text:
# Westinghouse Bankruptcy and the Future of Nuclear Power

Scana Corp the owner of South Carolina Gas and Electric has decided to scrap the reactors partially completed due to the bankruptcy of Westinghouse.

https://www.nytimes.com/2017/07/31/climate/nuclear-power-project-canceled-in-south-carolina.html

Included in the decision was the fact that electricity demand is not increasing and cheap natural gas. SC utilities still has coal fired plants which were to be closed on completion of this project.

russ_watters
Mentor
Scana Corp the owner of South Carolina Gas and Electric has decided to scrap the reactors partially completed due to the bankruptcy of Westinghouse.

https://www.nytimes.com/2017/07/31/climate/nuclear-power-project-canceled-in-south-carolina.html

Included in the decision was the fact that electricity demand is not increasing and cheap natural gas. SC utilities still has coal fired plants which were to be closed on completion of this project.
Disappointing, but instructive economics lesson: it shows that in a well integrated market where there is one final price for a product, Product A (natural gas electricity) can undercut both Products B (nuclear electricity) and C (coal electricity) while having the even bigger impact of altering the balance (or rather, reversing the trend in the balance) between B and C. Counter-intuitively, it means in this case that more future natural gas power leads to more future coal power (less of a drop).

Astronuc
Staff Emeritus
Scana Corp the owner of South Carolina Gas and Electric has decided to scrap the reactors partially completed due to the bankruptcy of Westinghouse.
It is more the case that SCANA and Santee abandoned the plant construction over the delays and coast.
Originally scheduled to come online by 2018, the V.C. Summer nuclear project in South Carolina had been plagued by disputes with regulators and numerous construction problems. This year, utility officials estimated that the reactors would not begin generating electricity before 2021 and could cost as much as $25 billion — more than twice the initial$11.5 billion estimate.
Another article in the NY Times mentions some of the problems that caused increase in cost.

The projects, more than three years late and billions over budget, are what pushed Westinghouse — one of the last private companies building nuclear reactors — and its parent, Toshiba, to the brink of financial ruin, resulting in Toshiba’s chairman stepping down.
Under the old system, companies received construction permits based on incomplete plans and then applied for an operating license, often leading to rebuilding and lengthy delays. The idea for the new system was that companies would submit much more complete design plans for approval, and then receive their operating licenses as construction started. That way, as long as they built exactly what they said they would, the process could move more quickly.
I know that one concern some in the nuclear industry had was that the industry had lost the expertise in manufacturing large components and larger projects.
Because nuclear construction had been dormant for so long, American companies lacked the equipment and expertise needed to make some of the biggest components, like the 300-ton reactor vessels. Instead, they were manufactured overseas, adding to expense and delays.

One reactor vessel, headed for Georgia Power’s Vogtle plant from the Port of Savannah, almost slipped off a specialized rail car. That led to a weekslong delay before a second attempt was made to deliver it.

And, in a separate snafu, while working on the plant’s basement contractors installed 1,200 tons of steel reinforcing bar in a way that differed from the approved design. That triggered a seven-and-a-half month delay to get a license amendment.
I certainly remember when the delivery of Vogtle's RPV went awry.

As far as I know, the regulatory issues stem from folks not doing the work according to plans, or changes made somewhere that conflict with the approved design basis. It takes time and money to resolve those issues, and they are cumulative.

I cited two articles:
https://www.nytimes.com/2017/07/31/climate/nuclear-power-project-canceled-in-south-carolina.html
https://www.nytimes.com/2017/02/18/...nment/nuclear-power-westinghouse-toshiba.html

While the first article mentions Summer is 40% complete, I read elsewhere that the plant or perhaps lead unit is 64% complete.

The liabilities for Westinghouse and Toshiba are astounding, and the cost overruns (the cost of finishing the project is another $14 billion) takes the plant from$11.5 billion to an estimated $25 billion. In the face of a decrease in the demand for electricity and relatively inexpensive natural gas, SCANA and Santee feel compelled to abandon the project. mheslep mheslep Gold Member I've not seen any satisfactory explanation for the amount of the cost increase reported for Summer, from$11 to $25B, for a 40% complete plant? mathman Science Advisor Two projects using AP1000 (in South Carolina) were just cancelled. Astronuc Staff Emeritus Science Advisor I've not seen any satisfactory explanation for the amount of the cost increase reported for Summer, from$11 to $25B, for a 40% complete plant? SCE&G and Santee apparently did a study. They've been dealing with delays and cost increases for years now, and perhaps the coup de grace is the Westinghouse bankruptcy. I posted about a Power Magazine article on the Summer NPP in the Nuclear Power Thread. According to the Power Magazine article, the Summer project is 64% complete based on a statement from SCE&G. http://www.powermag.com/vc-summer-project-64-complete-sceg-says/ http://www.powermag.com/scana-santee-cooper-abandon-v-c-summer-ap1000-units-citing-high-costs/ SCANA Corp. and Santee Cooper have ceased construction of Units 2 and 3 at the V.C. Summer Nuclear Station in South Carolina. The project owners said the decision, prompted by analysis of detailed schedule and cost data, would save customers nearly$7 billion. The project, which was about 64% complete, has been in limbo since key contractor Westinghouse filed for bankruptcy in March.
The escalating costs have been in the news for a while now.
http://www.northaugustastar.com/new...cle_05f1095c-160e-11e7-a2ea-2babdc28c02e.html

Part of the problem of estimating the cost will involve assumptions about idle workers and equipment and manhours needed to complete tasks.

Just prior to the beginning of construction, there was a concern that the available workforce had no prior experience in constructing nuclear power plants which have to be done under tighter quality control than non-nuclear plants. That makes changing the design costly, as well as remediating mistakes.

It was a costly mistake for Westinghouse to try to absorb CBI.
http://www.reuters.com/article/us-toshiba-accounting-westinghouse-cbi-idUSKBN17Z2AF
2017 - http://investors.cbi.com/news/press...Position-in-Westinghouse-Lawsuit/default.aspx

etudiant
Gold Member
I suspect there may be a domino effect, as the Vogtle plants are now the last two under construction in the US. That means every nuclear construction supervisor and regulator will have only Vogtle to focus on.
Ideally, this might mean the best and most experienced would now try 100% to help this project succeed. Realistically the more likely outcome is 'too many cooks'.
Nuclear may have a rebirth in this country, but it will need a very different form. We don't have the management or regulatory competency required to build big plants anymore. So the future if any is smaller units, probably centrally produced. Until that gets going, another long hiatus such as post TMI awaits us, at least imho.

Astronuc
Staff Emeritus
The situation with Westinghouse and the new AP1000 builds makes me wonder, if we can build nuclear powered aircraft carriers and submarines, why can't we be successful at building NPPs?

mheslep
russ_watters
Mentor
The situation with Westinghouse and the new AP1000 builds makes me wonder, if we can build nuclear powered aircraft carriers and submarines, why can't we be successful at building NPPs?
It's because ships are built in dedicated, purpose-built, high-tech factories and nuclear power plants are just buildings -- and any idiot can pour concrete or weld a pipe or lay a beam where a blueprint says to.....except that they can't. If a contractor screws up on building an apartment building, the engineer complains and makes him fix it if it is important or just accepts it and docks their fee if it isn't, and that's the end of it. If a contractor screws up building a nuclear plant, it is quite literally a federal case. In the non-nuclear world, nobody really cares about bad concrete pours until after it fails and a panel falls out of the ceiling of your \$30 billion tunnel in Boston and kills a family driving under it.

and any idiot can pour concrete or weld a pipe or lay a beam where a blueprint says to.....except that they can't.
or don't. Back in the beginning of NP in the 60's I read that a new plant under construction had to be stopped because the steel reinforcing rods where not be laid according to the schedule. The contractor unfamiliar with this new type of construction thought that it was way over designed at cut back on the steel.

In the case of the Scana plant they too did not follow the plans delaying the construction. What is it that they can't follow directions.

russ_watters
mheslep
Gold Member
The situation with Westinghouse and the new AP1000 builds makes me wonder, if we can build nuclear powered aircraft carriers and submarines, why can't we be successful at building NPPs?
US NRC. The NRC does not, can not, impose new requirements on US naval nuclear vessels after construction begins. Imagine 'aircraft impact' containment requirements imposed naval vessels.

mheslep
Gold Member
It's because ships are built in dedicated, purpose-built, high-tech factories and nuclear power plants are just buildings -- and any idiot can pour concrete or weld a pipe or lay a beam where a blueprint says to.....except that they can't.
Which does not explain how China, S. Korea builds large reactors in building mode in five to seven years for a fraction of the cost of new US reactors, or how the US itself built them cheaply 40 years ago. Differences in labor costs does not explain all.

Last edited:
US NRC. The NRC does not, can not, impose new requirements on US naval nuclear vessels after construction begins. Imagine 'aircraft impact' containment requirements imposed naval vessels.
It's my understanding that the NRC has no jurisdiction over government controlled reactors or nuclear material.

mheslep
Gold Member
except that they can't....
One such problem is loss of skills, like the near miss Astro reported. Another occurred with the forging of the big RPV. Apparently, in big pieces the carbon concentration goes wrong on one end of the piece. The solution, going back to cannon age, was to lop off the end. This time, they forgot.

russ_watters
Mentor
Which does not explain how China, S. Korea builds large reactors in building mode in five to seven years for a fraction of the cost of new US reactors, or how the US itself built them cheaply 40 years ago. Differences in labor costs does not explain all.
My response would probably be too political, though...
One such problem is loss of skills...
Yes, I think that is a lot of it.

jim hardy


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# The Azimuth Project Burgers' equation (Rev #4, changes)

Showing changes from revision #3 to #4: Added | Removed | Changed

# Contents

## Idea

Burger’s equation is a partial differential equation that was originally proposed as a simplified model of turbulence as exhibited by the full-fledged Navier-Stokes equations. It is a nonlinear equation for which exact solutions are known and is therefore important as a benchmark problem for numerical methods.

In one spatial dimension it is

$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} - \nu \frac{\partial^2 u}{\partial x^2} = 0$

in a spatial region $\Omega$, positive time $t \gt 0$ and with a real constant $\nu \gt 0$, with appropriate boundary and initial conditions.

We write for the initial condition:

$u(x, o) = u_0(x)$

It is possible to write Burgers’s equation in a conservation form (with a flow that is conserved):

$\frac{\partial u}{\partial t} + \frac{\partial F(u)}{\partial x} = 0$

with

$F(u) := \frac{1}{2} u^2 - \nu \frac{\partial u}{\partial x}$

## Details

### Transformation to the Heat Equation

It is possible to transform Burgers’ equation to the heat equation via

$u = -2 \nu \frac{\phi_x}{\phi}$

so that the function $\phi$ satisfies

$\frac{\partial \phi}{\partial t} - \nu \frac{\partial^2 \phi}{\partial x^2} = 0$

iff the function $u$ satisfies Burgers’ equation.

### Approximation via Fourier-Galerkin Spectral Method

We will use a spectral method for an approximate solution of Burgers’ equation on the domain $\Omega = (0, 2 \pi)$. “Fourier” means that we will use an approximation via a Fourier series, and “Galerkin” means that we will use the approximation functions also as test functions. As usual, we will use the spectral approximation for the spatial dimension only, not for the temporal. This results in an ansatz that is usually called “separation of variables”. One could indeed call “separation of variables” a special case of spectral methods.

Our ansatz for the approximate solution is:

$u_N (x, t) = \sum_{k = - \frac{N}{2}}^{k = \frac{N}{2} - 1} \hat{u}_n (t) e^{i k t}$

Since we choose as test functions our approximation functions, the conditions resulting from the requirement that $\langle \; M(f_{\alpha}), \; h_i \; \rangle = 0$ (see spectral methods for an explanation of the nomenclature) results in our case in the system of equations

$\int_0^{2 \pi} (\frac{\partial u_N}{\partial t} + u_N \frac{\partial u_N}{\partial x} - \nu \frac{\partial^2 u_N}{\partial x^2} ) \; e^{i k t} \; d x = 0$

for $k = - \frac{N}{2}...\frac{N}{2} - 1$ .

This results in a system of ordinary differential equations for the Fourier components $\hat{u}_n$.

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# The figure shows graphs of $f$, $f'$, $f''$, and $f'''$. Identify each curve, and explain your choices.

## $f$ is d, $f^{\prime}$ is $c, f^{\prime \prime}$ is $b,$ and $f^{\prime \prime \prime}$ is a.

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### Discussion

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### Video Transcript

in this problem. It is given that the figure shows graphs of F F dash double dash and F double dash identify each girl and explain your choices. Okay, so here we can see A D is the function of if because we can just see one thing in differentiation, what happens is to understand it we will just get into there should be a maximum and minimum points. Right? So maximum and minimum points are found in the CO D. You can see maximum and minimum points. Uh huh. Found in. Mhm Go deep. That means this means D is a function of that function graph Yeah. So, now um what is less than we are just we are just writing in the decreasing order. Right. Mhm. Mhm So now here we have two points. Right, two points. We are getting here for grab. So obviously where we can get f dash request to zero. So there must be two points again. So only two points are available in Kirby cough c so that's right. F dash zero. I'm sorry. This is f dash at dash. We are considering the function graph. Right? So F dash, mm hmm. Shown by graph. Okay, seeing you graph syria Okay. Mhm No, again, if we will differentiate again. So F double dish will be decided. Bye. Mhm Better. The function is one gave up. Yeah. What on game don't Yeah, why? Because this is in graph. See this is decreasing first and it is increasing first. So after differentiation, the graph should increase first, concave up. That is and it should decrease then we can say this uh F dash, double dash, therefore if double dash is represented us. Yeah, graph, uh huh, graphs go B. Sanjay. So moving ahead. Right, we can say the only remaining remaining graph fish E, which is differentiation of triple differentiation of F. Right? So here we can easily write it like this. He is, you can see you this must be written like this, he is nothing. But this is F. Triple dash B is is double dash, no, see is have dash and D is Mhm. Function that that is. So this is how we calculate this graph. Oh, that's good. I hope your industry the concept. Thanks for watching.

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# Using .Net link with fortran DLL created by Intel Fortran [closed]

I am following unstable's wonderful answer on .NET and fortran.

I can reproduce his result using gfortran without problem. But I want to use Intel Fortran, because gfortran generates slower code on my computer.

All the other steps are the same, only the dll is generated by the following command

ifort /dll testfunction.f90

After I install .NET and load the dll, running TestFunction[{1.0, 2.0}, 3.0] gives me errors.

NET::netexcptn: A .NET exception occurred: System.EntryPointNotFoundException: 无法在 DLL"D:\w\WORK\mathematica\NET\testfunction.dll"中找到名为"testfunction_"的入口点。 在 Wolfram.NETLink.DynamicDLLNamespace.DLLWrapper1.testfunction_(Double[] , Double& ).

The Chinese part seems to say that it can't find the entrance point named "testfunction_"

How can I solve this?

• Different Fortran compilers follow different name mangling specifications. This is most likely the problem you face here. – Oleksandr R. Jan 17 '16 at 16:45
• @OleksandrR. Hi,OleksandrR. I know intel fortran will uppercase all the subroutine name. But change the name also didn't work. How to solve this? – matheorem Jan 18 '16 at 5:32


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# Revision of omxRAMtoML from Tue, 11/01/2011 - 12:18

Revisions allow you to track differences between multiple versions of your content, and revert back to older versions.

# omxRAMtoML

### list: ol; title: Table of Contents; minlevel: 1; maxlevel: 4; attachments: 1;

#### Usage

This function accepts a model as input, and returns a model as output where all the RAM objective functions have been transformed to either ML or FIML objective functions, where the necessary algebras have been auto-generated in order to calculate the expected covariance and means matrices.

omxRAMtoML(model)

#### Arguments

 model a MxModel object.

#### Examples

# Read libraries and set options.

require(OpenMx)

# ----------------------------------
# Read the data and print descriptive statistics.

data(factorExample1)

# ----------------------------------
# Build an OpenMx single factor FIML model with fixed variance

indicators <- names(factorExample1)
latents <- c("F1")
uniqueLabels <- paste("U_", indicators, sep="")
meanLabels <- paste("M_", indicators, sep="")
factorVarLabels <- paste("Var_", latents, sep="")

oneFactorRaw1 <- mxModel("Single Factor FIML Model with Fixed Variance",
type="RAM",
manifestVars=indicators,
latentVars=latents,
mxPath(from=latents, to=indicators,
#           arrows=1, all=TRUE,
arrows=1, connect="all.pairs",
free=TRUE, values=.2,
mxPath(from=indicators,
arrows=2,
free=TRUE, values=.8,
labels=uniqueLabels),
mxPath(from=latents,
arrows=2,
free=FALSE, values=1,
labels=factorVarLabels),
mxPath(from="one", to=indicators,
arrows=1, free=TRUE, values=.1,
labels=meanLabels),
mxData(observed=factorExample1, type="raw")
)
oneFactorRawML <- omxRAMtoML(oneFactorRaw1)
oneFactorRawMLOut <- mxRun(oneFactorRawML, suppressWarnings=TRUE)

# See the results...
summary(oneFactorRawMLOut)
data:
$Single Factor FIML Model with Fixed Variance.data x1 x2 x3 x4 Min. :-2.99780 Min. :-1.579400 Min. :-2.13250 Min. :-3.00650 1st Qu.:-0.62555 1st Qu.:-0.365850 1st Qu.:-0.26977 1st Qu.:-0.69588 Median :-0.03170 Median : 0.007300 Median : 0.05055 Median :-0.04330 Mean :-0.01161 Mean :-0.006821 Mean : 0.02396 Mean :-0.03135 3rd Qu.: 0.59815 3rd Qu.: 0.333675 3rd Qu.: 0.33495 3rd Qu.: 0.68142 Max. : 2.54270 Max. : 1.800600 Max. : 1.26530 Max. : 2.88340 x5 x6 x7 x8 Min. :-3.20380 Min. :-3.54670 Min. :-4.15680 Min. :-2.05160 1st Qu.:-0.71252 1st Qu.:-0.98603 1st Qu.:-1.07967 1st Qu.:-0.64263 Median :-0.02015 Median :-0.07750 Median :-0.14610 Median :-0.05310 Mean :-0.04548 Mean :-0.09178 Mean :-0.06732 Mean :-0.03902 3rd Qu.: 0.62877 3rd Qu.: 0.77910 3rd Qu.: 0.91097 3rd Qu.: 0.58552 Max. : 2.85080 Max. : 3.26040 Max. : 3.74800 Max. : 2.63280 x9 Min. :-3.68950 1st Qu.:-0.83327 Median :-0.04285 Mean :-0.05999 3rd Qu.: 0.72447 Max. : 3.47750 free parameters: name matrix row col Estimate Std.Error lbound ubound 1 b_x1 A x1 F1 0.68395558 0.03517218 2 b_x2 A x2 F1 0.32481984 0.02238500 3 b_x3 A x3 F1 0.10886694 0.02076627 4 b_x4 A x4 F1 0.47440890 0.04457067 5 b_x5 A x5 F1 0.60180412 0.04221052 6 b_x6 A x6 F1 1.12063877 0.04569668 7 b_x7 A x7 F1 1.25933139 0.04883099 8 b_x8 A x8 F1 0.64739267 0.03057637 9 b_x9 A x9 F1 0.71872734 0.04926900 10 U_x1 S x1 x1 0.35279611 0.02484526 11 U_x2 S x2 x2 0.17619283 0.01193414 12 U_x3 S x3 x3 0.19353556 0.01230270 13 U_x4 S x4 x4 0.79987497 0.05201061 14 U_x5 S x5 x5 0.63305704 0.04272612 15 U_x6 S x6 x6 0.36762720 0.03207912 16 U_x7 S x7 x7 0.34023767 0.03483551 17 U_x8 S x8 x8 0.23403773 0.01730076 18 U_x9 S x9 x9 0.85441146 0.05777368 19 M_x1 M 1 x1 -0.01161303 0.04050289 20 M_x2 M 1 x2 -0.00682285 0.02373273 21 M_x3 M 1 x3 0.02396104 0.02026669 22 M_x4 M 1 x4 -0.03135672 0.04527295 23 M_x5 M 1 x5 -0.04548168 0.04460893 24 M_x6 M 1 x6 -0.09178376 0.05696533 25 M_x7 M 1 x7 -0.06732317 0.06204879 26 M_x8 M 1 x8 -0.03902037 0.03613431 27 M_x9 M 1 x9 -0.05999675 0.05235709 observed statistics: 4500 estimated parameters: 27 degrees of freedom: 4473 -2 log likelihood: 9706.388 saturated -2 log likelihood: NA number of observations: 500 chi-square: NA p: NA Information Criteria: df Penalty Parameters Penalty Sample-Size Adjusted AIC 760.3878 9760.388 NA BIC -18091.5542 9874.182 9788.483 CFI: NA TLI: NA RMSEA: NA  Please add material here as you learn... If you have questions not answers, then add those here: That's how a wiki works. ## Comments ### utilization of created matrices How can one use the matrices created by omxRAMtoML with the mxFIMLObjective? mraw is my model as RAM. When I use mrawML$covariance@name and mrawML$covariance@name as expected covariance and expected means, respectively, I get the error at the end of this post: > mrawML <- omxRAMtoML(mraw) > FIMLobjective <- mxFIMLObjective(covariance=mrawML$covariance@name,means=mrawML\$means@name,dimnames=Namen)
>
> mrawMLwithFIML <- mxModel('Modell SES FIML', mraw, FIMLobjective)
>
> FitrawFIML <- mxRun(mrawMLwithFIML)
Running Modell SES FIML
Error: Unknown expected covariance name 'covariance' detected in the objective function of model 'Modell SES FIML'
>

Thanks in advance for any help!

### In the future, I recommend

In the future, I recommend posting your questions to the forums. New forum posts are noticed much faster than new comments to a wiki page. With regards to the issue at hand, you'll need to include the original model as the 1st argument to the mxModel() function in order to modify the existing model. As currently specified, you are constructing a new model with the objective function and data but it is missing all of the other matrices and algebras. Use the function mxRename() to give the model a new name, it will correct any internal references from the old name to the new name. I would write it as follows:

mrawMLwithFIML <- mxModel(mrawML, mraw, FIMLobjective)
mrawMLwithFIML <- mxRename(mrawMLwithFIML, 'Modell SES FIML')

In playing around with this example, I found that omxRAMtoML() has a bug and incorrectly throws an error if the input model has data with type='cov'. This will be corrected in OpenMx 1.2.5.


text:
# One-sided Bound on Sum of Fourth Moments

I'm interested in methods for proving one-sided bounds of the form $$\mathbb{P}[\frac{1}{n}\sum_{i=1}^n X^4_i \geq 3+t]\leq Ce^{-nt}$$ where $X_i$ are standard normal random variables. I've run a few basic random experiments, it seems that this should be possible for t relatively large (but independent of $n$). Note that I don't care about concentration, just a loose upper bound.

Is this possible?

• Did you mean $\geq 3+t$? As it is written, the inequality doesn't make sense: consider very large $t$ for example. The LHS goes to $1$ whereas the RHS goes down to $0$. – Kim Jong Un Jul 23 '15 at 0:20
• Also as $n \to \infty$, since $\mathbb E[X_i^4] = 3$, the left side goes to $1$ for any fixed $t > 0$. $\mathbb P[n^{-1} \sum_{i=1}^n X_i^4 \ge 3 + t]$ does decay exponentially by Cramér's theorem from Large Deviations theory. – Robert Israel Jul 23 '15 at 0:40
• Hi @RobertIsrael, would you mind elaborating on the exponential decay? I'm not familiar with Cramer's Theorem, and the few things I found online don't state the theorem in a clear enough form (to me at least). – squattyroo Jul 23 '15 at 1:17
• @RobertIsrael The random variables $X^4$ have no finite exponential moments hence Cramér needs at least some tweaking before being an option. – Did Jul 23 '15 at 1:19
• The result is false as $t\to\infty$ even for $n=1$. – user940 Jul 23 '15 at 2:46


text:
# Editing custom XML part in word document sometimes corrupts document

We have a system that stores some custom templating data in a Word document. Sometimes, updating this data causes Word to complain that the document is corrupted. When that happens, if I unzip the docx file and compare the contents to the previous version, the only difference appears to be the expected change in the customXML\item.xml file. If I re-zip the contents using 7zip, it seems to work OK (Word no longer complains that the document is corrupt).

The (simplified) code:

void CreateOrReplaceCustomXml(string filename, MyCustomData data)
{
using (var doc = WordProcessingDocument.Open(filename, true))
{
var part = GetCustomXmlParts(doc).SingleOrDefault();
if (part == null)
{
}

var serializer = new DataContractSerializer(typeof(MyCustomData));
using (var stream = new MemoryStream())
{
serializer.WriteObject(stream, data);
stream.Seek(0, SeekOrigin.Begin);
part.FeedData(stream);
}
}
}

IEnumerable<CustomXmlPart> GetCustomXmlParts(WordProcessingDocument doc)
{
return doc.MainDocumentPart.CustomXmlParts
.Where(part =>
{
using (var stream = doc.Package.GePart(c.Uri).GetStream())
{
}
});
}

Any suggestions?

• Would it possible to post a reproducing sample? – Simon Mourier May 19 '14 at 6:46

Since re-zipping works, it seems the content is well-formed.

So it sounds like the zip process is at fault. So open the corrupted docx in 7-Zip, and take note of the values in the "method" column (especially for customXML\item.xml).

Compare that value to a working docx - is it the same or different? Method "Deflate" works.

• the same - everything is "Deflate" – Jon May 19 '14 at 0:42
• OK, good. I'd focus on your C# code then. Does the problem occur when the update makes the data shorter, or longer, or is there no pattern? Is the packed size (reported in 7-Zip) the same before and after re-zip? – JasonPlutext May 19 '14 at 7:53
• this helped me figure out my problem, so happy to award it with rep :) My actual problem was to do with writing the finished word document to file... – Jon May 27 '14 at 1:20
• thanks Jon :-) glad you worked it out! – JasonPlutext May 28 '14 at 5:05

I faced the same issue and it turned out it was due to encoding. Do you already specify the same encoding when serializing/deserializing?

Couple of suggestion a. Try doc.Package.Flush(); after you write the data back into the custom xml. b. You may have to delete all custom part and add a new custom part. We are using the following code and it seems working fine.

public static void ReplaceCustomXML(WordprocessingDocument myDoc, string customXML)
{

MainDocumentPart mainPart = myDoc.MainDocumentPart;
mainPart.DeleteParts<CustomXmlPart>(mainPart.CustomXmlParts);
using (StreamWriter ts = new StreamWriter(customXmlPart.GetStream()))
{
ts.Write(customXML);
ts.Flush();
ts.Close();
}
}

public static MemoryStream GetCustomXmlPart(MainDocumentPart mainPart)
{
foreach (CustomXmlPart part in mainPart.CustomXmlParts)
{
{
{
byte[] byteArray = Encoding.ASCII.GetBytes(str);
MemoryStream stream = new MemoryStream(byteArray);

return stream;
}
}
}

return null; //result;
}

using (WordprocessingDocument myDoc = WordprocessingDocument.Open(ms, true))
{


text:
# The area of a circle is 2826 square inches. What is the length of its radius?

Mar 19, 2018

#### Explanation:

A is area (D is diameter)

$A = \pi \times {D}^{2} / 4 = 2826$

${D}^{2} = \frac{11304}{\pi} ^$

${D}^{2} = 3598$ square inches

$D = 59.985$ inches


text:
# Migrating standalone Mongo to Atlas

Some time ago I worked on a very interesting project, which was a bit different from all the other things a was used to do. It was about migrating our production database from a standalone self-hosted MongoDB instance to one of the cloud providers.

### Evaluating our options

At first, I summarized different options that we could go with. They included continuing using a self-hosted cluster as well as moving to any of the well known cloud DBaaS providers. We chose to go with Mongo Atlas as it suited our needs and the price was acceptable.

One of the most important things I learned at this stage was that it is a real pain to run your own MongoDB cluster, especially without having a dedicated DevOps engineer.

### Preparing the application

Long before the actual migration began, I started figuring out how the process would go to have close to zero down time, and as little restarted jobs as possible.

The application was deployed in Docker containers on AWS ECS. Most of its components were very easy to restart without losing anything, but there also were some long running tasks (jobs) that we would like to keep running no matter what. A sudden disconnect from our database could mean failing some of them, if the new replica set would not become available in a very short time. To ensure that the connections will have enough time to restore, our system was adjusted to retry database connections for a much longer time before failing, than before. This should not be the case in a stable production environment since problems should become clear as fast as possible.

### Testing in staging environment

It was very clear that such a migration must be rehearsed. I learned a lot from migrating a standalone MongoDB instance in staging environment to Atlas, and surely felt more confident when doing the same things again in production. You should become very familiar with the process before you touch your production database.

### Cluster preparation

Migration process could be time consuming, especially if something went wrong. That’s why any task that could be completed separately in advance, should have been done as early as possible in the process. Setting up a cluster in Atlas is a pretty easy task, but there are a lot of things to test. Did we add all the addresses to IP white list in AWS? Did we create a user with enough permissions for Atlas Migration Tool? Are the new instances accessible from our app? Is VPC peering working properly?

Once the answers to all these, and some other questions, were positive, the process could be started.

### Migration from a standalone instance to Mongo Atlas replica set

At first, I needed to convert our standalone MongoDB deployment to a single node replica set, since it was impossible to migrate a standalone instance to Atlas directly. This step is covered in most of the guides:

# /etc/mongod.conf
replication:
replSetName: rs0

followed by service restart. This part is quick and causes very little down time. Our app knew how to handle short database hiccups, so that was fine. Since it was a new replica set, another command was required: rs.initiate(), when running MongoDB shell as admin (mongo -u admin --authenticationDatabase admin admin --password). It takes very little time as well, but you need to initiate a new replica set as close to service restart as possible.

At this point, everything was ready for Live Import. Our app had a fairly small database, so this step did not take too much time. While it was happening, we constantly monitored our application and made sure that it works properly.

Another test we decided to run while having some time during data import was to create a very small application using the same stack as our main app, and make sure the exact code we use in production was able to communicate with a new server. That’s how we ensured that drivers we used were compatible with the MongoDB version that Atlas had.

#### Cutoff

When cutoff option became available in Atlas, we updated the application to use a new connection path. All the tasks that started after this point were using Atlas. All the other tasks were still using our old database. It was OK, since live import was still running, and it was able to catch up with all the changes on the legacy instance almost instantly.

We restarted almost all containers with the new code. ECS made this part easy. There were still some jobs using the old database though. We could tell about some of them that they were supposed to take a long time to complete; these jobs were killed and restarted. Other jobs were given some time to complete.

To tell how many connections were left to the legacy instance, we used the following command:

netstat -tn 2>/dev/null | grep :27017 | awk '{print \$5}' | \
cut -d: -f1 | sort | uniq -c | sort -nr | head

We waited for some time until we saw only Atlas IP addresses in the output. It was the point where everything was connected to Atlas. We shut the old node down.

The final steps were to setup alarms in Atlas and create user credentials for whoever was granted access to the database. In the following days we rolled back the retry intervals that were added to deal with short database hiccups during the process.

This operation was successful mostly because of the extensive testing in staging and very methodical and cautious process in production.


text:
# Borel $\sigma$-Algebra generated by open intervall with rational end point

$$A$$ is the $$\sigma$$-Algebra generated by $$(p,\infty)$$ with $$p\in \mathbb{Q}$$. I need to show that $$(a,b)\in A$$ for $$a,b\in \mathbb{R}$$ and $$a so that $$A$$ is a Borel $$\sigma$$-Algebra on $$\mathbb{R}$$.

Proof (what I tried):

We now that rational numbers are dense in the reals, so for a we can find a series ('from the right') with $$\lim_{n\to\infty} q_n = a$$ with $$q_n>a$$. Then $$\bigcup\limits_{n=1}^{\infty} (q_n,\infty)=(a,\infty)$$ This is a countable Union of sub-sets of $$A$$ so it is itselfe in $$A$$ and we get $$(a,b)\in A$$. That is true for every open set in $$\mathbb{R}$$. Since every open set in $$\mathbb{R}$$ can be written as a a countable union of intervals from $$A$$, $$A$$ contains the Borel $$\sigma$$-Algebra of $$\mathbb{R}$$. Because we said $$q_n \downarrow a$$ it is the smallest $$\sigma$$-Algebra containing the open intervall by definition. So it is the Borel $$\sigma$$-Algebra of $$\mathbb{R}$$.

I hope this is not a duplicate and not too bad, I would appreciate corrections, help and tips.

You proved that for every $$a\in\mathbb R$$ the set $$(a,\infty)$$ is an element of $$\mathcal A$$, but your conclusion that $$(a,b)\in\mathcal A$$ is not justified.

You can solve this by proving that also sets like $$[a,\infty)$$ are elements of $$\mathcal A$$ on a similar way with $$q_n$$ approaching from the left and taking $$\bigcap_{n=1}^{\infty}(q_n,\infty)=[a,\infty)$$. This in the knowledge that a $$\sigma$$-algebra is also closed under countable intersections.

Then also $$(-\infty, b)=[b,\infty)^{\complement}$$ and $$(a,b)=(a,\infty)\cap(-\infty,b)$$ are elements of $$\mathcal A$$.

If that is done then you can indeed claim that all open sets - as countable unions of open intervals - are elements of $$\mathcal A$$ because a $$\sigma$$-algebra is closed under complements and countable intersections.

This ensures that the Borel-$$\sigma$$-algebra on $$\mathbb R$$ if it is equipped with its usual order topology is a subcollection of $$\mathcal A$$.

Conversely we know that this Borel-$$\sigma$$-algebra will contain all sets $$(p,\infty)$$ where $$p\in\mathbb Q$$ since these sets are open. This ensures that $$\mathcal A$$ which is generated by these sets is a subcollection of the Borel-$$\sigma$$-algebra.

• Thanks a lot for your help, really appreciate it. – KingDingeling Feb 18 at 14:23
• You are welcome. – drhab Feb 18 at 14:24

There are a couple of things that are a little inaccurate in your proof. That is, the general proof is fine, but it has a couple of issues.

Issue 1:

You claim that

I need to show that $$(a,b)\in A$$ for $$a,b\in\mathbb R$$ and $$a so that $$A$$ is a Borel $$\sigma$$-Algebra on $$\mathbb R$$

however, in your proof, you don't do this. That is, you prove that $$A$$ is a borel sigma algebra on $$\mathbb R$$, but you didn't prove that by first proving $$(a,b)\in A$$ for $$a,b\in\mathbb R$$. This, to a reader of the proof, can be confusing. In a proof, if you announce you will prove $$X$$ by doing $$Y$$, you should then actually do $$Y$$.

Issue 2:

You claim that

Because we said $$q_n \downarrow a$$ it is the smallest $$\sigma$$-Algebra containing the open interval by definition.

However, this claim is quite nonsensical. I don't really know what you were trying to prove here. Since $$A$$ is generated by a subset of the Borel $$\sigma$$-algebra, it is obvious that $$A$$ must also be a subset of the Borel $$\sigma$$-algebra.

• Thank you for your help, very kind. I will check your points. – KingDingeling Feb 18 at 14:24


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# Cotangent Calculator for Degrees or Radians

Written by:
PK

On this page is a cotangent calculator, for either angles or degrees. Input an angle's measurement, choose your units, and run the calculator to see the cotangent.

## Using the Cotangent Calculator

To use the tool to find the cotangent, enter the measurement of the angle in either degrees or radians, choose the units, and compute.

• Angle - measurement of the angle, in degrees or radians)

Once happy, click the "Compute Cotangent" button.

• Cotangent - the cotangent, or ratio of the adjacent side over the opposite side's measurement

## What is the cotangent?

For a given angle in a right triangle, cotangent is the ratio of the adjacent side to the opposite side. Cotangent is the reciprocal of the tangent; tangent is the ratio of the opposite side's length to the adjacent side.

Trigonometric functions like the cotangent can also be used to find the ratio between sides of a unit circle (a circle with radius 1).

### Cotangent as a formula

As a trigonometry formula, cotangent is usually abbreviated as "cot", as in the following equation:

\cot(\theta)

As the reciprocal of tangent, it's equivalent to:

\cot(\theta)=\frac{1}{\tan(\theta)}

If you know the cotangent, use the inverse cotangent or arccotangent to find the angle.

### Don't Quit Your Day Job...

DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.


text:
# What “is” a matrix in the context of a vector space?

I'm familiar with the definition of a vector space $V$ over a field $F$

I'm also comfortable with the notion that a matrix "represents" a linear map from one vector space $V$ to another vector space $W$.

The Wiki article on Vector Spaces says this:

Matrices are a useful notion to encode linear maps.

But this conflicts with what I think a matrix is. Is a matrix a "thing" or "element" that we can write down and multiply by other matricies/vectors? Or is it simply a "representation" of a linear map between spaces? Does it depend on the context?

If it is a "thing" - then how do we define the multiplication of objects from the vector space with objects of a different kind that do not belong to the vector space?

We can write a meaningful equation such as:

$M \cdot \textbf{x} = \textbf{y}$

Where x and y are vectors, and M is a matrix.

I know how to perform the multiplication, but it seems wrong to me that $M$ is of a different "species" (apologies - I'm not familiar enough with abstract algebra to know the correct classification - set/group/etc.) than the thing we are multiplying it with.

Sorry if this is vague, I'm just hoping someone might be able to clear up this slight confusion.

Thanks!

-
If you imagine a bijection $\phi$ which associates a matrix as a thing in itself with the linear transformation it represents, then you can think of $\mathbf{Mx}=\mathbf y$ as notation for $\phi(\mathbf M)(\mathbf x)=\mathbf y$. –  Rahul Jul 8 '13 at 17:47
Incidentally, you cannot write $M\cdot x =y$ where $x,y$ are vectors in arbitrary vector spaces of the correct dimension, absent a specific basis for each vector space. –  Thomas Andrews Jul 8 '13 at 17:51

To understand the connection between matrix multiplication and linear maps, it is important to understand that a linear map is uniquely determined by what happens to a basis.

To be more specific, consider a linear map $\phi: V \to W$. For the sake of example, let's imagine that $V$ and $W$ are finite dimensional. Say $\dim_{\ F} (V) = n$ and $\dim_{\ F}(W) = m$. Now choose a basis $B=\{v_1, \ldots, v_n\}$ for $V$. I claim that the linear map $\phi$ is uniquely determined by $\phi(v_1), \ldots, \phi(v_n)$. Now for an arbitrary $v \in V$, write $v = c_1 v_1 + \ldots + c_n v_n$ for uniquely determined coefficients $c_i \in F$. Then $\phi(v) = \phi(c_1 v_1 + \ldots + c_n v_n)=c_1\phi(v_1)+ \ldots + c_n\phi(v_n)$, which follows automatically from the linearity of $\phi$.

If you examine the last equation, you should find that this information can be encoded by the multiplication of an $m\times n$ matrix with an $n \times 1$ vector. The columns of the matrix are $\phi(v_1), \ldots, \phi(v_n)$, these expressed in some basis for $W$, while the vector is the coordinates of $v$ in the basis $B$.

Thus, once we have chosen bases for our finite dimensional spaces, there is a perfect correspondence between these linear maps and $m \times n$ matrices. For each choice of bases for $V$ and $W$, we obtain an isomorphism of vector spaces $L(V, W) \cong M_{m \times n}(F)$, where $L(V, W)$ is the space of linear maps from $V$ to $W$ and $M_{m \times n}(F)$ (which we might also call $F^{m\times n}$) is the space of $m \times n$ matrices with entries in $F$. When we view a matrix as a linear map without any other context, we may as well imagine $V = \mathbb{R}^n$ and $W = \mathbb{R}^m$ with the standard bases. In this picture, multiplication of matrices with suitable dimensions corresponds to composition of linear maps. In the special case $V = W$, we get square matrices, and the correspondences $L(V, V) \cong M_{n \times n}(F)$ are $F$-algebra isomorphisms.

In some sense, matrix multiplication is best understood as a window into the inner workings of a linear map. This is not the only way to view a matrix, but the connection between matrix operations and the operations available for linear maps is inescapable.

-
I think I understand. Since the space of ALL possible linear maps from V to W is isomorphic with the set of ALL m x n matrices, we can work with matrices and perform arithmetic on them and interpret our results in terms of linear mappings because of the isomorphism. So would it be reasonable then to view the "multiplication" of a vector by a matrix just an expression of the matrix's corresponsing map being applied to the vector? I think I'm splitting hairs at this point but I do feel more comfortable thanks to your post. –  Chase Meadors Jul 8 '13 at 18:10
Yes, this is a reasonable interpretation. The action of matrices on a vector space via matrix multiplication is in some sense the same as the action of linear maps on that vector space. The equivalence depends on choosing bases. –  James Staff Jul 8 '13 at 18:18
It might also be useful to think about things in this way: Choosing bases for $V$ and $W$ amounts to choosing coordinates for those spaces, and matrix multiplication in this sense is what a linear map looks like when expressed in terms of those coordinates. –  James Staff Jul 8 '13 at 18:25

Note, a matrix does not represent a linear transformation $V\to W$, unless you have fixed bases for $V,W$.

But you are stuck believing that a matrix must be "one" kind of thing. It can be thought of either as a thing of its own, or it can be thought of as a specific linear transformation from $\mathbb k^n\to\mathbb k^m$, and in particular, how it acts on their usual bases.

Picking one view is merely definitional - the two definitions are completely equivalent. Nothing is really gained by treating one of these ideas as "primary," and it is probably better to get used to dichotomy. Sometimes it is useful to think of matrices in their own rights, sometimes it is useful to think of them as linear transformations.

For example, in matrices, it makes much more obvious sense to investigate eigenvalues that are not in the base field $k$. It is harder, initially, to see what that would mean for a linear transformation $T:V\to V$. (It can certainly be defined, it is just not initially obvious.) The fact that these eigenvalues are the same no matter which basis we choose for $V$, however, is really interesting, and thus clearly is a property of the transformation, not the matrix.

On the other hand, a lot of the row calculations that you first learn for matrices are really about changes of bases in the vector space(s).

Sometimes, operations are really "about" the transformation, and sometimes, the operation is deeply tied to the basis. For example, the "transpose" of a matrix is deeply tied to a basis, but the determinant, trace, eigenvalues, rank, etc. of a matrix are independent of the basis. When we really don't care about the basis, we are often dealing with something that is "primarily" a linear transformation.

I suppose the dichotomy starts to have problems when dealing with infinite-dimensional vector spaces.

-

If you consistently look at matrices as linear transformations, there is a way to understand multiplication as a transformation as well so they are not "different species".

You may have seen multiplication connected to the AND operation in logic (distributivity), combinatorics (disjoint possibility), probability and the like. The common understanding for matrices is very similar to these (and for very similar reasons). Transformations multiply when you apply one transformation AND THEN the next. Notice that there is a temporal quality to this application.

So, if you follow a point $p$ (seen as a vector from some basis origin) through this transformation, the point after the first transformation $M_1$ is $M_1 p$ AND THEN, after applying another transformation $M_2$ the point is finally $M_2 M_1 p$.

The reason there is this inherent temporal part to the operation is that matrix multiplication is not commutative, nor is the application of linear transformations. The order of applying the transformations matter.

Part of the difficulty in understanding that matrices are associated with transformations is that you may be thinking that matrices need numbers in them and transformations do not need a basis to be meaningful. However, just as you can meaningfully talk about transformations without a basis by describing their properties, you can just as well speak of matrices through the equations they obey without referencing a specific basis or solution. The space of solutions to the equation is the same as the space of transformation representations for transformations with the property.

For example, you might call transformations that applied twice return the space to it's original form as "reflections". You can look at the space of matrices obeying $R^2 = I$ and study the same thing. Similarly, if you call transformations that applied twice are the same as applied only once as "projections", you can study $P^2 = P$ and determine properties of solutions there.

NOTE: the idea of looking at multiplication as AND THEN is taken quite seriously when mathematicians looks at describing logics of transformation. This is found quite a lot in the study of temporal logics, quantum logics, and other nonstandard logic research. The connection then can be seen as an extension of the existing one for standard logic. Addition has a slightly different understanding than one might naively expect from the OR case, but it too has a natural understanding.

-


text:
Match Fishtank is now Fishtank Learning!

# Addition and Subtraction of Fractions/Decimals

## Objective

Add fractions with unlike denominators whose sum is less than 1.

## Common Core Standards

### Core Standards

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• 5.NF.A.1 — Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)

• 5.NF.A.2 — Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.

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• 4.NF.A.1

• 4.NF.B.3

## Criteria for Success

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1. Understand that in order to add quantities, they must have the same unit (which in the case of fractions, are their denominators).
2. Find common units for fractions with unlike denominators by finding equivalent fractions using a number line and an area model (MP.5).
3. Understand that there is more than one possibility for the common unit used, and use that to optionally find the least common denominator.
4. Add two fractions with unlike denominators whose sum is less than 1 (and therefore does not require regrouping), simplifying the sum if applicable.
5. Simplify a fraction.
6. Solve one-step word problems involving the addition of two fractions with unlike denominators whose sum is less than 1 (MP.4).

## Tips for Teachers

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• The following material is needed for today's lesson: Square-sized paper
• The Progressions for the Common Core State Standards in Math state, “It is not necessary to find a least common denominator to calculate sums of fractions, and in fact the effort of finding a least common denominator is a distraction from understanding algorithms for adding fractions” (p. 11). Thus, throughout the unit, students are encouraged to find a common denominator but never forced to use the least common denominator.
• There are five Anchor Tasks in today’s lesson. However, because the first one just lays a conceptual foundation for adding like units and all the subsequent ones include just a single computation, it should be possible to fit it in a single lesson.
• The algorithm for adding fractions with unlike denominators is explicitly discussed in Lesson 6. However, students might uncover it before then, especially since students have used multiplication and division to find equivalent fractions in Grade 4 (4.NF.1) and in Lesson 1. If that does happen, you can discuss it as a class when it comes up naturally, and allow students to use it as a strategy before Lesson 6.
• For when students get to the generalized algorithm for adding fractions with unlike denominators in Lesson 6, it may be helpful for students to see multiple examples of previous problems they’ve completed. Thus, keep anchor charts from the Anchor Tasks in Lessons 4 and 5 up after their instruction.
• Before the Problem Set, you could have students play a game to practice adding fractions with unlike denominators, such as "Wacky Fractions" or "Fraction Addition War", from Games with Fraction Strips and Fraction Cards on The Max Ray Blog.

#### Remote Learning Guidance

If you need to adapt or shorten this lesson for remote learning, we suggest prioritizing Anchor Task 4 (benefits from worked example). Find more guidance on adapting our math curriculum for remote learning here.

#### Fishtank Plus

• Problem Set
• Student Handout Editor
• Vocabulary Package

?

### Problem 1

1. Solve.
1. 1 orange + 3 oranges = ___________
2. 1 child + 3 adults = ___________
2. What do you notice about #1 above? What do you wonder?

#### References

EngageNY Mathematics Grade 5 Mathematics > Module 3 > Topic B > Lesson 3Concept Development

Grade 5 Mathematics > Module 3 > Topic B > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by Fishtank Learning, Inc.

### Problem 2

Solve. Show your work with an area model.

${{1\over2}+{1\over4}}$

#### References

EngageNY Mathematics Grade 5 Mathematics > Module 3 > Topic B > Lesson 3Concept Development

Grade 5 Mathematics > Module 3 > Topic B > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by Fishtank Learning, Inc.

### Problem 3

Solve. Show your work with an area model and a number line.

${{1\over3}+{1\over2}}$

#### References

EngageNY Mathematics Grade 5 Mathematics > Module 3 > Topic B > Lesson 3Concept Development

Grade 5 Mathematics > Module 3 > Topic B > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by Fishtank Learning, Inc.

### Problem 4

Solve. Show your work with an area model and a number line.

${{2\over3}+{1\over4}}$

#### References

EngageNY Mathematics Grade 5 Mathematics > Module 3 > Topic B > Lesson 3Concept Development

Grade 5 Mathematics > Module 3 > Topic B > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by Fishtank Learning, Inc.

### Problem 5

1. Solve. Show your work with an area model and a number line.

${{2\over9}+{2\over3}}$

1. Is there another common unit that can be used to solve?

## Problem Set & Homework

#### Discussion of Problem Set

• Look at #1. What fractions are equivalent to $\frac{4}{9}$? What fractions are equivalent to $\frac{2}{6}$? What equivalent fractions did you use to add these fractions together?
• Look at #2. How did you use the number line to figure out the sum?
• Look at #6. How did you use an area model to find your answer?
• Look at #8. For part (a), what common denominators did you choose? Why do both of them work? For part (c), how did you solve? What did you get as the sum?
• Look at #9. What is the sum of the two fractions you came up with? Was anyone able to find a closer sum? Did anyone get an exact sum of $\frac{1}{2}$

?

### Problem 1

Solve. Show or explain your work.

${{1\over2}+{1\over5}}$

#### References

EngageNY Mathematics Grade 5 Mathematics > Module 3 > Topic B > Lesson 3Exit Ticket, Question #1

Grade 5 Mathematics > Module 3 > Topic B > Lesson 3 of the New York State Common Core Mathematics Curriculum from EngageNY and Great Minds. © 2015 Great Minds. Licensed by EngageNY of the New York State Education Department under the CC BY-NC-SA 3.0 US license. Accessed Dec. 2, 2016, 5:15 p.m..

Modified by Fishtank Learning, Inc.

### Problem 2

Solve. Show or explain your work.

Carpenters are laying down hardwood floors in a living room. On the first day, they laid down $\frac{2}{5}$ of the whole floor. On the second day, they laid down $\frac{1}{4}$ of the whole floor. The area model shown can be used to find how much of the whole floor they laid over the two days.

• How much of the floor did they lay down over the two days?
• Explain how the area model can be used to answer the question.

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text:
Gold Member

## Homework Statement

For the solution to a given problem, in the second to last step I had:
##-\frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}##

I stated next that the solution was ##-\frac{\sqrt{6}+\sqrt{2}}{4}##

I was told this was incorrect and that the correct solution is ##\frac{-\sqrt{6}+\sqrt{2}}{4}##

## The Attempt at a Solution

Could someone explain this error to me? Is it not true that ##-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}##?
Or is this case different because I have an expression on the top, not just a single variable?

Related Precalculus Mathematics Homework Help News on Phys.org
Gold Member
Screenshot attached for reference.

#### Attachments

• 10.6 KB Views: 369
ehild
Homework Helper

## Homework Statement

For the solution to a given problem, in the second to last step I had:
##-\frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}##

I stated next that the solution was ##-\frac{\sqrt{6}+\sqrt{2}}{4}##

I was told this was incorrect and that the correct solution is ##\frac{-\sqrt{6}+\sqrt{2}}{4}##

## The Attempt at a Solution

Could someone explain this error to me? Is it not true that ##-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}##?
##-\frac{a}{b}=\frac{-a}{b}## is true. So ##-\frac{\sqrt 6}{4} =\frac{-\sqrt 6}{4}##and ##\frac{-\sqrt 6}{4}+\frac{\sqrt2}{4}=\frac{-\sqrt 6+\sqrt2}{4}##

opus
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

For the solution to a given problem, in the second to last step I had:
##-\frac{\sqrt 6}{4} + \frac{\sqrt 2}{4}##

I stated next that the solution was ##-\frac{\sqrt{6}+\sqrt{2}}{4}##

I was told this was incorrect and that the correct solution is ##\frac{-\sqrt{6}+\sqrt{2}}{4}##

## The Attempt at a Solution

Could someone explain this error to me? Is it not true that ##-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}##?
Or is this case different because I have an expression on the top, not just a single variable?
You are guilty of sloppy use (or non-use) of parentheses:
$$-\frac{a}{b} + \frac{c}{b}$$
means that you subtract the fraction ##\frac{a}{b}## from the fraction ##\frac{c}{b}##, and since they have the same denominator, you can combine the numerators as ##c-a##. That is, you get
$$\frac{c-a}{b},$$
which can be written as $$\frac{-a + c}{b}.$$
If you write
$$-\frac{a+c}{b}$$
that means $$- \left( \frac{a+c}{b} \right),$$
which was not what you started with.

Remember the priority order of mathematical operations: reading from left to right, parentheses have precedence over everything, then powers have precedence over multiplication or division, and these have precedence over addition and subtraction. Therefore, when we parse the expression ##-\frac{a}{b} + \frac{c}{b}##, we do "##-##" first, then ##a/b## then "##+##", then ##\frac{c}{b}##. In other words, had we used parentheses we would have had ##(- \frac{a}{b}) + (\frac{c}{b})##.

opus
Gold Member
Ahh I see. So by pulling that negative all the way out front, I negated the entire expression which is not equal to what I had. The negative needed to stay with the ##\sqrt 6##

FactChecker
Ray Vickson
Homework Helper
Dearly Missed
Ahh I see. So by pulling that negative all the way out front, I negated the entire expression which is not equal to what I had. The negative needed to stay with the ##\sqrt 6##
Yes, exactly, because in ##-\frac{a+c}{b}## the division has prioity over the first "##-##", so first we do the fraction, then we change its sign.

See, eg.,
http://www.purplemath.com/modules/orderops.htm
or
https://en.wikipedia.org/wiki/Order_of_operations

opus
Gold Member
Excellent, thanks guys.

berkeman
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# A posteriori error estimates for mixed finite element discretizations of the Neutron Diffusion equations

1 POEMS - Propagation des Ondes : Etude Mathématique et Simulation
Inria Saclay - Ile de France, UMA - Unité de Mathématiques Appliquées, CNRS - Centre National de la Recherche Scientifique
Abstract : We analyse $a\ posteriori$ error estimates for the discretization of the neutron diffusion equations with mixed finite elements. We provide guaranteed and locally efficient estimators on a base block equation, the one-group neutron diffusion equation. We pay particular attention to AMR strategies on Cartesian meshes, since such structures are common for nuclear reactor core applications. We exhibit a robust marker strategy for this specific constraint, the $direction\ marker$ strategy. The approach is further extended to a Domain Decomposition Method, the so-called DD+$L^2$ jumps method, as well as to the multigroup neutron diffusion equation.
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https://hal-cea.archives-ouvertes.fr/cea-02893125
Submitted on : Wednesday, July 8, 2020 - 9:20:19 AM
Last modification on : Tuesday, July 28, 2020 - 9:02:11 AM

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CiDM20_submission.pdf
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• HAL Id : cea-02893125, version 1

### Citation

Patrick Ciarlet, Minh-Hieu Do, François Madiot. A posteriori error estimates for mixed finite element discretizations of the Neutron Diffusion equations. 2020. ⟨cea-02893125⟩

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## Algebra and Trigonometry 10th Edition

We are given the function: $f(x)=\sqrt[3]{4x}$ Graph the function: The graph shows that the function is symmetrical with respect to the origin, which means the function is odd. Check that the function is odd algebraically: $f(-x)=\sqrt[3]{-x}=-\sqrt[3]x=-f(x)$ As $f(-x)=-f(x)$, the function is odd.