text:
Performant subgraph isomorphism

# Grand Isomorphisms

Subgraph isomorphism is a resource-heavy (but branch-parallelizable) algorithm that is hugely impactful for large graph analysis. SotA algorithms for this (Ullmann, VF2, BB-Graph) are heavily RAM-bound, but this is due to a large number of small processes each of which hold a small portion of a traversal tree in memory.

Grand-Iso is a subgraph isomorphism algorithm that exchanges this resource-limitation for a parallelizable partial-match queue structure.

It performs favorably compared to other pure-python (and even some non-pure-python!) implementations:

See the wiki for more documentation.

## Example Usage

from grandiso import find_motifs
import networkx as nx

host = nx.fast_gnp_random_graph(10, 0.5)

motif = nx.Graph()

len(find_motifs(motif, host))

Directed graph support:

from grandiso import find_motifs
import networkx as nx

host = nx.fast_gnp_random_graph(10, 0.5, directed=True)

motif = nx.DiGraph()

len(find_motifs(motif, host))

## Counts-only

For very large graphs, you may use a good chunk of RAM not only on the queue of hypotheses, but also on the list of results. If all you care about is the NUMBER of results, you should pass count_only=True to the find_motifs function. This will dramatically reduce your RAM overhead on higher-count queries.

There are many other arguments that you can pass to the motif search algorithm. For a full list, see here.

## Hacking on this repo

### Running Tests

coverage run --source=grandiso -m pytest

## Citing

# https://doi.org/10.1038/s41598-021-91025-5
@article{Matelsky_Motifs_2021,
title={{DotMotif: an open-source tool for connectome subgraph isomorphism search and graph queries}},
volume={11},
ISSN={2045-2322},
url={http://dx.doi.org/10.1038/s41598-021-91025-5},
DOI={10.1038/s41598-021-91025-5},
number={1},
journal={Scientific Reports},
publisher={Springer Science and Business Media LLC},
author={Matelsky, Jordan K. and Reilly, Elizabeth P. and Johnson, Erik C. and Stiso, Jennifer and Bassett, Danielle S. and Wester, Brock A. and Gray-Roncal, William},
year={2021},
month={Jun}
}

## Project details

Uploaded source


text:
# nLab Stefan Schwede

## Selected writings

### Articles

• Stefan Schwede, Spectra in model categories and applications to the algebraic cotangent complex, Journal of Pure and Applied Algebra 120 (1997) 77-104 (pdf)

On orbispaces:

category: people

format: plain


text:
# Sensitivity Analysis of SVM

This notebook illustrates sensitivity analysis of data points in a Support Vector Machine (inspired from @matbesancon's SimpleSVMs.)

For reference, Section 10.1 of https://online.stat.psu.edu/stat508/book/export/html/792 gives an intuitive explanation of what it means to have a sensitive hyperplane or data point. The general form of the SVM training problem is given below (without regularization):

$$$\begin{split} \begin{array} {ll} \mbox{minimize} & \sum_{i=1}^{N} \xi_{i} \\ \mbox{s.t.} & \xi_{i} \ge 0 \quad i=1..N \\ & y_{i} (w^T X_{i} + b) \ge 1 - \xi[i]\\ \end{array} \end{split}$$$

where

• X, y are the N data points
• ξ is the soft-margin loss.

## Define and solve the SVM

Import the libraries.

using SCS, DiffOpt, LinearAlgebra, JuMP
import Random, Plots

Construct separable, non-trivial data points.

N = 100
D = 2
Random.seed!(62)
X = vcat(randn(N ÷ 2, D), randn(N ÷ 2, D) .+ [4.0, 1.5]')
y = append!(ones(N ÷ 2), -ones(N ÷ 2));

Let's initialize a special model that can understand sensitivities

model = Model(() -> diff_optimizer(SCS.Optimizer))
A JuMP Model
Feasibility problem with:
Variables: 0
Model mode: AUTOMATIC
CachingOptimizer state: EMPTY_OPTIMIZER
Solver name: SCS

@variable(model, l[1:N])
@variable(model, w[1:D])
@variable(model, b);

@constraint(
model,
1.0 * l ∈ MOI.Nonnegatives(N),
)
@constraint(
model,
cons,
y .* (X * w .+ b) + l .- 1 ∈ MOI.Nonnegatives(N),
);

Define the objective and solve

@objective(
model,
Min,
sum(l),
)

optimize!(model) # solve

We can visualize the separating hyperplane.

loss = objective_value(model)
wv = value.(w)
bv = value(b)

svm_x = [0.0, 3.0]
svm_y = (-bv .- wv[1] * svm_x )/wv[2]

p = Plots.scatter(X[:,1], X[:,2], color = [yi > 0 ? :red : :blue for yi in y], label = "")
Plots.yaxis!(p, (-2, 4.5))

## Experiment 2: Gradient of hyperplane wrt the data point coordinates

Similar to previous example, construct perturbations in data points coordinates X.

∇ = Float64[]
dX = zeros(N, D);

begin differentiating

for Xi in 1:N
dX[Xi, :] = ones(D)  # set

for i in 1:D
MOI.set(
model,
DiffOpt.ForwardInConstraint(),
cons,
MOI.Utilities.vectorize(dX[:,i] .* MOI.SingleVariable(w[i])),
)
end

DiffOpt.forward(model)

dw = MOI.get.(
model,
DiffOpt.ForwardOutVariablePrimal(),
w
)
db = MOI.get(
model,
DiffOpt.ForwardOutVariablePrimal(),
b
)
push!(∇, norm(dw) + norm(db))

dX[Xi, :] = zeros(D)  # reset the change made ago
end
normalize!(∇);

We can visualize point sensitivity with respect to the separating hyperplane. Note that the gradients are normalized.

p3 = Plots.scatter(
X[:,1], X[:,2],
color = [yi > 0 ? :red : :blue for yi in y], label = "",
markersize = ∇ * 20,
)
Plots.yaxis!(p3, (-2, 4.5))
Plots.plot!(p3, svm_x, svm_y, label = "loss = \$(round(loss, digits=2))", width=3)


text:
[–] 8 points9 points ago

sorry, this has been archived and can no longer be voted on

There is a few approaches you can do:

1. Put everything in one repository (repo) and create branches. Then after a while you can merge them completely or cherry pick some parts.

2. Have one big master source and use latex's \if to conditionally compile them.

Even though 1. is more pure and elegant I'd personally chooose 2. since it's just a lot easier. I do use git every day but having to switch branches a lot, worrying about where you are and then merging (a possibly messy) history can be a pain.

Check out bitbucket for some free hosting. I'd also recommend going for mercurial instead of git since you said you have no prior experience with version controlling.

[–] 5 points6 points ago

sorry, this has been archived and can no longer be voted on

I use git for all of my LaTeX papers, and I'm very happy with it. The "git rebase" command sounds like it's exactly what you're looking for. It allows you to make changes to the base version of your document, and then propagate these changes to all the forks. In other words, it moves the base of the branches -- hence "rebase".

[–] 5 points6 points ago

sorry, this has been archived and can no longer be voted on

An alternative is to break the content into separate files, and use \include and \input to build separate documents from appropriately-selected parts. The content files could have shared material, or general/specific presentations of some topic. Then any shared content would be present in only one file, and changes to this content would immediately be in each version.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

I have my resume/cv typeset exactly this way. Each element (education, papers published, work history, etc) live in their own file.

[–] 7 points8 points ago

sorry, this has been archived and can no longer be voted on

The sociologist Kieran Healy uses version control extensively with his Latex docs and offers suggestions about this in his Workflow document. I'm linking to his overall resources page because they're all good.

Here

[–][S] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

Thanks for the responses so far. Here's a silly question: are there any version control schemes that don't require me signing up for a website and storing everything online? It seems that's the case with Github, and I'm not thrilled about that...

[–] 2 points3 points ago

sorry, this has been archived and can no longer be voted on

If you have a .edu email you'll get 5 private repost on github. Bitbucket will offer you unlimited. Their security should be pretty good so I wouldn't worry about that. Of course, NEVER store any keys/passwords on git.
Git and Mercurial don't need a server. You just run a git init and have it locally. No need for a remote. Though, you will have to read into the whole DVCS thingy. No need for me here to repeat the hundreds of tutorials available on the web. :)
Again: Start with Mercurial. It's more intuitive and since you're free to choose (you don't have a team that dictates your workflow) there is no reason to go for git.

[–] 4 points5 points ago

sorry, this has been archived and can no longer be voted on

you can use git locally too, though i'm not sure why you wouldn't want to have it backed up on a server somewhere...

one of the beautiful things about git is that it's distributed, so the version control is done locally, and then you can push and pull changes to and from a server or other users, but it's not really necessary. I have used git locally for projects i'm not sharing with anyone (e.g. homework projects)

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

(this is a big benefit, even if you do use a server, because it means you can work and track changes and do version control even while away from the network for long periods of time)

[–] 8 points9 points ago

sorry, this has been archived and can no longer be voted on

Git is just a program, I don't know who told you that you need to sign up on a website to use it. Also check out mercurial its easy to learn and much like git.

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

Both git and svn can be used locally. People just store things online so that other people may access their files (or so they may be accessed from different computers). Github is just one option. Your company or university might allow you to set up a free public svn or git repository, which can be very convenient.

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

Any of the 'distributed version control systems' easily support local only repositories. Bazaar is probably the easiest to learn (but also has a less powerful branching model). Git has everything, including the kitchen sink, but has a nearly vertical learning curve. Mercurial is more like git than bazaar, but is a little easier to get started with. There's also darcs, but I do not know much about it.

The only reason to bother with an account somewhere for storing a copy of the repository is for backup purposes.

[–] 0 points1 point ago

sorry, this has been archived and can no longer be voted on

I'm not sure what is causing all of the git hate in this thread. If you can handle LaTeX, you can handle git just fine. You'll also probably be able to blow through a good starter book in an afternoon.

My only caution is that most version control systems diff and merge by line (ahem, mercurial here too), and LaTeX isn't always fun to merge since LaTeX code is prone to very long lines. Good luck, with whatever you choose!

[–] 1 point2 points ago

sorry, this has been archived and can no longer be voted on

TeX isn't always fun to merge since LaTeX code is prone to very long lines

Some people mitigate this by putting each sentence (or blocks of sentences) on a separate line.


text:
# Thread: Simple integration

1. ## Simple integration

\int_{0}^{1}\int_{0}^{1} min(x,y) dxdy
= \int_{0}^{1}\int_{y}^{1} y dxdy
= \int_{0}^{1} y(1-y)dy
= 1/6
What's wrong with it?
Thank you.

2. ## Re: Simple integration

Can you write this in LaTex? I can't seem to follow what you have there.

3. ## Re: Simple integration

Originally Posted by xiongs8421
\int_{0}^{1}\int_{0}^{1} min(x,y) dxdy
= \int_{0}^{1}\int_{y}^{1} y dxdy
= \int_{0}^{1} y(1-y)dy
= 1/6
What's wrong with it?
Thank you.
On the square you can write the function

$\text{min}(x,y)=\begin{cases} x, \qaud \text{ if } y \ge x \\ y, \qaud \text{ if } y \le x \end{cases}$

So you can break this into two different integrals over each triangle. Or you can do one of the integrals and multiply it by 2. Why does this work?


text:
Mathematics

# Classify the following angle as right, straight, acute, obtuse or reflex.

##### SOLUTION
If the angle is less than $90^o$ than it is a acute angle and as here we can see that the angles measure are less than $90^o$ so both are acute angle.

You're just one step away

Subjective Medium Published on 09th 09, 2020
Questions 120431
Subjects 10
Chapters 88
Enrolled Students 151

#### Realted Questions

Q1 Single Correct Medium
A closed figure having three sides is
• A. not a polygon
• B. a triangle
• D. none of these

Asked in: Mathematics - The Triangle and Its Properties

1 Verified Answer | Published on 23rd 09, 2020

Q2 Subjective Medium
Fill in the blanks with the correct word/symbol to make it a true statement:
A triangle whose all the angle are of measure less than $90^{o}$ is known as .........

Asked in: Mathematics - The Triangle and Its Properties

1 Verified Answer | Published on 23rd 09, 2020

Q3 Subjective Medium
$FACE$ is a parallelogram. $L$ and $R$ are mid-points of $EF$ and $AC$ respectively. Prove that the line segment $ER,\ AL$ triset the diagonal $FC$.

1 Verified Answer | Published on 09th 09, 2020

Q4 Single Correct Medium
Consider the following statements : When two straight lines intersect,
(i) adjacent angles are complementary      (ii) adjacent angles are supplementary       (iii) opposite angles are equal       (iv) opposite angles are supplementary.
• A. (i) and (iii)
• B. (ii) and (iii)
• C. (i) and (iv)
• D. (ii) and (iv)

Asked in: Mathematics - Lines and Angles

1 Verified Answer | Published on 09th 09, 2020

Q5 Subjective Medium
If two angles of a triangle are completely then what type of triangle will be formed.

Asked in: Mathematics - The Triangle and Its Properties

1 Verified Answer | Published on 23rd 09, 2020


text:
• # question_answer If you do not put an equal to (=) before a formula, _______. A) ###### is displayed in the cellB) #REF! is displayed in the cellC) The formula is displayed as it is in the cellD) The formula is still executed and its results are displayed in the cell

You need to login to perform this action.
You will be redirected in 3 sec


text:
Term for a momentary geometric pattern formed by astronomical objects

Few days back, the Pleiades, the Sun and the Moon were forming an almost perfect equilateral triangle.

Is there a term that describes such momentary geometrical patterns? Asterism and conjunction are close, but not exactly applicable here.

-

1 Answer

Momentary events in general (comets, supernovae, solar flares, etc.) can be lumped under the general heading of "transient phenomena". 'Transient asterisms' may be a candidate term for the concept you are describing.

Then again, we now know that on a large enough time scale all asterisms are eventually transient in nature. So one might justifiably take the point of view that the adjective "transient" is redundant. So perhaps calling the patterns you describe simply as "asterisms" is not as bad of a candidate term as you might suspect.

-
There's transient and then there's transient. Some of stars we see may be moving 100s of km/s with regard to our solar system. But they are so distant the angular velocity is small. In contrast the sun moves about a degree per day against the celestial sphere and the moon about 12 degrees. –  HopDavid Jun 1 at 15:49


text:
Bright - Maple Help

Bright Color Palette

Description

 • The Bright plot palette is a collection of bright variants of primary and secondary colors. It is designed for coloring collections of filled areas. There are six main colors in the palette as well as ten more named colors.
 > ColorTools:-DisplayPalette("Bright",noextend);

 1. Blue 2. Red 3. Green 4. Purple 5. Cyan 6. Orange 7. DarkBlue 8. BluishGreen 9. ReddishPurple 10. DarkRed 11. DarkGreenishBlue 12. GreenishYellow 13. DarkBlueGreen 14. DarkGreenishYellow 15. PurpleRed 16. VividGreen

Examples

You can reference colors in the palette by name or number in most commands that understand color names.

 > $\mathrm{plot}\left(\left[x,{x}^{3}\right],x=-1..1,\mathrm{color}=\left["Bright 2","Bright Blue"\right]\right)$
 > $C≔\mathrm{ColorTools}:-\mathrm{Color}\left("Bright 10"\right)$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Color}}{}\left({"RGB"}{,}\left[{0.60000000}{,}{0.039215686}{,}{0.}\right]\right)$ (1)

You can make Bright the default palette in plots using setcolors

 > $\mathrm{plots}:-\mathrm{setcolors}\left("Bright"\right):$

You can programatically access the colors using ColorTools commands

 > $P≔\mathrm{ColorTools}:-\mathrm{GetPalette}\left("Bright"\right)$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Palette}}{}\left(\left[{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}{,}{\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}\right]{,}\left[{"Blue"}{,}{"Red"}{,}{"Green"}{,}{"Purple"}{,}{"Cyan"}{,}{"Orange"}{,}{"DarkBlue"}{,}{"BluishGreen"}{,}{"ReddishPurple"}{,}{"DarkRed"}{,}{"DarkGreenishBlue"}{,}{"GreenishYellow"}{,}{"DarkBlueGreen"}{,}{"DarkGreenishYellow"}{,}{"PurpleRed"}{,}{"VividGreen"}\right]{,}\left[{}\right]{,}{\mathrm{palettename}}{=}{"Bright"}{,}{\mathrm{palettetype}}{=}{"fill"}\right)$ (2)
 > $P\left[1\right]$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Color}}{}\left({"RGB"}{,}\left[{0.20000000}{,}{0.40000000}{,}{0.80000000}\right]{,}{\mathrm{caption}}{=}{"Blue"}\right)$ (3)
 > $n≔\mathrm{numcolors}\left(P\right)$
 ${n}{≔}{16}$ (4)
 > $\mathrm{plot}\left(\left[\mathrm{seq}\left(\mathrm{sin}\left(x+\frac{i\mathrm{\pi }}{n}\right),i=1..n\right)\right],x=0..2\mathrm{\pi },\mathrm{axes}=\mathrm{boxed},\mathrm{color}=:-\mathrm{GetColors}\left(P\right)\left[1..n\right],\mathrm{legend}=:-\mathrm{GetColorNames}\left(P\right)\left[1..n\right],\mathrm{numpoints}=25,\mathrm{style}=\mathrm{point}\right)$

ColorTools algorithms will automatically generate new colors beyond the main 16 colors:

 > $P\left[20\right]$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Color}}{}\left({"RGB"}{,}\left[{0.1709078406}{,}{0.2427565597}{,}{0.2825290500}\right]\right)$ (5)

The Bright palette is designed for filled areas, but the Lookup command can provide slightly altered shades of its colors for use in coloring points or lines.

 > $\mathrm{Lookup}\left(P,1,\mathrm{shade}="line"\right)$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Color}}{}\left({"RGB"}{,}\left[{0.063687133}{,}{0.35704766}{,}{0.76481354}\right]\right)$ (6)
 > $\mathrm{Lookup}\left(P,1,\mathrm{shade}="point"\right)$
 ${\mathrm{ColorTools}}{:-}{\mathrm{Color}}{}\left({"RGB"}{,}\left[{0.13919861}{,}{0.35249922}{,}{0.72966107}\right]\right)$ (7)
 >


text:
Illuminate
Search…
⌃K

# YieldSpace AMM

### Secondary Markets

In order to provide easy on-chain access to what integrators can be confident is the best rate, an on-chain secondary market for iPTs is necessary.
In order to provide liquid secondary markets for Illuminate's iPTs, we've partnered with Yield Protocol to utilize their upcoming variant of the YieldSpace AMM design.

### Pricing Theta

The YieldSpace design is a variant of the traditional
$x*y=k$
AMM, with an additional a modifier applied in between trades that ensures the price of a PT increases as time passes.
This constantly increasing price ensures that even in the absence of trades, the PTs themselves can trade at a consistent rate. Though the price of the PT may change, the rate itself remains consistent.

## Choosing A YieldSpace Variant

After significant due diligence, we concluded that the Yield Protocol YieldSpace implementation was both the most gas efficient and simple integration offered on the market.
Element Protocol's YieldSpace implementation is then largely a direct implementation of YieldSpace, but in the Balancer ecosystem.
While Sense Protocol has innovated on YieldSpace with the SenseSpace AMM that accrues interest to the underlying asset while providing liquidity.
That said, YieldSpace is able to emulate the benefits of SenseSpace with minimal external or internal improvements that are likely to be implemented by Yield Protocol themselves.
With this context, the unique benefits of each are largely:
YieldSpace: Gas Efficient, Simple Integration
SenseSpace: Balancer Ecosystem Access
Element's YieldSpace: None? (Feel free to correct us!)
Last modified 3mo ago


text:
A Sudden Shake

Classical Mechanics Level 4

Estimate the recoil speed (in $$\frac{m}{s}$$) of the Earth if the entire population of China jumped simultaneously and as high as they could. Give your answer as an order of magnitude.

I created this problem for the 2013 UBC Physics Olympics (Problem 3)

×


text:
# Propagator.m

Calculates exponential propagators. Syntax:

           P=propagator(spin_system,L,timestep)

returns exp(-i*L*t). The following calculation methods are supported:

  'cpu'       - Taylor series with scaling and squaring
on CPU, spmd parallel if possible

  'gpu'       - Taylor series with scaling and squaring
on GPU

The propagator calculation method is chosen by setting the sys.enable parameter during the call to create.m function.

Notes: we did have Chebyshev and Newton series here at one point, as well as the Pade method. None of them has lived up to their marketing.


text:
lilypond-user
[Top][All Lists]

Re: Help with Tweaking of Volta Brackets - without repeat marks

 From: Phil Holmes Subject: Re: Help with Tweaking of Volta Brackets - without repeat marks Date: Fri, 14 May 2010 13:10:51 +0100

----- Original Message ----- From: "James Lowe" <address@hidden>
To: "Phil Holmes" <address@hidden>
Sent: Friday, May 14, 2010 12:39 PM
Subject: Re: Help with Tweaking of Volta Brackets - without repeat marks

Phil,

Phil Holmes wrote:

James,

I'm by no means an expert on LilyPond or music, but can offer the following
observations:

You can get rid of the error message by ending the first volta with a finish
marking.  If you replace your second \set Score command with:

\set Score.repeatCommands = #'((volta #f) (volta "Fine."))

It will compile cleanly.  However, you won't get the ending lines you're

looking for. I think that this is because you're not actually signalling a repeat from a musical perspective. My understanding from playing with this
in LilyPond is that the double bar is not a repeat mark, and therefore

LilyPond has not found anywhere to actually terminate that section of music. Replace your \bar "||" with \bar ":|" and you get the "terminated" bracket
you're looking for.  Ditto the second repeat isn't actually (in musical

terms) terminated - the music carries on past the 4th bar into the 5th. If
you want a "terminated" bracket on the second time, you need to align the

end of the bracket with where the music ends, and tell LilyPond explicitly
where this is.  I do this with an end bar.

My version below isn't what you want, because it uses repeat bars where you have double bars, but it is musically correct and ends the brackets as you
want.

\relative c' {
\time 4/4
d'4. c8 b a g f
\set Score.repeatCommands = #'((volta "1."))
e4 e' e r
\set Score.repeatCommands = #'((volta #f) (volta "Fine.") end-repeat)
e,4 e e r \bar "||"
c'4 c8 c c4 c8 c
\set Score.repeatCommands = #'((volta #f))
\bar "|."
}

--
Phil Holmes

Thanks for your time. That does help explains things; I don't really fully understand syntax of these 'tweaks', maybe there is an alternative to 'end-repeat' that just 'ends' the bracket regardless.

In this case the end bar is incorrect in this piece at the 'fine' which is why it has double bars, and as I said before the repeats are not relevant either.

Perhaps I could remove the repeat bar engraver or make it invisible so that it prints the voltas right but 'hides' the repeats?

Again that is something I need to sit and figure out as I rarely have to make such tweaks so have to re-learn it each time on the rare occasions I do it.

Thanks again.

James

As I said, James, I'm a bit dubious about the layout of the music. The voltas are alternative endings, and from my musical theory I would say don't indicate a repeat. The double dotted bar indicates a repeat, and so is required when a piece of music is repeated. That's what I would have thought, anyway.

--
Phil Holmes


text:
# Microbit Module¶

The microbit module gives you access to all the hardware that is built-in into your board.

## Functions¶

microbit.panic(n)

Enter a panic mode. Requires restart. Pass in an arbitrary integer <= 255 to indicate a status:

microbit.panic(404)

microbit.reset()

Restart the board.

microbit.sleep(n)

Wait for n milliseconds. One second is 1000 milliseconds, so:

microbit.sleep(1000)

will pause the execution for one second. n can be an integer or a floating point number.

microbit.running_time()

Return the number of milliseconds since the board was switched on or restarted.

microbit.temperature()

Return the temperature of the micro:bit in degrees Celcius.


text:
# Hello this is my health management class so i need 2 page paper for about health...

###### Question:

hello this is my health management class so i need 2 page paper for about health profession
thanks
HEALTH PROFESSIONAL INTERVIEW: Prepare a written 2-3 page summation of an interview with a current or retired health care provider or professional. Paper should describe the individual's role in health care, their thoughts on the industry as a profession, and their opinions about any future needs or trends occurring in the industry.

#### Similar Solved Questions

##### Verify the identity. (Simplify your answers completely.) 5 cot? t = 5 coS t(csc2 t 1) CSC tcot2 t5 cot3 t CSC tCSC t(csc2 t - 1)CSC t(csc2 t - 1)sin tsin t5 coS t(csc2 t - 1)
Verify the identity. (Simplify your answers completely.) 5 cot? t = 5 coS t(csc2 t 1) CSC t cot2 t 5 cot3 t CSC t CSC t (csc2 t - 1) CSC t (csc2 t - 1) sin t sin t 5 coS t(csc2 t - 1)...
##### 3.1.Q: How is Inditex financing the Business in 2017? Show the different ways of financing with...
3.1.Q: How is Inditex financing the Business in 2017? Show the different ways of financing with numbers 3.2.Q: Is Inditex having Profit/Loss, please explain how they operate? 3.3.Q: What are the most important costs for Inditex Accumulated YTD (Year to Date) 3.4.Q: Give me 2 examples of possible tan...
##### Large San Antonia school district ; thcre ur AP Statistics (cucher The tuble below provides some information about theae (eacher, Name Weckly Salun Amandh 740 Beth 6700 Jenmler S800 Sandra S7 i0 Tulip 5780List all I0 possihle SRSs of size culculate the meAn wcekly salury for each sumple Md display the sampling distribution olthe sumple mean on dotplot.
large San Antonia school district ; thcre ur AP Statistics (cucher The tuble below provides some information about theae (eacher, Name Weckly Salun Amandh 740 Beth 6700 Jenmler S800 Sandra S7 i0 Tulip 5780 List all I0 possihle SRSs of size culculate the meAn wcekly salury for each sumple Md display ...
##### Find the eigenvalues of the given matrix.102 36 297 105OA 3, -60 B. -3,6 Oc 3 0 D. 6
Find the eigenvalues of the given matrix. 102 36 297 105 OA 3, -6 0 B. -3,6 Oc 3 0 D. 6...
##### Part A If a scuba diver fills his lungs to full capacity of 5.3 L when...
Part A If a scuba diver fills his lungs to full capacity of 5.3 L when 9.0 m below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume he dives in the sea, thus the water is salt. Express your answer using two significant figures Submit Request Answer Part B ...
##### Estimate: GPA = Bo + BIGRE + ε, where GRE is a student's score on the...
Estimate: GPA = Bo + BIGRE + ε, where GRE is a student's score on the math portion of the Graduate Record Examination (GRE) score and GPA is the student's grade point average in graduate school. Click here for the Excel Data File Click here for the CSV Data File a. Use R's predic...
##### A 40.0 mL solution of 30.0 M of succinic acid (aq) is titrated with 0.300 M...
A 40.0 mL solution of 30.0 M of succinic acid (aq) is titrated with 0.300 M of KOH. Find the pH of the solution. (pKa=4.2). A) initially B) 1/2 way to the equivalence point C) at the equivalence point D) 1.00 mL past the equivalence post 2) sketch the titration curve that results from above titratio...
##### Tmihic 4For uosjad e observing another person in & car with decreasing moving velocity; the true statement from the following IS3lji 6 nl a The car 65" q The accelerating force acting on the car is equal to C. The zero car is a non-Newtonian frame of [ The car reference is a Newtonian auej} of reference
Tmihic 4 For uosjad e observing another person in & car with decreasing moving velocity; the true statement from the following IS 3lji 6 nl a The car 65" q The accelerating force acting on the car is equal to C. The zero car is a non-Newtonian frame of [ The car reference is a Newtonian aue...
##### How to calculate propagation uncertianty for Period T? My uncertianty for time measurment is +/- 0.2s...
How to calculate propagation uncertianty for Period T? My uncertianty for time measurment is +/- 0.2s and and +/- 0.005cm for length uncertianty. Here is some of the data: I counted the number of oscillations the pendulum made in 10s. Trial Length (cm) Time (s) Oscillations 1 1 5.00+ 0.05 10.2 0.2 ...
##### What is the limit as x approaches 0 from the right of 1/x?
What is the limit as x approaches 0 from the right of 1/x?...
##### McvlnzupdonLouccQuestion 30PpottDrtpuru-Eicda niya Ov Taenene0ro Ina & Mpfini ncilde C7ea Mieacio pln You can dernityely conciupeErak;07i709rdes1ew Mendel -nhenbnetths flontt Folor EatermicanDi inecmciamedeminanetIne Zene5) *OFiOt.er Cooe fs O0 Ta Cnrocoioctat{nttnjeOonnnT PureleMyoC7ot
Mcvlnz updon Loucc Question 30 Ppott Drt puru-Eicda niya Ov Taenene 0ro Ina & Mpfini ncilde C7ea Mieacio pln You can dernityely conciupe Erak;07i7 09rdes1 ew Mendel - nhenbnet ths flontt Folor EatermicanDi inecmciamedeminanet Ine Zene5) *OFiOt.er Cooe fs O0 Ta Cnrocoioctat {nttnje Oonnn T Purel...
##### Exercise 6 If p is a prime of the form 2k + 1 where k is a positive integer,; show that every quadratic nonresidue of p is a primitive root of p. Deduce the list of the primitive roots modulo 17 .
Exercise 6 If p is a prime of the form 2k + 1 where k is a positive integer,; show that every quadratic nonresidue of p is a primitive root of p. Deduce the list of the primitive roots modulo 17 ....
##### Question 23botuling company produces bottles that hold 10 Ckn 0f llquid Perodically: the company gets camplalnts that thoit bottles ara not tenouun Ilquid; To test Ho:H= 10 JGainst Hj p< 10 the comp ry randarnty sunpled 36 bottles and found thet Inc werago Arnount Moubole d eere8 bottles Ws Oulic a wth = sund-Ird dewationor0 T cunce whaecOticlu cn could Yol IuILE OuS velol eruficancor Do nat relect Ho Brtduse 257 Reject Ho becauSt-257 ~e Reject Ho because -2.,57 < -1.96 Do not reject Ho
Question 23 botuling company produces bottles that hold 10 Ckn 0f llquid Perodically: the company gets camplalnts that thoit bottles ara not tenouun Ilquid; To test Ho:H= 10 JGainst Hj p< 10 the comp ry randarnty sunpled 36 bottles and found thet Inc werago Arnount Moubole d eere8 bottles Ws Ouli...
##### Question 1 (1 point) ListenSavedProduction of a normal phenotype in the offspring of two recessive mutants due to completion of pathway in the offspring is called
Question 1 (1 point) Listen Saved Production of a normal phenotype in the offspring of two recessive mutants due to completion of pathway in the offspring is called...
##### List of Equations (Use only if the problem must be solved in terms of vectors) 3)...
List of Equations (Use only if the problem must be solved in terms of vectors) 3) d,-xf-Xi or 4) d=d1 + d2 + dy-yr-yi (Do not use if 2 applies) (Use only if the problem must be solved in terms of vectors) 7) uk dx/ t or vy dy/ t (Do not use if 6 applies) ,8) s=l/ht 9) [Cycle Length]/[Cycle Time] 1...
##### Part II: A 35-year-old woman is in the intensive care unit with sepsis following a cesarean...
Part II: A 35-year-old woman is in the intensive care unit with sepsis following a cesarean section.                                  ...
##### Solve each problem. A pulley rotates through $75^{\circ}$ in $1 \mathrm{~min}$. How many rotations does the pulley make in $1 \mathrm{hr} ?$
Solve each problem. A pulley rotates through $75^{\circ}$ in $1 \mathrm{~min}$. How many rotations does the pulley make in $1 \mathrm{hr} ?$...
##### What is self-serving bias, and how is it different from the actor-observer effect? What are the explanations given by researchers for this bias?
What is self-serving bias, and how is it different from the actor-observer effect? What are the explanations given by researchers for this bias?...
##### Find the area under the curve below from x = 1 to x = 4. Y = Zx + (2/x2)3. [~/1 Points]DETAILSFind the area between the two curves on [1,3]. e* (1/x)
Find the area under the curve below from x = 1 to x = 4. Y = Zx + (2/x2) 3. [~/1 Points] DETAILS Find the area between the two curves on [1,3]. e* (1/x)...
##### Find the radius of convergence R of the series_7(x + 2) VnFind the interva of convergence of the series (Enter your answer using interval notation:)15 13Need Help?Watch ItMastor ItTalk to Tutor
Find the radius of convergence R of the series_ 7(x + 2) Vn Find the interva of convergence of the series (Enter your answer using interval notation:) 15 13 Need Help? Watch It Mastor It Talk to Tutor...
##### 1Ihe Icngth ~orasumpIe PCIALU:L 11
1 Ihe Icngth ~orasumpIe PCIALU:L 1 1...
##### Easy problem, Timed assignment please answer within 30 mins Problem2 Simply F using Boolean algebra postulates...
Easy problem, Timed assignment please answer within 30 mins Problem2 Simply F using Boolean algebra postulates and theorems. Show all the details. (10 points)...
##### Circle A has a radius of 1  and a center of (1 ,2 ). Circle B has a radius of 2  and a center of (5 ,3 ). If circle B is translated by <-2 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Circle A has a radius of 1  and a center of (1 ,2 ). Circle B has a radius of 2  and a center of (5 ,3 ). If circle B is translated by <-2 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?...
##### 199.9 NH2 HNO3 H,SO Synthesize each of the following beginning with benzene. google good
199.9 NH2 HNO3 H,SO Synthesize each of the following beginning with benzene. google good...
##### Es lo 514. What is the news and Od? What type of disequilibrium situation exists, and...
es lo 514. What is the news and Od? What type of disequilibrium situation exists, and what is its maumitude Briefly describe what will happen to bring the market back to equilibrium. 2) Price Elasticity of Demand (5 points) When the price of Java Joe's lattes decreased from $5.00 to$4.50, the q...
##### Find the x, Y and componcnts of thc clectric ficld if the potential is givcn by the function V = (4 Volts/m-)x- (3 Volts m)y (IVolt/m)z
Find the x, Y and componcnts of thc clectric ficld if the potential is givcn by the function V = (4 Volts/m-)x- (3 Volts m)y (IVolt/m)z...
##### Second orderandlinear DilclentolequatonDroeucngninanr 0 Irrentujiequbdjn0 ninaenl ?
Second orderandlinear Dilclentolequaton Droeucngninanr 0 Irrentujiequbdjn 0 ninaenl ?...


text:
# Revolution of an area around y axis [closed]

I'm trying to plot an area that revolves around the y axis in 3D, but can't obtain what I need. I checked alot of related questions but can't really find a solution. I need to make revolve the area between:

y^2=x and 2y=x

So far , the closest I've got is :

RevolutionPlot3D[{y^2, 2 y}, {y, 0, 2}, PlotRange -> All,RevolutionAxis -> Z]

The form seems to be it, but it's not "filled", I have just the surface... So I'm looking to have the whole volume, Thanks in advance.

## closed as off-topic by Mr.Wizard♦Mar 9 '17 at 23:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Mr.Wizard
If this question can be reworded to fit the rules in the help center, please edit the question.

• @Kuba Yes, and with that I tried RegionPlot3D[ y^4 < x^2 + z^2 < (2 y)^2, {x, -2, 2}, {y, 0, 2}, {z, -2, 2}, PlotTheme -> "Classic", Mesh -> None, PlotStyle -> Opacity[0.5], BoxRatios -> Automatic, Boxed -> False, Axes -> False, Background -> GrayLevel[0]] But didn't work out as expected... – Sebastien Martinez Mar 9 '17 at 20:07
• Can't you just plot both surfaces? Show[RevolutionPlot3D[Sqrt[x], {x, 0, 4}], RevolutionPlot3D[x/2, {x, 0, 4}]] – Simon Woods Mar 9 '17 at 22:18
• @Simon Woods Oh yeah,didn't think at that, seems to be perfect, thanks alot ! – Sebastien Martinez Mar 9 '17 at 22:25

As suggested in the comments, I plotted both surfaces : Show[RevolutionPlot3D[Sqrt[x], {x, 0, 4}], RevolutionPlot3D[x/2, {x, 0, 4}]] Thanks for your help !


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

I found an R script in R-blogger that can be used to track PubMed trend. The script needs Perl package TGen-EUtils to perform query and it is not available now.

It’s not difficult to query Pubmed record in R. We can use RCurl package to fetch and use XML package to parse the downloaded record as shown in stackoverflow.

Before I write my own script, I found that there is a well written package, RISmed, that provides many functions to access the NCBI databases.

I write a wrapper function called getPubmedTrend, which import EUtilsSummary and QueryCount from RISmed, to track PubMed trend. Another function called plotPubmedTrend was also implemented for visualizing the trend. These two functions is available in my toy package, yplots.

I run these functions with two influenza strains and RSV (Respiratory Syncytial Virus) between 2001 and 2014.

?View Code RSPLUS
 1 2 3  term = c('"H7N9"', '"H5N1"', '"RSV"') pm=getPubmedTrend(term, year=2001:2014) plotPubmedTrend(pm)

And here is the plot:

Based on the document, the search terms should be double quoted.

As shown in the above figure, there is a sharp increase of publication of H5N1 in 2006 since the H5N1 outbreak in 2005. Nobody care about the H7N9 until it cause an outbreak in 2013. The increase of publication number of H7N9 in 2013 is dramatically.


text:
Free Version
Moderate

Empirical and Molecular

SATCHM-8TZ4YE

Which choice is BOTH an empirical formula and a molecular formula?

A

$H_2O$

B

$Fe_2O_3$

C

$2H_2O_2→ 2H_2O + O_2$

D

$Na_2Cl_2$

E

$N_2O_4$


text:
# Mathematical expression for SARIMA [duplicate]

• If you are having problems reproducing the forecast values from a software package, these posts may give you some helpful hints: here and here. Check the definition of the ARIMA model that is used by your sofware package, in particular check the signs of the AR and MA coefficients; check also whether a constat is included as an intercept, $y_t = \alpha + \phi y_{t-1} + \epsilon_t$, or as a mean $(y_t - \mu) = \alpha + \phi(y_{t-1} - \mu) + \epsilon_t$. – javlacalle Sep 22 '15 at 10:46


text:
134 views

### How to remedy the “freshman's dream”? [duplicate]

I am teaching a mid-level calculus course, and I see my students making the freshman's dream mistake of thinking that every function is a homomorphism. In particular, they think that exponents can be ...
2k views

### How to cure students from the idea that root and squaring are identity operators?

I tutor high school algebra and I’ve noticed that a lot of my students don’t seem to understand what they’re doing when they “convert” between different ways of writing numbers involving perfect ...
2k views

### How to Teach Adults Elementary Concepts

I've recently taken on the task of helping out in my school's Math Center. The courses I assist in range from Algebra to Calculus. While I'm younger (in my 20's), most of the students at the school ...
710 views

### What deficiencies are present in Precalculus curricula that causes so many students to fail Calculus I?

At our university we now require one semester of Pre-calculus instead of one semester of Algebra and one semester of Trigonometry before you take Calculus I (for those who do not test into Cal I). ...
624 views

### What are some common fallacies students make when they learn $X$ concept?

What are some common mistakes students often make, which may seem logical at first? I'm a student myself, but I'm curious of what some of the most frequent mistakes which happens. I'm thinking of ...
426 views

### A more rigorous approach to Precalculus

I am a pure mathematics PhD student and graduate teaching assistant at a major state university. During the summers here, teaching assistants are typically appointed to teach an entire course, rather ...
1k views

### Commonly taught method divides by zero

I just saw this question and the two answers. Now I am very sure that if many teachers see these answers they would accept them. Unfortunately they are deeply flawed. The second one is clearly invalid ...


text:
# Give an example about flatness.

Please give an example of a flat family {X_t} of closed subschemes of PP^n such that the family of projective cones of X_t is not a flat family in PP^{n+1}.

I still could not find...

-
Why do you believe there is such a thing? –  Steven Landsburg Aug 22 '11 at 14:34

As an example, let $X_t$ be three points in $\mathbb{P}^2$, at positions $(1:0:0)$, $(0:1:0)$ and $(1:1:t)$. This is certainly a flat family. Notice that the points are colinear when $t=0$ and not for $t \neq 0$.
For $t \neq 0$, the homogenous ideal of $X_t$ is $$\langle xz-txy, yz-txy, z^2-t^2 xy \rangle$$
Let $Y$ be the closure of $\langle xz-txy, yz-txy, z^2-t^2 xy \rangle$ in $\mathbb{A}^3 \times \mathbb{A}^1$. If I haven't made any errors, the ideal of $Y$ is $$\langle xz-txy, yz-txy, z^2-t^2 xy, xy(x-y) \rangle.$$ Since I defined $Y$ as the closure of a flat family over a smooth one dimensional base, it is flat. For $t \neq 0$, the fiber $Y_t$ is the cone on $X_t$.
However, at $t=0$, the fiber $Y_0$ is not the cone on $X_0$. They are the same away from the origin, but $Y_0$ has an extra nonreduced bit. Specifically, since $X_0$ lies on the line $z=0$, the cone on $X_0$ has $z$ in its defining ideal, while $Y_0$ has nothing in degree $1$. You can picture $Y_0$ as three coplanar lines, plus a nonreduced tangent vector pointing out the the $z=0$ plane; the cone on $X_0$ doesn't have that extra nonreduced piece.
What may be confusing you is that there isn't some non-flat family here for me to point to. The family $Y_t$, which exists, is flat, but its fibers are not the cones on the $X_t$. There is no closed subfamily of $\mathbb{A}^3 \times \mathbb{A}^1$ whose fibers are the cones on the $X_t$. This is the general picture whenever you have a flat family in $\mathbb{P}^N$ which does not lift to a flat family of projective cones.
I just noticed you asked for a failure in $\mathbb{P}^{n+1}$, not $\mathbb{A}^{n+1}$. You can take the projective closure of this example to get that; the interest is all at the origin. –  David Speyer Aug 22 '11 at 15:16


text:
# A candle weighs 12 oz. It burns for 5 hrs and 0.8 oz are....

#### mhb1959

##### New member
A friend and I are getting different answers for the same problem. I would like to know which method for solving is correct, and why the other one does not work.

The Problem:

A candle weighs 12 oz.
It burns for 5 hours and .8 oz are consumed.
How long will it take for the remainder of the candle to burn?

I set up my formula this way:

(12 / 12) - (.8 / 12) = (11.2 / 12), or .94, that's how much candle you still have left

therefore, .06 is the percentage which burned in 5 hours

so you set up a purportional equation

(5 hrs./ .06) = (x / .94), where x = how long it will take the remainder to burn, assuming the same rate

5 hrs. /.06 * .94 = x

78.333 hrs. = x

==================

My friend figures it this way:

Total estimated burn time = (initial mass)(duration of test burn)/(change in mass)

so,

(12 oz)(5 hours) = 60 divided by .8 = 75

[/u]

#### galactus

##### Super Moderator
Staff member
$$\displaystyle \L\\\frac{x+5}{12}=\frac{5}{0.8}$$

x=70 hrs for the rest of the candle to burn.

#### Subhotosh Khan

##### Super Moderator
Staff member
Both methods are correct - the problem is in execution.

You have a problem of rounding 0.8/12 = 0.0667

The remainder is then 0.9333

Now

x = 5 * 0.9333/0.0667 = 69.96

If you left everything in fraction - the answer would be a nice and even 70.

Your friend calculated the time to burn the whole candle (not remainder - as asked). He would get a 100% if he had subtracted 5 (the time to burn the initial part) from his answer.

#### TchrWill

##### Full Member
A candle weighs 12 oz.
It burns for 5 hours and .8 oz are consumed.
How long will it take for the remainder of the candle to burn?

If it tkes 5 hr to burn 8 oz., it will take x hr to burn 12 oz., or 5/8 = x/12 or x = 7.5 hr.

Therefore, it will take 7.5 - 5 = 2.5 hr to burn the remainder of the candle.

#### Subhotosh Khan

##### Super Moderator
Staff member
TchrWill said:
A candle weighs 12 oz.
It burns for 5 hours and .8 oz are consumed.

If it tkes 5 hr to burn 8 oz ., ......
TchrWill,

In 5 hours it burnt 0.8 oz - not 8 oz.

#### TchrWill

##### Full Member
My most humble apologies. This will teach me to wait for the effects of my eye injection to wear off before I tackle any problems. I completely missed the decimal point as in .8 oz. as opposed to 8 oz. which I assumed I was reading.

Clearly, the burn time remaining derives from

5/.8 = x/12 making x = 75 and the remaining burn time is 75 - 5 = 70 ours.

#### stapel

##### Super Moderator
Staff member
TchrWill said:
This will teach me to wait for the effects of my eye injection to wear off before I tackle any problems....
Ouch! You're getting stuff injected into your eye?!? :shock:

By the way, I nearly missed the decimal point, too. My only excuse, I supposed, is age and bifocals, but the confusion does point up the need for clarity: "0.8" is so much easier to "see" than is ".8"!

Eliz.

#### galactus

##### Super Moderator
Staff member
That's why it should be written as 0.8


text:
# Derivative of measure-valued function

I have a measure $\mu^x$ which is the law of a random variable and depends on $x$. The specific situation I am thinking of is $\mu^x$ is the law of $X_t$, the solution of an SDE with $X_0=x$. If I consider $x \mapsto \mu^x$ as a measure-valued function, is there a notion of differentiation (possibly in a Frechet sense) for such functions?

-
What do you want to do with such a derivative? –  Michael Greinecker May 24 '12 at 22:58
en.wikipedia.org/wiki/… –  Ricky Demer May 24 '12 at 23:05
@Michael: integrate, apparently? –  Ilya May 25 '12 at 8:49
To explain a little bit more: if we have a regular SDE $$X^x_t = x + \int_0^t b(X^x_s) ds + \int_0^t \sigma(X^x_s) dW_s$$ then under certain assumptions on the coefficients, we can prove that, for fixed $t$, $X^x_t$ is differentiable in $x$. We can actually still make sense of an SDE if the drift and diffusion parts are functions not only of $X_t$ but also $\mu_t:=$ the law of $X_t$ i.e. $$X^x_t = x + \int_0^t b(X^x_s, \mu^x_s) ds + \int_0^t \sigma(X^x_s, \mu^x_s) dW_s$$ I am now thinking about the differentiability of the solution to such an equation. –  mathman May 25 '12 at 9:07
Do you mean that $X^x_t(\omega)$ is differentiable in $x$ for fixed $t$ and $\omega$? Otherwise, what is the definition of the differentiability of a radnom variable $X^x_t(\cdot)$ w.r.t. parameter $x$? –  Ilya May 25 '12 at 10:06


text:
Introduction
The very basics
Infinite looping
Processing trees with recursive CTEs
Processing graphs with recursive ctes
Summary

## Instruction

Oops, our query failed. That was unexpected. What happened?

If you carefully read the error message, it will tell you that the types in the anchor and the recursive parts of our query do not match.

The data type for

SELECT
N''

is the data type for column X, i.e. Unicode String Literal.

The data type for

SELECT
X + W

is NVARCHAR of certain length. Unicode String Literal and NVARCHAR types do not match. The UNION ALL operator expects the types of corresponding columns to match; when they don't, the query fails.

What can we do about it? We have to explicitly cast the value N'' to match the type of the other expression, like this:

SELECT
CAST(N'' AS NVARCHAR(MAX))

The MAX argument makes sure that the type will be correct, because it forces the concatenation operator to return an NVARCHAR(MAX) type.

Unfortunately, there are no simple rules explaining what data type you must cast a column to get your query right.

## Exercise

Fix the query so that it runs correctly.

Cast the value N'' to type NVARCHAR(MAX).


text:
## Tohoku Mathematical Journal 2015 September SECOND SERIES VOL. 67, NO. 3

 Tohoku Math. J. 67 (2015), 465-479

Title BIHARMONIC HYPERSURFACES WITH THREE DISTINCT PRINCIPAL CURVATURES IN EUCLIDEAN SPACE

Author Yu Fu

(Received February 21, 2014, revised June 24, 2014)
Abstract. The well known Chen's conjecture on biharmonic submanifolds states that a biharmonic submanifold in a Euclidean space is a minimal one ([10--13, 16, 18--21, 8]). For the case of hypersurfaces, we know that Chen's conjecture is true for biharmonic surfaces in $\mathbb E^3$ ([10], [24]), biharmonic hypersurfaces in $\mathbb E^4$ ([23]), and biharmonic hypersurfaces in $\mathbb E^m$ with at most two distinct principal curvature ([21]). The most recent work of Chen-Munteanu [18] shows that Chen's conjecture is true for $\delta(2)$-ideal hypersurfaces in $\mathbb E^m$, where a $\delta(2)$-ideal hypersurface is a hypersurface whose principal curvatures take three special values: $\lambda_1, \lambda_2$ and $\lambda_1+\lambda_2$. In this paper, we prove that Chen's conjecture is true for hypersurfaces with three distinct principal curvatures in $\mathbb E^m$ with arbitrary dimension, thus, extend all the above-mentioned results. As an application we also show that Chen's conjecture is true for $O(p)\times O(q)$-invariant hypersurfaces in Euclidean space $\mathbb E^{p+q}$.

Mathematics Subject Classification. Primary 53D12; Secondary 53C40.

Key words and phrases. Chen's conjecture, biharmonic submanifolds.


text:
more useful options PerlMonks

### Re: Re: Re: Re: CGI Image on localhost

by eoin (Monk)
 on Dec 21, 2003 at 23:07 UTC ( #316258=note: print w/replies, xml ) Need Help??

in reply to Re: Re: Re: CGI Image on localhost
in thread CGI Image on localhost

NO I was not trying to get you to solve my problem.

I wrote the HTML myself and was only asking how do I link to it from a different folder

I.E. CGI SCRIPT "Apache2\cgi-bin\script.cgi"
HTML       "Apache2\data\home.html"
IMAGES     "Apache2\htdocs\images\image.jpg"
I wanted to know what should I put in the HTML to let me link to the IMAGES from the CGI SCRIPT. I knew that I could simply put in F:\foo\bar\Apache2\htdocs\images\image.jpg but I didn't WANT to do this as, on a remote server I wouldn't have access to the root directory e.g. F:\. I wanted to know how do you go link in the HTML down one directory and up two.

or

should I just generate the HTML using perl and CGI and maybe HTML::Template. or something like that.

I was just asking for help
Eoin.

Create A New User
Domain Nodelet?
Node Status?
node history
Node Type: note [id://316258]
help
Chatterbox?
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others exploiting the Monastery: (1)
As of 2021-09-25 14:37 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?

No recent polls found

Notices?


text:
# What is an antenna tuner? Why bother with resonant antennas in the first place?

There are a lot of antenna tuners on the market that have various capabilities in terms of how they can allow random radiators to work with our transceivers.

What exactly do they do? My understanding is that they can neutralize the imbalance between inductive and capacitive reactance so the transceiver can couple the most power out (or in) even though an antenna might not be resonant on a particular operating frequency, since overall impedance (Z) consists of three parts, two of which are reactance dependent on frequency: X(sub)L and X(sub)C, and R, the radiation resistance of the whole system. And the transmitters do not work as well into a non-resistive load.

With a resonant antenna these are in balance such that the characteristic radiation resistance of the antenna is presented to the transmitter. For a plain dipole that tends to be roughly 72 ohms. And there are many antenna analyzers out there that can assist in the physical pruning of antennas in order to reduce the SWR on the feedlines.

(I remember using light bulbs to tune up old tube-type transmitters because they were mostly resistive and non-reactive at HF)

300-ohm ladder line to a dipole requires a 4:1 balun to bring the impedance down to a range the transmitter can handle, which is typically 50-75 ohms, resistive, to keep the transmitters within their limits.

So what does an antenna tuner do? Does it reduce the SWR (loss) on the feedlines? Or does it simply ensure a match for the transmitter so it can operate at maximum efficiency regardless of whether a signal is coupled from the antenna to the outside world?

To further complicate the question, there is the matter of Q - the measure of the bandwidth for which the antenna will most efficiently accept transmitter power and radiate it. My understanding is that the best resonant antennas have a high Q which means a very narrow bandwidth for which they are most effective.

I've often used tuners when using 80-meter dipoles on the 75-meter phone section of the band, in order to reduce indicated SWR.

Similar to using 40-meter antennas on 15 meters, though that is more problematic since the harmonics don't line up as closely and it takes a better tuner than my current travel model.

So the question is in two parts:

What is an antenna tuner? If it is mounted at the transmitter is it actually tuning the antenna itself? Does it reduce the SWR in the feedline to a non-resonant antenna, for instance? Should the tuner be placed at the antenna instead of at the transmitter in order to reduce SWR loss (heating) in the feedline?

And if it is possible to use a tuner to make a transmitter feed power into a coat hanger then why do we worry about making resonant antennas in the first place?

• Great answers. This whole topic makes me wonder about cellphones. How in the world can they make an antenna that works when held inside a closed hand and also sitting on a table. That must be a major design triumph. – SDsolar Aug 13 '17 at 2:24
• – SDsolar Aug 29 '17 at 15:19
• cell phone: low power, wide band fractal antenna or multiple antennas – user10489 Jun 16 '19 at 12:48

And if it is possible to use a tuner to make a transmitter feed power into a coat hanger then why do we worry about making resonant antennas in the first place?

In a word - efficiency.

Consider that a full size 40 meter, 1/2 wavelength dipole is approximately 65 feet (19.8 meters) long. A dipole that is only 2 feet (0.6 meters) long, would have a gain of only 0.5 dB less than the 65 foot dipole. This would hardly be a noticeable difference on the air. But the problem is we cannot make the antenna system efficient enough to realize that minor 0.5 dB loss. The system inefficiencies often result in substantial losses compared to a full size dipole.

But we also should not be obsessed with resonant antennas - it makes little difference with practical antennas if we properly deal with their non-resonant effects. A 10/8 wave dipole has nearly double the gain of a 1/2 wave dipole. It is not resonant but can easily be matched to a transmission line with a simple matching network. A Yagi antenna is not resonant without the addition of a matching network. Also consider that many resonant antennas are not resonant at the impedance of the feedline anyway - for example, an ideal 1/4 wave ground plane vertical is resonant at approximately 34 ohms.

Here are some typical sources of antenna system inefficiencies:

Radiation resistance is an often misunderstood and misapplied term. The radiation resistance of an antenna is caused by the radiation of electromagnetic waves. For a free space, 1/2 wave, center fed dipole of reasonable construction, the radiation resistance is ~73 ohms. Any resistive losses (power that is not radiated) in the antenna add to this radiation resistance to contribute to the feed point impedance. In the case of this dipole, there will be a very small amount of RF resistance due to the wires that make up the dipole. If we use 14 gauge (1.45 mm) wire to construct the antenna, the RF resistance will be ~2.7 ohms. The total feed point impedance would then be 75.7 ohms in this example.

The efficiency of an antenna is given by the formula:

$$Efficiency = \left( \frac{R_r}{R_r+R_l} \right)$$

where Rr is the radiation resistance and Rl is the resistive losses.

So if we apply this to our dipole example above, the efficiency would be 96.4%. By contrast, the two foot dipole will have an Rr of approximately 0.04 ohms and a comparative Rl of 0.08 ohms resulting in an efficiency of 33%.

To complete this part of the efficiency picture, consider that:

$$Gain = Directivity \times Efficiency$$

A full size dipole has a directivity of 1.65. Multiply this times the 96.4% efficiency of the above example and the gain becomes 1.59 (2.02 dBi). For the short dipole, the directivity is 1.5. Multiply this times the 33% efficiency of the above example and the gain becomes 0.5 (-3.01 dBi). So we already have a 5 dB difference between the two antennas and there still are other system losses that must be taken into account.

Matching Networks / Tuners

In the example of our 1/2 wave dipole, the feed point impedance of the antenna is ~ 75 ohms. If we are attempting to drive this with a 50 ohm source, we may wish to have a tuner or matching network that does the transformation between the two largely resistive impedances. In this case, a well built tuner or matching network will have an efficiency in the 80-95% range (less than 1 dB of loss).

In the case of our short dipole example, the situation is a bit more complex. We need to both cancel the capacitive reactance of the dipole and match the very low feed point impedance. Without going through laborious calculations of matching network efficiencies, it would not be unreasonable to expect less than 10% efficiency (>10 dB loss) from the matching network. This means that at least 90% of the applied power or signal will be lost to heat in the matching network alone. Our short dipole system now has a gain of -13 dBi which is more than 15 dB down from our full size dipole example. This is the equivalent to comparing the signal strength of a 100 watt transmitter to a 3 watt transmitter.

Transmission Line Loss

All real world transmission lines exhibit loss. This in itself is a reduction in efficiency. The specification for the transmission line lists the loss for the transmission line at a given frequency assuming that the transmission line is terminated in its characteristic impedance. When the transmission line is not terminated in its characteristic impedance, the loss of the transmission line increases (the efficiency further decreases).

By placing a tuner or matching network close to the antenna to provide a match to the characteristic impedance of the transmission line, the additional losses due to a transmission line mismatch can be avoided. This maximizes the efficiency of a given transmission line.

There is a secondary effect when the transmission line is not connected to a load that matches its characteristic impedance - the transmission line will no longer exhibit its characteristic impedance. Another way of saying this is it becomes an impedance transformer. For example, our 75 ohm dipole impedance in the early example when connected to 23 feet of RG213 will be transformed to 34 ohms at the transmitter end of the transmission line. The total losses in the transmission line will be 0.119 dB, of which only 0.009 dB is due to the mismatched load. A free program such as TLDetails makes quick work of these calculations:

In some cases, this transformation can be used to our advantage by transforming the antenna impedance to something that is more usable from a system efficiency perspective. In other cases, it can worsen the system efficiency. But do consider this transforming effect in light of transmitter output efficiency.

Transmitter Output Efficiency

The final efficiency factor to be considered is the output power of the transmitter. Most amateur transmitters are designed to deliver their rated output power when connected to a 50 ohm resistive load. Any deviation from this load will typically result in lower output power from the transmitter. This loss of transmitter output power effectively reduces the efficiency of the system.

For example, a 100 watt, 50 ohm source impedance transmitter that is connected to the 23 feet of RG213 that is terminated with our 75 ohm dipole, will output 96 watts (if no protection circuits kick in). This is an output efficiency of 96% (a 0.17 dB loss). Note that this is most likely more efficient, in this example, than using an antenna tuner at the transmitter to match the transmitter impedance to the feedline impedance (even though this is a 1.5:1 SWR).

SWR

SWR is based on the relationship of the transmission line characteristic impedance to the impedance of the load on the transmission line. In the case of a transmitting system, the load is typically the antenna. In the case of a receiving system, the load is typically the receiver input impedance.

Since we have already dealt with the relationship of the transmission line characteristic impedance to the load impedance in the earlier transmission line section, we need not consider SWR as another source of inefficiency.

My understanding is that they can neutralize the imbalance between inductive and capacitive reactance so the transceiver can couple the most power out (or in) even though an antenna might not be resonant on a particular operating frequency, since overall impedance (Z) consists of three parts, two of which are reactance dependent on frequency: X(sub)L and X(sub)C, and R, the radiation resistance of the whole system. And the transmitters do not work as well into a non-resistive load.

It some cases, it's not a question of coupling power into the load. For some impedances, a mismatched load may enable more power to be coupled into the load, however this would destroy the transmitter! Thus, the purpose of the tuner is also to present an impedance to the transmitter that allows it to work within the limits of its maximum current and voltage.

Radiation resistance also depends on frequency, and the tuner doesn't necessarily just cancel reactance. Consider an extreme example: a 1-wavelength long wire dipole has a radiation resistance around 4500 ohms and no reactance. That doesn't mean there's nothing for the tuner to do: the tuner needs to step down that impedance to 50 ohms.

300-ohm ladder line to a dipole requires a 4:1 balun to bring the impedance down to a range the transmitter can handle, which is typically 50-75 ohms, resistive, to keep the transmitters within their limits.

That's not the only way to use 300-ohm ladder line. If it's used in 1/2 wavelength multiples of length, then the ladder line will have no effect on the load impedance seen by the transmitter. The impedance step-down can also be accomplished with a network of inductors or capacitors, either fixed, or variable as in a tuner.

Or if the antenna wasn't matched to start, the right length of 300-ohm ladder line might result in a matched impedance at the transmitter. Using controlled lengths of feedline to match loads is called stub matching. Given the ability to pick the transmission line length and impedance, any load can be matched. If the transmission line impedance can't be picked, then any load can be matched with no more than two lengths of transmission line.

there is the matter of Q - the measure of the bandwidth for which the antenna will most efficiently accept transmitter power and radiate it. My understanding is that the best resonant antennas have a high Q which means a very narrow bandwidth for which they are most effective.

Antenna losses reduce Q, so for example a vertical with a poor ground plane will have a lower Q and wider bandwidth. A dummy load has a Q of zero and infinite bandwidth.

But losses are not the only factor. A vertical shortened by loading coil has a higher Q than one a full 1/4 wavelength long, yet the shortened vertical is generally a poorer antenna. A 1/2 wave dipole made of thicker wires, or made of a cage of wires, or thickened as in a folded dipole, has a lower Q than one made of thin wire, and yet is just as effective.

I've often used tuners when using 80-meter dipoles on the 75-meter phone section of the band, in order to reduce indicated SWR.

Tuners are often required on 80/75-meters due to two factors:

• electrically shortened antennas (loading coils, for example) are frequently used to make the otherwise very large antennas manageable, and
• the fractional bandwidth of that band is very large.

What is an antenna tuner?

It's any network of components designed to match one impedance to another, with variable components to accommodate use at a number of frequencies or with a variety of antennas. Frequently, it's a PI network:

simulate this circuit – Schematic created using CircuitLab

If it is mounted at the transmitter is it actually tuning the antenna itself?

No. Arguably, if it's mounted at the antenna it's not actually tuning the antenna either, although such an arrangement can avoid additional losses in the feedline due to increased SWR.

It just depends on how you define "the antenna". As far as the transmitter is concerned, all that matters is the impedance appearing at its terminals. Some antennas could be said to contain feedlines, such as those with stub matches, or phased arrays using transmission lines for phasing delay. Are those transmission lines part of "the antenna"? What if a transmission line is an integral part of the antenna, such as in a zepp, or a gamma match, or a folded dipole?

Does it reduce the SWR in the feedline to a non-resonant antenna, for instance?

Not if it's mounted at the transmitter.

Should the tuner be placed at the antenna instead of at the transmitter in order to reduce SWR loss (heating) in the feedline?

Should the tuner be placed in the shack so the knobs can be adjusted without going outside? The answer will depend on individual circumstances. Automatic tuners are significantly more expensive. While placing the tuner at the antenna does reduce mismatch losses in the feedline, those losses may not be high enough to justify the cost of a remote tuner. See What is the actual loss in a feed line with high SWR? Reducing the SWR on the feedline also reduces the peak voltage and current on the line and any components along it such as surge supressors, which may become a limiting factor when operating at high power.

And if it is possible to use a tuner to make a transmitter feed power into a coat hanger then why do we worry about making resonant antennas in the first place?

A coat hanger at HF would be a challenging antenna to match, since its extremely small length relative to wavelength would present such an extreme mismatch that most tuners would not be able to match it.

Resonant antennas are just one of many possible antenna designs. Not requiring a tuner reduces costs.

Simply put:

If it is mounted at the transmitter is it actually tuning the antenna itself?

No, it is not tuning the antenna. It just matches the resistance and reactance at the tuner end of the feedline to your transmitter.

Does it reduce the SWR in the feedline to a non-resonant antenna?

No, it does not change the VSWR on the feedline.

Should the tuner be placed at the antenna instead of at the transmitter in order to reduce SWR loss (heating) in the feedline?

Ideally, yes. However, that is usually not practical. It's often better to use a low-loss feedline.

300-ohm ladder line to a dipole requires a 4:1 balun to bring the impedance down

That sometimes works for one (maybe two) bands, but not if you want to cover 80-10.

My understanding is that the best resonant antennas have a high Q which means a very narrow bandwidth for which they are most effective.

Not necessarily so. A dipole with less Q and a wider bandwidth can be very effective indeed! One way of doing this is to replace a single wire with a "cage", greatly increasing the effective diameter of the element(s).

Although the difference is buried in the answers so far, it is worth pointing out that tuning and matching are two entirely different things.

To state the obvious, tuning is ensures that the antenna is resonant at the desired frequency, matching is to achieve the desired impedance which the transmitter and/or receiver sees.

REPEAT AFTER ME; THERES NO SUCH THING AS AN ANTENNA TUNER. THERES NO SUCH THING AS AN ANTENNA TUNER. LABEL BE DAMNED, ITS A LIE, A PLOT TO OIVERTHROW THE HAM COMMUNITY.

BUT WHAT ARE WE DOING??? LET US TUNE: AT 14.25MHZ, YOUR 1/2 WAVE DIPOLE IS "TUNED" WHEN YOU CUT IT TO THE LENGTH OF 32.842'

IF YOU LIKE TO WORK C.W. AT THE BOTTOM OF THE BAND, SAY 14.05MHZ, YOUR DIPOLE WOULD NOW NEED TO BE 33.309' ABOUT 5 1/2" LONGER. ALSO ON THE 75 & 80 MTR BANDS: 3.975MHZ NEEDS 117.735'. - AT 3.6MHZ - 130' . ABOUT 12 ' LONGER.

THOSE WOULD BE TUNED ANTENNAS. NO TUNER DOES THAT. A "BROOMSTICK" TYPE WITH THE VARIABLE LOADED WHIP COMES PRETTY CLOSE IN THAT YOU ARE CHANGING THE ANTENNA'S PHYSICAL CHARACTERISTICS.

WHAT IS HE SAYING? THIS GUY IS NUTS YOU SAY. WELL HERE IS THE DIFFERENCE YOU ARE NOT USING A TUNER, YOU ARE USING A "MATCHING DEVICE". (OFTEN CALLED A MATCHBOX, TRANSMATCH, COUPLER, AMU) YOU ARE NOT DOING SQUAT TO THE ANTENNA, YOU ARE MATCHING THE ANTENNA - AND THIS IS KEY: SYSTEM - TO THE RIG. ALL YOU'RE DOING IS CHANGING THE DIFFERENT VALUE OF THAT SYSTEM (FEEDLINE, TRAPS, BALUNS, CHOKES ETC.) BETWEEN THE TX AND THE RADIATING ELEMENT. REMEMBER: THERES NO SUCH THING AS AN ANTENNA TUNER.

SO LET US MATCH.

HOPEFULLY I HAVEN'T BORED U W/ DETAILS, BUT AFTER 58 YRS AS A HAM, I HAD TO LEARN MANY THINGS THE HARD WAY. MOST OF THIS WAS FROM W1ICP LEW MCCOY. '73 RICH KF9F

• What's with the yelling? – mike65535 Apr 9 '19 at 23:22
• Seems to me, the ultimate purpose of an antenna tuner is to come close to maximizing the current through the radiation resistance of the antenna (at the antenna). – Cecil - W5DXP Apr 10 '19 at 20:11


text:
# zbMATH — the first resource for mathematics

## Found 2 Documents (Results 1–2)

London Mathematical Society Lecture Note Series. 228. New York: Cambridge University Press. viii, 484 p. (1996).


text:
Linear Algebra Proof

1. May 14, 2008

cscott

1. The problem statement, all variables and given/known data

A = A(t), B = constant

$$\frac{d}{dt} \left[A \cdot (\dot{A} \times B) \right] = \frac{d}{dt} \left[A \cdot (\ddot{A} \times B) \right]$$

3. The attempt at a solution

In Einstein notation I get

$$LHS = \frac{d}{dt} \left [ A_i (\dot{A} \times B)_i \right] = \frac{d}{dt} \left [ \epsilon_{ilm} A_i \dot{A}_l B_m \right] = \epsilon_{ilm} B_m \frac{d}{dt} \left [A_i \dot{A}_l \right]$$

Is this right so far? Product rule on A and A-dot from here?

Last edited: May 14, 2008
2. May 14, 2008

Dick

I think you have an extra d/dt on the right side of what you are trying to prove. But yes, now product rule. Then what?

3. May 14, 2008

cscott

I think I got it:

After using product rule I get

$$A \cdot ( \ddot{A} \times B) + \dot{A} \cdot ( \dot{A} \times B)$$

Where the second term would be 0

So the second term must expand as (A . A) x ( A . B) ?

4. May 14, 2008

cscott

Yeah the RHS time derivative shouldn't be there.

5. May 14, 2008

Dick

Yes, but you don't expand it like that A.A and A.B are scalars. How can you cross them? In terms of vectors axb is perpendicular to a and b. In terms of your tensor expansion the product of the two A dots is symmetric, the corresponding indices of the epsilon tensor are antisymmetric. So?

6. May 14, 2008

cscott

Ah. This way just seemed quicker but I see where it makes no sense.


text:
# What is the domain and range of y=cos^-1 x?

domain is $\left[- 1 , 1\right]$ and the
range is $\left[0 , \pi\right]$.


text:
# Two AC Sources with One Transformer

I want to program a little power grid game and have the problem of connecting things via transformers. Imagine the following setup:

You can see the AC voltage sources as generators (or together with the resistor as a "consumer", if the amplitude is 0).

What I need to figure out are the Voltages across the inductors L1 and L2 and the currents which flow through the resistors R1 and R2. So the only things given are U1(t), R1, L1, U2(t), R2 and L2. What I already know is that the voltage across resistor R1 is:
$$U_{R1} = U_{L1} - U_1$$
the same thing goes for the voltage across R2:
$$U_{R2}=U_{L2}-U_2$$

Through some research I also found that the voltage across an inductor is given by:
$$U_{L1}=L_1 \cdot \frac{dI_1}{dt}+ M \cdot \frac{dI_2}{dt}$$

From that we can conclude that:
$$U_{R1}=L_1 \cdot \frac{dI_1}{dt} + M \cdot \frac{dI_2}{dt} - U_1$$
which is equavilent to:
$$I_1 = \frac{L_1}{R_1} \cdot \frac{dI_1}{dt} + \frac{M}{R_1} \cdot \frac{dI_2}{dt} - \frac{U_1}{R_1}$$
and for the other side:
$$I_2 = \frac{L_2}{R_2} \cdot \frac{dI_2}{dt} + \frac{M}{R_2} \cdot \frac{dI_1}{dt} - \frac{U_2}{R_2}$$

Where $$M = \sqrt{L_1 \cdot L_2}$$ is the mutual inductance.

The problem is, that this differntial equation seems to be highly unstable in a simulation (using numerical differentiation). Is there any solution to that? Just having one side without the other, so setting $$M=0$$ works just fine, but with both sides everything just "explodes".

Hopefully someone can help me :D

#### Attachments

• 4.9 KB Views: 581
Last edited by a moderator:

Related Electrical Engineering News on Phys.org
anorlunda
Staff Emeritus
You can find what you need here https://en.wikipedia.org/wiki/Transformer#Ideal_transformer
The trick is this

the apparent load impedance referred to the primary circuit is derived in eq. (7) to be equal to the turns ratio squared times the secondary circuit load impedance.
Then you put the referred impecance in series with R1, completely removing the transformer from the circuit.

Be sure that you can control the phase angles of the two AC sources. Also do all your calculations in complex arithmetic, because you need to calculate real and imaginary voltages, currents and power (P=watts Q=VARS)

Good luck, what you're doing is good self-education.

anorlunda
Staff Emeritus
Whoops. I forgot to mention that you should not think of that transformer as two inductors. Again, that Wikipedia article explains more.

scottdave
Hmm, I'll try this and see if it works, but I have a concern: if U1 is larger than U2 and both are in phase, then wouldn't power from the left side be transferred to the right? If R1 = R2? This can't be modeled with additional impedance, can it? I thought of replacing the inductor with impedance and an additional AC source, but I can't work that out either.

I guess I have to first look into the complex arithmetic stuff. I'm familar with complex numbers, but not too much with electronics (yet).


text:
hw-balancedparens-0.4.1.1: Balanced parentheses

Synopsis

# Documentation

Find the position of the first unmatch parenthesis.

The digits 1 and 0 are treated as an open parenthesis and closing parenthesis respectively.

All positions are indexed from zero. If the search runs out of bits, then continue as if there remain an infinite string of zeros.

>>> import HaskellWorks.Data.Bits.BitRead
>>> import Data.Maybe

The following scans for the first unmatched closing parenthesis from the beginning of the bit string:

>>> findUnmatchedCloseFar 0 0 $fromJust$ bitRead "00000000"
0

The following scans for the first unmatched closing parenthesis after skipping one bit from the beginning of the bit string:

>>> findUnmatchedCloseFar 0 1 $fromJust$ bitRead "00000000"
1

The following scans for the first unmatched closing parenthesis from the beginning of the bit string. To find unmatched parenthesis, the scan passes over the first parent of matching parentheses:

>>> findUnmatchedCloseFar 0 0 $fromJust$ bitRead "10000000"
2

The following scans for the first unmatched closing parenthesis from the beginning of the bit string, but runs out of bits. The scan continues as if there an inifinite string of zero bits follows, the first of which is at position 64, which also happens to be the position of the unmatched parenthesis.

>>> findUnmatchedCloseFar 0 0 $fromJust$ bitRead "11110000"
8

The following scans for the first unmatched closing parenthesis from the beginning of the bit string, but runs out of bits. The scan continues as if there an inifinite string of zero bits follows and we don't get to the unmatched parenthesis until position 128.

>>> findUnmatchedCloseFar 0 0 $fromJust$ bitRead "11111111"
16

Following are some more examples:

>>> findUnmatchedCloseFar 0 0 $fromJust$ bitRead "11110000"
8
>>> findUnmatchedCloseFar 0 1 $fromJust$ bitRead "11110000"
7
>>> findUnmatchedCloseFar 0 2 $fromJust$ bitRead "11110000"
6
>>> findUnmatchedCloseFar 0 3 $fromJust$ bitRead "11110000"
5
>>> findUnmatchedCloseFar 0 4 $fromJust$ bitRead "11110000"
4
>>> findUnmatchedCloseFar 0 5 $fromJust$ bitRead "11110000"
5
>>> findUnmatchedCloseFar 0 6 $fromJust$ bitRead "11110000"
6
>>> findUnmatchedCloseFar 0 7 $fromJust$ bitRead "11110000"
7
>>> findUnmatchedCloseFar 0 8 $fromJust$ bitRead "11110000"
8


text:
## STNB2017(31è any)

### On Maeda's Conjecture

#### Ponents

Paloma Bengoechea

#### Resum

Let T_{n,k}(X) be the characteristic polynomial of the n-th Hecke operator acting on the space of cusp forms of weight k. A popular conjecture, called Maeda's Conjecture, asserts that, for every n, T_{n,k}(X) is irreducible in Q[X] and has full Galois group over Q. We will discuss some recent progress on the conjecture and will show the following: Let k be any positive integer congruent to 0 modulo 4. If one T_{n,k}(X) is irreducible and has full Galois group, then the same is true of T_{p,k}(X) for all primes p.

#### Fitxers

No hi ha fitxers per descarregar

Compte
Llengües:


text:
# Features#

## Annotate records#

The Argilla UI has a dedicated mode to quickly label your data in a very intuitive way, or revise previous gold labels and correct them. Argilla’s powerful search and filter functionalities, together with potential model predictions, can guide the annotation process and support the annotator.

You can access the Annotate mode via the sidebar of the Dataset page.

### Create labels#

For the text and token classification tasks, you can create new labels within the Annotate mode. On the right side of the bulk validation bar, you will find a “+ Create new label” button that lets you add new labels to your dataset.

### Annotate#

To annotate the records, the Argilla UI provides a simple and intuitive interface that tries to follow the same interaction pattern as in the Explore mode. As in the Explore mode, the record cards in the Annotate mode are also customized depending on the task of the dataset.

#### Text Classification#

When switching in the Annotate mode for a text classification dataset, the labels in the record cards become clickable and you can annotate the records by simply clicking on them. For multi-label classification tasks, you can also annotate a record with no labels by either validating an empty selection or deselecting all labels.

Once a record is annotated, it will be marked as Validated in the upper right corner of the record card.

Note

When the dataset contains a large list of labels (10 +) we encourage viewing the record one by one (from the footer) and using the arrow keys to move quickly to the next record.

#### Token Classification#

For token classification datasets, you can highlight words (tokens) in the text and annotate them with a label. Under the hood, the highlighting takes advantage of the tokens information in the Token Classification data model. You can also remove annotations by hovering over the highlights and pressing the X button.

After modifying a record, either by adding or removing annotations, its status will change to Pending and a Save button will appear. Once a record is saved, its status will change to Validated.

#### TextGeneration#

For TextGeneration datasets, you have a text box available, in which you can draft or edit an annotation. After editing or drafting your annotation, don’t forget to save your changes.

### Bulk annotate#

For all tasks, you can bulk validate the predictions of the records. You can either select the records one by one with the selection box on the upper left of each card, or you can use the global selection box below the search bar, which will select all records shown on the page. Then you can either Validate or Discard the selected records.

For the text classification task, you can additionally bulk annotate the selected records with a specific label, by simply selecting the label from the “Annotate as …” list.

### Validate predictions#

In Argilla you can pre-annotate your data by including model predictions in your records. Assuming that the model works reasonably well on your dataset, you can filter for records with high prediction scores, and simply validate their predictions to quickly annotate records.

#### Text Classification#

For this task, model predictions are shown as percentages in the label tags. You can validate the predictions shown in a slightly darker tone by pressing the Validate button:

• for a single label classification task, this will be the prediction with the highest percentage

• for a multi label classification task, this will be the predictions with a percentage above 50%

#### Token Classification#

For this task, predictions are shown as underlines. You can also validate the predictions (or the absence of them) by pressing the Validate button.

Once the record is saved or validated, its status will change to Validated.

#### TextGeneration#

You can validate or edit a prediction, by first clicking on the view predictions button, and then the Edit or Validate button. After editing or drafting your annotation, don’t forget to save your changes.

### Search and filter#

The powerful search bar allows you to do simple, quick searches, as well as complex queries that take full advantage of Argilla’s data models. In addition, the filters provide you a quick and intuitive way to filter and sort your records with respect to various parameters, including the metadata of your records. For example, you can use the Status filter to hide already annotated records (Status: Default), or to only show annotated records when revising previous annotations (Status: Validated).

You can find more information about how to use the search bar and the filters in our detailed search guide and filter guide.

Note

Not all filters are available for all tasks.

### Progress metric#

From the sidebar you can access the Progress metrics. There you will find the progress of your annotation session, the distribution of validated and discarded records, and the label distribution of your annotations.

## Weak labelling#

The Argilla UI has a dedicated mode to find good heuristic rules, also often referred to as labeling functions, for a weak supervision workflow. As shown in our guide and tutorial, these rules allow you to quickly annotate your data with noisy labels in a semiautomatic way.

You can access the Weak labelling mode via the sidebar of the Dataset page.

Note

The Weak labelling mode is only available for text classification datasets.

### Query plus labels#

A rule in Argilla basically applies a chosen set of labels to a list of records that match a given query, so all you need is a query plus labels. After entering a query in the search bar and selecting one or multiple labels, you will see some metrics for the rule on the right and the matches of your query in the record list below.

Warning

Filters are not part of the rule, but are applied to the record list. This means, if you have filters set, the record list does not necessarily correspond to the records affected by the rule.

If you are happy with the metrics and/or the matching record list, you can save the rule by clicking on “Save rule”. In this way it will be stored as part of the current dataset and can be accessed via the manage rules button.

Hint

If you want to add labels to the available list of labels, you can switch to the Annotation mode and create labels there.

### Rule Metrics#

After entering a query and selecting labels, Argilla provides you with some key metrics about the rule. Some metrics are only available if your dataset has also annotated records.

• Coverage: Percentage of records labeled by the rule.

• Annotated coverage: Percentage of annotated records labeled by the rule.

• Correct/incorrect: Number of labels the rule predicted correctly/incorrectly with respect to the annotations.

• Precision: Percentage of correct labels given by the rule with respect to the annotations.

Note

For multi-label classification tasks, we only count wrongly predicted labels as incorrect, not labels that the rule misses.

#### Overall rule metrics#

From the right sidebar you can access the Overall rule metrics. Here you will find the aggregated metrics, such as the coverages, the average precision and the total number of correctly/incorrectly predicted labels. You can also find an overview about how many rules you saved and how they are distributed with respect to their labels.

Hint

If you struggle to increase the overall coverage, try to filter for the records that are not covered by your rules via the Annotation filter.

### Manage rules#

Here you will see a list of your saved rules. You can edit a rule by clicking on its name, or delete it by clicking on the trash icon.

## Explore records#

If you want to explore your dataset or analyze the predictions of a model, the Argilla UI offers a dedicated Explore mode. The powerful search functionality and intuitive filters allow you to quickly navigate through your records and dive deep into your dataset. At the same time, you can view the predictions and compare them to gold annotations.

You can access the Explore mode via the sidebar of the Dataset page.

### Search and filter#

The powerful search bar allows you to do simple, quick searches, as well as complex queries that take full advantage of Argilla’s data models. In addition, the filters provide you a quick and intuitive way to filter and sort your records with respect to various parameters, including predictions and annotations. Both of the components can be used together to dissect in-depth your dataset, validate hunches, and find specific records.

You can find more information about how to use the search bar and the filters in our detailed search guide and filter guide.

Note

Not all filters are available for all tasks.

### Predictions and annotations#

Predictions and annotations are an integral part of Argilla’s data models. The way they are presented in the Argilla UI depends on the task of the dataset.

#### Text classification#

In this task the predictions are given as tags below the input text. They contain the label as well as a percentage score. Annotations are shown as tags on the right together with a symbol indicating if the predictions match the annotations or not.

#### Token classification#

In this task, predictions and annotations are displayed as highlights in the input text. To easily identify them at a glance, annotations are highlighted with the color of their corresponding label, while predictions are underlined with a solid line (see picture).

For datasets with available score, the solid line for predictions disappears when the score filter (in Predictions filter section) is applied.

#### TextGeneration#

In this task predictions and the annotation are given in a text field below the input text. You can switch between prediction and annotation via the “View annotation”/”View predictions” buttons. For the predictions you can find an associated score in the lower left corner. If you have multiple predictions you can toggle between them using the arrows on the button of the record card.

### Metrics#

From the side bar you can access the Stats metrics that provide support for your analysis of the dataset.

## Search Records#

The search bar in Argilla is driven by Elasticsearch’s powerful query string syntax.

It allows you to perform simple fuzzy searches of words and phrases, or complex queries taking full advantage of Argilla’s data model.

Check out the query guide for a comprehensive guide on how to search in the Argilla UI.

## Filter Records#

The Argilla UI provides a set of filters that help you to quickly and conveniently navigate your records. For example, you can filter records by their predictions, by a specific metadata, or choose to only view records that are missing annotations. The filters are available in all modes of the UI.

Hint

You can also filter the records through the search bar by means of queries.

Note

Not all filters listed below are available for all tasks.

### Predictions filter#

This filter allows you to filter records with respect of their predictions:

• Predicted as: filter records by their predicted labels

• Predicted ok: filter records whose predictions do, or do not, match the annotations

• Score: filter records with respect to the score of their prediction

• Predicted by: filter records by the prediction agent

### Annotations filter#

This filter allows you to filter records with respect to their annotations:

• Annotated as: filter records with respect to their annotated labels

• Annotated by: filter records by the annotation agent

• Only records not covered by rules: this option only appears if you defined rules for your dataset. It allows you to show only records that are not covered by your rules.

### Status filter#

This filter allows you to filter records with respect to their status:

• Default: records without any annotation or edition

• Validated: records with validated annotations

• Edited: records with annotations but still not validated

This filter allows you to filter records with respect to their metadata.

Hint

Nested metadata will be flattened and the keys will be joint by a dot.

### Sort records#

With this component you can sort the records by various parameters, such as the predictions, annotations or their metadata.

### Examples#

Here we will provide a few examples how you can take advantage of the filters for different use cases.

#### Missing annotations#

If you are annotating records and want to display only records that do not have an annotation yet, you can set the status filter to Default.

#### Low scores#

If you uploaded model predictions and want to check for which records the model still struggles, you can use the score filter to filter records with a low score.

#### High loss#

If you logged the model loss as a metadata for each record, you can sort the records by this loss in descending order to see records for which the model disagrees with the annotations (see this tutorial for an example).

## View dataset metrics#

Argilla metrics are very convenient in terms of assesing the status of the dataset, and to extract valuable information.

### How to use Metrics#

Metrics are composed of two submenus: Progress and Stats. These submenus might be different for Token and Text Classification tasks, as well as for the different modes (especially the Weak labelling mode).

#### Progress#

This submenu is useful when users need to know how many records have been annotated, validated and/or discarded.

##### Annotation and Explore modes#

When clicking on this menu, not only the progress is shown. The number of records is also displayed, as well as the number of labeled records or entities that are validated or discarded.

##### Weak labelling mode#

In this mode, progress is related to the coverage of the rules. It shows the model coverage and the annotated coverage, and also the precision average and the number of correct and incorrect results.

In the total rules section, users can find the number of rules related to the different categories.

#### Stats#

This submenu allows users to know more about the keywords of the dataset.

##### Explore and Weak labelling mode#

In both modes, the Keywords list displays a list of relevant words and the number of occurrences.

##### Annotation mode#

In the annotation mode, the stats show the mentions (this is, the entities) which are present in the records.

This submenu has the Predicted as (for predictions) and the Annotated as (for annotation) sections, and here users can see the number of entities predicted or annotated with a specific label. The number of occurrences is ordered from highest to lowest, and the labels are also ordered in that way.

#### Refresh button#

Users should click this button whenever they wanted to see the page updated. If any change is made, this button displays the updated page.