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Main

## Temperature Control with an Arduino

This lab is an application of feedback control for a temperature control device. Heat output is adjusted by modulating the voltage to a transistor. A thermistor measures the temperature. Energy from the transistor output is transferred by conduction and convection to the temperature sensor.

This lab teaches principles of system dynamics and control. In particular, this lab reinforces:

#### Step 1: Dynamic Modeling

The three important elements for a control loop are the measurement device (thermistor temperature sensor), an actuator (voltage to the transistor), and capability to perform computerized control (USB interface).

At maximum output the transistor dissipates 3.1 W of power with a voltage of 8.9 V and current of 0.35 A. The transistor is controlled with a 0 to 255 mV signal that linearly adjusts the power output between 0 and 3.1 W. The mass of the transistor and thermistor is about 2 gm with an approximate heat capacity of 460 J/{kg K}. The surface area of the heater and sensor is about 5e-4 m^2. A convective heat transfer coefficient for quiescent air is approximately 10 W/{m^2K}. The heat generated by the transistor transfers primarily by convection and conduction to the temperature sensor. Heat transfer is improved with a thermal coupling (silver epoxy) that connects the two components.

Create a dynamic model of the dynamic response between input power to the transistor and the temperature sensed by the thermistor. Use an energy balance to start the derivation.

$$m\,c_p\frac{dT}{dt} = \sum \dot h_{in} - \sum \dot h_{out} + Q$$

Expand or simplify terms that are needed for this application. The full energy balance includes convection and radiation terms.

$$m\,c_p\frac{dT}{dt} = h\,A\,\left(T_\infty-T\right) + \epsilon\,\sigma\,A\,\left(T_\infty^4-T^4\right) + \alpha\,I\,V$$

where m is the mass, c_p is the heat capacity, T is the temperature, h is the heat transfer coefficient, A is the area, T_\infty is the ambient temperature, \epsilon is the emissivity, \sigma is the Stefan-Boltzmann constant, I is the current, and V is the voltage. The parameter \alpha is the fraction of electrical energy converted into heat in the transistor. Use this equation to develop a dynamic simulation of the temperature response due to an impulse (off, on, off) in the heater output. Leave the heater on for sufficient time to observe nearly steady state conditions.

#### Step 2: Generate Data, Fit Physics-based Model

As done in simulation, test the same heater impulse response on the Arduino device and record the temperature. Adjust the uncertain parameters such as h, \alpha, and m from Step 1 to best match the dynamic data from the impulse response. A linearized version of the model can be used to compare the FOPDT model of step 3 to the physics-based model in this step. A linearized model is derived as:

$$m\,c_p\frac{dT}{dt} = f\left(T_\infty,T,V\right) \approx f \left(\bar T_\infty, \bar T, \bar V\right) \ldots \\ + \frac{\partial f}{\partial T_\infty}\bigg|_{\bar T_\infty,\bar T,\bar V} \left(T_\infty-\bar T_\infty\right) \ldots \\ + \frac{\partial f}{\partial T}\bigg|_{\bar T_\infty,\bar T,\bar V} \left(T-\bar T\right) \ldots \\ + \frac{\partial f}{\partial V}\bigg|_{\bar T_\infty,\bar T,\bar V} \left(V-\bar V\right)$$

where

$$\frac{\partial f}{\partial T_\infty}\bigg|_{\bar T_\infty,\bar T,\bar V} = h\,A\,\bar T_\infty + 4\epsilon\,\sigma\,A\,\bar T_\infty^3 = \beta$$

$$\frac{\partial f}{\partial T}\bigg|_{\bar T_\infty,\bar T,\bar V} = -h\,A\,\bar T - 4\epsilon\,\sigma\,A\,\bar T^3 = \gamma$$

$$\frac{\partial f}{\partial V}\bigg|_{\bar T_\infty,\bar T,\bar V} = \alpha \, \bar I$$

The final linearized equation is further simplified by replacing the partial derivatives with constants \beta, \gamma, and \alpha \bar I that are evaluated at the steady state conditions \bar T_\infty, \bar T, and \bar V.

$$m\,c_p\frac{dT}{dt} = \beta \left(T_\infty-\bar T_\infty\right) + \gamma \left(T-\bar T\right) + \alpha \bar I \left(V-\bar V\right)$$

This is placed into a time-constant form as by dividing by gamma and substituting the deviation variables T\prime_\infty = T_\infty -\bar T_\infty, T\prime = T -\bar T, and V\prime = V -\bar V.

$$\frac{m\,c_p}{\gamma}\frac{dT^\prime}{dt} = \frac{\beta}{\gamma} T^\prime_\infty + T^\prime + \frac{\alpha \bar I}{\gamma} V^\prime$$

#### Step 3: Fit FOPDT Model

Using the data from Step 2, determine the parameters of an FOPDT model that best match the dynamic temperature data including K_p, \tau_p, and \theta_p. A first-order linear system is:

$$\tau_p \frac{dT(t)}{dt} = -T(t) + K_p \, V\left(t-\theta_p\right)$$

where K_p is the process gain, \tau_p is the process time constant, and \theta_p is the process dead-time. A second order (SOPDT) model may also give insight on whether the system shows evidence of a higher order response.

#### Step 4: PID Control

Obtain PID tuning constants K_c, \tau_I, and \tau_D from IMC correlations. Use the tuning constants for PID control of temperature. Demonstrate step changes in temperature set point and comment on the performance of the controller using the calculated constants. Tune the controller by adjusting the constants to improve performance. Comment on the difference between IMC tuning constants and the improved tuning constants in terms of rise time, overshoot, decay ratio, heater fluctuations, or other relevant performance criteria.

#### Lab Sample Code

Below is sample code to help with steps of the project. It is a starting point that can be modified to investigate issues such as whether radiative heat transfer is significant, is the temperature response inherently first order or higher order, and what values of uncertain parameters in the physics based model help the predicted temperature agree with the data.

#### Step 1: Simulate Temperature Response

A first step is to simulate the Arduino temperature response with an energy balance. Just like the physical system, this simulated system responds to changes in the input voltage.

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

# define Arduino model
def arduino(x,t,mV,Ta):
## Parameters
h = 1.0               # W/m^2-K
m = 4.0/1000.0        # kg
Cp = 0.5 * 1000.0     # J/kg-K
A = 6.0 / 100.0**2    # Area in m^2
Amp = 1.0             # Amps to Transitor
T = x[0]

# acc = inlet - outlet
dTdt = (1.0/(m*Cp))*(-h*A*(T-Ta) + Amp*mV/1000.0)
return dTdt

# Ambient temperature
Ta = (80.0-32.0)*5.0/9.0 + 273.15  # 80 degF in K
# Starting temperature
T0 = Ta  # 80 degF in K

# Time Intervals (sec)
t = np.linspace(0,600,601)

# Manipulated variable
mV = np.ones(len(t))* 0.0 # millivolts input
mV[20:] = 100
# Store results for plotting
T = np.ones(len(t)) * T0

for i in range(len(t)-1):
ts = [t[i],t[i+1]]
y = odeint(arduino,T0,ts,args=(mV[i],Ta))
T0 = y[-1]
T[i+1] = T0

# Construct results and save data file
# Column 1 = time
# Column 2 = cooling temperature
# Column 3 = reactor temperature
data = np.vstack((t,mV,T)) # vertical stack
data = data.T             # transpose data
np.savetxt('data.txt',data,delimiter=',')

# Plot the results
plt.figure()
plt.subplot(2,1,1)
plt.plot(t,mV,'b--',linewidth=3)
plt.ylabel('Milli-Volts')
plt.legend(['Heater'],loc='best')

plt.subplot(2,1,2)
plt.plot(t,T,'r-',linewidth=3)
plt.ylabel('Temperature (K)')
plt.legend(['T'],loc='best')

plt.xlabel('Time (sec)')
plt.savefig('step1.png')

plt.show()

#### Step 2: Optimize Model Parameters

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from scipy.optimize import minimize

# Import data file
# Column 1 = time (t)
# Column 2 = input (u)
# Column 3 = output (yp)
mV0 = data[0,1]
Tmeas0 = data[0,2]
t = data[:,0].T
mV = data[:,1].T
Tmeas = data[:,2].T

# specify number of steps
ns = len(t)
delta_t = t[1]-t[0]

Cp = 0.5 * 1000.0     # J/kg-K
A = 6.0 / 100.0**2    # Area in m^2
Ta = Tmeas0           # Ambient temperature (K)

# define Arduino model
def arduino(x,t,mVi,p):
# uncertain parameters
h = p[0]
m = p[1]
Amp = p[2]

T = x[0]
# acc = inlet - outlet
dTdt = (1.0/(m*Cp))*(-h*A*(T-Ta) + Amp*mVi/1000.0)
return dTdt

def calc_T(p):
T = np.ones(len(t)) * Tmeas0
T0 = T[0]
for i in range(len(t)-1):
ts = [t[i],t[i+1]]
y = odeint(arduino,T0,ts,args=(mV[i],p))
T0 = y[-1]
T[i+1] = T0
return T

# define objective
def objective(p):
# simulate model
Tp = calc_T(p)
# calculate objective
obj = 0.0
for i in range(len(Tmeas)):
obj = obj + ((Tp[i]-Tmeas[i])/Tmeas[i])**2
# return result
return obj

# Parameter initial guess
h = 10.0               # W/m^2-K
m = 4.0/1000.0         # kg
Amp = 1.0              # Amps to transistor
p0 = [h,m,Amp]

# show initial objective
print('Initial SSE Objective: ' + str(objective(p0)))

# optimize parameters
# bounds on variables
bnds = ((1.0, 20.0),(1.0/1000.0,10.0/1000.0),(0.5,2.0))
solution = minimize(objective,p0,method='SLSQP',bounds=bnds)
p = solution.x

# show final objective
print('Final SSE Objective: ' + str(objective(p)))

h = p[0]
m = p[1]
Amp = p[2]
print('h: ' + str(h))
print('m: ' + str(m))
print('Amp: ' + str(Amp))

print('')
print('FOPDT Equivalent')
#dx/dt = (-1/taup) * x + (Kp/taup) * u
#dTdt = (1.0/(m*Cp))*(-h*A*(T-Ta) + Amp*mVi/1000.0)
dfdT = -h*A/(m*Cp)
dfdmV = Amp/(1000.0*m*Cp)
taup = -1.0/dfdT
Kp = dfdmV * taup
print('Kp: ' + str(Kp))
print('taup: ' + str(taup))

# calculate model with updated parameters
T1 = calc_T(p0)
T2 = calc_T(p)

# Plot the results
plt.figure()
plt.subplot(2,1,1)
plt.plot(t,mV,'b--',linewidth=3)
plt.ylabel('milliVolts')
plt.legend(['Heater'],loc='best')

plt.subplot(2,1,2)
plt.plot(t,T1,'b:',linewidth=3,label='Initial Guess')
plt.plot(t,Tmeas,'r-',linewidth=3,label='Measured')
plt.plot(t,T2,'k--',linewidth=3,label='Final Prediction')
plt.ylabel('Temperature (K)')
plt.legend(loc='best')
plt.xlabel('Time (sec)')
plt.savefig('step2.png')
plt.show()

#### Step 3: Fit FOPDT and SOPDT to Data

Determine the parameters of an FOPDT model that best match the dynamic temperature data including K_p, \tau_p, and \theta_p. A second order (SOPDT) can also be fit to investigate whether a higher order model is more accurate.

#### Step 4: Implement PID Controller

Simulate a PID Controller with one of the models determined in steps 2 or 3. It is suggested to tune the controller in simulation before implementing with an Arduino. Tuning on a device that takes 10-20 minutes per test is much slower than running a PID controller in simulation. Once optimized PID tuning values are obtained, demonstrate the performance with the physical device.

#### Solution Files

Additional approaches for modeling, estimation, and control are provided at the link below. These include:

• Dynamic modeling
• First principles
• Empirical / machine learning
• Dynamic Estimation
• Parameter regression
• Kalman filter
• Moving Horizon Estimation
• Process control
• ON/OFF control
• PID control
• Stability analysis


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# Equation of plane

1. May 12, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

Find the equation of the plane which passes through M(2,3,1) and the line

$$\left\{\begin{matrix} 2x-y+3z-1=0 & \\ 5x+4y-z-7=0 & \end{matrix}\right.$$

2. Relevant equations

$$\frac{x-x_1}{\begin{vmatrix} B_1 & C_1\\ C_1 & C_2 \end{vmatrix}}=\frac{y-y_1}{\begin{vmatrix} C_1 & A_1\\ C_2 & A_2 \end{vmatrix}}=\frac{z-z_1}{\begin{vmatrix} A_1 & B_1\\ A_2 & B_2 \end{vmatrix}}$$

3. The attempt at a solution

first I find the equation of the line:

$$\frac{x}{-1}=\frac{y+7}{-5}=\frac{z+3}{-3}$$

What should I do next?

2. May 12, 2008

### BrendanH

Are you familiar with cross products? (You must be, to some extent, as you have posted determinants. A solution of this form I have never seen before.)
If yes, then $$N1 \\ X \\ N2$$ will be the normal to your plane, where N1 and N2 are the normals to the lines given.

Last edited: May 12, 2008
3. May 12, 2008

### Physicsissuef

$n_1 x n_2$ will give me vector perpendicular to this ones, but how it will help me?

4. May 12, 2008

### BrendanH

from that, you will have a, b, and c of the following equation for a plane.

$$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=d$$

In other words, you will have the normal to the plane you're trying to solve for. Substitution of the point they give you will solve it.

5. May 12, 2008

$$n_1 x n_2=(-1,-5,-3)[/itex] A(x+1)+B(y+5)+c(z+3)=D Hmm... And what about x,y,z, should I substitute for M? 6. May 12, 2008 ### HallsofIvy Unfortunately, any line, R3 has an normals in an infinite number of directions- it doesn't make sense to talk about "the" normal to a line. 7. May 12, 2008 ### HallsofIvy Choose any two points on the line. For example, if you choose x= 0, you can solve the two equations 2(0)-y+3z-1=0 and 5(0)+4y-z-7=0 for y= 2 and z= 1: (0, 2, 1) is on the line. If you take x= 1, you can solve 2(1)- y+ 3z- 1= 0 and 5(1)+ 4y- z- 7= 0 for y= -23/11 and z= -2/11. The cross product of the vectors from (2, 3, 1) to those points will give you the normal to the plane. 8. May 12, 2008 ### BrendanH I retract having said 'normals' to the line. I got 'normal-happy' , what with the terminology for planes. 9. May 12, 2008 ### Physicsissuef But the plane don't necessarily need to intersect the line in 2 points. It can intersect it in one point... 10. May 12, 2008 ### dx The plane intersects the line at an infinite number of points, i.e. the line is contained in the plane. 11. May 12, 2008 ### dx Otherwise, you would need more information than you are given to determine the plane uniquely. 12. May 12, 2008 ### HallsofIvy What plane are you talking about? You were told to find the plane that contains that line. Every point on the line is in that plane. 13. May 13, 2008 ### Physicsissuef Yes, sorry. My fault. Will it be correct if I show it like this: [tex]a \circ n=0$$

$$(a_1,a_2,a_3) \circ (A,B,C)=0$$

So one of the conditions is:

-A-5B-3C=0

the second condition is:

2A+3B+C+D=0

and the third condition:

0*A-7B-3C+D=0

$$\left\{\begin{matrix} -A-5B-3C=0 & \\ 2A+3B+C+D=0 & \\ -7B-3C+D=0 & \end{matrix}\right.$$

The result is:

$$-5x-y-7=0$$

right?

14. May 13, 2008

### Physicsissuef

Just to mention that in the first post the line is:

$$\left\{\begin{matrix} x+y-2z+1=0 & \\ 2x-y+z-4=0 & \end{matrix}\right.$$

Sorry, for mistake, I type the wrong line...

15. May 14, 2008

### Physicsissuef

And can I ask you something?

Two planes to be parallel, must this condition be satisfied?

$$A_1x+B_1y+C_1z+D=0$$

$$A_2x+B_2y+C_2z+D=0$$

$$\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{C_1}{C_2}$$, since in my book doesn't look like that...

format: plain


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# Down in the delta family theory

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Down in the delta family theory
Rated 3/5 based on 37 review


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# Classical and quantum electromagnetic field

1. ### karlzr

131
In classical physics, charged particles induce electric field ##\vec{E}_c## around them. How do we interpret this classical electric field ##\vec{E}## in quantum mechanics. Is this just the vacuum expectation value ##\vec{E}_c=<0|\vec{E}|0>##? if so, it means ##<A>\neq 0##. This would lead to Lorentz violation. So I don't know what's going on.

Another thing, we always try to define quantum modes in vacuum (the state with minimum energy). For instance, Higgs field is defined in some arbitrary vacuum of the potential: ##\phi=v+h## where ##\phi##, ##v## and ##h## are the original scalar field, vev ##<\phi>## and the higgs field respectively. Do we always have to do in this way(define excitation in vacuum ? I know this is essential for fermions, and W/Z bosons in SM to get mass. But does this really make sense? this makes me wonder what is field on earth.

2. ### ChrisVer

2,378
For the first question I am not quiet sure for the answer... In general it's better to work with the 4-vector potentials rather than fields.
Also the expression that you give for the expectation value of the vacuum does not feel right. I mean it takes no account of the existence of particle you are talking about.

As for the Higgs field..... You have the vev of the field, that minimizes the potential. Now something moving in the minimum potential is not interesting - it corresponds to a Goldstone boson, and as such could not have masses. At least that you get from perturbing the vev.
In general the perturbated field, will not get only the
$Φ= v+h$
in fact the higgs field would be:
$Φ= e^{iθ ξ} \left( \begin{array}{c} 0 \\ v+h \\ \end{array} \right)$

for which ξ can be a goldstone boson field, that is being dropped out once you work in the unitary gauge.... It also can be seen as performing a perturbation along the potential minimum (the slightest push along the minimum will start circulating you around it). The $h$ is the perturbation that happens along the field's values and that's the higg's field (by the last term I mean that it's the higgs that when coupled to itself will give the mass of the higg's boson) since it's the field that excites the vacuum.

3. ### karlzr

131
Do you mean the virtual electron-position pairs or photons?

My doubt is whether it is legitimate to define a new field ##h## by a constant? At high energies before symmetry breaking, maybe we would quantize ##\Phi## canonically with some creation and annihilation operators. But after a constant translation, I suppose the creation and annihilation operators in ##h## would correspond to totally distinct particles. That's why I have this concern.


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# Efficient Algorithms for Computing Approximate Equilibria in Bimatrix, Polymatrix and Lipschitz Games

Deligkas, A
(2016) Efficient Algorithms for Computing Approximate Equilibria in Bimatrix, Polymatrix and Lipschitz Games. Doctor of Philosophy thesis, University of Liverpool.


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# Documentation:Soil pH

## What is Soil pH?

Soil reactions are described in terms of pH values. Soil pH is a measurement of the hydrogen ion (H+ ) concentration within the soil. It is an extremely important soil chemical property, which has a profound effect on the availability of nutrients and pollutants to plants and activity of soil microorganisms. The pH scale is defined by the equation:

${\displaystyle pH=-\log[H^{+}]}$

Therefore, the pH decreases (or becomes more acidic) as the concentration of hydrogen ions increases.

The pH scale ranges from 0 to 14, while soil pH values usually range from extremely acidic (pH 2-3) to extremely basic or alkaline (pH 11-12). Because pH is defined as the negative decimal logarithm of [H+ ], each pH increase of one unit is a 10-fold jump in actual H+ concentration. Most soils pH range from 4.5 (considered strongly acidic for soils) to 8.5 (medium alkaline). The best plant growth takes place within pH 6 to 7 (i.e., in slightly acidic to neutral soils); however, some plants thrive in an alkaline environment while others prefer more acidic soils. Extreme acidity and alkalinity are conditions that usually need to be corrected if satisfactory crop production is to be achieved.

## Types of Soil Acidity

1. Active acidity: amount of hydrogen ions (H+ ) present in the soil solution (i.e., soil water + solutes).
2. Exchangeable acidity: is represented by the quantity of hydrogen and aluminum ions present on exchange sites on colloids, easily replaced by a salt solution (such as 0.01 M CaCl2). Recall that cations in the soil solution are in equilibrium with adsorbed cations.
3. Residual or reserve acidity: amount of hydrogen and aluminum ions strongly bound to soil particles, not replaced by a salt solution but can be neutralized by limestone.

## Methods for Determining Soil pH

### Active Acidity (pH of the soil samples in water)

1) Prepare a 1:5 soil : water suspension by adding 50 mL of distilled water to 10 g soil in a clean paper cup. Mix thoroughly with stirring rod.

2) Let suspension equilibrate for 30 minutes, during which time most of the soil solids should settle out.

3) Insert pH electrodes into supernatant solution and when a stable reading is displayed (‘S’ appears on the screen) record the pH to one decimal place on the data collection table provided below.

4) Rinse the electrodes well using distilled water.

### Exchangeable Acidity (pH of the soil samples in CaCl2 solution)

1) Prepare a 1:5 soil : CaCl2 suspension by adding 50 mL of 0.01 M CaCl2 to 10 g soil in a clean paper cup. Mix thoroughly with stirring rod.

Note: the symbol M denotes moles of solute per liter of solution: M = mol/L

2) Let suspension equilibrate for 30 minutes, during which time most of the soil solids should settle out.

3) Insert pH electrodes into supernatant solution and when a stable reading is displayed (‘S’ appears on the screen) record the pH to one decimal place on the data collection table provided below.

4) Rinse the electrodes well using distilled water.

### Field Test Kit

Follow instructions printed on the colour comparison card in the test kit:

1) Obtain 1/2 teaspoon of soil sample and place into a ceramic or plastic well.

2) Wearing gloves, add just enough Reagent N to absorb into your soil sample. If there is excess Reagent N, sprinkle a small amount of soil until fully absorbed.

3) Ensure the fine mesh sieve is placed on top of Reagent M bottle. Shake out a fine layer of Reagent M over your sample.

4) Wait for a a couple of minutes until the colour has developed and compare to the colour comparison card in the test kit.

## References

1. Krzic, M., Grand, G., Ballard, T., Brown, S., and L. Dampier. (2020). Lab Manual for Introduction to Soil Science course. The University of British Columbia. Vancouver, BC. Retrieved from: http://lfs-sw.sites.olt.ubc.ca/files/2020/01/APBI200_Lab-Manual_2020_Jan2120_V2.pdf
2. Thomas, G.W. 1996. Soil pH and soil acidity. p. 475-490. In D.L. Sparks et al. (ed.) Methods of soil analysis. Part 3. Chemical Methods. Soil Sci. Soc. Am. Book Series No. 5. ASASSSA, Madison, WI


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Next: The secant equation Up: Annexes Previous: Gram-Schmidt orthogonalization procedure.   Contents

# Notions of constrained optimization

Let us define the problem:
Find the minimum of subject to constraints .

To be at an optimum point we must have the equi-value line (the contour) of tangent to the constraint border . In other words, when we have constraints, we must have (see illustration in Figure 13.2) (the gradient of and the gradient of must aligned):

In the more general case when , we have:

 (13.19)

Where E is the set of active constraints, that is, the constraints which have We define Lagrangian function as:

 (13.20)

The Equation 13.19 is then equivalent to:

 where (13.21)

In unconstrained optimization, we found an optimum when . In constrained optimization, we find an optimum point ( ), called a KKT point (Karush-Kuhn-Tucker point) when:

 is a KKT point (13.22)

The second equation of 13.22 is called the complementarity condition. It states that both and cannot be non-zero, or equivalently that inactive constraints have a zero multiplier. An illustration is given on figure 13.3.
To get an other insight into the meaning of Lagrange Multipliers , consider what happens if the right-hand sides of the constraints are perturbated, so that

 (13.23)

Let , denote how the solution and multipliers change as changes. The Lagrangian for this problem is:

 (13.24)

From 13.23, , so using the chain rule, we have

 (13.25)

Using Equation 13.21, we see that the terms and are null in the previous equation. It follows that:

 (13.26)

Thus the Lagrange multiplier of any constraint measure the rate of change in the objective function, consequent upon changes in that constraint function. This information can be valuable in that it indicates how sensitive the objective function is to changes in the different constraints.

Next: The secant equation Up: Annexes Previous: Gram-Schmidt orthogonalization procedure.   Contents
Frank Vanden Berghen 2004-04-19


text:
# zbMATH — the first resource for mathematics

The rationality problem for norm one tori. (English) Zbl 1223.14017
This paper is on the rationality problem of algebraic tori of norm type. It contains several very significant results with elegant proofs.
Let $$K/k$$ be a finite separable field extension, $$R_{K/k}(G_m)$$ be the Weil restriction of $$G_{m,K}$$ to $$k$$. Define the norm one torus $$R^1_{K/k}(G_m):= \ker \{ \phi : R_{K/k}(G_m) \rightarrow G_{m,k} \}$$ where $$\phi$$ is the norm map $$\phi(\alpha)=N_{K/k}(\alpha)$$ for any $$\alpha \in K^{\times}$$. The main question is to study under what situation the torus $$R^1_{K/k}(G_m)$$ is rational over $$k$$.
The function field of $$R^1_{K/k}(G_m)$$ may be represented as follow. Let $$L/k$$ be the Galois closure of $$K/k$$, $$G=\mathrm{Gal}(L/k)$$, and $$H$$ be the subgroup such that $$L^H=K$$. Consider the augmentation map of $$G$$-lattices $$\epsilon : Z[G/H] \rightarrow Z$$. Let $$I_{G/H}= \ker(\epsilon)$$, and $$J_{G/H}=\operatorname{Hom} (I_{G/H}, Z)$$. Then the character group and the function field of $$R^1_{K/k}(G_m)$$ are $$J_{G/H}$$ and $$L(J_{G/H})^G$$ respectively.
From the theory of flabby classes the rationality problem can be reduced to
(1) $$L(J_{G/H})^G$$ is stably rational over $$k$$ if and only if $$J_{G/H}$$ is quasi-permutation, i.e. there is a short exact sequence of $$Z[G]$$-modules $$0 \rightarrow J_{G/H} \rightarrow P_1 \rightarrow P_2 \rightarrow 0$$ where $$P_1,P_2$$ are permutation $$G$$-lattices, and
(2) $$L(J_{G/H})^G$$ is retract rational over $$k$$ if and only if $$J_{G/H}$$ is quasi-invertible, i.e. there is a short exact sequence of $$Z[G]$$-modules $$0 \rightarrow J_{G/H} \rightarrow P \rightarrow E \rightarrow 0$$ where $$P$$ is a permutation $$G$$-lattice and $$E$$ is an invertible $$G$$-lattice (see Theorem 1.1).
The case when $$K/k$$ is Galois has been solved by S. Endo and T. Miyata [Nagoya Math. J. 56, 85–104 (1975; Zbl 0301.14008)]; also see Theorem 1.2 of this paper. Thus only the non-Galois case remains, i.e. $$H$$ is a non-normal non-trivial subgroup of $$G$$ (in particular, $$G$$ is non-abelian).
The first two main results are the following.
Theorem 1. Let $$G$$ be a non-abelian nilpotent group and $$H$$ be a non-normal non-trivial subgroup of $$G$$. Then $$J_{G/H}$$ is not quasi-invertible (see Theorem 2.1).
Theorem 2. Let $$G$$ be a non-abelian group and $$H$$ be a non-normal non-trivial subgroup of $$G$$. Then $$J_{G/H}$$ is quasi-invertible if and only if all the Sylow subgroups of $$G$$ are cyclic. In this situation, $$J_{G/H}$$ is quasi-permutation if and only if $$G \simeq C_m \times D_n$$ ($$m,n$$ are odd, gcd$$\{m,n \}=1$$, $$n \geq 3$$) and $$H \subset D_n$$ is of order 2 (see Theorem 3.1).
The author then consider the generic tori, i.e. $$G=S_n$$ and $$H=S_{n-1}$$ where $$S_n$$ denotes the symmetric group of degree $$n$$.
Theorem 3. For $$n \geq 2$$, $$J_{S_n/S_{n-1}}$$ is quasi-invertible if and $$n$$ is a prime number. Moreover, it is quasi-permutation if and only if $$n=2,3$$ (see Theorem 4.3). For previous partial results of the above theorem, see Remark 2 of this paper and Section 8 of the paper by A. Cortella and B. Kunyavskii [J. Algebra 225, No. 2, 771–793 (2000; Zbl 1054.14507)]. Finally consider the alternating groups.
Theorem 4. For $$n \geq 3$$, $$J_{A_n/A_{n-1}}$$ is quasi-invertible if and $$n$$ is a prime number. Moreover, the direct sum $$[J_{A_n/A_{n-1}}]^{(t)}$$ is quasi-permutation for some $$t \geq 1$$ if and only if $$n=3, 5$$ (see Theorem 4.5).
{Remarks}: The reader is warned that the definition of a meta-cyclic group is different from the usual one. The usual definition of a meta-cyclic group is an extension of a cyclic group by another cyclic group (see, for example, [I. M. Isaacs, Finite Group Theory. Graduate Studies in Mathematics 92. Providence, RI: American Mathematical Society (AMS). (page 160) (2008; Zbl 1169.20001)]). In this paper, a group $$G$$ is called meta-cyclic if all the Sylow subgroups of $$G$$ are cyclic, equivalently, $$G \simeq C_m \cdot C_n$$ ($$m$$ is odd and gcd$$\{m,n \}=1$$), where $$C_m \cdot C_n$$ denotes the semi-direct product with $$C_m$$ being a normal subgroup there.

##### MSC:
 14E08 Rationality questions in algebraic geometry 20C10 Integral representations of finite groups
##### Keywords:
Rationality problem; algebraic torus; retract rational
Full Text:
##### References:
 [1] Y. Berkovich, Groups of Prime Power Order, I , de Gruyter, Berlin, 2008. [2] K. S. Brown, Cohomology of Groups , Grad. Texts in Math. 87 , Springer, New York, 1982. [3] J.-L. Colliot-Thélène and J.-J. Sansuc, La R-équivalence sur les tores , Ann. Sci. Éc. Norm. Supér. (4) 10 (1977), 175-230. · Zbl 0356.14007 [4] J.-L. Colliot-Thélène and J.-J. Sansuc, Principal homogeneous spaces under flasque tori: Applications , J. Algebra 106 (1987), 148-205. · Zbl 0597.14014 [5] A. Cortella and B. Kunyavskiĭ, Rationality problem for generic tori in simple groups , J. Algebra 225 (2000), 771-793. · Zbl 1054.14507 [6] C. W. Curtis and I. Reiner, Methods of Representation Theory, I , John Wiley, New York, 1981. [7] A. Dress, The permutation class group of a finite group , J. Pure Appl. Algebra 6 (1975), 1-12. · Zbl 0302.20010 [8] S. Endo, On the rationality of algebraic tori of norm type , J. Algebra 235 (2001), 27-35. · Zbl 1062.20505 [9] S. Endo and T. Miyata, Invariants of finite abelian groups , J. Math. Soc. Japan 25 (1973), 7-26. · Zbl 0245.20007 [10] S. Endo and T. Miyata, On a classification of the function fields of algebraic tori , Nagoya Math. J. 56 (1975), 85-104; Correction , Nagoya Math. J. 79 (1980), 187-190. · Zbl 0301.14008 [11] M. Florence, Non rationality of some norm one tori , preprint, 2006. [12] L. Le Bruyn, Generic norm one tori , Nieuw Arch. Wiskd. (5) 13 (1995), 401-407. · Zbl 0872.20043 [13] N. Lemire and M. Lorenz, On certain lattices associated with generic division algebras , J. Group Theory 3 (2000), 385-405. · Zbl 0998.16012 [14] M. Rosen, Representations of twisted group rings , Ph.D. dissertation, Princeton University, Princeton, N. J., 1963. · Zbl 0115.40602 [15] D. J. Saltman, Retract rational fields and cyclic Galois extensions , Israel J. Math. 47 (1984), 165-215. · Zbl 0546.14013 [16] V. E. Voskresenskiĭ, Algebraic Groups and Their Birational Invariants , Transl. Math. Monogr. 179 , Amer. Math. Soc., Providence, 1998. · Zbl 0974.14034
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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Encyclopedia > Horizontal cell
Plan of retinal neurons.

Horizontal cells are the laterally interconnecting neurons in the outer plexiform layer of the retina. Image File history File links Gray882. ... Image File history File links Gray882. ... Human eye cross-sectional view. ...

## Contents

There are three basic types of horizontal cells, designated HI, HII and HIII. The selectivity of these three horizontal cells, towards one of the three cone types, is a matter of debate. According to studies conducted by Boycott and Wassle neither HI cells nor HII cells were selective towards S,M, or L cones. By contrast, Anhelt and Kolb claim that in their observations HI cells connected to all three cone types indiscremenantly, however, HII cells tended to contact S cones the most. They also identified a third type of horizontal cell, HIII, which was identical to HI but did not make contact with S cones. Normalised absorption spectra of human cone (S,M,L) and rod (R) cells Cone cells, or cones, are cells in the retina of the eye which only function in relatively bright light. ...

The HII cells also make connections with rods, but do so far enough away from the horizontal cell's soma such that they do not interfere with the activities of the cones.

They span across cones and summate inputs from them all to control the amount of GABA released back onto the photoreceptor cells, which depolarizes them. Their arrangement together with the on-centre and off-centre bipolar cells that receive input from the photoreceptors constitutes a form of lateral inhibition, increasing spatial resolution at the expense of some information on absolute intensity. The eye is thus more sensitive to contrast and differences in intensity. Gaba may refer to: GabÃ¢ or gabaa (Philippines), the concept of negative karma of the Cebuano people GABA, the gamma-amino-butyric acid neurotransmitter GABA receptor, in biology, receptors with GABA as their endogenous ligand Gaba 1 to 1, an English conversational school in Japan Marianne Gaba, a US model... Photoreceptor cells are contained in the retina and are responsible for transducing, or converting, light into signals that can be ultimately transmitted to the brain via the optic nerve. ... As a part of the retina, the bipolar cell exists between photoreceptors (rod cells and cone cells) and ganglion cells. ... simply; your nose receives messages from the environment sending them to the olfactory centre which is present in the cerebrum the largest part of brain: then you either smell a good smell or a bad smell stink with parts of seconds. ...

## Functional Properties

When light is shone onto a photoreceptor, the photoreceptor hyperpolarizes and reduces the release of glutamate, when this happens, horizontal cells reduce the release of GABA, which has an inhibitory effect on the photoreceptors. This reduction of inhibition leads to a depolarization of the photoreceptors. We therefore have the following negative feedback

Illumination$to$photoreceptor hyperpolarization$to$horizontal cell hyperpolarization$to$photoreceptor depolarization

One proposed theory for facilitation by the horizontal cells proceeds as follows. Assume we have 11 photoreceptors, one hyperpolarizing (H) bipolar cell, and one horizontal cell. All ten photoreceptors connect to the horizontal cell, and the middle photoreceptor (Pm) connects to the bipolar cell. The surrounding cells, which represent the outer receptive field, will be designated Po then we can explain an off-centre arrangement as follows. If light is shown onto the Pm then

1. Pm is activated by light and therefore hyperpolarizes
2. Pm reduces release of glutamate
3. Reduction of glutamate depolarizes the H bipolar cell
4. Reduction of glutamate hyperpolarizes the horizontal cell and it reduces release of GABA
5. Since Pois still releasing glutamate, reduction in GABA is marginal

If the light is shone onto the surrounding area then

1. Po is activated and therefore hyperpolarizes
2. Po reduce release of glutamate
3. Reduction of glutamate hyperpolarizes the horizontal cell
4. Horizontal cell reduces release of GABA
5. Reduction of GABA depolarizes photoreceptors
6. Po not affected since they are strongly being hyperpolarized by activation
7. Pm is affected and therefore depolarizes
8. Pm releases glutamate
9. H Bipolar cell is hyperpolarized

To explain diffuse light, then we consider both cases together, and as it turns out, the two effects cancel each other out, and we get little to no affects

Photoreceptors are light-sensitive proteins involved in the function of photoreceptor cells. ... As a part of the retina, the bipolar cell exists between photoreceptors (rod cells and cone cells) and ganglion cells. ... This article or section is incomplete and may require expansion and/or cleanup. ... A Ganglion Cell (or sometimes called a gangliocyte) is a type of Neuron located in the retina that receives visual information from bipolar cells; its axons give rise to the optic nerve. ...

Results from FactBites:

 Journal of Vision - Receptive field structure of H1 horizontal cells in macaque monkey retina, by (10768 words) The horizontal cells of mammalian retinas, such as those in cat and rabbit, are also extensively coupled and consequently have uniformly large receptive fields (Mills and Massey, 1994; Bloomfield, Xin, and Persky, 1995). H1 horizontal cell receptive fields were characterized by measuring their responses to drifting sinusoidal gratings as a function of spatial frequency, to flashing spots as a function of spot diameter, and to flashing annuli as a function of annulus inner diameter. In cat retina, increased cell density has not been correlated with a reduction in horizontal cell receptive field size as a function of decreasing retinal eccentricity (Lankheet et al., 1990, 1992; Nelson, 1977), a correlation that is very strong in primate retina.
 Webvision: Physiology of Horizontal Cells (2854 words) During the horizontal cell response to light, the trans-membrane potential increases or 'hyperpolarizes', from ~-30 mV in darkness to -40 to -70 mV in the presence of the light stimulus. The inference from such models is that signals in cell bodies of either the A or B type horizontal cells are expected to reflect only the local synaptic inputs from cones while signals in axon terminals of the B-type horizontal cells are expected to reflect only the local synaptic inputs from rods. Horizontal cells are affected by light stimuli over a much wider area than one might suppose based on the lateral extent of process arborization, or 'dendritic field'.
More results at FactBites »

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# What Causes That Ephemeral Facial Flush

When alcohol is consumed, it is not processed by the stomach - instead, the liver is responsible for breaking it down into something easily expended by the body. The process by which the liver accomplishes this transformation is fairly simple, but surprisingly noxious.

## #H: Oxidation of Ethanol

Because ethanolEthanol is used in this discussion because it is the most common intoxicating agent. cannot be easily digested, the body converts it to acetic acid, which can be comfortably processed. This conversion is described by a two-step oxidation:

In organic chemistry, oxidation state refers to the heterolytic count of electrons on an atom. In these reactions, the central carbon first has an oxidation state of -1, then +1, then +3. Generally, adding electronegative substituents oxidizes a molecule.

This reaction requires two enzymes: alcohol dehydrogenase and aldehyde dehydrogenase. We will soon see why these are important! In the meantime, let us not ignore the intermediate of this reaction -

## #He: Acetaldehyde

As it turns out, acetaldehyde is responsible for many of the side effects experienced when consuming alcohol. Perhaps the most prominent and visible of these consequences is the alcohol flush responseOften referred to as Asian Glow or Asian Flush Syndrome., delineated by reddish blotches on the face and other parts of the body. Referral of this condition as Asian Glow actually comes from the fact that a disproportionate number of Eastern Asians experience the syndrome.

Why so?

Eastern Asians have a known genetic deficiency of aldehyde dehydrogenase, responsible for oxidizing toxic acetaldehydeNot only does significant amounts acetaldehyde cause facial flushing; it also induces nausea, headaches, is an irritant, and is a likely carcinogen. into expendable acetic acid. When excess acetaldehyde accumulates, either because of a deficiency of aldehyde dehydrogenase or because of an excess of alcohol dehydrogenaseAn excess of alcohol dehydrogenase causes ethanol to oxidize to acetaldehyde more rapidly than acetaldehyde can be oxidized to acetic acid., its toxic effects begin to impact the human body.

## #Li: Side-Side Effects

Ayaz,

You cry,

it can’t be all bad… can it?!

It should be noted that formal research of this theory is currently developing, and there are is notable counter-evidence: for one, Native Americans are known to have an inordinately high rate of alcoholism, despite there being no enzymic differences between Native Americans and Whites.

No, it can’t. Reports have shown that Eastern Asian populations also have a disproportionately low rate of alcoholism - the prevailing theory being that the unpleasant consequences due to a aldehyde dehydrogenase deficiency drives individuals against drinking more.


text:
open import Cat.Displayed.Cartesian
open import Cat.Functor.Equivalence
open import Cat.Diagram.Pullback
open import Cat.Displayed.Fibre
open import Cat.Instances.Slice
open import Cat.Displayed.Base
open import Cat.Functor.Base
open import Cat.Prelude

import Cat.Reasoning as CR

module Cat.Displayed.Instances.Slice {o ℓ} (B : Precategory o ℓ) where

# The canonical self-indexing🔗

There is a canonical way of viewing any category $\mathcal{B}$ as displayed over itself, given fibrewise by taking slice categories. Following , we refer to this construction as the canonical-self indexing of $\mathcal{B}$ and denote it $\underline{\mathcal{B}}$. Recall that the objects in the slice over $y$ are pairs consisting of an object $x$ and a map $f : x \to y$. The core idea is that any morphism lets us view an object $x$ as being “structure over” an object $y$; the collection of all possible such structures, then, is the set of morphisms $x \to y$, with domain allowed to vary.

Contrary to the maps in the slice category, the maps in the canonical self-indexing have an extra “adjustment” by a morphism $f : x \to y$ of the base category. Where maps in the ordinary slice are given by commuting triangles, maps in the canonical self-indexing are given by commuting squares, of the form

where the primed objects and dotted arrows are displayed.

record
Slice-hom
{x y} (f : Hom x y)
(px : /-Obj {C = B} x) (py : /-Obj {C = B} y)
: Type (o ⊔ ℓ)
where
constructor slice-hom
field
to      : Hom (px .domain) (py .domain)
commute : f ∘ px .map ≡ py .map ∘ to

open Slice-hom

The intuitive idea for the canonical self-indexing is possibly best obtained by considering the canonical self-indexing of ${{\mathbf{Sets}}}_\kappa$. First, recall that an object $f : {{\mathbf{Sets}}}/X$ is equivalently a $X$-indexed family of sets, with the value of the family at each point $x : X$ being the fibre $f^*(x)$. A function $X \to Y$ of sets then corresponds to a reindexing, which takes an $X$-family of sets to a $Y$-family of sets (in a functorial way). A morphism $X' \to Y'$ in the canonical self-indexing of ${{\mathbf{Sets}}}$ lying over a map $f : X \to Y$ is then a function between the families $X' \to Y'$ which commutes with the reindexing given by $f$.

It’s straightforward to piece together the objects of the (ordinary) slice category and our displayed maps Slice-hom into a category displayed over $\mathcal{B}$.

Slices : Displayed B (o ⊔ ℓ) (o ⊔ ℓ)
Slices .Ob[_] = /-Obj {C = B}
Slices .Hom[_] = Slice-hom
Slices .Hom[_]-set = Slice-is-set
Slices .id′ = slice-hom id id-comm-sym
Slices ._∘′_ {x = x} {y = y} {z = z} {f = f} {g = g} px py =
slice-hom (px .to ∘ py .to) $(f ∘ g) ∘ x .map ≡⟨ pullr (py .commute) ⟩≡ f ∘ (y .map ∘ py .to) ≡⟨ extendl (px .commute) ⟩≡ z .map ∘ (px .to ∘ py .to) ∎ Slices .idr′ {f = f} f′ = Slice-pathp (idr f) (idr (f′ .to)) Slices .idl′ {f = f} f′ = Slice-pathp (idl f) (idl (f′ .to)) Slices .assoc′ {f = f} {g = g} {h = h} f′ g′ h′ = Slice-pathp (assoc f g h) (assoc (f′ .to) (g′ .to) (h′ .to))  It’s only slightly more annoying to show that a vertical map in the canonical self-indexing is a map in the ordinary slice category which, since the objects displayed over $x$ are defined to be those of the slice category $\mathcal{B}/x$, gives an equivalence of categories between the fibre $\underline{\mathcal{B}}^*(x)$ and the slice $\mathcal{B}/x$. Fibre→slice : ∀ {x} → Functor (Fibre Slices x) (Slice B x) Fibre→slice .F₀ x = x Fibre→slice .F₁ f ./-Hom.map = f .to Fibre→slice .F₁ f ./-Hom.commutes = sym (f .commute) ∙ eliml refl Fibre→slice .F-id = /-Hom-path refl Fibre→slice .F-∘ f g = /-Hom-path (transport-refl _) Fibre→slice-is-ff : ∀ {x} → is-fully-faithful (Fibre→slice {x = x}) Fibre→slice-is-ff {_} {x} {y} = is-iso→is-equiv isom where isom : is-iso (Fibre→slice .F₁) isom .is-iso.inv hom = slice-hom (hom ./-Hom.map) (eliml refl ∙ sym (hom ./-Hom.commutes)) isom .is-iso.rinv x = /-Hom-path refl isom .is-iso.linv x = Slice-pathp refl refl Fibre→slice-is-equiv : ∀ {x} → is-equivalence (Fibre→slice {x}) Fibre→slice-is-equiv = is-precat-iso→is-equivalence$
record { has-is-ff = Fibre→slice-is-ff
; has-is-iso = id-equiv
}

## As a fibration🔗

If (and only if) $\mathcal{B}$ has all pullbacks, then the self-indexing $\mathcal{B}$ is a Cartesian fibration. This is almost by definition, and is in fact where the “Cartesian” in “Cartesian fibration” (recall that another term for “pullback square” is “cartesian square”). Since the total space $\int \underline{\mathcal{B}}$ is equivalently the arrow category of $\mathcal{B}$, with the projection functor $\pi : \int \underline{\mathcal{B}} \to \mathcal{B}$ corresponding under this equivalence to the codomain functor, we refer to $\underline{ca{B}}$ regarded as a Cartesian fibration as the codomain fibration.

Codomain-fibration
: (∀ {x y z} (f : Hom x y) (g : Hom z y) → Pullback B f g)
→ Cartesian-fibration Slices
Codomain-fibration pullbacks .has-lift f y′ = lift-f where
open Pullback (pullbacks f (y′ .map))

lift-f : Cartesian-lift Slices f y′
lift-f .x′ = cut p₁
lift-f .lifting .to = p₂
lift-f .lifting .commute = square
lift-f .cartesian .universal m h′ .to = limiting (assoc _ _ _ ∙ h′ .commute)
lift-f .cartesian .universal m h′ .commute = sym p₁∘limiting
lift-f .cartesian .commutes m h′ = Slice-pathp refl p₂∘limiting
lift-f .cartesian .unique m′ x = Slice-pathp refl $Pullback.unique (pullbacks f (y′ .map)) (sym (m′ .commute)) (ap to x)  Since the proof that Slices is a cartesian fibration is given by essentially rearranging the data of pullbacks in $\mathcal{B}$, we also have the converse implication: If $\underline{\mathcal{B}}$ is a Cartesian fibration, then $\mathcal{B}$ has all pullbacks. Codomain-fibration→pullbacks : ∀ {x y z} (f : Hom x y) (g : Hom z y) → Cartesian-fibration Slices → Pullback B f g Codomain-fibration→pullbacks f g lifts = pb where open Pullback open is-pullback the-lift = lifts .has-lift f (cut g) pb : Pullback B f g pb .apex = the-lift .x′ .domain pb .p₁ = the-lift .x′ .map pb .p₂ = the-lift .lifting .to pb .has-is-pb .square = the-lift .lifting .commute pb .has-is-pb .limiting {p₁' = p₁'} {p₂'} p = the-lift .cartesian .universal {u′ = cut id} p₁' (slice-hom p₂' (pullr (idr _) ∙ p)) .to pb .has-is-pb .p₁∘limiting = sym (the-lift .cartesian .universal _ _ .commute) ∙ idr _ pb .has-is-pb .p₂∘limiting = ap to (the-lift .cartesian .commutes _ _) pb .has-is-pb .unique p q = ap to$ the-lift .cartesian .unique
(slice-hom _ (idr _ ∙ sym p)) (Slice-pathp refl q)

Since the fibres of the codomain fibration are given by slice categories, then the interpretation of Cartesian fibrations as “displayed categories whose fibres vary functorially” leads us to reinterpret the above results as, essentially, giving the pullback functors between slice categories.


text:
# Compute $T_{E|\mathbb{Q}}(\alpha)$ and $N_{E|\mathbb{Q}}(\alpha)$ for every $\alpha \in E$.

Consider the splitting field $E$ of $X^4 -2$.

Compute $T_{E|\mathbb{Q}}(\alpha)$ and $N_{E|\mathbb{Q}}(\alpha)$ for every $\alpha \in E$.

My attempt:

The splitting field $E = \mathbb{Q}(i,\sqrt{2},\sqrt[4]{2})$ which is a degree 8 extension with basis $\mathcal{B} = \{1,i,\sqrt{2},\sqrt[4]{2},i\sqrt{2},i\sqrt[4]{2},\sqrt[4]{8}\}$

We can use the definition of the trace in terms of the roots of the minimal polynomial over $\mathbb{Q}$ which is $X^4 -2$, since its an anihilating polynomial of order 4, which is monic. This polynomial is irreducible over $\mathbb{Q}$ by Einsestein criterion, then its the minimal polynomial.

Since the trace is the sum of all the zeros, and every zero is a conjugate of the other, the trace is Also the norm, can be obtained similarly.

I am right ?

First of all you say that "Hence, all the basis elements are the roots", which isn't true, as the basis has $8$ elements, as noted above, while the polynomial has $4$ roots. Nevertheless the trace can be easily calculated in terms of the representation in basis elements. So we have that

$$T(\alpha) = T(a + b\sqrt[4]{2} + c\sqrt{2} + d\sqrt[4]{8} + ei + fi\sqrt[4]{2} + gi\sqrt{2} + fi\sqrt[4]{8}) = aT(1) = 8a$$

This is true as the trace function is linear and the traces of the basis elements are $0$, except for the element $1$.

On the other side calculating the norm is substantially harder, as the norm is multiplicative, instead of linear. In particular you need to compute:

$$N(\alpha) = \prod_{\sigma \in \text{Gal}(E/\mathbb{Q})}\sigma(\alpha)$$

I guess doing this by hand is almost impossible, as there would be $8^8$ terms if you want to get an expression with respect to the basis representation of $\alpha$.

Note: This answer comes after a comment in a related post, 2819799, although it was already "green" as i started. Well, the dialog is important, this is my only interest...

Let us denote by $$t=\sqrt[4]2$$ the generator of the subfield $$K=\Bbb Q(t)=\Bbb Q(\sqrt[4]2)$$ of the given splitting field, $$E=K(i)=\Bbb Q(\sqrt[4]2,i)\ ,$$ and since we will write this $t$ very often, so a short-hand was imperative above.

We will also identify it (agains my rules of always working algebraically) with the only complex root of the polynomial $X^4-2$, which is real, and positive. The conjugates of $t$ in $\Bbb C$ are $$t_1=t\ ,\ t_2=it\ ,\ t_3=i^2t\ ,\ t_4=i^3t\ ,$$ and corresponding to them we get embeedings $\sigma_j:\Bbb Q(t)\to\Bbb C$, $t\to t_j$.

The system $$\{\ 1,\ t,\ t^2,\ t^3\ \}$$ is a base of $K=\Bbb Q(t)$ over $\Bbb Q$, and computing the trace and the norm for the extension $K:\Bbb Q$ can be done in terms of it.

The trace is linear, so for computing it, it is enough to compute \begin{aligned} \operatorname{Trace}(1) &= \sum_{1\le 4\le j}\sigma_j(1) &&=1+1+1+1 &&=4\ ,\\ \operatorname{Trace}(t) &= \sum_{1\le 4\le j}\sigma_j(t) &&=t+it+-t-it&&=0\ ,\\ \operatorname{Trace}(t^2) &= \sum_{1\le 4\le j}\sigma_j(t^2) &&=t^2-t^2+t^2-t^2 &&=0\ ,\\ \operatorname{Trace}(t^3) &= \sum_{1\le 4\le j}\sigma_j(t^3) &&=t^3-it^3-t^3+it^3 &&=0\ . \end{aligned} So for a general $$\xi = a+bt+ct^2+dt^3\in K=\Bbb Q(t)\ ,\qquad a,b,c,d\in \Bbb Q\ ,$$ we have $$\operatorname{Trace}_{K:\Bbb Q}\xi = 4a\ .$$ The norm is relatively simple, too. First, we need an identity: \begin{aligned} &(X+Y+Z+T) (X+iY-Z-iT) (X-Y+Z-T) (X-iY-Z+iT) \\[3mm] &\qquad= \Big(\ (X+Z)^2-(Y+T)^2\ \Big) \Big(\ (X-Z)^2+(Y-T)^2\ \Big) \\[3mm] &\qquad= \Big(\ (X^2+Z^2-2YT)-(Y^2+T^2-2XZ)\ \Big) \\ &\qquad\ \phantom{=} \Big(\ (X^2+Z^2-2YT)+(Y^2+T^2-2XZ)\ \Big) \\[3mm] &\qquad= (X^2+Z^2-2YT)^2-(Y^2+T^2-2XZ)^2 \\[3mm] &\qquad= X^4-Y^4+Z^4-T^4 \\ &\qquad\qquad -2X^2Z^2 + 2Y^2T^2 \\ &\qquad\qquad\qquad - 4 X^2 YT + 4Y^2XZ - 4 Z^2 YT + 4T^2XZ\ . \end{aligned} Now we can simply compute the norm for the "smaller extension" of an element of the form $\xi=a+bt+ct^2+dt^3$ \begin{aligned} \operatorname{Norm}_{K:\Bbb Q}\xi &= \operatorname{Norm}_{K:\Bbb Q}(a+bt+ct^2+dt^3) \\[3mm] &= (a+bt+ct^2+dt^3) (a+ibt-ct^2-idt^3) (a-bt+ct^2-dt^3) (a-ibt-ct^2+idt^3) \\[3mm] &=a^4-b^4t^4+c^4t^8-d^4t^{12}\\ &\qquad -2a^2c^2t^4 + 2b^2d^2t^8\\ &\qquad\qquad-4a^2bdt^4 + 4b^2act^4-4c^2bdt^8+4d^2act^8 \\[3mm] &=a^4-2b^4+4c^4-8d^4\\ &\qquad -4a^2c^2 + 8b^2d^2\\ &\qquad\qquad-8a^2bd + 8b^2ac - 16c^2bd + 16d^2ac\ . \end{aligned} We can ask the computer for a confirmation of the above:

var('a,b,c,d')
t = 2^(1/4)
prod( [ a + b*i^k*t + c*i^(2*k)*t^2 + d*i^(3*k)*t^3 for k in [0,1,2,3] ] ).expand()
latex(_)

and we get the result:

a^4 - 2*b^4 + 8*a*b^2*c - 4*a^2*c^2 + 4*c^4 - 8*a^2*b*d
- 16*b*c^2*d + 8*b^2*d^2 + 16*a*c*d^2 - 8*d^4

(Result was manually broken in two lines.) In LaTeX: $$a^{4} - 2 \, b^{4} + 8 \, a b^{2} c - 4 \, a^{2} c^{2} + 4 \, c^{4} - 8 \, a^{2} b d - 16 \, b c^{2} d + 8 \, b^{2} d^{2} + 16 \, a c d^{2} - 8 \, d^{4}$$ Now each element in $E=K(i)$ can be written uniquely in the form $$\xi+i\xi'\ .$$ We can use the above computation and the "transitivity of the related norms". The norm for the extension $E:K$ for the above element is $$(\xi+i\xi')(\xi-i\xi')\ .$$ This brings in scene $\xi^2$ and ${\xi'}^2$. We have to isolate the many components of them w.r.t. the basis $1,t,t^2,t^3$, then apply the above formula.

(For the trace, the game is simpler, the trace over $E:K$ of $\xi+i\xi'$ is the sum of the elements $\xi\pm i\xi'$, so it is $2\xi$. From here we apply the already computed trace.

It is hard to say more in detail for "generic elements".

• I love how you transform the problem to avoid the roots. Thanks. – Richard Clare Jun 16 '18 at 20:02


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# 6.1: Inequality Expressions

Difficulty Level: Basic Created by: CK-12
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Practice Inequality Expressions
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Suppose you're having a party, and you know that the number of people attending will be greater than or equal to 25. How would you write this inequality? If you had to graph the solutions to this inequality on a number line, could you do it? After completing this Concept, you'll not only be able to express inequalities such as this one with a graph, but you'll also be able to look at a graph and determine what inequality it represents.

### Guidance

Verbs that translate into inequalities are:

\begin{align*}>\end{align*} “greater than”

\begin{align*}\ge\end{align*} “greater than or equal to”

\begin{align*} < \end{align*} “less than”

\begin{align*}\le\end{align*} “less than or equal to”

\begin{align*}\neq\end{align*} “not equal to”

Definition: An algebraic inequality is a mathematical sentence connecting an expression to a value, a variable, or another expression with an inequality sign.

Solutions to one-variable inequalities can be graphed on a number line or in a coordinate plane.

#### Example A

Graph the solutions to \begin{align*}t > 3\end{align*} on a number line.

Solution: The inequality is asking for all real numbers larger than 3.

You can also write inequalities given a number line of solutions.

#### Example B

Write the inequality pictured below.

Solution: The value of four is colored in, meaning that four is a solution to the inequality. The red arrow indicates values less than four. Therefore, the inequality is:

\begin{align*}x \le 4\end{align*}

Inequalities that “include” the value are shown as \begin{align*}\le\end{align*} or \begin{align*}\ge\end{align*}. The line underneath the inequality stands for “or equal to.” We show this relationship by coloring in the circle above this value on the number line, as in the previous example. For inequalities without the “or equal to,” the circle above the value on the number line remains unfilled.

Four Ways to Express Solutions to Inequalities

1. Inequality notation: The answer is expressed as an algebraic inequality, such as \begin{align*}d \le \frac{1}{2}\end{align*}.

2. Set notation: The inequality is rewritten using set notation brackets { }. For example, \begin{align*}\left \{ d | d \le \frac{1}{2} \right \}\end{align*} is read, “The set of all values of \begin{align*}d\end{align*}, such that \begin{align*}d\end{align*} is a real number less than or equal to one-half.”

3. Interval notation: This notation uses brackets to denote the range of values in an inequality.

1. Square or “closed” brackets [ ] indicate that the number is included in the solution
2. Round or “open” brackets ( ) indicate that the number is not included in the solution.

Interval notation also uses the concept of infinity \begin{align*}\infty\end{align*} and negative infinity \begin{align*}- \infty\end{align*}. For example, for all values of \begin{align*}d\end{align*} that are less than or equal to \begin{align*}\frac{1}{2}\end{align*}, you could use set notation as follows: \begin{align*} \left . \left (-\infty, \frac{1}{2}\right ] \right .\end{align*}.

4. As a graphed sentence on a number line.

#### Example C

Describe the set of numbers contained by the given set notation for the following:

a) (8, 24)

b) [3, 12)

Solution:

(8, 24) states that the solution is all numbers between 8 and 24 but does not include the numbers 8 and 24.

[3, 12) states that the solution is all numbers between 3 and 12, including 3 but not including 12.

### Guided Practice

Describe and graph the solution set expressed by \begin{align*}(\infty, 3.25)\end{align*}.

Solution:

The solution set contains all numbers less than 3.25, not including 3.25.

The graph on the number line is:

### Practice

1. What are the four methods of writing the solution to an inequality?

Graph the solutions to the following inequalities using a number line.

1. \begin{align*}x < -3\end{align*}
2. \begin{align*}x \ge 6\end{align*}
3. \begin{align*}x > 0\end{align*}
4. \begin{align*}x \le 8\end{align*}
5. \begin{align*}x < -35\end{align*}
6. \begin{align*}x > -17\end{align*}
7. \begin{align*}x \ge 20\end{align*}
8. \begin{align*}x \le 3 \end{align*}

Write the inequality that is represented by each graph.

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Date Created:
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Manipulate the Form of the Equation of a Line - Maple Help

Home : Support : Online Help : Tasks : Algebra : Equation of a Line : Task/ManipulateFormEquationOfLine

Manipulate the Form of the Equation of a Line

 Description Use the LineTutor to manipulate the form of the equation of a line.

 Manipulate the Form of the Equation of a Line Enter the equation of a line:

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