text:
# For an ideal gas, a constant pressure line and a constant volume line intersect at a point, in the Temperature (T) versus specific entropy (S) diagram. Cp is the specific heat at constant pressure and Cv is the specific heat at constant volume. The ratio of the slopes of the constant pressure and constant volume lines at the point of intersection is

1. $$\frac{{Cp - Cv}}{{Cp}}$$
2. $$\frac{{{C_p}}}{{{C_v}}}$$
3. $$\frac{{{C_p} - {C_v}}}{{{C_v}}}$$
4. $$\frac{{{C_v}}}{{{C_p}}}$$

Option 4 : $$\frac{{{C_v}}}{{{C_p}}}$$

## Temperature - Entropy Diagram MCQ Question 1 Detailed Solution

Concept:

Combined equations of first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

Calculation:

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}$$

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}$$

Therefore, ratio of slope of constant pressure and constant volume lines at point of intersection:

$$\frac{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{p = c}}}}{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{v = c}}}} = \frac{{\frac{T}{{cp}}}}{{\frac{T}{{{c_v}}}}} = \frac{{{C_v}}}{{{C_p}}}$$

# 'Absolute zero' refers to ______.

1. minus 273 °C
2. minus 295 °C
3. minus 300 °C
4. minus 255 °C

Option 1 : minus 273 °C

## Temperature - Entropy Diagram MCQ Question 2 Detailed Solution

The correct answer is minus 273 degrees Celsius.

Key Points

• Absolute zero is the lowest possible temperature and is a point at which fundamental particles of nature have minimal vibration motion resulting in retaining only quantum mechanical zero-point energy induced particle motion.
• It is equivalent to 0 Kelvin or minus 273 degrees Celsius
• It is practically impossible to reach out to the absolute zero temperature
• It is the temperature at which the particles in the substance are essentially motionless which means that there is no way to slow them down further.
• Nothing can be colder than absolute zero on the Kelvin scale.

# The thermal efficiency of the hypothetical cycle shown is

1. 0.6
2. 0.5
3. 0.4
4. 0.3

Option 3 : 0.4

## Temperature - Entropy Diagram MCQ Question 3 Detailed Solution

Concept:

The area under the curve on T-S diagram shows the heat interaction for a reversible process

The area enclosed by the cycle on the T-S diagram gives the work interaction for the cycle

In the cyclic process

Wnet = Qnet = QIn – QOut

Qout = Heat rejection at the lowest temperature in the cycle

Qout = Tmin∆S

Where ∆S is the entropy change

Calculation:

Net work is equal to the area enclosed by the cycle, i.e. area of the triangle

$${W_{net}} = \frac{1}{2} \times 4 \times 400 = 800\;kJ$$

∴ Wnet = 800 kJ

Qout = 300 × (5 - 1) = 1200 kJ

So, QIn = 800 + 1200 = 2000 kJ

The efficiency of the cycle $${\eta _{cycle}} = \frac{{{W_{net}}}}{{{Q_{in}}}} = \frac{{800}}{{2000}} = 0.4$$

# Two processes isobaric and isochoric are represented on T-s diagram. They are starting from the same point. Out of these process which shall have a higher slope?

1. isobaric
2. both have same slope
3. isochoric
4. depend on final point

Option 3 : isochoric

## Temperature - Entropy Diagram MCQ Question 4 Detailed Solution

Concept:

Combined equations of the first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}$$

Hence on the T-S diagram, the slope of the constant pressure line = T/Cp

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}$$

Hence on the T-S diagram, the slope of the constant pressure line = T/Cv

As, C> Cv , T/Cv > T/Cp , Hence the slope of the Isochoric curve will be more than the slope of the Isobaric Curve on TS plane.

# On T-S diagram, constant pressure line slope is expressed as:

1. Cp/γT
2. T/cp
3. T/S
4. S/T

Option 2 : T/cp

## Temperature - Entropy Diagram MCQ Question 5 Detailed Solution

Concept:

Combined equations of the first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}$$

Hence on the T-S diagram, the slope of the constant pressure line = T/Cp

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}$$

# What is the ratio of the efficiencies $$\left( {\frac{{{\eta _I}}}{{{\eta _{II}}}}} \right)$$ for the two cycles as shown in the T-s diagrams?

1. $$\frac{{{T_1} + {T_2}}}{{2{T_1}}}$$
2. $$\frac{{{T_1} + {T_2}}}{{2{T_2}}}$$
3. $$\frac{{2{T_2}}}{{{T_1} + {T_2}}}$$
4. $$\frac{{2{T_1}}}{{{T_1} + {T_2}}}$$

Option 4 : $$\frac{{2{T_1}}}{{{T_1} + {T_2}}}$$

## Temperature - Entropy Diagram MCQ Question 6 Detailed Solution

Concept:

The area under the T-S diagram gives the heat interaction by a reversible process, and the ara enclosed by a reversible cycle represents net work or net heat.

The efficiency of a cycle is given by $$\eta = 1 - \frac{{Heat\;Rejected}}{{Heat\;Supplied}}$$

Calculation:

Case (i)

Heat Supplied $${Q_s} = \frac{1}{2}\left( {{T_1} + {T_2}} \right)\left( {{S_1} - {S_2}} \right)$$

Heat Rejected $$\;{Q_R} = {T_2}\left( {{S_1} - {S_2}} \right)$$

$${\eta _I} = 1 - \frac{{{T_2}\left( {{S_1} - {S_2}} \right)}}{{\frac{1}{2}\left( {{T_1} + {T_2}} \right)\left( {{S_1} - {S_2}} \right)}}$$

$$\Rightarrow {\eta _I} = 1 - \frac{{2{T_2}}}{{{T_1} + {T_2}}}$$

$$\Rightarrow {\eta _I} = \frac{{{T_1} - {T_2}}}{{{T_1} + {T_2}}}$$

Case (ii)

Heat Supplied QS = T1 (S1 – S2)

Heat Rejected $${Q_R} = \frac{1}{2}\left( {{T_1} + {T_2}} \right)\left( {{S_1} - {S_2}} \right)$$

$${\eta _{II}} = 1 - \frac{{\frac{1}{2}\left( {{T_1} + {T_2}} \right)\left( {{S_1} - {S_2}} \right)}}{{{T_1}\left( {{S_1} - {S_2}} \right)}}$$

$${\eta _{II}} = 1 - \frac{{{T_1} + {T_2}}}{{2{T_1}}}$$

$$\Rightarrow {\eta _{II}} = \frac{{{T_1} - {T_2}}}{{2{T_1}}}$$

$$\therefore \frac{{{\eta _I}}}{{{\eta _{II}}}} = \frac{{\left( {{T_1} - {T_2}} \right)\;/{T_1} + {T_2}}}{{\left( {{T_1} - {T_2}} \right)/2{T_1}}} = \frac{{2{T_1}}}{{{T_1} + {T_2}}}$$

# For an ideal gas, a constant pressure line and a constant volume line intersect at a point, in the Temperature (T) versus specific entropy (S) diagram. Cp is the specific heat at constant pressure and Cv is the specific heat at constant volume. The ratio of the slopes of the constant pressure and constant volume lines at the point of intersection is

1. $$\frac{{Cp - Cv}}{{Cp}}$$
2. $$\frac{{{C_p}}}{{{C_v}}}$$
3. $$\frac{{{C_p} - {C_v}}}{{{C_v}}}$$
4. $$\frac{{{C_v}}}{{{C_p}}}$$

Option 4 : $$\frac{{{C_v}}}{{{C_p}}}$$

## Temperature - Entropy Diagram MCQ Question 7 Detailed Solution

Concept:

Combined equations of first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

Calculation:

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}$$

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

$${\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}$$

Therefore, ratio of slope of constant pressure and constant volume lines at point of intersection:

$$\frac{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{p = c}}}}{{{{\left( {\frac{{dT}}{{ds}}} \right)}_{v = c}}}} = \frac{{\frac{T}{{cp}}}}{{\frac{T}{{{c_v}}}}} = \frac{{{C_v}}}{{{C_p}}}$$

# 'Absolute zero' refers to ______.

1. minus 273 °C
2. minus 295 °C
3. minus 300 °C
4. minus 255 °C

Option 1 : minus 273 °C

## Temperature - Entropy Diagram MCQ Question 8 Detailed Solution

The correct answer is minus 273 degrees Celsius.

Key Points

• Absolute zero is the lowest possible temperature and is a point at which fundamental particles of nature have minimal vibration motion resulting in retaining only quantum mechanical zero-point energy induced particle motion.
• It is equivalent to 0 Kelvin or minus 273 degrees Celsius
• It is practically impossible to reach out to the absolute zero temperature
• It is the temperature at which the particles in the substance are essentially motionless which means that there is no way to slow them down further.
• Nothing can be colder than absolute zero on the Kelvin scale.

# The thermal efficiency of the hypothetical cycle shown is

1. 0.6
2. 0.5
3. 0.4
4. 0.3

Option 3 : 0.4

## Temperature - Entropy Diagram MCQ Question 9 Detailed Solution

Concept:

The area under the curve on T-S diagram shows the heat interaction for a reversible process

The area enclosed by the cycle on the T-S diagram gives the work interaction for the cycle

In the cyclic process

Wnet = Qnet = QIn – QOut

Qout = Heat rejection at the lowest temperature in the cycle

Qout = Tmin∆S

Where ∆S is the entropy change

Calculation:

Net work is equal to the area enclosed by the cycle, i.e. area of the triangle

$${W_{net}} = \frac{1}{2} \times 4 \times 400 = 800\;kJ$$

∴ Wnet = 800 kJ

Qout = 300 × (5 - 1) = 1200 kJ

So, QIn = 800 + 1200 = 2000 kJ

The efficiency of the cycle $${\eta _{cycle}} = \frac{{{W_{net}}}}{{{Q_{in}}}} = \frac{{800}}{{2000}} = 0.4$$

# The thermal efficiency of the hypothetical cycle shown is

1. 0.6
2. 0.5
3. 0.4
4. 0.3
5. 0.2

Option 3 : 0.4

## Temperature - Entropy Diagram MCQ Question 10 Detailed Solution

Concept:

The area under the curve on T-S diagram shows the heat interaction for a reversible process

The area enclosed by the cycle on the T-S diagram gives the work interaction for the cycle

In the cyclic process

Wnet = Qnet = QIn – QOut

Qout = Heat rejection at the lowest temperature in the cycle

Qout = Tmin∆S

Where ∆S is the entropy change

Calculation:

Net work is equal to the area enclosed by the cycle, i.e. area of the triangle

$${W_{net}} = \frac{1}{2} \times 4 \times 400 = 800\;kJ$$

∴ Wnet = 800 kJ

Qout = 300 × (5-1) = 1200 kJ

So, QIn = 800 + 1200 = 2000 kJ

The efficiency of the cycle $${\eta _{cycle}} = \frac{{{W_{net}}}}{{{Q_{in}}}} = \frac{{800}}{{2000}} = 0.4$$


text:
## Wednesday, July 1, 2015

### Determine the Two Best Fitting Parameters

#### Integrations

The concept that my Python script had to employ in order to integrate the distribution produced by my contour plot is expressed in the following equation,

#### CDF$$(a_i) = \int_{a_{i_{min}}}^{a_{i_{max}}}$$PDF$$(a_i^{\prime}) da^{\prime}$$

where CDF is the Cumulative Distribution Function and $$a_i$$ can be, in this case, $$a_0$$ or $$a_1$$.
The purpose of evaluating the CDF is to find where the area under the marginalization curves becomes half the total area under the curve.  At that point, or x-axis value, is where the peak of the marginalization curves of Figures 3 and 4 is located.  Thus, I will then have the x-axis value that yields the highest likeliness of being the correct value for that fitting parameter ($$a_0$$ or $$a_1$$).

My Python script has no integral calculations in it.  I avoided using integrals in my calculations because of the likelihood matrix I made early on.  However, I am still getting the integral of PDF.
Marginalizing that matrix (used to plot the contour) allows me to now add the likelihood-sums in order to obtain the integral of the PDF.
One way I like to think about this complex situation is by discerning the fact that my PDF actually is not a continuous distribution and thus an integral is not an apt operator for this method.  I have discrete data points representing likelihood - not a continuous line.  Therefore, I am actually doing the following evaluation: CDF$$= \sum_{a_{i_{min}}}^{a_{i_{max}}}$$ PDF$$(a_{i})$$.  (When considering the summation, this is similar to the Riemann sum technique for continuous line integral estimation.) The main difference between this equation and my method is that I am using a distribution of likelihoods instead of probabilities.  But is that really so different?   Well, when considering the fact that likelihood is simply the product of all probabilities (which is elaborated in my Exercise in Likelihood and $$\chi^2$$, continued... post), it is not different at all.
Seeing the plot will make this concept easier to comprehend.

If you are familiar with FOR loops then you can easily see my code to see my execution of this method as well.  For example, for each value of $$a_1$$ I will check its corresponding likelihood-sum value and then add that with the each other likelihood-sum value corresponding to only the $$a_1$$ values that were previously checked.  Once the sum of the likelihood-sums equate to 1/2 of the sum of all of the likelihood-sums, I have found the point at where the slope of the CDF plot is greatest and therefore have found the point where the PDF is at maximum.  A simpler way of thinking about this concept is that once I find the value that indicates 1/2 of the sum of all of the likelihood-sums, that value is where the Likelihoods curve is at the global maximum, i.e. the peak of the Likelihoods plot.

 Figure 5: Top: Marginalization of Matrix plot shown again.  Bottom: CDF plot revealing the $$a_1$$ value that yields the highest probability of being the correct value.  The green helps to see what value of $$a_1$$ was chosen.  The red lines represent the uncertainties of this calculation of $$a_1$$.

In my code, I added the "areas", or likelihood-sums, from the leftmost hypothesized value of $$a_1$$ to the rightmost.  This is why the CDF reaches 100% when on the right side of the plot.  This means that 100% of the area under the PDF curve has been evaluated.  The mark where 50% of the area was accounted for is indicated by the green line.  However, one must consider how I managed to find that optimal $$a_1$$ value when none of my originally hypothesized $$a_1$$ values yield a result at exactly the 0.50 mark on the y-axis.

#### Interpolations

Python has a beautiful command called "interp1d" that accepted the values in my marginalization plot and found a formula to describe those points.  With this equation, I could then interpolate to find the best value for the fitting parameter.  I needed to interpolate because none of my $$a_1$$ x-axis values produced a CDF value directly at the 0.50 y-axis point.  This is important because I want the point at which the integral of the PDF is precisely 50%.

By interpolation, I also found the length of my asymmetric error bars for this evaluation of $$a_1$$.  Because the PDF follows a normal distribution, I need to find the boundaries around that best $$a_1$$ value that encompasses 68.3% of the integral of the PDF.  That newfound best $$a_1$$ must lie in the middle of this region.  Therefore, I needed to find boundaries that encompass 68.3% symmetrically about the 0.5 mark.  These two boundaries are 0.158 and 0.842.  This is why I interpolated at those marks and found their corresponding $$a_1$$ values.  For my example, these values are 2.15 and 1.86, which is why my uncertainties are +0.14 and -0.15.

#### Nuisance Parameter

I neglected to show the CDF for $$a_0$$ because it is what most would call a nuisance parameter.  It is the y-intercept--a constant.  Terms that merely provide an offset for the data are typically irrelevant.  Despite this truth, it is still uncomfortable for me to simply drop constants during any calculations I do because I am still an undergraduate.  For homework, I get points deducted off for anything.  This fact is magnified especially because of the relentless professors at Embry-Riddle Aeronautical University.  (This is a good and bad thing for a variety of reasons.)  Maybe some day I'll feel more comfortable about dropping constants.  Until then, I still keep them close.

 Figure 6: Top: Marginalization of Matrix plot shown again.  Bottom: CDF plot revealing the $$a_0$$ value that yields the highest probability of being the correct value.  The green helps to see what value of $$a_0$$ was chosen.  The red lines represent the uncertainties of this calculation of $$a_0$$.
This is actually not the best example for finding the optimal $$a_0$$ value.  As you can see in the Likelihoods plot, the right end of the curve seems cut off and doesn't reach as low as the values on the left side of the curve.  Consequently, this does not resemble a normal distribution as well as it should.  Therefore, when using this CDF plot technique to find the point at which 50% of the integral of the PDF is accounted for, the chosen $$a_0$$ in the CDF plot will not be identical to the $$a_0$$ value in Likelihoods plot that indicates the peak of that curve.  Because my range of sampling $$a_0$$ values was not sufficient in producing a curve where both sides have similar local minima, I should re-run my code and make my $$a_0$$ range of samples extend past 2.0 so that I can get a more precise $$a_0$$ measurement.  Notice that the uncertainties for $$a_0$$ are almost 50% of the estimated value.  But look at how low the uncertainties are for $$a_1$$!  Experiencing this relatively not-so-great estimation for $$a_0$$ was a great way of learning the concept and small but necessary details behind statistics.  This exercise was especially helpful for a person like myself who has not taken a full course in any statistics yet.

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# ZEU-THAT

 2018-12-0313:23 [PUBDB-2018-05150] Report/Journal Article et al Generation of quasi continuous-wave electron beams in an L-band normal conducting pulsed RF injector for laboratory astrophysics experiments [arXiv:1803.05540]   We report on an approach to produce quasi continuous-wave (cw) electron beams with an average beam current of milliamperes and a mean beam energy of a few MeV in a pulsed RF injector. Potential applications are in the planned laboratory astrophysics programs at DESY. [...] 2018-11-0713:22 [PUBDB-2018-04279] Report/Journal Article et al Binary neutron star merger remnants as sources of cosmic-ray [DESY-18-097; arXiv:1806.01624] Astroparticle physics 106, 10 - 17 (2018) [10.1016/j.astropartphys.2018.10.007]   We investigate non-thermal electron and nuclei energy losses within the binary neutron star merger remnant produced by the event GW170817. The lack of a cooling feature within the detected synchrotron emission from the source is used to constrain the magnetic field at the mG level, assuming that this emission is electron synchrotron in origin, and that the accelerated spectrum in the electrons follows the form $dN/dE_e \propto E_e^{-2}$. [...] OpenAccess: NS-NS_Mergers_v2(2) - PDF PDF (PDFA); 1-s2.0-S092765051830166X-main - PDF PDF (PDFA); External link: Fulltext 2018-11-0513:36 [PUBDB-2018-04143] Report/Journal Article et al Inefficient Cosmic-Ray Diffusion around Vela X: Constraints from H.E.S.S. Observations of Very High-energy Electrons [arXiv:1807.04182] The astrophysical journal 866(2), 143 (2018) [10.3847/1538-4357/aadfed]   Vela X is a nearby pulsar wind nebula (PWN) powered by a ~104 year old pulsar. Modeling of the spectral energy distribution of the Vela X PWN has shown that accelerated electrons have largely escaped from the confinement, which is likely due to the disruption of the initially confined PWN by the supernova remnant reverse shock [...] OpenAccess: PDF PDF (PDFA); 2018-10-2908:01 [PUBDB-2018-03978] Preprint/Report Pohl, M. Modeling of the spatially resolved non-thermal emission from the Vela Jr. supernova remnant [DESY-18-039; arXiv:1807.10551]   Vela Jr. (RX J0852.0$-$4622) is one of just a few known supernova remnants (SNRs) with a resolved shell across the whole electromagnetic spectrum from radio to very-high-energy ($>100$ GeV; VHE) gamma-rays. [...] OpenAccess: PDF PDF (PDFA); 2018-09-0417:33 [PUBDB-2018-03368] Preprint/Report et al Data-driven model of the cosmic-ray flux and mass composition from 10 GeV to $10^{11}$ GeV [arXiv:1711.11432]   We present a new parametrization of the cosmic-ray flux and its mass composition over an energy range from 10 GeV to $10^{11}$ GeV. Our approach is data-driven and relies on theoretical assumptions as little as possible. [...] OpenAccess: PDF PDF (PDFA); 2018-09-0414:43 [PUBDB-2018-03367] Preprint/Report et al The hadronic interaction model SIBYLL 2.3c and Feynman scaling [arXiv:1709.07227]   The Monte Carlo model Sibyll has been designed for efficient simulation of hadronic multiparticle production up to the highest energies as needed for interpreting cosmic ray measurements. For more than 15 years, version 2.1 of Sibyll has been one of the standard models for air shower simulation. [...] OpenAccess: PDF PDF (PDFA); 2018-09-0414:26 [PUBDB-2018-03366] Master Thesis Heinze, J. Cosmogenic neutrinos as secondary messengers of Ultra-High Energy Cosmic Rays x (2016) = Masterarbeit, Humboldt-Universität zu Berlin, 2016 2018-09-0414:20 [PUBDB-2018-03365] Master Thesis Rudolph, A. Implications of collision dynamics in Gamma-Ray Burst internal shocks on the production of multiple astrophysical messengers x (2017) = Masterarbeit, KIT Karlruhe, 2017 2018-09-0414:12 [PUBDB-2018-03364] Dissertation / PhD Thesis Winter, W. Implications of physics beyond the Standard Model in the quark and lepton sectors Humboldt-Universität zu Berlin 154 pp. (2018) [10.18452/19174] = Dissertation, Humboldt-Universität zu Berlin, 2018   Das Standardmodell (SM) der Teilchenphysik hat sich in der Praxis als vielseitige Theorie bewährt, dennoch deuten nichtverschwindende Neutrinomassen, dunkle Materie und Baryonenasymmetrie auf Physik jenseits des SM (BSM) hin. Um also ein mit den Beobachtungen konsistentes Modell zu entwickeln, ist eine umfassendere Theorie nötig. [...] OpenAccess: PDF PDF (PDFA); 2018-09-0414:01 [PUBDB-2018-03363] Journal Article/Contribution to a conference proceedings et al The hadronic interaction model Sibyll – past, present and future XIX International Symposium on Very High Energy Cosmic Ray Interactions, ISVHECRI 2016, MoscowMoscow, Russia, 22 Aug 2016 - 27 Aug 2016   Sibyll is one of the first microscopic interaction models that was specifically developed for interpreting cosmic ray data. It combines non-perturbative concepts of simulating hadronic particle production with predictions derived from perturbative QCD calculations, focusing on forward particle production of relevance in studying cosmic ray interactions [...] OpenAccess: PDF PDF (PDFA);


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${\displaystyle 2\gamma _{a}\sigma T_{a}^{4}=\gamma _{s}\sigma T_{s}^{4}}$ (4j)
 ${\displaystyle {\sqrt[{4}]{\tfrac {S_{o}(1-\sigma )}{4\sigma (1-{\frac {\gamma _{a}}{2}})}}}=T_{s}}$ (4k)
With no atmosphere at all, ${\displaystyle \gamma _{a}=0}$ and the equation above just reduces to our original equation, giving us an answer of 255K. By plugging in the observed temperature at the Earth's surface (288K) and solving for ${\displaystyle \gamma _{a}}$, we obtain a value of ${\displaystyle \gamma =.76}$. With that value in hand, then, we can actually use this model to explore the response of the planet to changes in albedo or greenhouse gas composition—we can make genuine predictions about what will happen to the planet if our atmosphere becomes more opaque to infrared radiation, more energy comes in from the sun, or the reflective profile of the surface changes. This is a fully-developed ZDEBM, and while it is only modestly powerful, it is a working model that could be employed to make accurate, interesting predictions. It is a real pattern.


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.

## Tag: graphs

### Science things that are awesome…

(Also, all of these goofy pics are now being archived at a tumblr I just set up – scienceisawesomethatisall.tumblr.com)

O.K. Yesterday was our provincial elections (in British Columbia), and in the end, the Liberal party came out winning. There’s quite a few environmental issues that are in the forefront in my neck of the woods, not the least of which concerns the Northern Gateway pipeline.

The Liberals didn’t actually have the greatest platform on this (at least from an environmental or science policy standpoint), but here’s hoping the public continues to pressure them to do the “best” (re: what scientific expert peer review suggests) thing for the province, and indeed the planet at large.

Last Saturday, my lab opened up the entire ground floor of the Michael Smith Building to the public. This was in conjunction with Science Rendezvous, a cross Canada science festival, and in the case of UBC, organized by the Faculty of Science. In the house (so to speak) were folks from the Beaty Museum, Civil Engineering, Pathology, Physics and Astronomy, as well as the Engineering Physics Robotics lab (who also brought in their 3D printers). We also used the building as ground zero for a number of tours throughout campus.

All in all, a great day (and busy too!). In my space, I actually brought out about a dozen or dissecting scopes and collected a nice jar of pond scum. Kids (and their parents), with some basic instructions, were let loose to find whatever they could find in the pond water. Lots of cooties were found, protozoa and algae abound, but my favourite was this Hydra that I managed to get a decent picture of on my iPhone.

The scientific method – it’s not perfect, but it’s pretty much the best way out there on collecting your thoughts and information to make sound decisions. All the more so, if the decision is high stakes IMHO.

### Tupper’s self-referential formula #whoa

“Tupper’s self-referential formula is a self-referential formula defined by Jeff Tupper that, when graphed in two dimensions, can visually reproduce the formula itself. It is used in various math and computer science courses as an exercise in graphing formulae.

Specifically (From Wikipedia):

The formula is an inequality defined by:

${1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor$

where $\lfloor \cdot \rfloor$ denotes the floor function and mod is the modulo operation.

Let k equal the following:

48584506361897134235820959624942020445814005879832445494830930850619347047088099284506447698655243648499972470249151191
104116057391774078569197543265718554420572104457358836818298237541396343382251994521916512843483329051311931999535024137
58765239264874613394906870130562295813219481113685339535565290850023875092856892694555974281546386510730049106723058933586
052544096664351265349363643957125565695936815184334857605266940161251266951421550539554519153785457525756590740540157929
001765967965480064427829131488548259914721248506352686630476300

If one graphs the set of points (x,y-k) with $0 \le x \le 106$ and $k \le y \le k + 17$ such that they satisfy the inequality given above, the resulting graph looks like this:

### Globally relevant infographics via the Olympic Rings.

Gustavo Sousa “uses the five colorful rings, representing each of the five continents taking part in the games every four years, to display a series of informative graphs about the world we live in today. The topics range from general facts like the world’s population to staggering statistics that reveal the ratio of people living with HIV, as symbolized by the size of the circle representing their continental location. Key: Blue is Oceania (Australia and its proximate islands); Yellow is Africa; Black is Europe; Green is Asia; Red is the Americas.”

By Gustavo Sousa, text via My Modern Met.

### Kurt Vonnegut and the graph of infinite happiness

Classic Vonnegut on “The Shape of Stories.” The graphs only make it cooler.

Via @cogdog

### Function World looks awesome!

By Grant Snider, via Drawn.ca.

### Why I majored in Biology: The Pie Chart.

Via sugarglue.tumblr.com

Via i.imgur.com.

### Percent chance that each of these bars will reach the top of this graph. #funny via@BenGreenman

Via I Love Charts and Ben Greenman


text:
TheInfoList

Gaussian noise is statistical noise having a probability density function (PDF) equal to that of the normal distribution, which is also known as the Gaussian distribution.[1][2] In other words, the values that the noise can take on are Gaussian-distributed.

The probability density function ${\displaystyle p}$ of a Gaussian random variable ${\displaystyle z}$ is given by:

${\displaystyle p_{G}(z)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {(z-\mu )^{2}}{2\sigma ^{2}}}}}$

where ${\displaystyle z}$ represents the grey level, ${\displaystyle \mu }$ the mean value and ${\displaystyle \sigma }$ the standard deviation.[3]

A special case is white Gaussian noise, in which the values at any pair of times are identically distributed and statistically independent (and hence uncorrelated). In communication channel testing and modelling, Gaussian noise is used as additive white noise to generate additive white Gaussian noise.

In telecommunications and computer networking, communication channels can be affected by wideband Gaussian noise coming from many natural sources, such as the thermal vibrations of atoms in conductors (referred to as thermal noise or Johnson-Nyquist noise), shot noise, black body radiation from the earth and other warm objects, and from celestial sources such as the Sun.

### Gaussian noise in digital images

Principal sources of Gaussian noise in digital images arise during acquisition e.g. sensor noise caused by poor illumination and/or high temperature, and/or transmission e.g. electronic circuit noise.[3] In digital image processing Gaussian noise can be reduced using a spatial filter, though when smoothing an image, an undesirable outcome may result in the blurring of fine-scaled image edges and details because they also correspond to blocked high frequencies. Conventional spatial filtering techniques for noise removal include: mean (convolution) filtering, median filtering and Gaussian smoothing.[1][4]

## References

1. ^ a b Tudor Barbu (2013). "Variational Image Denoising Approach with Diffusion Porous Media Flow". Abstract and Applied Analysis. 2013: 8. doi:10.1155/2013/856876.
2. ^ Barry Truax, ed. (1999). "Handbook for Acoustic Ecology" (Second ed.). Cambridge Street Publishing. Retrieved 2012-08-05.
3. ^ a b Philippe Cattin (2012-04-24). "Image Restoration: Introduction to Signal and Image Processing". MIAC, University of Basel. Retrieved 11 October 2013.
4. ^ Robert Fisher; Simon Perkins; Ashley Walker; Erik Wolfart. "Image Synthesis — Noise Generation". Retrieved 11 October 2013.


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Increasing And Decreasing Functions

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If 0 lt alpha lt beta lt (pi)/...

Updated On: 27-06-2022

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Text Solution

(tan beta)/(tan alpha ) lt (alpha)/(beta)(tan beta)/(tan alpha) gt (alpha)/(beta)(tan beta )/(tan alpha ) gt (alpha)/(beta)(tan alpha)/(tan beta )le (alpha)/(beta)

Solution : Consider the function f(x) given by f(x)=x tan x, x in(0 , pi //2)
 we have ,
 f(x) = x sec^2 x + tan x gt 0 " for all " x In (0,pi//2)
 rArr f(x) is increasing on (0,pi//2)
 rArr f(alpha) lt f (beta) for 0 alpha lt beta lt (pi)/(2)
 rArr alpha tan alpha lt beta
 rArr /beta (tan beta)/(tan alpha)

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

hello students in this question we are given that if Alpha is greater than zero and less than B H and M S an informed decision prior to that we need to find the correct option among these for a given right she carefully in the options have to find the relation between alpha and beta 10 litre to find relation between these two quantities so let's assume function f x = 2 x 10 x right and rear axle belong 2022 right now to differentiate this function that is left X equals to 8 will be X into x square

plus derivative of X is 1126 night so it is x square x + 10 x 10 x now you can see that X is a positive 28 and sec Square X will also also be positive because I square of any number is positive rate so this hotel is a positive country and we know that 10x in first quadrant is also positive overall and this all about X is greater than 0 9 x is greater than 0 in explains 2025-2 so what we can say that we can say that FX filmi increasing function effects will be increasing in excellence 2025

right now it is given in the question that Alpha is less than Peter Wright Alpha is less than beta and if FX is an increasing function we can say that a for Alpha will be less than half of Beta because if an increasing function will be something like this it and if Alpha is here that is alpha is less than Pita induction CPC alpha and beta then obviously if the function is increasing then the act of Alpha will be less than half of Peter Wright show upon price of Peter now shows that f of X is equals to x 10 x right so what will be a free-form

nothing Alpha 1 to 10 mph with less than Peter into 10 beta right now if we write this in depend on then it will be 10 beta / 10 awful it will be greater than what a bi beta Koteswara answer now we compared with the options that 10 beta by Alpha is what alphabet M letter alphabet M so aur option to is the correct option that 10 beta B and Alpha is relevant to for my beta so this is your answer in


text:
# Jon Aaronson (TAU) On the bounded cohomology of actions of multidimensional groups.

Date:
Tue, 06/11/201814:15-15:15
Although each cocycle for a action of the integers is
specified by the sequence of Birkhoff sums of a function,
it is relatively difficult to specify cocycles for the actions of
multidimensional groups such as $\Bbb Z^2$.
We'll see that if $(X,T)$ is a transitive action of the finitely
generated (countable) group $\Gamma$ by homeomorphism of the polish space $X$,
and $\Bbb B$ is a separable Banach space, there is a cocycle
$F:\Gamma\times X \to\Bbb B$
with each $x\mapsto F(g,x)$ bounded and continuous
so that the skew product action $(X x \Bbb B,S)$ is transitive where
$S_g(x,b)=(T_gx,b+F(g,x))$.
Depending on time available, we'll also discuss measure theoretic analogues.
This result was shown for transitive actions of the integers by E.A.
Sidorov in 1973.
Joint work with Benjamin Weiss: arXiv:1712.05196


text:
NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Related rates (dy/dx=dy/dt x dt/dx)

## Interactive practice questions

Let $y=\left(x+5\right)^5$y=(x+5)5 be defined as a composition of the functions $y=u^5$y=u5 and $u=x+5$u=x+5.

a

Determine $\frac{dy}{du}$dydu.

b

Determine $\frac{du}{dx}$dudx.

c

Hence determine $\frac{dy}{dx}$dydx.

Easy
Approx 3 minutes

A sandbox is being filled at a local playground. The density of sand is given by $\frac{dm}{dV}=4.5$dmdV=4.5 g/cm3, where $m$m is the mass in grams and $V$V is the volume in cubic centimetres.

If sand is being added to the sandbox at a rate of $\frac{dV}{dt}=400$dVdt=400 cm3/min, find $\frac{dm}{dt}$dmdt, the rate at which the sandbox gets heavier over time.

A point moves along the curve $y=5x^3$y=5x3 in such a way that the $x$x-coordinate of the point increases by $\frac{1}{5}$15 units per second.

Let $t$t be the time at which the point reaches $\left(x,y\right)$(x,y).

Find the rate at which the $y$y-coordinate is changing with respect to time when $x=9$x=9.

A spherical hot air balloon, whose volume and radius at time $t$t are $V$V m3 and $r$r m respectively, is filled with air at a rate of $4$4 m3/min.

At what rate is the radius of the balloon increasing when the radius is $2$2 m?

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems


text:
# Math Help - A Little 3D Maths...

1. ## A Little 3D Maths...

Not really sure if this is the correct forum to help, but i'm having a couple a problems with 3d co-ordinates.

I am working on a project to visualise some data. I am being supplied with vectors of data (a magnitude (-9999 to 9999) plus an axis reading 0-180 degrees). I am trying to made a 3d visualisation of this.
I have the start coordinates of each vector (x y z), and the 2D information(magnitude, and direction in the 2D plane), but don't have a clue how to calculate the 3D vector for each point from this information. any ideas or links to information on this sort of thing?

2. ## Just an update

i am now working on the assumption that i can calculate a vector which is on the same plane as the 2d point, calculate the equation of the plane, and then put the point hat i need in 3d into this equation to get the cartesian co-ordinates. Does this sound correct (or at least feasible).
If so can somebody suggest how i can get the plane normal from a vector and a point on the plane, as i cannot quit understand that much yet...

3. Originally Posted by kevd
i am now working on the assumption that i can calculate a vector which is on the same plane as the 2d point, calculate the equation of the plane, and then put the point hat i need in 3d into this equation to get the cartesian co-ordinates. Does this sound correct (or at least feasible).
If so can somebody suggest how i can get the plane normal from a vector and a point on the plane, as i cannot quit understand that much yet...
I'm not following why and how you're switching between 2D and 3D. But I can tell you how to get a normal to a plane using the cross product.

If you have trouble understanding that, then post an example so that I or others here can understand what notation you're using. Then we can translate to that notation. By notation, I mean do you use the $\mathbf{v} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ notation for vectors, or the cartesian coordinates $\mathbf{v} = [2,1,3]$ or something else?

4. I am working on a project to visualise in a 3d manner stress in pipe. I am being provided with a stress vector (containing 2D polar coordinates, being the magnitude (-9999 to 9999) and and direction (0 to 180 degrees)). I am taking this data and mapping it onto a 3D visualisation i am about to create.

I will know the location of the centre of the pipe, A, (say (0,0,0)) and the location of the data point where the vector is to be mapped to, B (say (0,1,0) for sake of argument). From this, i'm hoping i can use the vector AB, to calculate the normal to the plane, and hence the equation of the plane.

And Thats where i need a way to get the 2d polar co-ordinates specified on that plane, to 3D so i can map the direction on the visualisation (maybe by using the equation of the plane and the one 3D point i already know on the plane, the Z co-ordinate of the second point may be obtainable), but this is where i need the help.

If my assumprions are way off please say, as i only was only given this information/project yesterday, and am only now beginning to look into how to achieve it.

5. Originally Posted by kevd
I am working on a project to visualise in a 3d manner stress in pipe. I am being provided with a stress vector (containing 2D polar coordinates, being the magnitude (-9999 to 9999) and and direction (0 to 180 degrees)). I am taking this data and mapping it onto a 3D visualisation i am about to create.

I will know the location of the centre of the pipe, A, (say (0,0,0)) and the location of the data point where the vector is to be mapped to, B (say (0,1,0) for sake of argument). From this, i'm hoping i can use the vector AB, to calculate the normal to the plane, and hence the equation of the plane.

And Thats where i need a way to get the 2d polar co-ordinates specified on that plane, to 3D so i can map the direction on the visualisation (maybe by using the equation of the plane and the one 3D point i already know on the plane, the Z co-ordinate of the second point may be obtainable), but this is where i need the help.

If my assumprions are way off please say, as i only was only given this information/project yesterday, and am only now beginning to look into how to achieve it.
I'm still not sure what you mean, but I'll try to state my interpretation so far.

Start with the pipe on the (x,y) plane. It lies along the x-axis starting at the origin. Let the direction of the pipe be denoted by the unit vector along the x-axis. In 3D coordinates this direction vector is (1,0,0). The stress vector translated to cartesian coordinates in the (x,y) plane is (s,t). The 3D coordinates are (s,t,0). The 3D unit normal to the (x,y) plane at the origin is just the unit vector (0,0,1) along the z-axis.

Now suppose we rotate the plane, the pipe, the stress vector and the unit normal vector all together in 3D space. We want to know what are the coordinates of the pipe, stress and unit normal vectors after the rotation.

We can represent the rotation with a matrix (see here and here)

where the angles $(\alpha,\beta,\gamma)$ are the angles of rotation, called the Euler angles. Then the rotated coordinates of the pipe, stress and unit normal vectors will just be their original coordinates multiplied by this matrix.

Postponing the question of how to calculate the Euler angles, is this anywhere near what you had in mind?

6. cheers,
Its definately given me somewhere to start. I'll probably be back posting when i get stuck again(or atl east when my head stops spinning from the confusion), but it seems to be exactly what i needed, and it even makes sense!

7. thanks Jake,

I'm sure you can guess the question thats coming now, but here goes. Euler angles.. i think i now understand the concept, but cannot get my head round how to calculate them for a given rotation. Could you possilbly explain using the following example?

the vector on x,y plane is (3,2,0), lets say i wanted to rotate everythin by 30 degrees on the x-axis, 20 degrees on the y-axis, and 45 degrees on the z-axis

Sorry if i'm missing something simple, but i'm new to eulers angles etc.

8. Originally Posted by kevd
thanks Jake,

I'm sure you can guess the question thats coming now, but here goes. Euler angles.. i think i now understand the concept, but cannot get my head round how to calculate them for a given rotation. Could you possilbly explain using the following example?

the vector on x,y plane is (3,2,0), lets say i wanted to rotate everythin by 30 degrees on the x-axis, 20 degrees on the y-axis, and 45 degrees on the z-axis

Sorry if i'm missing something simple, but i'm new to eulers angles etc.
You're not missing something simple. This is complicated stuff.

I'm new to this too. I liked your question because it gave me a chance to better understand what I had run across before. But after reading more links on the web and playing with a computer program I wrote, my conclusion is I don't really get it and I don't think it simple enough to be described on the web. The questions you and I are both asking have been asked before and simple answers just aren't there. It seems reading books and/or taking classes are required.

So I'm bailing out here. Framing the problem and pointing to a few links I think you need to understand to do 3D is about as far as I can go. Good luck!

9. thanks anyway. I'm sure i'll work it out (sooner or later)

10. Originally Posted by kevd
thanks Jake,

I'm sure you can guess the question thats coming now, but here goes. Euler angles.. i think i now understand the concept, but cannot get my head round how to calculate them for a given rotation. Could you possilbly explain using the following example?

the vector on x,y plane is (3,2,0), lets say i wanted to rotate everythin by 30 degrees on the x-axis, 20 degrees on the y-axis, and 45 degrees on the z-axis

Sorry if i'm missing something simple, but i'm new to eulers angles etc.
kevd, while I bailed on trying to explain this question to you, I continued to work on understanding the problem. I found out a few things that explain the confusion I had and may help you.

First off, the angles you are describing are strictly speaking not Euler angles. They are the roll, pitch and yaw angles. From this Wikipedia page:

• To add to the confusion, flight and aerospace engineers, when using yaw, pitch, and roll ... to refer to rotations about the z, y, x axes, respectively, often call these the Euler angles. These x-y-z angles are properly known as the Tait-Bryan angles...

This page gives to a cookbook I could follow to compute the rotation matrix for the rotation you describe. You just calculate separate rotation matrices for each of the 3 axes and multiply them together. It seems simple enough except for notice some things:

• The angles are called Euler angles.
• The meanings of roll and pitch are switched from above.
• A positive rotation angle means clockwise when viewing an axis from the positive numbers to the origin. The author actually doesn't say which direction you are viewing the axis from; I found it is from the positive side by trying it out.
• The order the rotations are applied matters. The order used there is y, then x, then z. Use a different order and the effect is different.

So these roll-pitch-yaw angles seem straightforward if you are careful about the definitions. I still don't know how these relate to Euler angles and why you would use those.

All I can say is, what a stinking mess!

11. lol, cheers jake, think i'm closer to getting something working now! i'll post here if i do just to satify your curiosity!
Thanks for all the help,
Kev


text:
# After school, Maun'ce walks 1/3 mile to the park and then walks 1/2 mile to his house. How far does Maurice walk from school to his house?

Haiden Meyer 2022-09-29 Answered
After school, Maun'ce walks $\frac{1}{3}$ mile to the park and then walks $\frac{1}{2}$ mile to his house. How far does Maurice walk from school to his house?
You can still ask an expert for help

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## Answers (1)

Emilia Boyle
Answered 2022-09-30 Author has 10 answers
$\frac{1}{3}+\frac{1}{2}$ change them into equivalent fractions
$\frac{2}{6}+\frac{3}{6}=\frac{5}{6}$
###### Did you like this example?

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# Why write plain text?

Published:

Here’s why:

• A writing environment that models an argument, not a sheet of paper. Plain-text documents decouple writing environment from distribution format. In my plain-text documents, each sentence gets a line of its own: sentences always begin at the left margin. This makes it easier to see and edit the sequence of argument within paragraphs. Lines of text are re-flowed into paragraphs for distribution — i.e., by the program that converts my plain-text document into pdf or docx or html.

• Persistent private comments. Like programming languages, plain-text markup languages have a syntax for comments — that is, for designating text that will not be read by the engine that processes the file. Plain-text comments can be used like the comment function in Microsoft Word or Google Docs. The difference is that comments in plain-text files are, by default, masked off and omitted from the pdf and docx files generated from them. Because the comments do not render in the publication formats, I can retain them in my working document throughout the whole life-cycle of the document, or as long as they remain useful to me. I can also comment out whole sentences or paragraphs if, for instance, I need to reduce the word count in one version of a document, but think I may re-instate the suppressed material in a later version.

• Persistent lightweight file history. Plain-text files can be placed under git version control, just like program code.

• Automated formatting of citations and bibliography. In a plain-text environment, I never keyboard citations. Citations are formatted by script, drawing on machine-readable bibliographies that I curate. The output must be proofed. This progressively improves the quality of my bibliographical data and debugs the scripts that format citations.

• Exposure to command-line utilities. As text files, my documents can be manipulated with the same command-line utilities and applications that I use to interact with any other text file in my system.

For an example of a simple plain-text document, view the Markdown source for this post.

Tags:


text:
# Generalizations of Fermat’s Little Theorem and combinatorial zeta functions

Everyone who studies elementary number theory learns two different versions of Fermat’s Little Theorem:

Fermat’s Little Theorem, Version 1: If $p$ is prime and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$.

Fermat’s Little Theorem, Version 2: If $p$ is prime and $a$ is any integer, then $a^{p} \equiv a \pmod{p}$.

as well as the following extension of Version 1 of Fermat’s Little Theorem to arbitrary positive integers $n$:

Euler’s Theorem: If $n$ is a positive integer and $(a,n)=1$, then $a^{\phi(n)} \equiv 1 \pmod{n}$, where $\phi$ is Euler’s totient function.

My first goal in this post is to explain a generalization of Version 2 of Fermat’s Little Theorem to arbitrary $n$. I’ll then explain an extension of this result to $m \times m$ integer matrices, along with a slick proof of all of these results (and more) via “combinatorial zeta functions”.

Continue reading

# An April Fools’ Day to Remember

Today is the 10th anniversary of the death of Martin Gardner. His books on mathematics had a huge influence on me as a teenager, and I’m a fan of his writing on magic as well, but it was only last year that I branched out into reading some of his essays on philosophy, economics, religion, literature, etc. In this vein, I highly recommend “The Night Is Large”, a book of collected essays which showcases the astonishingly broad range of topics about which Martin had something interesting to say. It’s out of print, but it’s easy to find an inexpensive used copy if you search online.

Thinking back on my favorite Martin Gardner columns, my all-time favorite has to be the April 1975 issue of Scientific American. In that issue, Martin wrote an article about the six most sensational discoveries of 1974. The whole article was an April Fools’ Day prank: among the discoveries he reported were a counterexample to the four-color problem and an artificial-intelligence computer chess program that determined, with a high degree of probability, that P-KR4 is always a winning move for white. The article also contained the following:

Continue reading

# A Very Meta Monday

Usually my blog posts are rather tightly focused, but today I’d just like to post a few stream-of-consciousness thoughts.

(1) My blog was recently featured in the AMS Blog on Math Blogs. Perhaps by mentioning this here I can create some sort of infinite recursion which crashes the internet and forces a reboot of the year 2020.

Continue reading


text:
# Game theory payoff table - small correction needed

I would like to create a payoff matrix that looks like this:

With the following code:

\begin{align*}
\begin{tabular}{cc|c|c|}
& \multicolumn{1}{c}{} & \multicolumn{2}{c}{Suspect 2} \\\cline{3-4}
& {} &{Quiet}  &{Fink}  \\\cline{2-4}
&{Quiet} & {4,4} & {0,6}  \\ \cline{2-4}
Suspect 1  &{Fink} & {6,0} & {2,2}  \\\cline{2-4}
\end{tabular}
\end{align*}

I've got it looking like this:

Help very much appreciated!

\documentclass[12pt]{article}
\usepackage{multirow}
\newcommand\ml[1]{\multicolumn{1}{|l|}{#1}}
\newcommand\mcc[1]{\multicolumn{2}{c}{#1}}

\begin{document}
\begin{center}
\renewcommand\arraystretch{1.3}
\begin{tabular}{lc|c|c|}
\mcc{}          &   \mcc{Suspect 2}         \\
\cline{3-4}
&           &   Quiet   &   Fink        \\
\cline{2-4}
\multirow{2}{*}{Suspect 1}
&   \ml{Quiet}  &   4,4     &   0,6     \\
\cline{2-4}
&   \ml{Fink}   &   6,0     &   2,2     \\
\cline{2-4}
\end{tabular}
\end{center}
\end{document}

edit: as suggested CarLaTeX in her comment below, table is now stretched by factor 1.3.

• Already awake? You were faster that me! I was just posting the same answer, the only difference is I would add \renewcommand*{\arraystretch}{1.2} :) – CarLaTeX Oct 23 '17 at 2:41
• Works like a charm - learned a few new tricks in the process! – Thevesh Theva Oct 23 '17 at 2:45
• @CarLaTeX, still awake ... :-(, so i'm bit tiered and forgot to suggest table more "open" ... – Zarko Oct 23 '17 at 2:45
• Still awake?! You should go to sleep, then! Good night :):):) – CarLaTeX Oct 23 '17 at 2:48


text:
The Annals of Probability

The Transition Function of a Fleming-Viot Process

Abstract

Let $S$ be a compact metric space, let $\theta \geq 0$, and let $\nu_0$ be a Borel probability measure on $S$. An explicit formula is found for the transition function of the Fleming-Viot process with type space $S$ and mutation operator $(Af)(x) = (1/2)\theta\int_S(f(\xi) - f(x))\nu_0(d\xi)$.

Article information

Source
Ann. Probab., Volume 21, Number 3 (1993), 1571-1590.

Dates
First available in Project Euclid: 19 April 2007

Permanent link to this document
https://projecteuclid.org/euclid.aop/1176989131

Digital Object Identifier
doi:10.1214/aop/1176989131

Mathematical Reviews number (MathSciNet)
MR1235429

Zentralblatt MATH identifier
0778.60038

JSTOR


text:
you are viewing a single comment's thread.

[–] -1 points0 points  (0 children)

sorry, this has been archived and can no longer be voted on

I got down voted for saying something like this before, but if you have an infinite number of strings of zero, and I say infinite because I assume the universe is infinite wouldn't that make the answer of the summation be 0*infinite, or 0/0, making the answer not 0? Sorry if I misunderstood what you meant by that, it just seemed to relate to my previous answer to this question.


text:
Michael Schapira: Internet Routing, Distributed Computation, Game Dynamics and Mechanism Design I

This post is authored by Michael Schapira. (It is the first in a series of two posts.)

In this post, I’ll outline work on Internet routing and sketch important areas for future work, both on routing itself and, more broadly, on mechanism design, game theory and distributed computation.

The Internet comprises administrative domains known as Autonomous Systems (ASes), which vary in size, from large (multi)national networks to small networks servicing schools or small businesses. The task of establishing routes between ASes, called interdomain routing, is currently handled by the Border Gateway Protocol (BGP)—the core routing protocol of the Internet. BGP is one of the most critical pieces of the Internet’s infrastructure and can be regarded as the glue that holds the Internet together.

Over the past decade, routing with BGP has been studied from engineering, computational, game theoretic and economic perspectives. Combining algorithmic and economic considerations in the study of interdomain routing is natural, because the many separate domains that make up the Internet are independent economic agents that must jointly execute a distributed algorithm in order to route traffic. In addition, interdomain routing motivates exciting new questions in computer science, game theory, and economics.

A (Simplified) Model of Interdomain Routing

The following is a simplified model of interdomain routing, based on the seminal work of Thimothy Griffin and Gordon Wilfong, and on the game-theoretic model of Hagay Levin, Michael Schapira, and Aviv Zohar   Continue reading

Joe Malkevitch: Why Planar Graphs are so Exceptional

Not only do interesting questions arise by considering the special class of planar graphs but additional special issues arise when one considers a specific plane drawing of a planar graph. This is because when a graph is drawn in the plane it becomes possible to order or number, say for example, the edges at a particular vertex of the graph, in an organized consistent way.

One example of results which started from this reality for plane graphs is this paper of Grunbaum and Motzkin: B. Grünbaum and T. S. Motzkin , The number of hexagons and the simplicity of geodesics on certain polyhedra. Canad. J. Math. 15 (1963), pp. 744–751. In this paper the following is explored (along with results about what today are called fullerenes). Suppose one has a 3-valent plane graph. If one picks any edge and moves along that edge (in either direction), when one gets to a vertex one has the choice of going left or right and moving on to another edge. Suppose one goes left, and at the next vertex in one’s “traversal” one goes right, alternating left, and right. Grunbaum and Motzkin refer to this as left-right path, and they prove that for a plane 3-valent graph with each of its faces having a multiple of 3 for its number of sides, that these left-right paths (starting on any edge) always generate “simple circuits.”

I was able to extend this result in several directions. Continue reading

(Eran Nevo) The g-Conjecture III: Algebraic Shifting

This is the third in a series of posts by Eran Nevo on the g-conjecture. Eran’s first post was devoted to the combinatorics of the g-conjecture and was followed by a further post by me on the origin of the g-conjecture. Eran’s second post was about the commutative-algebra content of the conjecture. It described the Cohen-Macaulay property (which is largely understood and known to hold for simplicial spheres) and the Lefshetz property which is known for simplicial polytopes and is wide open for simplicial spheres.

The g-conjecture and algebraic shifting

Squeezed spheres

Back to the question from last time, Steinitz showed that

any simplicial 2-sphere is the boundary of a convex 3-polytope.

However, in higher dimension

there are many more simplicial spheres than simplicial polytopes,

on a fixed large number of vertices. Continue reading

Ehud Friedgut: Murphy’s Law of Breastfeeding Twins

This post is authored by Ehud Friedgut. Congratulations to Keren, Ehud and Michal for the birth of Shiri and Hillel!

Murphy’s law of breastfeeding twins, like all of Murphy’s laws, is supported by strong empirical evidence.

The twins’ feeding rhythm depends on your approach. If you decide to try and synchronize them in order to minimize your hassle and maximize your rest periods then twin A’s hunger will be governed by $sine(t)$ and twin B’s by $cosine(t)$.

If you decide to respect their natural rhythm in hope of falling into a fixed predictable pattern then twin A will follow $sine(t)$ and twin B $sine(\phi\times t)$, where $\phi$, the golden ratio, is the number hardest to approximate by rationals

(Eran Nevo) The g-Conjecture I

This post is authored by Eran Nevo. (It is the first in a series of five posts.)

Peter McMullen

The g-conjecture

What are the possible face numbers of triangulations of spheres?

There is only one zero-dimensional sphere and it consists of 2 points.

The one-dimensional triangulations of spheres are simple cycles, having $n$ vertices and $n$ edges, where $n\geq 3$.

The 2-spheres with $n$ vertices have $3n-6$ edges and $2n-4$ triangles, and $n$ can be any integer $\geq 4$. This follows from Euler formula.

For higher-dimensional spheres the number of vertices doesn’t determine all the face numbers, and for spheres of dimension $\geq 5$ a characterization of the possible face numbers is only conjectured! This problem has interesting relations to other mathematical fields, as we shall see. Continue reading

Michal Linial: Vive La Technology

Nebi Samuel

It was a long night after crossing the Atlantic. I got to the airport and indeed the big signs of ‘Welcome’ and “TAAM SHEL BAIT” (taste of home) had a special appeal. It was 3:30 in the morning when I took the Taxi. The driver was very talkative and wanted to know where I am coming from, are the stories that he had heard on America true. Specifically he wanted to know if I went to the Twins to see the city from the tallest buildings etc, etc.
I replied shortly: “Yes, New York is as everybody said”. For the Twins, I replied: “No, there is no elevator there anymore”… I tried to be polite but at the same time to keep our touristy discussion to a minimum.  The Taxi driver turned out to be quite sensitive to my poor situation and told me very proudly: “Please tell me where, and I will take you, I have a new GPS, just tell me the address and the GPS will show me the way”. Indeed, I shared with him the importance of the GPS technology and even told him that I may soon buy one for my own car. My address is Nave Sha’anan 18, I told him, and 2 minutes later, I was asleep, this time very confident that in 45 min I am at home and to my bed. The next thing I know is Continue reading

1:0 to Italy: Michal Linial

Coming back from Hinxton, the most exciting genome center in the world and the most boring place after 5pm… At 5:30, a shuttle bus takes all the ‘workers’ to the big city – to Cambridge. The shuttle bus dropped us in the middle of town and without noticing, I remained alone. I quickly looked around; actually, there was no one on the street… OK I decided, the evening is not lost yet, I will go to London…

I waited 5 minutes and a Taxi stopped next to me. To the train station” I said. My taxi driver started the ride and then slowed down, stopped for a minute next to an open bar looked extremely worried and after 2 minutes speeded up again. We moved another 100 meters and again, he slows down and stop for 2 minutes next to a TV screen in the next open bar. He looks bothered and mumbled to himself, I hate him, Continue reading


text:
# Properties

 Label 475.2.j.b Level $475$ Weight $2$ Character orbit 475.j Analytic conductor $3.793$ Analytic rank $0$ Dimension $12$ CM no Inner twists $4$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$475 = 5^{2} \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 475.j (of order $$6$$, degree $$2$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$3.79289409601$$ Analytic rank: $$0$$ Dimension: $$12$$ Relative dimension: $$6$$ over $$\Q(\zeta_{6})$$ Coefficient field: 12.0.50712647503417344.1 Defining polynomial: $$x^{12} - 13 x^{10} + 119 x^{8} - 552 x^{6} + 1863 x^{4} - 2450 x^{2} + 2401$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 95) Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{11}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + \beta_{1} q^{2} -\beta_{11} q^{3} + ( 2 - \beta_{2} + 3 \beta_{3} - \beta_{4} - \beta_{6} ) q^{4} + ( 1 - 2 \beta_{2} + 3 \beta_{3} - 2 \beta_{4} + \beta_{5} - 2 \beta_{6} ) q^{6} + ( \beta_{8} + \beta_{10} + \beta_{11} ) q^{7} + ( \beta_{1} + 2 \beta_{8} - \beta_{9} - \beta_{10} - \beta_{11} ) q^{8} + ( 1 + \beta_{3} + \beta_{5} ) q^{9} +O(q^{10})$$ $$q + \beta_{1} q^{2} -\beta_{11} q^{3} + ( 2 - \beta_{2} + 3 \beta_{3} - \beta_{4} - \beta_{6} ) q^{4} + ( 1 - 2 \beta_{2} + 3 \beta_{3} - 2 \beta_{4} + \beta_{5} - 2 \beta_{6} ) q^{6} + ( \beta_{8} + \beta_{10} + \beta_{11} ) q^{7} + ( \beta_{1} + 2 \beta_{8} - \beta_{9} - \beta_{10} - \beta_{11} ) q^{8} + ( 1 + \beta_{3} + \beta_{5} ) q^{9} + ( -1 + \beta_{4} - \beta_{5} + \beta_{6} ) q^{11} + ( 2 \beta_{1} - \beta_{8} - 2 \beta_{9} - \beta_{10} - \beta_{11} ) q^{12} + ( 5 \beta_{7} - 5 \beta_{8} ) q^{13} + ( 2 \beta_{2} - 3 \beta_{3} ) q^{14} + ( -\beta_{2} + 2 \beta_{3} - 2 \beta_{6} ) q^{16} + ( -3 \beta_{1} - 2 \beta_{7} - 2 \beta_{11} ) q^{17} + ( -5 \beta_{8} - \beta_{10} - \beta_{11} ) q^{18} + ( -1 - \beta_{2} + \beta_{3} - 2 \beta_{4} + \beta_{5} - 3 \beta_{6} ) q^{19} + ( \beta_{2} - 4 \beta_{3} - \beta_{6} ) q^{21} + ( -\beta_{1} + 3 \beta_{7} + 2 \beta_{11} ) q^{22} + ( -3 \beta_{7} + 3 \beta_{8} - 3 \beta_{9} + 2 \beta_{10} ) q^{23} + 7 \beta_{3} q^{24} + ( -5 \beta_{5} + 5 \beta_{6} ) q^{26} + ( -\beta_{1} - 3 \beta_{8} + \beta_{9} ) q^{27} + ( -4 \beta_{7} + 4 \beta_{8} + \beta_{9} + 2 \beta_{10} ) q^{28} + ( \beta_{2} - \beta_{3} + \beta_{4} - \beta_{5} + \beta_{6} ) q^{29} + ( -\beta_{5} + \beta_{6} ) q^{31} + ( -3 \beta_{7} + 3 \beta_{8} - \beta_{9} + 2 \beta_{10} ) q^{32} + ( -\beta_{1} + 4 \beta_{7} + 2 \beta_{11} ) q^{33} + ( -10 - \beta_{2} - 9 \beta_{3} - \beta_{4} + 4 \beta_{5} - \beta_{6} ) q^{34} + ( -2 \beta_{2} + \beta_{3} + 2 \beta_{6} ) q^{36} + ( -2 \beta_{1} - 3 \beta_{8} + 2 \beta_{9} - 5 \beta_{10} - 5 \beta_{11} ) q^{37} + ( -8 \beta_{7} + 7 \beta_{8} - 2 \beta_{9} - 3 \beta_{11} ) q^{38} + ( 5 + 5 \beta_{4} - 5 \beta_{5} + 5 \beta_{6} ) q^{39} + ( -\beta_{2} - \beta_{3} + 2 \beta_{6} ) q^{41} + ( -8 \beta_{7} + 8 \beta_{8} + 2 \beta_{9} + 3 \beta_{10} ) q^{42} + ( 3 \beta_{1} + \beta_{7} + \beta_{11} ) q^{43} + ( -4 + 3 \beta_{2} - 7 \beta_{3} + 3 \beta_{4} - 3 \beta_{5} + 3 \beta_{6} ) q^{44} + ( -10 - \beta_{4} + 5 \beta_{5} - 5 \beta_{6} ) q^{46} + ( \beta_{7} - \beta_{8} - 2 \beta_{9} + \beta_{10} ) q^{47} + ( -2 \beta_{7} + 2 \beta_{8} - 3 \beta_{9} + 2 \beta_{10} ) q^{48} + ( 4 + 2 \beta_{4} - 3 \beta_{5} + 3 \beta_{6} ) q^{49} + ( 3 + 4 \beta_{2} - \beta_{3} + 4 \beta_{4} + \beta_{5} + 4 \beta_{6} ) q^{51} + ( 5 \beta_{1} + 15 \beta_{7} + 5 \beta_{11} ) q^{52} + ( -2 \beta_{7} + 2 \beta_{8} + \beta_{9} - 4 \beta_{10} ) q^{53} + ( \beta_{2} - 5 \beta_{3} + 4 \beta_{6} ) q^{54} + ( 4 - \beta_{4} + 2 \beta_{5} - 2 \beta_{6} ) q^{56} + ( 3 \beta_{1} - 7 \beta_{7} + 2 \beta_{8} - 3 \beta_{9} + 3 \beta_{10} + \beta_{11} ) q^{57} + ( 3 \beta_{8} + 2 \beta_{10} + 2 \beta_{11} ) q^{58} + ( -\beta_{2} - \beta_{3} - 2 \beta_{6} ) q^{59} + ( 3 - 2 \beta_{2} + 5 \beta_{3} - 2 \beta_{4} - 2 \beta_{5} - 2 \beta_{6} ) q^{61} + ( \beta_{1} + 5 \beta_{7} + \beta_{11} ) q^{62} + ( -4 \beta_{7} + 4 \beta_{8} + 3 \beta_{10} ) q^{63} + ( -5 \beta_{4} + 3 \beta_{5} - 3 \beta_{6} ) q^{64} + ( -6 + 5 \beta_{2} - 11 \beta_{3} + 5 \beta_{4} - 6 \beta_{5} + 5 \beta_{6} ) q^{66} + ( -8 \beta_{7} + 8 \beta_{8} - 2 \beta_{9} + 2 \beta_{10} ) q^{67} + ( -7 \beta_{1} - 14 \beta_{8} + 7 \beta_{9} - \beta_{10} - \beta_{11} ) q^{68} + ( 2 + 3 \beta_{4} + 2 \beta_{5} - 2 \beta_{6} ) q^{69} + ( 3 \beta_{2} - 11 \beta_{3} + \beta_{6} ) q^{71} + ( 6 \beta_{7} - 6 \beta_{8} + 3 \beta_{9} - 4 \beta_{10} ) q^{72} + ( -3 \beta_{1} - 9 \beta_{7} - 2 \beta_{11} ) q^{73} + ( -8 \beta_{2} + 5 \beta_{3} ) q^{74} + ( -3 - 2 \beta_{2} + 7 \beta_{3} - 2 \beta_{4} + 7 \beta_{5} - 7 \beta_{6} ) q^{76} + ( \beta_{1} - 6 \beta_{8} - \beta_{9} - 3 \beta_{10} - 3 \beta_{11} ) q^{77} + ( 5 \beta_{1} + 15 \beta_{7} + 10 \beta_{11} ) q^{78} + ( 2 \beta_{2} + 6 \beta_{3} + 4 \beta_{6} ) q^{79} + ( -\beta_{2} - 6 \beta_{3} - 2 \beta_{6} ) q^{81} + ( 13 \beta_{7} - 13 \beta_{8} + 4 \beta_{9} - 4 \beta_{10} ) q^{82} + ( 3 \beta_{1} + 2 \beta_{8} - 3 \beta_{9} ) q^{83} + ( 5 - 6 \beta_{4} + 7 \beta_{5} - 7 \beta_{6} ) q^{84} + ( 11 - \beta_{2} + 12 \beta_{3} - \beta_{4} - 2 \beta_{5} - \beta_{6} ) q^{86} + ( -\beta_{1} + 4 \beta_{8} + \beta_{9} + \beta_{10} + \beta_{11} ) q^{87} + ( -2 \beta_{1} + 3 \beta_{8} + 2 \beta_{9} + 2 \beta_{10} + 2 \beta_{11} ) q^{88} + ( -6 - 4 \beta_{2} - 2 \beta_{3} - 4 \beta_{4} - 4 \beta_{6} ) q^{89} + ( -5 \beta_{2} + 5 \beta_{3} - 5 \beta_{4} + 5 \beta_{5} - 5 \beta_{6} ) q^{91} + ( -8 \beta_{1} - 17 \beta_{7} - 2 \beta_{11} ) q^{92} + ( \beta_{1} + 3 \beta_{7} + 2 \beta_{11} ) q^{93} -7 q^{94} + ( 4 - \beta_{4} + 4 \beta_{5} - 4 \beta_{6} ) q^{96} + ( 4 \beta_{1} + 3 \beta_{11} ) q^{97} + ( 5 \beta_{1} + 11 \beta_{7} + 5 \beta_{11} ) q^{98} + ( -2 + 3 \beta_{2} - 5 \beta_{3} + 3 \beta_{4} - 4 \beta_{5} + 3 \beta_{6} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$12 q + 14 q^{4} + 12 q^{6} + 8 q^{9} + O(q^{10})$$ $$12 q + 14 q^{4} + 12 q^{6} + 8 q^{9} - 20 q^{11} + 14 q^{14} - 6 q^{16} + 24 q^{21} - 42 q^{24} - 20 q^{26} - 4 q^{29} - 4 q^{31} - 50 q^{34} - 6 q^{36} + 20 q^{39} + 4 q^{41} - 36 q^{44} - 96 q^{46} + 28 q^{49} + 12 q^{51} + 20 q^{54} + 60 q^{56} + 12 q^{59} + 18 q^{61} + 32 q^{64} - 58 q^{66} + 20 q^{69} + 58 q^{71} - 14 q^{74} - 38 q^{76} - 48 q^{79} + 42 q^{81} + 112 q^{84} + 64 q^{86} - 28 q^{89} + 20 q^{91} - 84 q^{94} + 68 q^{96} - 26 q^{99} + O(q^{100})$$

Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{12} - 13 x^{10} + 119 x^{8} - 552 x^{6} + 1863 x^{4} - 2450 x^{2} + 2401$$:

 $$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$736 \nu^{10} - 31772 \nu^{8} + 290836 \nu^{6} - 1755683 \nu^{4} + 4553172 \nu^{2} - 5987800$$$$)/1328929$$ $$\beta_{3}$$ $$=$$ $$($$$$850 \nu^{10} - 9867 \nu^{8} + 90321 \nu^{6} - 370073 \nu^{4} + 1414017 \nu^{2} - 1859550$$$$)/1328929$$ $$\beta_{4}$$ $$=$$ $$($$$$169 \nu^{10} - 1547 \nu^{8} + 14161 \nu^{6} - 24219 \nu^{4} + 31850 \nu^{2} + 467838$$$$)/189847$$ $$\beta_{5}$$ $$=$$ $$($$$$-200 \nu^{10} + 3917 \nu^{8} - 21252 \nu^{6} + 87076 \nu^{4} + 37412 \nu^{2} + 124852$$$$)/189847$$ $$\beta_{6}$$ $$=$$ $$($$$$9 \nu^{10} - 26 \nu^{8} + 238 \nu^{6} + 289 \nu^{4} + 3726 \nu^{2} - 4900$$$$)/5131$$ $$\beta_{7}$$ $$=$$ $$($$$$52 \nu^{11} - 476 \nu^{9} + 2271 \nu^{7} - 7452 \nu^{5} + 9800 \nu^{3} - 164812 \nu$$$$)/189847$$ $$\beta_{8}$$ $$=$$ $$($$$$-100 \nu^{11} + 859 \nu^{9} - 10626 \nu^{7} + 43538 \nu^{5} - 200461 \nu^{3} + 62426 \nu$$$$)/251419$$ $$\beta_{9}$$ $$=$$ $$($$$$-850 \nu^{11} + 9867 \nu^{9} - 90321 \nu^{7} + 370073 \nu^{5} - 1414017 \nu^{3} + 1859550 \nu$$$$)/1328929$$ $$\beta_{10}$$ $$=$$ $$($$$$8973 \nu^{11} - 159328 \nu^{9} + 1458464 \nu^{7} - 7813716 \nu^{5} + 22832928 \nu^{3} - 30027200 \nu$$$$)/9302503$$ $$\beta_{11}$$ $$=$$ $$($$$$273 \nu^{11} - 2499 \nu^{9} + 18703 \nu^{7} - 39123 \nu^{5} + 51450 \nu^{3} + 328061 \nu$$$$)/189847$$
 $$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$-\beta_{6} - \beta_{4} + 5 \beta_{3} - \beta_{2} + 4$$ $$\nu^{3}$$ $$=$$ $$-\beta_{11} - \beta_{10} - 5 \beta_{9} + 2 \beta_{8} + 5 \beta_{1}$$ $$\nu^{4}$$ $$=$$ $$-8 \beta_{6} + 28 \beta_{3} - 7 \beta_{2}$$ $$\nu^{5}$$ $$=$$ $$-6 \beta_{10} - 29 \beta_{9} + 19 \beta_{8} - 19 \beta_{7}$$ $$\nu^{6}$$ $$=$$ $$-13 \beta_{6} + 13 \beta_{5} + 41 \beta_{4} - 122$$ $$\nu^{7}$$ $$=$$ $$28 \beta_{11} - 147 \beta_{7} - 176 \beta_{1}$$ $$\nu^{8}$$ $$=$$ $$232 \beta_{6} + 119 \beta_{5} + 232 \beta_{4} - 964 \beta_{3} + 232 \beta_{2} - 732$$ $$\nu^{9}$$ $$=$$ $$113 \beta_{11} + 113 \beta_{10} + 1083 \beta_{9} - 1059 \beta_{8} - 1083 \beta_{1}$$ $$\nu^{10}$$ $$=$$ $$2255 \beta_{6} - 5754 \beta_{3} + 1309 \beta_{2}$$ $$\nu^{11}$$ $$=$$ $$363 \beta_{10} + 6700 \beta_{9} - 7348 \beta_{8} + 7348 \beta_{7}$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/475\mathbb{Z}\right)^\times$$.

 $$n$$ $$77$$ $$401$$ $$\chi(n)$$ $$-1$$ $$-1 - \beta_{3}$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
49.1
 −2.17114 − 1.25351i −1.97899 − 1.14257i −1.05818 − 0.610938i 1.05818 + 0.610938i 1.97899 + 1.14257i 2.17114 + 1.25351i −2.17114 + 1.25351i −1.97899 + 1.14257i −1.05818 + 0.610938i 1.05818 − 0.610938i 1.97899 − 1.14257i 2.17114 − 1.25351i
−2.17114 1.25351i −1.05818 0.610938i 2.14257 + 3.71104i 0 1.53163 + 2.65287i 0.221876i 5.72889i −0.753509 1.30512i 0
49.2 −1.97899 1.14257i −2.17114 1.25351i 1.61094 + 2.79023i 0 2.86445 + 4.96137i 3.50702i 2.79216i 1.64257 + 2.84502i 0
49.3 −1.05818 0.610938i 1.97899 + 1.14257i −0.253509 0.439091i 0 −1.39608 2.41808i 1.28514i 3.06327i 1.11094 + 1.92420i 0
49.4 1.05818 + 0.610938i −1.97899 1.14257i −0.253509 0.439091i 0 −1.39608 2.41808i 1.28514i 3.06327i 1.11094 + 1.92420i 0
49.5 1.97899 + 1.14257i 2.17114 + 1.25351i 1.61094 + 2.79023i 0 2.86445 + 4.96137i 3.50702i 2.79216i 1.64257 + 2.84502i 0
49.6 2.17114 + 1.25351i 1.05818 + 0.610938i 2.14257 + 3.71104i 0 1.53163 + 2.65287i 0.221876i 5.72889i −0.753509 1.30512i 0
349.1 −2.17114 + 1.25351i −1.05818 + 0.610938i 2.14257 3.71104i 0 1.53163 2.65287i 0.221876i 5.72889i −0.753509 + 1.30512i 0
349.2 −1.97899 + 1.14257i −2.17114 + 1.25351i 1.61094 2.79023i 0 2.86445 4.96137i 3.50702i 2.79216i 1.64257 2.84502i 0
349.3 −1.05818 + 0.610938i 1.97899 1.14257i −0.253509 + 0.439091i 0 −1.39608 + 2.41808i 1.28514i 3.06327i 1.11094 1.92420i 0
349.4 1.05818 0.610938i −1.97899 + 1.14257i −0.253509 + 0.439091i 0 −1.39608 + 2.41808i 1.28514i 3.06327i 1.11094 1.92420i 0
349.5 1.97899 1.14257i 2.17114 1.25351i 1.61094 2.79023i 0 2.86445 4.96137i 3.50702i 2.79216i 1.64257 2.84502i 0
349.6 2.17114 1.25351i 1.05818 0.610938i 2.14257 3.71104i 0 1.53163 2.65287i 0.221876i 5.72889i −0.753509 + 1.30512i 0
 $$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 349.6 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner
19.c even 3 1 inner
95.i even 6 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 475.2.j.b 12
5.b even 2 1 inner 475.2.j.b 12
5.c odd 4 1 95.2.e.b 6
5.c odd 4 1 475.2.e.d 6
15.e even 4 1 855.2.k.g 6
19.c even 3 1 inner 475.2.j.b 12
20.e even 4 1 1520.2.q.j 6
95.i even 6 1 inner 475.2.j.b 12
95.l even 12 1 1805.2.a.g 3
95.l even 12 1 9025.2.a.ba 3
95.m odd 12 1 95.2.e.b 6
95.m odd 12 1 475.2.e.d 6
95.m odd 12 1 1805.2.a.h 3
95.m odd 12 1 9025.2.a.z 3
285.v even 12 1 855.2.k.g 6
380.v even 12 1 1520.2.q.j 6

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
95.2.e.b 6 5.c odd 4 1
95.2.e.b 6 95.m odd 12 1
475.2.e.d 6 5.c odd 4 1
475.2.e.d 6 95.m odd 12 1
475.2.j.b 12 1.a even 1 1 trivial
475.2.j.b 12 5.b even 2 1 inner
475.2.j.b 12 19.c even 3 1 inner
475.2.j.b 12 95.i even 6 1 inner
855.2.k.g 6 15.e even 4 1
855.2.k.g 6 285.v even 12 1
1520.2.q.j 6 20.e even 4 1
1520.2.q.j 6 380.v even 12 1
1805.2.a.g 3 95.l even 12 1
1805.2.a.h 3 95.m odd 12 1
9025.2.a.z 3 95.m odd 12 1
9025.2.a.ba 3 95.l even 12 1

## Hecke kernels

This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{12} - 13 T_{2}^{10} + 119 T_{2}^{8} - 552 T_{2}^{6} + 1863 T_{2}^{4} - 2450 T_{2}^{2} + 2401$$ acting on $$S_{2}^{\mathrm{new}}(475, [\chi])$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$2401 - 2450 T^{2} + 1863 T^{4} - 552 T^{6} + 119 T^{8} - 13 T^{10} + T^{12}$$
$3$ $$2401 - 2450 T^{2} + 1863 T^{4} - 552 T^{6} + 119 T^{8} - 13 T^{10} + T^{12}$$
$5$ $$T^{12}$$
$7$ $$( 1 + 21 T^{2} + 14 T^{4} + T^{6} )^{2}$$
$11$ $$( -1 + 2 T + 5 T^{2} + T^{3} )^{4}$$
$13$ $$( 625 - 25 T^{2} + T^{4} )^{3}$$
$17$ $$2401 - 95550 T^{2} + 3798139 T^{4} - 173452 T^{6} + 5971 T^{8} - 89 T^{10} + T^{12}$$
$19$ $$( 6859 - 133 T^{3} + T^{6} )^{2}$$
$23$ $$5764801 - 4593113 T^{2} + 3433875 T^{4} - 175020 T^{6} + 6923 T^{8} - 94 T^{10} + T^{12}$$
$29$ $$( 1 - 5 T + 23 T^{2} - 12 T^{3} + 9 T^{4} + 2 T^{5} + T^{6} )^{2}$$
$31$ $$( -7 - 6 T + T^{2} + T^{3} )^{4}$$
$37$ $$( 51529 + 15069 T^{2} + 242 T^{4} + T^{6} )^{2}$$
$41$ $$( 1369 - 1591 T + 1923 T^{2} + 12 T^{3} + 47 T^{4} - 2 T^{5} + T^{6} )^{2}$$
$43$ $$214358881 - 31888098 T^{2} + 3440635 T^{4} - 164560 T^{6} + 5743 T^{8} - 89 T^{10} + T^{12}$$
$47$ $$5764801 - 1529437 T^{2} + 285719 T^{4} - 27048 T^{6} + 1863 T^{8} - 50 T^{10} + T^{12}$$
$53$ $$9354951841 - 832380926 T^{2} + 54235431 T^{4} - 1570788 T^{6} + 33419 T^{8} - 205 T^{10} + T^{12}$$
$59$ $$( 2401 - 343 T + 343 T^{2} - 56 T^{3} + 43 T^{4} - 6 T^{5} + T^{6} )^{2}$$
$61$ $$( 2401 - 2401 T + 2842 T^{2} + 343 T^{3} + 130 T^{4} - 9 T^{5} + T^{6} )^{2}$$
$67$ $$59969536 - 63067136 T^{2} + 64899840 T^{4} - 1483008 T^{6} + 25712 T^{8} - 184 T^{10} + T^{12}$$
$71$ $$( 218089 - 110212 T + 42153 T^{2} - 5910 T^{3} + 605 T^{4} - 29 T^{5} + T^{6} )^{2}$$
$73$ $$35153041 - 61074629 T^{2} + 104628351 T^{4} - 2563392 T^{6} + 52199 T^{8} - 250 T^{10} + T^{12}$$
$79$ $$( 61504 - 28768 T + 19408 T^{2} + 3280 T^{3} + 460 T^{4} + 24 T^{5} + T^{6} )^{2}$$
$83$ $$( 5929 + 2454 T^{2} + 117 T^{4} + T^{6} )^{2}$$
$89$ $$( 3136 + 2016 T + 2080 T^{2} - 392 T^{3} + 232 T^{4} + 14 T^{5} + T^{6} )^{2}$$
$97$ $$214358881 - 88578050 T^{2} + 33952479 T^{4} - 1065768 T^{6} + 26711 T^{8} - 181 T^{10} + T^{12}$$


text:
# Wang Xiaotong

### Quick Info

Born
about 580
China
Died
about 640
Chian

Summary
Wang Xiaotong was a Chinese mathematician and astronomer who was involved in calendar reform and the solution of cubic equations.

### Biography

Wang Xiaotong is also known as Wang Hs'iao-t'ung. He lived the early part of his life under the Sui dynasty (581-618). Although it was only a short-lived dynasty, during these years the north and south of China were unified. Wang is famed as the author of the Jigu suanjing (Continuation of Ancient Mathematics) one of The Ten Classics. He presented his treatise to the Emperor of China, Li Yuan the first emperor of the T'ang dynasty, who came to power in 618. Li Yuan, who had been an official in the Sui dynasty, did not control the whole country at first. He had full control of the eastern part by 621 and of the north and south as well by 624. We know a little about Wang Xiaotong's life because when he presented his treatise to Emperor Li Yuan he gave the emperor a brief biography at the same time. This biography has survived, becoming attached to the treatise.

Wang Xiaotong's biography tells us that he became interested in mathematics at a young age. He made a careful study of the Nine Chapters on the Mathematical Art and was particularly impressed with the commentary on that text written by Liu Hui. Wang went on to become a teacher of mathematics, and later he was appointed as deputy director of the Astronomical Bureau.

It was known that the Chinese calendar at that time was in need of reform since, although only in operation for a few years, already predictions of eclipses were getting out of step. In 623 Wang and Zu Xiaosun, a Civil Servant, were given the task of reporting on the problems with the calendar and making recommendations. There was disagreement between Wang and another calendar expert Fu Renjun about certain aspects of the calendar and in fact Wang's ideas were not particularly good since he wished to ignore the irregularity of the sun's motion and he also wanted to ignore the precession of the equinoxes which had first been incorporated in calendar calculations by Zu Chongzhi in the fifth century.

One might ask why we have included Wang in this archive when his achievements on calendar reform seem retrograde. Well, he has not been included for this work but rather for the remarkable contribution he made to Chinese mathematics in the Jigu suanjing (Continuation of Ancient Mathematics). The book contains 20 problems. The first is a pursuit problem of a dog chasing a hare, but Wang tells us that this is really a problem about movements of astronomical bodies. The next 13 problems concern engineering constructions and the volume of granaries. The final six problems concern right angled triangles.

The important innovation which is incorporated in most of these problems is that they reduce to a cubic equation which Wang solves numerically. We do not know of any earlier Chinese work on cubic equations. Of course one has to understand that when we say that the text is concerned with cubic equations, we do not see expressions with $x, x^{2}$ and $x^{3}$ in them. Rather the equations are expressed in words and Wang thinks in a geometrical way. For example where we might say "Let the height be $x$" and then produce an equation in $x$, Wang writes:-
Let the height be the side of a cube.
He then goes on to set up a cubic equation for the height. Again he will write:-
Let the depth be the side of a cube
when he is about to set up a cubic equation for the depth. It is interesting to compare this geometrical way of thinking about cubics with that of Cardan 900 years later. Many see this work by Wang as the first steps towards the "tian yuan" or "coefficient array method" or "method of the celestial unknown" of Li Zhi. For example Ruan Yuan writes in his Biographies of astronomers and mathematicians (1799) (see [5]):-
Really his work is the true source of the later "tian yuan" method.
Mikami writes [3]:-
In setting up cubic equations Wang Xiaotong utilised a rule which is the same as the "tian yuan". Thus the rule is new in form, but, in fact, it is the continuation of the ancient style handed down from earlier generations.
The problems in the Jigu suanjing (Continuation of Ancient Mathematics) are stated in a very complicated form. Here is a flavour of the style with a translation of Problem 3 as given in [1]:-
Suppose that a dyke is to be built. The difference between the lower and upper widths of the west face is 6 zhang 8 chi 2 cun ..., the height of the east face is 3 zhang 1 chi less than the height of the west face .... Sub-prefecture A gives 6724 men, sub-prefecture B gives 16677 men ... Each man of the four sub-prefectures excavates 9 dan 9 dou 2 sheng of earth per day and packs down on average 11 chi 4 cun and 6/12 cun per day ... Previously a man was able to cover a horizontal road of length 192 bu 62 times per day carrying 2 dou 4 sheng and 8 he of earth on his back. But now, it is necessary to climb a hill and cross a watercourse in order to fetch the earth. A climb of 3 bu is equivalent to 4 bu of a flat road, 1 bu for a ford is equivalent to 2 bu for a flat road, ... What is the daily task of a man who digs, transports and constructs; what are the heights and the upper and lower widths of the dyke ...
If we strip away some of the complications then we are left with workers from four countries, $A, B, C,$ and $D$ who cooperate to build a dyke. The east and west ends of the dyke are trapezoidal. The length, upper width, lower widths of the east and west ends and the height of the west end are all known as functions of the height $x$ of the east end. This allows the volume to be calculated as a function of $x$. Data given for the work done by the workers allows the volume to be calculated, and a cubic equation is arrived at for $x$.

To be able to solve this problem Wang has not only to be able to set up a cubic equation and solve it, but he also needs to know a formula for the volume of his dyke with trapezoidal ends and varying cross-section. One of Wang's achievements was calculating this formula and in the preface to the treatise he gives a reference to ideas which had led him to the volume formula (see for example [6]):-
People much appreciated Zu Geng's Appended Rules. It is a pity Zu did not clearly explain the problems on the frustum. Here I give alternative methods for extending them.
Note that Zu Geng was Zu Chongzhi's son.

Let us now look at the problems on right angled triangles, namely problems 15 to 20 inclusive. We use letters to name the sides of the right angled triangle where Wang uses names:-
Problem 15: Let a right angled triangle have sides a, b, c where c is the hypotenuse. If a times b is seven hundred and six and one fiftieth, and if c is thirty six and nine tenths more than a. What are the values of the three sides.
Suppose we forget the numbers for a moment and look at the data we are given. We know $ab, c - a$ and of course that $a^{2} + b^{2} = c^{2}$ by the Gougu rule (Pythagoras). Wang calls $a$ the unknown and finds a cubic equation in terms of $a$. Let us see how this can be done (we do not know exactly how Wang reasoned).
$(ab)^{2} = a^{2}b^{2} = a^{2}(c^{2}- a^{2}) = a^{2}(c - a)(c + a)$

Hence, dividing by $2(c - a)$ we get
$(ab)^{2}/2(c - a) = a^{2}(c + a)/2 = a^{2}(2a + (c - a))/2 = a^{2}(a + (c - a)/2) = a^{3} + a^{2}(c - a)/2$.

Writing $x$ for the unknown $a$, we have the cubic equation
$x^{3} + x^{2}(c - a)/2 = (ab)^{2}/2(c - a)$

and we know $ab$ and $c - a$ so the coefficients are known. Isn't that clever! Remember that Wang did not have our symbolic notation which makes the whole process so much easier today.

Other problems here are of the same type. In Problem 17 we are given $bc$ and $a + b$, while in Problem 19 we are given $bc$ and $a$. Try to set up the necessary equations in these two cases in a similar way to our solution to Problem 15 above.

In 656, after editing by Li Chunfeng, the treatise Jigu suanjing (Continuation of Ancient Mathematics) became a text for the Imperial examinations and it became one of The Ten Classics when reprinted in 1084. Not only did Wang's work influence later Chinese mathematicians, but it is said that it was his ideas on cubic equations which Fibonacci learnt, probably first transmitted into the Islamic/Arabic world, and then brought to Europe.

### References (show)

1. J-C Martzloff, A history of Chinese mathematics (Berlin-Heidelberg, 1997).
2. J-C Martzloff, Histoire des mathématiques chinoises (Paris, 1987).
3. Y Mikami, The Development of Mathematics in China and Japan (New York, 1974).
4. B Qian (ed.), Ten Mathematical Classics (Chinese) (Beijing, 1963).
5. Y Ruan, Biographies of Mathematicians and Astronomers (Chinese) 1 (Shanghai, 1955).
6. K Shen, J N Crossley and A W-C Lun, The nine chapters on the mathematical art : Companion and commentary (Beijing, 1999).
7. S Guo, Wang Xiaotong, in Du Shiran (ed.), Zhongguo Gudai Kexuejia Zhuanji (Biographies of Ancient Chinese Scientists) (Beijing, 1992), 317 -319.

### Cross-references (show)

Written by J J O'Connor and E F Robertson
Last Update December 2003