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What is Reward Estimation?

One of the biggest issues faced when solving prescriptive problems is the question of how to evaluate the quality of a proposed prescription policy. The key issue is that a new policy will almost certainly prescribe treatments that are different to those observed in the data, and so we do not know what would have happened under the new treatments. Reward estimation is an approach that seeks to address this problem, by estimating the reward that should be attributed to a policy for assigning a given treatment to a given observation.

Throughout this guide we use the following notation:

• $\mathbf{x}_i$ is the vector of observed covariates for the $i$th point
• $T_i$ is the observed treatment for the $i$th point
• $y_i$ is the observed outcome for the $i$th point (corresponding to treatment $T_i$)
• $p_i(t)$ is the predicted probability that the $i$th point was assigned treatment $t$ in the original data, i.e., the propensity score
• $f_i(t)$ is the predicted outcome for the $i$th point under treatment $t$
• $\Gamma_i(t)$ is the estimated reward realized if the $i$th point is assigned treatment $t$

Reward Estimation Process

The reward estimation process consists of three steps.

1. Estimating Propensity Scores

The propensity score$\ p_i(t)$, is the probability that a given point $i$ would be assigned treatment $t$ based on its features $\mathbf{x}_i$.

In a randomized experiment where treatments are randomly assigned to points, the propensity scores would be equal for all treatments.

In observational data, there is typically some non-random treatment assignment process that assigns treatments based on the observed features. In these cases, we can estimate the propensity scores by treating the problem as a multi-class classification problem, where we try to predict the assigned treatments using the features. This trained model can then estimate the probabilities of treatment assignment for each point.

2. Estimating Outcomes

The predicted outcome $f_i(t)$ is the outcome that we would predict to occur if point $i$ with features $\mathbf{x}_i$ were assigned treatment $t$. These are often referred to as counterfactuals or potential outcomes.

We can estimate the outcomes by training a separate regression model for each treatment to predict the outcomes under that treatment. For a given treatment, we train the model on the subset of the data that received this treatment. We can then apply each of the trained regression models in order to predict the outcomes under each treatment for each point.

3. Estimating Rewards

Given estimates of the propensity scores and outcomes, there are different approaches for estimating the reward associated with assigning a given observation a given treatment. For more information on these approaches, you can refer to papers in the causal inference literature such as Dudik et al. (2011).

Warning

The inverse propensity weighting and doubly robust estimators divide by the predicted propensity as part of calculating the reward, meaning that small propensity predictions can lead to very large or infinite reward predictions. If this occurs, you should use the propensity_min_value parameter to control the minimum propensity value that can be predicted.

Direct Method

Argument name: :direct_method

The direct method estimator simply uses the predicted outcome under each treatment as the estimated reward:

$\Gamma_i(t) = f_i(t)$

The quality of the reward estimation is thus determined solely by the quality of the outcome prediction model. One potential issue with this approach is that the outcome prediction model is trained holistically, rather than being trained with prescription in mind. This means the model might focus attention on predicting outcomes in areas that are irrelevant for comparing the relative quality of treatments.

Inverse Propensity Weighting

Argument name: :inverse_propensity_weighting

The inverse propensity weighting estimator predicts the rewards by adjusting the observed treatments using estimates of the propensity score:

$\Gamma_i(t) = \begin{cases} \frac{y_i}{p_i(t)}, & \text{if } t = T_i\\ 0, & \text{otherwise} \end{cases}$

This can be interpreted as taking the observed outcomes and correcting for any shift in treatment assignment probabilities between the observed data and the policy being evaluated. The quality of this method is tied to the probability estimation model. In particular, small probability estimates can cause large reward estimates. It may also require a larger number of observations with good coverage of the possible treatments to ensure that the probability estimates are non-zero in most places.

Doubly Robust

Argument name: :doubly_robust

The doubly robust estimator uses both the predicted probabilities and predicted outcomes to estimate the rewards:

$\Gamma_i(t) = \begin{cases} f_i(t) + \frac{y_i - f_i(t)}{p_i(t)}, & \text{if } t = T_i\\ f_i(t), & \text{otherwise} \end{cases}$

It can be seen as a combination of the direct method and the inverse propensity weighting approaches. This estimator is shown to be accurate as long as either the probability and/or outcome predictions are accurate, hence the name doubly robust.

Evaluating Prescriptive Policies Using Rewards

Given a policy $\pi(\mathbf{x})$ that assigns treatments to points, we can evaluate its quality using the estimated rewards in multiple ways.

Note that for observational data, the current policy is given by the treatment assignments. We can evaluate the quality of this policy to gain a standard of performance against which we can compare alternative policies.

Aggregate Policy Reward

The simplest way to evaluate the quality of a policy is by summing the rewards of the assigned treatments for each point:

$\sum_i \Gamma_i(\pi(\mathbf{x}_i))$

Comparison to Oracle

If the rewards have been estimated using Direct Method, you can also evaluate the performance of the policy by comparing to an oracle that picks the treatment with the best reward for each point. You might consider the following metrics:

• the proportion of points where the policy prescription matches the oracle prescription
• an assessment of the difference in rewards under the policy prescription and the oracle prescription, e.g. mean-squared error or mean-absolute error of the difference
Warning

You can only evaluate performance by comparing to an oracle if you estimated the rewards using the Direct Method. The other reward estimation methods only make sense when comparing the aggregate policy values

Training and Testing Setup

Evaluating prescriptive policies is more complicated than evaluating predictive problems. In particular, we recommend the following steps to ensure that you get a fair and useful out-of-sample evaluation when evaluating prescriptive policies.

Ensure Sufficient Test Data

Warning

Make sure you include enough data in the test set to train high-quality reward estimation models

Unlike a normal predictive model evaluation, reward estimation requires training models on the test set as part of the evaluation. If there is not enough data in the test set, the quality of the reward estimation may suffer, leading to an evaluation of the prescriptive policy that may be inaccurate and unreasonable.

We recommend using much more data for the test set than usual for a prescriptive problem. Allocating 50% of the data for testing is a good starting point.

Proper Separation of Training and Test Sets

Warning

When using reward estimation for evaluating prescription policies, you should estimate the rewards using fit_predict! on the test data.

In order to simulate a fair out-of-sample evaluation, we need to prevent any information leaking from the training data into the estimated rewards. This ensures that the rewards used for evaluation are purely a function of the testing data.

In this circumstance, it is important that fit_predict! is used to fit the model and estimate the rewards in a single step, rather than using fit! followed by predict. This is because fit_predict! contains additional measures that reduce overfitting in the reward estimation process.


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# Why is hydrogen the most abundant element in the Universe?

Hydrogen is the most abundant element in nature. Does cosmological nucleosynthesis provide an explanation for why is this the case? Is the explanation quantitatively precise?

• I don't think my statement qualifies as a total answer, but it's worth considering that Hydrogen-1 nuclei is just a proton. Quarks naturally become paired together under a certain temperature (2x10^12 K) and the most stable hadron configuration is a proton, so an enormous amount of protons were created (either straightforward or from decay). Other nuclei require penetrating the coulomb barrier and often rely on the weak interaction to to be formed which requires much more demanding circumstances. Jan 24, 2019 at 16:26

The short answer is that (i) protons (hydrogen nuclei) are produced abundantly in the early universe, but only a small fraction of these are able to engage in nuclear reactions leading to heavier elements, either during primordial nucleosynthesis or later inside stars. This means that hydrogen ends up being the most abundant element in the universe. (ii) Big bang nucleosynthesis makes very precise predictions about the fraction of baryons that end up as protons vs other nuclei.

Further details

In the early phases of the big bang only the constituent parts of nucleons (quarks and anti-quarks) plus leptons (e.g. electrons, positrons) and light existed. As the universe expanded and cooled, quarks were able to combine and form the basic building blocks of nuclei - the neutrons and protons. A proton is of course a hydrogen nucleus; any heavier elements needed to be built by fusing together protons.

Protons are positively charged and repel each other strongly. In order to fuse protons and make helium requires high energies/temperatures and the intermediate stage of forming deuterons - a proton plus neutron bound pair.

The reason that most protons do not fuse in this way is threefold.

1. The small excess mass of the neutron compared to the proton means that the equilibrium reactions that produce nucleons end up producing about six times as many protons as neutrons.

2. Free neutrons then decay into protons on timescales of minutes, which increases this ratio further to about seven.

3. The deuteron is weakly bound, so there is only a limited window of time between when it is too hot to remain bound (prior to 10s after the big bang) and when it is too cool to get deuterons to fuse to become helium (at about 10 minutes after the big bang).

Of these things, by far the most important factor is #1 and this is fundamentally the answer to your question. It is because the neutron is more massive than the proton. All heavier nuclei must contain neutrons because the Coulomb repulsion between two or more protons is too strong to form stable nuclei without them. By the end of big bang nucleosynthesis, essentially all of the available free neutrons end up in helium nuclei (with traces of deuterons and lithium nuclei) and thus the H/He ratio ends up at 12 protons for every He nucleus (2 protons, 2 neutrons).

Is this quantitatively precise? Well, my figure of 12 protons for every He nucleus, or 75% by mass, is a bit handwaving. A detailed model must take into account in a more accurate way the temperature evolution of the universe, the various (more minor) nucleosynthetic pathways, photodisintegration and so-on, but yes, the models do provide a very precise prediction of the H/He (and the minor species) at the end of cosmological nucleosynthesis. There are small uncertainties in the neutron lifetime, the various reaction cross-sections and so-on, but the most important uncertainty is that there is an important "free parameter" - the baryon to photon ratio that must be fixed. This can be constrained by demanding that one consistent value of this number can explain all the measured primordial abundance ratios (He/H, Li/H, D/H, $$^3$$He/H) or it can be found from other cosmological measurements (such as from the cosmic microwave background).

The He/H ratio is not very sensitive to this free parameter and hydrogen is always by far the most dominant nucleus for the reasons I explained above. Current levels of uncertainty on the mass fraction of hydrogen produced in the (standard) big bang are significantly smaller than 1% (e.g. Peimbert 2008). Coc et al. (2013) use the Planck cosmic microwave background constraints on the baryon-to-photon ratio and estimated a standard big-bang He/H mass ratio of $$0.2463\pm 0.0003$$. One could consider additions to this standard model - e.g. change the number of neutrino families, have decaying dark matter particles in the early universe and so-on, but it seems hard to change the mass fraction of hydrogen by more than 1% without upsetting the concordance with other observations.

A final point to make is that very little of this cosmological material has so far found its way into stars (perhaps 10%), and of that, much of it is still in the same (low-mass) stars that were formed. The amount of "processed" material made of heavier elements, fused from hydrogen in stars, that has enriched the cosmological material is therefore comparatively small - of order 1-2%. So the predominance of hydrogen has hardly changed since the big bang.

This latter property can be used to test the whole model. By looking at the He/H ratio as we go "back in time" we can see if the primordial ratio matches that predicted by the big-bang. In practice this can be done by estimating the He/H ratio in the oldest stars or by estimating He/H in the interstellar medium of the most metal-poor galaxies. These measurements are more uncertain than the predictions above, but are in reasonable agreement with them. An example would be Izotov & Thuan (2010), who estimate a primordial He/H mass ratio of $$0.2565 \pm 0.005$$ from metal-poor galaxies - about two (small) error bars higher than the prediction above.

• +1 I was waiting for someone to mention factor numero uno. :-) Jul 20, 2018 at 8:36
• How many gigayears (or terayears or yottoyears or whatever) will it take for hydrogen to not be the most abundant element? Jul 23, 2018 at 14:18
• @CurtF. Could be a new question though there may be a duplicate. Jul 23, 2018 at 16:46
• @RobJeffries Points 1 and 2 of your answer imply that the number density of protons in equilibrium is almost seven times the number density of neutrons. How does that disfavour the fusion of protons? Mar 2, 2019 at 6:48
• @mithusengupta123 It doesn't, but that fusion reaction is neligibly slow compared with the fusion of protons and neutrons, where there is no coulomb barrier. Mar 2, 2019 at 8:39

Between about t=$10^{-12}$ and t=$10^{-6}$ seconds, the universe was filled with a quark-gluon plasma. Temperatures were too high for mesons and baryons (like protons and neutrons) to remain bound, should they briefly form.

Then, between about t=$10^{-6}$ and 1 second, during the "Hadron epoch," the temperature was low enough that quarks that stuck together could remain together. Statistically, the simplest stable configurations were protons and neutrons, so that is how the majority of quarks ended up bound.

During the next 200 seconds, temperatures were still high enough for nuclear reactions, so about one-quarter of baryons (neutrons and protons) were bound together into helium (and traces of other elements).

During the next hours, any remaining neutrons decayed into protons and electrons (and neutrinos).

And that's roughly where we're at. Solo protons are H$^+$ (hydrogen ions) and those that combined with electrons are H (hydrogen atoms).

• that figure is an excellent visualisation of early nuclei transformation. Did you make it, or do you have a citation for it?
– craq
Jul 24, 2018 at 1:34
• I too found the figure very apt and would appreciate a reference or link to follow it up. Aug 2, 2021 at 8:14
• The image is from this Lawrence Berkeley Laboratory page: universeadventure.org/big_bang/elemen-formation.htm There is no specific copyright notice attached to the image, so my guess is that it was created by LBL, and as such, should be public domain. I'm not a lawyer, so don't take my word for it Aug 2, 2021 at 14:46

Cosmological nucleosynthesis furnishes a remarkably accurate accounting of the amount of both hydrogen and helium in the observable universe. The satisfactory correspondence between the amounts observed and the amounts predicted by the standard big bang model is one of the main reasons for the general acceptance of the validity of the big bang model.

Weinberg's book "The First Three Minutes" provides an excellent description of these processes and is highly recommended.

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# How do you add two fractions? I have a fraction I am trying to solve. I know the answer already, as

How do you add two fractions?
I have a fraction I am trying to solve. I know the answer already, as Wolfram says it is $\frac{143}{300}$.
The fraction is:

You can still ask an expert for help

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Immanuel Glenn
Null has given you a good way. Here's a way without worrying about the LCM:
$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$
In the example,
$\frac{5}{12}+\frac{3}{50}=\frac{\left(5\right)\left(50\right)+\left(12\right)\left(3\right)}{\left(12\right)\left(50\right)}=\frac{286}{600}=\frac{143}{300}$
The price of not worrying about the LCM is that you get an answer $286/600$ , that isn't in lowest terms, so you have the extra step at the end of reducing the fraction.
###### Did you like this example?
Rebecca Villa
The fractions need to have the same denominator in order to add them together. The denominator must be the least common multiple of the two denominators, or a multiple of it. The least common multiple of 12 and 50 is 300.
First multiply $5/12$ by $25/25$. Since $25/25=1$ the product is the same as the original fraction. You simply multiply the numerators together and the denominators together:
$\frac{5}{12}×\frac{25}{25}=\frac{125}{300}$
Then multiply $3/50$ by $6/6$ . Again, the product is the same as the original fraction:
$\frac{3}{50}×\frac{6}{6}=\frac{18}{300}$
Now both fractions have the same denominator and you can simply add the numerators:
$\frac{125}{300}+\frac{18}{300}=\frac{143}{300}$
###### Did you like this example?

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• PhoenixFire
A uniform rod of length 'L' sits at rest vertically. The base of the rod is connected to a frictionless pivot. It then falls over under the influence of gravity. What portion of the rod has a tangential acceleration greater than $$gsin(\theta)$$ where $$\theta$$ is the angle the rod makes with the vertical axis.
Physics

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This choice is one of the main questions to be ... [more ▼]Since 10 years topology optimisation has been trying to bring an efficient answer to the problem of automatic choice of morphology of mechanical components. This choice is one of the main questions to be addressed during the preliminary design phase of mechanical and structural components. By topology or morphology of a mechanical or structural component one means here all the basic data that touch the layout. So topology covers for example the number and the relative positions of the wholes in the structural domains, the number and the nature of the structural members, their connectivity and the character of the connecting joints. Before having topology optimisation tool, the selection of the mechanical morphology had been let to engineers’ experience or (even worse sometimes) to their intuition. For example it was a common use in industry to take the topology of an existing product and to use it as it is for the new design. With topology optimisation the choice of morphology can now rely on rational arguments and can be made in order to fit to the product characteristics. Furthermore mathematical tools, because of the optimisation formulation of the design problem, drive the determination of the structural layout. This has two advantages. At first topology optimisation can facilitate the automation of preliminary design steps. Then it can improve substantially the performance of new mechanical products. This means that topology optimisation can propose original and innovative solutions to engineering problems. Some authors suggested that in some problems topology could lead to a gain of performance that could grow up to 50 percents. This paper reports some novel contributions to topology optimisation techniques. Two areas will be addressed. The first one is concerned with recent progress related to the perimeter method of topology optimisation. The perimeter method, which was originally introduced by Haber et al (1996) in topology optimisation, consists in bounding the perimeter of the material distribution in addition to its area. At first recent research focussed on extending the method to 3-D structures. Then other work was made to new quasi-isotropic measures of the perimeter that are nearly insensitive to the mesh. The second axis of our work has been devoted to the treatment of stress constraints. We have continued along the initial work of Duysinx and Bendsøe (1998). The new developments were made to consider stress constraints in practical (industrial) design problems. Firstly we investigated the formulation of the problem in terms of global (i.e. integrated) stress constraints instead of the local stress constraints which can be very cumbersome for practical applications. A second research was devoted to extend the classic von Mises equivalent stress criterion to other kinds of criteria. Indeed in many cases such as in structures made of a material with unequal stress limits, the von Mises criterion is unable to predict a correct topology design. In order to include the effect of different behaviours in tension and compression, we are going to show that Raghava and Ishai quadratic criteria can be used. Finally in the final stage of the paper we will discuss the position of topology optimisation in the design chain. Usual answers in accordance with the current state of the art consider this topology tool as a preliminary design tool. However our experience lead us to a more complicated answer. In a similar way to stress constraints, the ‘optimal’ topology can be dependent on all the design constraints and not only stiffness performance. These constraints can come from the structural (or functional) behaviour, but they can also be related to the manufacturing aspects. Our experience showed that perimeter constraint is quite efficient to limit the design complexity in same cases, especially for planar structures. However this perimeter constraint can lead designs that are totally impossible to manufacture especially in 3-D. For example, perimeter constraint never prevents included wholes that would be impossible to carve out with some fabrication techniques. Thus we come to the conclusion that new progress in topology optimisation should be oriented towards a simultaneous approach of the design problem including most of the functional requirements as well as of the manufacturing restrictions. [less ▲]Detailed reference viewed: 44 (3 ULg) Modeling of Electro-mechanical coupling in MEMSRochus, Véronique ; Duysinx, Pierre ; Golinval, Jean-Claude Scientific conference (2002)Detailed reference viewed: 23 (4 ULg) Topology optimization with self-weight loading: (un-expected) problems and solutionsBruyneel, Michaël ; Duysinx, Pierre in Bendsoe, M. P.; Olhoff, Niels; Rassmussen, John (Eds.) Proceedings of the 2nd Max Planck Workshop on Engineering Design Optimization (2001, October)Detailed reference viewed: 36 (0 ULg) A family of MMA approximations for structural optimizationBruyneel, Michaël ; Duysinx, Pierre ; Fleury, Claude Conference (2001, July)This paper deals with the approximation concepts approach applied to structural optimization. In 1987, Svanberg proposed the method of moving asymptotes for solving structural optimization problems. This ... [more ▼]This paper deals with the approximation concepts approach applied to structural optimization. In 1987, Svanberg proposed the method of moving asymptotes for solving structural optimization problems. This approximation is monotonous and can efficiently treat problems characterized by such a behavior. Svanberg (1995) proposed a modification of the MMA approximation by making it non monotonous. This property is based on an heuristically updated parameter. The resulting approximation is called GCMMA. In Bruyneel et al. (1999), it is shown that it is possible to generate non monotonous GCMMA based approximations, called GBMMA, by using the gradient and/or the function values from previous iteration. It was shown that such approximations improve the convergence speed of the optimization process. In many optimization problems (for example composite structures optimization or simultaneous sizing and optimal configuration of truss structures), the structural responses present both monotonous and non monotonous behaviors. A mixed monotonous/non monotonous approximation scheme has to be used for approximating in the best way the optimization problem (Zhang et al., 1998 and Bruyneel and Fleury, 2000). In this paper, we propose to show that it is possible to derive a very general approximation of the MMA family based on gradients and/or functions values at two successive design step, that present a mixed monotonous/non monotonous behavior. This approximation scheme is based on the GMMA approximation of Dusyinx et al. (1995) and on the non monotonous GBMMA approximations of Bruyneel and Fleury (1999). As the approximation scheme proposed in this paper is general, it contains all the approximations of the MMA family described above, that is MMA, GCMMA, GBMMA and GMMA. According to the characteristics of the problem under consideration, one of those approximations or of a mix of them is used to solve the optimization problem. This selection can be automatic or based on the designer’s knowledge. Numerical applications will show that the derived GMMA/GBMMA approximation scheme is efficient for solving structural optimization problems. Results will be compared with the ones obtained with the other approximations of the MMA family. [less ▲]Detailed reference viewed: 90 (3 ULg) Estimating diagonal second order terms in structural approximations with quasi-Cauchy techniquesDuysinx, Pierre ; Nguyen, Van Hien; Bruyneel, Michaël et alin CHENG, Gen Dong (Ed.) Proceedings of the 4th World Congress of Structural and Multidisciplinary Optimization WCSMO4 (2001, June)This paper reports preliminary results obtained when estimating diagonal second order terms to be used in structural approximations with the quasi-Cauchy updates which was recently proposed by Zhu ... [more ▼]This paper reports preliminary results obtained when estimating diagonal second order terms to be used in structural approximations with the quasi-Cauchy updates which was recently proposed by Zhu, Nazareth, and Wolkowicz (SIAM J. of Optimization, 9 (4), 1192-1204, 1999). At first, the theory of quasi-Cauchy updates is presented. Main characteristics of the developments that were necessary to use quasi-Cauchy updates in the context of structural optimization are drawn. The available numerical results allow comparing quasi-Cauchy second order term estimations with other estimation procedures. [less ▲]Detailed reference viewed: 20 (0 ULg) Composite structures design for strength and stiffness with respect to ply thickness and/or fibers orientationBruyneel, Michaël ; Duysinx, Pierre ; Fleury, Claude in CHENG, Gen Dong (Ed.) Proceedings of the 4th World Congress of Structural and Multidisciplinary Optimization WCSMO4 (2001, June)This paper presents an optimization approach that allows the design of composite structures under stiffness and strength criteria, with fibers orientations and plies thickness taken into account in the ... [more ▼]This paper presents an optimization approach that allows the design of composite structures under stiffness and strength criteria, with fibers orientations and plies thickness taken into account in the same optimization loop. The approach is based on the sequential convex programming. An industrial application is presented. [less ▲]Detailed reference viewed: 41 (3 ULg) L’Optimisation Topologique en MécaniqueDuysinx, Pierre Article for general public (1999)Detailed reference viewed: 24 (3 ULg) Compressor and Turbine Blade Design by OptimizationLéonard, Olivier ; Rothilde, André; Duysinx, Pierre in Bloebaum, C. (Ed.) Proceedings of the 3rd World Congress of Structural and Multidisciplinary Optimization WCSMO3 (1999, May)Compressor and turbine blade design involves thermodynamical, aerodynamical and mechanical aspects, resulting in an important number of iterations. Inverse methods and optimization procedures help the ... [more ▼]Compressor and turbine blade design involves thermodynamical, aerodynamical and mechanical aspects, resulting in an important number of iterations. Inverse methods and optimization procedures help the designer in this long and eventually frustrating process. In this paper an optimization procedure is presented which solves two types of two-dimensional or quasi-three-dimensional problems: the inverse problem, for which a target velocity distribution is imposed, and a more global problem, in which the aerodynamic load is maximized. [less ▲]Detailed reference viewed: 70 (2 ULg) Almost Isotropic Perimeters in Topology Optimization: Theoretical and Numerical AspectsPetersson, Joakim; Beckers, Muriel; Duysinx, Pierre in Bloebaum, C. (Ed.) Proceedings of the 3rd World Congress of Structural and Multidisciplinary Optimization WCSMO3 (1999, May)We consider topology optimization of elastic continuum structures including a bound on the perimeter of the structural domain. Such a bound is known to ensure existence of solutions and it stabilizes the ... [more ▼]We consider topology optimization of elastic continuum structures including a bound on the perimeter of the structural domain. Such a bound is known to ensure existence of solutions and it stabilizes the behavior of numerical finite element (FE) solutions. However the straightforward way of calculating the perimeter is rotationally mesh-dependent. In this paper we present new perimeter formulae with weaker rotational dependence i.e. perimeters that are almost isotropic. An overview of some theoretical results as well as numerical tests are given. [less ▲]Detailed reference viewed: 37 (2 ULg) Topology Optimization with Different Stress Limit in Tension and CompressionDuysinx, Pierre in Bloebaum, C. (Ed.) Proceedings of the 3rd World Congress of Structural and Multidisciplinary Optimization WCSMO3 (1999, May)This report presents new progresses in topology optimization of continuum structures with stress constraints. One principal contribution consists in the consideration of equivalent stress criteria which ... [more ▼]This report presents new progresses in topology optimization of continuum structures with stress constraints. One principal contribution consists in the consideration of equivalent stress criteria which can generalization of von Mises criterion and which are able to take into account non equal stress limits in tension and compression. A literature review led us to consider Raghava and Ishai criteria, which include a contribution of hydrostatic pressure. With the help of these criteria topology optimization can predict more realistic designs in which structural members are able to withstand better tension loads than compression loads, or vice-versa, as it is sometimes encountered in civil engineering or in composite material design. The implementation and sensitivity analysis aspects of Raghava and Ishai criteria in the Finite Element context are presented. We also present recent advanced developments to the solution of topology problems with stress constraints like the stress constraint relaxation technique and the numerical optimization procedure based on convex approximations and dual optimizers. Finally numerical applications will show the original character of the stress based topology designs ad versus compliance designs when there are unequal stress limits or when there are more than one load case. [less ▲]Detailed reference viewed: 146 (2 ULg) Enhanced Approximation Schemes for Classical Optimization AlgorithmBruyneel, Michaël ; Vermaut, O.; Duysinx, Pierre et alReport (1998)The report assesses the capabilities of the algorithms CONLIN, MDQA, an GCM available in Boss Quattro to solve composite structure optimization. The algorithms are evaluated with respect to the ... [more ▼]The report assesses the capabilities of the algorithms CONLIN, MDQA, an GCM available in Boss Quattro to solve composite structure optimization. The algorithms are evaluated with respect to the optimization of sandwich panel parameters (ply thickness and orientation, and foam core thicknesss). [less ▲]Detailed reference viewed: 18 (0 ULg) Topology Optimization of Continuum Structures with Local Stress ConstraintsDuysinx, Pierre ; Bendsoe, Martin Philipin International Journal for Numerical Methods in Engineering (1998), 43We introduce an extension of current technologies for topology optimization of continuum structures which allows for treating local stress criteria. We rst consider relevant stress criteria for porous ... [more ▼]We introduce an extension of current technologies for topology optimization of continuum structures which allows for treating local stress criteria. We rst consider relevant stress criteria for porous composite materials, initially by studying the stress states of the so-called rank 2 layered materials. Then, on the basis of the theoretical study of the rank 2 microstructures, we propose an empirical model which extends the power penalized sti ness model (also called SIMP for Solid Isotropic Microstructure with Penalization for intermediate densities). In a second part, solution aspects of topology problems are considered. To deal with the so-called 'singularity' phenomenon of stress constraints in topology design, an -constraint relaxation of the stress constraints is used. We describe the mathematical programming approach that is used to solve the numerical optimization problems, and show results for a number of example applications. [less ▲]Detailed reference viewed: 63 (3 ULg) New Developments in Handling Stress Constraints in Optimal Material DistributionsDuysinx, Pierre ; Sigmund, Olein Proceedings of 7th AIAA/USAF/NASA/ISSMO symposium on Multidisciplinary Design Optimization (1998)There is a general interest to consider stress constraints in topology optimization of continuum structures. By their very nature stress constraints are local constraints which result in large scale ... [more ▼]There is a general interest to consider stress constraints in topology optimization of continuum structures. By their very nature stress constraints are local constraints which result in large scale optimization problems that are often expensive to solve. Here in order to reduce the computing the effort we explore an alternative technique based on equivalent global (that is integrated) constraints. We define two global stress constraints based on the p-norm and p-mean of the epsilon-relaxed overall stress criteria in the finite elements. We present a new formulation of the epsilon-relaxation technique which is better suited to topology optimization of continuum structures and which makes the relaxation process automatic. The "p-mean" and "p-norm" functions bound by lower and upper value the maximum value of the epsilon-relaxed overall stress criterion. Based on numerical experiments this study compares the global and the local constraint formulations. Even if the use of integrated constraints leads a reduction of the computing time by one or two orders of magnitude, they definitely give a weaker control of local stress level. This sometimes can lead to solutions that are a bit different. [less ▲]Detailed reference viewed: 255 (1 ULg) Topology Optimization of Continuum Structures with Stress ConstraintsDuysinx, Pierre ; Bendsoe, Martin Philipin Gutkowski, W.; Mroz, Z. (Eds.) Proceedings of the Second World Congress of Structural and Multidisciplinary Optimization (WCSMO2) (1997, May)We introduce an extension of topology optimization of continuum structures to deal with local stress criteria. We first consider relevant stress criteria for porous composite materials, initially by ... [more ▼]We introduce an extension of topology optimization of continuum structures to deal with local stress criteria. We first consider relevant stress criteria for porous composite materials, initially by studying the stress states of the so-called rank 2 layered materials. Then, an empirical model is proposed for the power law materials (also called SIMP materials). In a second part, solution aspects of topology problems are considered. To deal with the so-called 'singularity' phenomenon of stress constraints in topology design, an $\epsilon$ constraint relaxation of the stress constraints is used. We describe the mathematical programming approach that is used to solve the numerical optimization problems. The proposed strategy is applied to illustrative applications. [less ▲]Detailed reference viewed: 46 (1 ULg) Layout Optimization : A Mathematical Programming ApproachDuysinx, Pierre Report (1997)This paper presents applications of specially tailored methods of the mathematical programming approach for solving topology design problems formulated as an optimal material distribution, providing a way ... [more ▼]This paper presents applications of specially tailored methods of the mathematical programming approach for solving topology design problems formulated as an optimal material distribution, providing a way to enlarge the scope of potential applications of topology optimization. The first feature of the methodology is to resort to dual maximization to solve optimization problems with a huge number of design variables, using the concept of Sequential Convex Programming (SCP). Here a central theme is the choice of convex approximations suited to the problems. First order approximations are firstly considered and compared. However, to increase the performances of the procedures (for example, to reduce the number of steps to arrive to a stationary design), a new approach to build un-expensive second approximation schemes, especially relevant for large scale problems, have been developed and validated. To illustrate the efficiency of the mathematical programming approach, the final part of the paper deals with an efficient treatment of perimeter constraints. Such constraints are essential to regularize material distribution with non optimal microstructures, but perimeter constraints are very difficult to handle in a numerical procedure for practical designs. The paper presents a solution to this problem that has been implemented and validated on a wide range of applications. [less ▲]Detailed reference viewed: 77 (0 ULg) Convex Approximation Methods for Large Scale Structural ProblemsDuysinx, Pierre in Proceedings of the 9th Nordic Seminar on Computational Mechanics (NSCM IX) (1996, October)Detailed reference viewed: 24 (0 ULg) Layout Optimization : A Mathematical Programming ApproachDuysinx, Pierre Conference (1996, August)Detailed reference viewed: 6 (1 ULg) Introduction à l'optimisation topologiqueDuysinx, Pierre Report (1996)Detailed reference viewed: 28 (3 ULg) OPTIMISATION TOPOLOGIQUE : DU MILIEU CONTINU A LA STRUCTURE ELASTIQUEDuysinx, Pierre Doctoral thesis (1996)Detailed reference viewed: 86 (11 ULg) A Generalized Method of Moving Asymptotes (GMMA) Including Equality ConstraintsZhang, Weihong; Fleury, Claude ; Duysinx, Pierre et alin Structural Optimization (1996), 12(2/3), 143-146Detailed reference viewed: 65 (3 ULg)


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# Skip one step of the loop if there is an NIntegrate::eincr message

Is there a possibility of skipping one step of a loop if in this step there appears an error like NIntegrate::eincr? After skipping, the loop should go to the next step and continue evaluation.

-
Have you looked at Check and Continue in the documentation? –  rm -rf Oct 5 '12 at 15:50
As rm-rf remarks, this can be done with Check and Abort (and Quiet to keep things tidy):


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# Mysql – Import or restore MySQL databases from var directory

MySQLosx-snow-leopard

I recently backed up my entire laptop hard drive with SuperDuper and reinstalled Snow Leopard. Because I made an image clone, I was able to restore all my files without losing any data. Go hard drive cloning.

However, the one thing I forgot to do was to do a dump of all my MySQL databases. Eek. Because of the nature of MySQL's storage, I can't just drag and drop the databases and tables—all the data is locked up in the server.

When I mount the cloned image I can navigate to /Volumes/laptop/usr/local/mysql/var (since I had installed MySQL to /usr/local/mysql) and I can see all my databases listed as directories with random files inside, which seems promising.

What should I do to restore all those databases I forgot to dump? Is there a way to restore the var/ folder to my new, fresh installation of MySQL and (hopefully) maintain all the users and privileges? Can I just copy that directory and have everything work?

I've done similar restorations (under Windows, though) by simply copying the files (the whole /usr/local/mysql directory...shouldn't that be under /var/mysql anyway?). Of course you'll need to stop the MySQL daemon and restart if after you're done.


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Volume 284 - QCD Evolution 2016 (QCDEV2016) - Tuesday, May 31, 2016 - Afternoon1
Towards the three loop evolution equation for GPDs
A. Manashov
Full text: pdf
Pre-published on: February 10, 2017
Published on: March 28, 2017
Abstract
QCD in $d=4-2\epsilon$ dimensions and large number of flavors possesses a nontrivial infrared stable critical point.
At the critical point the theory is invariant under the conformal transformations. It implies that the evolution kernel of
leading-twist light-ray operators commutes with the generators of conformal transformations. Taking into account that the evolution
kernels in the \rm{MS}-scheme are not sensitive to the space-time dimensions one concludes that
QCD evolution equations in \rm{MS}-schemes have a hidden
symmetry. Namely, the evolution kernel commutes with the generators of the conformal algebra. The explicit form of the generators differs
from their canonical (classical) form and can be derived by studying conformal Ward identities.
Invariance with respect to the conformal transformations imposes nontrivial constraints on
the form of evolution kernels and allows one to restore the nonforward evolution kernels for the nonsinglet operators from
the known NNLO anomalous dimensions.
We present here the two loop expressions for the generators of conformal algebra.
DOI: https://doi.org/10.22323/1.284.0019
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


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Smooth population counts in 5-year age groups.

smooth_age_5_cf(Value, Age, OAG = TRUE, start_on = 0)

Arguments

Value numeric vector of counts in single, abridged, or 5-year age groups. numeric vector of ages corresponding to the lower integer bound of the counts. logical. Whether or not the top age group is open. Default TRUE. integer. Either 0 or 5. Do we constrain decades starting on the 0s or 5s?

Value

numeric vector of smoothed counts in 5-year age groups.

Details

This method does not account for ages < 10 nor for the 10 year age interval prior to the open age group. These are returned imputed with NA. Age classes must be cleanly groupable to 5-year age groups. Smoothed counts are constrained to sum to original totals in 10-year age groups.

References

Carrier NH, Farrag AM (1959). “The reduction of errors in census populations for statistically underdeveloped countries.” Population Studies, 12(3), 240--285. United States Census Bureau (2017). “Population Analysis System (PAS) Software.” https://www.census.gov/data/software/pas.html, https://www.census.gov/data/software/pas.html.

Examples

# from PASEX AGESMTH
Ages         <- seq(0, 80, by = 5)
KKNtest <- c(NA,NA,354871,278502,285508,261429,236513 ,
204233,162138,126555,90094,65988,54803,41041,NA,NA,NA)

CFmales <- smooth_age_5_cf(pop5m_pasex, Ages, TRUE)
CFtest <- c(NA,NA,346290,287083,285855,261082,237937,
202809,162973,125720,88730,67352,55187,40657,NA,NA,NA)
all(round(CFmales) - CFtest == 0, na.rm = TRUE)
#> [1] TRUEif (FALSE) {
plot(Ages, pop5m_pasex)
lines(as.integer(names(CFmales)),CFmales)
}


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# Suppose $p$, $q$ are distinct odd primes, $a\in\mathbb{Z}$, and $q|a^p-1$ but $q\nmid a-1$

From the assumptions above, I am trying to prove that $q=1+kp$ for some integer $k$ and that $k$ is even.

My thoughts thus far: Since $a^p\equiv 1$ mod $q$, I know that by a corollary of Fermat's little theorem that $a^p\equiv a$ mod $q$. So $q|a$ or $q|a^{p-1}-1$, but $q\nmid a$, so $q|a^{p-1}-1$.

And this is where I'm not sure where to go. If I can prove that $q\equiv 1$ mod $p$, then the conclusion would follow, and hopefully $k$ being even would follow as well. But I'm not sure.

Thoughts are greatly appreciated!

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This is homework, so I'll get you started.

We must have $q\not\mid a$, as otherwise $a^p\equiv 0\not\equiv 1\mod{q}$. Let $d$ be the order of $a$ modulo $q$, that is, the least positive integer such that $a^d\equiv 1\mod{q}$.

From here, what can we say about $d$? We have two conditions: the given one $a^p\equiv 1\mod{q}$ and Fermat's Little Theorem $a^{q-1}\equiv 1\mod{q}$. What do these tell us about $d$?

Once we know that $q=kp+1$ for some integer $k$, what happens if $k$ is odd?

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Excellent, thank you! –  Domonic Mei Nov 26 '12 at 3:37

Let $b$=$a$mod$q$. By considering the order of $a$ in $U_q$, we find that $|a|$ divides $q-1$. Since, |a| is different from 1 and divides p. Thus, |a|=p. Hence p|(q-1). Now it follows that q=1+kp

K has to be even. If k is odd we will find that 1+kp is even. This contradicts the fact that q is odd.

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Giáo trình

# College Physics

Science and Technology

## Quantum Numbers and Rules

Tác giả: OpenStaxCollege

Physical characteristics that are quantized—such as energy, charge, and angular momentum—are of such importance that names and symbols are given to them. The values of quantized entities are expressed in terms of quantum numbers, and the rules governing them are of the utmost importance in determining what nature is and does. This section covers some of the more important quantum numbers and rules—all of which apply in chemistry, material science, and far beyond the realm of atomic physics, where they were first discovered. Once again, we see how physics makes discoveries which enable other fields to grow.

The energy states of bound systems are quantized, because the particle wavelength can fit into the bounds of the system in only certain ways. This was elaborated for the hydrogen atom, for which the allowed energies are expressed as ${E}_{n}\propto 1/{n}^{2}$, where $n=1, 2, 3, ...$. We define $n$ to be the principal quantum number that labels the basic states of a system. The lowest-energy state has $n=1$, the first excited state has $n=2$, and so on. Thus the allowed values for the principal quantum number are

$n=1, 2, 3, ....$

This is more than just a numbering scheme, since the energy of the system, such as the hydrogen atom, can be expressed as some function of $n$, as can other characteristics (such as the orbital radii of the hydrogen atom).

The fact that the magnitude of angular momentum is quantized was first recognized by Bohr in relation to the hydrogen atom; it is now known to be true in general. With the development of quantum mechanics, it was found that the magnitude of angular momentum $L$ can have only the values

$L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left(l=0, 1, 2, ...,\phantom{\rule{0.25em}{0ex}}n-1\right)\text{,}$

where $l$ is defined to be the angular momentum quantum number. The rule for $l$ in atoms is given in the parentheses. Given $n$, the value of $l$ can be any integer from zero up to $n-1$. For example, if $n=4$, then $l$ can be 0, 1, 2, or 3.

Note that for $n=1$, $l$ can only be zero. This means that the ground-state angular momentum for hydrogen is actually zero, not $h/2\pi$ as Bohr proposed. The picture of circular orbits is not valid, because there would be angular momentum for any circular orbit. A more valid picture is the cloud of probability shown for the ground state of hydrogen in [link]. The electron actually spends time in and near the nucleus. The reason the electron does not remain in the nucleus is related to Heisenberg’s uncertainty principle—the electron’s energy would have to be much too large to be confined to the small space of the nucleus. Now the first excited state of hydrogen has $n=2$, so that $l$ can be either 0 or 1, according to the rule in $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ . Similarly, for $n=3$, $l$ can be 0, 1, or 2. It is often most convenient to state the value of $l$, a simple integer, rather than calculating the value of $L$ from $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$. For example, for $l=2$, we see that

$L=\sqrt{2\left(2+1\right)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }=0\text{.}\text{390}h=2\text{.}\text{58}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s.$

It is much simpler to state $l=2$.

As recognized in the Zeeman effect, the direction of angular momentum is quantized. We now know this is true in all circumstances. It is found that the component of angular momentum along one direction in space, usually called the $z$-axis, can have only certain values of ${L}_{z}$. The direction in space must be related to something physical, such as the direction of the magnetic field at that location. This is an aspect of relativity. Direction has no meaning if there is nothing that varies with direction, as does magnetic force. The allowed values of ${L}_{z}$ are

${L}_{z}={m}_{l}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left({m}_{l}=-l,-l+1, ...,\phantom{\rule{0.25em}{0ex}}-1, 0, 1, ...\phantom{\rule{0.25em}{0ex}}l-1,\phantom{\rule{0.25em}{0ex}}l\right)\text{,}$

where ${L}_{z}$ is the $z$-component of the angular momentum and ${m}_{l}$ is the angular momentum projection quantum number. The rule in parentheses for the values of ${m}_{l}$ is that it can range from $-l$ to $l$ in steps of one. For example, if $l=2$, then ${m}_{l}$ can have the five values –2, –1, 0, 1, and 2. Each ${m}_{l}$ corresponds to a different energy in the presence of a magnetic field, so that they are related to the splitting of spectral lines into discrete parts, as discussed in the preceding section. If the $z$-component of angular momentum can have only certain values, then the angular momentum can have only certain directions, as illustrated in [link].

What Are the Allowed Directions?

Calculate the angles that the angular momentum vector $\mathbf{\text{L}}$ can make with the $z$-axis for $l=1$, as illustrated in [link].

Strategy

[link] represents the vectors $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ as usual, with arrows proportional to their magnitudes and pointing in the correct directions. $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ form a right triangle, with $\mathbf{L}$ being the hypotenuse and ${\mathbf{\text{L}}}_{z}$ the adjacent side. This means that the ratio of ${\mathbf{L}}_{z}$ to $\mathbf{L}$ is the cosine of the angle of interest. We can find $\mathbf{\text{L}}$ and ${\mathbf{\text{L}}}_{z}$ using $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$ and ${L}_{z}=m\frac{h}{2\pi }$.

Solution

We are given $l=1$, so that ${m}_{l}$ can be +1, 0, or −1. Thus $L$ has the value given by $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$.

$L=\frac{\sqrt{l\left(l+1\right)}h}{2\pi }=\frac{\sqrt{2}h}{2\pi }$

${L}_{z}$ can have three values, given by ${L}_{z}={m}_{l}\frac{h}{2\pi }$.

${L}_{z}={m}_{l}\frac{h}{2\pi }=\left\{\begin{array}{cccc}\phantom{\rule{0.25em}{0ex}}\frac{h}{2\pi },& \phantom{\rule{0.25em}{0ex}}{m}_{l}& =& +1\phantom{\rule{0.25em}{0ex}}\\ \phantom{\rule{0.25em}{0ex}}0,\phantom{\rule{0.25em}{0ex}}& {m}_{l}& =& 0\phantom{\rule{0.25em}{0ex}}\\ -\frac{h}{2\pi },& \phantom{\rule{0.25em}{0ex}}{m}_{l}& =& -1\end{array}$

As can be seen in [link], $cos\phantom{\rule{0.25em}{0ex}}\theta ={\text{L}}_{z}\text{/L,}$ and so for ${m}_{l}=+1$, we have

$\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{1}=\frac{{L}_{Z}}{L}=\frac{\frac{h}{2\pi }}{\frac{\sqrt{2}h}{2\pi }}=\frac{1}{\sqrt{2}}=0\text{.}\text{707.}$

Thus,

${\theta }_{1}={\text{cos}}^{-1}\text{0.707}=\text{45}\text{.}0º.$

Similarly, for ${m}_{l}=0$, we find $\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{2}=0$; thus,

${\theta }_{2}={\text{cos}}^{-1}0=\text{90}\text{.}0º.$

And for ${m}_{l}=-1$,

$\text{cos}\phantom{\rule{0.25em}{0ex}}{\theta }_{3}=\frac{{L}_{Z}}{L}=\frac{-\frac{h}{2\pi }}{\frac{\sqrt{2}h}{2\pi }}=-\frac{1}{\sqrt{2}}=-0\text{.}\text{707,}$

so that

${\theta }_{3}={\text{cos}}^{-1}\left(-0\text{.}\text{707}\right)=\text{135}\text{.}0º.$

Discussion

The angles are consistent with the figure. Only the angle relative to the $z$-axis is quantized. $L$ can point in any direction as long as it makes the proper angle with the $z$-axis. Thus the angular momentum vectors lie on cones as illustrated. This behavior is not observed on the large scale. To see how the correspondence principle holds here, consider that the smallest angle (${\theta }_{\text{1}}$ in the example) is for the maximum value of ${m}_{l}=0$, namely ${m}_{l}=l$. For that smallest angle,

$\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{L}_{z}}{L}=\frac{l}{\sqrt{l\left(l+1\right)}}\text{,}$

which approaches 1 as $l$ becomes very large. If $\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =1$, then $\theta =0º$. Furthermore, for large $l$, there are many values of ${m}_{l}$, so that all angles become possible as $l$ gets very large.

# Intrinsic Spin Angular Momentum Is Quantized in Magnitude and Direction

There are two more quantum numbers of immediate concern. Both were first discovered for electrons in conjunction with fine structure in atomic spectra. It is now well established that electrons and other fundamental particles have intrinsic spin, roughly analogous to a planet spinning on its axis. This spin is a fundamental characteristic of particles, and only one magnitude of intrinsic spin is allowed for a given type of particle. Intrinsic angular momentum is quantized independently of orbital angular momentum. Additionally, the direction of the spin is also quantized. It has been found that the magnitude of the intrinsic (internal) spin angular momentum, $S$, of an electron is given by

$S=\sqrt{s\left(s+1\right)}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\text{(}s=1/2\phantom{\rule{0.25em}{0ex}}\text{for electrons),}$

where $s$ is defined to be the spin quantum number. This is very similar to the quantization of $L$ given in $L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }$, except that the only value allowed for $s$ for electrons is 1/2.

The direction of intrinsic spin is quantized, just as is the direction of orbital angular momentum. The direction of spin angular momentum along one direction in space, again called the $z$-axis, can have only the values

${S}_{z}={m}_{s}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left({m}_{s}=-\frac{1}{2},+\frac{1}{2}\right)$

for electrons. ${S}_{z}$ is the $z$-component of spin angular momentum and ${m}_{s}$ is the spin projection quantum number. For electrons, $s$ can only be 1/2, and ${m}_{s}$ can be either +1/2 or –1/2. Spin projection ${m}_{s}\text{=+}1/2$ is referred to as spin up, whereas ${m}_{s}=-1/2$ is called spin down. These are illustrated in [link].

To summarize, the state of a system, such as the precise nature of an electron in an atom, is determined by its particular quantum numbers. These are expressed in the form $\left(\mathrm{n, l,}\phantom{\rule{0.25em}{0ex}}{m}_{l},\phantom{\rule{0.25em}{0ex}}{m}_{s}\right)$ —see [link] For electrons in atoms, the principal quantum number can have the values $n=1, 2, 3, ...$. Once $n$ is known, the values of the angular momentum quantum number are limited to $l=1, 2, 3, ...,n-1$. For a given value of $l$, the angular momentum projection quantum number can have only the values ${m}_{l}=-l,\phantom{\rule{0.25em}{0ex}}-l\phantom{\rule{0.25em}{0ex}}+1, ...,-1, 0, 1, ...,\phantom{\rule{0.25em}{0ex}}l-1,\phantom{\rule{0.25em}{0ex}}l$. Electron spin is independent of $\mathrm{n, l,}$ and ${m}_{l}$, always having $s=1/2$. The spin projection quantum number can have two values, ${m}_{s}=1/2\phantom{\rule{0.25em}{0ex}}\text{or}\phantom{\rule{0.25em}{0ex}}-1/2$.

 Name Symbol Allowed values Principal quantum number $n$ $1, 2, 3, ...$ Angular momentum $l$ $0, 1, 2, ... n − 1$ Angular momentum projection $m l$ $− l , − l + 1, ... , − 1, 0, 1, ... , l − 1, l ( or 0, ±1, ±2, ... , ± l )$ SpinThe spin quantum number s is usually not stated, since it is always 1/2 for electrons $s$ $\text{1/2}\left(\text{electrons}\right)$ Spin projection $m s$ $-1/2,\phantom{\rule{0.25em}{0ex}}+1/2$

[link] shows several hydrogen states corresponding to different sets of quantum numbers. Note that these clouds of probability are the locations of electrons as determined by making repeated measurements—each measurement finds the electron in a definite location, with a greater chance of finding the electron in some places rather than others. With repeated measurements, the pattern of probability shown in the figure emerges. The clouds of probability do not look like nor do they correspond to classical orbits. The uncertainty principle actually prevents us and nature from knowing how the electron gets from one place to another, and so an orbit really does not exist as such. Nature on a small scale is again much different from that on the large scale.

We will see that the quantum numbers discussed in this section are valid for a broad range of particles and other systems, such as nuclei. Some quantum numbers, such as intrinsic spin, are related to fundamental classifications of subatomic particles, and they obey laws that will give us further insight into the substructure of matter and its interactions.

# Section Summary

• Quantum numbers are used to express the allowed values of quantized entities. The principal quantum number $n$ labels the basic states of a system and is given by
$n=1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}3,\text{.}\text{.}\text{.}.$
• The magnitude of angular momentum is given by
$L=\sqrt{l\left(l+1\right)}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left(l=0, 1, 2, ...,\phantom{\rule{0.25em}{0ex}}n-1\right),$
where $l$ is the angular momentum quantum number. The direction of angular momentum is quantized, in that its component along an axis defined by a magnetic field, called the $z$-axis is given by
${L}_{z}={m}_{l}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left({m}_{l}=-l,-l+1, ...,\phantom{\rule{0.25em}{0ex}}-1, 0, 1, ...\phantom{\rule{0.25em}{0ex}}l-1,\phantom{\rule{0.25em}{0ex}}l\right)\text{,}$
where ${L}_{z}$ is the $z$-component of the angular momentum and ${m}_{l}$ is the angular momentum projection quantum number. Similarly, the electron’s intrinsic spin angular momentum $S$ is given by
$S=\sqrt{s\left(s+1\right)}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\text{(}s=1/2\phantom{\rule{0.25em}{0ex}}\text{for electrons),}$
$s$ is defined to be the spin quantum number. Finally, the direction of the electron’s spin along the $z$-axis is given by
${S}_{z}={m}_{s}\frac{h}{2\pi }\phantom{\rule{1.00em}{0ex}}\left({m}_{s}=-\frac{1}{2},+\frac{1}{2}\right)\text{,}$
where ${S}_{z}$ is the $z$-component of spin angular momentum and ${m}_{s}$ is the spin projection quantum number. Spin projection ${m}_{s}\text{=+}1/2$ is referred to as spin up, whereas ${m}_{s}=-1/2$ is called spin down. [link] summarizes the atomic quantum numbers and their allowed values.

# Conceptual Questions

Define the quantum numbers $\mathrm{n, l,}\phantom{\rule{0.25em}{0ex}}{m}_{l}\mathrm{, s}$, and ${m}_{s}$.

For a given value of $n$, what are the allowed values of $l$?

For a given value of $l$, what are the allowed values of ${m}_{l}$? What are the allowed values of ${m}_{l}$ for a given value of $n$? Give an example in each case.

List all the possible values of $s$ and ${m}_{s}$ for an electron. Are there particles for which these values are different? The same?

# Problem Exercises

If an atom has an electron in the $n=5$ state with ${m}_{l}=3$, what are the possible values of $l$?

$l=4, 3$ are possible since $l and $\mid {m}_{l}\mid \text{≤}l$.

An atom has an electron with ${m}_{l}=2$. What is the smallest value of $n$ for this electron?

What are the possible values of ${m}_{l}$ for an electron in the $n=4$ state?

$n=4⇒l=3, 2, 1, 0⇒{m}_{l}=±3,\phantom{\rule{0.25em}{0ex}}±\text{2,}\phantom{\rule{0.25em}{0ex}}±\text{1, 0}$ are possible.

What, if any, constraints does a value of ${m}_{l}=1$ place on the other quantum numbers for an electron in an atom?

(a) Calculate the magnitude of the angular momentum for an $l=1$ electron. (b) Compare your answer to the value Bohr proposed for the $n=1$ state.

(a) $1\text{.}\text{49}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s$

(b) $1\text{.}\text{06}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s$

(a) What is the magnitude of the angular momentum for an $l=1$ electron? (b) Calculate the magnitude of the electron’s spin angular momentum. (c) What is the ratio of these angular momenta?

Repeat [link] for $l=3$.

(a) $3\text{.}\text{66}×{\text{10}}^{-\text{34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s$

(b) $s=9\text{.}\text{13}×{\text{10}}^{-\text{35}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot s$

(c) $\frac{L}{S}=\frac{\sqrt{\text{12}}}{\sqrt{3/4}}=4$

(a) How many angles can $L$ make with the $z$-axis for an $l=2$ electron? (b) Calculate the value of the smallest angle.

What angles can the spin $S$ of an electron make with the $z$-axis?

$\theta =\text{54.7º, 125.3º}$


text:
# Lesson 14

Representing Fractions on a Line Plot

## Warm-up: Which One Doesn’t Belong: Line Plot (10 minutes)

### Narrative

The purpose of this warm-up is for students to recall the line plots with fractional measurements which they have studied in prior courses. This prepares them to do more arithmetic with fractions using the data from line plots in the next two lessons.

### Launch

• Groups of 2
• Display the image.
• “Pick one that doesn’t belong. Be ready to share why it doesn’t belong.”
• 1 minute: quiet think time

### Activity

• 2–3 minutes: partner discussion
• Share and record responses.

### Student Facing

Which one doesn’t belong?

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

• “Why doesn’t B belong?” (There are no fractions marked and there is no data with fractional values.)
• “Today we are going to work with line plots with fractional data and use what we have learned about fractions to solve problems.”

## Activity 1: Sums of Fractions (15 minutes)

### Narrative

The purpose of this activity is for students to make a line plot and answer questions about the data collected. The numbers that students plot come from spinning a spinner twice and adding the fractions on the spinner. The denominators are chosen so that 8 can be used as a common denominator. Students observe and think about patterns and then discuss them during the synthesis.

Engagement: Provide Access by Recruiting Interest. Synthesis: Revisit math community norms to prepare students for the whole-class discussion.
Supports accessibility for: Attention, Social-Emotional Functioning

### Required Materials

Materials to Gather

### Required Preparation

• Each group of 2 needs 1 paper clip and one pencil.

### Launch

• Groups of 2
• Display the number line image from student workbook.
• “You are going to play a game with your partner. You will use a paper clip and a pencil to make a spinner out of the image in your workbook. Let’s practice.”
• Demonstrate how to use the pencil and paper clip as a spinner. Spin twice and record the fractions you landed on for all to see.
• “I need to find the value of the sum of these two fractions.”
• Demonstrate how to record the sum on the number line with an X.
• “Which number do you think will have the most Xs if you spin the spinner a lot of times? Why?”
• 1–2 minutes: partner discussion

### Activity

• 1–2 minutes: quiet think time
• 6–8 minutes: partner work time
• Monitor for students who:
• partition the number line into eighths
• use common denominators to convert fractions with unlike denominators to fractions with like denominators

### Student Facing

1. Play Sums of Fractions with your partner.

• Take turns with your partner.
• Spin the spinner twice.
• Record the sum on the line plot.
• Play the game until you and your partner together have 12 data points.

2. How did you know where to plot the sums of eighths?
3. What is the difference between your highest and lowest number?
4. What do you notice about the data you collected?

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

• Ask previously identified students to share their thinking.
• “Did anyone record a one? What did you spin?” (I got $$\frac{1}{2}$$ on both spins.)
• “Is there any other way to get 1 as a sum?” (No. I would need to add $$\frac{3}{4}$$ to$$\frac{1}{4}$$, $$\frac{7}{8}$$ to $$\frac{1}{8}$$, and $$\frac{3}{8}$$ to $$\frac{5}{8}$$ and none of those is possible.)
• “What is the largest number you recorded?” (Sample responses: $$1\frac{2}{8}$$, $$1 \frac{1}{8}$$)
• “Is it possible to get more than $$1\frac{2}{8}$$?” (No, the biggest number is $$\frac{5}{8}$$ and two of those is $$1\frac{2}{8}$$.)
• “Name a fraction that would have made the game more challenging if it were on the number mat. Why would this have made the game more challenging?” ($$\frac{2}{3}$$, $$\frac{4}{5}$$, or any other fraction with a denominator that is not a factor or multiple of 8. It would be more challenging because we could not use 8 as a common denominator to easily add the fractions.)

## Activity 2: Lots of Eggs (20 minutes)

### Narrative

The purpose of this activity is for students to use measurement data to make a line plot and then solve problems about the data presented in the line plot (MP2). The line plot is blank so students will choose which whole numbers to label and which fractions to label in between. They will use their understanding of equivalent fractions (halves, quarters, and eighths) to accurately make the line plot. Jada’s statement about the eggs that weigh $$1\frac{7}{8}$$ ounces is interesting because it uses two fractions referring to different quantities: $$\frac{1}{4}$$ is a fraction of the eggs and $$1\frac{7}{8}$$ is their weight in ounces. The focus of the synthesis is on how students reason about Jada’s statement. As students reason through Jada’s statement, they critique the reasoning of others (MP3).

MLR8 Discussion Supports. Prior to solving the problems, invite students to make sense of the situations and take turns sharing their understanding with their partner. Listen for and clarify any questions about the context.

### Launch

• Groups of 2
• Display the image.
• “What do you notice?” (It is a chicken. There is an egg. There are numbers and units.)
• “What do you wonder?” (What is it for? Is it a scale? Why does it say small, medium, large, X-Large?)
• “This is an egg scale. It is used to weigh eggs. What can you say about the egg on the scale?” (It's small. It weighs less than 2 ounces.)
• “Now you will make a line plot and answer questions about the egg weights.”

### Activity

• 5 minutes: independent work time
• 5 minutes: partner work time
• Monitor for students who use either of these expressions to determine if Jada is correct:
• $$\frac{1}{4}\times12=3$$
• $$12\div4=3$$

### Student Facing

1. Use the egg weights below to make a line plot.

$$1\frac{7}{8}$$, $$2 \frac{1}{2}$$, $$2\frac{3}{8}$$, $$1 \frac{3}{4}$$, $$2\frac{1}{4}$$, $$2\frac{4}{8}$$, $$2 \frac{1}{8}$$, $$1\frac{7}{8}$$, $$2\frac{1}{4}$$, $$1\frac{6}{8}$$, $$2\frac{1}{8}$$, $$1\frac{7}{8}$$

2. Jada said that $$\frac{1}{4}$$ of the eggs weigh $$1\frac{7}{8}$$ ounces. Do you agree? Explain or show your reasoning.
3. How much heavier is the heaviest egg than the lightest egg? Explain or show your reasoning.

### Student Response

For access, consult one of our IM Certified Partners.

### Activity Synthesis

• Ask previously selected students to share their response to Jada's statement.
• “How did you know how many eggs Jada weighed?” (I could count the measurements or count the marks on the graph.)
• “How did you decide if her statement was correct?” (I found $$\frac{1}{4}$$ of 12 and then counted the number eggs that weighed $$1\frac{7}{8}$$ ounces. They were both 3 so Jada is correct.)
• If students do not write an equation matching Jada’s response, consider displaying the equations $$\frac{1}{4} \times 12 = 3$$ and $$12 \div 4 = 3$$
• “How do these equations represent Jada’s reasoning?” (They both say that $$\frac{1}{4}$$ of the 12 egg measurements is 3 eggs. One uses multiplication and the other one uses division.)

## Lesson Synthesis

### Lesson Synthesis

“How are the line plots different?” (The numbers for the game are just numbers. The other ones are weights. The numbers for the game only go up to $$\frac{10}{8}$$. There is more variation in the weights of the eggs.)