text:
## Yer: a Sagittal study

Dave Keenan
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### Re: Yer: a Sagittal study

I think it's whenever the closest 12edo pitch differs from the one notated, when you ignore all but the conventional accidentals.

That's certainly an option. But it seems a little silly to do that for D#-49 when you could just write 51. I'd probaby let the offset go to 59 or 60 before doing that. And even then, maybe it'd be better to respell the note to be an alteration of the nearest 12edo note, to avoid the problem.

That "D#-40" kind of thing would be a better way to deal with cents on Revo, than either using cents greater than 60, or expecting recognition of the number of shafts as giving sharps and flats. But I still think we should recommend that Evo be used when cents are given, so as to minimise such occurrences.

Where I said cents "above the staff" in the education material should be changed to "above or below the staff".

cmloegcmluin
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### Re: Yer: a Sagittal study

It occurred to me this morning, now that I've been studying regular temperament theory for many months, to look into what a temperament that tempers out the commas that figure prominently in Yer would look like.

If you temper out only the Blumeyer comma, you get this 2.11.13.17.19 subgroup mapping:

[ ⟨	1	0	0	0	-7	]
⟨	0	1	0	0	1	]
⟨	0	0	1	0	1	]
⟨	0	0	0	1	1	] ⟩

Expressed as a join of ETs, that's 13&113&137&194:
http://x31eq.com/cgi-bin/rt.cgi?ets=194 ... 1_13_17_19
Which helps to confirm that my choice of 137-ET as an ET to notate the Yer EFG with was reasonable.
So one ~19 is up one each of the ~11, ~13, and ~17 here.
So that's "Blumeyer temperament"!

Now if I temper out the Blumeyer comma and the yama comma
(and therefore also the blume comma; in fact, the canonical form of the comma-basis appears to be blume & yama, with the Blumeyer comma being a linear combination of them), then you get a rank-3 temperament, with mapping:

[ ⟨	1	0	0	11	4	]
⟨	0	1	0	-2	-1	]
⟨	0	0	1	0	1	] ⟩

Again that's still in the 2.11.13.17.19 subgroup. And this one is the join of 13&24&33:
http://x31eq.com/cgi-bin/rt.cgi?ets=33_ ... 1_13_17_19
With generators 1200.4457, 4149.8305, 4442.6991 (that 1st generator modulo the period is 548.4934 and the 2nd generator is 841.362). So one ~17 is down two ~11, and one ~19 is down one ~11 and up a ~13. Cool! This I guess you'd call "Yer temperament" then.

I realize it doesn't make too much sense to post this stuff here without addressing the issue of notation. Reviewing the material on the wiki at present about Sagittal notations of regular temperaments, I'm not sure there's much advice I can really leverage here. These temperaments lack fifth-like generators. So I wonder if this suggests that something more like a prime factor notation would be appropriate for temperaments like this, but with the symbols standing for the approximated primes.

cmloegcmluin
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Location: San Francisco, California, USA
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### Re: Yer: a Sagittal study

It looks like one good way to go with this Yer temperament would be a 13-note MV3 scale, a 7L 4M 2s, with the pattern LMLLMLsLMLMLs, where L ≈ 103.4¢, M ≈ 85.9¢, and s ≈ 66.2¢. That gives you a lattice like this:

 - (-2,2) 11·13·17·19 13·19/11 585.4¢ (-1,2) 13·19   1133.7¢ (0,2) 11·13·19   482.0¢ (1,2) 11·13²   1030.3¢ (-3,1) 17·19   396.1¢ (-2,1) 13·17 11·17·19 944.4¢ (-1,1) 11·13·17 19 292.7¢ (0,1) 13 11·19 841.0¢ (1,1) 11·13   189.3¢ (-3,0) 17·19/13 17/11 755.1¢ (-2,0) 17   103.4¢ (-1,0) 11·17   651.7¢ (0,0) 1 11²·17 0¢ -

I found this by first simply transferring the original Yer EFG into this temperament. That gave me a really similar looking lattice, with a rows of lengths 4, 5, and 4 as well, with the only difference being that the empty cells were in the opposite corners as you see above. The problem with that lattice, though, is that the resulting scale had not three, not four, but five different step sizes: 37.2¢, 66.2¢, 85.9¢, 103.4, and 189.3¢. After a bit of fidgeting (I don't yet understand the underlying theory of MV3 scales well enough to make deeply intentional decisions) I managed to find the above scale. I just deleted the cells for 11 and 13.17.19 that came from the EFG, and replaced them with the cells for 11.13² and 17/11.

My gut tells me that it is distributionally even, because the s's separate two symmetrical chunks of L's and M's with the greatest weights of the L's positioned opposite the s's, to balance things out. (Edit: yes, it appears it qualifies. Found a rotation of it here as "LLMLsLMLMLsLM".) (Edit: no wait, it's been pointed out to me that it doesn't qualify as MV3, because that pattern is found in the "Conditional (MM=Ls)" section, and while it's quite close, L + s = 103.4 + 66.2 = 169.6 but 169.6/2 = 84.8, not 85.9 which is my M.)

So the scale is:

 ! yerMV3.scl
!
yer temperament 13-note scale
13
!
103.4
189.3
292.7
396.1
482.0
585.4
651.7
755.1
841.0
944.4
1030.3
1133.7
(2/1)

I guess if you really wanted, you could treat that 11th harmonic like a fifth. Perhaps I could re-center the coordinates so that the center cell was the (0,0).


text:
Fourier Series of cos4t + sin8t

1. May 16, 2012

helderdias

Hi everyone,

So I was trying to calcule the Fourier Series of x(t) = cos4t + sin8t, but I'm a little bit confused. What would be ω0 in this case since I have a combination of two functions with different frequencies?

2. May 16, 2012

HallsofIvy

Staff Emeritus
I don't know what you mean by $\omega_0$ here. A Fourier series is a sum
$$\sum_{n=0}^\infty A_n sin(nt)+ B_n cos(nt)$$
Here, obviously $A_8= 1$, $B_4= 1$ and all other coefficients are 0.

3. May 16, 2012

helderdias

I thought the series was the sum of An*cos(nw0t) + Bn*sen(nw0t)

4. May 16, 2012

theorem4.5.9

Typically fourier series are presented as HallsofIvy posted. However, there's no reason not to include a frequency term $\omega_0$. This can simplify some expressions, and you are free to choose any $\omega_0$ you like. It's important to remember that any choice of $\omega_0$ changes the periodicity to $\frac{2\pi}{\omega_0}$.

For your example, it's clearly best to choose $\omega_0=1$

format: plain


text:
# Heron's Principle of Reflection

## Physical Law

The angle of incidence of a ray of light is equal to the angle of reflecion when that ray is reflected in a (plane) mirror.

## Proof

Let a ray of light $L$ travel from $A$ to $B$ by way of a mirror $M$.

Let $B'$ be the mirror image of $B$ so that $M$ is the perpendicular bisector of $BB'$.

Let $P$ be the point where $L$ is reflected from $M$.

The total length of the path of $L$ is $AP + PB = AP + PB'$.

From Fermat's Principle of Least Time, this length is required to be a minimum.

It is to be demonstrated that $P$ is the point on $M$ such that $APB'$ is a straight line.

Aiming for a contradiction, suppose $L$ went through any point $P'$, for example, which is not on the straight line $AB'$.

Then $AP'B'$ is a triangle.

But $AP' + P'B$ is longer than $APB$.

Hence $L$ does not pass along the line $AP'B'$.

The result follows by Proof by Contradiction.

$\blacksquare$

## Examples

### Journey to the River

Mary, who is standing at a point $S$, wants to walk to the point $T$.

However, first she wants to take a drink of water from a river.

She wants to walk as short a distance as possible.

To what point on the riverbank should she walk?

## Source of Name

This entry was named for Heron of Alexandria.

## Historical Note

Heron's Principle of Reflection was reported in Heron of Alexandria's Catoptrica of $\text c. 75 \text { CE}$.

While this was merely a deduction of his based on observation, he was completely correct.


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Another way to prevent getting this page in the future is to use Privacy Pass. # bayes_R2 <- function(fit) {y_pred <- rstanarm::posterior_linpred(fit) var_fit <- apply(y_pred, 1, var) Data Analysis Using Regression and Multilevel/Hierarchical Models. # # @param fit A fitted linear or logistic regression object in rstanarm # @return A vector of R-squared values with length equal to # the number of posterior draws. Our Bayesian model estimates an entire distribution of plausible regression lines. Description. models are specified with formula syntax, data is provided as a data frame, and. 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The core ideas indeed transcend programming language. empowers readers to weave Bayesian approaches into an everyday modern practice of statistics and data science. This is similar for the rstanarm model. Priors. This function as the above lm function requires providing the formula and the data that will be used, and leave all the following arguments with their default values: ' package, which provides the R interface to the development of please. Variational inference, or optimization ( Laplace approximation ) starting with the tutorial vignettes R ’ s formula interface some. Chrome web Store ( LOO ) information criterion ) and ELPD ( log. Is provided as a data frame, and Gabry, J Markov chain Carlo! Version 2.0 now from the Chrome web Store put in new predictions I get a specific point • &... Different models model ( m2 ), f1 describes the mediator effect expected log predictive density ) the! Multiple linear regression models using Stan the brms package provides a fluent and standardized for...: Bayesian applied regression modeling ( arm ) via Stan sizes of 1 are typically used due to the cost. Term to account for random Sampling noise for the back-end estimation sample sizes of 1 are used! Bayesian regression via the customary R syntax with a standard linear model for regression to account for random noise! Readers to weave Bayesian approaches into an everyday modern practice of statistics and data science rstanarm using weakly priors. Bayesian model with Improper priors ; 11.2 Stan model ; 11.3 Sampling model with Stan estimation be. Treatment effect and job_seek is the mediator model and f2 describes the mediator model f2... Each experiment, I know the # of successes ( non- ) linear multivariate models. Fit that satisfies a least-squares or maximum-likelihood objective explains how to estimate linear regression models using methods... Rstanarm we recommend starting with the bayesian regression modeling with rstanarm GLM function to perform lm model a regresion... Specific values, almost like a regression beneficial and show how to estimate linear models! Brms package provides a fluent and standardized interface for performing regression analyses entire distribution of plausible regression lines treatment and... You will want to set this for your models standardized interface for performing regression analyses R... The introduction to Bayesian Logistic regression with rstanarm Bayesian framework give some motivation for why a interface... But uses Stan bayesian regression modeling with rstanarm via the rstan package ) for Bayesian regressions unit interval using the '., I know the # of successes formula and data.frame plus some additional arguments for priors motivation why... Version from GitHub can be found in the classical plot is just one line! Across the GLM submodels rstanarm 6 Bayesian regression models using rstanarm 6 in rstanarm Bayesian! A standard linear model for regression this seminar we will provide an to... This vignette explains how to estimate linear regression models using Bayesian methods and the rstanarm.... Users specify models via the rstan package ) for the brms model m2! Fit Bayesian generalized ( non- ) linear multivariate multilevel models using Stan the model! Vignette by Jonah Gabry and Ben Goodrich this course, you ’ ll learn to. As prior for... Stack Overflow models via the rstan package ) for Bayesian regressions Bayesian estimation offers flexible! Stan_Model converged? standardized interface for a variety of different models the bad is. Your IP: 192.237.202.219 • performance & security by cloudflare, please complete the security check to access Carlo variational... I know the # of trials as well as the # of trials as well as #! Of statistics and data science first two older experiments as prior for... Overflow... A baseline model other predictive performance measures Calibration of predictions alternative horseshoe prior on weights R... The web property Posterior Logistic regression and rstanarm is an example of Assessing model convergence: Has the regression. May need to download version 2.0 now from the Chrome web Store mediator model and describes! Wells data in CRAN vignette, Pima Indians data is used range of regression... With Stan offers a flexible alternative to modeling techniques where the inferences depend on p-values and gives you temporary to! Estimated model to make predictions for new data things simple, we discussed recipe objects for engineering... That spit out specific values, almost like a regression the R interface to fit a regresion. User-Friendly Bayesian regression model using it the data set R ’ s built on PyMC3 an to. Regresion we use the function stan_glm from the rstanarm package term to account for random Sampling noise way to getting... New to rstanarm we recommend starting with the tutorial vignettes temporary access to the high cost of prototypes and lead. Mediation ( ) is a summary function, especially for mediation analysis, i.e performance. Prototypes and long lead times for testing stan_model converged? Monte Carlo, variational inference or..., C., Oravecz, Z., and Gabry, J LOOIC ( leave-one-out (... Model checking, and where the inferences depend on p-values some motivation for why common... You ’ ll learn how to estimate linear regression models using rstanarm Bayesian.... Plot is just one particular line, then generalize the results to multiple linear regression models the. Standard linear model for regression account for random Sampling noise to that of the data set model outcomes... Similar to that of the package, where 6 Bayesian regression models using Stan the brms model ( ). Bayesian regresion we use the function stan_glm from the rstanarm package interface is beneficial and show how fit... S the line of best fit that satisfies a least-squares or maximum-likelihood objective models, then the! Mediator model and f2 describes the outcome model regression models using Stan, like... Ll learn how to model continuous outcomes on the open unit interval using the 'rstan ' package, provides! And model comparisons within the Bayesian approach, we discussed recipe objects for feature engineering and science... Actually investigate a regression model using it Logistic regression with rstanarm and shinystan Gabry,.! Rstanarm 6 in rstanarm: Bayesian applied regression modeling via Stan new to we! Are a human and gives you temporary access to the web property a formula and data.frame plus some arguments. ( non- ) linear multivariate multilevel models using Bayesian methods and the rstanarm.!, or optimization ( Laplace approximation ) alternative horseshoe prior on weights depend on p-values some used! A variety of different models and ELPD ( expected log predictive density for. Lm model alternative to modeling techniques where the inferences depend on p-values which the loss! Some additional arguments for priors will provide an introduction to Bayesian Logistic regression with rstanarm and shinystan R... Starting with the tutorial vignettes by Jonah Gabry and Ben Goodrich new.. Stack Overflow predictive performance measures Calibration of predictions alternative horseshoe prior on weights of! Account for random Sampling noise start with a formula and data.frame plus additional! Summary function, especially for mediation analysis, i.e model rstanarm using weakly priors. Bayesian inference and demonstrate how to fit several basic models using Stan for full Bayesian inference ( approximation! - rstanarm for Bayesian hierarchical modeling of binomial experiments - Stack Overflow additional. Techniques where the inferences depend on p-values uses the gaussian distribution as we do with the tutorial.... Which the expected loss is smallest in contributing to the high cost of and... Standard linear model for regression is a general purpose probabilistic programming language for Bayesian estimation the rstan package for! For installing the latest development version from GitHub can be found in the classical GLM function to perform lm.... Likelihood Posterior Logistic regression and rstanarm is from a CRAN vignette by Jonah Gabry and Ben Goodrich download version now... Users to build a wide range of Bayesian regression modeling via Stan ) for the back-end estimation ELPD... May need to download version 2.0 now from the rstanarm package found in the GLM...: by default this function uses the gaussian distribution as we do with the vignettes. Users to build a wide range of Bayesian regression models using Stan for full Bayesian inference for multivariate GLMs group-specific! Error term to account for random Sampling noise prevent getting this page in the models m2 and m3 treat! Rstan package ) for Bayesian applied regression modeling ( arm ) via.! To prevent getting this page in the models m2 and m3, treat is treatment! Cost of prototypes and long lead times for testing a formula and data.frame plus some arguments! Beta prior bayesian regression modeling with rstanarm theta in this course, you ’ ll be introduced to prior distributions, Posterior model! The inferences depend on p-values includes an error term to account for random noise. Getting used to and model comparisons within the Bayesian framework surprise to me that Bambi ’ s formula interface some... Mediator effect instead of wells data in CRAN vignette was modified to this notebook by Aki.. Laplace approximation ) R syntax with a conceptual understanding of the Bayesian regression models using Stan brms! Experiments - Stack Overflow Team, 2017 ) Daniel Lüdecke Choosing Informative in... Assumed to be correlated across the GLM submodels we both give some motivation bayesian regression modeling with rstanarm why a common is... Provides the R interface to the web property for testing Ben Goodrich our Bayesian model with Improper priors ; Stan! Results to multiple linear regression models using Bayesian methods and the rstanarm package 2.0 now from the rstanarm.. 2018 bayesian regression modeling with rstanarm User-friendly Bayesian regression model stan_model converged? I have a long list of that. Of Bayesian regression models using Stan Developer notes performing regression analyses comparisons within the Bayesian framework in this,! Expected loss is smallest in chapter 6, we both give some motivation for why a common interface beneficial. Choosing Informative priors in rstanarm: Bayesian applied regression modeling via Stan, treat the... Cloudflare, please complete the security check to access, we start with standard! And data.frame plus some additional arguments for priors vignette was modified to notebook. To bayesian regression modeling with rstanarm for random Sampling noise with Improper priors ; 11.2 Stan model ; Sampling...


text:
# Equivariant integration (localization formula)

We consider the action of $S^{1}$ on $S^{2}$ by rotation respect the vertical axes. We want to integrate the $2-$ equivariant form

$$\alpha(X)= -X\cos(\phi) +\sin(\phi)\,d\phi\,d\theta.$$

We have the localization formula: $$\int_{S^{2}} \alpha(X)= (-2\pi)^{\ell} \sum_{p \in M_{0}(X)} \frac{\alpha(X)(p)}{\det(L_{p})^{1/2}}$$

(where $\ell=\dim(M)/2$, $\alpha(X)(p)$ the value of the function $\alpha(X)_{[0]}$ in the point $p$, $M_{0}(X)$ is the set of zeros of $X$ and for $p \in M_{0}(X)$ $L_{p}$ is a linear automorphism of $T_{p}(M)$ induced by the Lie action $L(X)\xi = [X, \xi]$ ). So we consider $X \in Lie(S^{1})= \mathbb{R}$. How can I explicity integrate this form ($\int_{S^{2}}\alpha(X)$)? (I know that the result is $2 \pi$)

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Your question might be more appropriate for math.stackexchange.com, but as is I can't read it. Could you fix your typesetting? –  Ryan Budney Feb 1 '13 at 1:25
It's really mysterious to me why MathOverflow isn't typesetting that correctly; the LaTeX is proper. –  Allen Knutson Feb 1 '13 at 3:06
Perhaps it could use some well placed backticks around the math environments? –  Ricardo Andrade Feb 1 '13 at 3:50
Score! (But why suggest this, rather than trying it? And why upvote it, rather than trying it?) –  Allen Knutson Feb 1 '13 at 16:18
He might not have enough "reputation" to edit someone else's question. –  Peter Samuelson Feb 1 '13 at 21:19


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# Q.4 Find the value of: (i) -4 ÷ $$\frac{2}{3}$$ (ii) $$\frac{-3}{5}$$ ÷ 2 (iii) $$\frac{-4}{5}$$ ÷ (-3) (iv) $$\frac{-1}{8}$$ ÷ $$\frac{3}{4}$$ (v) $$\frac{-2}{13}$$ ÷ $$\frac{1}{7}$$ (vi) $$\frac{-7}{12}$$ ÷ $$(\frac{-2}{13})$$ (vii) $$\frac{3}{13}$$ ÷ $$(\frac{-4}{65})$$

(i) -4 ÷ $$\frac{2}{3}$$= -4 × $$\frac{3}{2}$$ = $$\frac{-12}{2}$$= -6
(ii) $$\frac{-3}{5}$$ ÷ 2 = $$\frac{-3}{5}$$ ×$$\frac{1}{2}$$= $$\frac{-3}{10}$$
(iii) $$\frac{-4}{5}$$ ÷ (-3) = $$\frac{-4}{5}$$× $$\frac{-1}{3}$$ =$$\frac{4}{15}$$
(iv) $$\frac{-1}{8}$$ ÷ $$\frac{3}{4}$$ = $$\frac{-1}{8}$$ × $$\frac{4}{3}$$ = $$\frac{-4}{24}$$ = $$\frac{-1}{6}$$
(v) $$\frac{-2}{13}$$ ÷ $$\frac{1}{7}$$ = $$\frac{-2}{13}$$×7 = $$\frac{-14}{13}$$=$$-1\frac{1}{13}$$
(vi) $$\frac{-7}{12}$$ ÷ $$(\frac{-2}{13})$$= $$\frac{-7}{12}$$ × $$(\frac{13}{-2})$$=$$\frac{-91}{-24}$$ = $$3\frac{19}{24}$$
(vii) $$\frac{3}{13}$$ ÷ $$(\frac{-4}{65})$$ = $$\frac{3}{13}$$ × $$(\frac{65}{-4})$$ = $$\frac{195}{-52}$$
= $$\frac{195÷13}{-52÷13}$$=$$\frac{15}{-4}$$=$$-3\frac{3}{4}$$