose traction. A better analogy for horizontal positioning would be using GPS to do the long-term estimate and inertial sensors for the short-term integrations.) | 3,348 | 14,826 | {"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} | 3.03125 | 3 | CC-MAIN-2019-39 | latest | en | 0.899428 |
http://toughstem.com/problem-answer-solution/2458/soccer-ball-mass-initially-moving-speed-soccer-player-kicks-ball-exerting-constant | 1,638,320,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00291.warc.gz | 80,755,914 | 19,076 | ToughSTEM
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A soccer ball with mass .42 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of 40N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?
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Using the work energy theorem:
m = 0.42 kg
vi = 2 m/s
vf = 6 m/s
F = 40 N
work = F.d = delta_KE
delta_KE = (1 / 2) * m * vf ^ 2 - (1 / 2) * m * vi ^ 2
delta_KE = 6.72 J
d = delta_KE / F
d=6.72/40
d = 0.168 m
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Almost done! | 370 | 1,078 | {"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} | 3.4375 | 3 | CC-MAIN-2021-49 | longest | en | 0.888127 |
https://www.physicsforums.com/threads/derivative-of-an-operator-valued-function.576671/ | 1,527,072,456,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00142.warc.gz | 822,200,541 | 19,636 | # Derivative of an operator valued function
1. Feb 11, 2012
### giova7_89
If I have a function
$$f:R\rightarrow L$$
where L is the space of linear operators from an hilbert space to itself, how can i define the derivative of f at a particular point of R? I mean, it is "obvious" that one should try:
$$f'(s_0)=lim_{\Delta s\rightarrow0}\frac{|f(s_0 + \Delta s) -f(s_0)|}{\Delta s}$$
but i'm concerned with the fact that when in the numerator i use ||, i mean the norm on L, which as i said is the space of linear operators from an hilbert space to itself. Since in physics one often deals with operators that have infinite operator norm, I wanted to know if the norm of that difference of operators is finite or not, even if we take unbounded operators.
It is a mathematical question, but since in QM one encounters these things all the time, I wanted some clarifications..
thanks!
2. Feb 11, 2012
### Staff: Mentor
Yea I was going to suggest to define the norm on the space of operators by the following inner product of operators (A,B) = Trace (Abar B) - here Abar means the adjoint which reduces to Trace (AB) for the usual Hermtian operators in QM.
The above is called the Frobenius norm (older texts refer to it as the Hilbert–Schmidt norm or the Schur norm).
But ouch - even the identity operator has an infinite norm.
Its probably better to use the operator norm:
http://en.wikipedia.org/wiki/Operator_norm
But even then I don't know how to guarantee it being finite.
Thanks
Bill
Last edited: Feb 12, 2012
3. Feb 12, 2012
### Bill_K
I would think the "obvious" thing to do is to take the ||'s off and just let f'(s) be an operator.
4. Feb 12, 2012
### giova7_89
Ok, one doesn't need to put || in THAT definition, and I know that f'(s_0) is an operator, what I meant was that when i use the usual $$\epsilon,\delta$$
definition of limit, the norm on L will surely pop out: I mean that f'(s_0) is the operator that does this thing:
$$\forall\epsilon>0\exists\delta>0:if\,\,0<|s-s_0|<\delta\Rightarrow\bigg|\frac{f(s) - f(s_0) - (s-s_0)f'(s_0)}{s-s_0}\bigg|_L<\epsilon$$
where the $$||_L$$ is the norm on L. Here I wanted to know if there are any problems if i'm dealing with unbounded operators. I admit I made a mistake, and in my first post the || were not needed, but they surely are in this definition (which is that of the differential of a function from a vector space to another, i think)
5. Feb 12, 2012
### giova7_89
In the original post, things should look like this:
$$f'(s_0)$$
is an operator such that:
$$lim_{\Delta s\rightarrow 0}\frac{|f(s_0+\Delta s) - f(s_0) - \Delta s\,f'(s_0)|_L}{\Delta s} = 0$$
The oroginal post was wrong since it compared a number to an operator
6. Feb 12, 2012
### strangerep
For bounded operators, one can use the supremum norm (aka operator norm), which bhobba mentioned already: http://en.wikipedia.org/wiki/Operator_norm
For unbounded operators, one can talk in terms of densely-defined operators on an ordinary Hilbert space, or else pass to a rigged Hilbert space and use the theory of generalized functions (distributions). But for that, one must first understand how operators on (say) the Schwartz space are extended to operators on its dual space of tempered distributions via the dual pairing. (If that all sounds like gobble-de-gook, I could probably post some introductory references. It depends what you're really trying to do.)
7. Feb 13, 2012
### giova7_89
My knowledge about this is very limited.. So i'd be glad if you could post some references. I just started a course on quantum field theory and since (but it also happens in the mathematical formalism of quantum mechanics) we're dealing with operator valued functions, I wanted to know more about the mathematics behind that..
So, if you know an introductory book about the mathematics of QFT, feel free to tell me about that, too
8. Feb 13, 2012
### Fredrik
Staff Emeritus
If you want to learn this stuff, you have to start with topology and functional analysis. Kreyszig's book on functional analysis covers the topology you need as well, so it's much easier to read than many of the others. (I haven't read it myself, but I know how crazy hard some other books are).
9. Feb 13, 2012
### lugita15
I second the recommendation of Kreyszig. It's probably the only functional analysis book that assumes so little background and yet covers so much ground.
10. Feb 13, 2012
### Staff: Mentor
I also recommend Kreyszig and I am thinking of actually purchasing a copy.
I also recommend the following book on Distribution Theory that I have in my library. It helped me a lot with this sort of stuff:
https://www.amazon.com/Theory-Distr...=sr_1_1?s=books&ie=UTF8&qid=1329180440&sr=1-1
You can usually define what you want on a test space then use distribution theory extend it. And if you are doing QFT then the above IMHO is a must - they are really distributions rather than functions.
Thanks
Bill
11. Feb 14, 2012
### strangerep
Yeah, I was going to say something about how basic stuff on distributions and generalized functions is probably more important to understand clearly (for QM/QFT purposes) before tackling Functional Analysis books (which usually don't cover generalized functions anyway). I didn't know a good basic book on distributions and generalized functions, so I'll take at look at the one you suggested (Richards & Youn) and see what it's like.
Cheers.
12. Feb 14, 2012
### Tarantinism
The space of BOUNDED operators on a Hilbert space is itself a Banach space, so it is easy to define the usual Fréchet derivative:
http://en.wikipedia.org/wiki/Fréchet_derivative
I wouldn't know how to define this derivative, on the other hand, if you map into some unbounded operators, typical in physics.
13. Feb 14, 2012
### strangerep
Gel'fand triples, i.e., using topological duals of a suitable nuclear space.
14. Feb 15, 2012
### lugita15
Can you spell this out in more detail? Are you taking the Frechet derivative of an unbounded operator on the dense nuclear space, and then doing ... something with the dual pairing between the nuclear space and the space of continuous antilinear functionals?
15. Feb 15, 2012
### strangerep
Well, it can't be the standard Frechet derivative, since that's for Banach spaces, but the small nuclear space $\Omega$ in a Gel'fand triple
$$\Omega ~\subset~ \bar{\Omega} ~\subset~ \Omega' ~,~~~~ \Big(\mbox{where}~ \Omega' ~\mbox{is the topological dual of}~ \Omega\Big) ~.$$
is not complete, hence not Banach.
But here's my (quick) understanding of what's going on. (Warning: I could be wrong about parts of this, in which case I hope someone more knowledgeable will say so.)
Let $\phi$, $\Psi$ be arbitrary vectors in $\Omega, \Omega'$ respectively. Let A be an operator defined everywhere on $\Omega$. Then by standard theorems (cf. Gelfand & Vilenkin vol 4) it can be extended uniquely to an operator A' on $\Omega'$ via the dual pairing
$$(\phi, A'\Psi) ~:=~ (A\phi, \Psi)$$
such that A' coincides with A when acting on elements of $\Omega$.
Now let A(t) be an operator-valued function of a real parameter t such that A(t) is defined everywhere on $\Omega$. Then, by the above construction it can be extended to $\Omega'$.
If, in addition, there is an operator B such that
$$\frac{A(t+h) - A(t) - B}{h} ~\rightarrow~ 0 ~~~~~~~~ (1)$$
in weak topology, we can call B a derivative of A at t. By duality, we can extend B to all of $\Omega'$.
The subtleties lie in what it means for the LHS of (1) to approach 0 weakly. The standard meaning for weak operator topology is that a sequence of operators $\{T_n\}$ converges weakly to $T$ iff $T_n \phi \to T \phi$ for all $\phi$.
(Once again: caveat emptor.)
16. Feb 15, 2012
### Staff: Mentor
He is talking about Rigged Hilbert Spaces which is an extension of distribution theory - probably best to get the book I recommended on distribution theory first but do check out:
http://www.abhidg.net/RHSclassreport.pdf [Broken]
Thanks
Bill
Last edited by a moderator: May 5, 2017 | 2,150 | 8,060 | {"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} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.940853 |
https://teamtreehouse.com/videos/111023/captions | 1,638,294,811,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-49/segments/1637964359065.88/warc/CC-MAIN-20211130171559-20211130201559-00044.warc.gz | 639,456,678 | 2,647 | 1 00:00:00,372 --> 00:00:02,543 Great job getting set up. 2 00:00:02,543 --> 00:00:06,343 Now it's time to challenge you to make some scatter plots. 3 00:00:06,343 --> 00:00:10,956 What is the relationship between Level and Attack Points and 4 00:00:10,956 --> 00:00:13,043 Level and Defense Points? 5 00:00:13,043 --> 00:00:20,518 Bonus: use sns.relplot(kind='scatter') to solve one of these questions. 6 00:00:20,518 --> 00:00:24,661 The seaborn library has many functions to achieve the same plots. 7 00:00:24,661 --> 00:00:26,865 For each of the following challenges, 8 00:00:26,865 --> 00:00:30,630 I'll drop a bonus hint into which ones you can use. 9 00:00:30,630 --> 00:00:35,732 You can find all of these functions in the seaborn API documentation. 10 00:00:35,732 --> 00:00:38,102 Now it's time to take the challenge. 11 00:00:38,102 --> 00:00:41,944 Pause me and try it out on your own. 12 00:00:41,944 --> 00:00:46,685 What is the relationship between Level and Attack Points? 13 00:00:46,685 --> 00:00:50,019 To draw a scatterplot, 14 00:00:50,019 --> 00:00:57,192 we can call sns.scatterplot(data=monsters, 15 00:00:57,192 --> 00:01:03,374 x='Level', y='Attack Points'). 16 00:01:06,365 --> 00:01:10,889 Remember that these parameters can be found in 17 00:01:10,889 --> 00:01:16,342 the previous cell when we called monsters.head(). 18 00:01:16,342 --> 00:01:19,886 They are the headings to each of these columns. 19 00:01:22,532 --> 00:01:24,581 And it looks like I'll have a typo here. 20 00:01:24,581 --> 00:01:29,355 I need Attack_Points. 21 00:01:34,223 --> 00:01:39,914 Nice, there seems to be a general positive correlation between Level and 22 00:01:39,914 --> 00:01:41,211 Attack Points. 23 00:01:41,211 --> 00:01:46,402 As Level gets higher, so do Attack Points. 24 00:01:46,402 --> 00:01:48,433 How about for Defense Points? 25 00:01:48,433 --> 00:01:54,662 Let's do one with the bonus hint, sns.relplot. 26 00:01:54,662 --> 00:01:59,297 In Jupyter Lab, you can use the shift tab keyboard combo on a function to open 27 00:01:59,297 --> 00:02:04,383 the method signature documentation to get more information about that function. 28 00:02:04,383 --> 00:02:08,514 It looks like this function is set up similarly to 29 00:02:08,514 --> 00:02:13,369 the scatterplot function where we have data, x, and y. 30 00:02:15,513 --> 00:02:20,831 When I scroll down, there's documentation about what relplot is used for. 31 00:02:20,831 --> 00:02:24,151 Feel free to read through this. 32 00:02:24,151 --> 00:02:29,322 It's the same as what is in the seaborn API documentation from their website. 33 00:02:29,322 --> 00:02:33,262 Let's use this function, 34 00:02:33,262 --> 00:02:40,800 sns.relplot(data=monsters, x='Level', 35 00:02:40,800 --> 00:02:48,347 y='Defense_Points'), with an underscore. 36 00:02:52,873 --> 00:02:58,105 The default kind for the relplot is a scatter plot, 37 00:02:58,105 --> 00:03:03,714 but we can be explicit by setting kind='scatter'. 38 00:03:03,714 --> 00:03:05,931 Let's run the cell. 39 00:03:05,931 --> 00:03:09,487 And we've got a scatterplot that shows a general 40 00:03:09,487 --> 00:03:13,746 positive correlation between Level and Defense Points. 41 00:03:13,746 --> 00:03:20,213 Generally, as Level increases, so do the Defense Points. 42 00:03:20,213 --> 00:03:23,993 Awesome, are you ready for the next challenge? 43 00:03:23,993 --> 00:03:24,920 I'll catch you there. | 1,254 | 3,429 | {"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} | 2.796875 | 3 | CC-MAIN-2021-49 | latest | en | 0.888322 |
https://www.groundai.com/project/euclidean-triangles-have-no-hot-spots/ | 1,558,328,373,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255562.23/warc/CC-MAIN-20190520041753-20190520063753-00114.warc.gz | 778,658,077 | 66,007 | No hot spots
# Euclidean Triangles have no hot spots
## Abstract.
We show that a second Neumann eigenfunction of a Euclidean triangle has at most one (non-vertex) critical point , and if exists, then it is a non-degenerate critical point of Morse index . Using this we deduce that
1. the extremal values of are only achieved at a vertex of the triangle, and
2. a generic acute triangle has exactly one (non-vertex) critical point and that each obtuse triangle has no (non-vertex) critical points.
This settles the ‘hot spots’ conjecture for triangles in the plane.
###### Key words and phrases:
Hot spots, Laplace operator, Neumann eigenfunctions
###### 1991 Mathematics Subject Classification:
35P05, 35B38, 35J05, 35J25, 58J50
The work of C.J. is partially supported by a Simons collaboration grant.
## 1. Introduction
Let be a domain in Euclidean space with Lipschitz boundary. The second Neumann eigenvalue, , is the smallest positive number such that there exists a not identically zero, smooth function that satisfies
(1) Δu = μ2⋅u \ and \ ∂u∂n∣∣∣∂Ω≡ 0
where denotes the outward pointing unit normal vector field defined at the smooth points of . A function that satisfies (1) will be called a second Neumann eigenfunction for , or simply a -eigenfunction.
One variant of the ‘hot spots’ conjecture, first proposed by J. Rauch at a conference in 1974,1 asserts that a second Neumann eigenfunction attains its extrema at the boundary. The main result of this paper implies the hot spots conjecture for triangles in the plane.
###### Theorem 1.1.
If is a second Neumann eigenfunction for a Euclidean triangle , then has at most one critical point.2 Moreover, if has a critical point , then lies in a side of and is a nondegenerate critical point with Morse index equal to .
In Theorem 12.4, we show that if is a generic acute triangle, then has exactly one critical point, and that if is an obtuse triangle, then has no critical points. Earlier, Bañuelos and Burdzy showed that if is obtuse, then has no interior maximum, and, in particular, the maximum and minimum values of are achieved at the acute vertices [Bnl-Brd99]. We extend the latter statement to all triangles (see Theorem 12.1). Unlike [Bnl-Brd99], our proof of Theorem 1.1 does not rely on probabilistic techniques.
For a brief history and various formulations of the ‘hot spots’ conjecture, we encourage the reader to consult [Bnl-Brd99]. We provide some highlights. The first positive result towards this conjecture was due to Kawohl [Kwl85] who showed that the conjecture holds for cylinders in any Euclidean space. Burdzy and Werner in [Brd-Wrn99] (and later Burdzy in [Brd05]) showed that the conjecture fails for domains with two (and one) holes. In the paper [Brd05] Burdzy made two separate (‘hot spot’) conjectures for ‘convex’ and ‘simply connected’ domains. We believe that the conjecture is true for all convex domains in the plane.
The conjecture has been settled for certain convex domains with symmetry. In 1999, under certain technical assumptions, Bañuelos and Burdzy [Bnl-Brd99] were able to handle domains with a line of symmetry. A year later Jerison and Nadirashvili [Jrs-Ndr00] proved the conjecture for domains with two lines of symmetry. In a different direction, building on the work in [Bnl-Brd99], Atar and Burdzy [Atr-Brd04] proved the conjecture for lip domains (a domain bounded by the graphs of two Lipschitz functions with Lipschitz constant 1). In 2012, the hot spots conjecture for acute triangles became a ‘polymath project’ [Polymath]. In 2015 Siudeja [Sdj15] proved the conjecture for acute triangles with at least one angle less than by sharpening the ideas developed by Miyamoto in [Mym09, Mym13]. Notably, in the same paper, Siudeja proved that the second Neumann eigenvalue of an acute triangle is simple unless is an equilateral triangle. An earlier theorem of Atar and Burdzy [Atr-Brd04] gave that the second Neumann eigenvalue of each obtuse and right triangle is simple.
Our approach to the conjecture differs from most of the previous approaches (but has some features in common with the approach in [Jrs-Ndr00]). For each acute or obtuse triangle, , we consider a family of triangles that joins to a right isosceles triangle . Using the simplicity of (due to [Atr-Brd04], [Mym13] and [Sdj15]) we then consider a family of second Neumann eigenfunctions associated to .3 Because is the right isosceles triangle, the function is explicitly known up to a constant, and a straightforward computation shows that has no critical points (see equation (21)). Therefore, if were to have a critical point, then it would have to somehow ‘disappear’ as tends to 1. Each nondengenerate critical can not disappear immediately, that is, it is ‘stable’. On the other hand, a degenerate critical point can instantaneously disappear, that is, it could be ‘unstable’. Thus, as varies from to , either a critical point of converges to a vertex or is or becomes degenerate and then disappears. Understanding the first case, among the last two possibilities, is more or less straightforward, and we do it by studying the expansion of in terms of Bessel functions near each vertex. Understanding the second case is more complicated. One particular reason for this complication is that disappearance of this type probably does occur for perturbations of general domains.
The study of how eigenvalues and eigenfunctions vary under perturbations of the domain is a classical topic (see for example [Kato]). Jerison and Nadirashvilli [Jrs-Ndr00] considered one-parameter families of domains with two axes of symmetry and studied how the nodal lines of the directional derivatives of the associated eigenfunctions varied. In particular, they used the fact that each constant vector field commutes with the Laplacian, and hence if is an eigenfunction, then is also an eigenfunction with the same eigenvalue. The eigenfunctions were also used in [Sdj15] and implicitly in [Bnl-Brd99] and [Atr-Brd04]
In the current paper, we consider the vector field , called the rotational vector field, that corresponds to the counter-clockwise rotational flow about a point . To be precise if , then
Rp = −(y−p2)⋅∂x + (x−p1)⋅∂y.
We will call the angular derivative of about . Each rotational vector field commutes with the Laplacian, and hence the angular derivative is an eigenfunction. By studying the nodal sets of where is a vertex of the triangle, one finds that if has an interior critical point, then also has a critical point on each side of the triangle (see Corollary 6.2). Moreover, we show that each of these three critical points is stable under perturbation even though might not be stable (see Proposition 9.4). We also use the nodal sets of both and , where is parallel to the side of , to show that, although a degenerate critical point of might not be stable under perturbation, there are at least two other critical points that are stable under perturbation (see Proposition 9.5).
### Outline of the paper
In §2, we recall Cheng’s [Chn76] theorem concerning the structure of the nodal set of an eigenfunction on surfaces. From a result of Lojasiewicz [Ljs59] it follows that the critical set of each eigenfunction is a disjoint union of isolated points and analytic one-dimensional manifolds. In §3, we consider domains obtained from a triangle via reflecting about its sides. By applying Cheng’s structure theorem to the extension of an eigenfunction to these extended polygonal domains, we obtain a qualitative result concerning nodal arcs whose endpoints lie in a side of the triangle. In §4, we consider the Bessel expansion of a Neumann eigenfunction on a sector. Using the radial and angular derivatives of this expansion, we obtain a qualitative description of the critical set of a Neumann eigenfunction on a sector. We use this lower estimate in §5 to prove that the critical set of a second Neumann eigenfunction on a triangle is finite. There we also (re)prove the fact that the nodal set of is a simple arc, and use this fact to obtain information about the first two Bessel coefficients of at the vertices of . For example, we deduce that can vanish at only one vertex of . In §6, we study the nodal set of both the angular derivatives, , about vertices and the directional derivatives, , parallel to an edge . We show that each component of each of these nodal sets is a finite tree, and use this to obtain information about the critical set of . For example, we show that if has an interior critical point then it has at least three more critical points, one critical point per side (Corollary 6.2), and if has a degenerate critical point on a side , then has a critical point on a side distinct from (Theorem 6.5).
In §8, we begin the proof of Theorem 1.1. Given an obtuse or (non-equilateral) acute triangle , we consider a ‘straight line path’ of triangles that joins to a right isosceles triangle and an associated path of second Neumann eigenfunctions. In §8, we suppose converges to and consider the accumulation points of a sequence where each is a critical points of . Using the Bessel expansion of , we find that if each lies in the interior of , then a vertex is not an accumulation point of . We also show that if each lies in a side and a vertex is an accumulation point of , then there does not exist a sequence of critical points lying in a distinct side that has as an accumulation point.
In §9, we address the issue of the stability of critical points. We regard a critical point of as ‘stable’ if for each neighborhood of the function has a critical point in for sufficiently close to . Non-degenerate critical points are stable, but, in general, degenerate critical points are not. Nonetheless, we use the results of §6 to show that if is a degenerate critical point of , then has at least two stable critical points. In §10, we use the existence of these two stable critical points to show that, if has an interior critical point, then also has at least two critical points for each that is near . In contrast, the eigenfunction for the right isosceles triangle has no critical points, and thus, to prove Theorem 1.1 for acute and obtuse triangles, it suffices to show that the number of critical points can not drop from two to zero in the limit as tends to one. This is accomplished by using the results of §8 and certain elementary properties of .
To make the exposition of the proof of Theorem 1.1 easier, we use the known simplicity of the second Neumann eigenvalue [Bnl-Brd99] [Atr-Brd04] [Mym13], [Sdj15]. However, we indicate in §11 how to avoid this assumption.
In §12, we use the topology of the nodal sets of the extension of to the double of the triangle to show that has a critical point if and only if each vertex is an isolated local extremum of . In particular, if is associated to an acute triangle, then has a critical point if and only if does not vanish at any of the vertices. In the final part of §12, we consider the parameter space of all labeled triangles up to homothety. Using analytic perturbation theory and Hartog’s separate analytic theorem we deduce that the Bessel coefficients of a second Neumann eigenfunction (at a labeled vertex) can be thought of (in a suitable sense) as analytic functions on a dense open subset of . Using this fact, we deduce that a generic acute triangle has exactly one critical point and obtuse triangles have no critical points.
###### Notation and terminology.
For notational convenience, we will regard the Euclidean plane as the complex plane. That is, we will use to represent a point in the plane. In particular, and , and if then and . We will also use to denote the set of such that , and to denote the interior of a set . For us, a Laplace eigenfunction is a smooth real valued solution to the equation where and . We will sometimes call such a solution a -eigenfunction.
###### Acknowledgments.
We thank Neal Coleman for producing contour plots of eigenfunctions in triangles. In particular, he created a very inspirational animation of the ‘straight-line’ family of triangles joining a triangle with labeled angles to a triangle with labeled angles . See https://youtu.be/bO50jFOxCAw. He created these contour plots with his ‘fe.py’ python script [Clm16]. We also thank David Jerison and Bartlomiej Siudeja for comments on the first version of the paper.
## 2. The nodal set and the critical set of an eigenfunction
Let be an open set, and let be an eigenfunction of the Laplacian. In this section, we recall some facts about the nodal set and the set, , of critical points of . The intersection is the set of nodal critical points.
The following is a special case of the stratification of real-analytic sets due to Lojasiewicz [Ljs59]. An elementary proof can be found in the proof of Proposition 5 in [Otl-Rss09].
###### Lemma 2.1.
Let be an open subset of . If is a real-analytic function, then each has a neighborhood such that is either equal to or is homeomorphic to a properly embedded finite graph. Moreover, if , then is a real-analytic arc.
Because the Laplacian is a constant coefficient elliptic operator, the eigenfunction is real-analytic function. Therefore, it follows from Lemma 2.1 that is a locally finite graph whose vertices are the nodal critical points, and the complement of these vertices is a disjoint union of real-analytic loops and arcs. Cheng observed [Chn76] that (in dimension 2) the nodal set has a special structure in a neighborhood of each nodal critical point.
###### Lemma 2.2 (Theorem 2.5 in [Chn76]).
Let be an eigenfunction of the Laplacian on an open set . If is a nodal critical point, then there exist a neighborhood of , a positive integer , a real number , and simple arcs , such that
1. ,
2. equals , and
3. for each , the arc is tangent at to the line .
###### Remark 2.3.
Arcs satisfying condition (3) of Lemma 2.2 are called equiangular.
###### Sketch of proof.
Without loss of generality . The Taylor series of about may be regarded as a sum of homogeneous polynomials of degree in and . Because is a nodal critical point, and vanish identically. Since is an eigenfunction and maps homogeneous polynomials of degree to homogeneous polynomials of degree , we have . In particular, if is the smallest such that , then . Thus, where is a harmonic polynomial of degree at least and denotes a sum of terms of degree at least . The restriction of the harmonic polynomial to the unit circle centered at is a Laplace eigenfunction with eigenvalue , and so since is homogeneous, the nodal set of equals the union of lines for some . One obtains the claim by applying the method of [Kuo69]. See Lemma 2.4 in [Chn76]. ∎
As a consequence of Lemma 2.2, the nodal set is the union of loops and proper4 arcs. We will call these the Cheng curves of .
We shall be interested in whether certain Cheng arcs cross a line or not. To make this precise, we note the following.
###### Lemma 2.4.
Let be a Cheng curve in and let be an intersection point of and a line . There exists an open neighborhood of and a parameterization of such that and either
1. the sets and lie in different components of ,
2. the sets and lie in the same component of or
3. the curve lies in the component of that contains .
In case (1), we say that the curve crosses the line .
###### Proof.
The restriction of to is a real-analytic function on . We have case (3) if and only if this restriction vanishes identically on the component containing . If it it does not vanish identically, then there exists a neighborhood of such that contains no zeros of the restriction other than . Choose a parameterization of so that and do not intersect . ∎
The set, , of critical points has the following description parallel to that of the nodal set .
###### Proposition 2.5.
Let be a Laplace eigenfunction on an open set . Each connected component of is either
1. an isolated point,
2. a proper real-analytic arc, or
3. a real-analytic curve that is homeomorphic to a circle.
###### Proof.
The function is analytic, and hence by Lemma 2.1 each critical point is either isolated or lies in a component of that is a locally finite graph.
Let be a component of the graph . If for some , then since is connected and on , it would follow that . By Lemma 2.2, the set of nodal critical points is discrete, and hence would consist of an isolated point.
If , either or . Without loss of generality, we may assume that , and hence . Therefore, the analytic implicit function theorem provides a neighborhood of such that is a real-analytic arc . The set lies in .
By Lemma 2.1, the set is either finite, and hence is an isolated point, or is a proper finite graph. In the latter case . Since is arbitrary, the component is a real-analytic 1-manifold (without boundary). If is compact, then is homeomorphic to a circle. Otherwise, there exists a possibly infinite open interval and a real-analytic unit speed parameterization . Since is closed in , the map is proper. ∎
## 3. Eigenfunctions on triangles, kites, and hexagons
In this section we consider eigenfunctions on the triangle that satisfy Neumann conditions along at least one of the sides of . Let be a side of , and let denote the reflection across the line containing . Following [Sdj15], we define the kite to be the closed set . If is an eigenfunction of the Laplacian that satisfies Neumann conditions along , then extends uniquely to a real-analytic Neumann eigenfunction on the kite such that . Note that whenever we refer to the nodal set of an extended eigenfunction, we are speaking of the nodal set in the extended domain.
If is an eigenfunction that satisfies Neumann conditions on all three sides, then we will find it useful to reflect about all three sides simultaneously. If some angle of is greater than , then one might not be able to extend to the union of the three kites unambiguously. But one may use a ‘smaller’ extension. For example, the bisectors of each angle of the triangle meet at the centroid to form a tripod. This tripod divides the triangle into three smaller triangles each of which contains exactly one edge of . Reflect each of these smaller triangles about the corresponding edge to obtain a ‘hexagon’ containing . The Neumann eigenfunction on extends uniquely via the reflection principle to a Laplace eigenfunction on .
Let be an eigenfunction of the Laplacian on the interior of that extends continuously to the vertices of . In this article will equal or where is either a constant or rotational vector field. By Lemma 2.2, the nodal set of is a union of curves where each curve is either homeomorphic to a circle (a ‘loop’) or is a proper arc. Recall that each such curve is called a Cheng curve of .
###### Definition 3.1.
Let be a Cheng curve of . The closure of a component of the intersection will be called a maximal subset of the nodal set of the restriction of to .
The nodal set of the restriction is a union of maximal subsets. Each maximal subset in the nodal set of is either a point, a loop5, or a arc with distinct endpoints in . Each intersection of such loops/arcs is equiangular (see Remark 2.3). If a maximal subset is homeomorphic to an interval, then we will call it a maximal arc.
If a maximal subset consists of a single point, then this point lies in a side of . Indeed, the nodal set of an eigenfunction defined on an open set has no isolated points. For the same reason, if satisfies Neumann or Dirichlet conditions along a side of , then contains no singleton maximal subsets. In particular, the nodal set of the Neumann eigenfunction is a union of maximal loops and maximal arcs. Each vertex of the graph is thus either a critical point of , an endpoint of a maximal arc, or an isolated point of .
The following should be compared to Lemma 5 in [Sdj15].
###### Lemma 3.2.
Let be a triangle. Let be an eigenfunction on that satisfies Neumann conditions along the side . If a piecewise smooth arc in has both endpoints in , then the eigenvalue of is strictly greater than the second Neumann eigenvalue of .
###### Proof.
The maximal arc and the side together bound a topological disc . Define by setting if and otherwise. The function satisfies Neumann conditions along and Dirichlet conditions along the other two sides of . Hence the eigenvalue, , of is larger than the first eigenvalue of the mixed eigenvalue problem on corresponding to Neumann conditions on and Dirichlet conditions on the other two sides. In turn, by Theorem 3.1 in [Ltr-Rhl17], the first eigenvalue of the mixed problem is greater than the second Neumann eigenvalue of . ∎
## 4. Neumann eigenfunctions on sectors
Let be a sector of angle and radius , that is
Ω: = {z:0≤arg(z)≤β and |z|<ϵ}.
In this section, is a (real) eigenfunction of the Laplacian on with eigenvalue that satisfies Neumann boundary conditions along the boundary edges corresponding to respectively. (We impose no conditions on the circle of radius .) We will use the expansion of in Bessel functions near the ‘vertex’ , to derive information about both the nodal set and the critical set of .
Separation of variables leads to the following expansion valid near :
(2) u(reiθ) = ∞∑n=0cn⋅Jnπβ(√μ⋅r)⋅cos(nπθβ).
Here and denotes the Bessel function of the first kind of order [Lbv72]. The series converges uniformly on compact sets that miss the origin. The Bessel function has the expansion [Lbv72]
Jν(r) = rν⋅∞∑k=0(−1)k⋅r2k22k⋅Γ(k+ν)⋅Γ(k+ν+1)
where is the Gamma function. In particular, for each , there exists an entire function so that .6 Note that none of the Taylor coefficients of vanish. In particular, neither nor vanishes in a neighborhood of for each . With this notation, the expansion in (2) takes a more compact form:
(3) u(reiθ) = ∞∑n=0cn⋅rn⋅ν⋅gn⋅ν(r2)⋅cos(n⋅ν⋅θ)
where .
We will be interested in the level set, , that contains the vertex . In particular, if , then is the nodal set of .
###### Lemma 4.1.
There exists a neighborhood of such that either equals or equals the union of real-analytic arcs such that the pairwise intersection of and equals for each .
###### Proof.
By expanding each , the expansion in (3) becomes
(4) u(reiθ) = ∞∑n=0∞∑j=0cn⋅an,j⋅rn⋅ν+2j⋅cos(n⋅ν⋅θ)
where each is nonzero. We have . Let and let . Then
(5) u(z)−u(0)rη = ∑(n,j)∈Acn⋅an,j⋅cos(n⋅ν⋅θ) + h(z)
where is a real-analytic function and both and are of order as tends to zero for some . The claim follows from the implicit function theorem. ∎
We will require more specialized information about the level sets that contain a vertex of a triangle when the vertex angle is acute or obtuse.
###### Lemma 4.2.
If the angle , then there exists a neighborhood of such that
1. if , then equals , and
2. if and , then is a simple arc containing .
If , then there exists a neighborhood of such that
1. if , then is a simple arc containing , and
2. if and , then
If , then there exists a neighborhood of such that consists of at least two arcs.
###### Proof.
Suppose . If , then defined in Lemma 4.1 equals and . In particular, the trigonometric polynomial appearing on the right hand side of (5) is a constant and hence . On the other hand, if and , then and . In this case, the trigonometric polynomial of the right hand side of (5) equals , and hence is a simple arc.
Suppose . If , then and . Thus, the trigonometric polynomial equals , and is an arc. On the other hand, if and , then the trigonometric polynomial is a constant and hence .
Finally, if , then each term in the trigonometric polynomial in (5) is the product of a constant and where and . Such a function has at least two roots. ∎
###### Proposition 4.3.
If is not an integer multiple of , then there exists a deleted neighborhood of that contains no critical points of . If , then there exists a neighborhood of such that is either empty, equals , or equals exactly one edge of the sector.
###### Remark 4.4.
The conditions on are necessary. For example, on the square we have the Neumann eigenfunction . In this case, the set lies in the critical set of .
###### Proof.
The point is a critical point of if and only if both the radial derivative and the angular derivative vanish at . Let be the first nonzero coefficient in the Bessel expansion in (3). By differentiating term-by-term we obtain
(6) ∂θu(z) = −∞∑n=mcn⋅rn⋅ν⋅gn⋅ν(r2)⋅n⋅ν⋅sin(n⋅ν⋅θ)
and
(7) ∂ru(z) = ∞∑n=mcn⋅rn⋅ν−1⋅(n⋅ν⋅gn⋅ν(r2)+2r2⋅g′n⋅ν(r2))⋅cos(n⋅ν⋅θ).
In particular,
(8) ∂θu(z) = −cm⋅rm⋅ν⋅gm⋅ν(r2)⋅m⋅ν⋅sin(m⋅ν⋅θ) + O(r(m+1)ν)
and
(9) ∂ru(z) = cm⋅rm⋅ν−1⋅(m⋅ν⋅gm⋅ν(r2)+2r2⋅g′m⋅ν(r2))⋅cos(m⋅ν⋅θ) + O(r(m+1)ν−1)
where represents a function defined in a neighborhood of that is bounded by a constant times .
Suppose that and . Then since , we find from (8) that . It follows that there exists so that
(10) ∣∣∣θ−km⋅β∣∣∣ = O(rν).
Suppose that . If , then , and so from (9) we find that . It follows that there exists so that
∣∣∣θ−2k+12m⋅β∣∣∣ = O(rν).
Therefore, if , there exists such that if then and can not both be zero.
If , then the term associated to in (7) might not be dominant and so (9) might not be useful. Which term is dominant depends on the value of .
If , then the term associated to is dominant, and thus does not vanish for small . If , then the term associated to is dominant, and we find that there exists so that for some . Comparison with (10) where then gives that and can not both vanish near .
If , then since satisfies Neumann conditions along the edges, we may use the reflection principle to extend to a smooth eigenfunction on the disk . By Proposition 2.5, if lies in the critical locus of , then there exists a disk neighborhood of zero such that or is a real-analytic arc . Because the extended eigenfunction is invariant under reflection across both the real and imaginary axes, the arc is also invariant under these reflections and hence lies either in the real or imaginary axis. ∎
###### Remark 4.5.
If , then the sector is a half-disk. One can apply the reflection principle to extend to the disk. Using Proposition 2.5, we find that if is a critical point, then there exists a neighborhood of such that is either , equals the real-axis, or is an arc that is orthogonal to the real-axis.
###### Remark 4.6.
If , then there exists , such that if and , then is not a critical point of . Indeed, for each , defines an analytic function on , and hence from (6) we have
∂θu(z) = −sin(ν⋅θ)⋅rν(c1⋅ν⋅gν(0)⋅sin(ν⋅θ) + O(rν′))
where . Thus, if is a critical point and then there exists such that
(11) |c1|⋅ν⋅gν(0) ≤ C⋅rν′.
and therefore .
## 5. A second Neumann eigenfunction on a Euclidean triangle
In this section, is a Euclidean triangle, and is a second Neumann eigenfunction for .
We will use the following well-known fact many times in the sequel.
###### Lemma 5.1.
Let be a subset of with piecewise smooth boundary, and let that satisfies Dirichlet boundary conditions on , that is . Then the Rayleigh quotient . In particular, if itself is a -eigenfunction on with Dirichlet boundary condition, then .
###### Proof.
By the variational characterization of the first Dirichlet eigenvalue we have . By the domain monotonicity of the first Dirichlet eigenvalue, and by a result of Polya [Ply52] we have , giving the first assertion. ∎
The following fact is also well-known.
###### Theorem 5.2.
The nodal set of consists of one simple maximal arc.
###### Proof.
By Lemma 2.2, the nodal set is a collection of loops and maximal arcs. Lemma 5.1 implies that there are no loops. By Courant’s nodal domain theorem, the complement has exactly two components. The claim follows. ∎
If is a vertex of the triangle , then an -neighborhood of can be identified with a subset of a sector. For each vertex , we consider the Bessel expansion of about , and we let denote the associated Bessel coefficient.
###### Corollary 5.3.
Let be a vertex of . The first two Bessel coefficients, and , can not both equal zero.
###### Proof.
If both were both zero, then by Lemma 4.2, there would exist (at least) two arcs in that emanate from the vertex. They could not form a loop by Lemma 5.1, and so they would have to be distinct, but this would contradict Theorem 5.2. ∎
###### Corollary 5.4.
If and are two distinct vertices of , then and can not both equal zero.
###### Proof.
Suppose to the contrary that and are both zero. Then by Corollary 5.3, the coefficients and are both nonzero. Thus, by Lemma 4.2, there would exist an arc in emanating from and an arc in emanating from . By Theorem 5.2 these arcs would belong to the same maximal arc in that joins and . This would contradict Lemma 3.2. ∎
The following is a consequence of a more general result of [Ndr86], but it follows easily from the previous corollary.
###### Corollary 5.5.
The dimension of the space of second Neumann eigenfunctions is at most two.
###### Proof.
Define the linear map by where and are distinct vertices of . By Corollary 5.4, the map has no kernel, and so the dimension of is at most two. ∎
###### Proposition 5.6.
The critical set of a second Neumann eigenfunction is finite.
###### Proof.
The Neumann eigenfunction extends via reflection to an eigenfunction on the interior of the ‘hexagon’ described in §3. By Proposition 2.5, each component of is either an isolated point, a proper analytic arc, or an analytic loop. It follows that each component of the critical set of is either an analytic arc with points in the boundary of , a loop in , or an isolated point in where is the set of vertices.
If were a loop, then each directional derivative, for example , would be a Dirichlet eigenfunction on the region bounded by the loop, contradicting Lemma 5.1.
If were an arc, then the endpoints of the arc lie in the union of two sides. If is the rotational vector field about the common vertex of these two sides, then is a Dirichlet eigenfunction on a subdomain of . This would contradict Lemma 5.1.
Thus each component of is an isolated point in . If each vertex angle is not equal to , then Proposition 4.3 implies that there is a neighborhood of the set, , that contains no critical points. Therefore is finite if is not a right triangle.
If is a right triangle, then Proposition 4.3 gives that either a deleted neighborhood of contains no critical points or one of the sides is a component of . In the former case, the preceding argument still applies. The latter case is impossible. Indeed, the other endpoint of the side is a vertex with angle strictly less than , contradicting Proposition 4.3. ∎
## 6. Derivatives of a second Neumann eigenfunction
In this section, is a second Neumann eigenfunction for a triangle . Here, we consider the nodal sets of the angular and directional derivatives of .
By ‘directional derivative’ we mean the result of applying a (real) constant vector field . Each such vector field commutes with the Laplacian and so if is an eigenfunction of the Laplacian, then is also an eigenfunction with the same eigenvalue. We are particularly interested in the unit vector field, , that is parallel to a side of a triangle such that a counterclockwise rotation of points into the half-plane containing . We will let denote the unit vector field that is outward normal to the side . Note that satisfies Neumann conditions if and only if for each side of .
By ‘angular derivative’ we mean the result of applying the rotational vector field that corresponds to the counter-clockwise rotational flow about a point . To be precise if , then
Rp = −(y−p2)⋅∂x + (x−p1)⋅∂y.
The vector field commutes with the Laplacian, and so if is a Laplace eigenfunction, then is also an eigenfunction with the same eigenvalue. We are particularly interested in the case where is a vertex of a triangle.
Recall that a tree is a simply connected graph. The degree of a vertex is the number of edges that contain the vertex. By the interior of a side of the triangle , we will mean the complement where are the vertices of . We will denote the interior with .
According to §3, the nodal sets of both and are locally finite graphs whose vertex set consists of the critical points of , endpoints of maximal arcs, and isolated points in the boundary of . We will now show that each of these graphs is finite and each component is a tree.
###### Lemma 6.1.
Let be a vertex of and let denote the side opposite to . The nodal set of is a finite disjoint union of finite trees, and it contains the sides adjacent to . If the nodal set of intersects the interior of , then the nodal set has a degree 1 vertex that lies in the interior of . Each point that lies in the intersection of and is a critical point of .
###### Proof.
The simple connectedness of the nodal set follows from Lemma 5.1.
If is a side that is adjacent to , then the restrictions of the vector fields and to agree up to a non-zero factor. Thus, since vanishes along , so does .
On the other hand, for each in the side opposite to , the vector is independent of the vector . Hence, if belongs to the interior of , then is a critical point of . Therefore by Lemma 5.6, the intersection is finite.
Suppose that lies in the intersection of and the interior of . Since is simply connected, the component of that contains is a maximal arc that has two distinct endpoints. Since the sides adjacent to are contained in , one of these endpoints lies in the interior of the side opposite to . This endpoint is a critical point of .
In sum, the set is the union of the sides adjacent to and the maximal arcs that have at least one endpoint in the interior of , and each such endpoint is a critical point of . Since there are only finitely many critical points and the degree of each vertex of is finite, there are finitely many maximal arcs. It follows that the set is a finite disjoint union of finite trees.
Each finite tree contains at least two degree 1 vertices. Let be a (tree) component of that intersects the interior of . If does not contain the union, , of the sides adjacent to , then each degree 1 vertex of lies in . Otherwise, note that the closure of is a finite union of trees, and let be a component that intersects the interior. Exactly one vertex of lies in , and hence at least one degree 1 vertex of lies in . ∎
###### Corollary 6.2.
If has a critical point that lies in the interior of , then for each vertex of , the nodal set of has a degree 1 vertex that lies in the interior of the side opposite to . In particular, if has a critical point that lies in the interior of , then has at least three more critical points each lying in a distinct side of .
###### Proof.
If is a critical of , then for each vertex of . Lemma 6.1 implies the claim. ∎
###### Lemma 6.3.
Let be a side of . The nodal set of is a finite union of finite trees. If a maximal arc of intersects the interior of , then one endpoint of the arc lies in , and if is not the vertex opposite to , then is a critical point of . If the nodal set of intersects the interior of , then the nodal set has a degree 1 vertex that lies in .
###### Proof.
Lemma 5.1 implies that the nodal set is simply connected.
Since satisfies Neumann conditions, the function satisfies Neumann conditions along . If , then there exists a maximal arc of containing that has distinct endpoints. The endpoints can not both lie in as a consequence of Lemma 3.2. Hence at least one endpoint lies in .
If is a side of that meets at an angle equal to , then . Thus, it follows from Lemma 3.2, that if one of the endpoints of the latter maximal arc lies in then the other can not lie on and hence lies on . If meets at an angle not equal to , then at each , the vectors and are independent. In particular is a critical point of .
Thus, if is a maximal arc that intersects the interior of , then at least one endpoint of is a critical point that lies in . By Lemma 5.6, the set of such points is finite. Each vertex of has finite degree and so the number of maximal arcs in is finite. It follows that is a finite disjoint union of finite trees.
The remainder of the argument is similar to that given at the end of the proof of Lemma 6.1. ∎
###### Lemma 6.4.
Let be a side of . The intersection equals the set of critical points of that lie in , and is hence finite. Each point in is an endpoint of at least one maximal arc of that intersects .
###### Proof.
The first assertion follows from the fact that and are independent. By Lemma 5.6, the function has only finitely many critical points.
Since satisfies Neumann conditions along , we may extend uniquely to an eigenfunction on the interior of the kite that is invariant under the reflection associated to . Since is parallel to , we find that is also invariant under , and hence the nodal set is also invariant. No Cheng arc of equals , and therefore there exists a maximal arc that intersects the interior of . ∎
The following theorem plays a prominent role in the proof Theorem 1.1.
###### Theorem 6.5.
Let be a side of . If has a degenerate critical point that lies in and does not have a critical point that lies in the interior of , then either
1. for each of the vertices adjacent to , the nodal set of has a degree 1 vertex that belongs to the edge opposite to , or
2. the nodal set of has a degree 1 vertex that belongs to the interior of a side distinct from , and the nodal set of has a degree 1 vertex that belongs to the interior of a side distinct from .
###### Proof.
Without loss of generality, the critical point is located at the origin, and the edge lies in the real axis. Let be the extension of to the interior of the kite obtained by reflecting about . The real-analytic Taylor expansion at has the form
(12) ˜u(z) = a00 + a20⋅x2 + a11⋅xy + a02⋅y2 + O(3).
where indicates a sum of terms in and of order at least . Since is a Neumann eigenfunction along , we have . Therefore, since by assumption is a degenerate critical point, either or .
The case leads to alternative (1). Indeed, if , then from (12) we find that the angular derivative of about an endpoint of equals
Rv˜u(z) = −2a20⋅xy + O(2) = O(2).
It follows that is a nodal critical point of . By Lemma 2.2, at least two Cheng curves of intersect at and the intersection is equiangular. In particular, since one of these arcs is , some other curve is transverse to at and hence intersects . Hence by Lemma 6.1, the nodal set of has a degree 1 vertex that lies in the interior of the side opposite to . (See Figure 1.) Letting , the two endpoints of , we obtain alternative (1).
The case leads to alternative (2). Indeed, in this case, from (12) we find that . Thus at least two Cheng curves of meet at . Note that is invariant under the reflection about , and hence these Cheng curves are also invariant. By Lemma 6.4 none of these Cheng curves is a subset of . It follows that each curve intersects the interior of . By Lemma 6.3, each of two of the corresponding maximal arcs, , has an endpoint . If is a vertex of , then the other endpoint, , lies in the interior of a side . If and lie in the interiors of distinct sides , then choose and . If and lie in the interior of the same side , then by relabeling if necessary, we may assume that and the vertex opposite to are separated by , and we choose . In fact, in each case the curve separates from the vertex opposite to . By Lemma 6.3, is a critical point of , and moreover, there exists a degree vertex of that lies in the interior of . (See Figure 1.)
Since is a critical point of , by Lemma 6.4, there exists a maximal arc of that intersects the interior of and has as an endpoint. The other endpoint of cannot be the vertex opposite to . Indeed, since the vectors and are independent at each , an intersection point would be a critical point. By considering a subtree containing the arc one finds a degree 1 vertex of that lies in the interior of a side distinct from . (See Figure 1.) ∎
###### Corollary 6.6.
If has exactly one critical point , then is a nondegenerate critical point.
###### Proof.
By Corollary 6.2, the point lies in . By Theorem 6.5, the point is nondegenerate. ∎
We will also use the next two results in the proof of Theorem 1.1, but Lemma 6.8 will not be used in the acute case.
###### Proposition 6.7.
Let be a side of . If contains at least two critical points of , then has a third critical point that lies in .
###### Proof.
Let and be critical points of that lie on . By Lemma 6.4, there exists a maximal arc of that intersects the interior of and has as an endpoint in . The other endpoints of | 9,971 | 40,537 | {"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} | 2.6875 | 3 | CC-MAIN-2019-22 | latest | en | 0.842288 |
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## To play this game, everyone stands in a circle. One player calls out ‘one’. The next player says ‘two’ and so on. A player who has to call out 3 or a number which can be divided by 3 has to say ‘Meow’ instead of the number. One who forgets to say ‘Meow’ is out of the game. The last player left is the winner. - class 5 math
(a) Which numbers did you replace with ‘Meow’?
(b) Now, which numbers did you replace with ‘Meow’?
(c) Write any ten multiple of 5.
By:milan-ransingh
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(a) The number which will be replaced with ‘Meow’ are:
3,6, 9, 12, 15, 18, 21, 24, 27, 30…
(b) The numbers which will be replaced with ‘Meow’ are:
4, 8, 12, 16, 20, 24, 28, 32…
(c) Ten multiple of 5 are: 5, 10, 15, 220, 25, 30, 35, 40, 45 and 50.
pankaj
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https://www.squareyards.com/blog/1-trillion-in-crores-stoc | 1,716,239,854,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00188.warc.gz | 898,827,369 | 81,187 | # 1 Trillion in Crores Converter – Trillion to Crores Conversion
1 Trillion is equal to 100000.00 Crore
• Million
• Hundred
• Thousand
• Lakh
• Crore
• Billion
• Arab
• Kharab
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In this blog we will talk about how to convert 1 Trillion to Crores in a detailed and simplified way, so as to not put too much pressure on you. The units of measurement are decided in accordance with the International System of Units, also known as SI. This system lays down the basis on which the physical dimensions of matter are measured. The International System of Units finds a wide range of use in education and industrial applications.
The SI system is widely used in many day to day activities like banking, grocery shopping, and cooking. You will come across a number of daily situations where you will have to rely on the SI system. Which makes it imperative that you should know how to convert one unit to the other. Let’s take a look at how to convert Trillion to Crore.
## What is a Trillion?
Trillion is a natural number in the International System of Units that has two different definitions. They are as follows:
• According to short scale system which is used in both American and UK English, 1 Trillion can be defined as 1 million million or 1,000,000,000,000
• According to the long scale definition, 1 Trillion can be defined as 1,000,000,000,000,000,000 or 10 raised to the power eighteen. This makes it million time larger than the short scale definition of 1 Trillion
The long scale variation of this definition was widely used in the UK up until 1974. The short scale definition is widely used in the field of technical writing, journalism, science, etc.
## What is a Crore?
Crore is a unit of measurement that is prominently used in the Indian Numbering System. In the International System of Units, this is known as ten million or 1,00,00,000.
## Converting 1 Trillion to Crores
To convert 1 Trillion to Crores, we will apply the following formula:
1 Trillion = 100000 x Crore
From this formula, we can see that a Trillion is equal to 1,00,000 Crores. Which means that you can convert trillion to crore by multiplying the value of trillion with 1,00,000. Let’s take a look at an example:
5 Trillion = 5 x 1,00,000 Crore
Therefore, 5 Trillion is equal to 5,00,000 crore
## Trillion to Crore Conversion
If you wish to convert trillion to crore, you will be required to multiply the number in the trillion position with 1,00,000. The following example will help you understand this a little better:
Example:
Converting 5 trillion to Crores:
1 Trillion = 1,00,000 Crores
Which means,
4 Trillion = 4 x 1,00,000 Crore or 4,00,000 Crore
Therefore, 4 Trillion is equal to 4,00,000 Crore
## Trillion to Crore Conversion Table
In Trillion Formula Applied In Crore I Trillion 1 x 1,00,000 1,00,000 Crores 2 Trillion 2 x 1,00,000 2,00,000 Crores 3 Trillion 3 x 1,00,000 3,00,000 Crores 4 Trillion 4 x 1,00,000 4,00,000 Crores 5 Trillion 5 x 1,00,000 5,00,000 Crores 6 Trillion 6 x 1,00,000 6,00,000 Crores 7 Trillion 7 x 1,00,000 7,00,000 Crores 8 Trillion 8 x 1,00,000 8,00,000 Crores 9 Trillion 9 x 1,00,000 9,00,000 Crores 10 Trillion 10 x 1,00,000 10,00,000 Crores
## How Many Zeroes are There in a Trillion?
As previously discussed,
1 Trillion = 1,00,000 Crore
Which means that 1 Trillion has 12 zeros
## Difference Between Trillion and Crore
The following table shows the difference between Trillion and Crore:
Basis of Comparison Crore Trillions Definition It is a part of the Indian Number System. It is defined as 10,000,000 A natural number in the international number system denoted by 1,000,000,000,000 Symbol Cr T Use Used in the Indian Number System Used in the international system of numbers Value 1 Crore = 10 Million 1 Trillion = 1000 Billion Origin It is derived from the Sanskrit word ‘koti’, which means ten millions The term trillion is a combination of two French words, namely tri (meaning three) illion which means a million million.
## You Might Also Like
### How many Crores are there in a Trillion?
There are 1,00,000 Crores in 1 Trillion.
### How much is a trillion?
One trillion is equal to 1000 Billion.
### How many zeros are in a trillion?
There are 12 zeros in one trillion.
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Contact Our Real Estate Experts | 1,319 | 4,903 | {"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} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.941055 |
https://it.mathworks.com/help/simulink/slref/designing-a-guidance-system-in-matlab-and-simulink.html | 1,719,298,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00406.warc.gz | 273,134,475 | 21,708 | # Design a Guidance System in MATLAB and Simulink
This example shows how to use the model of the missile airframe presented in a number of published papers on the use of advanced control methods applied to missile autopilot design. See [1], [2], and [3]. The model represents a tail-controlled missile traveling between Mach 2 and Mach 4, at altitudes ranging between 10,000 ft (3,050 m) and 60,000 ft (18,290 m), with typical angles of attack ranging between +/- 20 degrees.
### Airframe Dynamics Model
The core element of the model is a nonlinear representation of the rigid body dynamics of the airframe. The aerodynamic forces and moments acting on the missile body are generated from coefficients that are non-linear functions of both incidence and Mach number.
Create the model using Simulink® and the Aerospace Blockset™. The Aerospace Blockset provides reference components, such as atmosphere models, that are common to all models irrespective of the airframe configuration.
Open the model.
The airframe model consists of four principal subsystems, controlled through the acceleration-demand autopilot. The `Atmosphere & Incidence, Airspeed Computation` subsystem calculates the change in atmospheric conditions with changing altitude. The `Fin Actuator` and `Sensors` subsystems couple the autopilot to the airframe. The `Aerodynamics & Equations of Motion` subsystem calculates the magnitude of the forces and moments acting on the missile body and integrates the equations of motion.
### International Standard Atmosphere Model
The `Atmosphere & Incidence Airspeed Computation` subsystem approximates the International Standard Atmosphere (ISA) and is split into two regions. The troposphere region lies between sea level and 11 km and has a linear temperature drop that corresponds to changing altitude. The lower stratosphere region ranges between 11 km and 20 km and has a constant temperature.
### Aerodynamic Coefficients for Forces and Moments
The `Aerodynamics & Equations of Motion` subsystem generates the forces and moments applied to the missile in body axes and integrates the equations of motion that define the linear and angular motion of the airframe.
The aerodynamic coefficients are stored in datasets. During the simulation, the value at the current operating condition is determined by interpolation using 2-D Lookup Table blocks.
### Classical Three-Loop Autopilot Design
The missile autopilot controls acceleration normal to the missile body. In this example, the autopilot structure is a three-loop design that uses measurements from an accelerometer placed ahead of the center of gravity. A rate gyro provides additional damping. The controller gains are scheduled on incidence and Mach number and are tuned for robust performance at an altitude of 10,000 ft.
Designing the autopilot using classical design techniques requires that linear models of the airframe pitch dynamics be derived about a number of trimmed flight conditions. MATLAB® can determine the trim conditions, and derive linear state space models directly from the nonlinear Simulink model. This method saves time and helps to validate the model. The functions provided by the MATLAB Control System Toolbox™ and Simulink® Control Design™ allow you to visualize the behavior of the airframe open-loop frequency (or time) responses. To see how to trim and linearize the airframe model, see Airframe Trim and Linearize.
### Airframe Frequency Response
Autopilot designs are carried out on a number of linear airframe models derived at varying flight conditions across the expected flight envelope. To implement the autopilot in the nonlinear model, you must store the autopilot gains in two-dimensional lookup tables and incorporate an anti-windup gain to prevent integrator windup when the fin demands exceed the maximum limits. You can then test the autopilot in the nonlinear Simulink model to show satisfactory performance in the presence of nonlinearities such as actuator fin and rate limits and with the gains dynamically varying with changing flight conditions. The `Autopilot` subsystem implements the gain schedule.
### Homing Guidance Loop
The homing guidance loop contains a `Seeker/Tracker` subsystem that returns measurements of the relative motion between the missile and target. The `Guidance` subsystem generates normal acceleration demands which are passed to the autopilot. The autopilot becomes part of an inner loop within the overall homing guidance system. For information on the differing forms of guidance and background information on the analysis techniques that are used to quantify guidance loop performance, see [4].
### Guidance Subsystem
The `Guidance` subsystem generates demands during closed-loop tracking and perform an initial search to locate the target position. A Stateflow® chart controls the transfer between these differing modes of operation. Switching between modes is triggered by events generated either in Simulink or internal to the Stateflow chart. You can control the behavior of the Simulink model by changing the value of Mode. This variable is used to switch between differing control demands. During target search, the Stateflow chart controls the tracker directly by sending demands to the seeker gimbals (Sigma). The tracker flags target acquisition once the target lies within the beamwidth of the seeker (Acquire). After a short delay, closed-loop guidance starts. Stateflow functionality enables the system to rapidly define all operational modes, for example, actions to take should when a loss of lock on the target occurs or when a target is not acquired during target search.
Once the seeker has acquired the target, a proportional navigation guidance (PNG) law is used to guide the missile until impact. This form of guidance law has been used in guided missiles since the 1950s and can be applied to radar, infrared, or television guided missiles. The navigation law requires these data:
• Measurements of the closing velocity between the missile and target, which can be obtained using a Doppler tracking device for a radar-guided missile
• An estimate for the rate of change of the inertial sightline angle
### Seeker/Tracker Subsystem
The `Seeker/Tracker` subsystem drives the seeker gimbals to keep the seeker dish aligned with the target and provides the guidance law with an estimate of the sightline rate. The tracker loop time constant tors is set to 0.05 s and represents a compromise between maximizing speed of response and keeping noise transmission within acceptable levels. The stabilization loop compensates for body rotation rates. The gain Ks, which is the loop crossover frequency, is set as high as possible subject to the limitations of the bandwidth of the stabilizing rate gyro. The sightline rate estimate is a filtered value of the sum of the rate of change of the dish angle measured by the stabilizing rate gyro and an estimated value for the rate of change of the angular tracking error (e) measured by the receiver. In this example, the bandwidth of the estimator filter is set to half the bandwidth of the autopilot.
For radar-guided missiles, a commonly modeled parasitic feedback effect is radome aberration. Radome aberration occurs when the shape of the protective covering over the seeker distorts the returning signal and gives a false reading of the look angle to the target. Generally, the amount of distortion is a nonlinear function of the current gimbal angle. However, the model approximates a linear relationship between the gimbal angle and the magnitude of the distortion using a Gain block named `Radome Aberration`. You can also model other parasitic effects, such as sensitivity in the rate gyros to normal acceleration, to test the robustness of the target tracker and estimator filters.
### Run Guidance Simulation
You can now test the performance of the overall system. The target is defined to be traveling at a constant speed of 328 m/s on a course reciprocal to the initial missile heading and 500 m above the initial missile position. Simulation results show that acquisition occurs at 0.69 seconds into the engagement, with closed-loop guidance starting after 0.89 seconds. Impact with the target occurs at 3.46 seconds, and the range to go at the point of closest approach is 0.265 m.
The `aero_guid_plot.m` script runs a performance analysis.
The animation block provides a visual reference for the simulation.
## References
[1] "Bennani, Samir, Dehlia M. C. Willemsen, and Cartsen W. Scherer. "Robust LPV Control with Bounded Parameter Rates." AIAA-97-3641 (August 1997). https://doi.org/10.2514/6.1997-3641.
[2] Mracek, Curtis P., and James R. Cloutier. "Full Envelope Missile Longitudinal Autopilot Design Using the State-Dependent Riccati Equation Method." AIAA-97-3767 (December 1994). https://doi.org/10.2514/6.1997-3767.
[3] Shamma, Jeff S., and James R. Cloutier. "Gain-Scheduled Missile Autopilot Design Using Linear Parameter Varying Transformations.” Journal of Guidance, Control, and Dynamics 16, no. 2 (March 1993): 256–63. https://doi.org/10.2514/3.20997.
[4] Ching-Fang Lin. Modern Navigation, Guidance, and Control Processing. Vol. 2. Englewood Cliffs, NJ: Prentice Hall, 1991. | 1,933 | 9,250 | {"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} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.854742 |
https://www.answers.com/Q/How_many_square_feet_in_3.1_acres | 1,566,319,348,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00239.warc.gz | 691,561,353 | 14,841 | # How many square feet in 3.1 acres?
Answer: 3.1 acres = 135,036 ft² | 27 | 69 | {"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} | 2.828125 | 3 | CC-MAIN-2019-35 | latest | en | 0.874884 |
https://inches-to-meters.appspot.com/1/et/75.4-toll-et-meeter.html | 1,618,801,420,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038863420.65/warc/CC-MAIN-20210419015157-20210419045157-00194.warc.gz | 420,843,965 | 6,016 | Toll Et Meeter
# 75.4 in m75.4 Tolli Meetrit
in
=
m
## Kuidas teisendada 75.4 tolli meetrit?
75.4 in * 0.0254 m = 1.91516 m 1 in
## Convert 75.4 in ühistele pikkused
MõõtühikPikkusühiku
nanomeeter1915160000.0 nm
Mikromeeter1915160.0 µm
Millimeeter1915.16 mm
Sentimeeter191.516 cm
Toll75.4 in
Jalg6.2833333333 ft
Õu2.0944444444 yd
Meeter1.91516 m
Kilomeeter0.00191516 km
Miil0.0011900253 mi
Meremiil0.0010341037 nmi
## Alternatiivsed õigekirja
75.4 Tolli m, 75.4 in Meetrit, 75.4 in m, 75.4 Tolli Meetrit, 75.4 Toll Meetrit, 75.4 Tolli Meeter, 75.4 in Meeter | 269 | 566 | {"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} | 2.703125 | 3 | CC-MAIN-2021-17 | latest | en | 0.260326 |
http://tex.stackexchange.com/questions/107745/why-does-pgfplots-plot-functions-only-until-x-5-and-y-5 | 1,469,556,364,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825048.60/warc/CC-MAIN-20160723071025-00268-ip-10-185-27-174.ec2.internal.warc.gz | 241,184,061 | 18,979 | # Why does pgfplots plot functions only until x = 5 and y = 5?
Why does pgfplots plot functions only until x = 5 and y = 5, but not any further?
\documentclass{article}
\usepackage{tikz}
\usepackage{pgfplots}
%%%<
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{5pt}
%%%>
\begin{document}
\begin{tikzpicture}
\begin{axis}[samples=100,ymin=0,ymax=10,xmin=0,xmax=20]
\addplot [thick] plot (\x, {1/(1 + exp(-0.6*(\x - 12)))});
\end{axis}
\end{tikzpicture}
\end{document}
-
Because that's the default domain. You can define over what range to plot the function by setting domain=0:20, for example. xmin and xmax only set the axis range (what you can see) but not the calculation range (what is computed). By the way, instead of saying \addplot plot (\x, {\x});, you can also say \addplot {x};. – Jake Apr 9 '13 at 15:03
Thank you, that was it! – Eekhoorn Apr 9 '13 at 15:07
There are a lot of "domain" options in pgfplots. The one which you are asking about is simply domain, which specifies what values of x (you don't need the backslash if you're using pgfplots) are used in plotting; by default, we have domain = -5:5, which is what the author apparently thinks is reasonable for typical graphs. There is a corresponding y domain for two-variable functions.
This domain is quite different from the limits established by xmin and xmax. While domain is set per-plot, these keys are set per-axis and just confine the actual drawing to these limits. There are corresponding ymin and ymax. These will be computed automatically by pgfplots if they are not given, but it is necessary for a really polished picture to set them yourself. Note that y domain does not have anything to do with ymin and ymax in a plot of one-variable functions, because it determines the inputs of the non-existent variable y. Instead, ymin and ymax, if they were determined automatically, would be calculated from the values output by your plotted functions across the domain.
There are more! My favorites are restrict x to domain and restrict y to domain, which are filters with the same input syntax as domain. These don't determine what numbers are used in the variables; they determine what values are used in the plot. They are enormously helpful with parametric or uncontrollable functions; i.e. \addplot {1/x}; will, with the default domain = -5:5, produce a rather hideous asymptote at x = 0 as well as (with the default ymin and ymax) a badly distorted view of the axes. But setting restrict y to domain = -5:5 in this plot will simply eliminate the large values, removing the asymptote and also scaling the picture back to a proportional square.
Or, alternatively, a parametric plot such as \addplot ({exp(x)},{exp(-x)}); (a funny way of drawing the same thing in the first quadrant alone), which is hard to tune directly because of the logarithmic connection between values on the plot and values of the variable. For this, both the default domain and the default axis sizes are inappropriate; I usually leave domain as it is (which gives numbers that are too huge in both coordinates) and then set restrict x to domain and restrict y to domain accordingly to trim the picture nicely. This is not to say it's a good idea to be completely oblivious to domain, since those points are computed...just not used.
These filter keys are different from the min and max keys because they actually ignore the filtered-out values, rather than simply cutting them out of the picture. This is essential if these values are larger than TeX is capable of computing with.
Finally, there are samples or sample at, which latter exists mutually exclusively with domain and say how many, or even exactly at which values of x to compute values. This can be an alternative to the restrict to domain keys, if you choose the samples carefully to avoid exceptional inputs. They are also useful for tuning the plot around rapidly-varying places in the graph, which would otherwise look rather choppy. These also interact with the restrict to domain keys in the sense that with, say,
\addplot[
domain = -5:5,
samples = 11,
restrict x to domain = -1:1,
restrict y to domain = -1:1
] ({exp(x)},{exp(-x)});
there will be exactly 11 points evaluated, namely ({exp(-5)},{exp(5)}) through ({exp(5)},{exp(-5)}), but only those with both coordinates in the interval [-1,1] will be plotted. Unfortunately, the only point with that property is ({exp(0)},{exp(0)}) = (1,1), so your plot will be rather vacant. The un-plotted points are not even used to anchor interpolating curves! So the filter keys are not a panacea.
My pictures tend to set all of these keys, since they each affect the drawing differently.
-
Nice overview! Two things: 1. While sample at and domain are mutually exclusive, samples and domain are not (I guess you didn't actually mean that, but that's how I read it). 2. Discarding points using restrict x to domain is probably okay in most situations, but it does introduce unnecessary computations. Setting the domain (or samples at) to appropriate values can sometimes be required if you have functions that are very expensive to evaluate. – Jake Apr 9 '13 at 15:29 | 1,244 | 5,202 | {"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} | 2.515625 | 3 | CC-MAIN-2016-30 | latest | en | 0.883596 |
http://nrich.maths.org/public/leg.php?code=-63&cl=3&cldcmpid=6207 | 1,448,838,447,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398460263.61/warc/CC-MAIN-20151124205420-00334-ip-10-71-132-137.ec2.internal.warc.gz | 170,647,276 | 6,347 | # Search by Topic
#### Resources tagged with Arithmetic sequence similar to Silver Line:
Filter by: Content type:
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### There are 16 results
Broad Topics > Sequences, Functions and Graphs > Arithmetic sequence
### Shifting Times Tables
##### Stage: 3 Challenge Level:
Can you find a way to identify times tables after they have been shifted up?
### Days and Dates
##### Stage: 4 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Coordinate Patterns
##### Stage: 3 Challenge Level:
Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead?
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Cellular
##### Stage: 2, 3 and 4 Challenge Level:
Cellular is an animation that helps you make geometric sequences composed of square cells.
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Investigating Pascal's Triangle
##### Stage: 2 and 3 Challenge Level:
In this investigation, we look at Pascal's Triangle in a slightly different way - rotated and with the top line of ones taken off.
### Prime AP
##### Stage: 4 Challenge Level:
Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference is divisible by 6. What if one of the terms is 3?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Slick Summing
##### Stage: 4 Challenge Level:
Watch the video to see how Charlie works out the sum. Can you adapt his method?
### More Number Pyramids
##### Stage: 3 and 4 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Odds, Evens and More Evens
##### Stage: 3 Challenge Level:
Alison, Bernard and Charlie have been exploring sequences of odd and even numbers, which raise some intriguing questions...
### Series Sums
##### Stage: 4 Challenge Level:
Let S1 = 1 , S2 = 2 + 3, S3 = 4 + 5 + 6 ,........ Calculate S17.
### Sums of Powers - A Festive Story
##### Stage: 3 and 4
A story for students about adding powers of integers - with a festive twist. | 739 | 3,048 | {"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} | 3.96875 | 4 | CC-MAIN-2015-48 | longest | en | 0.856813 |
https://conklinfangmanbuickgmckc.com/auto-parts/your-question-how-do-you-calculate-the-lifting-capacity-of-a-motor.html | 1,632,348,127,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00028.warc.gz | 244,838,994 | 19,150 | # Your question: How do you calculate the lifting capacity of a motor?
Contents
## How do you calculate motor load capacity?
It is combination from Amperage and Voltage value. Basic formula for Power is P (watt) =I(ampere) x V (volt). Power also measured in Horsepower (hp) unit. For common conversion from electrical Horsepower to Watt is 1hp = 746 watt.
## How do you calculate lifting power?
Let us calculate the work done in lifting an object of mass m through a height h. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh.
## How much weight can a 1 hp motor lift?
Most engines use horsepower to describe how much work they can do in a given amount of time. The constant 1 horsepower equals 550 foot-pounds per second. In other words, 1 horsepower is the amount of work required to move a load of 550 pounds over 1 foot, in 1 second.
## How do you calculate motor output?
Calculations
1. I = V / R. where I – current, measured in amperes (A); …
2. Pin = I * V. where Pin – input power, measured in watts (W); …
3. Pout = τ * ω where Pout – output power, measured in watts (W); …
4. ω = rpm * 2π / 60. …
5. E = Pout / Pin. …
6. Pout = Pin * E. …
7. τ * ω = I * V * E.
8. τ * rpm * 2π / 60 = I * V * E.
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## How is motor FLC calculated?
The motor FLA calculator uses the following formulas:
1. Single Phase AC Motor FLA (Amperes) = (P [kW] × 1000) / (V × cos ϕ)
2. Single Phase AC Motor FLA (Amperes) = (P [HP] × 746) / (V × cos ϕ × η)
3. Three Phase AC Motor FLA (Amperes) = (P [kW] × 1000) / (V × 1.732 × cos ϕ)
## What is work formula?
We can calculate work by multiplying the force by the movement of the object. W = F × d. Unit. The SI unit of work is the joule (J)
## What is the lift formula?
The lift equation states that lift L is equal to the lift coefficient Cl times the density r times half of the velocity V squared times the wing area A. For given air conditions, shape, and inclination of the object, we have to determine a value for Cl to determine the lift.
## How much power is required to lift a 30 N?
The power needed is 3 watts.
## How much weight can a 1/2 hp motor lift?
To find the weight each opener can lift simply divide 550 lbs by the manufacturer HP rating and then multiply by (2/3). For example, a 1/2 horsepower garage door opener should be able to lift 275 pounds while a 1 HP opener can lift 550 pounds.
## How fast is 10 horsepower?
That can be anything from 1 km/h (0.6 mph) to 300 km/h and above (200 mph). I have personally driven several vehicles that just by chance were powered by a 90hp engine. VW Golf, I would drive it routinely at 180km/h (110 mph) but it could go a bit faster.
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## What is the rpm of 1 hp motor?
1HP 750 W 1 HP 3000 RPM Three Phase AC Induction MotorPower1HP 750 WSpeed (RPM)1440,2880Voltage230
## How do I calculate RPM?
How to Calculate Motor RPM. To calculate RPM for an AC induction motor, you multiply the frequency in Hertz (Hz) by 60 — for the number of seconds in a minute — by two for the negative and positive pulses in a cycle. You then divide by the number of poles the motor has: (Hz x 60 x 2) / number of poles = no-load RPM.
## How many watts is 23000 rpm?
The Salton Compact Power Blender has a Vortex Jar with 6 stainless steel, Japanese blades and a 2HP, 23,000RPM, 1000 watt motor.
## What is the efficiency of motor?
For an electric motor, efficiency is the ratio of mechanical power delivered by the motor (output) to the electrical power supplied to the motor (input). Thus, a motor that is 85 percent efficient converts 85 percent of the electrical energy input into mechanical energy. | 1,053 | 3,866 | {"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} | 4.40625 | 4 | CC-MAIN-2021-39 | latest | en | 0.869981 |
https://www.coursehero.com/file/5971025/2-Variables-and-Their-Distributions/ | 1,529,936,291,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00135.warc.gz | 771,520,327 | 371,166 | 2 Variables and Their Distributions
# 2 Variables and Their Distributions - BUAD 310 Applied...
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Outline for Today Descriptive statistics Variables and their distributions Finding the center Measuring the spread Relationship between variables 2
Data Consists of values of some variables measured or observed for some individuals . E.g., the data set from the class survey (13 variables for the students in our class). Variable is a characteristic of an individual. Qualitative or Categorical : places an individual into one of several groups or categories (finitely many possible) Quantitative : measurements represent quantities, e.g. “how many” (infinitely many possible), could be discrete or continuous 3
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Scales of Measurement Quantitative Ratio variable, almost all quantitative variables are ratio variables Ratios meaningful, inherently defined zero (e.g., salary, weight, distance) Interval variable, very few quantitative variables are interval variables Ratios not meaningful, no inherently defined zero (e.g., temperature) Categorical Ordinal variable Meaningful ordering or ranking of categories (e.g., class) Nominative variable No meaningful ordering or ranking of categories (e.g., gender, color) 4 R a t i o ( a l m o s t a l l ) I n t e r v a l ( v e r y f e w ) Q u a n t i t a t i v e O r d i n a l N o m i n a t i v e Q u a l i t a t i v e ( C a t e g o r i c a l ) D a t a
Data Analysis Initial examination of the data is called Exploratory Data Analysis (EDA). Strategy for EDA: Begin by examining each variable and then move on to the study of relationships At each stage start with graphs and then add numerical summaries 5
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Distribution of a Variable Describes what possible values the variable takes and how frequently it takes those values. We describe the overall pattern of a distribution by its shape, center, and spread. There are many ways to describe and display distributions. Graphs Word descriptions Numerical measures 6
Describing Categorical Data Describing Categorical Data 7 Motivating Example: Everyday, more than 100 million shoppers visit amazon.com, either by typing amazon.com or by clicking an ad. Which hosts send the most visitors to Amazon’s Web site? Data set consists of 188,996 visits Host is a categorical variable To answer this question we must describe the variation in Host
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Data Table Frequency and Relative Frequency Tables The distribution of a categorical variable is a list of values with its associated count (frequency) A frequency table summarizes the distribution of a categorical variable A relative frequency table shows the proportion (or percentage) in each category 8
9 Looking At Data
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10 Charts of Categorical Data Bar Charts and Pie Charts Unless you need to know exact counts, charts are better than tables for summarizing more than five categories The two most common displays of a categorical variable are a bar chart and a pie chart
11 Charts of Categorical Data The Bar Chart
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Jill Tulane University ‘16, Course Hero Intern | 979 | 4,678 | {"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} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.831388 |
https://www.mathsatsharp.co.za/august-2018-maths-at-sharp-newsletter | 1,623,682,185,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00533.warc.gz | 804,187,242 | 33,273 | # August 2018 Maths at Sharp Newsletter
### Dear Friend of Maths at Sharp,
It’s suddenly August and the weather is turning to sunshine again. We hope you have enjoyed your break and that the IEB teachers enjoy their break now.
### I am very excited to share some personal news with you –
my husband and I are expecting a little girl for Christmas. What has amazed me is all the maths that is involved in a pregnancy. For one, you need to countdown to your due date, and calculate how many weeks you are, or how many you have left before you are due (I don’t know which is scarier 😉). When you have an ultrasound or scan (which we recently had the privilege of doing in 3D), the machine bounces sound waves around your insides. A very clever computer then uses mathematical algorithms to plot a picture of the little one inside. The algorithms have gotten so advanced you can now do it in 3D and HD! If you would like more information – here is a great article.
### The 2018 AMESA
that took place at the University of the Free State at the end of June was not quite as chilly as we were all expecting. The workshops were stimulating with a lot of good discussion. The one that stands out is the first one held on Tuesday afternoon. In it, we were discussing the applications of the unlimited table mode on the Sharp EL-W535SAB. One of the applications was to use the table mode to teach a particular factorising method. If you recall from the previous newsletter, there was a method called airplane factorising which helped students to factorise trinomials with a value for “a” other than one. (Please take a look here for the method). We encountered a problem with the use of the “a” value in both brackets because it was mathematically incorrect.
For example, when we initially write down the brackets as
(3x )(3x )
Instead of writing the brackets as (3x )(x ) as we would usually do.
This additional 3 is “lost” in one of the last steps when we divide one of the brackets by 3 again.
In our discussion, this additional “3” or “a” in the second bracket was a problem because it wasn’t mathematically correct, as the entire trinomial was not multiplied by 3 or “a”. Only the c value was.
What are your thoughts? Do you think that if the method is easier it shouldn’t matter that it isn’t perfectly correct? Or do you think that the method should be mathematically correct? Or do you think that the method doesn’t matter at all, as long as the student is able to get the correct answer and they can justify their steps?
For all the Sharp AMESA workshops click here and for the competition results click here. Congratulations to Emmly Sibanyoni from the North West for winning the 34 litre Sharp grill microwave!
A picture of Thabo (last year’s winner) handing over the prize to Emmly.
### I came across a very interesting article
about using knitting to explain different mathematical concepts. The author of the article opens with the observation that students see maths as “computation” and “equation”, while mathematicians see mathematics as logical thinking and problem-solving. This changes a person’s perception of maths from boring and rule bound to something that is used for finding solutions to interesting problems. This article goes on to explain how they explored mathematics without writing out “equations”, but exploring it through using knitting, drawing pictures and creation. There are many great ways to share the passion of maths without bringing in the rules that students find so restrictive.
For a different take, here are 23 random maths facts that all students enjoy. I confess that I didn’t believe that 10! seconds is 6 weeks but it is in fact true!
### We will be at several events over the next couple of weeks,
from the Maths conference in Sebokeng, to the Sasol Techno X the following week (13th of August – 17th of August) in Sasolburg. This is a great place to bring your students as most of the universities are there, as well as several other companies that demonstrate different career options. Some of the exhibitors include CSIR, Eskom, Jumpstart Foundation, Mintek, National Zoological Gardens, Plastics SA, Sci-Bono, SAPS, and many more. Come pop in and take a look at the new Sharp scientific calculators and get a taste of what you can do in the classroom.
### I am also excited to share the progress we are making with the #FutureFund.
We have partnered with MAHLE, an engineering firm in Durban that makes filtration systems (amongst other things) for Volkswagen and BMW. On the 21st of September we will be part of an East Coast Radio breakfast show where we will be taking pledges for the #FutureFund. If you are in the Durban area keep your radio on and find out whether we reach our target of R1 million! Seartec has also agreed to match every donation with a second calculator! That means that for every R200 donated, there will be two students that benefit.
Alternatively, you can purchase your EL-W506 calculators directly from Seartec for R200 each, and we will match each of these purchases with the same number of calculators, to be sent to your school, or another school of your choice.
### Calculator tip:
On the new Sharp EL-W535SAB there are two new functions for lowest common multiple and highest common factor. To find the highest common factor of 25 and 30, press 25
then press 30 . You should get:
To find the lowest common multiple type in 25
30 and this will give you:
### Last month’s riddle in the newsletter was:
How can you add eight 8’s to make 1000?
The answer is 888 + 88 + 8 + 8 + 8 = 1000. Hope you enjoyed playing around with all the 8’s.
This month’s riddle is:
If 1 + 9 + 8 = 1 then what is 2 + 8 + 9 = ?
Wishing you a fantastic women’s month and great third term,
Best maths wishes
Tal (mini-me), and the Seartec team
And because its women’s month, here are two 😉 | 1,334 | 5,900 | {"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} | 3.8125 | 4 | CC-MAIN-2021-25 | latest | en | 0.975468 |
mrpredict.com | 1,582,430,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145746.24/warc/CC-MAIN-20200223032129-20200223062129-00006.warc.gz | 483,284,936 | 42,836 | # Odds of upcoming football fixtures, dropping odds and sure bets
rotate for better view
### Odds of winning are a numerical expression
Odds of winning are a numerical expression, usually expressed as a pair of numbers, used in both gambling and statistics. In statistics, the odds for an event or the odds of an event reflect the probability of the occurrence of the event, while the odds against the event reflect the probability that this is not the case. In gambling, the chances of winning are the ratio of payout to bet and do not necessarily accurately reflect the probabilities. Opportunities are expressed in several ways (see below), and sometimes the term is misused to simply indicate the likelihood of an event. Traditionally, game odds are expressed in the form "X to Y", where X and Y are numbers, and it is implied that the odds are odds against the event to which the player is considering betting. In both gambling and statistics, the 'odds' are a numerical expression of the probability of a possible event.
Need football betting tips for free ? Or place free bet on premier league football fixture ? MrPredict gives you the guidance in your football betting. Find the best football odds, read premier league news or latest premier league videos. Will Chelsea or Arsenal win the game ? Who will be the goalscorer, Abraham ? MrPredict provides daily updated football betting tips and predictions.
If you bet on rolling one of the six sides of a fair die, with a probability of one in six, chances are five to one against you (5 to 1) and you would win five times as much as your bet. If you bet six times and win once, you win five times your bet and lose your bet five times. The chances of winning offered by the bookmaker thus reflect the probabilities of the cube.
In gambling, odds represent the ratio between the amounts bet or wagered by the parties. A quota of 5 to 1 means that the first party (usually a bookmaker) is six times the second party's bet. In the simplest sense, 5 to 1 odds means when you put a dollar (the "1" in the term), and you win, you get five dollars (the "5" in the term) or five times 1. If you bet two dollars you would paid ten dollars or five times. 2. If you bet three dollars and win, you would pay fifteen dollars or five times. 3. If you bet one hundred dollars and win, you would get five hundred dollars. or 5 times 100. If you lose one of these bets, you lose the dollar or two dollars or three dollars or one hundred dollars. | 544 | 2,493 | {"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} | 3.09375 | 3 | CC-MAIN-2020-10 | longest | en | 0.947667 |
https://www.woodworkingtalk.com/threads/crazy-angle-cutting-question.52599/ | 1,653,173,583,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00691.warc.gz | 1,276,518,976 | 19,619 | 1 - 5 of 5 Posts
#### Anguspapa
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Why when I set my chop saw to cut a 50 degree angle, is the angle on the wood @ 40? If I set it to cut a 30 degree angle is the angle on the wood @ 60. The angle is always opposite but always totaling 90. Why does it not cut the angle I set it for, 60 cut 60. Or if I want a 60 degree cut do I and everyone out have to set the saw at 30? Everyone probably thinks I'm a nut and this is driving me nuts. Thanks once again for everyone's help.
#### tc65
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Forget all those other angles, just set it to 45 and you'll never have those problems again:laughing::laughing:
Sorry, couldn't resist
#### BZawat
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It all comes back to your point of reference. The fence is @180 degrees (straight line). The blade set @0 is actually @90degrees (perpendicular) to the reference of the fence. So when you set your saw to 30 degrees left, the 30 is actually measuring the angle between the right side of the blade and the "zero" reference mark (perpendicular to the 180 fence). What remains is its complementary angle of 60 degrees. Make any sense?
#### Anguspapa
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##### Registered
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I think I just figured out how I'm going nuts! I'm looking at my cuts based on the horizontal axes and the saw is pulling its cuts off of the vertical axes. Am I right?
#### Anguspapa
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BZawat said:
It all comes back to your point of reference. The fence is @180 degrees (straight line). The blade set @0 is actually @90degrees (perpendicular) to the reference of the fence. So when you set your saw to 30 degrees left, the 30 is actually measuring the angle between the right side of the blade and the "zero" reference mark (perpendicular to the 180 fence). What remains is its complementary angle of 60 degrees. Make any sense?
I guess when you were typing it bit me in the a**! Thanks. Things like that will drive me nuts till I figure it out. Then I kick my self.
1 - 5 of 5 Posts | 556 | 2,133 | {"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} | 3.28125 | 3 | CC-MAIN-2022-21 | latest | en | 0.910054 |
https://www.geeksforgeeks.org/rearrange-given-array-place/amp/ | 1,576,494,193,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00326.warc.gz | 712,381,431 | 20,155 | # Rearrange an array so that arr[i] becomes arr[arr[i]] with O(1) extra space
Given an array arr[] of size n where every element is in range from 0 to n-1. Rearrange the given array so that arr[i] becomes arr[arr[i]]. This should be done with O(1) extra space.
Examples:
```Input: arr[] = {3, 2, 0, 1}
Output: arr[] = {1, 0, 3, 2}
Input: arr[] = {4, 0, 2, 1, 3}
Output: arr[] = {3, 4, 2, 0, 1}
Input: arr[] = {0, 1, 2, 3}
Output: arr[] = {0, 1, 2, 3}```
If the extra space condition is removed, the question becomes very easy. The main part of the question is to do it without extra space.
## We strongly recommend that you click here and practice it, before moving on to the solution.
The credit for following solution goes to Ganesh Ram Sundaram . Following are the steps.
1) Increase every array element arr[i] by (arr[arr[i]] % n)*n.
2) Divide every element by n.
```Let us understand the above steps by an example array {3, 2, 0, 1}
In first step, every value is incremented by (arr[arr[i]] % n)*n
After first step array becomes {7, 2, 12, 9}.
The important thing is, after the increment operation
of first step, every element holds both old values and new values.
Old value can be obtained by arr[i]%n and new value can be obtained
by arr[i]/n.
In second step, all elements are updated to new or output values
by doing arr[i] = arr[i]/n.
After second step, array becomes {1, 0, 3, 2}
```
Following is the implementation of the above approach.
`#include ` `using` `namespace` `std; ` ` ` `// The function to rearrange an array in-place so that arr[i] ` `// becomes arr[arr[i]]. ` `void` `rearrange(``int` `arr[], ``int` `n) ` `{ ` ` ``// First step: Increase all values by (arr[arr[i]]%n)*n ` ` ``for` `(``int` `i=0; i < n; i++) ` ` ``arr[i] += (arr[arr[i]]%n)*n; ` ` ` ` ``// Second Step: Divide all values by n ` ` ``for` `(``int` `i=0; i
`class` `Rearrange ` `{ ` ` ``// The function to rearrange an array in-place so that arr[i] ` ` ``// becomes arr[arr[i]]. ` ` ``void` `rearrange(``int` `arr[], ``int` `n) ` ` ``{ ` ` ``// First step: Increase all values by (arr[arr[i]]%n)*n ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``arr[i] += (arr[arr[i]] % n) * n; ` ` ` ` ``// Second Step: Divide all values by n ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``arr[i] /= n; ` ` ``} ` ` ` ` ``// A utility function to print an array of size n ` ` ``void` `printArr(``int` `arr[], ``int` `n) ` ` ``{ ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``System.out.print(arr[i] + ``" "``); ` ` ``System.out.println(``""``); ` ` ``} ` ` ` ` ``/* Driver program to test above functions */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``Rearrange rearrange = ``new` `Rearrange(); ` ` ``int` `arr[] = {``3``, ``2``, ``0``, ``1``}; ` ` ``int` `n = arr.length; ` ` ` ` ``System.out.println(``"Given Array is :"``); ` ` ``rearrange.printArr(arr, n); ` ` ` ` ``rearrange.rearrange(arr, n); ` ` ` ` ``System.out.println(``"Modified Array is :"``); ` ` ``rearrange.printArr(arr, n); ` ` ``} ` `} ` ` ` `// This code has been contributed by Mayank Jaiswal `
`# Python3 program to Rearrange ` `# an array so that arr[i] becomes ` `# arr[arr[i]] ` ` ` `# The function to rearrange an ` `# array in-place so that arr[i] ` `# becomes arr[arr[i]]. ` `def` `rearrange(arr, n): ` ` ` ` ``# First step: Increase all values ` ` ``# by (arr[arr[i]] % n) * n ` ` ``for` `i ``in` `range``(``0``, n): ` ` ``arr[i] ``+``=` `(arr[arr[i]] ``%` `n) ``*` `n ` ` ` ` ``# Second Step: Divide all values ` ` ``# by n ` ` ``for` `i ``in` `range``(``0``, n): ` ` ``arr[i] ``=` `int``(arr[i] ``/` `n) ` ` ` `# A utility function to print ` `# an array of size n ` `def` `printArr(arr, n): ` ` ` ` ``for` `i ``in` `range``(``0``, n): ` ` ``print` `(arr[i], end ``=``" "``) ` ` ``print` `("") ` ` ` `# Driver program ` `arr ``=` `[``3``, ``2``, ``0``, ``1``] ` `n ``=` `len``(arr) ` ` ` `print` `(``"Given array is"``) ` `printArr(arr, n) ` ` ` `rearrange(arr, n); ` `print` `(``"Modified array is"``) ` `printArr(arr, n) ` ` ` `# This code is contributed by shreyanshi_arun `
`// C# Program to rearrange an array ` `// so that arr[i] becomes arr[arr[i]] ` `// with O(1) extra space ` `using` `System; ` ` ` `class` `Rearrange ` `{ ` ` ` ` ``// Function to rearrange an ` ` ``// array in-place so that arr[i] ` ` ``// becomes arr[arr[i]]. ` ` ``void` `rearrange(``int` `[]arr, ``int` `n) ` ` ``{ ` ` ` ` ``// First step: Increase all values ` ` ``// by (arr[arr[i]] % n) * n ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``arr[i] += (arr[arr[i]] % n) * n; ` ` ` ` ``// Second Step: Divide all values by n ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``arr[i] /= n; ` ` ``} ` ` ` ` ``// A utility function to ` ` ``// print an array of size n ` ` ``void` `printArr(``int` `[]arr, ``int` `n) ` ` ``{ ` ` ``for` `(``int` `i = 0; i < n; i++) ` ` ``Console.Write(arr[i] + ``" "``); ` ` ``Console.WriteLine(``""``); ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``Rearrange rearrange = ``new` `Rearrange(); ` ` ``int` `[]arr = {3, 2, 0, 1}; ` ` ``int` `n = arr.Length; ` ` ` ` ``Console.Write(``"Given Array is :"``); ` ` ``rearrange.printArr(arr, n); ` ` ` ` ``rearrange.rearrange(arr, n); ` ` ` ` ``Console.Write(``"Modified Array is :"``); ` ` ``rearrange.printArr(arr, n); ` ` ``} ` `} ` ` ` `// This code has been contributed by Nitin Mittal. `
` `
Output:
```Given array is
3 2 0 1
Modified array is
1 0 3 2```
Time Complexity: O(n)
Auxiliary Space: O(1)
The problem with above solution is, it may cause overflow.
Please see below post for a better solution:
Rearrange an array such that ‘arr[j]’ becomes ‘i’ if ‘arr[i]’ is ‘j’ | 2,243 | 6,142 | {"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} | 3.734375 | 4 | CC-MAIN-2019-51 | latest | en | 0.770056 |
http://www.javaist.com/rosecode/problem-226-Meanderings-askyear-2015 | 1,519,467,854,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00197.warc.gz | 480,132,333 | 6,955 | RoseCode
Problem #226
Meanderings
Public ★(x19) 03/20/15 by Philippe_57721 4xp Programming 55.9%
A ant walks exploring each lattice point in the first quadrant of the plane with the following pattern:
37 → 38 → 39 → 40 → 41 → 42 → 43 ↑ ↓ 36 ← 35 ← 34 ← 33 ← 32 ← 31 44 ↑ ↓ 17 → 18 → 19 → 20 → 21 30 45 ↑ ↓ ↑ ↓ 16 ← 15 ← 14 ← 13 22 29 46 ↑ ↓ ↑ ↓ 5 → 6 → 7 12 23 28 47 ↑ ↓ ↑ ↓ ↑ ↓ 4 ← 3 8 11 24 27 48 ↑ ↓ ↑ ↓ ↑ ↓ 1 → 2 9 → 10 25 → 26 49
The 1st point visited (index = 1) is (0,0)
The 10th point is (3,0)
The 100000th point is (143,316)
What is the 10^17-th point?
What is the index of the point (718281828,141592653)?
Example: 143,316/51594074 // For 10^5 and (7182,1415)
[My timing: < 100 ms]
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https://resource.flexrule.com/knowledge-base/decision-graph/ | 1,701,576,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00222.warc.gz | 550,096,993 | 19,528 | # Decision Graph
Contents
## What is a Decision Graph?
Decision Graph is an orchestration model that inter-connects decisions such as Decision Requirement Diagram (DRD), reusable sub-DRD, and other types of graphs.
Decision Graph can also be created following Decision Model and Notation (DMN), which is an open standard notation for modeling decisions that are based on logic, data, and processes. It was developed by Object Management Group (OMG) in 2014. It provides meta-model, notation, and semantics for decision modeling. This is a complimentary notation to BPMN for modeling operational decisions and business rules.
### What is a Decision
decision is an act of determining an output value, based on a number of input values, using logic defining how the output is determined from the inputs. A decision model is a model in a specific area that defines decision requirements and decision logic.
In DMN, A decision is modeled at two levels:
1. Decision Requirement
2. Decision Logic
A Decision Graph is a high-level model that defines one or more Decision Graphs, which makes a Decision Requirement Graph (DRG). At the Decision Logic level, the different details of the implementation of logically related to each node (e.g., Decision) in the DRG is defined. This logic can have any implementation, but as far as DMN is concerned the Decision Table is a standard way of logic implementation.
### Sub-Decision Graph
Decision Graph allows connecting sub-decision graphs in the decision hierarchy. This also ensures that the reusable decision graphs can be connected as necessary, without having to duplicate the decisions.
For example, in the following figure, you can see that we have reused the Post-Bureau Affordability graph in both Bureau Strategy Decision and Approval Decision.
### Knowledge Source
One of the benefits of modeling in DMN is that, not only can it model the actual decision-making process at different levels, but also that you can model (link) the documentation, references, and authorities around these. The Knowledge Source is for that purpose. Also, it can be used to check the impact of change if these sources are changed.
### Decision Logic Level
At this level, the implementation of decisions and business knowledge can be linked to each node at the decision requirement level. These implementations can vary based on your needs, product capabilities, and other factors (e.g., your familiarity with a specific type of logic: Decision Table, Tree, etc.).
#### Decision Table
In DMN, a Decision Table is a common way to present business knowledge in a tabular form. It has its own characteristics compared to a generic decision table.
#### Other Logic Types
In FlexRule, we support other types of logic as well as Decision Tables, so you can simply link this logic to business knowledge as the implementation of the logic. This logic is:
1. Decision Table (DMN standard method)
2. Tree (i.e. Validation, Decision)
3. Natural Language
### Decision Graph Elements
In the table below, all of the elements on a Decision Graph are illustrated:
ElementNotationDescription
DecisionThe act of determining an output from a number of inputs, using decision logic which may reference one or more business knowledge models.
Business Knowledge ModelA function encapsulating business knowledge, in the form of business rules, decision tables or an analytic model. Some of the tools may not support this element. In such cases, the decision logic is directly linked to the Decision rather than the business knowledge model.
Knowledge SourceThe authority for a business knowledge model or decision.
Input DataInformation used as an input by one or more decisions. It also denotes the parameters of a Business Knowledge Model.
Information RequirementInformation – input data or decision output – required for a decision.
Knowledge RequirementThe invocation of a business knowledge model.
Authority RequirementShowing the knowledge source of an element or the dependency of a knowledge source on input data.
To add a new Decision Graph,
`Document Types --> Business Logic --> Decision Graph`
Once you added a term, this will be the window.
### Decision Graph Properties
1. Name: Name of the flow
2. Node Type: Type of the node
3. Types Definition: List of types and functions
4. Variable Definition: Define variables to be used in the flow
6. Resources: List of resources to reference
### Toolbox
The toolbox has the following items.
Complete list of commands
Check the DRD Commands for the complete list.
At the top menu, you can see the following items.
1. Open the related logic document for this node: Open the logic document of a selected node
2. Create a logic document for selected node: Create a logic document for a selected node
3. Delete item: Delete a selected item
4. Highlight the impacts of the selected node: Highlight the impacts of a selected node
5. Properties: Properties of the page or a node
6. Variable properties: Define/ update/ delete variables
7. Compile to execution plan: Shows the XML version of the diagram
9. Overview of entire model: See the entire model
10. Zooming: Zoom in or zoom out
11. Export to image: Export the document to an image
12. Alignments: Align nodes
13. Resize: Resize selected nodes to a single size
14. Layout arrangements: Automatically arrange and layout the nodes
15. Change connections: Change connection styles to straight/ Bezier or Right Angle
### Node/ Connection Properties
If you click on a node/ connection, its property window will appear. The properties vary depending on the node type.
## Video Description
Updated on October 13, 2022 | 1,164 | 5,655 | {"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} | 2.515625 | 3 | CC-MAIN-2023-50 | longest | en | 0.919208 |
https://www.cfd-online.com/Forums/openfoam/74742-interfoam-shear-stress-water-air-interface.html | 1,685,752,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00716.warc.gz | 737,321,609 | 19,783 | # (interfoam) shear stress at water air interface !
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April 7, 2010, 14:58 (interfoam) shear stress at water air interface ! #1 Member Join Date: Dec 2009 Posts: 46 Rep Power: 15 hi guys I'm using interfoam solver for my 2D multiphase problem (water - air) My goal on my simulation is to calculate the shear stress on the water-air interface Because alpha_1 is not a discreet variable (range from 1 for water to 0 for air) so i can't really define the interface Can any one have experience in this matter tell me how to calculate the shear stress on the interface using some utility in openfoam or even share an effective idea.. thank you very much .. best regards.. Last edited by openfoam1; April 7, 2010 at 15:35.
April 7, 2010, 15:03 #2 Senior Member Sebastian Gatzka Join Date: Mar 2009 Location: Frankfurt, Germany Posts: 729 Rep Power: 19 I think in gerneral the interface is defined as . May this be a good asumption in your case? __________________ Schrödingers wife: "What did you do to the cat? It's half dead!"
April 7, 2010, 15:48
#3
Member
Join Date: Dec 2009
Posts: 46
Rep Power: 15
Quote:
Originally Posted by sega I think in gerneral the interface is defined as . May this be a good asumption in your case?
yes i know that is very good ,, but do you know a way to automatically calculate shear stress using a utility in openfoam or in paraview
i think if i didn't find any of those ,, the standard way is to calculate the velocity gradient in the cells of alpha1=0.5 ,, then i must get the gradient in the direction of the interface ( because the shear force must be tangent to the interface ),, which is a very complicated issue (i must also know the direction of the interface at every cell )
best regards
April 8, 2010, 13:31
#4
Senior Member
Sebastian Gatzka
Join Date: Mar 2009
Location: Frankfurt, Germany
Posts: 729
Rep Power: 19
Quote:
Originally Posted by openfoam1 thank you for your fast reply yes i know that is very good ,, but do you know a way to automatically calculate shear stress using a utility in openfoam or in paraview i think if i didn't find any of those ,, the standard way is to calculate the velocity gradient in the cells of alpha1=0.5 ,, then i must get the gradient in the direction of the interface ( because the shear force must be tangent to the interface ),, which is a very complicated issue (i must also know the direction of the interface at every cell ) best regards
Oh, I'm not aware if such a tool exists. I'm afraid you wil have to write it yourself.
But - good for you - the normal vector of the interface (its direction) is already used in interFoam to calculate the force due to surface tension and will point into the direction of the gradient of !
So concider this part solved. The rest should be up to you.
Get back to us!
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April 8, 2010, 16:10
#5
Member
Patricio Bohorquez
Join Date: Mar 2009
Location: Jaén, Spain
Posts: 95
Rep Power: 16
Quote:
Originally Posted by openfoam1 hi guys I'm using interfoam solver for my 2D multiphase problem (water - air) My goal on my simulation is to calculate the shear stress on the water-air interface Because alpha_1 is not a discreet variable (range from 1 for water to 0 for air) so i can't really define the interface Can any one have experience in this matter tell me how to calculate the shear stress on the interface using some utility in openfoam or even share an effective idea.. thank you very much .. best regards..
It is quite elegant to add a function to the interfaceProperties in order to do so automatically during the numerical simulation. There you could gain access to the interface curvature and to the velocity field.
Application
interfaceProperties
Description
Properties to aid interFoam :
1. Correct the gamma boundary condition for dynamic contact angle.
2. Calculate interface curvature.
3. Calculate the shear stress.
April 10, 2010, 10:46
#6
Member
Join Date: Dec 2009
Posts: 46
Rep Power: 15
Quote:
Originally Posted by sega Oh, I'm not aware if such a tool exists. I'm afraid you wil have to write it yourself. But - good for you - the normal vector of the interface (its direction) is already used in interFoam to calculate the force due to surface tension and will point into the direction of the gradient of ! So concider this part solved. The rest should be up to you. Get back to us!
Hi sega
do you know how can i get the velocity gradient in the output data of my final time like U , p , and alpha1
thank you
April 10, 2010, 11:28
#7
Senior Member
Sebastian Gatzka
Join Date: Mar 2009
Location: Frankfurt, Germany
Posts: 729
Rep Power: 19
Quote:
Originally Posted by openfoam1 Hi sega do you know how can i get the velocity gradient in the output data of my final time like U , p , and alpha1 thank you
If you have luck, you can create a velocity gradient field with foamCalc.
But from the information at slice 34 I'm not so sure anymore.
Otherwise you will have to do it yourself.
You can create the explicit gradient from the finite volume calculus class:
Code:
fvc::grad(U)
Have a look at the Programmers Guide (Page 37)!
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April 11, 2010, 02:20
#8
Member
Join Date: Dec 2009
Posts: 46
Rep Power: 15
Quote:
Originally Posted by sega If you have luck, you can create a velocity gradient field with foamCalc. But from the information at slice 34 I'm not so sure anymore. Otherwise you will have to do it yourself. You can create the explicit gradient from the finite volume calculus class: Code: fvc::grad(U) Have a look at the Programmers Guide (Page 37)!
if i know that this member functin grad(U) for fvc class is used to get the gradient of the velocity , how can i use it to get the gradient of the last time of the simulation
I'mvery sorry cause I'm quite new to C++ programming
best regards
Last edited by openfoam1; April 11, 2010 at 02:46.
April 11, 2010, 05:15
#9
Senior Member
Sebastian Gatzka
Join Date: Mar 2009
Location: Frankfurt, Germany
Posts: 729
Rep Power: 19
Quote:
Originally Posted by openfoam1 unfortunately foamCalc only have gradient magnitude magGrad , but i want the gradient itself if i know that this member functin grad(U) for fvc class is used to get the gradient of the velocity , how can i use it to get the gradient of the last time of the simulation I'mvery sorry cause I'm quite new to C++ programming best regards
You will have to loop though all the timesteps which have been written to the folders. So you will be able to get the gradient for each timestep.
I suggest you have a look into the code of the foamCalc function which caluclates magGrad or magU. I suppose it will do the same loop.
__________________
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January 13, 2014, 11:25 #10 Member Join Date: Aug 2011 Posts: 89 Rep Power: 13 Hello, I know this thread is a little bit old but nevertheless I hope someone could help me: I also want to calculate the interfacial shear stress. Did anyone find a good way to calculate it? Thanks a lot.
July 16, 2014, 12:31 #11 New Member Nara Shikamaru Join Date: Apr 2012 Posts: 22 Rep Power: 13 I am working on the same problem and almost done with the equations to be calculated. I got a lot of help from this post. I will suggest anyone working on this problem reads this post first. http://www.cfd-online.com/Forums/ope...bend-pipe.html I understand all the formulations given in that post. I am working with two incompressible immiscible fluids (VOF formulation) with interFoam solver. The only question I have is regarding the stress tensor that should be used for the calculation of traction. Should the viscous stress tensor be used for calculating the Traction vector? This is what I would do if I was doing solids or, The total stress tensor including the pressure term should be used? Last edited by shikamaru; July 16, 2014 at 19:22.
October 27, 2014, 03:59 #12 Member Join Date: Aug 2011 Posts: 89 Rep Power: 13 Hello shikamaru, are you still interested in this topic? I would use the first equation you wrote in the last post for i is not equal to j. Did you calculate the shear stress on the wall or at the interface? See you idefix | 2,139 | 8,378 | {"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} | 2.5625 | 3 | CC-MAIN-2023-23 | longest | en | 0.92848 |
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Nov 25, 2022, 15 tweets
Encyclopedia of Freemasonry:
On the topic of the "Number Nine"
#mason #masonic #freemason
Number Seven
Number Three
Twenty Seven and Other Suspicions
Comments: 9 is the number of the moon's completion being composed of a triangle whose points are 3 (3x3x3). Again this triangular cycle of 3 sets of 9 nakshatras gives the 27 day lunar cycle. 9 also gives birth to the svbc & moons magic square of 9x9=81.
Comments: 7 is composed of the square 4 of the Sun & the triangle 3 of the Moon. However it is rooted in the Sun (the totality of the 7 visible planets). We can also see this in the approx correlation to the solar revolution (-1.25) 7×52= 364. 52 is also a digitized 5+2=7.
Twelve:
Just as 7 is composed of the Sun's 4 and Moon's 3, so too is the number 12. And just as 7 is predominantly Solar, so too is the number 12.
5 elements 7 archetypes
MAGIC SQUARES
Likewise we can see the numerics behind all the planetary forces through their magic squares. In which the Sun square of 36 (6by6) had a Magic sum of 111=3 and a total sum of 666=9. The Moons square of 81 (9by9) gives a magic sum of 369=9 and total sum 3321=9.
Digitized Sums:
The total Sum of Magic squares will be either 1 or 9, while the magic Sums will be unique to each Planet:
Mars 2 & 1 (5by5)
Sun 3 & 9 (6by6)
Venus 4 & 1 (7by7)
Saturn 6 & 9 (3by3)
Jupiter 7 & 1 (4by4)
Mercury 8 & 1 (8by8)
Moon 9 & 9 (9by9)
1 and 5 are absent.
Magic square Dimensions:
You can easily see that the dimensions follow the Descending order:
Saturn 3by3
Jupiter 4by4
Mars 5by5
Sun 6by6
Venus 7by7
Mercury 8by8
Moon 9by9
2 groups of planets
Mars, Venus, Jupiter, Mercury having total magic square digitized summations of 1. These set up 2 pairs of opposites (Ven-Mars & Jup-Mer)
This creates the Square of 4
&
Saturn, Sun, Moon having total digitized summations of 9. (Su-Mo opp Sat). Triangle of 3
In the system of 7:
1 and 5 are absent as these are the two odd numbers which are secondary numbers in creation, whereas 7 and 9 are primary creators.
Su-Mo opposite Saturn
give the 3,6,9 triangle.
2-4 Mars-Venus opposition &
7-8 Jup-Mer opposition
give the Square of 2,4,7,8
In the system of 9 the odd numbers are:
Sun, Rahu, Ketu, Saturn, Mars
(1,3,5,7,9 respectively)
Evens are:
Mercury, Jupiter, Moon, Venus
(2,4,6,8 respecitvely)
(Symmetric vs. Asymmetric)
In the end astrology is just simple math of even and odd vibrations and repeating patterns of mathematical symmetry
All the best 🙏 | 812 | 2,592 | {"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} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.847405 |
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# Chap10onSol_partial - CHAPTER 10 Making Capital Investment...
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1 CHAPTER 10 Making Capital Investment Decisions 1. (LO1) The \$5 million acquisition cost of the land six years ago is a sunk cost. The \$5.3 million current aftertax value of the land is an opportunity cost if the land is used rather than sold off. The \$11.6 million cash outlay and \$425,000 grading expenses are the initial fixed asset investments needed to get the project going. Therefore, the proper year zero cash flow to use in evaluating this project is \$5,300,000 + 11,600,000 + 425,000 = \$17,325,000 2. (LO1) Sales due solely to the new product line are: 19,000(\$12,000) = \$228,000,000 Increased sales of the motor home line occur because of the new product line introduction; thus: 4,500(\$45,000) = \$202,500,000 in new sales is relevant. Erosion of luxury motor coach sales is also due to the new mid-size campers; thus: 900(\$85,000) = \$76,500,000 loss in sales is relevant. The net sales figure to use in evaluating the new line is thus: \$228,000,000 + 202,500,000 – 76,500,000 = \$354,000,000 3. (LO1) We need to construct a basic income statement. The income statement is: Sales \$ 740,000 Variable costs 444,000 Fixed costs 173,000 Depreciation 75,000 EBT \$ 48,000 16,800 Net income \$ 31,200 4. (LO3) To find the OCF, we need to complete the income statement as follows: Sales \$ 876,400 Costs 547,300 Depreciation 128,000 EBIT \$ 201,100 68,374 Net income \$ 132,726 The OCF for the company is: OCF = EBIT + Depreciation – Taxes OCF = \$201,100 + 128,000 – 68,374 OCF = \$260,726 The depreciation tax shield, also called the CCA tax shield, is the depreciation times the tax rate, so: Depreciation tax shield = t c Depreciation Depreciation tax shield = .34(\$128,000) Depreciation tax shield = \$43,520 The depreciation tax shield shows us the increase in OCF by being able to expense depreciation.
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2 5. (LO3) To calculate the OCF, we first need to calculate net income. The income statement is: Sales \$ 96,000 Variable costs 49,000 Depreciation 4,500 EBT \$ 42,500 14,875 Net income \$ 27,625 Teaching note: focus on the first approach and the tax shield approach. Using the most common financial calculation for OCF, we get: OCF = EBIT + Depreciation – Taxes OCF = \$42,500 + 4,500 – 14,875 OCF = \$32,125 The top-down approach to calculating OCF yields: OCF = Sales – Costs – Taxes OCF = \$96,000 – 49,000 – 14,875 OCF = \$32,125 The tax-shield approach is: OCF = (Sales – Costs)(1 – t C ) + t C Depreciation OCF = (\$96,000 – 49,000)(1 – .35) + .35(4,500) OCF = \$32,125 And the bottom-up approach is: OCF = Net income + Depreciation OCF = \$27,625 + 4,500 OCF = \$32,125 All four methods of calculating OCF should always give the same answer. 6.
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Ask a homework question - tutors are online | 940 | 3,256 | {"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} | 2.796875 | 3 | CC-MAIN-2018-05 | latest | en | 0.873838 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/How-to-find-the-list-of-the-K-nearest-neighbors/td-p/24256?nobounce | 1,516,665,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891546.92/warc/CC-MAIN-20180122232843-20180123012843-00577.warc.gz | 646,640,626 | 27,322 | New Contributor
Posts: 2
# How to find the list of the K nearest neighbors
Hello,
I’m searching a proc in sas to find the K nearest neighbors in a set of points.
In the following data step, I generate a random set of 100 observations.
DATA observations;
do ID=1 TO 100;
x1 = RAND('NORMAL',0,1);
x2 = RAND('NORMAL',0,1);
x3 = RAND('NORMAL',0,1);
output;
end;
run;
For example, with this set of 100 observations, is there a proc to search the 10 nearest neighbor (Euclidian distance) of the point [ 0.5 ; 0.5 ; 0.5 ]?
Please note that I already found proc discrim to apply a KNN classification. But I want the list of the K nearest neighbor and not to make a classification.
Thx!
Super User
Posts: 7,269
## Re: How to find the list of the K nearest neighbors
Why not just code it yourself?
`data observations; do ID=1 TO 100; x1 = RAND('NORMAL',0,1); x2 = RAND('NORMAL',0,1); x3 = RAND('NORMAL',0,1); output; end;run;proc sql noprint ; create table distance as select distinct id,x1,x2,x3,sqrt((x1-y1)**2 + (x2-y2)**2 + (x3-y3)**2) as distance from observations , (select 0.5 as y1,0.5 as y2, 0.5 as y3 from observations(obs=1) ) order by distance desc ;quit;data top10; set distance (obs=10);run;proc print; run;`
Posts: 5,055
## Re: How to find the list of the K nearest neighbors
One quick way to find nearest neighbors in a large set of observations is PROC MODECLUS. Your specific problem could be solved by:
`DATA observations;do ID=1 TO 100; x1 = RAND('NORMAL',0,1); x2 = RAND('NORMAL',0,1); x3 = RAND('NORMAL',0,1);output;end;run;DATA position;ID = -999;x1 = 0.5;x2 = 0.5;x3 = 0.5;run;proc sql;create table test as(select * from position)union all(select * from observations)order by ID;ods _all_ close;proc modeclus data=test dk=11 /* = 10 observations + 1 position */ neighbor;var x1 x2 x3;id ID;ods output Neighbor=nTest;run;ods listing;proc sql inobs=10;select nBor as ID, distance from nTest;`
PG
New Contributor
Posts: 2
## Re: How to find the list of the K nearest neighbors
Hello Tom & PGStats,
many thanks for your help !
based on the message from Tom, I have buil the following macro "listOfKNN":
data observations;
do ID=1 TO 100000;
x1 = RAND('NORMAL',0,1);
x2 = RAND('NORMAL',0,1);
x3 = RAND('NORMAL',0,1);
output;
end;
run;
%macro listOfKNN (obser , target, K, nameOutput);
%Let dim = %eval(%SYSFUNC(count(&target,%NRSTR( )))+1) ;
%let ttt =;
%DO ii = 1 %TO &dim;
%let ttt = &ttt.(x&ii.-%scan(&target,&ii," "))**2;
%IF &ii NE &dim %THEN
%let ttt = &ttt.+;
%END;
proc sql noprint ;
create table &nameOutput as
select distinct *, sqrt(&ttt)as distance
from &obser
order by distance;
quit;
data &nameOutput;
set &nameOutput (obs=&K);
drop distance;
run;
%mend listOfKNN;
/*make a test*/
%listOfKNN( obser = observations,
target = 0.5 1 1,
K = 5,
nameOutput = out
);
B.R.
Olivier
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https://www.quantifiedstrategies.com/zweig-breath-thrust-indicator/ | 1,674,770,922,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00875.warc.gz | 992,374,482 | 45,840 | Last Updated on October 22, 2022
Zweig Breadth Thrust Indicator is a market breadth indicator that helps you understand the internal strength of the market, at least that’s what the theory says. There are many breadth indicators – among them the Zweig Breadth Thrust Indicator (one of many).
The best way to measure market breadth is by looking at the number of advancers and decliners. This article looks at The Zweig Breadth Thrust Indicator and shows you how you can use the ratio between advancing and declining stocks to make an indicator. Does the Zweig Market Breadth Thrust Indicator work? We backtest Zweig Breadth Thrust Indicator.
## What is the Zweig Market Breadth Thrust?
It’s an overbought/oversold indicator that oscillates up and down and is applied to the stock market. The indicator is named after its inventor – Martin Zweig. Mr. Zweig (1942 -2013) was an investor, advisor, and analyst and held a Ph.D. in finance. (The famous writer Jason Zweig is NOT related to Martin Zweig.)
Zweig Breadth Thrust Indicator = A 10-day moving average of the breadth
The formula creates an indicator that goes up and down, but Zweig was original when he looked at the indicator (as we understand how he interpreted readings):
A buy signal happens when it goes from an oversold market to overbought within any ten-day period. To be precise: An oversold position is a reading below 0.4 and an overbought position is when the breadth thrust indicator is above 0.615.
We assume that Zweig’s reasoning is that such a sudden change in market sentiment over a short period of time signals a bullish reversal.
## A visual look at the Zweig Market Breadth Thrust
Let’s type the formula in Amibroker and see what the indicator looks like on the S&P 500:
As you can see, the Zweig Breadth Thrust Indicator goes up and down in rapid moves.
## Does the Zweig Breadth Thrust Indicator work?
Let’s test the indicator as Zweig recommended. We test based on the following criteria and use SPY as a proxy for the S&P 500:
1. Today’s reading must be at least 0.615.
2. At any time during the last ten days, the reading must have been less than 0.4.
3. If 1 and 2 are true, we buy the close and sell x days later.
As you might have already imagined, the buy signal doesn’t happen often: Since the year 2000 we have only witnessed 20 observations, some of the signals “overlapping” or happening very close to each other.
The table below shows the results when we exit a trade (time stop) after 5 to 50 days with 5 days intervals (we like to optimize in trading to get a better understanding of the results):
The first row shows the number of days when we exit our position. It’s ten trades and all trades are profitable – the ones holding for the longest time are the most profitable.
The few trades make the setup for most traders pretty uninteresting, we would assume.
## Let’s test Zweig Breadth Trust Indicator as an oversold and overbought indicator
Let’s test by going long the S&P 500 when the Zweig indicator is below 0.4 and sell when it crosses a higher reading. Unfortunately, no matter what we tested we couldn’t come up with anything that looked promising. We even tested by changing the length of the moving average of the indicator, but to no avail.
However, if we use the Zweig Breadth Thrust Indicator as an indicator together with another indicator we get some really good results:
The number of trades is 266, the average gain per trade is 0.73%, the win ratio is 77%, the profit factor is 2.55, and the max drawdown is 23%. We don’t want to publish the strategy because we might consider making it a Trading Edge for our subscribers: | 849 | 3,662 | {"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} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.937075 |
https://archinfos.com/library/lecture/read/105962-how-do-you-sort-numbers-from-smallest-to-largest-in-python | 1,670,012,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00802.warc.gz | 132,119,118 | 4,610 | # How do you sort numbers from smallest to largest in Python?
## How do you sort numbers from smallest to largest in Python?
It's a built-in function that is available in a standard installation of Python. sorted() , with no additional arguments or parameters, is ordering the values in numbers in an ascending order, meaning smallest to largest.
## How do I merge two sorted lists?
Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list. The new list should be made by splicing together the nodes of the first two lists.
## How do you write Lambda?
Log in to your AWS Account, and navigate to the Lambda console. Click on Create function. We'll be creating a Lambda from scratch, so select the Author from scratch option. Enter an appropriate name for your Lambda function, select a Python runtime and define a role for your Lambda to use.
## What is lambda in programming?
Lambda expressions (or lambda functions) are essentially blocks of code that can be assigned to variables, passed as an argument, or returned from a function call, in languages that support high-order functions. They have been part of programming languages for quite some time.
## What is lambda in math?
Lambda calculus (also written as λ-calculus) is a formal system in mathematical logic for expressing computation based on function abstraction and application using variable binding and substitution. ... Function definition (M is a lambda term). The variable x becomes bound in the expression.
## What does λ mean?
Lambda (uppercase/lowercase Λ λ) is a letter of the Greek alphabet. It is used to represent the "l" sound in Ancient and Modern Greek. In the system of Greek numerals, it has a value of 30. Letters that came from it include the Roman L and Cyrillic Л.
## What is lambda in English?
1 : the 11th letter of the Greek alphabet — see Alphabet Table. 2 : an uncharged unstable elementary particle that has a mass 2183 times that of an electron and that decays typically into a nucleon and a pion. | 448 | 2,156 | {"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} | 2.765625 | 3 | CC-MAIN-2022-49 | latest | en | 0.904231 |
http://us.metamath.org/mpeuni/mmtheorems165.html | 1,643,455,682,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00228.warc.gz | 62,666,265 | 20,869 | Home Metamath Proof ExplorerTheorem List (p. 165 of 424) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
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Theorem List for Metamath Proof Explorer - 16401-16500 *Has distinct variable group(s)
TypeLabelDescription
Statement
Definitiondf-sect 16401* Function returning the section relation in a category. Given arrows 𝑓:𝑋𝑌 and 𝑔:𝑌𝑋, we say 𝑓Sect𝑔, that is, 𝑓 is a section of 𝑔, if 𝑔𝑓 = 1‘𝑋. If there there is an arrow 𝑔 with 𝑓Sect𝑔, the arrow 𝑓 is called a section, see definition 7.19 of [Adamek] p. 106. (Contributed by Mario Carneiro, 2-Jan-2017.)
Sect = (𝑐 ∈ Cat ↦ (𝑥 ∈ (Base‘𝑐), 𝑦 ∈ (Base‘𝑐) ↦ {⟨𝑓, 𝑔⟩ ∣ [(Hom ‘𝑐) / ]((𝑓 ∈ (𝑥𝑦) ∧ 𝑔 ∈ (𝑦𝑥)) ∧ (𝑔(⟨𝑥, 𝑦⟩(comp‘𝑐)𝑥)𝑓) = ((Id‘𝑐)‘𝑥))}))
Definitiondf-inv 16402* The inverse relation in a category. Given arrows 𝑓:𝑋𝑌 and 𝑔:𝑌𝑋, we say 𝑔Inv𝑓, that is, 𝑔 is an inverse of 𝑓, if 𝑔 is a section of 𝑓 and 𝑓 is a section of 𝑔. Definition 3.8 of [Adamek] p. 28. (Contributed by FL, 22-Dec-2008.) (Revised by Mario Carneiro, 2-Jan-2017.)
Inv = (𝑐 ∈ Cat ↦ (𝑥 ∈ (Base‘𝑐), 𝑦 ∈ (Base‘𝑐) ↦ ((𝑥(Sect‘𝑐)𝑦) ∩ (𝑦(Sect‘𝑐)𝑥))))
Definitiondf-iso 16403* Function returning the isomorphisms of the category 𝑐. Definition 3.8 of [Adamek] p. 28, and definition in [Lang] p. 54. (Contributed by FL, 9-Jun-2014.) (Revised by Mario Carneiro, 2-Jan-2017.)
Iso = (𝑐 ∈ Cat ↦ ((𝑥 ∈ V ↦ dom 𝑥) ∘ (Inv‘𝑐)))
Theoremsectffval 16404* Value of the section operation. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & · = (comp‘𝐶) & 1 = (Id‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑𝑆 = (𝑥𝐵, 𝑦𝐵 ↦ {⟨𝑓, 𝑔⟩ ∣ ((𝑓 ∈ (𝑥𝐻𝑦) ∧ 𝑔 ∈ (𝑦𝐻𝑥)) ∧ (𝑔(⟨𝑥, 𝑦· 𝑥)𝑓) = ( 1𝑥))}))
Theoremsectfval 16405* Value of the section relation. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & · = (comp‘𝐶) & 1 = (Id‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝑋𝑆𝑌) = {⟨𝑓, 𝑔⟩ ∣ ((𝑓 ∈ (𝑋𝐻𝑌) ∧ 𝑔 ∈ (𝑌𝐻𝑋)) ∧ (𝑔(⟨𝑋, 𝑌· 𝑋)𝑓) = ( 1𝑋))})
Theoremsectss 16406 The section relation is a relation between morphisms from 𝑋 to 𝑌 and morphisms from 𝑌 to 𝑋. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & · = (comp‘𝐶) & 1 = (Id‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝑋𝑆𝑌) ⊆ ((𝑋𝐻𝑌) × (𝑌𝐻𝑋)))
Theoremissect 16407 The property "𝐹 is a section of 𝐺". (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & · = (comp‘𝐶) & 1 = (Id‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝐹(𝑋𝑆𝑌)𝐺 ↔ (𝐹 ∈ (𝑋𝐻𝑌) ∧ 𝐺 ∈ (𝑌𝐻𝑋) ∧ (𝐺(⟨𝑋, 𝑌· 𝑋)𝐹) = ( 1𝑋))))
Theoremissect2 16408 Property of being a section. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & · = (comp‘𝐶) & 1 = (Id‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐻𝑌)) & (𝜑𝐺 ∈ (𝑌𝐻𝑋)) (𝜑 → (𝐹(𝑋𝑆𝑌)𝐺 ↔ (𝐺(⟨𝑋, 𝑌· 𝑋)𝐹) = ( 1𝑋)))
Theoremsectcan 16409 If 𝐺 is a section of 𝐹 and 𝐹 is a section of 𝐻, then 𝐺 = 𝐻. Proposition 3.10 of [Adamek] p. 28. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐺(𝑋𝑆𝑌)𝐹) & (𝜑𝐹(𝑌𝑆𝑋)𝐻) (𝜑𝐺 = 𝐻)
Theoremsectco 16410 Composition of two sections. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & · = (comp‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝑍𝐵) & (𝜑𝐹(𝑋𝑆𝑌)𝐺) & (𝜑𝐻(𝑌𝑆𝑍)𝐾) (𝜑 → (𝐻(⟨𝑋, 𝑌· 𝑍)𝐹)(𝑋𝑆𝑍)(𝐺(⟨𝑍, 𝑌· 𝑋)𝐾))
Theoremisofval 16411* Function value of the function returning the isomorphisms of a category. (Contributed by AV, 5-Apr-2017.)
(𝐶 ∈ Cat → (Iso‘𝐶) = ((𝑥 ∈ V ↦ dom 𝑥) ∘ (Inv‘𝐶)))
Theoreminvffval 16412* Value of the inverse relation. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑆 = (Sect‘𝐶) (𝜑𝑁 = (𝑥𝐵, 𝑦𝐵 ↦ ((𝑥𝑆𝑦) ∩ (𝑦𝑆𝑥))))
Theoreminvfval 16413 Value of the inverse relation. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑆 = (Sect‘𝐶) (𝜑 → (𝑋𝑁𝑌) = ((𝑋𝑆𝑌) ∩ (𝑌𝑆𝑋)))
Theoremisinv 16414 Value of the inverse relation. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑆 = (Sect‘𝐶) (𝜑 → (𝐹(𝑋𝑁𝑌)𝐺 ↔ (𝐹(𝑋𝑆𝑌)𝐺𝐺(𝑌𝑆𝑋)𝐹)))
Theoreminvss 16415 The inverse relation is a relation between morphisms 𝐹:𝑋𝑌 and their inverses 𝐺:𝑌𝑋. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐻 = (Hom ‘𝐶) (𝜑 → (𝑋𝑁𝑌) ⊆ ((𝑋𝐻𝑌) × (𝑌𝐻𝑋)))
Theoreminvsym 16416 The inverse relation is symmetric. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝐹(𝑋𝑁𝑌)𝐺𝐺(𝑌𝑁𝑋)𝐹))
Theoreminvsym2 16417 The inverse relation is symmetric. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑(𝑋𝑁𝑌) = (𝑌𝑁𝑋))
Theoreminvfun 16418 The inverse relation is a function, which is to say that every morphism has at most one inverse. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → Fun (𝑋𝑁𝑌))
Theoremisoval 16419 The isomorphisms are the domain of the inverse relation. (Contributed by Mario Carneiro, 2-Jan-2017.) (Proof shortened by AV, 21-May-2020.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) (𝜑 → (𝑋𝐼𝑌) = dom (𝑋𝑁𝑌))
Theoreminviso1 16420 If 𝐺 is an inverse to 𝐹, then 𝐹 is an isomorphism. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) & (𝜑𝐹(𝑋𝑁𝑌)𝐺) (𝜑𝐹 ∈ (𝑋𝐼𝑌))
Theoreminviso2 16421 If 𝐺 is an inverse to 𝐹, then 𝐺 is an isomorphism. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) & (𝜑𝐹(𝑋𝑁𝑌)𝐺) (𝜑𝐺 ∈ (𝑌𝐼𝑋))
Theoreminvf 16422 The inverse relation is a function from isomorphisms to isomorphisms. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) (𝜑 → (𝑋𝑁𝑌):(𝑋𝐼𝑌)⟶(𝑌𝐼𝑋))
Theoreminvf1o 16423 The inverse relation is a bijection from isomorphisms to isomorphisms. This means that every isomorphism 𝐹 ∈ (𝑋𝐼𝑌) has a unique inverse, denoted by ((Inv‘𝐶)‘𝐹). Remark 3.12 of [Adamek] p. 28. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) (𝜑 → (𝑋𝑁𝑌):(𝑋𝐼𝑌)–1-1-onto→(𝑌𝐼𝑋))
Theoreminvinv 16424 The inverse of the inverse of an isomorphism is itself. Proposition 3.14(1) of [Adamek] p. 29. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) (𝜑 → ((𝑌𝑁𝑋)‘((𝑋𝑁𝑌)‘𝐹)) = 𝐹)
Theoreminvco 16425 The composition of two isomorphisms is an isomorphism, and the inverse is the composition of the individual inverses. Proposition 3.14(2) of [Adamek] p. 29. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) & · = (comp‘𝐶) & (𝜑𝑍𝐵) & (𝜑𝐺 ∈ (𝑌𝐼𝑍)) (𝜑 → (𝐺(⟨𝑋, 𝑌· 𝑍)𝐹)(𝑋𝑁𝑍)(((𝑋𝑁𝑌)‘𝐹)(⟨𝑍, 𝑌· 𝑋)((𝑌𝑁𝑍)‘𝐺)))
Theoremdfiso2 16426* Alternate definition of an isomorphism of a category, according to definition 3.8 in [Adamek] p. 28. (Contributed by AV, 10-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & (𝜑𝐶 ∈ Cat) & 𝐼 = (Iso‘𝐶) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐻𝑌)) & 1 = (Id‘𝐶) & = (⟨𝑋, 𝑌⟩(comp‘𝐶)𝑋) & = (⟨𝑌, 𝑋⟩(comp‘𝐶)𝑌) (𝜑 → (𝐹 ∈ (𝑋𝐼𝑌) ↔ ∃𝑔 ∈ (𝑌𝐻𝑋)((𝑔 𝐹) = ( 1𝑋) ∧ (𝐹 𝑔) = ( 1𝑌))))
Theoremdfiso3 16427* Alternate definition of an isomorphism of a category as a section in both directions. (Contributed by AV, 11-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & 𝐼 = (Iso‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐻𝑌)) (𝜑 → (𝐹 ∈ (𝑋𝐼𝑌) ↔ ∃𝑔 ∈ (𝑌𝐻𝑋)(𝑔(𝑌𝑆𝑋)𝐹𝐹(𝑋𝑆𝑌)𝑔)))
Theoreminveq 16428 If there are two inverses of an morphism, these inverses are equal. Corollary 3.11 of [Adamek] p. 28. (Contributed by AV, 10-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → ((𝐹(𝑋𝑁𝑌)𝐺𝐹(𝑋𝑁𝑌)𝐾) → 𝐺 = 𝐾))
Theoremisofn 16429 The function value of the function returning the isomorphisms of a category is a function over the square product of the base set of the category. (Contributed by AV, 5-Apr-2017.)
(𝐶 ∈ Cat → (Iso‘𝐶) Fn ((Base‘𝐶) × (Base‘𝐶)))
Theoremisohom 16430 An isomorphism is a homomorphism. (Contributed by Mario Carneiro, 27-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐻 = (Hom ‘𝐶) & 𝐼 = (Iso‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝑋𝐼𝑌) ⊆ (𝑋𝐻𝑌))
Theoremisoco 16431 The composition of two isomorphisms is an isomorphism. Proposition 3.14(2) of [Adamek] p. 29. (Contributed by Mario Carneiro, 2-Jan-2017.)
𝐵 = (Base‘𝐶) & · = (comp‘𝐶) & 𝐼 = (Iso‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝑍𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) & (𝜑𝐺 ∈ (𝑌𝐼𝑍)) (𝜑 → (𝐺(⟨𝑋, 𝑌· 𝑍)𝐹) ∈ (𝑋𝐼𝑍))
Theoremoppcsect 16432 A section in the opposite category. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑂 = (oppCat‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑆 = (Sect‘𝐶) & 𝑇 = (Sect‘𝑂) (𝜑 → (𝐹(𝑋𝑇𝑌)𝐺𝐺(𝑋𝑆𝑌)𝐹))
Theoremoppcsect2 16433 A section in the opposite category. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑂 = (oppCat‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑆 = (Sect‘𝐶) & 𝑇 = (Sect‘𝑂) (𝜑 → (𝑋𝑇𝑌) = (𝑋𝑆𝑌))
Theoremoppcinv 16434 An inverse in the opposite category. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑂 = (oppCat‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Inv‘𝐶) & 𝐽 = (Inv‘𝑂) (𝜑 → (𝑋𝐽𝑌) = (𝑌𝐼𝑋))
Theoremoppciso 16435 An isomorphism in the opposite category. See also remark 3.9 in [Adamek] p. 28. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑂 = (oppCat‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝐼 = (Iso‘𝐶) & 𝐽 = (Iso‘𝑂) (𝜑 → (𝑋𝐽𝑌) = (𝑌𝐼𝑋))
Theoremsectmon 16436 If 𝐹 is a section of 𝐺, then 𝐹 is a monomorphism. A monomorphism that arises from a section is also known as a split monomorphism. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑀 = (Mono‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹(𝑋𝑆𝑌)𝐺) (𝜑𝐹 ∈ (𝑋𝑀𝑌))
Theoremmonsect 16437 If 𝐹 is a monomorphism and 𝐺 is a section of 𝐹, then 𝐺 is an inverse of 𝐹 and they are both isomorphisms. This is also stated as "a monomorphism which is also a split epimorphism is an isomorphism". (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝑀 = (Mono‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑁 = (Inv‘𝐶) & (𝜑𝐹 ∈ (𝑋𝑀𝑌)) & (𝜑𝐺(𝑌𝑆𝑋)𝐹) (𝜑𝐹(𝑋𝑁𝑌)𝐺)
Theoremsectepi 16438 If 𝐹 is a section of 𝐺, then 𝐺 is an epimorphism. An epimorphism that arises from a section is also known as a split epimorphism. (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐸 = (Epi‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹(𝑋𝑆𝑌)𝐺) (𝜑𝐺 ∈ (𝑌𝐸𝑋))
Theoremepisect 16439 If 𝐹 is an epimorphism and 𝐹 is a section of 𝐺, then 𝐺 is an inverse of 𝐹 and they are both isomorphisms. This is also stated as "an epimorphism which is also a split monomorphism is an isomorphism". (Contributed by Mario Carneiro, 3-Jan-2017.)
𝐵 = (Base‘𝐶) & 𝐸 = (Epi‘𝐶) & 𝑆 = (Sect‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & 𝑁 = (Inv‘𝐶) & (𝜑𝐹 ∈ (𝑋𝐸𝑌)) & (𝜑𝐹(𝑋𝑆𝑌)𝐺) (𝜑𝐹(𝑋𝑁𝑌)𝐺)
Theoremsectid 16440 The identity is a section of itself. (Contributed by AV, 8-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Id‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) (𝜑 → (𝐼𝑋)(𝑋(Sect‘𝐶)𝑋)(𝐼𝑋))
Theoreminvid 16441 The inverse of the identity is the identity. (Contributed by AV, 8-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Id‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) (𝜑 → (𝐼𝑋)(𝑋(Inv‘𝐶)𝑋)(𝐼𝑋))
Theoremidiso 16442 The identity is an isomorphism. Example 3.13 of [Adamek] p. 28. (Contributed by AV, 8-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Id‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) (𝜑 → (𝐼𝑋) ∈ (𝑋(Iso‘𝐶)𝑋))
Theoremidinv 16443 The inverse of the identity is the identity. Example 3.13 of [Adamek] p. 28. (Contributed by AV, 9-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Id‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) (𝜑 → ((𝑋(Inv‘𝐶)𝑋)‘(𝐼𝑋)) = (𝐼𝑋))
Theoreminvisoinvl 16444 The inverse of an isomorphism 𝐹 (which is unique because of invf 16422 and is therefore denoted by ((𝑋𝑁𝑌)‘𝐹), see also remark 3.12 in [Adamek] p. 28) is invers to the isomorphism. (Contributed by AV, 9-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Iso‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) (𝜑 → ((𝑋𝑁𝑌)‘𝐹)(𝑌𝑁𝑋)𝐹)
Theoreminvisoinvr 16445 The inverse of an isomorphism is invers to the isomorphism. (Contributed by AV, 9-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Iso‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) (𝜑𝐹(𝑋𝑁𝑌)((𝑋𝑁𝑌)‘𝐹))
Theoreminvcoisoid 16446 The inverse of an isomorphism composed with the isomorphism is the identity. (Contributed by AV, 5-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Iso‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) & 1 = (Id‘𝐶) & = (⟨𝑋, 𝑌⟩(comp‘𝐶)𝑋) (𝜑 → (((𝑋𝑁𝑌)‘𝐹) 𝐹) = ( 1𝑋))
Theoremisocoinvid 16447 The inverse of an isomorphism composed with the isomorphism is the identity. (Contributed by AV, 10-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝐼 = (Iso‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) & 1 = (Id‘𝐶) & = (⟨𝑌, 𝑋⟩(comp‘𝐶)𝑌) (𝜑 → (𝐹 ((𝑋𝑁𝑌)‘𝐹)) = ( 1𝑌))
Theoremrcaninv 16448 Right cancellation of an inverse of an isomorphism. (Contributed by AV, 5-Apr-2017.)
𝐵 = (Base‘𝐶) & 𝑁 = (Inv‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝑍𝐵) & (𝜑𝐹 ∈ (𝑌(Iso‘𝐶)𝑋)) & (𝜑𝐺 ∈ (𝑌(Hom ‘𝐶)𝑍)) & (𝜑𝐻 ∈ (𝑌(Hom ‘𝐶)𝑍)) & 𝑅 = ((𝑌𝑁𝑋)‘𝐹) & = (⟨𝑋, 𝑌⟩(comp‘𝐶)𝑍) (𝜑 → ((𝐺 𝑅) = (𝐻 𝑅) → 𝐺 = 𝐻))
8.1.5 Isomorphic objects
In this subsection, the "is isomorphic to" relation between objects of a category 𝑐 is defined (see df-cic 16450). It is shown that this relation is an equivalence relation, see cicer 16460.
Syntaxccic 16449 Extend class notation to include the category isomorphism relation.
class 𝑐
Definitiondf-cic 16450 Function returning the set of isomorphic objects for each category 𝑐. Definition 3.15 of [Adamek] p. 29. Analogous to the definition of the group isomorphism relation 𝑔, see df-gic 17696. (Contributed by AV, 4-Apr-2020.)
𝑐 = (𝑐 ∈ Cat ↦ ((Iso‘𝑐) supp ∅))
Theoremcicfval 16451 The set of isomorphic objects of the category 𝑐. (Contributed by AV, 4-Apr-2020.)
(𝐶 ∈ Cat → ( ≃𝑐𝐶) = ((Iso‘𝐶) supp ∅))
Theorembrcic 16452 The relation "is isomorphic to" for categories. (Contributed by AV, 5-Apr-2020.)
𝐼 = (Iso‘𝐶) & 𝐵 = (Base‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝑋( ≃𝑐𝐶)𝑌 ↔ (𝑋𝐼𝑌) ≠ ∅))
Theoremcic 16453* Objects 𝑋 and 𝑌 in a category are isomorphic provided that there is an isomorphism 𝑓:𝑋𝑌, see definition 3.15 of [Adamek] p. 29. (Contributed by AV, 4-Apr-2020.)
𝐼 = (Iso‘𝐶) & 𝐵 = (Base‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) (𝜑 → (𝑋( ≃𝑐𝐶)𝑌 ↔ ∃𝑓 𝑓 ∈ (𝑋𝐼𝑌)))
Theorembrcici 16454 Prove that two objects are isomorphic by an explicit isomorphism. (Contributed by AV, 4-Apr-2020.)
𝐼 = (Iso‘𝐶) & 𝐵 = (Base‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑋𝐵) & (𝜑𝑌𝐵) & (𝜑𝐹 ∈ (𝑋𝐼𝑌)) (𝜑𝑋( ≃𝑐𝐶)𝑌)
Theoremcicref 16455 Isomorphism is reflexive. (Contributed by AV, 5-Apr-2020.)
((𝐶 ∈ Cat ∧ 𝑂 ∈ (Base‘𝐶)) → 𝑂( ≃𝑐𝐶)𝑂)
Theoremciclcl 16456 Isomorphism implies the left side is an object. (Contributed by AV, 5-Apr-2020.)
((𝐶 ∈ Cat ∧ 𝑅( ≃𝑐𝐶)𝑆) → 𝑅 ∈ (Base‘𝐶))
Theoremcicrcl 16457 Isomorphism implies the right side is an object. (Contributed by AV, 5-Apr-2020.)
((𝐶 ∈ Cat ∧ 𝑅( ≃𝑐𝐶)𝑆) → 𝑆 ∈ (Base‘𝐶))
Theoremcicsym 16458 Isomorphism is symmetric. (Contributed by AV, 5-Apr-2020.)
((𝐶 ∈ Cat ∧ 𝑅( ≃𝑐𝐶)𝑆) → 𝑆( ≃𝑐𝐶)𝑅)
Theoremcictr 16459 Isomorphism is transitive. (Contributed by AV, 5-Apr-2020.)
((𝐶 ∈ Cat ∧ 𝑅( ≃𝑐𝐶)𝑆𝑆( ≃𝑐𝐶)𝑇) → 𝑅( ≃𝑐𝐶)𝑇)
Theoremcicer 16460 Isomorphism is an equivalence relation on objects of a category. Remark 3.16 in [Adamek] p. 29. (Contributed by AV, 5-Apr-2020.)
(𝐶 ∈ Cat → ( ≃𝑐𝐶) Er (Base‘𝐶))
8.1.6 Subcategories
Syntaxcssc 16461 Extend class notation to include the subset relation for subcategories.
class cat
Syntaxcresc 16462 Extend class notation to include category restriction (which is like structure restriction but also allows limiting the collection of morphisms).
class cat
Syntaxcsubc 16463 Extend class notation to include the collection of subcategories of a category.
class Subcat
Definitiondf-ssc 16464* Define the subset relation for subcategories. Despite the name, this is not really a "category-aware" definition, which is to say it makes no explicit references to homsets or composition; instead this is a subset-like relation on the functions that are used as subcategory specifications in df-subc 16466, which makes it play an analogous role to the subset relation applied to the subgroups of a group. (Contributed by Mario Carneiro, 6-Jan-2017.)
cat = {⟨, 𝑗⟩ ∣ ∃𝑡(𝑗 Fn (𝑡 × 𝑡) ∧ ∃𝑠 ∈ 𝒫 𝑡X𝑥 ∈ (𝑠 × 𝑠)𝒫 (𝑗𝑥))}
Definitiondf-resc 16465* Define the restriction of a category to a given set of arrows. (Contributed by Mario Carneiro, 4-Jan-2017.)
cat = (𝑐 ∈ V, ∈ V ↦ ((𝑐s dom dom ) sSet ⟨(Hom ‘ndx), ⟩))
Definitiondf-subc 16466* (Subcat‘𝐶) is the set of all the subcategory specifications of the category 𝐶. Like df-subg 17585, this is not actually a collection of categories (as in definition 4.1(a) of [Adamek] p. 48), but only sets which when given operations from the base category (using df-resc 16465) form a category. All the objects and all the morphisms of the subcategory belong to the supercategory. The identity of an object, the domain and the codomain of a morphism are the same in the subcategory and the supercategory. The composition of the subcategory is a restriction of the composition of the supercategory. (Contributed by FL, 17-Sep-2009.) (Revised by Mario Carneiro, 4-Jan-2017.)
Subcat = (𝑐 ∈ Cat ↦ { ∣ (cat (Homf𝑐) ∧ [dom dom / 𝑠]𝑥𝑠 (((Id‘𝑐)‘𝑥) ∈ (𝑥𝑥) ∧ ∀𝑦𝑠𝑧𝑠𝑓 ∈ (𝑥𝑦)∀𝑔 ∈ (𝑦𝑧)(𝑔(⟨𝑥, 𝑦⟩(comp‘𝑐)𝑧)𝑓) ∈ (𝑥𝑧)))})
Theoremsscrel 16467 The subcategory subset relation is a relation. (Contributed by Mario Carneiro, 6-Jan-2017.)
Rel ⊆cat
Theorembrssc 16468* The subcategory subset relation is a relation. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝐻cat 𝐽 ↔ ∃𝑡(𝐽 Fn (𝑡 × 𝑡) ∧ ∃𝑠 ∈ 𝒫 𝑡𝐻X𝑥 ∈ (𝑠 × 𝑠)𝒫 (𝐽𝑥)))
Theoremsscpwex 16469* An analogue of pwex 4846 for the subcategory subset relation: The collection of subcategory subsets of a given set 𝐽 is a set. (Contributed by Mario Carneiro, 6-Jan-2017.)
{cat 𝐽} ∈ V
Theoremsubcrcl 16470 Reverse closure for the subcategory predicate. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝐻 ∈ (Subcat‘𝐶) → 𝐶 ∈ Cat)
Theoremsscfn1 16471 The subcategory subset relation is defined on functions with square domain. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻cat 𝐽) & (𝜑𝑆 = dom dom 𝐻) (𝜑𝐻 Fn (𝑆 × 𝑆))
Theoremsscfn2 16472 The subcategory subset relation is defined on functions with square domain. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻cat 𝐽) & (𝜑𝑇 = dom dom 𝐽) (𝜑𝐽 Fn (𝑇 × 𝑇))
Theoremssclem 16473 Lemma for ssc1 16475 and similar theorems. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻 Fn (𝑆 × 𝑆)) (𝜑 → (𝐻 ∈ V ↔ 𝑆 ∈ V))
Theoremisssc 16474* Value of the subcategory subset relation when the arguments are known functions. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝐽 Fn (𝑇 × 𝑇)) & (𝜑𝑇𝑉) (𝜑 → (𝐻cat 𝐽 ↔ (𝑆𝑇 ∧ ∀𝑥𝑆𝑦𝑆 (𝑥𝐻𝑦) ⊆ (𝑥𝐽𝑦))))
Theoremssc1 16475 Infer subset relation on objects from the subcategory subset relation. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝐽 Fn (𝑇 × 𝑇)) & (𝜑𝐻cat 𝐽) (𝜑𝑆𝑇)
Theoremssc2 16476 Infer subset relation on morphisms from the subcategory subset relation. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝐻cat 𝐽) & (𝜑𝑋𝑆) & (𝜑𝑌𝑆) (𝜑 → (𝑋𝐻𝑌) ⊆ (𝑋𝐽𝑌))
Theoremsscres 16477 Any function restricted to a square domain is a subcategory subset of the original. (Contributed by Mario Carneiro, 6-Jan-2017.)
((𝐻 Fn (𝑆 × 𝑆) ∧ 𝑆𝑉) → (𝐻 ↾ (𝑇 × 𝑇)) ⊆cat 𝐻)
Theoremsscid 16478 The subcategory subset relation is reflexive. (Contributed by Mario Carneiro, 6-Jan-2017.)
((𝐻 Fn (𝑆 × 𝑆) ∧ 𝑆𝑉) → 𝐻cat 𝐻)
Theoremssctr 16479 The subcategory subset relation is transitive. (Contributed by Mario Carneiro, 6-Jan-2017.)
((𝐴cat 𝐵𝐵cat 𝐶) → 𝐴cat 𝐶)
Theoremssceq 16480 The subcategory subset relation is antisymmetric. (Contributed by Mario Carneiro, 6-Jan-2017.)
((𝐴cat 𝐵𝐵cat 𝐴) → 𝐴 = 𝐵)
Theoremrescval 16481 Value of the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) ((𝐶𝑉𝐻𝑊) → 𝐷 = ((𝐶s dom dom 𝐻) sSet ⟨(Hom ‘ndx), 𝐻⟩))
Theoremrescval2 16482 Value of the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) & (𝜑𝐶𝑉) & (𝜑𝑆𝑊) & (𝜑𝐻 Fn (𝑆 × 𝑆)) (𝜑𝐷 = ((𝐶s 𝑆) sSet ⟨(Hom ‘ndx), 𝐻⟩))
Theoremrescbas 16483 Base set of the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) & 𝐵 = (Base‘𝐶) & (𝜑𝐶𝑉) & (𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝑆𝐵) (𝜑𝑆 = (Base‘𝐷))
Theoremreschom 16484 Hom-sets of the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) & 𝐵 = (Base‘𝐶) & (𝜑𝐶𝑉) & (𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝑆𝐵) (𝜑𝐻 = (Hom ‘𝐷))
Theoremreschomf 16485 Hom-sets of the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) & 𝐵 = (Base‘𝐶) & (𝜑𝐶𝑉) & (𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝑆𝐵) (𝜑𝐻 = (Homf𝐷))
Theoremrescco 16486 Composition in the category restriction. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐻) & 𝐵 = (Base‘𝐶) & (𝜑𝐶𝑉) & (𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝑆𝐵) & · = (comp‘𝐶) (𝜑· = (comp‘𝐷))
Theoremrescabs 16487 Restriction absorption law. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐶𝑉) & (𝜑𝐻 Fn (𝑆 × 𝑆)) & (𝜑𝐽 Fn (𝑇 × 𝑇)) & (𝜑𝑆𝑊) & (𝜑𝑇𝑆) (𝜑 → ((𝐶cat 𝐻) ↾cat 𝐽) = (𝐶cat 𝐽))
Theoremrescabs2 16488 Restriction absorption law. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐶𝑉) & (𝜑𝐽 Fn (𝑇 × 𝑇)) & (𝜑𝑆𝑊) & (𝜑𝑇𝑆) (𝜑 → ((𝐶s 𝑆) ↾cat 𝐽) = (𝐶cat 𝐽))
Theoremissubc 16489* Elementhood in the set of subcategories. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐻 = (Homf𝐶) & 1 = (Id‘𝐶) & · = (comp‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝑆 = dom dom 𝐽) (𝜑 → (𝐽 ∈ (Subcat‘𝐶) ↔ (𝐽cat 𝐻 ∧ ∀𝑥𝑆 (( 1𝑥) ∈ (𝑥𝐽𝑥) ∧ ∀𝑦𝑆𝑧𝑆𝑓 ∈ (𝑥𝐽𝑦)∀𝑔 ∈ (𝑦𝐽𝑧)(𝑔(⟨𝑥, 𝑦· 𝑧)𝑓) ∈ (𝑥𝐽𝑧)))))
Theoremissubc2 16490* Elementhood in the set of subcategories. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐻 = (Homf𝐶) & 1 = (Id‘𝐶) & · = (comp‘𝐶) & (𝜑𝐶 ∈ Cat) & (𝜑𝐽 Fn (𝑆 × 𝑆)) (𝜑 → (𝐽 ∈ (Subcat‘𝐶) ↔ (𝐽cat 𝐻 ∧ ∀𝑥𝑆 (( 1𝑥) ∈ (𝑥𝐽𝑥) ∧ ∀𝑦𝑆𝑧𝑆𝑓 ∈ (𝑥𝐽𝑦)∀𝑔 ∈ (𝑦𝐽𝑧)(𝑔(⟨𝑥, 𝑦· 𝑧)𝑓) ∈ (𝑥𝐽𝑧)))))
Theorem0ssc 16491 For any category 𝐶, the empty set is a subcategory subset of 𝐶. (Contributed by AV, 23-Apr-2020.)
(𝐶 ∈ Cat → ∅ ⊆cat (Homf𝐶))
Theorem0subcat 16492 For any category 𝐶, the empty set is a (full) subcategory of 𝐶, see example 4.3(1.a) in [Adamek] p. 48. (Contributed by AV, 23-Apr-2020.)
(𝐶 ∈ Cat → ∅ ∈ (Subcat‘𝐶))
Theoremcatsubcat 16493 For any category 𝐶, 𝐶 itself is a (full) subcategory of 𝐶, see example 4.3(1.b) in [Adamek] p. 48. (Contributed by AV, 23-Apr-2020.)
(𝐶 ∈ Cat → (Homf𝐶) ∈ (Subcat‘𝐶))
Theoremsubcssc 16494 An element in the set of subcategories is a subset of the category. (Contributed by Mario Carneiro, 6-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & 𝐻 = (Homf𝐶) (𝜑𝐽cat 𝐻)
Theoremsubcfn 16495 An element in the set of subcategories is a binary function. (Contributed by Mario Carneiro, 4-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝑆 = dom dom 𝐽) (𝜑𝐽 Fn (𝑆 × 𝑆))
Theoremsubcss1 16496 The objects of a subcategory are a subset of the objects of the original. (Contributed by Mario Carneiro, 4-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝐽 Fn (𝑆 × 𝑆)) & 𝐵 = (Base‘𝐶) (𝜑𝑆𝐵)
Theoremsubcss2 16497 The morphisms of a subcategory are a subset of the morphisms of the original. (Contributed by Mario Carneiro, 4-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝐽 Fn (𝑆 × 𝑆)) & 𝐻 = (Hom ‘𝐶) & (𝜑𝑋𝑆) & (𝜑𝑌𝑆) (𝜑 → (𝑋𝐽𝑌) ⊆ (𝑋𝐻𝑌))
Theoremsubcidcl 16498 The identity of the original category is contained in each subcategory. (Contributed by Mario Carneiro, 4-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝐽 Fn (𝑆 × 𝑆)) & (𝜑𝑋𝑆) & 1 = (Id‘𝐶) (𝜑 → ( 1𝑋) ∈ (𝑋𝐽𝑋))
Theoremsubccocl 16499 A subcategory is closed under composition. (Contributed by Mario Carneiro, 4-Jan-2017.)
(𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝐽 Fn (𝑆 × 𝑆)) & (𝜑𝑋𝑆) & · = (comp‘𝐶) & (𝜑𝑌𝑆) & (𝜑𝑍𝑆) & (𝜑𝐹 ∈ (𝑋𝐽𝑌)) & (𝜑𝐺 ∈ (𝑌𝐽𝑍)) (𝜑 → (𝐺(⟨𝑋, 𝑌· 𝑍)𝐹) ∈ (𝑋𝐽𝑍))
Theoremsubccatid 16500* A subcategory is a category. (Contributed by Mario Carneiro, 4-Jan-2017.)
𝐷 = (𝐶cat 𝐽) & (𝜑𝐽 ∈ (Subcat‘𝐶)) & (𝜑𝐽 Fn (𝑆 × 𝑆)) & 1 = (Id‘𝐶) (𝜑 → (𝐷 ∈ Cat ∧ (Id‘𝐷) = (𝑥𝑆 ↦ ( 1𝑥))))
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posted by .
A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?
I had some ideas as to how to solve this problem.
First, I thought of using the Specific Heat Formula to find a value, that could resemble a known element.
Q = m CP delta T
1 = m(2.550)(373.15)
1 = m(951.53)
m = 0.00105 g
At this point, I'm really stuck. I know that to find the molecular formula of a compound, I need to divide the empirical mass over the molecular mass. However, I am not sure how to find both masses. I was thinking to find the empirical mass I could calculate CH.
C = 12.011 g
H = 1.007 g
CH = 12.011 + 1.007 = 13.018 g
• Chemistry -
The easiest way to find the molar mass is
to use a rearranged PV = nRT to
P*M = density*RT
M = molar mass
P in atm.
T in kelvn.
Then go through your procedure to change the empirical formula to the molecular formula.
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# 1.3 Digital signals and modulation
Radio waves are naturally sinusoidal, with frequencies covering a wide range. They are capable of travelling through space, and are widely used for communication. This is a brief explanation of how they are able to carry information. Many of the same principles apply to other communication media, such as optical signals and electric currents.
## Activity 1.3 Exploratory
Radio waves cover a wide range of frequencies, some of which are more suitable than others for a particular service. You can explore some uses of radio with this interactive chart.
Click on the image of the electromagnetic spectrum below to learn more about the highlighted part of the spectrum (radio and microwave frequencies). You will see that this part of the spectrum is conventionally divided into bands, each covering a decade in frequency (or wavelength). Make a note of the frequencies and wavelengths and the typical uses of each band.
Interactive feature not available in single page view (see it in standard view).
Generally a medium used for communication (such as radio waves) needs to be processed in some way to carry information. The process is called modulation. Two signals are combined in modulation:
• The message signal, called the modulating signal. (Often this is non-periodic.)
• A signal of the right frequency for transmission, called the carrier signal.
When they are combined, the modulating signal changes the carrier signal in some way, such as by changing its amplitude or frequency. This creates a new signal that contains the message information and is also at the correct transmission frequency. Note that although modulation of some kind is essential for wireless transmission, it is also used in much wired transmission, for example broadband and optical fibre.
In the next section, assume that the message to be sent is in the form of a digital signal (that is, a signal that is interpreted as a sequence of discrete values). In fact, most communications fall into this category; computer networks and almost all telephony, as well as digital TV and radio. Analogue signals such as speech are converted to digital form at one end of a communications link and back to analogue at the other. When the message signal is digital, modulation produces distinct states of the carrier wave that can be distinguished by the receiver and can be used to represent ones and zeros, or groups of ones and zeros. Next you will see some basic digital modulation schemes. | 529 | 2,768 | {"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} | 3.03125 | 3 | CC-MAIN-2023-14 | longest | en | 0.928228 |
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# Topic: Extended Euclidean algorithm
PlanetMath: fast Euclidean algorithm The algorithm can also be used to compute any particular pair of coefficients from the Extended Euclidean Algorithm, although computing every pair of coefficients would involve The full algorithm, and a comprehensive runtime analysis is given in “Modern Computer Algebra” by von zur Gathen and Gerhard. This is version 4 of fast Euclidean algorithm, born on 2004-07-22, modified 2005-04-14. www.planetmath.org /encyclopedia/FastEuclideanAlgorithm.html (222 words)
PlanetMath: Berlekamp-Massey algorithm The Berlekamp-Massey algorithm is used for finding the minimal polynomial of a linearly recurrent sequence. The algorithm itself is presented at the end of this article. This is version 4 of Berlekamp-Massey algorithm, born on 2004-07-22, modified 2005-04-14. planetmath.org /encyclopedia/BerlekampMasseyAlgorithm.html (0 words)
Random Works of the Web » Blog Archive » Extended Euclidean algorithm (Site not responding. Last check: ) The extended Euclidean algorithm is an algorithm used to calculate the greatest common divisor (gcd, or also highest common factor, HCF) of two integers a and b, as well as integers x and y such that To illustrate the extension of the Euclid’s algorithm, consider the computation of gcd(120, 23), which is shown on the table on the left. This method attempts to solve the required equation with the sum replaced by the remainders in each step of the algorithm, which is larger than their gcd, but are decreasing in magnitude, and so will eventually become the required equation. random.dragonslife.org /extended-euclidean-algorithm/4414 (0 words)
Euclidean algorithm - Wikipedia, the free encyclopedia In number theory, the Euclidean algorithm (also called Euclid's algorithm) is an algorithm to determine the greatest common divisor (GCD) of two integers or elements of any Euclidean domain (for example, polynomials over a field) by repeatedly dividing the two numbers and the remainder in turns. This is known as the extended Euclidean algorithm. The quotients that appear when the Euclidean algorithm is applied to the inputs a and b are precisely the numbers occurring in the continued fraction representation of a/b. en.wikipedia.org /wiki/Euclidean_algorithm (0 words)
Extended Euclidean algorithm - Wikipedia, the free encyclopedia The extended Euclidean algorithm is an extension to the Euclidean algorithm for finding the greatest common divisor (GCD) of a and b: it also finds the integers x and y in Bezout's identity The extended Euclidean algorithm is particularly useful when a and b are coprime, since x is the multiplicative inverse of a modulo b. The extended Euclidean algorithm can also be used to calculate the multiplicative inverse in a finite field. en.wikipedia.org /wiki/Extended_Euclidean_algorithm (0 words)
Extended Euclidean Algorithm You repeatedly divide the divisor by the remainder until the remainder is 0. This is known as the extended Euclidean Algorithm. Before presenting this extended Euclidean algorithm, we shall look at a special application that is the most common usage of the algorithm. www-math.cudenver.edu /~wcherowi/courses/m5410/exeucalg.html (718 words)
Euclidean algorithm used to decode Reed-Solomon codes? | Comp.DSP | DSPRelated.com Euclidean algorithm is based, or perhaps give some pointers in Yes, it is the same algorithm or more precisely, the extended Euclidean algorithm for finding the GCD of two polynomials www.dsprelated.com /showmessage/22571/1.php (636 words)
Math 413 Lecture 2 - Divisibility & Euclidean Algorithm This is useful both because the Euclidean algorithm is the primary computational tool in number theory and because the existence of a Euclidean algorithm has strong consequences for the structure of the ring. The existence of a Euclidean algorithm is sufficiently important that we call a ring with such an algorithm a Euclidean ring. Careful analyses of the Euclidean algorithm and faster variants of the algorithm can be found in [Knuth2] and [BS96]. www.math.umbc.edu /~campbell/Math413Spr01/Lectures/lecture2.html (455 words)
PlanetMath: fast Euclidean algorithm The algorithm can also be used to compute any particular pair of coefficients from the Extended Euclidean Algorithm, although computing every pair of coefficients would involve The full algorithm, and a comprehensive runtime analysis is given in “Modern Computer Algebra” by von zur Gathen and Gerhard. This is version 4 of fast Euclidean algorithm, born on 2004-07-22, modified 2005-04-14. planetmath.org /encyclopedia/FastEuclideanAlgorithm.html (0 words)
[No title] (Site not responding. Last check: ) Use the extended Euclidean algorithm to solve g = mx + ny for a pair of integers m and n. Describe an algorithm to convert a base 10 fraction of the form m / n where m and n are small positive integers to base b. Describe an algorithm to convert a fraction with denominator that is a power of two to base two. www.cbu.edu /~yanushka/j0/r.0 (0 words)
Euclidean Algorithm -- from Wolfram MathWorld There are even principal rings which are not Euclidean but where the equivalent of the Euclidean algorithm can be defined. The Euclidean algorithm is an example of a P-problem whose time complexity is bounded by a quadratic function of the length of the input values (Bach and Shallit 1996). remainders, so the algorithm can be easily applied by hand by repeatedly computing remainders of consecutive terms starting with the two numbers of interest (with the larger of the two written first). mathworld.wolfram.com /EuclideanAlgorithm.html (0 words)
re: brain teaser #64 I imagine that Nate's approach is closely related to the approach using the extended euclidean algorithm. The trick to the extended euclidean algorithm is to remember the steps you performed while computing the gcd. The first step of the euclidean algorithm for computing this gcd is to subtract from 109 a multiple of 37, preferably the largest multiple of 37 that is less than 109, which is 74. www.physicsforums.com /showthread.php?p=96495 (0 words)
S2.html First of all the algorithm terminates since by the division equation thereom, the remainders form a decreasing sequence of non-negative integers. Since these equations form a k-step Eucldean algorithm for inputs r[0] and r[1], use the induction assumption to assume that they can be back-solved for the integers x and y satiifying the equation x*r[0]+y*r[1]=r[k]=g=gcd(r[0],r[1]). We use the extended Euclidean algorithm on 48 and 37. www.math.sfu.ca /~gfee/Math342/A2/S21.html (0 words)
Euclidean algorithm and its applications The Euclidean algorithm is simply a sequence of long divisions (repeated until the remainder becomes zero). The following extended Euclidean algorithm (EEA) also expresses each remainder as a linear combination of the original polynomials. EEA is fine to use for finite fields, since the coefficients of the intermediate polynomials never grow big. www.math.clemson.edu /faculty/Gao/calg/node5.html (306 words)
Cryptography Tutorial - The Euclidean Algorithm finds the Greatest Common Divisor of two Integers 1) Understand the Extended Euclidean Algorithm to determine the inverse of a given integer. Consequently, if a and b have a greatest common divisor different from 1 (that is the gcd(a,b) is not 1) a does not have an inverse mod b. Again, the Extended Euclidean Algorithm should be performed by a computer as it is very easy to implement and it yields the answer quickly. www.antilles.k12.vi.us /math/cryptotut/extended_euclidean_algorithm.htm (0 words)
Extended Euclidean algorithm The extended Euclidean algorithm is a version of the Euclidean algorithm; its input are two integers a and b and the algorithm computes their greatest common divisor (GCD) as well as integers x and y such that ax + by = gcd(a,b). The extended Euclidian algorithm can also be used to calculate the multiplicative inverse in a finite field. Given the irreducible polynomial f(x) used to define the finite field, and the element a(x) whose inverse is desired, then a form of the algorithm suitable for determining the inverse is given by the following pseudocode: www.fact-index.com /e/ex/extended_euclidean_algorithm.html (0 words)
[No title] (Site not responding. Last check: ) CS 236 Extended Euclidean & RSA Algorithms 20 Feb. For positive integers M and N, find the greatest common divisor G of M and N and find integers x and y of opposite signs with Mx + Ny = G efficiently. RSA Algorithm RSA denotes the first letters of the last names of the inventors of the algorithm, namely, Ronald L. Rivest, Adi Shamir and Leonard Adleman. The RSA algorithm depends on difficulty of factoring a large integer into two primes and on the ease of finding large primes. www.cbu.edu /~yanushka/od/n.0 (0 words)
Euclidean algorithm The Euclidean algorithm is a way to find the greatest common divisor of two positive integers, a and b. The greatest common divisor of 210 and 45 is 15, and we have written 15 as a sum of integer multiples of 210 and 45. The extended Euclidean algorithm is easy to implement on a computer and the amount of memory needed is not large. www.math.rutgers.edu /~greenfie/gs2004/euclid.html (793 words)
Polynomials - HowTo: Euclidean algorithm for polynomials In this HowTo we will describe the analogue of the euclidean algorithm to compute the greatest common divisor of any two polynomials in The euclidean algorithm then works in the same way. One such condition is that the gcd be monic; that is, that the coefficient of highest degree be xmlearning.maths.ed.ac.uk /lecture_notes/polynomials/howto_euclidean_algorithm_polynomials/howto_euclidean_algorithm_polynomials.php (202 words)
Hidden Small Exponent Method Information (Site not responding. Last check: ) However, it is a simple matter to decrypt messages that have been encrypted using pseudo keys generated by the hidden exponent algorithm. We invite you to learn about this algorithm, try out the 128-bit key generator (details on implementation provided) and try the official challenge yourself. Now that the RSA algorithm is public domain, companies no longer need to license it from RSA, leaving the purity of the implementation open to "interpretations". crypto.cs.mcgill.ca /~crepeau/RSA (294 words)
A Matrix Interpretation of the Extended Euclidean Algorithm (ResearchIndex) (Site not responding. Last check: ) The extended Euclidean algorithm for polynomials and formal power series that is used for the recursive computation of Pade approximants can be viewed in various ways as a sequence of successive matrix multiplications that are applied to a Sylvester matrix with the original data. Here we present this result in a general version that includes the treatment of the Cabay{Meleshko look-ahead algorithm, which generalizes the extended Euclidean algorithm and yields a weakly stable (forward stable)... 1 A weakly stable algorithm for Pade approximants and the inv.. citeseer.ist.psu.edu /310864.html (0 words)
[No title] CS 122 Extended Euclidean algorithm 30 Mar. I feel some of you did not take notes yesterday. Find the greatest command divisor of the positive integers a and b as g and then solve the equation g = x * a + y * b for integers x and y of opposite signs. Modify Euclid's algorithm to find the greatest common divisor to compute and save the quotients as an array. www.cbu.edu /~yanushka/j1/n.0 (0 words)
The Laws of Cryptography: Cryptographers' Favorite Algorithms It's possible to code the extended gcd algorithm following the model above, first using a loop to calculate the gcd, while saving the quotients at each stage, and then using a second loop as above to work back through the equations, solving for the gcd in terms of the original two numbers. An essential part of many of the algorithms involved is to raise an integer to another integer power, modulo an integer (taking the remainder on division). It is very similar to the previous algorithm, but differs in processing the binary bits of the exponent in the opposite order. www.cs.utsa.edu /~wagner/laws/fav_alg.html (0 words)
[No title] (Site not responding. Last check: ) The Extended Euclidean Algorithm -------------------------------- There are two prevailing versions of expositions of the extended Euclidean algorithm. This is where we extend the Euclidean algorithm. Because this process is just the plain old Euclidean algorithm when you just look at the values of w,z, eventually one of them, say w, will become 0, and z will become gcd(a,b): a*u + b*v = 0 a*x + b*y = gcd(a,b) Then you take the second equation and you are done! www.cs.toronto.edu /~trebla/ExtendedEuclid.txt (0 words)
PHP Bugs: #15835: Problem on function gmp_invert () I implemented the extended Euclides algoritm that do exactly the same that this function and, it returned me the right value. The Euclidean Algorithm computes the greatest common divisor, gmp_invert doesn't seem to do that... The "Euclidean Algorithm" computes the greatest common divisor but, the "Extended Euclidean Algorithm" give you the multiplicative inverse. bugs.php.net /15835 (0 words)
Euclidean algorithm algorithm wszystko o google- cybersocjologia (Site not responding. Last check: ) The Euclidean algorithm (also called Euclid's algorithm) is an algorithm to determine the greatest common divisor (gcd) of two integers. The proof of this algorithm is not difficult. When analyzing the runtime of Euclid's algorithm, it turns out that the inputs requiring the most divisions are two successive Fibonacci numbers, and the worst case requires ¥È(n) divisions, where n is the number of digits in the input (see Big O notation). www.ifb.pl /~pneuma/old/archiwum-google.php?lng=en&pg=322 (0 words)
On the Complexity of the Extended Euclidean Algorithm (Extended Abstract) (ResearchIndex) (Site not responding. Last check: ) On the Complexity of the Extended Euclidean Algorithm (Extended Abstract) (ResearchIndex) On the Complexity of the Extended Euclidean Algorithm (Extended Abstract) 2 A new algorithm and re ned bounds for extended GCD computati.. citeseer.ist.psu.edu /641687.html (0 words)
Footnotes (Site not responding. Last check: ) The naming convention is lower case sans-serif for example generators of cryptosystems and initial capitals for actual cryptosystems e.g. This refers to using the wrong result from performing the extended euclidean algorithm on K This refers to using the wrong result from performing the extended euclidean algorithm on weight w www.csse.monash.edu.au /~skcho5/Thesis/footnode.html (0 words)
An Example Using the Extended Euclidean Algorithm (Site not responding. Last check: ) If you're using the algorithm by hand (as opposed to using a computer), you can use a table to keep the numbers straight and remember what to do. I'll illustrate by applying the algorithm to find the greatest common divisor of 187 and 102. The statement and proof of the Extended Euclidean Algorithm marauder.millersville.edu /~bikenaga/absalg/exteuc/exteucex.html (209 words)
The Extended Euclidean Algorithm for the Euclidean algorithm for m/n are printed. The length l of the algorithm is printed. The continued fraction for -m/n is also printed. www.numbertheory.org /php/euclid.html (96 words)
Extended Euclidean Algorithm Enter 2 positive integers to find their G.C.D using the extended Euclidean Algorithm Please note : This script has been written to show the Extended Euclidean Algorithm. No guarantee can be given over the validity of the answers. homepages.inf.ed.ac.uk /s0563270/maths/php/euclidean.php (73 words)
The Euclidean algorithm The division algorithm (theorem 1.2.7) tells us that there are integers The proof of this is constructive and follows an algorithm called the Euclidean algorithm. Use the division algorithm (theorem 1.2.7) to compute integers web.usna.navy.mil /~wdj/book/node13.html (217 words)
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About us | Why use us? | Reviews | Press | Contact us | 3,699 | 16,257 | {"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} | 3.4375 | 3 | CC-MAIN-2019-22 | latest | en | 0.834209 |
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Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics1.76 Division In Decimals - (Divide by 10,100, and 1000)
Directions: Solve the following problems. Also write at least 10 examples of your own.
Q 1: 276.7/1000 =Answer: Q 2: 256.28/10 =Answer: Q 3: 0.76/10 =Answer: Q 4: 780.89/10 =Answer: Q 5: 386.78/100 =Answer: Q 6: 976.3/100 =Answer: Q 7: 89.7/1000 =Answer: Q 8: 789.67/1000 =Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 228 | 669 | {"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} | 3.015625 | 3 | CC-MAIN-2020-16 | latest | en | 0.805741 |
https://classplayground.com/ten-frame/ | 1,726,583,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00831.warc.gz | 150,202,336 | 18,925 | Categories
Ten Frame
Printables Interactive
What are Ten Frames?
A ten frame is made up of ten boxes arranged into two rows of five columns. Each box can be filled with a counter or left blank, allowing the representation of numbers up to ten. The concept of ten frames is rooted in the idea of benchmark numbers, specifically five and ten. By arranging counters in various configurations, students can form mental images of numbers. The ultimate goal is for students to develop automaticity, enabling them to recognize the number of counters without having to count them individually.
Why Use Ten Frames in the Elementary Classroom?
Ten frames are not only easy and cost-effective to implement into math instruction but also highly effective. They offer hands-on manipulatives that visually represent numbers less than, equal to, or more than ten. The concept of ten is fundamental to understanding place value, a critical skill in fostering a robust comprehension of number sense. It’s essential that students understand what a number consists of, rather than merely identifying a number as a digit.
Strategies for Using Ten Frames
The successful integration of ten frames into your classroom revolves around a few key strategies:
• Consistent Use: Regularly incorporating ten frames into your daily math lessons is crucial for students to become comfortable with them. This usage can be as simple as starting each math lesson with a quick ten frame activity or using ten frames during independent work time. The more they are exposed to and use ten frames, the more instinctive their understanding will become.
• Modeling: Before you expect students to use ten frames independently, it’s essential to model how to use them. Start by demonstrating how to represent different numbers on the ten frame. Then, show them how to use ten frames for addition and subtraction. For example, if you are teaching the concept of “4 + 3”, start with four counters on the ten frame, then add three more and count the total. Modeling these processes helps students understand how to use ten frames for mathematical operations.
• Differentiate Instruction: Ten frames can be a versatile tool for differentiating instruction to meet diverse learning needs. For students who find certain numerical concepts challenging, ten frames can simplify complex ideas into manageable parts. You can use a single ten frame to help these students master numbers up to ten. For more advanced students, introduce double ten frames or multiple ten frames to represent and work with larger numbers. This differentiation allows students to work at their own pace and ability level, fostering a more inclusive learning environment.
• Interactive Learning: Ten frames can also be used to promote interactive learning. Encourage students to explain their thinking while using ten frames. For instance, ask them why they placed counters in a particular arrangement or how they used the ten frame to solve a problem. This strategy promotes mathematical discourse, enhances students’ communication skills, and deepens their understanding of numerical concepts.
• Integration with Other Learning Areas: Don’t limit the use of ten frames to math lessons. Integrate them into other learning areas. For example, in a science lesson about seasons, use ten frames to count and represent the number of months in each season. In a language arts lesson, use them to count syllables in words. This cross-curricular integration reinforces students’ understanding of ten frames and their applicability across different contexts.
Activities for Using Ten Frames
Ten frame activities can make the learning process interactive and enjoyable:
• Flash and Tell: This activity is designed to improve students’ subitizing skills – their ability to quickly identify the number of objects in a small group without counting. To conduct this activity, prepare a ten frame with a specific number of counters. Show it to the students for a few seconds and then hide it. Ask students to recall the number of counters they saw and describe how they visualized them. For instance, if there were six counters, did they see two rows of three? Or perhaps five counters in one row and one in the other? Repeating this activity with different numbers of counters can help students develop a range of mental images for numbers.
• Ten Frame Memory Game: This game requires a set of ten frame cards and corresponding number cards. To prepare, create pairs of cards: one card showing a number as a numeral (e.g., “4”) and another showing that number on a ten frame. Spread all the cards face down. Students take turns flipping two cards at a time, trying to find a match between the numeral and the ten frame representation. When a match is found, the student keeps the pair of cards. The student with the most pairs at the end of the game wins. This activity reinforces number recognition and the visual representation of numbers.
• Addition and Subtraction Stories: This activity uses ten frames to visualize addition and subtraction word problems. First, present a word problem to the students. For example, “Billy has 7 apples. He eats 2. How many does he have left?” Then, use counters on a ten frame to represent the problem. Start with 7 counters, remove 2, and ask students to count how many are left. This activity helps students understand the process of addition and subtraction in a concrete, visual way.
• Fill the Frame: In this activity, give each student a blank ten frame and a set of counters. Call out a number and ask students to represent that number on their ten frame using the counters. For early learners, start with numbers less than or equal to 10. For more advanced students, introduce larger numbers and use multiple ten frames. This activity can be made competitive by timing it and seeing who can fill their frame correctly the fastest. It helps students practice number representation and reinforces the concept of ten as a benchmark number. | 1,146 | 6,009 | {"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} | 4 | 4 | CC-MAIN-2024-38 | latest | en | 0.923935 |
https://www.proofwiki.org/wiki/Definition:Stereographic_Projection | 1,679,375,735,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00146.warc.gz | 1,037,321,912 | 11,306 | # Definition:Stereographic Projection
## Definition
Let $\PP$ be a the plane.
Let $\mathbb S$ be a sphere which is tangent to $\PP$ at the origin $\tuple {0, 0}$.
Let the diameter of $\mathbb S$ perpendicular to $\PP$ through $\tuple {0, 0}$ be $NS$ where $S$ is the point $\tuple {0, 0}$.
Let the point $N$ be referred to as the north pole of $\mathbb S$ and $S$ be referred to as the south pole of $\mathbb S$.
Let $A$ be a point on $P$.
Let the line $NA$ be constructed.
Then $NA$ passes through a point of $\mathbb S$.
Thus any point on $P$ can be represented by a point on $\mathbb S$.
With this construction, the point $N$ on $\mathbb S$ maps to no point on $\mathbb S$. | 209 | 686 | {"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} | 2.890625 | 3 | CC-MAIN-2023-14 | latest | en | 0.758456 |
https://socratic.org/questions/how-do-you-use-the-remainder-theorem-to-determine-the-remainder-when-the-polynom-9 | 1,586,199,138,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00434.warc.gz | 691,063,634 | 6,134 | # How do you use the remainder theorem to determine the remainder when the polynomial (n^4-3n^2-5n+2)/(n-2)?
Dec 30, 2016
$R = - 4$
#### Explanation:
The remainder theorem states that if a polynomial $\text{ "P(x)" }$ is divided by the linear factor $\left(x - a\right) \text{ }$then the remainder is $\text{ } P \left(a\right)$
proof;
let $\text{ "P(x) " }$be divided by $\text{ "(x-a)" }$ to give quotient $\text{ "Q(x)" }$and remainder $\text{ } R$
then:$\text{ } P \left(x\right) = \left(x - a\right) Q \left(x\right) + R$
so:$\text{ } P \left(a\right) = \cancel{\left(a - a\right) Q \left(a\right)} + R$
$\text{ ":.P(a)=R" }$as required.
In this case we have :$\left({n}^{4} - 3 {n}^{2} - 5 n + 2\right) \div \left(n - 2\right)$
$P \left(n\right) = {n}^{4} - 3 {n}^{2} - 5 n + 2$
the divisor is $\text{ } n - 2 \implies a = 2$
$R = P \left(2\right) = {2}^{4} - 3 \times {2}^{2} - 5 \times 2 + 2$
$R = P \left(2\right) = 16 - 12 - 10 + 2 = - 4$ | 395 | 962 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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} | 4.75 | 5 | CC-MAIN-2020-16 | longest | en | 0.618163 |
https://codecogs.com/library/maths/discrete/number_theory/euler.php | 1,642,667,531,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00376.warc.gz | 246,544,802 | 8,307 | I have forgotten
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# Euler
viewed 3024 times and licensed 45 times
Calculates Euler numbers by means of recurrent relation
Controller: CodeCogs
C++
## Overview
Euler numbers, which are also called secant numbers or 'zig' numbers are defined for $\inline&space;&space;|z|&space;<\pi/2$ by
$sech(x)&space;-&space;1&space;\equiv&space;-\frac{E_1^*&space;z^2}{2!}&space;+\frac{E_2^*&space;z^4}{4!}&space;-\frac{E_3^*&space;z^6}{6!}$
The Euler number give the number of odd alternating permutations and are related to Genocchi numbers.
A slightly different convention, which is also used here, defines
$E_{2n}&space;=&space;(-1)^n&space;E_n^*$
$E_{2n+1}&space;=&space;0$
Thus, the Euler numbers $\inline&space;E_n$ and polynomials $\inline&space;E_n(x)$ are defined through equations
$sech(x)&space;=&space;\frac{2{E^z}}{{E^{2z}}+1}&space;=&space;\sum&space;_{n=0}^{\infty}{E_n}\frac{{z^n}}{n!},&space;|z|<&space;\frac{\pi&space;}{2}$
$\frac{2&space;E^{xz}&space;}{{E^z}+1}&space;=&space;\sum_{n=0}^{\infty&space;}{E_n}(x)\frac{{z^n}}{n!},&space;|z|&space;\pi$
The relation between Euler polynomials and numbers:
${E_n}&space;=&space;{2^n}{E_n}\big(\frac{1}{2}\big)$
## References:
• Higher Transcendental Functions, vol.1, (1.14) by H.Bateman and A.Erdelyi (Bateman Manuscript Project), 1953
• M.Abramowitz and I.A.Stegun, Handbook of Mathematical Functions, 1964 chapt.23
### Authors
Will Bateman (September 2005)
## EulerA
voidEulerA( int iMax double* daEuler )
Calculates Euler numbers using the recurrent relation
${E_{2n}}&space;=&space;-\sum_{r=0}^{n-1}\big(\begin{array}{c}&space;2n&space;\\&space;2r&space;\end{array}\big){E_{2r}},&space;{e_0}=1$
${E_{2n+1}}=0,&space;n=0,1,2,...$
### Example 1
#include <stdio.h>
#include <codecogs/maths/discrete/number_theory/euler.h>
#define MAX_INDEX 16
int main()
{
double dEulerA[MAX_INDEX+1];
double dEulerB[MAX_INDEX+1];
printf( "%10s %20s %20s\n", "2n", "E_A(2n) (recurrent)", "E_B(2n) (series)" );
printf( "%8s", " " );
for( int i = 0; i < 52; i++ )
printf( "%c", '-' );
printf( "\n" );
// array calculation
Maths::NumberTheory::EulerA( MAX_INDEX, dEulerA );
Maths::NumberTheory::EulerB( MAX_INDEX, dEulerB );
// results output
for( int i = 0; i <= MAX_INDEX; i+=2 )
printf( "%10d %20.6f %20.6f\n", i, dEulerA[i], dEulerB[i]);
return 0;
}
Output:
2n E_A(2n) (recurrent) E_B(2n) (series)
----------------------------------------------------
0 1.000000 1.000000
2 -1.000000 -1.000000
4 5.000000 5.000000
6 -61.000000 -61.000000
8 1385.000000 1385.000000
10 -50521.000000 -50521.000000
12 2702765.000000 2702765.000000
14 -199360981.000000 -199360981.000000
16 19391512145.000000 19391512145.000027
### Parameters
iMax input maximal index requested. Must be a power of 2. daEuler array into which to place the result, with iMax parameters.
### Returns
pointer on the resulting array of type double
### Authors
Anatoly Prognimack (Mar 19, 2005)
Developed & tested with: Borland C++ 3.1 for DOS Microsoft Visual C++ 5.0, 6.0
Updated by Will Bateman (March 2005)
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login.
## EulerB
voidEulerB( int iMax double* daEuler )
Calculates Euler numbers using the recurrent series expansion
$\begin{array}{l}&space;{E_{2n}}={{(-1)}^n}2(2n)!{{\big(\frac{2}{\pi&space;}\big)}^{2n+1}}\sum&space;&space;&space;_{r=0}^{\infty&space;}\frac{{{(-1)}^r}}{{{(2r+1)}^{2n+1}}},&space;&space;\\&space;n=0,1,2,...\\&space;\end{array}$
### Parameters
iMax input maximal index requested. Must be a power of 2. daEuler array into which to place the result, with iMax parameters.
### Returns
pointer on the resulting array of type double
### Authors
Anatoly Prognimack (Mar 19, 2005)
Developed & tested with: Borland C++ 3.1 for DOS Microsoft Visual C++ 5.0, 6.0
Will Bateman (March 2005)
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login. | 1,471 | 4,263 | {"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": 12, "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} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.62062 |
http://mathhelpforum.com/number-theory/117300-fibonacci-proof-print.html | 1,529,407,401,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862248.4/warc/CC-MAIN-20180619095641-20180619115641-00306.warc.gz | 204,518,242 | 3,485 | # Fibonacci Proof
• Nov 28th 2009, 09:54 PM
redsoxfan325
Fibonacci Proof
I hate to say this, but...I'm stuck on an induction proof. I'm trying to prove that $\displaystyle \gcd(F_m,F_n)=F_{\gcd(m,n)}$, where $\displaystyle F_j$ is the $\displaystyle j$th Fibonacci number. I have it reduced to proving that $\displaystyle F_m\,|\,F_{qm}$ for all $\displaystyle m$. I have the rest of the proof.
So I tried proving it with induction.
It's trivial for $\displaystyle m=1$, so I assumed $\displaystyle F_m\,|\,F_{qm}$. Then I tried to prove that $\displaystyle F_{m+1}\,|\,F_{qm+q}$.
Using an identity I know and have proved,
$\displaystyle F_{qm+q}=F_qF_{qm+1}+F_{q-1}F_{qm}=F_qF_{qm+1}+kF_{q-1}F_m$
for some $\displaystyle k\in\mathbb{N}$, but now I'm stuck.
Any hints?
• Nov 28th 2009, 10:00 PM
aman_cc
Quote:
Originally Posted by redsoxfan325
I hate to say this, but...I'm stuck on an induction proof. I'm trying to prove that $\displaystyle \gcd(F_m,F_n)=F_{\gcd(m,n)}$, where $\displaystyle F_j$ is the $\displaystyle j$th Fibonacci number. I have it reduced to proving that $\displaystyle F_m\,|\,F_{qm}$ for all $\displaystyle m$. I have the rest of the proof.
So I tried proving it with induction.
It's trivial for $\displaystyle m=1$, so I assumed $\displaystyle F_m\,|\,F_{qm}$. Then I tried to prove that $\displaystyle F_{m+1}\,|\,F_{qm+q}$.
Using an identity I know and have proved,
$\displaystyle F_{qm+q}=F_qF_{qm+1}+F_{q-1}F_{qm}=F_qF_{qm+1}+kF_{q-1}F_m$
for some $\displaystyle k\in\mathbb{N}$, but now I'm stuck.
Any hints?
Hint: Do induction on 'q' rather than m. That should do it.
• Nov 28th 2009, 10:08 PM
redsoxfan325
I can do that, but does that prove the same thing as induction on $\displaystyle m$? (Because the idea was supposed to be that $\displaystyle q$ is constant and $\displaystyle m$ is the variable.)
• Nov 28th 2009, 10:13 PM
aman_cc
Quote:
Originally Posted by redsoxfan325
I can do that, but does that prove the same thing as induction on $\displaystyle m$? (Because the idea was supposed to be that $\displaystyle q$ is constant and $\displaystyle m$ is the variable.)
I guess you want to prove that the above is true for any +ve integers m,q
So you can induct on m and under the induction hypothesis prove it is true for all q.
Or you can induct on q and under the induction hypothesis prove it is true for all m.
If you think carefully both prove the original statement. Atleast I'm pretty convinced.
The reason to chose 2nd approach is that the result is closely tied to induction on q rather than induction on m. It's almost trivial that way.
• Nov 28th 2009, 10:16 PM
aman_cc
Quote:
Originally Posted by aman_cc
I guess you want to prove that the above is true for any +ve integers m,q
So you can induct on m and under the induction hypothesis prove it is true for all q.
Or you can induct on q and under the induction hypothesis prove it is true for all m.
If you think carefully both prove the original statement. Atleast I'm pretty convinced.
The reason to chose 2nd approach is that the result is closely tied to induction on q rather than induction on m. It's almost trivial that way.
You may want to look at a similar post I did a few days ago -
http://www.mathhelpforum.com/math-he...tml#post410668
Your original problem follows directly from the identity, I wrote there. This is just FYI
• Nov 28th 2009, 10:19 PM
redsoxfan325
Yes, I agree that the second method is quite easy. I'll think about this until I'm convinced that inducting on q gives the same result.
• Nov 28th 2009, 10:21 PM
aman_cc
Quote:
Originally Posted by redsoxfan325
Yes, I agree that the second method is quite easy. I'll think about this until I'm convinced that inducting on q gives the same result.
Sure. I can help you prove that if you are not convinced. Please note the 'for all' in my post above. That would be critical in the proof I will give (for induction on 'q') | 1,140 | 3,928 | {"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} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.867111 |
http://www.xent.com/pipermail/fork/Week-of-Mon-20070402/044400.html | 1,529,570,197,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864110.40/warc/CC-MAIN-20180621075105-20180621095105-00135.warc.gz | 547,765,710 | 4,105 | # [FoRK] Data Structures Question
Stephen D. Williams <sdw at lig.net> on Wed Apr 4 06:59:54 PDT 2007
```The original question was trying to optimize when there were _any_
number of variables. I took any to be very large (thousands or 10's of
thousands), so O(n Log(n)) isn't very good compared to what can probably
be done. In many cases, these kinds of problems will have mostly false
variables or rules with limited interdependency which can be optimized for.
sdw
Nalin Savara wrote:
>> No, that doesn't do the job. I'm not looking to rewrite a boolean
>> calculator. I'm looking for a structure, such as a tree, that has inherent
>> properties that provide loading of an expression with and's and or's, but
>> enables to you go to a specific node, or use an algorithmic method of
>> navigation through the nodes, so that getting the subexpressions costs, at
>> the worst case, n * log(n)
>>
>>
> Reza; Andy; Joe ;
>
> (and I say this as someone who was once co-author of a algorithm design book
> that had a whole chapter on this):-- (Joe, Andy et al, feel free to comment)
>
> you can do the following:-- (the below are the exact steps--- hopefully
> this'll help you get it to code and skip the complicated proofs in Dragon
> book):--
>
> (1) First input the boolean expression in infix notation
> (2) Convert above boolean expression to postfix notation-- use the
> stack-based infix to postfix conversion algo found in any elementary
> data-structures book.
> Note: this algo is very efficient; because it involves peeking at the
> stack-top-- which is the equivalent of LALR(1) parsing-- because peeking at
>
> (3) Take the postfix expression and generate a tree out of it--> involves
> walking through the expression pushing nodes with each node having a single
> operand from the expression onto the stack until you 'see' a operator--> and
> at that point, you pop two nodes with operands, and replace those with a
> parent node with one operator and two terminal nodes which contain the
> operands (or a single child node with a single operand incase of a unary
> operator).
>
> (4) When there are no more operators or operands left in the expression; at
> that point the stack will automatically contain just one pushed node--> and
> that will be the root node of the expression tree representing the
> expression.
>
> The order of the above expression tree generation will be O(n Log(n) ) -->
> and assuming that nodes have pointers to parent nodes; and also references
> to symbols in original expression, you can go from any node to any other
> node as desired by you.
>
> If you want I can dig around for my old code-- see if i can find
> somethingthat does almost exactly what you want.
>
> Hope that helps...
>
> Nalin
>
>
> ----- Original Message -----
> From: "Andy Armstrong" <andy at hexten.net>
> To: <reza at voicegenesis.com>; "Friends of Rohit Khare" <fork at xent.com>
> Sent: Tuesday, April 03, 2007 4:20 PM
> Subject: Re: [FoRK] Data Structures Question
>
>
>
>> On 3 Apr 2007, at 05:33, Reza B'Far wrote:
>>
>>> However, I purposefully am trying not to treat it as a "logic"
>>> question. I
>>> want to load n expressions into a structure, such as a tree, that
>>> has the
>>> properties that support the decomposition that I want.
>>> Essentially, what I
>>> want is to explode the boolean (Andy refers to this) but with some
>>> constraints... Get all the "anded" terms together in one single
>>> expression.
>>> So, the subexpressions should only have AND's.
>>>
>> That /is/ a sum of products then... If you rewrite your original
>> expression in SoP form you get
>>
>> (e1 && e2 && e5) || (e3 && e5) || (e4 && e5)
>>
>> which is just your three AND terms ORed together.
>>
>>
>>> So, to summarize, I'm trying to find a structure that has this
>>> inherently
>>> built in... which by several degrees of indirection means
>>> significantly
>>> improved performance if you have n expressions and need to find the
>>> subexpressions and the path that was necessary to navigate to each
>>> randomly... basically, looking for the canonical solution... don't
>>> know if
>>> one exists or not.
>>>
>> I'm obviously having YASD[1] but I'm still not really clear what you
>> mean. For your example does the tree look like this?
>>
>> ( or )
>> / | \
>> / | \
>> / (and) \
>> / / \ \
>> (and) e3 e5 (and)
>> / | \ / \
>> e1 e2 e5 e4 e5
>>
>> It might be worth looking at this:
>>
>> http://www.dei.isep.ipp.pt/~acc/bfunc/
>>
>> Unfortunately it's binary-only but, if nothing else, some of the
>> links might be useful. I recently used it in conjunction with a
>> little Perl wrapper to semi-automatically refactor a bunch of complex
>>
>> [1] Yet Another Stupid Day
>>
>> --
>> Andy Armstrong, hexten.net
>>
>> _______________________________________________
>> FoRK mailing list
>> http://xent.com/mailman/listinfo/fork
>>
>
> _______________________________________________
> FoRK mailing list
> http://xent.com/mailman/listinfo/fork
>
``` | 1,301 | 5,203 | {"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} | 3.015625 | 3 | CC-MAIN-2018-26 | latest | en | 0.942906 |
https://axiumlearn.co.za/question-5-52/ | 1,723,473,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00673.warc.gz | 89,379,062 | 21,378 | # Question 5
A satellite of mass 1 000 kg is orbiting the earth at a distance of 200 km from the surface of the earth as shown on the diagram below.
5.1 State Newton’s Law of Universal Gravitation in words. (2)
5.2 Calculate the magnitude of the force that the earth exerts on the satellite to
keep it in orbit. (5)
5.3 Calculate the weight of the satellite on the earth surface. (3)
5.4 The same satellite is now orbiting the earth at a distance twice the radius of the earth from the centre of the earth. Without any further calculations, determine the force that the earth exerts on the satellite at the new distance. Explain how you arrived at your answer. (3)
[13] | 167 | 677 | {"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} | 2.765625 | 3 | CC-MAIN-2024-33 | latest | en | 0.878214 |
https://convertoctopus.com/9-9-ounces-to-kilograms | 1,627,493,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153739.28/warc/CC-MAIN-20210728154442-20210728184442-00149.warc.gz | 205,697,018 | 7,961 | ## Conversion formula
The conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:
1 oz = 0.028349523125 kg
To convert 9.9 ounces into kilograms we have to multiply 9.9 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:
1 oz → 0.028349523125 kg
9.9 oz → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 9.9 oz × 0.028349523125 kg
M(kg) = 0.2806602789375 kg
The final result is:
9.9 oz → 0.2806602789375 kg
We conclude that 9.9 ounces is equivalent to 0.2806602789375 kilograms:
9.9 ounces = 0.2806602789375 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 3.5630264595536 × 9.9 ounces.
Another way is saying that 9.9 ounces is equal to 1 ÷ 3.5630264595536 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that nine point nine ounces is approximately zero point two eight one kilograms:
9.9 oz ≅ 0.281 kg
An alternative is also that one kilogram is approximately three point five six three times nine point nine ounces.
## Conversion table
### ounces to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms
ounces (oz) kilograms (kg)
10.9 ounces 0.309 kilograms
11.9 ounces 0.337 kilograms
12.9 ounces 0.366 kilograms
13.9 ounces 0.394 kilograms
14.9 ounces 0.422 kilograms
15.9 ounces 0.451 kilograms
16.9 ounces 0.479 kilograms
17.9 ounces 0.507 kilograms
18.9 ounces 0.536 kilograms
19.9 ounces 0.564 kilograms | 498 | 1,778 | {"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} | 4.15625 | 4 | CC-MAIN-2021-31 | longest | en | 0.735237 |
https://www.coursehero.com/file/1650245/Experiment7/ | 1,524,485,227,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00356.warc.gz | 750,573,656 | 108,893 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Experiment7
# Experiment7 - Experiment#7 Diffraction and Interference 1...
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Experiment #7 Diffraction and Interference 1 Lab report for experiments 7 and 8 You will write one, formal, report for experiments 8 and 9 combined. It will differ from the reports for the previous experiments in that it will include the essential components of the type of report that is submitted to management or to a Journal for publication. A general outline of what should be included is provided in the page “Introduction to the course” on this web site. 2 Interference and diffraction In experiment 6 you measured the velocity of sound by finding the frequency of standing waves. Standing waves are a consequence of the phenomenon of constructive interference. In this experiment you will use another consequence of interference; the creation of a diffraction pattern due to the constructive and destructive interference of two or more waves of the same wavelength. You will measure the diffraction pattern due to the interference of identical waves emanating from two closely spaced sources. The two sources are obtained by placing a single source behind a barrier with two slots cut into it. The waves from the two slots behave as if they are two identical sources. Figure 1. Illustration of the waves from a plane wave incident on two slits in a barrier.
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The existence of a null or, more realistically a minimum, in the diffraction pattern is due to destructive interference between waves from the two apertures. Consider two narrow slits in a barrier with a plane wave incident on its back side. If the width of the slits, a, is very small compared to the wavelength, you will learn in experiment 8 that the wave will radiate uniformly in all directions from the slit. F or points of observation very far from the screen, strong intensity maxima will occur when the waves from both slits arrive in phase, so that they will add. This occurs when the differences in distance between the two slits and the point of observation are an integer number of wavelengths; λ θ n d = sin 2.1 Acoustic waves All wave phenomena display the properties of interference and diffraction. In this experiment you will measure the phenomenon using sound waves. Sound waves are obvious in our everyday experience and these phenomena have a strong influence on what we hear. They also play an important role in the design of acoustic devices such as speakers. A disadvantage of using sound waves for this experiment is that they reflect off of virtually all surfaces, including you. As a consequence you will see some interference between the waves you are interested in measuring and waves reflected off of many other surfaces. That makes this experiment more like a real experiment in which you are looking for one effect in the presence of many others. You will need to develop your own measuring technique and exercise some judgment in the way you obtain and analyze your data.
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{[ snackBarMessage ]} | 654 | 3,270 | {"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} | 3.046875 | 3 | CC-MAIN-2018-17 | latest | en | 0.924273 |
http://cs.stackexchange.com/questions/7473/propositional-formula-in-dnf-can-be-decided-in-polynomial-time | 1,467,039,080,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00002-ip-10-164-35-72.ec2.internal.warc.gz | 71,693,837 | 17,613 | # Propositional formula in DNF can be decided in polynomial time?
For a given propositional formula f in DNF, one can decide in polynomial time, if the formula is satisfiable: Just walk through all subformulas (l_1 and ... and l_k) and check, wheter it has NO complementary pair of literals. Formula f is satisfiable iff such subformula exists.
Is my approach above correct ?
If yes, I'm wondering why all modern SAT solvers get a CNF as input format, and don't just use DNF.
-
The conversion from CNF to DNF can come at an exponential cost. For example $(a_1 \lor b_1) \land \cdots \land (a_n \lor b_n)$ expands to $2^n$ many terms. As you comment, for DNF satisfiability is easy - it is falsifiability which is hard. If the problem is trivial, you don't input it to a SAT solver, and that's why SAT solvers accept CNFs instead of DNFs.
If you believe that P is different from NP, then this implies that there is no polynomial time way to convert CNF satisfiability to DNF satisfiability, since the former is NP-complete while the latter is in P.
-
Thank you. I was just wondering: When the SAT problem for DNF formulas is in P, why not just describe problems directly into DNF (instead of CNF) ? Because translating a formula into CNF might take exponential cost as well as translating it into DNF. – John Threepwood Dec 17 '12 at 22:26
@JohnThreepwood most problems aren't easily expressed in DNF. And, the conversions to DNF usually take exponential time, and necessarily so, for example if there are exponential satisfiable solutions. – Realz Slaw Dec 18 '12 at 2:57
@RealzSlaw Thank you. So most real-life problems are easier encoded in CNF. Is this because most problems are 'naturally' structured as constraints (in CNF each clause is a constraint) ? – John Threepwood Dec 18 '12 at 16:10
@JohnThreepwood If a problem can be expressed as DNF, a satisfiable assignment is there naturally. There would be nothing to solve. And intuitively yes, I think the problems solved using SAT solvers are naturally described via constraints which CNF is useful for. – Realz Slaw Dec 18 '12 at 17:29
@RealzSlaw Good point. That helped understanding. – John Threepwood Dec 18 '12 at 18:05 | 564 | 2,188 | {"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} | 3.1875 | 3 | CC-MAIN-2016-26 | latest | en | 0.931029 |
https://graphicmaths.com/gcse/geometry/area-perimeter-circles/ | 1,725,847,833,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00263.warc.gz | 266,194,846 | 7,779 | # Area and circumference of circles
By Martin McBride, 2022-11-04
Tags: triangle area perimeter
Categories: gcse geometry
A circle is a flat shape that has no corners or straight lines. Every point in the edge of a circle is a fixed distance (called the radius) from the centre.
In this article, we will look at the area and circumference of a circle. We will also look at:
• Arcs
• Segments
• Sectors
The radius of a circle is the length of a line drawn from the centre of the circle to any point on its edge. It is the distance r on this diagram:
The diameter of a circle is the length of a line drawn between two points on the edge of a circle that goes through the centre of the circle. It is the distance d on this diagram:
The diameter is twice as long as the radius.
## Circumference of a circle
The perimeter of a circle is the distance all the way around the edge of the circle. For circles, the perimeter is usually called the circumference.
The circumference of a circle, C, is given by the formula:
Here, d is the diameter of the circle, and π is a number that is approximately equal to 3.14159.
Since the diameter of a circle is twice its radius, we can write this equation as:
## Area of a circle
The circumference of a circle, A, is given by the formula:
The equation for the area of a circle also uses the same number, π, that is used in the equation for the circumference.
The number π (sometimes written as pi) can be defined as the ratio of the circumference of a circle to its diameter.
Mathematicians have been studying π since ancient times. At first, its value was probably estimated by measuring circles and would have been known to be about 3. Later on, geometric constructions were used to estimate π to several decimal places. Later still, various other methods were discovered, and the advent of modern computers has allowed us to calculate its value to over a billion digits.
We know that π is an irrational number. It cannot be expressed as an exact fraction. If it is expressed as a decimal number (3.141592653589793...) the digits go on forever and never repeat. This means that we can never write down the exact numerical value of π.
## Length of an arc
An arc is part of the circumference of a circle. Here are some examples. The thick red line shows the arc:
The arc is defined by the angle it makes at the centre of the circle.
The first example shows an arc of 90°. This takes up a quarter of the circle (because there are 360° in a full circle). Since the circumference of the circle is 2πr, the length of a 90° arc is a quarter of that, which is πr/2.
The second example shows an arc of 180°. This takes up half of the circle. The length of the arc is half of the circumference, which is equal to πr.
The general formula, for the length L of an arc with angle a, is:
This can be simplified by cancelling out the 2:
For an arc of angle 240°, this formula gives an arc length of 4πr/3, and for an angle of 72° the length is 2πr/5.
## Area of a sector
A sector is a part of the circle enclosed by an arc and two radii. Here are some examples:
The first example shows a sector with an angle of 60°. This shape is like a pie slice.
This sector represents a sixth of the area of the triangle (since 60° is a sixth of 360°). This means that the area of the sector is πr^2/6.
In general, a sector with an angle a has an area of:
The second example shows a sector with an angle of 240°.
## Minor and major sectors
Notice that in each case the circle is actually divided into 2 sectors. In the first example, we have created a sector of angle 60°, but the remainder of the circle is also a sector of angle 300°. One sector is less than 180°, and we call that the minor sector. The other sector is greater than 180° and we call that the major sector.
In the second example, we have created a major sector of 240°, which leaves a minor sector of 120°.
The exception is if the angle is 180°. In that case, the circle is divided into 2 semicircles. A semicircle is a sector, but it is neither minor nor major.
## Perimeter of a sector
The perimeter of a sector is the length of the arc plus the lengths of the 2 radii. Using the previous formula for the arc length and adding 2r gives us this formula for the perimeter of a sector:
## Area of a segment
A segment is formed when a circle is divided by a chord. A chord is a line joining any two points on a circle. Here is an example segment:
A segment is defined by the angle it makes at the centre, just like a sector.
To calculate the area of a segment, we need to calculate the area of the equivalent sector (shown on the left, below), and then subtract the area of the triangle formed by the chord and the 2 radii (shown on the left):
The sector area, as we saw earlier, is:
The area of the triangle, from the article Area and perimeter of triangles, is:
Subtracting the triangle area from the segment area gives:
This can be tidied up to give:
## Minor and major segments
The chord divides the circle into two parts:
The smaller area, in blue, is called the minor segment. The larger area, in green, is called the major segment.
The major segment is the one that contains the centre of the circle.
The area calculation above applies to the minor segment. To find the area of the major segment we simply subtract the area of the minor segment from the area of the full circle.
If the angle of the segment is 180°, the circle will be divided into two semicircles. They could be considered segments, but they are neither major nor minor segments. | 1,309 | 5,575 | {"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} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.954803 |
https://youvan.com/Examples/Example_2._Pseudocoloring_Ordinary_Photographs_to_Highlight_Grayscale_Changes.htm | 1,712,923,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00486.warc.gz | 953,530,984 | 2,508 | Example 2. Pseudocoloring Ordinary Photographs to Highlight Grayscale Changes
We have developed several hundred algorithms for converting grayscale images to color. Pseudocoloring schemes are seen everyday in a variety of applications ranging from radar maps to visualizing heat loss from houses. Often these colorization schemes are haphazard, and no mathematical basis is considered. It would seem rather obvious that the operator of a pseudocoloring scheme should have at his disposal many different methods for colorization. By visualizing such schemes within (Schrödinger) color cubes, as seen below, one can begin to understand why particular colorization schemes are better than others. In addition, through a process call orthonormalization, one color scheme can be converted to many others by employing tensor algebras.
The image that is analyzed next was captured at night during a thunder storm. There is a very large dynamic range of light intensity that varies from an actual bolt of lightening, to a background of clouds, to structures on the ground - including a church steeple. This image was captured using a Fuji FinePix S2, whereafter what little color detail was present in the original digital image was flattened to gray by averaging all three RGB channels into one gray channel.
Grayscale image (2.1) with a grayscale bar at left. The pseudocoloring scheme shown below was originally developed in the Youvan laboratory at MIT ca. 1990 as a 'hotmap' to colorize images of bacterial colonies taken with a digital imaging spectrophotometer.
'Hotmapped' pseudocolored image (2.2). Low grayscale values are mapped to black and blue hues whereas brighter grays are mapped progressively through green, yellow, red, pink, and white as seen in the color cube, below.
This figure (2.3) shows our 'hotmap' coloring screen traced inside a Schrödinger cube. Increasing values of R, G, and B are shown on the x, y, and z axis, respectively. The lower left corner of the cube is black (0,0,0), and the upper right corner is white (255,255,255). The paradigm for the hotmap arrows, follows: 1-42 ++ B, 43-84 --B; ++G, 85- 128 ++R , 129-170 --G, 171-212 ++B, 213-256 ++G . A line from 255,255,255 to 0,0,0 (not shown) is a trace-back of the hottest color (i.e., white) to the beginning of the color map (black) that can be seen as a black dot capping the hotmap scale in other figures on this page. Often, programmers work in collinear spatial coordinates (x, y, z) with coincident color vales (R, G, B) that are given on a scale of 0.0 to 1.0 in 256 increments of 1/256.
Source code in .nb and .PDF formats. | 655 | 2,650 | {"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} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.924208 |
https://proofwiki.org/wiki/Leigh.Samphier/Sandbox/Set_Difference_of_Matroid_Circuit_with_Element_is_Independent | 1,603,759,234,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107892710.59/warc/CC-MAIN-20201026234045-20201027024045-00048.warc.gz | 485,211,180 | 9,560 | # Leigh.Samphier/Sandbox/Set Difference of Matroid Circuit with Element is Independent
## Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $x \in C$.
Then:
$C \setminus \set x$ is an independent subset of $C$
## Proof
$C \setminus \set x \subseteq C$
Because $x \in C$ and $x \notin C \setminus \set x$:
$C \setminus \set x \ne C$
Hence:
$C \setminus \set x \subsetneq C$
By definition of circuit:
$C$ is a minimal dependent subset
Hence:
$C \setminus \set x \in \mathscr I$
$\blacksquare$ | 187 | 558 | {"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} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.482162 |
http://www.edhelper.com/visual_skills1002.htm | 1,493,374,814,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00177-ip-10-145-167-34.ec2.internal.warc.gz | 522,295,797 | 3,572 | edHelper subscribers - Create a new printable
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Sample edHelper.com - Visual Perceptual Skills Worksheet
Name _____________________________ Date ___________________
Review
(Answer ID # 0369653)
Match the picture on top with one of the four choices.
1.
Match the picture on top with one of the four choices.
2.
Complete.
3.
How many times is the number 76 in the above picture?
7 times 1 time 3 times 5 times
Match the complete picture on top with one of the four choices. Lines are missing from each of the four choices.
4.
Match the complete picture on top with one of the four choices. Lines are missing from each of the four choices.
5.
What is added to the first picture to make the second picture?
6.
First Picture to Second Picture
Copy the lines shown on the left to the blank grid on the right.
7.
Copy the lines shown on the left to the blank grid on the right.
8.
Copy the lines shown on the left to the blank grid on the right.
9.
One picture is different from the other three pictures. Circle the picture that is different from the others.
10.
Circle the picture that is exactly the same as the picture on top.
11.
One picture is different from the other three pictures. Circle the picture that is different from the others.
12.
Match the picture on top with one of the four choices.
13.
Complete.
14.
How many times is the word step in the above picture?
3 times 9 times 6 times 2 times
Complete.
15.
Count the numbers greater than 798 in the above picture. How many times did you count a number greater than 798?
12 times 5 times 13 times 20 times
Draw a line from start to finish. Do not cross any lines.
16
Connect the dots. Start from the number 1 and count by ones. Draw a line from start to finish.
17
Complete.
18.
Draw a triangle in box A2.
Draw a trapezoid in box A1.
Draw a square in box B1.
1 2 A B
19.
Draw an octagon in box B2.
Draw an oval in box A2.
Draw a circle in box A1.
A B 1 2
Pick the choice that is a wrong statement.
20.
HNPXV
P is third X is to the left of P X is between V and N N is to the right of H
Pick the choice that is a correct statement.
21.
the rectangle is between the oval and the trapezoid the oval is to the right of the rectangle the trapezoid is to the left of the oval the rectangle is to the right of the oval
Draw the missing parts of the picture on the right. Color the picture on the left.
22.
Left Right
23.
Left Right
Color all of the shapes.
24.
Color red Color pink Color gray Color purple
edHelper subscribers - Create a new printable | 641 | 2,605 | {"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} | 2.640625 | 3 | CC-MAIN-2017-17 | latest | en | 0.882217 |
https://questions.kunduz.com/math/application-of-derivatives/changes-saved-2-abail-is-thrown-into-the-air-with-an-upward-velocey-of-48-tis-is-height-hin-feetattert-seconds-is-given-by-the-21178579 | 1,656,310,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00090.warc.gz | 535,446,715 | 20,853 | Question:
# A ball is thrown into the air with an upward velocity of 48 ft/s. Is height h in feet after t seconds is given by the function h= -16t²+42t+6. How long does it take the ball to reach its maximum heigh
A ball is thrown into the air with an upward velocity of 48 ft/s. Is height h in feet after t seconds is given by the function h= -16t²+42t+6. How long does it take the ball to reach its maximum height? What is the ball's maximum height? Round to the nearest hundredth, if necessary. | 133 | 498 | {"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} | 3.59375 | 4 | CC-MAIN-2022-27 | latest | en | 0.94076 |
http://bedtimemath.org/fun-math-sharks/ | 1,550,395,952,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481832.13/warc/CC-MAIN-20190217091542-20190217113542-00606.warc.gz | 32,770,765 | 15,328 | # A Big Bite of Shark
Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.
# A Big Bite of Shark
August 6, 2014
Normally we think of sharks as those scary fish that might bite us. But don’t worry, we humans get to bite back — when we eat shark for dinner. Shark steaks are pretty tasty to a lot of folks, so people do catch and sell sharks for their meat, and handling those sharks can get pretty extreme. A fisherman in China caught a 16-foot long whale shark in his fishing net, weighing more than 2 tonnes. It was so huge that he had to drape it over a truck to drive it to the fish market! Whale sharks are endangered, so fishermen are not allowed to fish for them and sell them…but since this one had sadly already died in his fishing net, he’s allowed to sell it without being punished with a fine. Most people can’t eat more than a half-pound steak at a time, so that shark is going to feed a lot of hungry fish lovers.
Wee ones: Do you know how tall you are? Who’s longer, you or a 16-foot shark?
Little kids: If the shark is 16 feet long and they wanted a truck that was 1 foot longer to carry it, how long would that truck have been? Bonus: If instead they got this 14-foot truck, how much longer was the shark?
Big kids: The fisherman had to drive with that huge fish on his truck for a half hour. If he started the trip at 2:39 pm in the afternoon, when did he and his smelly passenger reach the fish market? Bonus: The shark weighs 2 tonnes, which means 2000 kg…which is about 4,400 pounds they had to lift on and off the truck. If a person can carry 100 pounds, how many people would it have taken to pick up that crazy fish? | 443 | 1,810 | {"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} | 3.46875 | 3 | CC-MAIN-2019-09 | latest | en | 0.979997 |
http://sourceforge.net/p/gdalgorithms/mailman/attachment/CAGWOt2-DSKyR%2Be8oo-20ikHbGq9hp3k4_deoBpx-LB74rCHcGw%40mail.gmail.com/1/ | 1,406,972,140,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510280868.21/warc/CC-MAIN-20140728011800-00184-ip-10-146-231-18.ec2.internal.warc.gz | 281,843,593 | 2,341 | There are also a number of line rasterizing algorithms that do 2, even 3 pixels per iteration, i.e. with them you will be able to fill the arrays faster. Lookup for double-step rasterizing algorithms.
Also, in my experience the extrema arrays are not needed: for convex polygons there are only two active edges at a time, therefore you just need to traverse the poly in clockwise and anti-clockwise order to switch the active edges until they meet at the maximum Y coordinate.
Hope that helps further.
Cheers,
Martin
On Wed, Aug 10, 2011 at 6:29 PM, Graham Rhodes ARA/SED wrote:
Some helpful Google search terms are "polygon rasterization" and "polygon scan conversion." Someone else mentioned the Foley/van Dam book/et al. book, which is a good reference that describes the classic algorithm. The book Graphics Gems I also has a discussion, though it is brief. You can also find this stuff online, for free. For example, two sets of lecture slides from University of Virginia that look good:
http://www.cs.virginia.edu/~gfx/courses/2004/Intro.Fall.04/handouts/11-polyscan.pdf
www.cs.virginia.edu/~asb/teaching/cs445-fall06/slides/09-rasterization.ppt
I even found this YouTube video:
Graham
-----Original Message-----
From: Joel B [mailto:onephatcat@earthlink.net]
Sent: Wednesday, August 10, 2011 7:49 AM
To: Game Development Algorithms
Subject: [Algorithms] Draw & fill regular polygon?
I should know how to do this, but i don't. I only have the ability in my system to draw lines (x1,y1,x2,y2)' to a bitmap, or write to the bitmap as a byte array. Can someone tell me the stepwise procedure to draw a polygon of n sides and then fill it? Having trouble finding anything online that doesn't use pre-existing primitives or libraries.
Thanks,
Joel
Sent from my iPhone
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