text:
## gold specific gravity formula

#### Specific Gravity of Gold Calculation Learn how to easily ...

Jan 02, 2017  Step 7. Match your calculation with the specific gravity to find out the metal purity in the chart below. Once you determine the specific gravity, it is important to note that tungsten gold bars are almost the same specific gravity as 24k gold bars. Tungsten’s specific gravity is 19.25 and 24k gold

#### Specific Gravity Formula - Softschools

Specific Gravity Formula Questions: 1) The density of gold is ρ gold = 19300 kg/m 3.What is the specific gravity of gold? Answer: The specific gravity can be found by dividing the density of gold by the density of water at 4.0°C: SG = 19.3. The specific gravity of gold is 19.3.

#### Specific gravity test for gold - Goldbay

Sep 30, 2020  Specific gravity Test for Gold and Quartz Specimens: Figure out how much gold is in a quartz gold specimen. The simplest way to test specific gravity is weigh the quartz gold specimen in “air” and multiply the weight by 1.9. Then weigh the quartz gold specimen fully submerged in “distilled water” and multiply that weight by 3.1.

#### Specific Gravity Chart for Gold and Metals - GoldHog Gold ...

95 行  Metal or Alloy Density; Actinium: 10.070: Admiralty Brass: 8.525: Aluminum: 2.712: Aluminum

#### Gold in Quartz - How to Calculate the Gold Content

The formula for calculating the specific gravity or Density of a Specimen rock. However the specimen that you want to estimate the gold content in also contains other material such as quartz, quartzite or ironstone or other substances so this needs to be taken into consideration in the calculation.

#### SPECIFIC GRAVITY FORMULA - Nugget Shooter

To do this, find some ironstone (free of gold or quartz) around where you find the specimen, and find out its density following steps (1) to (3) above and plug it in. This is necessary with ironstone specimens for accuracy because different iron stones have different densities (not like say different colored quartz which is all 2.65g/cubic cm ...

#### How to Calculate the Specific Gravity of a Substance - dummies

In physics, a substance’s specific gravity is the ratio of that substance’s density to the density of water at 4 degrees Celsius. Because the density of water at 4 degrees Celsius is 1,000 kg/m3, that ratio is easy to find. For example, the density of gold is 19,300 kg/m3, so its specific gravity is the []

#### How to Test Gold Using Specific Gravity. - YouTube

Feb 26, 2015  This video demonstrates how a specific gravity test device can be used to determine the purity of your metals.Get the specific gravity table at the bottom of...

#### Gold Mineral Properties - Geology

Specific gravity decreases as gold naturally alloys with silver, copper or other metals. Diagnostic Properties Color, hardness, streak, specific gravity, ductility, malleability.

#### Gold in Quartz Table - A simple way to calculate the gold ...

This gold in quartz table it has been calculated using the Specific Gravity of gold as 19.3 and Specific Gravity of Quartz as 2.65. In reality of course there will be impurities in the gold and in the quartz so as usual we caution you to use it as a quick guide, not a definitive analysis of the amount of gold

#### Specific Gravity Chart for Gold and Metals - GoldHog Gold ...

Metal or Alloy Density; Actinium: 10.070: Admiralty Brass: 8.525: Aluminum: 2.712: Aluminum – melted: 2.560 – 2.640: Aluminum – 1100: 2.720: Aluminum – 6061

#### Specific Gravity Calculations for Gold Content on 3 ...

May 11, 2015  Using the methodology shown by goldnuggetsales, and the standard equation of specific gravity to calculate the amount of gold in a the 3 specimens I rece...

#### SPECIFIC GRAVITY FORMULA - Nugget Shooter

To do this, find some ironstone (free of gold or quartz) around where you find the specimen, and find out its density following steps (1) to (3) above and plug it in. This is necessary with ironstone specimens for accuracy because different iron stones have different densities (not like say different colored quartz which is all 2.65g/cubic cm ...

#### Specific Gravity - Definition, Calculation, Solved ...

Example 2: If the density of gold is 19300 kg/m 3, what is the specific gravity of gold? ... Substituting the values in the above equation, we get $$\rho _{substance}=0.92\cdot 1000\,kg/m^3 =920\,kg/m^3$$ Specific Gravity of Gas. For gases, the specific gravity is normally calculated with reference to air. Specific gravity for gases is defined ...

#### Gold Specimen Calculations - Geocities.ws

The Specific Gravity Test (used to calculate the gold content of gold nuggets that contain a mixture of gold and other materials) In this example of the specific gravity test, we use gold mixed with quartz. Simply put, the formula is 3.1 x the weight in water, minus 1.9 x the weight in air: Wet nugget weighs 74.5 grams x 3.1 = 230.95

#### How to Test Gold Using Specific Gravity. - YouTube

Feb 26, 2015  This video demonstrates how a specific gravity test device can be used to determine the purity of your metals.Get the specific gravity table at the bottom of...

#### Calaverite - Wikipedia

Specific gravity: 9.1–9.3: Optical properties: Anisotropic: Pleochroism: Weak: Ultraviolet fluorescence: None: References: Calaverite, or gold telluride, is an uncommon telluride of gold, a metallic mineral with the chemical formula AuTe 2, with approximately 3% of the gold replaced by silver. It was first discovered in Calaveras County, ...

#### Minerals and some other materials - Specific Gravities

Related Topics . Material Properties - Material properties for gases, fluids and solids - densities, specific heats, viscosities and more ; Related Documents . Concrete Mixtures - Cement, sand and gravel mixtures; Density of Selected Solids - Density of selected solids; Density, Specific Weight and Specific Gravity - An introduction to density, specific gravity and specific

#### Specific Gravity Test - Gold and Quartz - Gold Nugget ...

Jan 17, 2019  Specific Gravity Test - Gold and Quartz Specific Gravity Test - Gold and Quartz. By nuggethunting, January 13, 2019 in Gold Nugget Hunting ... did the short math came to 77.30 grams of gold, the long formula and more accurate according to the video came to 72.4 grams. Regardless if its off, plus or minus its nice to have an idea of amount. ...

#### Relative density - Wikipedia

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. Specific gravity for liquids is nearly always measured with respect to water at its densest (at 4 °C or 39.2 °F); for gases, the reference is air at room temperature (20 °C or 68 °F). The term "relative density" is often preferred in ...

#### Specific Gravity Calculator - Calculator Academy

Since specific gravity is directly related to buoyancy, you can use the force of buoyancy to calculate specific gravity and that’s exactly what this type of device does. There are over 50 other instruments that can be used to calculated specific gravity, but these are the main forms used in science.

#### Conversion Factors and the Specific Gravity of Minerals ...

Aug 18, 2020  SG might also be used (as Archimedes did) to distinguish pure gold (SG = 19.3) from 14K gold (SG =13.3) or pyrite (SG = 5.1). Some specific gravities of minerals are shown below. Online mineral vendors sell equipment for determining specific gravity, but it is simple to do in the lab. Specific gravity is related to density, but it is unitless.

#### specific gravity Formula, Units, Equation Britannica

Specific gravity, ratio of the density of a substance to that of a standard substance. Solids and liquids are often compared with water at 4 C, which has a density of 1.0 kg per liter. Gases are often compared with dry air, having a density of 1.29 grams per liter (1.29 ounces per cubic foot) under standard conditions.

#### Specific Gravity - Recreational Gold Prospecting

What is specific gravity? Specific gravity is simply the weight in grams of one cubic centimeter of a metal. The Specific Gravity Test is used to

#### Calculate :: Specific Gravity :: Coins :: Gold :: Silver

if for example we use ethanol at 95% that have a specific gravity of 0,81 the formula to use get the following: Ps = Pm * 0,81 / Pq A different solution is to wet the coin in alcohol, the adherence of alcohol to coin is, we would dare to say, far less than volume of the thread that hangs the coin.

#### Conversion Factors and the Specific Gravity of Minerals ...

Aug 18, 2020  SG might also be used (as Archimedes did) to distinguish pure gold (SG = 19.3) from 14K gold (SG =13.3) or pyrite (SG = 5.1). Some specific gravities of minerals are shown below. Online mineral vendors sell equipment for determining specific gravity, but it is simple to do in the lab. Specific gravity is related to density, but it is unitless.

#### How to Determine Approximate Specific Gravity of Alloys

Divide 1,000 by this total‑the answer is the specific gravity of the alloy. Example. Find the specific gravity of 14K yellow gold having . 583 parts gold, 104 parts silver and 313 parts copper. First figure the reciprocals of the specific gravities. Fine Gold ‑ 1 divided by 19.32=.051759. Fine Silver ‑

#### Gold Mineral Properties - Geology

Specific gravity decreases as gold naturally alloys with silver, copper or other metals. Diagnostic Properties Color, hardness, streak, specific gravity, ductility, malleability.

#### Specific Gravity - The Gemology Project

Dec 13, 2016  Consider two balls of equal weight but of different specific gravity, for instance a 10 gram gold ball and a 10 gram silver ball. The gold ball has a sg of 19.3, while silver has a sg of 10.5. Because gold has a sg that is almost twice as much as silver, you can imagine that the 10 gram gold ball will be smaller than the 10 gram silver ball.

#### Gold testing using specific gravity - Quicktest

- the specific gravity method tells us whether gold is "about 20ct" or "about 22ct" but it's not quite accurate enough to measure to the nearest 1ct. THE FOLLOWING TESTS WERE CARRIED OUT ON A PURE (.999 / 24ct) BAR OF GOLD, which we know, from the chart above, should read 19.3 .

#### Minerals and some other materials - Specific Gravities

Related Topics . Material Properties - Material properties for gases, fluids and solids - densities, specific heats, viscosities and more ; Related Documents . Concrete Mixtures - Cement, sand and gravel mixtures; Density of Selected Solids - Density of selected solids; Density, Specific Weight and Specific Gravity - An introduction to density, specific gravity and specific weight - formulas ...

#### How to Find and Determine the Specific Gravity of a Rock

In the method we use to How to Find and Determine Specific Gravity of Rocks (some call it Specific Density because they don’t know better!), we’ll use simple equipment. What is required to make this measurement is a weight scale, and the list of the different minerals SPECIFIC GRAVITIES. What specific gravity is, is the mass of a solid in relationship to its weight. An extreme example of ...

#### Percent Solids to Specific Gravity Conversion of Slurry

Here is the formula for specific gravity of a slurry, S is: S = specific gravity of slurry Si= specific gravity of liquid phase Ss= specific gravity of solids phase Cw = concentration of solids by weight Cv= concentration of solids by volume EXAMPLE: if the liquid has a specific gravity of 1.2 and the concentration of solids by weight is 35% with the solids having a specific gravity of 2.2 ...

#### specific gravity Formula, Units, Equation Britannica

Specific gravity, ratio of the density of a substance to that of a standard substance. Solids and liquids are often compared with water at 4 C, which has a density of 1.0 kg per liter. Gases are often compared with dry air, having a density of 1.29 grams per liter (1.29 ounces per

#### Density and Specific Gravity - Practice Problems

Jun 19, 2020  You know the mass (40 g), but the volume is not given. To find the volume, use the formula for the volume of a box. volume = length x width x height. The volume of the cube is 2cm x 2cm x 2cm = 8cm 3. The density then is the mass divided by the volume: Thus the cube is NOT gold, since the density (5.0 g/cm 3) is not the same as gold (19.3g/cm 3).

#### specific gravity - calculator.org

The specific gravities of gases usually are compared to dry air which generally has a density of 1.29 g per litre. The specific gravity of all other materials is compared to water as a fraction heavier lighter or heavier density. Calculating Specific Gravity. The following formula is used to calculate the specific gravity of a material. SG = ρ ...

#### Gold converter Gold Volume vs. Weight Size ...

Rod from gold with D sizes of 0.2043 in = 5.189 mm = AWG gauge-4. Gold rods of sizes: Diameter 0.2043 inch and 5.189 millimeters (which is Gauge #4 AWG), and with lengths of 3 feet (1.18 in^3) or 1 meter (21147.41 mm^3), weights exactly 13.17 oz/3ft length and 408.15 g/m in pure 24 karat gold.. The gold calculator can be applied for many other units and measures not only in ounces and grams of ...

#### MINERAL GALLERY - SPECIFIC GRAVITY

Specific Gravity is a measure of the density of a mineral. At times it is such a useful property that it is the only way to distinguish some minerals without laboratory or optical techniques. Gold (pictured) can easily be distinguished from "fool's gold" by specific gravity alone, although there are many other ways. Specific gravity is a ...


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Top

# Coordinate System

Coordinate systems identify locations by making measurements on a framework of intersecting lines that resemble a complex net. On a map, the lines are straight and the measurements are done in terms of distance but on a circular surface (like a football or a sphere) the lines become circles and the measurements are done in terms of angle.
These measurements are based on Cartesian coordinates.

 Related Calculators Polar Coordinates to Cartesian Coordinates Calculator Rectangular Coordinates to Polar Coordinates Calculator Cartesian Coordinates Calculator Graphing Inequalities on a Coordinate Plane Calculator

## Coordinate System

Cartesian Coordinate System:
The following are some important concepts to know about the coordinate system.
A two dimensional surface on which a coordinate system has been set up, is known as Cartesian plane, or graph or coordinate grid.
The numbers in an ordered pair that locate a point in the coordinate plane is known as coordinates.
X – axis is the horizontal number line on the coordinate grid.
Y – axis is the vertical number line on the coordinate grid.
The four parts made by the above X and Y axes are called quadrants.
The intersection of the X and Y axis is known as the origin.
For plotting any point of the coordinate system, we have to write first the value of x and then y, like (2, 3), here x = 2 and y = 3.

If the coordinates of any group of points are given to us then we can:
1.    Find out the distance between those two points.
2.    Determine their mid point, their slopes and their equation of lines.
3.    Check if those lines are perpendicular lines or the parallel lines.
4.    Calculate the perimeter and determine the area of the polygon which is defined by its points.
5.    Do the transformation by either rotation, reflection or by moving it.
6.    Determine the equations of different conic sections.
7.    Determine the point of intersection of two lines etc.

Some Common Formulas:
The distance formula:
The distance D between any two points is given by the following equation:
D = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$, where the coordinates  are (x1, y1) and (x2, y2) respectively.
The mid point formula:
If a line segment has the end points ((x1, y1) and (x2, y2), the midpoint is given by the following formula: $[(\frac{x_{1}+x_{2}}{2}),(\frac{y_{1}+y_{2}}{2})]$.

Solved Examples:
Example1:  Find the point of intersection of the lines:
2x + y = 3    (a)
and    y = 3x – 1.    (b)
Solution: To find the point of intersection we need to solve the equations (a) and (b) simultaneously. We can substitute (b) into equation (a):
2x + (3x – 1) = 3
i.e.    5x – 1 = 3
i.e.    x = $\frac{4}{5}$
Substituting this into equation (b):
y = 3($\frac{4}{5}$) – 1 = $\frac{7}{5}$.
Therefore the lines intersect at the point
($\frac{4}{5}$, $\frac{7}{5}$).

Example 2: Calculate the difference between the two points (15, 20) and (35, 5).
Solution: For A is at (15, 20) its x-coordinate is the first number or 15. The x-coordinate of B is 35. So the difference is 20. Now, the difference in the point's y-coordinates would be 20 - 5 = 15.
Now using the distance formula for finding distance between two points which is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$, we will get, distance = $\sqrt{(20)^{2}+(15)^{2}}$ = 25.

Example 3: Find the coordinates (x, y) of the midpoint of the segment that connects the points (-4, 6) and (3, -8).
Solution: Applying the midpoint formula, we get, (x, y) = $[(\frac{x_{1}+x_{2}}{2}),(\frac{y_{1}+y_{2}}{2})]$.

(x, y) = $[(\frac{-4+3}{2}),(\frac{6-8}{2})]$

format: plain


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#### Volume 10, issue 3 (2010)

 Recent Issues
Author Index
 The Journal About the Journal Editorial Board Subscriptions Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 To Appear Other MSP Journals
Continuous interval exchange actions

### Christopher F Novak

Algebraic & Geometric Topology 10 (2010) 1609–1625
##### Abstract

Let $\mathsc{ℰ}$ denote the group of all interval exchange transformations on $0\le x<1$ Given a suitable topological group structure on $\mathsc{ℰ}$, it is possible to classify all one-parameter interval exchange actions (continuous homomorphisms $ℝ\to \mathsc{ℰ}$). In particular, up to conjugacy in $\mathsc{ℰ}$, any one-parameter interval exchange action factors through a rotational torus action.

##### Keywords
interval exchange, group action, one-parameter subgroup
##### Mathematical Subject Classification 2000
Primary: 37E05, 54H15
Secondary: 57S05, 37A10, 57M60


text:
# [scribus] Bulk Image Path modification / Relative Image Paths

john Culleton john at wexfordpress.com
Wed Aug 17 13:41:50 UTC 2011

On Wed, 17 Aug 2011 07:35:20 +0200
Martin Kempf <baubiologie at gmail.com> wrote:

>
> a.l.e. and John, I tried to change the "geeky way" with notepad++
> (handling big files), but inside the .sla opened as text file I
> didn't find any path in clear... did I something wrong?
>
> So I re-changed the path on the HD as before and created the
> "éléments pour la sortie", but I never manipulated those files, can I
> use them afterward to reopen the project and modify it like before?
>
> John, I just try more logical structure in my new organization of my
> paths :-)
>
> Martin Kempf
>
>
>
>
>
> Le 15/08/2011 17:39, john Culleton a écrit :
> > On Mon, 15 Aug 2011 16:49:37 +0200
> > Martin Kempf<baubiologie at gmail.com>  wrote:
> >
> >> Hello,
> >>
> >> I just cleaned up my HD, and changed the absolute paths of my
> >> scribus files and image folders on my main working HD.
> >> Now, of course, I have to change image per image the path inside my
> >> nearly hundred scribus files...
> >> I'm using 1.3.3.14 portable on Win7
> >>
> >> Is there any way to bulk change all paths inside a scribus
> >> document? Or if I could just click inside of the path-text in the
> >> image manager and copie-past the new path...
> >>
> >> And, which would make the future file manipulations much easier, is
> >> there any way to put the image path in a relative way? for example
> >> all images in a sub-folder, pointed like "../Images/" or somehow,
> >> and I could find my images once the whole  docs folder structure
> >> placed elsewhere...
> >>
> >> Thanks for your help,
> >> Martin
> >>
> > On my Linux system I have a subfolder:
> >
> > /usr/local/active
> >
> > And each project has its own folder within active:
> >
> > /usr/local/active/grammar
> >
> > for a typesetting and indexing project for an academic with the last
> > name of Grammar (sic).
> >
> > Within each project folder are all the pertinent files, including
> > graphics, source and pdf. If things get too crowded I may have yet
> > another subfolder for the graphics, e.g.:
> >
> > /usr/local/active/grammar/illos
> >
> >
> >
> > I recognize that fiddling with folder structures is more automatic
> > to Linux users than it is to Windows users. But the same mechanisms
> > exist.
> >
> > As for bulk changes within a Scribus .sla file this also is easy for
> > me. I am used to editing plain text documents in Gvim. Again this
> > may not be as common in the Windows world. Gvim has a change
> > statement like:
> >
> > :% s/\\myfolder\\foo/\\myfolder\\foo\\bar/g
> >
> > this string translates into
> >
> > : do a command
> > % apply it to all lines
> > s substitute
> > / / / the from-to string dividers
> > \\myfolder = \myfolder
> > \\foo = \foo
> > \\bar = \bar
> > g apply it to all instances in a line (i.e. globally)
> >
> > The g is necessary because Scribus sla files tend to run many
> > statements together without intervening end-of-line characters. Now
> > the above command looks very complicated to the first time user but
> > to the experienced Gvim user it is automatic. I have used similar
> > substitution strings literally thousands of times in the past two
> >
> > Another popular text editor is Emacs.
> >
> > The default text editor Notepad that comes with Windows is very
> > limited and tends to blow up on big files etc. So I suggest to all
> > users that they learn a more capable editor. Vim/Gvim is free as is
> > Emacs.
> >
> > John Culleton
> >
> > ___
> > Scribus Mailing List: scribus at lists.scribus.net
> > Edit your options or unsubscribe:
> > http://lists.scribus.net/mailman/listinfo/scribus
> > http://wiki.scribus.net
> > http://forums.scribus.net
> >
> -------------- next part --------------
> An HTML attachment was scrubbed...
> URL:
> <http://lists.scribus.net/pipermail/scribus/attachments/20110817/8697a7fc/attachment.html>
> ___ Scribus Mailing List: scribus at lists.scribus.net
> Edit your options or unsubscribe:
> http://lists.scribus.net/mailman/listinfo/scribus
> http://wiki.scribus.net
> http://forums.scribus.net

really an inferior product. For windows users Vim three icons will
appear: Gvim, the full product, Evim, a simplified user interface for
newcomers, and Rvim, a read-only version. I always delete the last one.

Using the full product, Gvim, you can add extra endlines to an .sla file
yet still have a functional file. For example, to append a newline
after each string consisting of "(space) I used the command:

<esc>:% s/" /" \r/g

The escape puts one in command line mode,
the :  is a command prefix,
the %  means apply this command to all lines,
the s  means substitute,
the    / / / enclose the before and after strings,
the "  followed by a space is the original string,
the " \r  is the new string (with end of line attached)
and the g  means change all occurrences in a line, not just the first
one.

The new .sla file is executable by scribus (I just checked) but the .sla
file is much more readable, and changeable.

If in your .sla file the addresses are relative rather than absolute
then my earlier suggestion of putting all the bits and pieces of the
document in one dedicated directory applies. The collect command will
pull all those bits and pieces together for transport, as someone (ALE?)
suggested previously.


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## Essential University Physics: Volume 1 (4th Edition)

$\vec{v} = -.8\hat{j} \ cm/h$ $\vec{d}=-4.8\hat{j} \ cm$
The displacement is $-4.8\hat{j} \ cm$. We know this because the hand goes from positive 2.4 cm on the y axis to -2.4 cm on that axis. The velocity is given by the displacement over the change in time, which is 6 hours. This gives: $\vec{v} = -4.8/6\hat{j} = -.8\hat{j} \ cm/h$


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# 2015 Countdown Problem #21: Buffon's Clean Tile Game Revisited

A rectangular tile has area $$2015$$ $$cm^2$$ and perimeter $$192$$ $$cm$$. A coin of diameter $$2$$ $$cm$$ is randomly tossed onto a floor laid with such rectangular tiles which tessellated in the following arrangement as shown (checkerboard arrangement).

Let $$P$$ be the probability of the coin landing on exactly 2 tiles (i.e. such that when it lands it covers part of exactly two tiles, as shown). Find $$2015P$$.

This problem is part of the set 2015 Countdown Problems.

×


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## volumetric analysis

PCl5 + 4H2O -> H3PO4 + 5HCl
If $5 * 10^{-3}$ mol of PCl5 is reacted completely with water and the solution made up to $100 cm^3$, what is the volume of $0.1 mol/dm^3$ sodium hydroxide required to react with $25 cm^3$ of this solution? (ans: $100cm^3$)

second question:
A 1 g sample of a mixture of sodium carbonate and sodium hydrogencarbonate was titrated with $0.2 mol/dm^3$ HCl using phenolphthalein as indicator. If $17.5 cm^3$ HCl were required, what is the percentage by mass of sodium hydrogencarbonate in the mixture? (ans: 62.9%)

how do i go about solving such questions where they're mixtures reacting with HCl? my ans differ a lot from the suggested answers for both.


text:
SQR-056: Guidelines for gated updates for SQuaRE services (including Science Platform)

• Frossie Economou and
• Russ Allbery

Latest Revision: 2021-09-15

Note

The aim of this document is to expose the heuristics by which we consider what the appropriate process is for various types of service updates. We also describe our extant processes around Science Platform and Science Platform adjacent services, as these are likely to be of most interest to developers outside our team.

Note

This is a descriptive (not prescriptive) document describing details of current practice of a specific team in a specific area. For the actual developer guide, see developer.lsst.io

1   Background¶

SQuaRE uses Configuration as Code (aka “gitops”) practices for service deployment whenever practical. What this means is that by merging code or configuration to a special branch (typically master), it is possible to trigger an action (via an agent or supervisory system) that automatically results in the deployment of the new version of the service.

Whether the new version of the service is actually deployed automatically (a practice known as continuous deployment) or whose deployment is gated behind a human decision making process is a matter of policy. The barrier to getting through that gate is itself a function of where a system lies on an axis of formality, with one end maximizing developer agency and the other maximizing organisational control.

Where any one particular instance of any one particular service lies in the Axis of Formality is a function of:

• The service’s maturity, including the number of users depending on the service
• The service environment, including whether it’s production, staging or development environment
• The service’s userbase, including whether it is a feature-seeking or risk averse population
• The actual and reputational impact of disruptions to the service
• The nature of the change being made.

It’s easy to see that a major change in a mature high visibility service that many users depend on is on the opposite side of the axis that a cosmetic change in a service under development deployed on a sandbox. The trickier situations lie in between these extremes, and the aim of this document is to expose the heuristics by which we consider these questions. We also describe our extant processes around the update of Science Platform and Science Platform adjacent services, as these are likely to be of most interest to developers outside our team and since they exercise a lot of the decision space described above.

2   Kubernetes services and Argo CD¶

SQuaRE develops Kubernetes-based services, and uses the ArgoCD platform to manage their deployment. The ArgoCD UI provides a valuable way both to assess the state of deployed services and to allow basic maintenance operations on deployed services even without being particularly intimate with the details of their deployments, and irrespective whether the underlying service is configured with Helm or kustomize.

ArgoCD continuously monitors running services and compares their state with that of their git repo. When it detects that a service is “out of sync” (ie there is drift between its deployed state and the repository state) it can sync automatically (the continuous deployment model) or as in Figure 2, indicate visually the out-of-sync state until an engineer pushes the sync button to resolve it (the gated deployment model). Generally unless working in a dev or bleed environment, we do not allow ArgoCD to sync automatically.

It follows then that for gated deployments there are two aspects to getting a new version of a service deployed into production:

1. Mechanics: Getting the right version of the service and its configuration merged into the deployment branch (typically master);
2. Process: Going through the process that allows it to be synced into production.

3   Mechanics¶

Assume that there is a new release of Gafaelfawr, the authentication service used by the Rubin Science Platform. The following steps notify Argo CD that this upgrade is ready to be performed:

1. Update the Helm chart in the [charts](https://github.com/lsst-sqre/charts) repository under charts/gafaelfawr. Do this by updating the appVersion key in Chart.yaml and the image.tag key in values.yaml to point to the new release. Also update the version of the Helm chart in Chart.yaml. When this PR is merged, a new version of the chart will be automatically published via GitHub Actions.
2. Update the Argo CD configuration to deploy the new version of the chart. The Chart.yaml file in services/gafaelfawr in the [phalanx](https://github.com/lsst-sqre/phalanx) repository determines what version of the Gafaelfawr chart is installed by Argo CD. (Phalanx is the Argo CD configuration repository for the Rubin Science Platform.) Under the stanza for the gafaelfawr chart in the dependencies key, change the version to match the newly-released chart version. Once this PR is merged, Argo CD will notice there is a pending upgrade for Gafaelfawr.

4   Process¶

The process used to deploy a specific instance of a service depends on where it lies on the axis of formality above. We use the following terminology for the various deployment environments, in decreasing formality:

• Prod: Production, the user-facing deployment. For historical reasons sometimes referred to as “stable” where the Science Platform is concerned.
• Int: Integration, a deployment for developers, typically from different teams, to converge upon to test the integration of services prior to deployment on Prod. This is the environment somethings referred to as “staging”. For telescope deployments, this is currently the NCSA Test Stand [NTS]
• Dev: Development, an instance that is primarily for developer testing. It may not be always available, or there may be several of them.
• Bleed: An environment that is left uncontrolled, either by being continuously deployed from master or by letting otherwise pinned versions of components float up

In some cases work is corralled in routine maintenance windows. This is both to minimize the potential of disruption in high availability environments and to allow co-ordination of work on multiple services and/or infrastructure; and also to issue a “hold off reporting problems” to users. Maintenance windows do not imply fixed downtime. Downtime (complete service unavailability) is extremely rare and we design our processes to avoid it. Routine maintenance work involves transient service unavailability at most and in most cases users are not barred from using the system during that time, though they are given notice to save their work and there is always a small chance of unforseen problems. A co-ordinator is assigned to announce work start and work end and field any questions.

Current fixed maintenance windows (for applicable services/deployments) are:

• All telescope deployments: 1st Wednesday of every month, late afternoon in Chile (approx 2pm PT), confirmed with the telescope software configuration manager.
• Any other deployment subject to maintenance window: Weekly, Thursday afternoons (approx 3pm PT).

Again, these are not scheduled downtimes. In the event that extended service downtime is needed in a production service (extremely rare), work would be scheduled out of normal hours with ample notice and co-ordination with stakeholders.

Here is a chart showing the current settled-upon practice in select areas:

Service Instance Bleed Dev Int Prod
Science Platform - Science Deployments (IDF, LDF, USDF) Automatic Developer-at-will, no heads-up required  #rsp-team  channel announcement, courtesy 1hr notice Release manager has to okay on  #rsp-team  , 24 hour notice minimum
Science Platform - Telescope Deployments(summit, base, NTS) N/A Developer-at-will, no heads-up required  #com-square  notice, in maintenance window  #com-square  notice,  #summit-testing-announcements  notice, in maintenance window
EFD N/A Developer-at-will, no heads-up required  #com-square  notice, in maintenance window  #com-square  notice,  #summit-testing-announcements  notice, in maintenance window
Roundtable (LSST-the-docs and other non-data services), SQuaSH N/A Developer-at-will, no heads-up required Developer-at-will, no heads-up required If service disruption is expected, in maintenance window, notice in relevant service channels (#dm-docs, etc). Otherwise at-will.
community.lsst.org N/A N/A N/A If service disruption is expected, out-of-hours in co-ordination with #rubinobs-forum-team . Otherwise in maintenance window.

5   Container Environments in the Science Platform¶

The above discussion pertains to services, ie codebases where an error could affect a service’s availability. When it comes to containers made available _by_ a service (eg in nublado), we are less risk averse as users, by design, can always fall back on a previously usable container in case of problems. We recognize that the Science Platform is a primary user environment and as such users do not wish to wait a week for some process to take its course in order to get a requested package or feature. We currently (and anticipate continuing to do so) provide containers labeled “experimental” to rapidly service ad-hoc user requests, for example.

How to reconcile this user-first orientation to the issue of scientific reproducibility is a matter for a future technote.


text:
# Refraction and reflection

1. Mar 15, 2005

### figs

Prove mathematically that when a plane mirror is rotated an angle theta about an axis through its center, the angle of deflection of a light is equal to 2theta. How am i to show draw this out? I have already drawn my points and lined up my angles with and without the mirror rotated. I dont know what direction to take my proof now.

2. Mar 15, 2005

it depends on what you mean by the angle of deflection. Do you mean the angle of incidense + the angle of reflection. If this is the case, rotationg the mirror by an angle $$\theta$$ would increase both the angle of incidence and the angle of reflection by $$\theta$$, resulting in an angle difference of $$2\theta$$.


text:
# Find the closest point along a rectangle, given an another point and direction

Given a rectangle, and a point with a vector direction towards the rectangle. How can I find the closest point on the outside of that rectangle to the point in question?

• Can you explain more what you're asking? I'm really not getting it. – user9471 Sep 5 '11 at 5:46
• Let's say I have a character facing an object (a rectangle), and I draw an imaginary line from that character to the rectangle. I would like to know how I can tell which point along the outside of the rectangular object the line is touching. – onedayitwillmake Sep 5 '11 at 5:49
• They keyword to look up in Google is "AABB" (axis-aligned bounding box). If your "box" (rectangle) isn't axis-aligned yet, a simple transformation matrix - used on all items you care about, obviously - can be used first to change it into one. – Martin Sojka Sep 5 '11 at 7:30

One technique which you could use is called "ray casting". It is commonly used for rendering graphics, but has other applications such as line-of-sight (as you are wanting to do) and path-finding. In general terms it works by finding the intersection of a ray and an object. In your example the ray is the vector for the character's direction.

A useful reference for ray/object intersections (and incidentally other object/object intersections) is www.realtimerendering.com/intersections.html (look under the references for ray/aabb and ray/obb).

The rectangle has four sides. Each side is a line segment.

Test each of the four sides for intersection with the ray. Track the closest hit.

Here's some code to find out where on the segment the ray hits:

bool intersect(const ray& ray, const segment& segment,point& hit) {
// where do we intersect this line?
float t = ((ray.direction.x * ray.origin.y + ray.direction.y *
(segment[0].x - ray.origin.x)) -
(ray.direction.x * segment[1].y)) /
(ray.direction.y * (segment[0].x + segment[1].x) -
ray.direction.x * (segment[0].y + segment[1].y));
if(t >= 0.0 && t<=1.0) { // in the segment
hit = segment[0] + (segment[1]-segment[0]*t);  // lerp
return true;
}
return false; // no hit
}

Well, you can use just linear algebra (analytic geometry, to be more specific) to solve this. It depends on how you modeled the rectangle.

Here's a general case: http://paulbourke.net/geometry/lineline2d/

If your box is axis aligned, you just have to clamp each coordinate axis to the box if the point is outside the box.

From RTCD pg 130:

// Do this for all 3 axes
if( point.x < min.x )  point.x = min.x ;
else if( point.x > max.x )  point.x = max.x ;

If you do this for x, y, z axes, then the point will be slammed to the nearest wall of the box, if it is outside the box to begin with. if it is already inside the box, it will be left alone (where it is).


text:
# A model for reasoning about JavaScript promises

As an antidote to callback-hell, ECMAScript 6 introduces Promises. Promises represent the value of an asynchronous computation, and the functions resolve and reject are used to settle the promise. Promises can be chained using then.

However, the semantics of JavaScript promises are quite complex, and since the feature is implemented by way of ordinary function calls, there are no static checks to ensure correct usage. As a result, programmers often make mistakes in promise-based code that leads to pernicious errors, as is evident from many reported issues on forums such as StackOverflow.

This paper introduces the notion of a promise graph, which can be used to visualise the flow in a promises-based program and help to detect a range of bugs. In order to have formal basis for such reasoning and program analysis the paper also introduces $\lambda_p$ , a calculus for the behaviour of promises, which is expressed as an extension to $\lambda_{JS}$.

### Promises essentials

A promise is an object that represents the result of an asynchronous operation.

A promise can be in one of three different states. Initially a promise is Pending, indicating that the operation has not yet completed and the promise holds no value. A Fulfilled promise is one in which the operation has succeeded and the promise holds a result value. A Rejected promise is one in which the operation has failed and the promise holds an error value. A promise that is either fulfilled or rejected is said to be settled.

A promise is constructed by passing a single callback function with two parameters: a resolve function that can be called to fulfil the promise, and a reject function that can be called to reject the promise. These parameters are named resolve and reject by convention.

[code lannguage=”javascript”]
var p = new Promise((resolve, reject) => {
resolve(42); // immediate resolution
});
[/code]

The then function can be used to register resolve reactions and reject reactions with promises, and creates in turn a dependent promise with the result computed by the resolve (reject) reaction.

### Promises calculus

Section 3 of the paper defines a formal $\lambda_p$ calculus for all of this. The runtime state has five components:

• A heap $\sigma$ that maps addresses to values
• A promise state map $\psi$ that maps addresses to promise values (Pending, Fulfilled(value), or Rejected(value)).
• A map $f$ that maps addresses to a list of fulfil reactions
• A map $r$ that maps addresses to a list of reject reactions
• And a queue $\pi$ that holds scheduled reactions – promised that have been settled and their reactions are awaiting asynchronous execution by the event loop.

A promisify expression in this language turns an object into a promise. Then there a bunch of reduction rules such as this one:

They can look a bit intimidating on first glance, but that’s mostly down to the syntax. For example, all this one says is that:

• given $a$ is an address in the heap, that is not currently mapped to promise values, and an expression promisify(a)
• then the system reduces to a new state where the expression value becomes undefined (the return value of a promisify expression),
• and the promise state map is updated with a new entry for this address that holds the value Pending, the fulfil reaction and reject reaction maps get an empty entry for the address, and the rest of the state of the system is unaltered

The fun part of the paper for me though, is what happens when you create promise graphs off of the back of this formalism. So let’s get straight into that…

### Promise graphs

The promise graph captures control- and dataflow in a promise-based program to represent the flow of values through promises, the execution of fulfil and reject reactions, and the dependencies between reactions and promises.

Promise graphs contain three different types of nodes:

• Value nodes represent value allocation sites in the program
• Promise nodes represent promise allocation sites in the program,
• Function nodes represent every lambda or named function in the program.

There are four different types of edges between nodes as well:

• A settlement edge (which can be labelled either ‘resolve’ or ‘reject’) from a value node $v$ to a promise node $p$ indicates that the value $v$ is used to resolve or reject the promise $p$.
• A registration edge (with label ‘resolve’ or ‘reject’) from a promise node $p$ to a function node $f$ indicates that the function $f$ is registered as a fulfil or reject reaction on the promise $p$.
• A link edge from a promise $p_1$ to a dependent promise $p_2$ represents the dependency that when the parent promise $p_1$ is resolved or rejected, the dependent promise will be resolved or rejected with the same value.
• A return edge from a function node $f$ to a value node $v$ represents the function $f$ returning the value allocated at $v$.

In the examples that follow, the subscript numbers (e.g., $v_{21}$ indicate the line number in the program on which the corresponding object is declared. At this point on first read I admit to thinking it would be easier just to read the code, but bear with me, it will all make more sense very soon.

Here’s a small example of a program and its corresponding promise graph:

You can read the graph forward: the value on line 3 (42) is used to resolve the promise declared on line 1 (p1), and flows into the fulfil reaction function declared on line 2. Inside that function, the newly computed value ($v_2$) is then used to resolve the second promise $p_2$. You can also read it backwards to trace the resolution of $p_2$ to understand how and why it was resolved.

Here’s an example with branching:

### A real world example

Here’s some code from a StackOverflow question (Q42408234) and it’s corresponding promise graph. The poster had spend a full two hours trying to figure out why the promise from mongoose.find (on line 15) was returning undefined.

Immediately the promise graph shows us two disconnected chains. The value returned on line 25 is used to resolve the promise on line 21, but then doesn’t go anywhere else. The function declared on lines 16-31 does not explicity return a value, so it will implicitly return undefined. The fix is to explicitly return the the value of bcrypt.compare as computed on line 20.

### A catalog of Promise-related bugs

Armed with the ability to create promise graphs, we can now detected a variety of promise-related bugs.

• A Dead Promise occurs when an object enters the Pending state and never transitions to either a fulfil or reject state. A promise node with no resolve, reject, or link edges is dead.

• A Missing resolve or reject reaction occurs when a promise is resolved with a non-undefined value, and the promise lacks a fulfil (or reject) reaction. This is a promise node with no outgoing edges.

• A Missing exceptional reject reaction occurs when a promise is implicitly rejected by throwing an exception. These are promise nodes with a reject edge, but to reject registration edge.

• A Missing return occurs when a fulfil or reject reaction unintentionally returns undefined and this value is used to resolve or reject a dependent promise.

• A Double resolve or reject occurs when a promise enters the fulfilled or rejected state, and later the promise is again resolved (or rejected). This can be detected in the graph by multiple resolve (or reject) edges leading to the same promise.

• An Unnecessary promise is a promise whose sole purpose is to act as an intermediary between two other promises.

• A Broken promise chain occurs when a programmer inadvertently creates a fork in a promise chain.

### Promise graphs and StackOverflow

The authors looked at the most recent 600 StackOverflow questions tagged with both promise and javascript (out of over 5,000 total). They focused on those with code fragments of at most 100 lines, a genuine promises-related issue, and an answer. This boiled things down to quite a small list, show below:

The fourth column in the table above shows the additional information — above and beyond the promise graph — that was needed to pinpoint the bug (if any).

• HB indicates that happens-before information is needed – i.e., the knowledge that one statement in the program always executes before another.
• LU indicates that loop unrolling is necessary to debug the root cause (i.e., what happens in one iteration affects a promise created in a subsequent one).
• EV indicates that event reasoning is needed to debug the root cause – e.g., the knowledge that an event handler for a button can fire multiple times.

Here’s an example analysis:

Q41268953. The programmer creates a promise, but never resolves or rejects it. As a result, the reactions associated with the promise are never executed. The programmer gets lost in the details and mistakenly believes the bug to be elsewhere in the program. He or she then proceeds to add additional logging and reject reactions in all the wrong places and these are never executed. In this scenario, the promise graph immediately identifies that the initial promise in the promise chain is never resolved or rejected, and hence that no reaction on chain is ever executed.

There are plenty more examples in section 5.3, and in appendix A.


text:
# proof that $e^x$ is a transcendental function of $x$?

Let a function $f(x)$ be algebraic if it satisfies an equation of the form $$c_n(x)(f(x))^n + c_{n-1}(x)(f(x))^{n-1} + \cdots + c_0(x)=0,$$ for $c_k(x)$ rational functions of $x$, and let $f$ be called transcendental if it is not algebraic. Is it possible to use this definition directly to show that $e^x$ is transcendental?

One way I have been considering for any complex number is this:

Let $x_0\in\mathbb{C}$ and $x_n=x_0+2\pi i n$, where $n\in\mathbb{Z}$. Hence $x_n\neq x_m$ for all $n\neq m$, but we do have $e^{x_n}=e^{x_m}$ for all $n,m\in\mathbb{Z}$ (since $e^{2\pi i n} = 1$ for all $n\in\mathbb{Z}$). But since the Implicit Function Theorem suggests there exists an exact algebraic formula for $x$ using the above definition of an algebraic function, then $e^x$ can not be algebraic since there are an infinite number of representations $x_n$ of $x$.

-
Does $f^n$ mean $f(x)^n$ or $\underbrace{f(f(\cdots f}_n (x)))$ here? –  Henning Makholm Aug 25 '12 at 14:05
No, it just means $(f(x))^n$ in this case. –  pbs Aug 25 '12 at 14:08
A slight nitpick: $f$ is algebraic if it satisfies an equation of the given form, not just one particular equation. –  Cameron Buie Aug 25 '12 at 14:38

One could use the growth at infinity of the function $f:x\mapsto\mathrm e^x$.

Assume that that $f$ is algebraic and choose a real number $x\geqslant0$. Then $|f(x)|\geqslant1$ and $$|c_n(x)|\,|f(x)|^n\leqslant b(x)|f(x)|^{n-1},\qquad b(x)=\sum\limits_{k=0}^{n-1}|c_k(x)|.$$ Hence, for every real number $x\geqslant0$ such that $c_n(x)\ne0$, $|f(x)|\leqslant b(x)/|c_n(x)|$. But indeed, $c_n(x)\ne0$ for every real number $x$ large enough and $b(x)/|c_n(x)|$ can grow at most polynomially when the real number $x$ goes to $+\infty$ while $|f(x)|=\mathrm e^x$ grows... well, exponentially. This is a contradiction.

-

Here is a purely algebraic proof using only the fact that if $y=e^x$, then $y'=y$, but no analytic facts about the exponential function. In other words it should hold in any differential field.

Let $K$ be a differential field, such that $K=k(x)$ where $k$ is the field of constants of $K$ (ie those numbers whose derivative is 0) and with $x'=1$ (think field of rational functions in $x$). And $K(y)$ is a differential extension with $y'=y.$

First observe that $y\notin K$. For elements of $K$ are of the form $c p/q$ with $p,q$ monic polynomials in $x$ over $k$ and $c$ a constant. If $(p/q)'=p/q,$ then $p'q-pq'=pq.$ But the polynomial on the right has degree $\deg p+\deg q$, while the polynomial on the left has degree $\deg p+\deg q-1,$ and both sides are monic, so they are not equal.

Now suppose $y$ were algebraic over $K$, then let the minimal polynomial of $y$ be $y^n+a_{n-1}y^{n-1}+\dotsb+a_0=0.$ Taking the derivative, we have $ny^n+(n-1)a_{n-1}y^{n-1}+a_{n-1}'y^{n-1}+\dotsb+a_0'=0.$ Subtracting from $n$ times the first equation gives $(a_{n-1}-a_{n-1}')y^{n-1}+\dotsb+na_0-a_0'=0.$ We already observed that $y'=y$ has no non-zero solutions in $K$, hence $a_{n-1}-a_{n-1}'\neq 0,$ so we have a monic polynomial of degree $n-1$ annihilating $y$, which contradicts the minimality of our polynomial.

-


text:
# Reference request: Reflection as a single axiom interpreting the whole of ZFC?

Working in first order languge $$\mathcal L(\in, W)$$, where $$W$$ is a constant symbol.

Reflection: if $$\varphi$$ is a formula in $$\mathcal L(\in)$$, in which $$x$$ is free, and $$\vec{p}$$ is the string of all of its parameters, and if $$\psi$$ is a formula in which $$z$$ is free and $$y$$ not free, then all closures of: $$\vec{p} \in W \wedge \exists x (\varphi) \to \exists x [\varphi\wedge \exists y \in W \ \forall z (z \in y \leftrightarrow z \in x \wedge \psi) ]$$; are axioms.

Now with Extensionality this would interpret all axioms of $$\text{ZFC}$$ since it would trivially interpret Harvey Friedman $$\text{K(W)}$$ theory (page 3). Although I'm not sure but I think even Extensionality is interpretable using the above scheme only.

Has there been prior attempts to non-trivially shortly axiomatize ZFC with a single comprehension schema? and had this schema been studied before specially in absence of Extensionality?

• Well, the Replacement scheme implies the Separation scheme, so the answer to your first question is immediately "yes." – Noah Schweber Jan 25 at 18:51
• To me all of the axioms of set union, power and infinity, all are individual comprehension axioms. What is not a comprehension axiom is Extensionality, Foundation and Choice. And by the way Replacement doesn't interpret all the others – Zuhair Jan 25 at 19:10


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# In the figure shown, QS is an altitude of right triangle PQR, PS = 16,

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In the figure shown, QS is an altitude of right triangle PQR, PS = 16,  [#permalink]

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25 Mar 2018, 23:17
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In the figure shown, QS is an altitude of right triangle PQR, PS = 16, and RS = 4. Find the perimeter of PQR.

A. $$20 + 4\sqrt{5}$$

B. $$20 + 6\sqrt{5}$$

C. $$20 + 8\sqrt{5}$$

D. $$20 + 12\sqrt{5}$$

E. $$32\sqrt{5}$$

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Re: In the figure shown, QS is an altitude of right triangle PQR, PS = 16,  [#permalink]

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26 Mar 2018, 01:34
Bunuel wrote:

In the figure shown, QS is an altitude of right triangle PQR, PS = 16, and RS = 4. Find the perimeter of PQR.

A. $$20 + 4\sqrt{5}$$

B. $$20 + 6\sqrt{5}$$

C. $$20 + 8\sqrt{5}$$

D. $$20 + 12\sqrt{5}$$

E. $$32\sqrt{5}$$

As the answers are close together we can't estimate and need to rely on specific geometric rules for an exact solution.
This is a Precise approach.

Since the larger triangle is right-angle then angle P and angle R sum to 90.
Since the two smaller triangle are also right-angle then angle SQP and angle P sum to 90 as do angles R and angle SRQ.
That is, angle SQP = angle R and angle SQR = angle P.
So our small triangles are similar both to each other and to the larger triangle.
Then using similarity of the two smaller triangles: SQ/SP = SR/SQ --> SQ^2 = 16*4 = 64 --> SQ = 8.
Then the hypotenuse of the small internal triangle can be calculated as $$\sqrt{8^2 + 4^2}=\sqrt{4^2(2^2+1)}=4\sqrt{5}$$
Then the hypotenuse of the large internal triangle can be calculated as the $$\sqrt{8^2 + 16^2}=\sqrt{8^2(1+2^2)}=8\sqrt{5}$$
Then our total perimeter is $$16 + 4 + 4\sqrt{5} + 8\sqrt{5} = 20 +12\sqrt{5}$$

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In the figure shown, QS is an altitude of right triangle PQR, PS = 16,  [#permalink]

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26 Mar 2018, 01:53
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In Triangle PQS, $$PQ^2 = PS^2 + QS^2$$ -> $$PQ^2 = 16^2 + x^2 = 256 + x^2$$

In Triangle RQS, $$RQ^2 = SR^2 + QS^2$$ -> $$RQ^2 = 4^2 + x^2 = 16 + x^2$$

In Triangle PQR, $$PR^2 = PQ^2 + RQ^2$$ -> $$20^2 = 256 + x^2 + 16 + x^2$$ -> $$2x^2 = 400 - 272 = 128$$ -> $$x^2 = 64$$ -> $$x = 8$$

Perimeter = $$20 + \sqrt{256 + 64} + \sqrt{16 + 64} = 20 + \sqrt{320} + \sqrt{80} = 20 + 2\sqrt{80} + \sqrt{80} = 20 + 3\sqrt{2^4*5} = 20 + 12\sqrt{5}$$

Therefore, the perimeter of triangle PQR is $$20 + 12\sqrt{5}$$(Option D)
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In the figure shown, QS is an altitude of right triangle PQR, PS = 16, &nbs [#permalink] 26 Mar 2018, 01:53
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# Things I like and things I don't

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Well I upgraded my Main Development Box to Office 2007 when Microsoft sent me the update. And I've been happy with it. That's to say that I have no complaints about it. It's backward compatible. It'll even write out PDF files directly (which isn't much of an issue as I have Acrobat, but hey). All in all, it's a pretty capable update. And I do rather like the new ribbon-toolbar-thingy. They managed to package up all the old functionality in an easier-to-navigate package.

But it is SLOW!

Actually it's just slow to load. Once it's loaded, it performs just fine. When I needed some office tools on my laptop, I considered Office for a bit, but then I decided to try OpenOffice again just for snorts and giggles because I shuddered to think how Office 2007 would perform on my cheapo laptop. I'd installed OpenOffice a few years ago to see if it'd be good for the company, and it really wasn't ready for prime-time. It had loads of ragged edges and it had its own weird desktop metaphor that was confusing. And it didn't have an analogue for Outlook. Quite simply, it wasn't as good.

Fast-forward to 2008, and I no longer need an analogue for Outlook, as I use Google Calendar for all my schedules and Yahoo Deluxe Mailer ($20 a year and well worth it) for all my email. So I tried OpenOffice again. And it's gotten quite a bit better. It was actually about as capable as Office and no longer had the ragged edges and it loaded up quite quickly even on my$350 Wal-Mart brand laptop.

Okay, there was one really ragged edge. Upon running it last week, it popped up a box that said "There's a new version out now. You want I should upgrade?". I pressed "yes" expecting a quick seamless Office-esque (or Firefox-esque for you who don't use Office) upgrade process. It then downloaded the complete installer for the new version, uninstalled the previous version, then re-installed the new version (after I re-told it what options I wanted) until it got up to about 80% complete at which point it said that something wasn't right and it was gonna uninstall itself and that I should probably try to run it again.

So its "upgrade" basically consisted of removing itself entirely from my machine. Thankfully I re-ran the installer as administrator and it worked the second time. Still, that's an ugly upgrade process they have there.

So OpenOffice is finally nice. Sorry if I'm late to that party.

Oh, and two pieces of anti-Kudos for Adobe. First is for their silly attempt to prevent Office 2007 from writing PDF files directly. If you remember a year or so ago, they were threatening to sue sue sue Microsoft for putting a "save as PDF" function into Office. Mind you, several other products already had this functionality at the time (including OpenOffice). They just didn't want Office to have it. They claimed it was because Microsoft would destroy their open standard, notwithstanding that PDF is an open standard over which MS has no control.

What was obvious and what Adobe wasn't saying was that Adobe had a nice cash-cow selling Acrobat Deluxe to Office users, and if Office 2007 could save to PDF, that market would dry up.

And Adobe succeeded. Okay, actually they didn't. They were successful in preventing MS from installing PDF support from the install CD. That means that if you install Office, you still have a "PDF" option in the save-as menu, and when you choose this for the first time, Office says "Oh, that didn't install by default. You mind if I run out to the website and grab the PDF-saver-thingy?" and then it pops up a loading box, downloads whatever PDF DLL it needs, and from then on you can save PDF files just fine.

So, in short, Adobe succeeded in preventing Office 2007 users from saving files as PDF for about the first two minutes of their first attempt to use it. Quite a victory there.

The second anti-Kudos for Adobe is that Fireworks CS3 is now the only piece of software on my machine that is STILL incompatible with Vista Aero. Mind you, they've only known about this problem since Vista was in early beta, and Fireworks CS3 has only been on the market for almost a year now, so it's understandable that they couldn't fix the problem yet.

Not. I'm now using Paint Shop Pro 5 for most of my image editing. It loads quickly and works with Aero. Amazing, eh?

Did you file a bug report about the update issue? It's possible they, like every other moron on the face of the planet, forgot to secure admin permissions before making system changes.

That's what I don't like about office 2007 also; the product's UI is nice but it's resource heavy. Even with it loaded it can be very slow.

For example, right now I'm doing a build using VS 2005. I alt-tabbed to Outlook 2007 and timed how long it took until I could interact with the UI: 27 seconds. I used to use Thunderbird and the same operation took only a few seconds. It's not like the machine is a slouch: it's a 2.4GHz Core Duo 2 with 2Gig of 800MHz memory. The odd thing is that the machine is not memory starved; the process meter is currently showing ~700M free.

I was just thinking I needed a calendar app the other day - cheers for the Google Calendar heads up - didn't know it even existed :)

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# Understanding a “Trivial” Result In Optimal Control Theory

In Lawrence C. Evans' online notes : Optimal control theory, page 33, Evans makes a very trivial looking statement,which doesn't seem trivial to me. I shall elaborate, giving necessary details, so that this reference is only for further reading.

Let $\alpha : [0, \infty) \to [-1,1]^m$ be a measurable function. Consider the differential equation: $$\dot x(t) = M x(t) + N\alpha(t) \\ x(0) = x_0$$

Where $M,N$ are constant real matrices of appropriate dimension, so that $x(t)$ has codomain $\mathbb{R}^n$. Now, we can solve this equation, with the solution: $$x(t) = X(t)x_0 + X(t) \displaystyle\int_0^t X^{-1}(s) N\alpha(s) \operatorname{ds}$$

Now, fix a time $T$, and define the following set: $$K(T,x_0) = \left\{ X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha(s) \operatorname{ds}\right\}$$ where the $\alpha$ can vary over all measurable functions possible from $[0,\infty) \to [-1,1]^m$.

In words, we are trying to find all reachable points at time $T$. If $x_1 \in K(T,x_0)$, there exists $\alpha_{x_1}$ such that $x_1 = X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha_{x_1}(s) \operatorname{ds}$.

One can prove that $K(T,x_0)$ is convex and closed, the former using an obvious candidate, and the latter via an application of the Banach-Alaoglu theorem.

Now, suppose that $0 \in K(\tau, x_0)$ for some $\tau$, but it is not true that $0 \in K(t, x_0)$ for all $t < \tau$. I want to prove that $0$ is a boundary point of $K(\tau,x_0)$.

In words: "If $0$ is reachable in time $\tau$, but is not reachable in any time smaller than $\tau$, then $0$ is a boundary point of the set of all points reachable in time $\tau$".

It did not seem trivial, so I tried letting $0$ be an interior point. It is clear that in a neighbourhood of $0$, there are points which can be reached in time $< \tau$ (when we take the trajectory from $x_0$ to $0$, it enters the neighbourhood before time $\tau$, so all points falling on that trajectory are reachable in time $< \tau$, naturally).However, I am unable to show why there must be points which are not reachable in time $\tau$ in a neighbourhood of $0$. This is because I am unable to use the nature of the given set to my benefit.

Surely if the question is trivial, then I am missing something. Do guide me across this one.

• Introduce an additional state variable, $y$, which behaves like time: $$\dot{y} = 1.$$ Now your set of all states reachable in time $T$ is simply the subset $y=\tau$ of the new state space. Now show that $y=\tau$ is a sufficiently smooth hypersurface in your (new) state space (hint: it is a hyperplane:), which "cuts" the neighborhood of $0$ you mentioned into 2 parts. – avs Feb 7 '17 at 0:32
• @avs Yes, I agree with you. However, I am unable to see why it should be a hyperplane. $y = \tau$ is cleary a hyperplane in the state space, however the transformation back to the original space may not be linear, right?(If what you are saying is right,which it should be, then we are done). Please excoriate me if I am making any comical/obvious statements in this comment. – астон вілла олоф мэллбэрг Feb 7 '17 at 0:38
• I may be mistaken (sorry, don't have time to get into it), but I was thinking, the transformation back to the original space is the orthogonal projection onto the $x$-coordinates. If so, this projection is a linear transformation. – avs Feb 7 '17 at 0:53
• @avs I'm unable to prove that's true, because the nature of "reachability" doesn't give us the flexibility to say the transformation is linear, because $\alpha$ is only given to be measurable. If you can come up with a more definitive explanation (that is, whenever you have the time to get to it) then I would be happy to hear. I do not wish to distract you, and sometimes a discussion is more enlightening than an answer. Thank you. – астон вілла олоф мэллбэрг Feb 7 '17 at 0:57
• I would experiment with some examples with $n \leq 2$ and $m \leq 2$, just to get insight. $\alpha$ parameterizes the vector fields on the state space, not the state space itself (sorry if I am saying what you already knew). Good luck. – avs Feb 7 '17 at 1:15

It is easier to argue if we introduce the following set $$R_T=\left\{\int_0^T X^{-1}(s) N\alpha(s)\,ds\right\}.$$ Then $$K(T,x_0)=X(T)(x_0+R_T)$$ and it is quite obvious that \begin{align} 0\not\in K(t,x_0)\quad&\Leftrightarrow\quad -x_0\not\in R_t,\\ 0\in \partial K(\tau,x_0)\quad&\Leftrightarrow\quad -x_0\in \partial R_{\tau}. \end{align} The sets $R_t$ are convex and closed as well, but they are ordered by inclusion as $$t_1\le t_2\quad\Rightarrow\quad R_{t_1}\subset R_{t_2}$$ since $\alpha=0$ is allowed. The sets are also continuous: for any $\epsilon>0$ there exists $t<\tau$ such that $$R_t\subset R_{\tau}\subset R_t+\epsilon B_1$$ where $B_1$ is the unit ball. It is because the integral $$\int_0^t X(s)^{-1}N\alpha(s)\,ds$$ is continuous in $t$ (actually equicontinuous for all $\alpha$ in our class).

To prove that $-x_0$ is the boundary point we can take a ball $B$ around it of an arbitrary radius $\mu>0$ and show that there are points in it that are outside of $R_{\tau}$.

The sets $R_t$ are convex and closed. Moreover, the center point $-x_0\not\in R_t$ for $t<\tau$, so one can separate the center of the ball from $R_t$ by a hyperplane. Thus the volume of $R_t$ inside the ball is less than half of the ball volume. By continuity we can find $t<\tau$ such that $$R_{\tau}\subset R_t+\frac{\mu}{4}B_1$$ and $R_t+\frac{\mu}{4}B_1$ not filling up the whole ball $B$, leaving the top quarter for points outside $R_{\tau}$.

• Thank you very much for your answer! I shall get back to you if I have any doubts. – астон вілла олоф мэллбэрг Mar 9 '17 at 23:26
• Here is my first doubt: Why is it obvious that $0 \in \partial K(\tau, x_0) \iff -x_0 \in \partial R_\tau$? – астон вілла олоф мэллбэрг Mar 10 '17 at 3:07
• The mapping $x\mapsto X(x_0+x)$ with an invertible $X$ is a homeomorphism, so a neighbourhood of $-x_0$ maps to a neighbourhood of $0$ and vice versa. It gives that if $-x_0$ is an interior point of $R$ then $0$ is an interior point of $K$, if $-x_0$ is a interior point of ${\Bbb R}^n\setminus R$ then $0$ is an interior point of ${\Bbb R}^n\setminus K$ and so on. – A.Γ. Mar 10 '17 at 5:09
• Yes, that's right. Thank you for your excellent response to this question, in honesty I have been struggling with it for some time. I shall get back to you in some time if I have further questions. – астон вілла олоф мэллбэрг Mar 11 '17 at 3:10
• Your questions are welcome. I'm used to work with control, but it has passed some time. – A.Γ. Mar 11 '17 at 7:37

Sorry, I'm not an expert. However, I try to give you my insight with the hope that it helps.
In order to check whether $0$ is a boundary point of $K(\tau,x_0)$, it suffices to show that any neighbourhood of $0$ contains a point in $K(\tau,x_0)$ and a point outside. Let's fix $\epsilon>0$ and take the open ball $B:=\{x\in\mathbb R^n \,:\, \|x\|<\epsilon\}$. Since $0\in K(\tau,x_0)$, there exists $\alpha:[0,\tau]\to[-1,1]^m$ so that $$0 = X(\tau)x_0 + X(\tau)\int_0^\tau X^{-1}(s)N\alpha(s)ds$$ For any $T<\tau$ take the state $$x(T) = X(T)x_0 + X(T)\int_0^T X^{-1}(s)N\alpha(s)ds$$ Then $x(T)$ is in $K(\tau,x_0)$, since it is in $K(T,x_0)$.
Let's choose $T$ so that $x(T)\in B$. That is possible, since, once fixed $x_0$, $t\mapsto x(t)$ is differentiable and thus locally Lipschitz. Hence we can bound $\|x(\tau)-x(T)\|$ as $$\|x(\tau)-x(T)\|\le L\|\tau-T\|$$ with $L$ that depends on $\tau$. We can thus chose $T$ such that $\|x(\tau)-x(T)\|<\epsilon$. This proves that in the $\epsilon$-neighbourhood of $x(\tau)$ there exists a point in $K(\tau,x_0)$. Take now $T>\tau$ and $\alpha'(s)$ that is any properly defined non-zero function that agrees with $\alpha$ on $[0,\tau]$. then $x(T)\notin K(t,x_0)$, for all $t\le \tau$, otherwise $x(\tau)$ would be reachable in less that $\tau$ seconds, that violates the hypothesis. Using the same argument as before you can chose $T>\tau$ so that $x(T)$ is in the $\epsilon$-neighbourhood of $x(\tau)$, thus proving that in such neighbourhood there exists a point that does not belong to $K(\tau,x_0)$. At this point, the claim should follow from the arbitrariness of $\epsilon$.
Maybe this is not a rigorous proof, but I hope that it can be an idea.

• It may not be rigorous, but I think the answer qualifies as an excellent one nevertheless. Thank you very much for your input. – астон вілла олоф мэллбэрг Feb 8 '17 at 23:16
• Cool. Glad it helped – LJSilver Feb 8 '17 at 23:17
• You are welcome! – астон вілла олоф мэллбэрг Feb 8 '17 at 23:18
• It is not necessarily true that $K(T,x_0)$ is a subset of $K(τ,x_0)$ for $T<τ$, However $X(T)^{-1}K(T,x_0)\subset X(τ)^{-1}K(τ,x_0)$. – LutzL Feb 9 '17 at 16:16
• @LutzL I'm responding late, my apologies. Why is it not true that $K(T,x_0) \subset K(\tau,x_0)$? We can just extend the control which steers to $0$ at time $T$, by a constant function zero. From the equation described, you can see that this doesn't affect the final constant. – астон вілла олоф мэллбэрг Feb 16 '17 at 6:47


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# A local firm manufactures LED products that have a lifespan that is approximately normally distributed with a std. dev. of 30 hours. If a sample of 30 LED products has an average lifespan of 780 hours, find a 96\% confidence interval for the population mean of all LED products produced by this firm.

A local firm manufactures LED products that have a lifespan that is approximately normally distributed with a std. dev. of 30 hours. If a sample of 30 LED products has an average lifespan of 780 hours, find a $96\mathrm{%}$ confidence interval for the population mean of all LED products produced by this firm.
Choose 2 answers in nearest unit (ones) or in whole number.
Example, if your answer is $888.83\le \mu \le 899.56$, choose 889 and 900.
$\begin{array}{|cccccccccccc|}\hline 775& 773& 807& 797& 791& 769& 789& 768& 805& 763& 771& 792\\ \hline\end{array}$

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Step 1
It is given that, and $\sigma =30$
At $96\mathrm{%}$ significance level, the value of z is 2.0537.
Then, the $96\mathrm{%}$ confidence interval is
$\stackrel{―}{x}±z\left(\frac{\sigma }{\sqrt{n}}=780±2.0537\left(\frac{30}{\sqrt{30}}\right)$
$=780±2.0537\left(5.4772\right)$
$=780±11.2486$

Rounded
Step 2
Therefore, 769 and 791 are the correct options.