text:
# Diagnostics in DHARMa for glmmTMB model

I am fitting models to a data set with 370 observations. It is ecological data with overdispersed counts as a response variable.

I have used the DHARMa package to show this overdispersion from a poisson distribution visually. As such I have been trying to get a well fit model using the nbinom1 and nbinom2 family arguments in glmmTMB. like so:

> gNB4<-glmmTMB(count ~ scale(Dichrostachys)*Season
+ scale(Acacia.sp)*scale(size)
+ scale(Commiphora)
+ (1|YEAR)
+ (1|TransectID)
, family="nbinom2", verbose=F, data=GIRAFFE.NB, ziformula=~0)

However, I have not managed to get a 'non-significant' deviation of the simulations from my predicted model as shown in the image below. Even after trying up to 5000 simulations.

I notice from the graphs that there is a peak in the histogram on the left side, that I interpreted as the bad fit of the QQ-plot close to zero. I also note that this shape (QQ-Plot) seems somehow similar to an underdispersed data set as shown in this vignette. I have not found a way of dealing with underdispersion (if that is the issue) in glmmTMB since I don't see an option for quasi-poisson distributions. Plus I thought the middle graph indicates the dispersion is more or less how it should be.

I have scoured the internet for weeks for other info on how to improve the fit, but either it is tacit knowledge or I am looking in all the wrong places.

Extra notes:

• Changing the variable combinations influences the QQ plot but I have not managed to get it any closer to passing the KS test than in the image above.
• I have tried using quadratic and square root adjustments to predictor variables (where it makes sense)
• I have tried fitting the most parsimonious models first (AIC selection) before testing
• I will happily share the data set if it is required but recreating a data set with the same nuances would be difficult (for me)

Question

Should I walk away at this point and in the write up basically void this model by saying it was not robust (i.e. too much variation between predictions from the fitted model) or do I still have some options? Ideally, I am looking for practical advice.

• Do you really want to treat YEAR as a grouping variable for which you include a random intercept? By doing that you assume that measurements on the same year are correlated, even if when they come from different TransectID`s.

format: plain


text:
# Is there a selective wet etch of Bi2O3 Bismuth Oxide?

I'm looking for a wet etch of Bismuth Oxide that does not etch $\ce{Bi}$ itself.

There are obviously chemicals that etch $\ce{Bi2O3}$, but $\ce{Bi}$ is pretty fragile, so almost everything I've found that can etch $\ce{Bi2O3}$ also etches Bi, and usually more rapidly than the $\ce{Bi2O3}$ itself.

However, someone recently mentioned to me in passing that there was a chemical that could do this, but they didn't remember what it was.

Searching Google, I found this, but that's RIE -- not wet etching.

I've also read about people using $\ce{H2}$ gas to reduce the oxide back to $\ce{Bi}$, but again, not wet etching.

Does anyone have any idea? I know it's a specific and uncommon question.

• I saw on wikipedia that bismuth oxide is transformed to bismuth trioleate by a mixture of acetic anhydride and oleic acid. I'm not 100% sure but I doubt either of those materials does much to bismuth metal. Maybe that could be the basis of a wet etching? – Curt F. Jul 7 '15 at 21:36


text:
# What is the Cartesian form of ( -5 , ( - 16pi)/3 ) ?

$\left(\frac{5}{2} , \frac{- 5 \sqrt{3}}{2}\right)$

#### Explanation:

$x = r \cos \theta$

$x = - 5 \cdot \cos \left(- 16 \frac{\pi}{3}\right)$

$x = \frac{5}{2}$

and

$y = r \cdot \sin \theta$

$y = - 5 \cdot \sin \left(- 16 \frac{\pi}{3}\right)$

$y = \frac{- 5 \sqrt{3}}{2}$


text:
Search for lepton-flavour-violating decays of the $Z$ boson into a $\tau$-lepton and a light lepton with the ATLAS detector

Submission summary

 As Contributors: Wing Sheung Chan Preprint link: scipost_201812_00035v1 Date accepted: 2019-01-17 Date submitted: 2018-12-07 Submitted by: Chan, Wing Sheung Submitted to: SciPost Physics Proceedings Proceedings issue: The 15th International Workshop on Tau Lepton Physics (Amsterdam, 2018-09) Domain(s): Experimental Subject area: High-Energy Physics - Experiment

Abstract

Lepton flavour violation in the charged lepton sector is an unambiguous signal of physics beyond the Standard Model. Searches for lepton flavour violation in decays of the $Z$ boson with the ATLAS detector is reported, focusing on decays into an electron or muon and a hadronically decaying $\tau$-lepton, using $pp$ collisions data with a centre-of-mass energy of 13 TeV. Upper limits on the branching ratios of lepton-flavor-violating decays are set at the 95\% confidence level: $\mathcal{B}(Z\to e\tau) < 5.8\times10^{-5}$ and $\mathcal{B}(Z\to \mu\tau) < 2.4\times10^{-5}$. When combined with a previous ATLAS result based on $pp$ collisions data with a centre-of-mass energy of 8 TeV, an upper limit of $\mathcal{B}(Z\to \mu\tau) < 1.3\times10^{-5}$ is obtained.

Submission & Refereeing History

Submission scipost_201812_00035v1 on 7 December 2018


text:
# Minimal dominating subsets in infinite graphs

Let $G=(V,E)$ be any simple, undirected graph. A dominating set is a set $D\subseteq V$ such that for all $v\in V\setminus D$ there is $d\in D$ such that $\{v,d\}\in E$.

Is there an infinite graph $G=(V,E)$ such that there is a dominating subset $D\subseteq V$ such that for any dominating subset $D_1\subseteq D$ there is a dominating subset $D_2\subseteq D_1$ with $D_2\neq D_1$?

• $G=(\mathbb N,\le)$ ?
– saf
Jan 30 '17 at 16:02
• @saf That doesn't work, since any singleton is dominating, but the empty set is not. Jan 30 '17 at 16:10

Let $V(G)$ be the set of non-empty subsets of $\mathbb N$ and join two sets by an edge whenever they intersect. Let $D$ be the set of initial segments of $\mathbb N$. Then subsets of $D$ are dominating if and only if they are infinite.


text:
x

## GENERAL

2022, 71 (22): 220201. doi: 10.7498/aps.71.20220765
Abstract +

## GENERAL

2022, 71 (22): 220301. doi: 10.7498/aps.71.20220871
Abstract +
In the application of quantum communication networks, it is an important task to realize the optimal allocation of resources according to the current situation. For example, We need to select the optimal quantum key distribution (QKD) protocol and parameters. Traditionally, the most commonly implemented method is the local search algorithm (LSA), which costs a lot of resources. Here in this work, we propose a machine learning based scheme, in which the regression machine learning is used to simultaneously select the optimal protocol and corresponding parameters. In addition, we make comparisons among a few machine learning models including random forest (RF), K-nearest neighbor (KNN) and logistic regression. Simulation results show that the new scheme takes much less time than the LSA scheme, and the RF achieves the best performance. In addition, through the RF residual analysis, we find that the machine learning scheme has good robustness. In conclusion, this work may play an important role in promoting the practical application of quantum communication networks.

## GENERAL

2022, 71 (22): 220302. doi: 10.7498/aps.71.20221125
Abstract +
Quantum communication is a frontier hotspot of current research, and it has ideal information security. In order to enable quantum systems in arid and desertified areas to work almost under all-weather condition, it is necessary to study the attenuation of free-space quantum signal transmission and the influence of the turbulence atmosphere carrying sand and dust on communication performance. Using Mie scattering theory, multiple scattering simulation method, and atmospheric turbulence theory, the attenuation of optical wave transmission in sand and dust turbulent atmospheric channels with different visibility, and the influence of multiple scattering and turbulence on attenuation are studied. The results show that the effect of multiple scattering increases with the decrease of visibility, the turbulence effect gradually strengthens with the increase of distance. According to the quantum amplitude damped channel model, the effects of multiple scattering and turbulence in the sand and dust turbulent atmosphere with different visibility on the quantum channel capacity, fidelity and bit error rate are analyzed. The results show that as the visibility decreases, the multiple scattering effect increases, resulting in the decrease of attenuation and bit error rate, but an increase in channel capacity, fidelity and the boundaries of security key rate. The existence of turbulence in the dust atmosphere will increase the attenuation and bit error rate, but reduce the channel capacity, fidelity and security key rate. It can be seen that the influence of multiple scattering and turbulence on the communication performance, when the visibility of the sand and dust atmosphere are both low, cannot be ignored. In practical applications, the relevant parameters of quantum communication should be adaptively adjusted according to the visibility and turbulence intensity to improve the probability, efficiency and reliability of quantum communication.

## GENERAL

2022, 71 (22): 220303. doi: 10.7498/aps.71.20220760
Abstract +
All open quantum systems are affected by environmental noises due to their interactions with the external environment and inevitably suffer from decoherence. Hence, it is fundamentally important and necessary to investigate decoherence suppression for open quantum systems via proper control strategies. Inspired by feed-forward control in the classical control theory, this paper proposes a novel decoherence suppression scheme via weak measurement and environment-assisted measurement. We first take the single-qubit system as an example to illustrate steps of the proposed scheme. To be specific, the single-qubit system is transferred to a state that is more robust to environmental noises via pre-weak measurement operators and feed-forward control operators before the decoherence channel, a measurement is performed on the environment coupled to the protected qubit during the decoherence channel, and the initial state is recovered via reversed feed-forward control operators and post-weak measurement operators after the decoherence channel. The optimum post-weak measurement strength is derived by setting the normalized final state equal to the initial state. By considering the optimum post-weak measurement strength, analytical formulas of the total success probability and the total fidelity are deduced. The proposed scheme is applicable for protecting quantum states from arbitrary decoherence channels with at least one invertible Kraus operator although only the amplitude damping channel and the phase damping channel are taken into account. Provided that the decay rate of the amplitude or phase damping channel is completely known, one can always achieve unit fidelity even for heavy damping cases, which is the biggest advantage of the proposed scheme. Influences of several parameters including strengths of weak measurements, the initial state and the decay rate of the decoherence channel on the performance of decoherence suppression are analyzed, and detailed procedures of a single-qubit pure and mixed state protection are presented on the Bloch sphere, respectively. Subsequently, the Kronecker product is employed to construct operators of dimension $2^N \times 2^N$, the proposed scheme is extended to the general N-qubit case, and unified analytical formulas of the total success probability and the total fidelity are deduced. By applying the proposed scheme to the protection of two-qubit entangled states, it is demonstrated that post-weak measurement operators are not necessary sometimes because of the particular structure of two-qubit entangled states. Furthermore, two numerical simulations are designed to enhance the concurrence of two-qubit entangled states and improve the average fidelity of the standard quantum teleportation in a noisy environment. Analytical formulas of the improvement of concurrence and the average teleportation fidelity are deduced, and the superiority of the proposed scheme is highlighted in comparison with unprotected scenarios.

## GENERAL

2022, 71 (22): 220304. doi: 10.7498/aps.71.20221312
Abstract +
Bose-Einstein condensate (BEC) is essentially a macroscopic quantum effect with quantum volatility, macroscopic quantum coherence and artificial controllability. Owing to its unique controllability, it becomes a new ideal platform for quantum simulations and studies of interacting quantum systems.In this paper, the generation of vortices and the formation of vortex chains, as well as characteristics of vortex chains in rotating two-dimensional BEC in a potential composed of harmonic potential and optical lattice are studied numerically. Firstly, the generation of vortices, the formation and distribution of vortex chains and the effects of different physical parameters on the vortex chains in two-dimensional BEC are investigated by using the multigrid preconditioned conjugate gradient method. Secondly, the evolution of the vortex chains with time is studied by using the time-splitting spectral method. The results show that the generation of vortices in BEC trapped in the compound potential corresponds to the minimum value of the potential. When the depth of the optical lattice increases to a certain value, vortex chains are formed in the BEC. With the further increase of the depth of the optical lattice, the vortex depth in the vortex chain in the BEC decreases continuously, and finally the vortex chain disappears completely. When the interaction strength between atoms increases, the distribution range of the condensate expands, and the number of vortices and the number of vortex chains in the condensate also increase. When the interaction strength between atoms increases to a certain value, the symmetry of the vortex chains is broken. As the rotation frequency of the condensate increases, the distribution range of the condensate expands, and the number of vortices and the number of vortex chains in the condensate also increase. When the rotation frequency is close to the external trapping potential frequency, the linear alignment of the vortex chains is disrupted. It is also found that there are three stages in the evolution of the vortex chains in the BEC: in the first stage, vortex chains rotate together with the condensate, and the original chain distribution keeps unchanged; in the second stage, the phenomenon of vortex space extrusion appears, and the vortex chain is destroyed; in the third stage, the phenomenon of vortex space expansion occurs, and finally the vortex chains disappear. The results above show that the depth of the optical lattice, the interaction strength between atoms, and the rotation frequency of the condensate have important effects on the vortices and vortex chains in the condensate. By adjusting these physical quantities, the number of vortices and the shape of vortex chains in the BEC can be effectively manipulated. This may provide some theoretical reference and guidance for future experiments and applications.

## GENERAL

2022, 71 (22): 220701. doi: 10.7498/aps.71.20220775
Abstract +
The spin relaxation time of alkali atoms in nuclear magnetic resonance gyroscope is usually on the order of 10–5 s, which is much less than that in atomic magnetometers. The response of electron paramagnetic resonance signals of atoms with short relaxation time is asymmetric in different directions under oscillating magnetic fields, which makes the measurement results of atomic transverse relaxation time and Larmor frequency different. In this work this phenomenon is analyzed based on Bloch equation theory and the theoretical correction is given. The shorter the relaxation time, the greater the differences of the response intensity and resonance frequency of the electron paramagnetic resonance signal under different magnetic field directions will be. Using this property, the transmission delay time of the system can be measured accurately. In this paper proposed is a method of measuring transverse relaxation time based on the difference between signal phases in X-axis direction and Y-axis direction, which can accurately and quickly measure very short transverse relaxation time. The difference between the half-width fitting method and the phase measurement method is compared by measuring the transverse relaxation times of 87Rb atoms under different magnetic field intensities. The half-width fitting method is greatly affected by the transmission delay time and has its measurement limit. The phase measurement method is greatly affected by the angle of the probe light, but the measurement range is wider and the anti-magnetic interference ability is stronger.

## NUCLEAR PHYSICS

2022, 71 (22): 222101. doi: 10.7498/aps.71.20220795
Abstract +
We investigate the thermodynamical properties of strange quark matter (SQM) at zero/finite temperature and under constant magnetic field within quasiparticle model. The quark matter symmetry energy, energy per baryon, free energy per baryon, anisotropic pressures are also studied and the result indicates that both the effects of temperature and magnetic field can significantly influence the thermodynamical properties of quark matter and proto-quark stars (PQSs). Our result also indicates that the maximum mass and the core temperature of PQSs not only depends on the heating process at the isentropic stages, but also but also the magnetic field strength and orientation distribution inside the magnetar within quasiparticle model.

## NUCLEAR PHYSICS

2022, 71 (22): 222901. doi: 10.7498/aps.71.20221193
Abstract +
In order to improve the temporal resolution of photomultiplier tubes, our research group has conducted the in-depth research on photomultiplier tubes based on microchannel plates that are widely used at present. The time resolution of photomultiplier tube based on microchannel plate is limited by the transit time of photoelectric signal in each part, including the transit time of photoelectric signal in the transmission process of photocathode to microchannel plate, the transit time of photoelectric signal in microchannel plate time, the transit time of the photoelectric signal from the microchannel plate to the detector anode, and the transit time of the photoelectric signal on the anode to the electrode port. The transit time of the whole process has a certain degree of influence on the time information measurement of the optoelectronic signal. In this study, various parameters affecting the time resolution of the photomultiplier tube are analyzed, and it is found that the different positions of the photoelectron signal on the anode will bring errors to the measurement of the arrival time of the signal at the anode, and the photoelectric signal is transmitted to the electrode port at the affected point of the anode The spent time will cause the signal measurement time to lag behind the real time, which indirectly affects the time resolution of the system. Therefore, a specific study is carried out on the time measurement error of the signal on the anode, and it is determined that the difference of the photoelectron signal on the anode position is an important factor causing the signal time measurement error, and a simple and effective method of compensating for error is proposed. In the research process, the delay line anode is used, and the positional resolution principle of the photoelectric signal is used to obtain the position information of the photoelectron signal on the anode, and the position information is converted into the time information transmitted from the position to the electrode port. The theoretical value of the transit time on the anode is offset, eliminating unnecessary time in the time-of-arrival measurement of the photoelectron signal. The time measurement error of the optoelectronic signal is compensated for by this time information. The experimental results show that the error compensation method can effectively improve the time measurement accuracy of optoelectronic signals, and provide solutions and theoretical basis for improving the time resolution of photomultiplier tubes based on microchannel plates.

## ATOMIC AND MOLECULAR PHYSICS

2022, 71 (22): 223101. doi: 10.7498/aps.71.20221052
Abstract +
The electronic structures and single point energy of 14 lowest electronic states of 88Sr79Br molecule are optimized by using the internal contraction multi-reference configuration interaction method and relativistic effective core pseudo-potential basis. Because 88Sr79Br molecule belongs to heavy element system, the single point energy must be corrected to obtain more accurate spectral parameters. Therefore, Davidson is introduced to correct the energy inconsistency, nuclear valence correlation is used to correct the electron correlation effect of inner shell and valence shell, and the relativistic scalar effect is corrected by calculating the third-order Douglas-Kroll-Hess Hamilton single electron integral. According to the single point energy calculated by the modified optimization, the potential energy curves, electric dipole moments, and transition dipole moments of 14 lowest electronic states are obtained. Using the latest LEVEL8.0 program to fit the modified potential energy curve, the spectral constants, molecular constants and vibration energy levels of 5 lowest bound states of 88Sr79Br molecule are given. In order to explain the changing trend of spectral constants of homologous compounds, the spectral parameters of each compound are compared and analyzed in this paper. At the same time, the vibration energy levels and molecular constants of 88Sr81Br molecule are also fitted and calculated for analyzing the influence of isotopes. The comparative analysis shows that the results of 88Sr79Br molecule are in better agreement with the experimental values. Finally, the Franck-Condon factors are gained by fitting the optimized single point energy and transition dipole moment of 88Sr79Br molecule. The transition band with the largest factor and obvious diagonalization is selected by analyzing the Franck-Condon factor of each transition band, and whether it meets the conditions for selecting laser cooling molecular system is judged. The radiation lifetimes of the transitions from the lowest two excited states to the ground state are calculated by combining the transition dipole moment, Franck-Condon factor, single point energy and vibration energy level of each electronic state. The results of this paper are in good agreement with the experimental values, which shows that the method in this paper is reliable. These spectral characteristic parameters provide theoretical support for further experimental measurement and construction of molecular laser cooling scheme of 88Sr79Br molecule.

## ELECTROMAGNETISM, OPTICS, ACOUSTICS, HEAT TRANSFER, CLASSICAL MECHANICS, AND FLUID DYNAMICS

2022, 71 (22): 224101. doi: 10.7498/aps.71.20221284
Abstract +
In this paper, cavity-excited Huygens’ metasurface is proposed for high-efficiency wavefront manipulation. By adjusting the length of electric dipole and magnetic dipole , the proposed Huygens’ metasurface meta unit can provide nearly 360° phase coverage with sufficiently high transmission efficiency. Based on the analysis of the resonance mode of the cavity, the Huygens’ metasurface has successfully performed its function by adopting integrated feeding method. According to the generalized Snell’s law, metasurfaces with different phase gradients are designed. Combined with the cavity structure, one-dimensional Huygens’ metasurfaces excited by cavity is realized, which can directionally emit the electromagnetic waves from the cavity. Both the simulation and experimental results show that the proposed cavity excited metasurfaces can effectively manipulate the direction of the emitted beam. Such a kind of cavity-excited metasurface can flexibly control the emission angle of the electromagnetic wave, reduce the energy loss and improve the efficiency of the electromagnetic wave. These designs have the advantages of compact, light and easy integration.
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text:
CJS Modeling in MRA' In mra: Mark-Recapture Analysis

This vignette repeats and embellishes examples in help(F.cjs.estim).

Time-varying Models

The following demonstrates two methods for fitting a time-varying capture and survival model, the so-called "small t" model. First, we attach the mra library and obtain access to the example dipper data.

library(mra)
data("dipper.histories")
dim(dipper.histories)

Method 1: Using factors

The following code constructs a factor variable containing one level for each capture occasion. The attribute of this factor tells mra the "other"" dimension of the problem. Internally, mra will use this attribute to replicate the factor into matrices that are the appropriate size.

ct <- as.factor( paste("T",1:ncol(dipper.histories), sep=""))
attr(ct,"nan")<-nrow(dipper.histories)
ct

Next, call F.cjs.estim and specify that ct is a time-varying vector covariate using the tvar function. When the vector given to tvar is a factor, there are additional options which allow the user to drop certain levels of the factor. This is useful when coefficients for some levels are not estimable, as in the case of a completely time-varying CJS model. Here, there are 7 levels in factor ct, but only 6 capture and survival parameters are defined (recall, 1st capture parameter is not estimable, and 7th survival parameter between occasions 7 and 8 does not exist). Consequently, we tell tvar to drop the first two levels of ct from the capture model. We drop the level 1st because only 6 parameters exist. We drop the 2nd to break the colinearity of levels and define $p_2$ as the reference level. In the survival model, we drop the first, sixth, and seventh levels of ct. The first level is dropped to break the colinearity of levels and define $\phi_1$ as the reference level. We drop level 6 because the last survival and capture parameters are confounded in CJS models. We drop the 7th level because there are only 6 survival parameters. The call the F.cjs.estim is:

dipper1.cjs <- F.cjs.estim( ~tvar(ct,drop=c(1,2)), ~tvar(ct,drop=c(1,6,7)), dipper.histories )
dipper1.cjs

Method 2: Using explicit 2-D matricies

While using factors (Method 1 above) produces the most economical code, it does not adequately illuminate the covariate matrices which are at the heart of CJS modeling. To illustrate covariates as explicit matricies, this method constructs one 2-D matrix for each paramter, then estimates the same model as Method 1.

First, we construct 6 matricies containing 1's in a single column only. In Method 1, this construction was performed behind-the-scenes by tvar. Note that only 6 matricies are required due to the number of parameters, breaking of colinearity, and confounding of CJS paramters mentioned above.

x2 <- matrix(c(0,1,0,0,0,0,0), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)
x3 <- matrix(c(0,0,1,0,0,0,0), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)
x4 <- matrix(c(0,0,0,1,0,0,0), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)
x5 <- matrix(c(0,0,0,0,1,0,0), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)
x6 <- matrix(c(0,0,0,0,0,1,0), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)
x7 <- matrix(c(0,0,0,0,0,0,1), nrow(dipper.histories), ncol(dipper.histories), byrow=TRUE)

Each of the above matrices have a column of 1's corresponding to the effect they estimate. The first six rows of x3 and x4 are:

head(x3)

We now call F.cjs.extim without aid of tvar by explicitely specifying the matricies in each model. Note that x2 is not included in the capture model, and that x6 and x7 are not included in the survival model.

dipper2.cjs <- F.cjs.estim( ~x3+x4+x5+x6+x7, ~x2+x3+x4+x5, dipper.histories )
dipper2.cjs

Note that parameter estimates produced by Method 1 and Method 2 are identical.

Plot: $\hat{N}_j$ estimates

Following is a plot of the Horvitz-Thomson population size estimates.

plot(dipper1.cjs)

Plot: $\hat{\phi}_j$ estimates

Following is a plot of survival estimates containing one line per individual.

plot(dipper1.cjs,type="s",ci=FALSE)

Try the mra package in your browser

Any scripts or data that you put into this service are public.

mra documentation built on May 1, 2019, 6:50 p.m.


text:
# How do you solve 2s - 3t = 13 and 2s + 2t = -2?

Aug 22, 2015

The solutions are
color(blue)(s=2,t=-3

#### Explanation:

color(blue)(2s)−3t=13 ....equation $1$
 color(blue)(2s)+2t=−2.....equation $2$

Solving by elimination

Subtracting equation $1$ from $2$

cancelcolor(blue)(2s)−3t=13
$\cancel{\textcolor{b l u e}{- 2 s}} - 2 t = + 2$

$- 5 t = 15$
color(blue)(t=-3

Finding $s$from equation $2$
2s−3t=13
$2 s = 13 + 3 t$
$2 s = 13 - 9$
color(blue)(s=2


text:
# Poinsot's spirals

Poinsot's spirals

In mathematics, Poinsot's spirals are two spirals represented by the polar equations:$r=a,operatorname\left\{csch\right\}\left(n heta\right)$:$r=a,operatorname\left\{sech\right\}\left(n heta\right)$where csch is the hyperbolic cosecant, and sech is the hyperbolic secant.

Wikimedia Foundation. 2010.

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text:
# Volcano Watch — More submersible work on Lō‘ihi

Release Date:

Scientists from the University of Hawaii at Manoa are once again engaged in a research cruise over Loihi, the submarine volcano southeast of the Island of Hawaii.

Scientists from the University of Hawaii at Manoa are once again engaged in a research cruise over Loihi, the submarine volcano southeast of the Island of Hawaii. This research cruise, and another research cruise next week, were scheduled much earlier. However, with the intense earthquake swarm that began in mid-July, and with the findings of the specially arranged cruise in response to the July earthquake swarm, the focus of the September and October cruises has changed somewhat.

The Hawaiian Volcano Observatory seismographic network continues to register earthquakes originating from Loihi. In recent weeks, the rate of earthquake activity has decreased: from the 1052 Loihi earthquakes located in July, to 142 in August and 29, to this date, in September. This decrease in activity has allowed the scientists to use the manned submersible vessel Pisces V for their reconnaissance work.

The scientists have been looking for evidence that will suggest probable causes of the July-August earthquake swarm. They are looking for indications of recent eruptive activity and for new volcanic vents. They are also exploring the new features visible at the summit of Loihi, where a crater dubbed Pele's Pit was formed. It is quite likely that Pele's Pit formed in connection with the earthquake swarm.

Besides trying to determine the causes of the earthquake swarm, a very important element of this research cruise is to develop a new map of the ocean bottom at Loihi. With sophisticated sonar mapping tools onboard the Pisces V mother ship Kaimikai o Kanaloa, detailed bathymetric maps can be drawn. Like a topographic map of the island, the bathymetric maps will show the pit craters and other features that scientists will have to incorporate into their interpretations of the Loihi data.

The bathymetric maps will also help the Hawaii Undersea Geo Observatory, or HUGO, project. This project is being spearheaded by UH-Manoa scientists and will involve HVO in the data acquisition, analysis, and transmission. Prior to the July earthquake swarm, the HUGO plan was to place their base station near the place where Pele's Pit formed. Needless to say, that plan has changed, and the newly acquired bathymetry will help the HUGO team determine a new location for their equipment and allow the project to move ahead.

### Volcano Activity Update

The eruption from flank vents on the western side of Puu Oo continues unabated, with lava flowing through the six-mile long tube system and entering the ocean at Lae`apuki.

During the past week, the HVO seismographic network recorded and located 140 earthquakes. None of these earthquakes were reported as felt by anyone on the island.


text:
**Update** July 12, 2012 Surgery. Healed from slight necrosis/scar revision

I have been dealing with back/neck pain for as...

I have been dealing with back/neck pain for as long as I can remember. Currently I am a 38G, at least that's what one of my braas says, others are DDD, the sizes are confusing. Anyways, I always thought to myself, I'll have a BR after I've had kids. Well, I am now 35, single, and no kids. Back in September I went through a pretty substantial break up with someone I thought I was going to spend the rest of my life with. I had brought up the idea of a BR to him and he was not thrilled with the idea (Typical man). I wound up going to a new chiropractor who suggested that they start documenting in case I ever decided I wanted to get BR. After doing x-rays and finding bone spurs on my ribs and buldging disks in my lower back, I decided it was time to do this for me. I've lost 30 pounds, but my chest doesn't seem affected by the weight loss.

I went to my fist consultation with a doctor who was recommended. He told me that if my insurance covered the BR, it would not cover the lift and handed me a paper that had a $2,500 additional fee. I was pretty upset thinking that it would take me a long time to come up with the extra money. After much thought, I contacted some other doctors who said the lift is included in the surgery. So, I went to two other consultations to find the doctor that I liked. I was approved by insurance within a month and my only delay was that I am a teacher and waiting until summer to have plenty of time to recoup. Once I met my deductable, my out of pocket for the surgery is going to be$302, plus the surgery center fee, which I think is $100. I was amazed at how little I am going to be paying and feel very fortunate. This leads me to where I am now. Over 6 months ago I picked my doctor, spoke to the nurse about my desire to do my BR at the beginning on the summer (JUNE) and was assured there would be no problems because it is the slow time of the year for them. I had my pre-op on May 24th and am still waiting to have my surgery date scheduled. When I arrived to my pre-op, they told me they were suddenly booked up until mid-end of July. I go back to work on August 1st and will have to put my classroom back together (move desks, get textbooks, staple into walls, etc). I'm getting kind of anxious. The nurse is still working on trying to get me in earlier than July 18th. (She said possibly July 5th) So, I am now sitting in limbo waiting. I made no plans for my summer break other than to recover and maybe enjoy my new image a little before going back to work. Now, I am just working out a lot to get in as good of shape as I can before my surgery. Yesterday the nurse finally called and my surgery... Yesterday the nurse finally called and my surgery date is July 12. I ordered a bra on-line and it came today also. My PS asked me to stop taking my birth control now and two weeks before surgery, no vitamins and such. With about a month to go, I am focusing on getting into as good of shape as I can before my surgery. I'm a teacher, so I have a lot of extra time right now. Happy 4th of July! Yesterday the surgery center... Happy 4th of July! Yesterday the surgery center called with the details. 7am arrival time, which means I am the first of the day! I'm so ready to get this over with and be on my way to recovery! I seem to think about boobs all of the time. lol Constantly on here reading and researching. I've been to the chiropractor twice this week because of the pain I am having in my back and neck. I was running last week and then...bam! Sore back and neck. I have 7 days to go and can't believe that it is just next week, I'm so over this back and neck pain from these huge boobs!!! Today I am watching my very lively 7 month old nephew ( who is also a bit of a chunky monkey). I love him so much and it makes me a little sad that I won't be able to lift him up and hold him for a while. I wonder how others with kids have dealt with this. He is 20 pounds and definately over my lifting capacity. I don't think I can even risk holding him in my lap because he moves so much that he could probably hurt me on accident. Other than that, I've done almost everything I need to get done prior to surgery. At least I hope so. I plan to go grocery shopping on Monday and stock up, but I'll be at my parents house for about a week. Looking for some good books to read during my down time. I already read 50 shades of Grey series. lol I have a feeling next Thursday will be here in no time. So today I am having a little bit of a freak-out... So today I am having a little bit of a freak-out session. Last night I had some mango that a coworker gave me. Within 10 minutes of eating it, my throat was scratchy. Woke up this am and it still hurts. I'm hoping this is an allergy thing (never had a prob w mango before). I'm less than 5 days away from surgery and I CAN'T be sick! No fever, so that's a plus. On a side note, I was thinking about the new clothes I'll be buying soon. ( as long as I'm not sick and can still have surgery). It's going to be so great to buy some new work clothes. Maybe a suit that has a jacket that fits the way it should. Can't wait! I now have 2 days until surgery!!! I feel great... I now have 2 days until surgery!!! I feel great and got over whatever that cold/allergic reaction was. I feel like I should be doing more to get ready. I already did a big cleaning of my place last week and will probably just do a once over before I leave tomorrow. I'm going to spend the night at my parents house and my mom is going to take me in the morning. I'll stay there for however long I need to. I'm still going between the excitement and nervousness of the whole thing. I've been approved for surgery since January, so I've had all of this time to prepare. It still doesn't feel quite real yet. I'm hoping I don't wake up on Thursday with an gian anxiety attack when the reality really sinks in. It's really strange thinking that I'll only be wearing these bras for a couple more days. I'll have an entire empty drawer that will be waiting to fill with pretty, smaller ones. I can't wait! Surgery was yesterday. I mostly slept the day away... Surgery was yesterday. I mostly slept the day away. I have some pain/discomfort mostly on the sides because my PS did some Lipo there. Went for post-op appointment this morning and they said everything looks good. There are no drains, but I am wrapped in something called Tegaderm. It looks like suran wrap. Yesterday I had an ace bandage, and today they put my surgical bra on. The nurse had to poke some holes In the Tegaderm to drain the fluids. Mostly had fluid on the right side. Then she added another layer of the Tegaderm, put my bra on and I was on my way. They said I could shower today, so that's a plus. Also, I feel very bloated. I've been taking a laxative pill every time I take my Percocet. Overall, I'm doing pretty good. The PS said I should be a full C, but to me they look do small especially ring bound. Today was day 2. I showered last night and today... Today was day 2. I showered last night and today wore a regular pull over shirt. My mom had to help me put it on. Physically, I am doing great. Not much pain except for the sides where the lipo was done. My emotions are a bit all over the place too. I'm worried that my chest is too small. I know that I am still bound and I need to not worry, but I am. I look at what I have and do not see a full C. My stomach is bloated and sticks out and I just feel like I look terrible. A family friend came over tonight and told me that I look great, skinnier. So why am I not happy? I am just going to try to be patient and see what happens. I'm sure I'm just being self conscious. Hopefully tomorrow will be better. Today was not the best day. It's been 5 days... Today was not the best day. It's been 5 days since surgery, I had my second appointment today. Yesterday I noticed my right nipple was looking a bit dark in spots. I started to get really concerned but tried to keep it under control knowing that my appointment was today. I still had the tegaderm on and I wasn't all cleaned up yet. So, I go to my appointment and the nurse removes the tegaderm (basically looks like suran wrap). It's not really painful, but was glued on pretty well. I could see her comparing my nipples, but she didn't say anything. I waited... She said she'd be right back and another nurse, one who saw me on Friday, came in to check it out. She walked right up and was like... they were both a good pink last time. I started to cry, fearing my right nip was dying. She told me that it wasn't black, but more dark red. My PS was in right after that and took a look. He assured me that he's "never lost a nipple" and that it was very much alive. I sat up and he poked some holes in my nipple to make sure that it would bleed, thus proving it is still viable. I'm still freaked out. From what I can see and what he sees, it looks almost like raw skin. Like the top layer somehow was removed. (possibly by the tegaderm as the swelling went down. It also looks almost like a burn. It's very disturbing to me because I feel like I have one pretty good looking breast and one that looks terrible. The left side looks round and normal and the right side, aside from the odd colored nipple, has an intersting shape to it. I guess it's kind of the boxy shape people have talked about. So in the mean time, I got a Rx for some nitrogliceride cream to put on it 3x a day. Apparently it helps to bring bloodflow to the surface. Then also add vaslene to keep it moist, oh and a warm compress 3x a day. I go back on Thursday. So I have 2 days till I see him again. Overall, I'm really not in pain. I'm getting a stabbing pain once in a while, again mostly on my right side. I think it hates me. haha I was itching like crazy this afternoon and took a shower to see if that would help. It didn't. I am getting used to the smaller size. I must admit that I was really freaked out at first. Could have sworn to anyone that I was an A cup because they looked that small to me. Then, I thought mayb a B cup. My sister, who never recieved the family curse of overly large breasts, was over tonight. After having her first child, she is now a B cup. To cheer me up she took her bra off and we kind of put it over my sports bra and come to find out I definately would not fit into her bra because it would be too small. So there's a relief. If my Right breast looked more like my Left breast, I'd be so happy. Now I just have to wait and see what happens. I'm going to try to post some pictures too. Feeling a lot better today. Went for another... Feeling a lot better today. Went for another appointment and they removed every 3rd stitch. Not really sure why, but whatever. My PS has assured me that my right nipple is fine. It looks like the top layer of skin on part of it may have come off with the tegederm and of course the trauma from it being moved. So he believes it will all scab over and them I will have a normal color once again. :). I was shopping with a friend today and tried on a few dresses and love the way they fit. You know, the summer dresses that nobody with huge breasts could ever wear. It was a great feeling to know that I can wear them. Just not with this sports bra. Lol It's been a week today and am looking forward to being completely healed. I feel really good physically and mentally. I understand what other ladies have said that your mind and your body aren't on the same page. I did start getting a weird shock-like feeling today in the sides. Its more uncomfortable than painful. I don't know if it's from the incisions or the lipo. Seems to come and go, probably nerves healing. It took me a week to adjust to the size, but I now believe that I am a C cup and I really think that on e everything settles, I'm going to be very happy. I'm very happy to have found this wonderful site. It has helped so much! Seriously, I feel like I'm the only one having an... Seriously, I feel like I'm the only one having an issue. I just knew that my right areola was going to be an issue once I saw the change in color last week. I called today because I thought I may have an infection due to some brownish oozing from the edge of my areola. They had me come in this afternoon and to my surprise, it was actually necrotic fat tissue.The oozing was the tissue that liquifies. Gross! My PS immediately started cutting it out. It didn't hurt, it was just awkward. He said the fatty tissue did not survive and it's important to removie it so it doesn't get infected. Apparently I'm a first for him of with his 20-something years of experience. Lucky Me! He said he's seen it before in other cases, and types of surgeries. I know that he had just come out of surgery, so as soon as he finished, he said some stuff and was out the door. I was fine until he left and then the nurse was finishing fixing me up and asked if I was ok and I just started crying. He said the hole was very small and he stiched it up. The nurse then removed some more of my stiches, but I still have a bunch. I have to put some cream on it they gave me and take more antibiotics. So, I cried all the way home not really understanding why this has happend to me. I'm posting a picture of it today, it's really hard to tell exactly where the hole is because he stiched it up, but it's on the top right side kind of between the patches of brown. I got home and was talking to a friend about my ordeal when my PS called me. I was actually really relieved that he called me himself. It actually impressed me that he took the time to do so. Apparently my nurse told him about my little crying incident and he wanted to talk to me to make sure I understood everything was ok and that I knew exactly what was going on. He appologized for having to rush out of the room earlier and he spent about 10 minutes on the phone with me. He was super nice, answered all of my questions, and told me that everything I was feeling was normal. He also tried to reassure me that within 6 months or so it will improve greatly and will eventually look normal. He mentioned possibly needing to revise the areola (I don't know what that entails), but that it is done in the office and it wouldn't cost me anything. He said that eventually, if the pigment doesn't come back, that I can have it tattoed. I left the office upset and wondering why I did this to myself. Questioning if it was worth it. Of course, I have to be one the low percentage of people that has a problem. But, I know that if I can get past this, it will get better. The thing that scares me the most ,and I shouldn't worry about it now, is the fact that I am single and I'm afraid that it will look terrible to a potential partner (I know my PS said differently, but I still fear this.) However, on the plus side, my neck and back do not hurt anymore, and everyone tells me how amazing and skinny I look. I just wish I could beam and say how wonderful and smooth everything went. I was scheduled to go in on Tuesday morning, but my PS said that since I was upset, he'd rather see me on Monday. I'll update after that appointment. Today is exactly 2 weeks post-op. Feeling a little... Today is exactly 2 weeks post-op. Feeling a little better today. The nurse called to check on me today, and that made me happy. I had a couple of questions. She told me she she was in the office tomorrow if I needed anything. Even with the issues I'm having, I'm so happy with my choice because I really feel like Dr. Kim and his staff care. That means a lot. I go in on Monday. I'll update later. Thanks to everyone and their kind words and support. It means a lot! So today I was laying here on the couch and I... So today I was laying here on the couch and I moved a certan way and I heard a sloshing sound. I heard the same sound a day or so ago, but thought it was just liquid in my stomach from drinking water. I did some more investigating and it is definately coming fro my left breast. Lefty has been so good during this entire process, how dare she act up now. lol I did a little research on the topic and it is something called a seroma, fluid buildup. Yes, there are other possibilities, but I'm going with that one for now, because it apparently is normal. I see my PS tomorrow and will get a certified oppinion. It mostly just sounds really weird. Righty looks maybe a little better, but about the same. I feel more prepared now to go in tomorrow and possibly have him remove more necrotic tissue. Again, I'm no doctor so I could be wrong. I'm just trying to think long term once all the wounds are healed that everything will finally look normal even if I need a slight revision. I'll update tomorrow. 18 Days Post-op. Went to PS today and the... 18 Days Post-op. Went to PS today and the first thing Dr. Kim asked me was "how are you feeling...mentally, not physically". He said my right nipple looked a little better. However, he did more cutting (debriding) of the area. Apparently, I am an anomaly. He has "never" had anyone with this issue. SERIOUSLY???? NEVER???? Lucky me! He went on to explain about how exteremely large breast sizes have greater risk for loss of blood flow, etc. (Stuff I already knew). Maybe that's why there was a new face in the room with me. I'm not sure if she was another Doctor or not (she had on a lab coat) but was mostly observing. I think she was the nurse for another doctor there who deals with breast reconstruction. She asked what size I was prior to surgery and before I could answer, he said "HUGE". I couldn't help but laugh! He told her he took 1000 grams from my right side and I think the other side was about 1/2 of that. The sloshing sound I heard in my left breast was exactly what I expected, seroma and it will absorb (hopefully). I asked him some questions about working out again (would this delay me). He said that I could maybe start some light stuff in a week or so. Running after about 6 weeks total. I was training for a 5K and want to get back into it. When he left, she finished cleaning up and bandaging me. Mentally, I felt much more at ease going in and leaving today. I now have to use diluted hydrogen peroxide, then air it out a while, then put on the antibiotic ointment, then gauze. Dr. Kim said in 3-6 months he will probably do a revision of the scarred areas and briefly mentioned something about nipple reconstruction. It's isn't dead, but it isn't responding to anything either. So, I'm not sure about that one. No tears leaving today, so that's a plus! On my way out, he said to me "hang in there". lol I go back on Thursday where Dr. Kim is going to stay later so I can see him after work and hopefully get the rest of these stiches out. I'll try to do an updated picture later tonight. Not really wanting to look right now :) 3 weeks post-op Went to my PS today. My... 3 weeks post-op Went to my PS today. My necrosis was worse than he thought it would be. I guess the hope was that it would slough off and the damage wouldnt be very deep. The reality was that it didn't get any better since Monday and was deeper than we hoped. He numbed me up a lot and then cut out all of the dead tissue. After that, the nurse cleaned the open wound and prepped the area kind of like a surgery. She brought in her iPhone and started some music (maybe to calm me, it helped). Dr. Kim came back in, explained again what he was going to do and then began sewing the area back up. I need to keep the area clean and use whatever ointment they gave me and keep it covered. They gave me an antibiotic pill in the office too. I won't post a picture because I don't know if it's too gross to post. Maybe in a couple of weeks after its healed up. I will definitely need some sort of revision. Dr. Kim is determined that he will make it look great when all of this is over. 3-6 months before a revision. My actual nipple may have to have some work now because it might not be centered now due to the skin being removed. You would think that I've known all of this for a couple of weeks, but I feel so numb right now. Dr. Kim and his staff were so great with me today too. I was the last patient of the day. I wish my friends and family could understand what I'm going through. I just want to have no open wounds and be trying on cute little tops and dresses without the need for a bra. I go back on Monday afternoon. I added a photo of what my right areola looks like... I added a photo of what my right areola looks like today after yesterday's procedure. It wasn't as gruesome as I thought it would be. It actually looks a lot better with all that dead tissue gone now. The nipple is definately in the wrong spot now, not that it really has looked like much of a nipple lately anyways. But I can tell where it's technically is and it's new location is now very close the the edge of the areola. I don't think it will ever actually work again correctly anyways. So, I guess it doesn't really matter. They said the skin may stretch more and it may go sort of back to a normal location. Just have to wait it out. If anything, I'll have one recreated. Just wondering how that works. Do you have one nipple that is always at attentention? Seems strange. The nurse called this morning to check on me. The positive again in all this is that I know I have a team that cares. 25 days post-op Went to my PS today for a... 25 days post-op Went to my PS today for a follow up since last weeks procedure. I'm finally on the upswing here. Yay! Other than the new stitches from last week, all the other stitches were removed. Dr. Kim seemed pleased with the way my right areola is now healing. It was kind of funny the way he responded when he looked at it, almost surprised by his own good work. Lol. He said that there is a possibility that I won't need a revision. ( I don't know how that is possible, but he's the expert). He said that the skin of my areola should stretch some (hoping it puts my nipple back in the correct spot). I left today feeling the best I've felt so far. I'm still having some drainage and need to keep a bandage on, but I'm hoping that it will scab up and heal super fast. I go back on Thursday for another check up because he wants to see me before he goes away for a week. I'm just happy that he is being so diligent with all of this. Ill try to update a picture tomorrow. Went to my now weekly visit to my PS today and... Went to my now weekly visit to my PS today and again everything looks good. There was one little cabby spot left that was a superficial area of necrotic tissue that came off earlier. Dr. Kim kind of cleaned it up a little with some more snipping and everything underneath it was a good pink color. Everything is healing well. I still have the stiches from last weeks procedure, but expected that because it's only been a week. He gave me some silicone sheet strips to put on my scars underneath and on the sides which I'm supposed to use at night. Overall, nothing new to really report...THANKFULLY! School started on Wed for kids and it's only been two days and I am exhausted already. Thank GOD it's a short week! My PS is going out of town next week, but I'm going in next Thursday so the nurses can take a look and then I see him the following Thursday again to probably have the stiches removed. I'll post new pics this weekend. Updated a couple of pictures today to show my... Updated a couple of pictures today to show my nipple/areola improvement. I noticed an internal stich comming up today on my left areola. I'm not really worried about it, but I go in on Thursday, so maybe they will remove it. Went for my all to familiar weekly update today.... Went for my all to familiar weekly update today. My PS is on vacation, so I saw the nurse. Nothing really new to report. I'm happy with that. She removed a couple of stiches that were starting to poke out. Said everything looked good. (no infection, etc) Back at the same time next week to see my PS. I don't know what it would be like to go a week without going there. lol I'm updating with a couple of new pics today. ... I'm updating with a couple of new pics today. It's been just over two weeks since the removal of a large section of necrotic tissue on part of my areola. Things are healing well. I just want to keep adding pics each week to show improvement for those who may go through something similar. From my experience, I was completely devastated at first and now I see so much improvement. I noticed a few days ago that my nipple is actually working (It reacts to cold), however, it's just not really in the right location because of the removal of tissue and then being stiched up. Will have to wait and see if it lowers a little. I've pretty much decided that I don't want to mess with it if I don't have to. It'll just be off center. I was expressing my concern about this with a friend over dinner the other night. Weird? Anyways, she held up a french fry that was a little curved and said, " you know most guys aren't perfectly straight either, if you know what I mean, but it doesn't mean it doesn't work just as well." It was a good laugh. Nobody is perfect! My pigment is starting to come back and I still have an area where tissue is still filling back in. Stiches should be out next week. 6 weeks post-op ***added photos*** Went for my... 6 weeks post-op ***added photos*** Went for my weekly appointment to see Dr. Kim today. Everything is healing wonderfully! He was very pleased with how everything looks. Thinks there is a very high possibility that I will need NO revision. My pigment is comming back and fingers are crossed that it will in ALL areas. I have been cleared to do whatever I want! Running is in the very near future, swimming, and so are whatever friggin bras I want to wear. YAY! I had the last 3 stitches removed today and there has been a little drainage from them. There is some very hard tissue right around my proplem areola that, from what Dr. Kim said, is probably some necrotic tissue. The expectation is that it will soften up and re-absorb. So, my Thursday afternoons are now free. I don't have to go back for an entire month. I almost feel like it is bitter-sweet. The nurses and I were joking about how weird it will be not to see each other every week! I asked what the heck I'm supposed to do now with my time. haha. Seriously, I've been there once or twice a week for the last 6 weeks. I think when I go next month, I'm going to bring all of the nurses something to show my appreciation to them. They have all been so wonderful to me. Just feeling happy overall. I went through some ups and downs, but pretty happy with how everything is going. My daily dose of of being told how skinny and amazing I look really helps. I love my co-workers! I also had the first comments from a couple of former students today (8th graders whom I had in 6th grade). They just complimented me on loosing weight and looking really great. They were very sweet! I hope that if anyone else has to go through some of the partial areola loss that I went through, they can see that all is not lost. Went for my first run today! YAY! I wore a super... Went for my first run today! YAY! I wore a super tight sports bra, and went slower than I did before, but it was great! AND.... my back didn't hurt from running! I'm probably going to be sore tomorrow though! Today we were off from school due to TS Issac. No... Today we were off from school due to TS Issac. No worries! We just got some wind and rain. Since I was off, I figured I would get into my PS office to get the one stich that was poking out pulled. Turns out it is what is called a running stich. The nurse just trimmed it and it's supposed to dissolve at some point. No worries. After that, I went and found a couple of bras. One is a wonder bra and one is a push up bra. lol I never thought, ever, that I would own those types of bras. So, I decided to take a couple of pics. Oh, I forgot the best part...$15.00 for both bras...

Oh, I forgot the best part...\$15.00 for both bras today! Unbelievable!

Went shopping today and was actually measured for...

Went shopping today and was actually measured for a bra. The lady measured me at 38 C, but I tried on a 38 B and it fit. I almost cried bc I really wanted to be a full C/ small D. And it appears that I am a full B/ small C. I know that all bras vary, but I wasn't trilled w that B cup. The others ,C's, I was fine with. I guess I get what I get. Lol.
In healing news, everything is great. Nothing looks really much different. Still not sure what my right areola is going to wind up looking like. Time will tell :)

4/20/13 It's been 9 months now. Everything has...

4/20/13

It's been 9 months now. Everything has healed very nicely. On my last visit with Dr.Kim, I expressed a desire to to have the dog ear fixed and the scar slightly adjusted to fix the uneven part of the scar. So on 4/18, I had a scar revision surgury. It brings back the memories of the pain from surgery, but definately not as bad. Just uncomfortable. The revision took less than an hour and was done in the office with a local anesthetic. Once I get the stiches out, I'll update with new pictures. You can't really see much now because it's covered with steri strips. I did have issues early in recovery with necrotic (dead) tissue on my areola. Some of the tissue was removed and then re-stiched. Although my nipple is now slightly off center, I am not going to have any corrective surgery to fix it. It is noticable by me, but does not seem to be an issue for the guy I am dating. Also, \it would require a lot to fix. Both nipples have complete sensitivity back and they seem to work noramally. Overall, I'm pretty happy with the procedure and am looking forward to the end results of the revision surgery.
Michael K. Kim, MD

Dr. Kim and his staff have been wonderful. I have had a minor complication, but trust that everything will be great. Dr. Kim called me several times in the days after my surgery to check on me. After a follow up appointment where I had to have some tissue removed, I was a bit emotional after Dr. Kim left the room. When I got home, Dr. Kim personally called me to make sure I was ok. He spent about 10 mins on the phone with me going over the issues and resolutions, as well as comforting me with how I was feeling. That call just made me feel so great about my having Dr. Kim as my PS.

5 out of 5 stars Overall rating
5 out of 5 stars Doctor's bedside manner
5 out of 5 stars Answered my questions
5 out of 5 stars After care follow-up
5 out of 5 stars Time spent with me
5 out of 5 stars Phone or email responsiveness
5 out of 5 stars Staff professionalism & courtesy
5 out of 5 stars Payment process
4 out of 5 stars Wait times


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Active contributors today

## What protects us from harmful radiation of the sun?

Hunaid L. Hanfee
Featured 1 year ago

The OZONE LAYER which lies in Stratosphere (16 to 48 kilometers above earth's surface) protects us from harmful UV radiation.

#### Explanation:

Ozone layer help us by trapping harmful radiation from the sun. Ultraviolet(UV) rays are harmful rays which come from the sun.

Ozone layer is made up of ${O}_{3}$ or three atoms of oxygen. When UV rays of 230 to 320 $\mu m$ hit the ozone layer, ${O}_{3}$ absorbs its energy and gets converted into ${O}_{2}$ & then $O$. ${O}_{2}$ molecule combine with this $O$ molecule and form ${O}_{3}$ by this ozone is formed again. This happens continuously in the whole layer.

The use of Chlorofluorocarbons (which is used in refrigerator) stops the ozone to form again & it damage the ozone layer

When UV rays strike chlorofluorocarbon($C F C {l}_{3}$) its carbon-chlorine bond breaks. This produces a Chlorine atom & this atom reacts with ${O}_{3}$ breaking it apart, which causes a hole in ozone layer.

Gamma ray and X-ray are also harmful radiation which comes from space. Just a single high-energy photon of gamma rays can cause significant damage to a living cell. But our earth's atmosphere protect us by absorbing it.

.

## In a star if helium fuses to make carbon, what makes the elements in between?

Phillip E.
Featured 1 year ago

The elements Lithium, Beryllium and Boron are made by cosmic rays.

#### Explanation:

The elements Hydrogen and Helium were made during the Big Bang. Helium is also made by fusion reactions in stars.

The elements Lithium Beryllium and Boron can't be made in any quantity is stars as they are intermediate steps in other fusion reactions. Any of these elements present at the birth of a star will get consumed by later reactions.

These intermediate elements (Lithium, Beryllium and Boron) are produced by cosmic rays. These are high energy protons and other particles. These collide with nuclei of Carbon and Oxygen which breaks them down and produces Lithium, Beryllium and Boron.

Carbon is produced by reactions such as the triple Helium reaction. This reaction explains what happens to Beryllium.

${\text{_2^4He+""_2^4He rarr }}_{4}^{8} B e$
${\text{_4^8Be+""_2^4He rarr }}_{6}^{12} C$

The diagram shows where all of the elements are created.

## What are nuclear fusion reactions?

Hunaid L. Hanfee
Featured 1 year ago

Nuclear fusion is the process of two small atomic nuclei coming together to make a larger nucleus which is stable.*

#### Explanation:

Nuclear fusion is the reverse process of nuclear fission. This process also results in the conversion of mass into energy.

Nuclear fusion is the energy source which causes stars to "shine", and hydrogen bombs to explode.

Nuclear fusion in star
The enormous luminous energy of the stars comes from nuclear fusion processes in their centers. Depending upon the age and mass of a star, the energy may come from proton-proton fusion, helium fusion, or the carbon cycle

Fusion releases energy. The energy released is related to Einstein's famous equation, $E = m {c}^{2.}$

To start nuclear fusion, protons and neutrons must be exposed to temperatures approaching 10 MK Celsius (180 million degrees Fahrenheit). In these reactions too much energy is released.

When nuclei fuse together, part of their mass is converted into energy. The sun and other stars convert this energy into light.

$\implies$ For example - Fusion of deuterium with tritium creating helium-4, freeing a neutron, and releasing 17.59 MeV of energy, as an appropriate amount of mass changing forms to appear as the kinetic energy of the products, in agreement with kinetic E = Δmc2, where Δm is the change in rest mass of particles. (Source : Wikipedia)

## Why hasn't the discovery in 1998 of the increasing rate of the expansion of the universe had a bigger effect on the philosophies of science and astronomy?

Pythagoras
Featured 5 months ago

It did have a huge effect on the philosophies of science and astronomy!

#### Explanation:

First of all, what do you mean by "bigger effect"? It did have quite a huge impact in cosmology and astrophysics that we had to rewrite entire libraries with the new discovery and its implications. Nowadays, all astronomy books include this unexplained "dark energy" and the mass-energy density pie-chart of the universe had to be revised.

Perhaps you felt it wasn't as big as the paradigm-shift that occurred in the early-1900s when Edwin Hubble proved that Andromeda and other spiral clouds were in fact entire galaxies similar to our own (which made the known universe to substantially increase in size, plus its expansion meaning that the universe had a beginning), or as big as the discovery of the Cosmic Microwave Background in the 1960s (which proved once and for all that the Big Bang really occurred and that the universe had a beginning).

The omega value, or critical density of the universe, is very close to 1 (i.e. flat universe), and since the expansion rate of the universe seems to be accelerating, it will keep expanding forever and faster and thus its geometry MAY be slightly open (but we don't know that for sure and it is up to debate, and if that is the case, the extent of the universe beyond the observable one could be infinite).

I realize that I need to explain more in my answer about another question asking for the volume of the universe, to which many of us replies that the "universe is infinite", as it would only make sense if the geometry is either really flat or open.

The current standard model of cosmology that best describes the universe is called the lambda CDM model and takes into account the cosmological constant (first introduced by Einstein but to have the opposite effect of what it is used for today, the dark energy). There are many references in the literature (perhaps not popular science), but I find the WMAP and Planck websites to be quite good to get a grasp of the current model. That's why, it still had a great effect.

David Drayer
Featured 3 months ago

The decadal oscillations cause temporary changes in world-wide climate, both global warming and global cooling.

#### Explanation:

One theory is that the present observed trend in global warming is at least partially the result of the decadal oscillations. The effects of the oscillations can be seen in the arctic ocean and the glaciers of Greenland.

During the 1930s and 1940s there was a global warming effect due to the decadal oscillations. During this time the Arctic Ice melted and the Northwest passage was open for a time in 1939 and 1940. Planes landed on the glaciers of Greenland were later covered with very thick layers of ice. These planes have recently been uncovered by the present trend in global warming.

Similar oscillations have been seen in the past. Vikings settled in Greenland, raised crops, grapes and cattle during a warming trend.
A later cooling trend forced the Vikings to abandon their settlements on Greenland and closed the Northwest passage that had allowed the Vikings to settle in what is now Russia.

Another theory is that the oscillations have nothing to do with the present trends in global warming. The oscillations are by definition are always moving back and forth. While the oscillations have a temporary affect on world climate the present trends are thought to be independent of the temporary affect of the oscillations. The theory is that the present trend in global warming is linked to the increased production of Carbon Dioxide.

## Why are planetary orbits elliptical and why do bodies in a solar system orbit the center of mass and nit the star itself?

Phillip E.
Featured 3 months ago

Planets orbits are defined by conservation laws.

#### Explanation:

Johannes Kepler discovered by observation that planets follow elliptical orbits. A few decades later Isaac Newton proved that by applying the law of conservation of energy that a planet's orbit is an ellipse.

When two bodies orbit around each other, they both always orbit about the centre of mass. This centre of mass is called the barycentre. The Moon doesn't orbit around the Earth. In fact both the Earth and Moon orbit around the Earth-Moon Barycentre (EMB).

When it comes to something more complex like the solar system a similar principle applies. None of the planets etc actually orbit around the Sun. In fact the Sun, planets, asteroids, comets and other bodies all orbit around the centre of mass of the solar system which is called the Solar System Barycentre (SSB).

The SSB is in constant motion and can be anywhere from near the centre of the Sun to over a Su radius outside of the Sun. So, everything in the solar system is orbiting around a point which is in constant motion.

The diagram shows the path of the SSB over several decades. The points where the SSB is furthest from the Sun occur when the planets are aligned.

##### Questions
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## Thursday, 7 August 2014

### chapter-23 Interesting Probability Questions

The subject of probability gives rise to some very interesting questions, which we believe everyone should be exposed to, even if to a minor extent. In this section, we’ll do precisely that. The examples will clearly show you the depth of this subject.
 Example: 23
 A drunk mathematician called $Pi$ is performing a random walk along the number line as follows. His original position is $x = 0$. He takes a step forward or a step backward with equal probability. Thus, for example, if he is at $x = 1$, he can go to $x = 2$ or $x = 0$ with equal probability. His home is at $x = 3$, while there is a pit at $\,x = - 2$. What is the probability that he’ll reach home before falling into the pit?
 Solution: 18
Assume the required probability to be $p$. Thus, $p$ denotes the probability $P(E)$ where
 $E$ : $Pi$ reaches home before the pit
Now, let us draw the probability tree that will make $E$ happen:
What should the events ${A_1}$ and ${A_2}$ be and what should be their probabilities? Once $Pi$ has reached $x = 1$, the situation is as follows:
${A_1}$ should therefore be the event that $Pi$ reaches $x = 3$ before $x = - 2$. Now comes the crucial insight. Compare Fig-$20$ with Fig-$18$. The situations have been symmetrically reversed. In Fig -$18$$p$ was the probability that $Pi$ reaches $x = 3$ before$x = - 2$, that is, $Pi$ travels $3$ units in one direction before travelling $2$ in the other. In Fig-$20$$P({A_1})$ is the probability that $Pi$ travels $2$ units in one direction before travelling $3$ in the other, so we simply have
 $P({A_1}) = 1 - p$
Make sure you understand this argument carefully.
Coming to ${A_2}$ , $Pi$ has reached $x = - 1$, he must return to $x = 0$, because he must not reach $x = - 2$. Once he is at $x = 0$, he comes back to the starting configuration, which means that now his probability of reaching $x = 3$ before $x = - 2$ is again $p$.
So let us complete the probability tree of Fig-$19$ now:
Thus (pay attention to how we write this!),
 $p\,\, = \,\,\dfrac{1}{2}\,\, \times \,\,\mathop {(1 - p)}\limits_{\scriptstyle{\rm{Reaching}}\;\;x = 3\;\hfill\atop \scriptstyle{\rm{ from}}\;x = 1\hfill} \,\, + \,\,\dfrac{1}{2}\,\,\, \times \,\,\,\dfrac{1}{2}\,\, \times \mathop p\limits_{\scriptstyle{\rm{Back}}\;{\rm{to}}\hfill\atop \scriptstyle\,\,\,\,x = 0\hfill}$ $\ldots(1)$
This relation is a form of recursion. In the definition (or expression) of $p$, we are using $p$ again. Thus, recursion is, in plain words, the process a ‘procedure’ goes through when one of the steps of the ‘procedure’ involves rerunning the procedure. In this example, the ‘procedure’ of calculating $p$ involved the step$(x = 0) \to (x = - 1) \to (x = 0),$ at which point we have to rerun the ‘procedure’ of calculating $p$.
From $(1)$
 $p = \dfrac{1}{2} - \dfrac{1}{2}p + \dfrac{1}{4}p$ $\Rightarrow\,\,\,\, \dfrac{5}{4}p = \dfrac{1}{2}$ $\Rightarrow\,\,\,\, p = \dfrac{2}{5}$
Thus, there is a $40\%$ chance that $Pi$ will return home safely!
Recursions can also take the following form: suppose we have to calculate ${T_n}$, the ${n^{th}}$term of some sequence, and suppose we are able to express ${T_n}$ as some function of ${T_{n - 1}}$, the previous term:
 ${T_n} = f({T_{n - 1}})$
This is also a recursion relation, for now we can use this relation repeatedly to descend, that is, to reach some ${T_m}$ that is calculable:
 ${T_n} = f({T_{n - 1}}) = f(f({T_{n - 2}})) = f(f(f({T_{n - 3}}))) = \ldots = f(f(f(\ldots {T_m}))))\ldots )$
For example, ${T_m}$ could be the first term of the sequence. Thus, here we are rerunning the ‘procedure’ $f$.


text:
C++ Fast-track for Games Programming Part 16: Physics

C++ Fast-track for Games Programming Part 16: Physics

In this tutorial, you will be introduced to game physics. This is an incredibly broad field, and quite complex too, so you’ll really only get a teaser. This will however produce some pretty neat effects.

Previous Part: File I/ONext Part: AI

Getting Started

For this tutorial, we’ll be using the same template from the 2nd tutorial:

TheTemplate.zip

Physics

Many recent games use highly advanced physics engines. A physics engine is a piece of software that gets linked to a game. It takes care of natural behaviour of objects in the game, taking into account forces, collisions, momentum, rotation and torque. Writing a good physics engine is hard for a number of reasons:

• Some game objects move fast, which makes accurate calculations tricky;
• Simulation in a game happens in steps, rather than continuously, like in nature.

For this tutorial, we will therefore focus on a simplified physics model, called Verlet Integration.

Verlet Integration

It works like this. If you have an object that moves along a straight line at a fixed speed, for each frame of your game, you can calculate its position like this:

$p=o+v\Delta{t}$

Where:

• $$p$$ is the new position of the object
• $$o$$ is the origin of the object
• $$v$$ is the speed (or velocity) the object is moving
• $$\Delta{t}$$ is the change in time since it was at the origin

That’s one way. Another way just looks at subsequent frames, and calculates the position incrementally:

$p\mathrel{+}=v\Delta{t}$

If the time it takes to render one frame is roughly the same each time, you can simplify this one step further:

$\begin{array}{rcl} \Delta{p} & = & (p-p_0) \\ p_0 & = & p \\ p & \mathrel{+}= & \Delta{p} \\ \end{array}$

Where:

• $$p_0$$ is the position of the object in the previous frame
• $$\Delta{p}$$ is the change of position from the previous frame

In other words: If we moved an object from $$p_0$$ to $$p$$ during the previous frame and both $$\Delta{t}$$ and $$v$$ remain the same, then we will move the same distance for the current frame. The distance between two snapshots now effectively becomes the object’s velocity.

Now imagine that there is a force that is pulling on our object (e.g. gravity). To apply that force, we adjust the position (by adding $$\Delta{p}$$) and then we move the object some more, to account for this force. If the object was already moving down, it will now move down faster, because in the next frame, $$\Delta{p}$$ includes the applied gravity.

Let’s put this into some code. Replace game.cpp with:

#include "game.h"
#include "surface.h"
#include "template.h"
#include "windows.h"

namespace Tmpl8
{
void Game::Init() {}

void Game::Shutdown() {}

void Game::Tick(float deltaTime)
{

}
};

First, I need a piece of code to draw a circle. That’s because circles (and spheres, in 3D) are excellent for physics. You’ll soon enough find out why.

void Circle( Surface* s, float x, float y, float r )
{
for ( int i = 0; i < 64; i++ )
{
float r1 = (float)i * PI / 32, r2 = (float)(i + 1) * PI / 32;
s->Line( x - r * sinf( r1 ), y - r * cosf( r1 ),
x - r * sinf( r2 ), y - r * cosf( r2 ), 0xff0000 );
}
}

No need to scrutinise that code, for now it’s fine to simply use it.

Then, the falling ball code, using Verlet integration:

    float x[2] = { 300, 320 }, y[2] = { 0, ScreenHeight - 50 };
float px[2] = { 300, 320 }, py[2] = { 0, ScreenHeight - 50 };

void Game::Tick(float deltaTime)
{
screen->Clear(0);
Circle(screen, x[0], y[0], 16);
Circle(screen, x[1], y[1], 49);
// compute deltas.
float dx = x[0] - px[0], dy = y[0] - py[0];
// store previous position.
px[0] = x[0], py[0] = y[0];
// Verlet integration
x[0] += dx;
y[0] += dy;
// add force due to gravity.
y[0] += 0.5f;
// delay
Sleep(50);
}

As you can see, the ball starts falling, and its velocity is increasing. It does fall through the floor though… And let’s not forget the other sphere.

Verlet Constraints

So, we now have subsequent positions (the difference of which is the velocity), and forces (implemented by changing positions directly on line 37). We need one more thing: constraints. In nature, a ball cannot fall through another ball, or through the floor. Using Verlet integration, we handle this using constraints.

Here’s one example: Since a ball cannot fall through the floor, we need to test for that. As soon as we detect that a ball does intersect the floor, we correct the situation, by moving the ball out of the floor. Note that this effectively changes the speed of the ball, if we keep interpreting the difference between subsequent positions as speed!

Apart from avoiding collisions using constraints, we can also keep objects together, using simple springs. This time, the constraint is: Two objects may not be further apart than a certain distance. When they are, we move both objects closer together to simulate the spring.

So, let’s add a first constraint. Here’s the code:

    void Game::Tick(float deltaTime)
{
screen->Clear(0);
Circle(screen, x[0], y[0], 16);
Circle(screen, x[1], y[1], 49);
// compute deltas.
float dx = x[0] - px[0], dy = y[0] - py[0];
// store previous position.
px[0] = x[0], py[0] = y[0];
// Verlet integration
x[0] += dx;
y[0] += dy;
// add force due to gravity.
y[0] += 0.5f;
// constraints
if (y[0] > (ScreenHeight - 17))
y[0] = ScreenHeight - 17;
// delay
Sleep(50);
}

The result may be a bit unsatisfying. No bouncing? We can improve it a little bit:

    void Game::Tick(float deltaTime)
{
screen->Clear(0);
Circle(screen, x[0], y[0], 16);
Circle(screen, x[1], y[1], 49);
// compute deltas.
float dx = x[0] - px[0], dy = y[0] - py[0];
// store previous position.
px[0] = x[0], py[0] = y[0];
// Verlet integration
x[0] += dx;
y[0] += dy;
// add force due to gravity.
y[0] += 0.5f;
// constraints
if (y[0] > (ScreenHeight - 17))
py[0] = y[0], y[0] = ScreenHeight - 17;
// delay
Sleep(50);
}

But that’s a hack. In general, Verlet integration is not very good for bounces, sadly. Let’s see what it is good at: The other sphere.

Sphere-Sphere Collision

How can we determine if the two balls are colliding? We need some math for that. If ball $$1$$ is located at $$(x_1,y_1)$$, and ball $$2$$ is located at $$(x_2,y_2)$$, then the distance is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$. If this distance is less than the sum of their radii (radiuses), then they are intersecting.

The distance between two spheres is $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

When the circles intersect, we have a number of options.

1. We could move the smallest ball, so that it doesn’t intersect anymore;
2. We could move both balls;
3. We could move both balls, but the lightest ball more than the heavier one.

Option 3 is clearly the most realistic one. But how do we move the balls away from each other? Maths to the rescue again: If the circles intersect, their distance is smaller than their combined radii. The combined radii minus their distance is the penetration depth, and thus the distance they need to be moved apart. Now, if the small ball is ten times lighter than the large one, the small one will do 90% of this distance, and the big one only 10%. Here we go:

    void Game::Tick(float deltaTime)
{
screen->Clear(0);
Circle(screen, x[0], y[0], 16);
Circle(screen, x[1], y[1], 49);
// compute deltas.
float dx = x[0] - px[0], dy = y[0] - py[0];
// store previous position.
px[0] = x[0], py[0] = y[0];
// Verlet integration
x[0] += dx;
y[0] += dy;
// add force due to gravity.
y[0] += 0.5f;
// constraints
if (y[0] > (ScreenHeight - 17))
py[0] = y[0], y[0] = ScreenHeight - 17;
float xdist = x[1] - x[0];
float ydist = y[1] - y[0];
float distance = sqrtf(xdist * xdist + ydist * ydist);
if (distance < (16 + 49))
{
float fix = (16 + 49) - distance;
float fraction = fix / distance;
x[0] -= fraction * xdist;
y[0] -= fraction * ydist;
}
// delay
Sleep(50);
}

In words: for two balls with radii $$16$$ and $$49$$, the distance should be at least $$16 + 49$$. If it isn't, then they are intersecting, by $$(16 + 49) – d$$, where $$d$$ is the distance between the two balls. They need to be moved apart, and not just in any direction, but along the line that connects their centres.

Once that is done, the small ball responds surprisingly realistic to the collision.

Assignment

There is clearly a very large amount of experiments that you could carry out with this code, so today's assignment is a somewhat random selection.

1. Fix the code so that the big ball moves. Assume that it is heavier than the small ball.
2. Add vertical walls, so that the small ball doesn't leave the screen anymore.
3. Add some extra small balls. Make sure that all balls collide with all other balls. Do not forget that a ball should not collide with itself.
4. Add a spring: Connect a heavy ball to a fixed point, and if it gets further than a preset distance from that point, pull it towards the point using Verlet physics.

Let it rain... At some point, you will bring C/C++ to its knees. How to make code like this faster is a topic for another day...

Previous Part: File I/ONext Part: AI

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text:
# How to derive van 't Hoff equation for Henry's law constant?

While studying about Henry's law, I encountered this equation:

$$\frac{\mathrm d\ln H}{\mathrm d(1/T)}= \frac{-\Delta_\text{sol}H }{R},$$

where $$\displaystyle\mathrm d(1/T) = -\frac{\mathrm dT}{T^2}$$ and the $$\displaystyle\frac{\Delta H}{RT^2}$$ reminds me of some kind of $$\Delta G$$ equation.

I didn't know how to derive it, so I searched about it and got nothing. I tried to substitute $$H$$ (Henry's law constant) by $$\gamma\cdot f_i$$ where $$\gamma$$ is the activity coefficient at infinite dilution and $$f_i$$ is the fugacity of pure species.

But still I have no idea what to do to obtain the formula that I'm familiar with, so I welcome any idea about how to derive or how to reverse engineer the formula.

Assuming Henry's coefficient $$\mathcal{H}_i$$ is defined by the following relation:

$$\mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat}$$

Where $$\gamma_i^\inf$$ is the activity coefficient for an infinite diluted solution and $$p_i^\mathrm{sat}$$ is the vapor pressure of pure component $$i$$.

One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):

$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)}$$

The RHS maybe rearranged in order to have the well know form of ratio of mixed equilibrium constants $$K_i$$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).

$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)}$$

That is, we are stating that ratio of Henry's constants approximates to the ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.

Where the concerned reaction is the converse of gas solubilization:

$$\ce{X_{i,(aq)} <=> X_{i,(\mathrm{g})}}{\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H}$$

And its mixed equilibrium constant:

$$K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i}$$

Then one may apply van 't Hoff relation using enthalpy $$\Delta_\mathrm{sol}H$$ of the concerned reaction, from its integrated form, it comes:

$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right]$$

This formulae is generally a good approximation around $$\pu{20°C}$$ over a small range of temperature. Do not forget that $$\Delta_\mathrm{sol}H$$ is not a constant but rather a function of temperature.

If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:

$$\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2}$$

Or equivalently: $$\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R}$$

• Can you please explain how you changed the right side of second equation? – Mo Samani May 24 '16 at 16:15
• assuming that ratio of $x's$ and ratio of activity coefficients are 1 right? so what's the formula used for $K$ here ? – Mo Samani May 24 '16 at 16:28
• How the equation for $Ki$ has obtained ? I know $K$ is the ratio of $fi$ to $fi$ (standard condition)، I understand the numerator,But Can't understand the denominator. – Mo Samani May 25 '16 at 10:17
• This is the expression of a mixed (polyphasic) equilibrium constant. $x_i$ stands for the bulk dissolved gas concentration at equilibrium and $\eta_i$ is its activity coefficient. It is just the ratio of product/reactant as you always write them for aqueous solutions. – jlandercy May 25 '16 at 11:35
• If you feel uncomfortable with this mixed constant, try convert it into $K_c$ or $K_p$. You will get some $RT$ terms that simply because of the ration of constants. – jlandercy May 25 '16 at 11:44


text:
# Application of Compound Interest Formula

There are some situations, we use CI formula. Here are a few :

• In a production unit, rate of growth of a product
• To determine population growth (increase or decrease)

Example

The present population of a town is 28000. If it increases at the rate of 5% per annum, what will be its population after 2 years?

Solution

Example

The cost of a machine is Rs. 175000. After one year the value of that machine was depreciated by 5%. Find the value of machine after one year.

Solution

Principal = Rs. 175000

Reduction = 5% of Rs. 1,75000 per year

Value at the end of 1 year = Rs. 1,75000 - Rs. 8750 = Rs. 1,66,250

Alternate method:
Value at the end of 1 year

= Rs. 1,66,250


text:
# Delta Jensen inequality

Let $a,b \in \mathbb{T}$ and $c,d \in \mathbb{R}$. Suppose $g \colon [a,b]\cap \mathbb{T} \rightarrow (c,d)$ is rd-continuous and $F \colon (c,d) \rightarrow \mathbb{R}$ is convex. Then $$F \left(\dfrac{\displaystyle\int_a^b g(t) \Delta t}{b-a}\right) \leq \dfrac{\displaystyle\int_a^b F(g(t))\Delta t}{b-a}.$$


text:
int_0^pi[cotx]dx, where [.] denotes the greatest integerfunction, | Filo

class 12

Math

Calculus

Integrals

551
150

, where [.] denotes the greatest integerfunction, is equal to

1. 1
551
150

Connecting you to a tutor in 60 seconds.


text:
Whilst reading Mícheál Ó Searcóid’s book Metric Spaces, I found out about a nuance in the definition of continuity that I was not previously aware of, and something which may be taught incorrectly at high schools. M. Searcóid states that a function such as tan(x) is continuous (read the page here). The definition of continuity at a point is based on the fact that the function has a value at that point (if a function is continuous at x = a, then f(a) has a value in the expression |f(x)-f(a)|<ϵ). However, following M. Searcóid’s line of thought about continuous functions, it does not make sense to consider points at which a function is not defined. If we were asked to prove that the function is ‘discontinuous’ at a point, we would need to show that the condition for continuity at that point is false. And the negation* of the condition for continuity would not make sense at a point where the function value is not defined. In this way, I think that high school mathematics may often overlook this nuance in the definition of continuity. One usually thinks of continuity in a setting where a function is mapped from a set of real numbers, but let’s quickly extend this idea to a more abstract setting. If X is a metric space, and b is not an element of X, then is a function f:X→X, that is otherwise continuous on X, discontinuous simply because it is not defined at b? One can clearly see the absurdity in this, and in the same way, we should only consider the subset of the real numbers that is the domain of a function when asking questions about continuity.

Here are some examples:

I argue that

$f(x) = \frac{(x - 1)(x + 2)}{x + 2}$

is a continuous function, because x = -2 is not in the domain of f, the domain being $\mathbb R \setminus \{-2\}$. There is no discontinuity at x = -2 nor is the function continuous at x = -2. The function is simply not defined at x = -2. It is as if the ‘world’ that f ‘knows’ does not include the number -2 in its domain—how could it ‘imagine’ otherwise?

A similar example, and much like the argument for the tan function,

$g(x) =\frac{x - 1}{x + 2}$

is continuous.

Now, in high school we are taught that there are two types of discontinuities: removable and jump. f (above) would have a removable discontinuity if f was defined at x = -2 and f(-2) ≠ -3. Similarly, g would have a discontinuity at x = -2 if $g(-2) \in \mathbb R$. Lastly, a jump discontinuity only exists if the function is defined at the jump:

is a continuous function, whereas

are discontinuous functions.

What do you think?

*The negation of the condition for continuity at a point a, otherwise known as the condition for discontinuity, is that there exists a real number ϵ>0 such that, for every real number δ>0, both |x-a|<δ and |f(x)-f(a)|≥ϵ.

 How clear is this post?


text:
# Bibliography after each chapter from single bib file

I am trying to compile a thesis which has a main.tex file which includes different chapters. I need to have references after each chapter. All the references are in a single .bib file. But when I compile I do not get references list and in place of reference number I get [?].

\documentclass[phd,tocprelim]{CU}
\usepackage[sectionbib]{chapterbib}

%\bibliographystyle{plain}
%\bibliographystyle{IEEEbib}

\begin{document}

\include{chapter1/thesis-chapter1}
\include{chapter2/thesis-chapter2}
\include{chapter3/thesis-chapter3}

\end{document}

The individual chapters are stored in separate sub-folders. An example for chapter 1 is (All_citations is bibliography file):

\chapter{abc}{\label{1}}
\section{Introduction}
...
...
\bibliographystyle{ieeetr}
\bibliography{All_citations}

Thanks

• I found threads similar to your question in the forum: tex.stackexchange.com/questions/282779/bibliography-by-chapter tex.stackexchange.com/questions/374546/… tex.stackexchange.com/questions/229846/… do you think they can help you? – Annie Fubini Feb 17 '20 at 14:41
• Thanks Annie for answering. But I have already seen these threads and unfortunately, they are not solving the issue. – Zulqar Feb 17 '20 at 14:57
• I'm sorry I can't help. Unfortunately it goes beyond my skills. I try to involve more experienced users. – Annie Fubini Feb 18 '20 at 11:52
• If the bibliography style is in the main directory, you should use \bibliography{../All_citations} --- does bibtex find the file or is it complaining? – Rmano Feb 18 '20 at 12:06
• did you run bibtex on all the aux-files from the \include commands? You will needed bibtex thesis-chapter1 (or perhaps bibtex chapter1/thesis-chapter1, I have forgotten and didn't check). – Ulrike Fischer Feb 18 '20 at 12:28


text:
Brain Teaser 5 ?

1. Jul 3, 2003

drag

Greetings !

What's all that stuff about (0! + 0! + 0! + 0!)! = 24 !?

Thanks !

Live long and prosper.

2. Jul 3, 2003

Greg Bernhardt

(0)! (pronounced as Zero -Factorial), is equal to 1. The factorial of the sum of the factorials of four zeros, each taken separately, giving 4!, which is equal to 24.

3. Jul 3, 2003

drag

Eeh... factorial ?

4. Jul 3, 2003

FZ+

Consider this:

4! = 4 * 3 * 2 * 1

So n!, where n is a natural number =

n * (n-1) * ... * 1

It can be mathematically proven that 0! = 1

Don't ask me how, because I forgot.

5. Jul 3, 2003

Hmmmm . . .

there is no 0! proof

actually there is no proof that 0! = 1. look at it from this perspective:

disregarding the order in which we choose, how many ways we can choose k objects from a set of n objects? when finding ways to choose we use the following: nCk = n!/(k!(n-k)!). since there is only one way that one can choose 0 objects, we define 0! = 1 so that n!/(0!(n-0)!) = n!/0!n! = 1/0! = 1.

bascially 0! = 1 is defined as such because we simply need it to be. try to choose k = 0 objects from a set of n objects if 0! = 0. now that's chaos.

i hope that helps.

6. Jul 4, 2003

drag

Re: there is no 0! proof

Aah...so 0! = 1 , in that case I knew what a factorial is but
I never heard of this equality so I thought this may be something else.

Thanks !

Live long and prosper.

7. Jul 4, 2003

Andy

4*3*2*1=24 how does that work?

8. Jul 5, 2003

dav2008

What do you mean? By definition 4!=4*3*2*1 which is 24. THats not the answer to the brain teaser, its just an example

9. Jul 5, 2003

Andy

Sarcasm gets lost online.

10. Jul 5, 2003

dav2008

What are you talking about?!?!?!...I was not being sarcastic. I was answering your question. If I didnt understand your question, that isnt my fault.

11. Jul 5, 2003

Andy

I was trying to be sarcastic, most probably just a piss poor attempt at sarcasm, but i was trying.

12. Jul 7, 2003

Imperial

Basically, (0! + 0! + 0! + 0!)! = (1 + 1 + 1 + 1)!, which equals 4! which evaluates to 24. That is all there is to it!