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#### Model 1 Mark Book Back Questions (New Syllabus) 2020

12th Standard EM

Reg.No. :
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Maths

Time : 00:20:00 Hrs
Total Marks : 28

Pat A

28 x 1 = 28
1. If A =$\left( \begin{matrix} cosx & sinx \\ -sinx & cosx \end{matrix} \right)$ and A(adj A) =$\lambda$ $\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$ then $\lambda$ is

(a)

sinx cosx

(b)

1

(c)

2

(d)

none

2. If $\rho$(A) = $\rho$([A/B]) = number of unknowns, then the system is

(a)

consistent and has infinitely many solutions

(b)

consistent

(c)

inconsistent

(d)

consistent and has unique solution

3. If $\rho$(A) ≠ $\rho$([AIB]), then the system is

(a)

consistent and has infinitely many solutions

(b)

consistent and has a unique solution

(c)

consistent

(d)

inconsistent

4. If, i2 = -1, then i1 + i2 + i3 + ....+ up to 1000 terms is equal to

(a)

1

(b)

-1

(c)

i

(d)

0

5. The complex number z which satisfies the condition $\left| \frac { 1+z }{ 1-z } \right|$ =1 lies on

(a)

circle x2+y2 =1

(b)

x-axis

(c)

y-axis

(d)

the lines x+y=1

6. If x =cosθ + i sinθ, then the value of xn+$\frac { 1 }{ { x }^{ n } }$ is

(a)

2 cosθ

(b)

2i sin nθ

(c)

2i sin nθ

(d)

2i cos nθ

7. If x is real and $\frac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 }$ then

(a)

$\frac{1}{3}$ ≤k≤

(b)

k≥5

(c)

k≤0

(d)

none

8. lf the root of the equation x3 +bx2+cx-1=0 form an lncreasing G.P, then

(a)

one of the roots is 2

(b)

one of the rots is 1

(c)

one of the rots is -1

(d)

one of the rots is -2

9. If ax2 + bx + c = 0, a, b, c E R has no real zeros, and if a + b + c < 0, then __________

(a)

c>0

(b)

c<0

(c)

c=0

(d)

c≥0

10. ·If $\alpha ={ tan }^{ -1 }\left( \cfrac { \sqrt { 3 } }{ 2y-x } \right) ,\beta ={ tan }^{ -1 }\left( \cfrac { 2x-y }{ \sqrt { 3y } } \right)$ then $\alpha -\beta$

(a)

$\cfrac { \pi }{ 6 }$

(b)

$\cfrac { \pi }{ 3 }$

(c)

$\cfrac { \pi }{ 2 }$

(d)

$\cfrac { -\pi }{ 3 }$

11. If tan-1(3)+tan-1(x)=tan-1(8)then x=

(a)

5

(b)

$\cfrac { 1 }{ 5 }$

(c)

$\cfrac { 5 }{ 14 }$

(d)

$\cfrac { 14 }{ 5 }$

12. If $4{ cos }^{ -1 }x+{ sin }^{ -1 }x=\pi$ then x is

(a)

$\cfrac { 3 }{ 2 }$

(b)

$\cfrac { 1 }{ \sqrt { 2 } }$

(c)

$\cfrac { \sqrt { 3 } }{ 2 }$

(d)

$\cfrac { 2 }{ \sqrt { 3 } }$

13. y2 - 2x - 2y + 5 = 0 is a

(a)

circle

(b)

parabola

(c)

ellipse

(d)

hyperbola

14. If the distance between the foci is 2 and the distance between the direction is 5, then the equation of the ellipse is

(a)

6x2 + 10y2 = 5

(b)

6x2 + 10y2 = 15

(c)

x2 + 3y2 = 10

(d)

none

15. The area of the circle (x - 2)2 + (y - k)2 = 25 is

(a)

25ㅠ

(b)

5ㅠ

(c)

10ㅠ

(d)

25

16. If $\overset { \rightarrow }{ d } =\overset { \rightarrow }{ a } \times \left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) +\overset { \rightarrow }{ b } \times \left( \overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \right) +\overset { \rightarrow }{ c } \times \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right)$, then

(a)

$\left| \overset { \rightarrow }{ d } \right|$

(b)

$\overset { \rightarrow }{ d } =\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c }$

(c)

$\overset { \rightarrow }{ d } =\overset { \rightarrow }{ 0 }$

(d)

a, b, c are coplanar

17. The area of the parallelogram having diagonals $\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k }$ and $\overset { \rightarrow }{ b } =\overset { \wedge }{ i } -3\overset { \wedge }{ j } +\overset { \wedge }{ 4k }$ is

(a)

4

(b)

2$\sqrt { 3 }$

(c)

4$\sqrt { 3 }$

(d)

5$\sqrt { 3 }$

18. The value of ${ \left| \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right| }^{ 2 }+{ \left| \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right| }^{ 2 }$ is

(a)

$2\left( { \left| \overset { \rightarrow }{ a } \right| }^{ 2 }+{ \left| \overset { \rightarrow }{ b } \right| }^{ 2 } \right)$

(b)

$\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b }$

(c)

$2\left( { \left| \overset { \rightarrow }{ a } \right| }^{ 2 }-{ \left| \overset { \rightarrow }{ b } \right| }^{ 2 } \right)$

(d)

${ \left| \overset { \rightarrow }{ a } \right| }^{ 2 }{ \left| \overset { \rightarrow }{ b } \right| }^{ 2 }$

19. Equation of the normal to the curve y=2x2+3 sin x at x=0 is

(a)

x + y = 0

(b)

3y = 0

(c)

x + 3y = 7

(d)

x + 3y = 0

20. The curve y = ex is ________

(a)

convex

(b)

concave

(c)

convex upwards

(d)

concave upwards

21. If f(x,y) = 2x2 - 3xy + 5y + 7 then f(0, 0) and f(1, 1) is

(a)

7,11

(b)

11,7

(c)

0,7

(d)

1,0

22. If u = sin-1 $\left( \frac { { x }^{ 4 }+{ y }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right)$ and f= sin u then f is a homogeneous function of degree ..................

(a)

0

(b)

1

(c)

2

(d)

4

23. $\int _{ 0 }^{ \infty }{ { e }^{ -mx } } { x }^{ 7 }$ dx is

(a)

(b)

(c)

(d)

24. $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sin \ x-cos \ x }{ 1+sin \ x cos \ x } dx= }$ ............

(a)

$\frac { \pi }{ 2 }$

(b)

0

(c)

$\frac { \pi }{ 4 }$

(d)

$\pi$

25. The I.F of $\frac{dy}{dx}-y$ tan x=cos x is

(a)

sec x

(b)

cos x

(c)

etan x

(d)

cot x

26. The number of binary operations that can be defined on a set of 3 elements is

(a)

32

(b)

33

(c)

39

(d)

31

27. If p is true and q is unknown, then _________

(a)

~ p is true

(b)

p v (~p) is false

(c)

p ∧ (~p) is true

(d)

p v q is true

28. In (N, *), x * y = max(x, y), x, y $\in$ N then 7 * (-7)

(a)

7

(b)

-7

(c)

0

(d)

-49


text:
# How to control pH in a fermenter based on the acid produced by mircoorganism

I have got a bioreactor with an aqueous medium which produces biomass (yeast; S.cerevisiae) after inoculation. I am observing and collecting data over the whole process via probes/etc. on different process parameters such as pH, temperature, offgas of $$\ce{O_2}$$ and $$\ce{CO_2}$$, base uptake ($$\ce{NaOH}$$), weight of bioreactor and some little more like stirrer speed and so on.

How can one calculate or make a model on adding base to control pH based on the produced acid
(for now I've got no information on which one that is exactly) of the yeast or other factors?
(Is there a mechanistic or only a deterministic solution?)

Edit:
I can indirectly calculate how much acid is produced because that corresponds to the biomass whose growth data I collect. I don't know the exact acid that is produced, f.ex. $$\ce{HCl}$$ or so, but I think that the molar ratio between my used base and the produced acid is either 1:1 or 2:1, based on my assumption that yeasts will not produce acids with 3 or more protons in their metabolism.

The medium used for growing:
I am currently using glucose as the main substrate, the fermentation itself lasts only about 5-7 hours which means the $$\ce{EtOH}$$ that is produced can be "neglected" as second substrate. Furthermore I use $$\ce{(NH4)2SO4}$$ and $$\ce{(NH4)H2PO4}$$ as N-source and P-source respectively.
[Also some trace elements but those are in minor concentrations/ negligible for my pH model.]

My question is if there is a way to make a model on the acid production to control the base feed to stabilize the pH.

• So let me get this straight: you don't currently have a way of measuring how much acid the yeast produces, but wonder if, knowing this, you can predict how much base you'd need to add to keep the pH at a given value? Jul 23 '19 at 8:28
• No, I can indirectly calculate how much acid is produced because that corresponds to the biomass whose growth data I collect. I meant that I don't know the exact acid that is produced, f.ex. $\ce{HCl}$ or so. But I think that the molar ratio between my used base and the produced acid is either 1:1 or 2:1, based on my assumption that yeasts will not produce acids with 3 or more protons in their metabolism. My question is if there is a way to make a model on the acid production to control the base feed to stabilize the pH. Jul 23 '19 at 9:06
• What is the medium on which the yeast grows? Does it contain sucrose or starch or something else? Jul 23 '19 at 10:07
• You should always edit in the additional information, not just include it in the comment sections. Aug 6 '21 at 20:28
• That said, simple feedback control connected to a continuous pH probe is much more reliable and is standard practice for well controlled fermentation. Aug 9 '21 at 20:44

format: plain


text: Patterning of the rib cage is complex, not least because it derives from two different tissues - the somitic mesoderm (which forms the vertebrae and ribs) and the lateral plate mesoderm (which forms the sternum). On p. 2981, Deneen Wellik and colleagues shed light on the Hox code' that orchestrates development of the thoracic skeleton. They made mutant mice lacking whole groups of paralogous Hox genes - Hox5, Hox6 and Hox9 - which had dramatic effects on rib cage morphology. Surprisingly, the phenotypes do not fit with the simplest version of the model of posterior prevalence', in which posterior groups of Hox genes are functionally dominant over the next-most anterior group. Consistent with Hox genes in other tissues, the phenotypes in the somite-derived skeleton do have consistent colinearity - the genomic distribution of the genes is reflected in their spatial and temporal expression and function during development. However, no such colinearity is seen in the sternum. The authors conclude that Hox patterning of the thoracic skeleton occurs through different mechanisms depending on the mesodermal tissue from which it derives.


text:
# Semiconductors

Diode - standard recovery rectifier, 3A, 1000V, 1N5408
1N5408 Diode. Lead mounted standard recovery rectifiers are designed for use in power supplies and other applications having need of a device with the following features:
• High Current to Small Size
• High Surge Current Capability
• Low Forward Voltage Drop
$0.38 LED - 3mm diameter, diffused lens Package of 5 Colors Available: Lens Diameter: 3 mm Overall Height: 5.4 mm Lens Type: Diffused Multiple color options available.$0.65
Diode - 1N34A Germanium
The 1N34A Germanium diode is an old standby in electronics. Widely used for detecting the rectifying efficiency or for switching on a radio, TV or stereo etc.
$1.75 Switch - Vibrato LDR, for Fender® Amps Optocoupler. Vibrato LDR for Fender Amps. Fix the dead LDR style tremolo on your vintage Fender amps as well as others. • Non- polarized optical side is indicated by silver dot and slightly longer leads • Resistive side is indicated only by slightly shorter leads This information can also be found within the diagram located in the Specifications, Files, and Documents section on this page.$8.95
Transistor - Peavey, SJE 5331, 120V, 3A
SJE 5331, 120V, 3A, Case-77.
$1.50 Transistor - Peavey, SJE 5332, 120V, 3A SJE 5332, 120V, 3A, Case-77.$1.65
Rectifier - Single-phase Bridge, 3A 400V
A full wave rectifier with 4 elements in a bridge circuit so that DC voltage is obtained from one pair of junctions when an AC voltage is applied to another pair of junctions.
$0.95 Triac - Q6025P5 - Alternistor, 400V, 25A, Fastpack TO-3 Base Alternistors are specifically designed for applications which are required to switch highly inductive loads. To accomplish this, a special chip has been designed which effectively offers the same performance as two thyristors (SCRs) wired inverse parallel (back to back); hence the alternistor has better turn-off behavior than a standard triac.$14.95
Integrated Circuit - Marshall, TDA7293 amplifier
120V - 100W DMOS audio amplifier with mute/standby. The TDA7293 is a monolithic integrated circuit in Multiwatt15 package, intended for use as audio class AB amplifier in Hi-Fi field applications, such as home stereo, self powered loudspeakers and Topclass TV. Thanks to the wide voltage range and to the high output current capability it is able to supply the highest power into both 4-Ω and 8-Ω loads.
$11.95 Transistor - MPSA13, Darlington 500mA 30V NPN NPN Darlington transistor (TO-92 package) - designed for applications requiring extremely high current gain at collector currents to 1.0A. • Max collector to emitter voltage = 30 V • Max collector to base voltage = 30 V • Max emitter to base voltage = 10 V • Max collector current = 1.2 A (continuous)$0.30
Op-Amp - TL072, Dual, Low-Noise, JFET Input, 8-Pin DIP
The TL072is a high speed JFET input dual operational amplifier incorporating well matched, high voltage JFET and bipolar transistors in a monolithic integrated circuit. The device features high slew rates, low input bias and offset current, and low offset voltage temperature coefficient.
$0.70 Diode - General purpose rectifier, 1A, 1N4005, 1N4006, 1N4007 A semiconductor device which passes current in the forward direction(from anode to cathode), and blocks current in the opposite direction. Starting at$0.15
Op-Amp - 4558, Dual, 8-Pin DIP
Dual Operational Amplifier. The 4558 integrated circuit is a dual high-gain operational amplifier internally compensated and constructed on a single silicon chip using an advanced epitaxial process. Combining the features of the NJM741 with the close parameter matching and tracking of a dual device on a monolithic chip results in unique performance characteristics. Excellent channel separation allows the use of the dual device in single NJM741 operational amplifier applications providing density. It is especially well suited for applications in differential-in, differential-out as well as in potentiometric amplifiers and where gain and phase matched channels are mandatory. Features:
• Operating Voltage (±4V$±18V) • High Voltage Gain (100dB typ.) • High Input Resistance (5MΩ typ.) • Bipolar Technology $0.75 Diode - small signal fast switching, 1N4148A A semiconductor device which passes current in the forward direction(from anode to cathode), and blocks current in the opposite direction. For extreme fast switches.$0.50 Optocoupler - VTL5C1, Vactrol, for Channel Switching VTL5C1 vactrols for use in amplifiers, effect pedals, synthesizers and more. The VTL5C1 was commonly used for channel switching circuits in amps made by Crate and others. $7.95 Transistor - Mosfet BS170, Small Signal MOSFET BS170 N-Channel Transistor. Used in MOD® Kits. Note - this is a fragile component! Use caution not to overheat or break leads.$1.25 Diode - Silicon Rectifier, 3A, 400V, 1N5404 A semiconductor device which passes current in the forward direction(from anode to cathode), and blocks current in the opposite direction. $0.35 Diode - Zener, 1W, 1N4741A, 1N4748A A semiconductor device that, above a certain reverse voltage(the zener value) has a sudden rise in current. If forward biased, the diode is an ordinary rectifier. When reverse biased, it exhibits a sharp break in its current-voltage graph. Voltage across the diode remains essentially constant for any further increase of reverse current.$0.15 Transistor - 2N3055, High Power NPN Silicon, TO-3 Case 15 A, 60 V, 115 W EPIBASE - Designed for power amplifier and switching regulator applications. The best choice where high frequency response, low switching losses and good safe operating area are required. • Current Gain - Bandwidth Product - fT = 2.5 MHz (Min) @ Ic = 1.0 Adc • Safe Operating Area - Full Power Rating to 40 V $2.25 Regulator - 5 Volt, used in MOD® Pedal Kits 5V Linear Regulator. Used in MOD® kits. This fixed-voltage monolithic integrated circuit voltage regulator is suitable for applications that required supply current up to 100 mA.$1.25 Diode - Hexfred, ultrafast soft recovery, 8A, 600V High performance, ultrafast diodes. Improve efficiency, reduce noise, allow higher frequency operation. $2.75 Op-Amp - NE5532, Dual, Low-Noise, 8-Pin DIP This a high performance dual low noise operational amplifier. Compared to the standard dual operational amplifiers, such as the NJM1458, it shows better noise performance, improved output drive capability, and considerably higher small-signal and power bandwidths. it is compensated internally for voltage follower circuit. This makes the device especially suitable for application in high quality and professional audio equipment, instrumentation, control circuits, and telephone channel amplifiers Features: • Operating Voltage = ±3V$±22V • Small Signal Bandwidth = 10MHz typ. • Output Drive Capability = 600Ω,10Vrms typ. • Input Noise Voltage = 5nV/√Hz typ. • Power Bandwidth = 140kHz typ. • Slew Rate = 8V/µs typ. • Bipolar Technology$1.15
Diode - Fast Recovery, 1A, 600V, 1N4937
A semiconductor device that, above a certain reverse voltage(the zener value) has a sudden rise in current. If forward biased, the diode is an ordinary rectifier. When reverse biased, it exhibits a sharp break in its current-voltage graph. Voltage across the diode remains essentially constant for any further increase of reverse current.
$0.20 Transistor - JFET N-Ch Transistor Lo Freq / Lo Noise, PF5102 PF5102 N-Channel Switch (TO-92 package) - designed for low level analog switching, sample and hold circuits and chopper stabilized amplifiers. • Max drain to gate voltage = 40 V • Max gate to source voltage = -40 V • Max forward gate current = 50 m$0.75
On Backorder
Don't see what you're looking for? Send us your product suggestions!


text:
# Making Use of the Cube Root Table, Find the Cube Root 833 . - Mathematics

Sum

Making use of the cube root table, find the cube root
833 .

#### Solution

We have: $830 < 833 < 840 \Rightarrow \sqrt[3]{830} < \sqrt[3]{833} < \sqrt[3]{840}$

From the cube root table, we have: $\sqrt[3]{830} = 9 . 398 \text{ and } \sqrt[3]{840} = 9 . 435$

For the difference (840 - 830), i.e., 10, the difference in values
$= 9 . 435 - 9 . 398 = 0 . 037$

∴  For the difference (833 - 830), i.e., 3, the difference in values

$= \frac{0 . 037}{10} \times 3 = 0 . 0111 = 0 . 011$  (upto three decimal places)
∴  $\sqrt[3]{833} = 9 . 398 + 0 . 011 = 9 . 409$ .
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 4 Cubes and Cube Roots
Exercise 4.5 | Q 21 | Page 36


text:
## Latest Posts

### Building an instrument to map the universe in 3-D

Wednesday, May 27th, 2015

The future Dark Energy Spectroscopic Instrument will be mounted on the Mayall 4-meter telescope. It will be used to create a 3-D map of the universe for studies of dark energy. Photo courtesy of NOAO

Dark energy makes up about 70 percent of the universe and is causing its accelerating expansion. But what it is or how it works remains a mystery.

The Dark Energy Spectroscopic Instrument (DESI) will study the origins and effects of dark energy by creating the largest 3-D map of the universe to date. It will produce a map of the northern sky that will span 11 billion light-years and measure around 25 million galaxies and quasars, extending back to when the universe was a mere 3 billion years old.

Once construction is complete, DESI will sit atop the Mayall 4-Meter Telescope in Arizona and take data for five years.

DESI will work by collecting light using optical fibers that look through the instrument’s lenses and can be wiggled around to point precisely at galaxies. With 5,000 fibers, it can collect light from 5,000 galaxies at a time. These fibers will pass the galaxy light to a spectrograph, and researchers will use this information to precisely determine each galaxy’s three-dimensional position in the universe.

Lawrence Berkeley National Laboratory is managing the DESI experiment, and Fermilab is making four main contributions: building the instrument’s barrel, packaging and testing charge-coupled devices, or CCDs, developing an online database and building the software that will tell the fibers exactly where to point.

The barrel is a structure that will hold DESI’s six lenses. Once complete, it will be around 2.5 meters tall and a meter wide, about the size of a telephone booth. Fermilab is assembling both the barrel and the structures that will hold it on the telescope.

“It’s a big object that needs to be built very precisely,” said Gaston Gutierrez, a Fermilab scientist managing the barrel construction. “It’s very important to position the lenses very accurately, otherwise the image will be blurred.”

DESI’s spectrograph will use CCDs, sensors that work by converting light collected from distant galaxies into electrons, then to digital values for analysis. Fermilab is responsible for packaging and testing these CCDs before they can be assembled into the spectrograph.

Fermilab is also creating a database that will store information required to operate DESI’s online systems, which direct the position of the telescope, control and read the CCDs, and ensure proper functioning of the spectrograph.

Lastly, Fermilab is developing the software that will convert the known positions of interesting galaxies and quasars to coordinates for the fiber positioning system.

Fermilab completed these same tasks when it built the Dark Energy Camera (DECam), an instrument that currently sits on the Victor Blanco Telescope in Chile, imaging the universe. Many of these scientists and engineers are bringing this expertise to DESI.

“DESI is the next step. DECam is going to precisely measure the sky in 2-D, and getting to the third dimension is a natural progression,” said Fermilab’s Brenna Flaugher, project manager for DECam and one of the leading scientists on DESI.

These four contributions are set to be completed by 2018, and DESI is expected to see first light in 2019.

“This is a great opportunity for students to learn the technology and participate in a nice instrumentation project,” said Juan Estrada, a Fermilab scientist leading the DESI CCD effort.

DESI is funded largely by the Department of Energy with significant contributions from non-U.S. and private funding sources. It is currently undergoing the DOE CD-2 review and approval process.

“We’re really appreciative of the strong technical and scientific support from Fermilab,” said Berkeley Lab’s Michael Levi, DESI project director.

Diana Kwon

### Drell-Yan, Drell-Yan with Jets, Drell-Yan with all the Jets

Monday, May 18th, 2015

All those super low energy jets that the LHC cannot see? LHC can still see them.

Hi Folks,

Particle colliders like the Large Hadron Collider (LHC) are, in a sense, very powerful microscopes. The higher the collision energy, the smaller distances we can study. Using less than 0.01% of the total LHC energy (13 TeV), we see that the proton is really just a bag of smaller objects called quarks and gluons.

This means that when two protons collide things are sprayed about and get very messy.

One of the most important processes that occurs in proton collisions is the Drell-Yan process. When a quark, e.g., a down quark d, from one proton and an antiquark, e.g., an down antiquark d, from an oncoming proton collide, they can annihilate into a virtual photon (γ) or Z boson if the net electric charge is zero (or a W boson if the net electric charge is one). After briefly propagating, the photon/Z can split into a lepton and its antiparticle partner, for example into a muon and antimuon or electronpositron pair! In pictures, quark-antiquark annihilation into a lepton-antilepton pair (Drell-Yan process) looks like this

By the conservation of momentum, the sum of the muon and antimuon momenta will add up to the photon/Z boson  momentum. In experiments like ATLAS and CMS, this gives a very cool-looking distribution

Plotted is the invariant mass distribution for any muon-antimuon pair produced in proton collisions at the 7 TeV LHC. The rightmost peak at about 90 GeV (about 90 times the proton’s mass!) is a peak corresponding to the production Z boson particles. The other peaks represent the production of similarly well-known particles in the particle zoo that have decayed into a muon-antimuon pair. The clarity of each peak and the fact that this plot uses only about 0.2% of the total data collected during the first LHC data collection period (Run I) means that the Drell-Yan process is a very useful for calibrating the experiments. If the experiments are able to see the Z boson, the rho meson, etc., at their correct energies, then we have confidence that the experiments are working well enough to study nature at energies never before explored in a laboratory.

However, in real life, the Drell-Yan process is not as simple as drawn above. Real collisions include the remnants of the scattered protons. Remember: the proton is bag filled with lots of quarks and gluons.

Gluons are what holds quarks together to make protons; they mediate the strong nuclear force, also known as quantum chromodynamics (QCD). The strong force is accordingly named because it requires a lot of energy and effort to overcome. Before annihilating, the quark and antiquark pair that participate in the Drell-Yan process will have radiated lots of gluons. It is very easy for objects that experience the strong force to radiate gluons. In fact, the antiquark in the Drell-Yan process originates from an energetic gluon that split into a quark-antiquark pair. Though less common, every once in a while two or even three energetic quarks or gluons (collectively called jets) will be produced alongside a Z boson.

Here is a real life Drell-Yan (Z boson) event with three very energetic jets. The blue lines are the muons. The red, orange and green “sprays” of particles are jets.

As likely or unlikely it may be for a Drell-Yan process or occur with additional energetic jets, the frequency at which they do occur appear to match very well with our theoretical predictions. The plot below show the likelihood (“Production cross section“) of a W or Z boson with at least 0, 1, 2, 3, or 4(!) very energetic jets. The blue bars are the theoretical predictions and the red circles are data. Producing a W or Z boson with more energetic jets is less likely than having fewer jets. The more jets identified, the smaller the production rate (“cross section”).

How about low energy jets? These are difficult to observe because experiments have high thresholds for any part of a collision to be recorded. The ATLAS and CMS experiments, for example, are insensitive to very low energy objects, so not every piece of an LHC proton collision will be recorded. In short: sometimes a jet or a photon is too “dim” for us to detect it. But unlike high energy jets, it is very, very easy for Drell-Yan processes to be accompanied with low energy jets.

There is a subtlety here. Our standard tools and tricks for calculating the probability of something happening in a proton collision (perturbation theory) assumes that we are studying objects with much higher energies than the proton at rest. Radiation of very low energy gluons is a special situation where our usual calculation methods do not work. The solution is rather cool.

As we said, the Z boson produced in the quark-antiquark annihilation has much more energy than any of the low energy gluons that are radiated, so emitting a low energy gluon should not affect the system much. This is like massive freight train pulling coal and dropping one or two pieces of coal. The train carries so much momentum and the coal is so light that dropping even a dozen pieces of coal will have only a negligible effect on the train’s motion. (Dropping all the coal, on the other hand, would not only drastically change the train’s motion but likely also be a terrible environmental hazard.) We can now make certain approximations in our calculation of a radiating a low energy gluon called “soft gluon factorization“. The result is remarkably simple, so simple we can generalize it to an arbitrary number of gluon emissions. This process is called “soft gluon resummation” and was formulated in 1985 by Collins, Soper, and Sterman.

Low energy gluons, even if they cannot be individually identified, still have an affect. They carry away energy, and by momentum conservation this will slightly push and kick the system in different directions.

If we look at Z bosons with low momentum from the CDF and DZero experiments, we see that the data and theory agree very well! In fact, in the DZero (lower) plot, the “pQCD” (perturbative QCD) prediction curve, which does not include resummation, disagrees with data. Thus, soft gluon resummation, which accounts for the emission of an arbitrary number of low energy radiations, is important and observable.

In summary, Drell-Yan processes are a very important at high energy proton colliders like the Large Hadron Collider. They serve as a standard candle for experiments as well as a test of high precision predictions. The LHC Run II program has just begun and you can count on lots of rich physics in need of studying.

Happy Colliding,

Richard (@bravelittlemuon)

### Anti-Matters: The Latest and Greatest from the Alpha Magnetic Spectrometer

Friday, May 8th, 2015

This past month in Geneva a conference took place bringing together the world’s foremost experiments in cosmic ray physics and indirect dark matter detection: “AMS Days at CERN”. I took a break from thesis-writing, grabbed a bag of popcorn, and sat down to watch a couple of the lectures via webcast. There was a stellar lineup, including but not limited to talks from IceCube, the Pierre Auger Observatory, H.E.S.S. and CTA, Fermi-LAT, and CREAM. The Alpha Magnetic Spectrometer (AMS) experiment was, of course, the star of the show. It is the AMS and its latest results that I’d like to focus on now.

But first, I’d like to give a brief introduction to cosmic rays, since that’s what AMS studies.

It turns out that space is not as empty as one might think. The Earth is constantly being bombarded by extremely-high-energy particles from all directions.  These cosmic rays were discovered in the early twentieth century by the Austrian physicist Victor Hess. Hess made several balloon-borne measurements of the Earth’s natural radiation at various altitudes and observed that the incidence of ionizing radiation actually increased with ascent, the exact opposite of what you would expect if all radioactivity came from the earth.

Fig. 1: An artist’s rendition of cosmic rays . Image from http://apod.nasa.gov.

The word “ray” is actually something of a misnomer – Cosmic rays are primarily charged matter particles rather than electromagnetic radiation. Their makeup goes as follows: approximately 98% are nuclei, of which 90% of are protons, 9% are alpha particles (helium nuclei), and only a small proportion heavier nuclei; and approximately 2% electrons and positrons. Only very small trace amounts (less than one ten-thousandth the number of protons) of antimatter are present, and of this, it is all positrons and antiprotons – not a single antihelium or heavier anti-nucleus has been discovered. There are two types of cosmic rays: primary rays, which come directly from extrasolar sources, and secondary rays, which come from primary rays crashing into the interstellar medium and forming new particles through processes such as nuclear spallation. Particles resulting from cosmic ray collisions with the Earth’s atmosphere are also considered secondary cosmic rays – these include particles like pions, kaons, and muons, and their decay products.

Fig. 2: Cosmic ray flux vs. particle energy. Image from http://science.nasa.gov/science-news/science-at-nasa/2001/ast15jan_1/

Despite being discovered over a hundred years ago, cosmic rays remain in a lot of ways a big mystery. For one thing, we don’t know exactly where they come from. Because cosmic rays are generally electrically charged, they don’t travel to us straight from the source. Rather, they are accelerated this way and that by magnetic fields in space so that when they finally reach us they could be coming from any direction at all. Indeed, the cosmic ray flux that we see is completely isotropic, or the same in all directions.

Not only do they not come straight from the source, but we don’t even know what that source is. These particles move orders of magnitude faster than particles in our most powerful accelerators on Earth. Astronomers’ best guess is that cosmic rays are accelerated by magnetic shocks from supernovae. But even supernovae aren’t enough to accelerate the highest-energy cosmic rays. Moreover, there are features in the cosmic ray energy spectrum that we just don’t understand (see Fig. 2). Two kinks, a “knee” at about 1016 eV and an “ankle” at about 1018 eV could indicate the turning on or off of some astrophysical process. Experiments like the Pierre Auger Observatory were designed to study these ultra-high-energy particles and hopefully will tell up a little bit more about them in the next few years.

The AMS is primarily interested in lower-energy cosmic rays. For four years, ever since its launch up to the International Space Station, it’s been cruising the skies and collecting cosmic rays by the tens of billions. I will not address the experimental design and software here. Instead I refer the reader to one of my previous articles, “Dark Skies II- Indirect Detection and the Quest for the Smoking Gun”.

In addition to precision studies of the composition and flux of cosmic rays, the AMS has three main science goals: (1) Investigating the matter-antimatter asymmetry by searching for primordial antimatter. (2) Searching for dark matter annihilation products amidst the cosmic rays. And (3), looking for strangelets and other exotic forms of matter.

The very small fraction of cosmic rays made up of antimatter is relevant not just for the first goal but for the second as well. Not many processes that we know about can produce positrons and antiprotons, but as I mention in “Dark Skies II”, dark matter annihilations into Standard Model particles could be one of those processes. Any blips or features in the cosmic ray antimatter spectrum could indicate dark matter annihilations at work.

Fig. 3. The positron fraction measured by AMS. Image from L. Accardo et al. (AMS Collaboration), September 2014.

On April 14 at “AMS Days at CERN”, Professor Andrei Kounine of MIT presented the latest results from AMS.

The first part of Kounine’s talk focused on a precise characterization of the positron fraction presented by the AMS collaboration in September 2014 and a discussion of the relevant systematics. In the absence of new physics processes, we expect the positron fraction to be smooth and decreasing with energy. As you can see in Fig. 3, however, the positron fraction starts rising at approximately 8 GeV and increases steadily up to about 250 GeV. The curve hits a maximum at about 275 GeV and then appears to begin to turn over, although at these energies the measurements are limited by statistics and more data is needed to determine exactly what happens beyond this point. Models of dark matter annihilation predict a much steeper drop-off than do models where the positron excess is produced by, say, pulsars. Five possible sources of systematic error were identified, all of which have been heavily investigated. These included a small asymmetry in positron and electron acceptance due to slight differences in some of the bits of the tracker; variations in efficiency with respect to energy of the incoming particle; binning errors, which are mitigated due to high experimental resolution; low statistics at the tails of the electron and positron distributions; and “charge confusion”, or the misidentification of electrons as positrons, which happens only in a very small number of cases.

Kounine also presented a never-before-seen, not-yet-published measurement of the antiproton-proton ratio as measured by AMS, which you can see in Fig. 4. This curve represents a total of 290,000 antiprotons selected out of total of 54 billion events collected by AMS over the past 4 years. Many of the same systematics (acceptance asymmetry, charge confusion, and so on) as in the positron measurement are relevant here. Work on the antiproton analysis is ongoing, however, and according to Kounine it’s too soon to try to match models to the data.

Fig. 4. AMS’s latest antiproton-proton ratio measurement, from Prof. Andrei Kounine’s presentation at “AMS Days at CERN”.

As a dark matter physicist, the question in my mind is, do these measurements represent dark matter annihilations? Professor Subir Sarkar of Oxford and the Niels Bohr Institute in Copenhagen thinks not. In his talk at “AMS Days”, Sarkar argues that the dark matter annihilation cross-section necessary to match the positron flux seen by AMS and other experiments such as Fermi-LAT and PAMELA needs to be so large that by all rights the dark matter in the universe should have all annihilated away already. This is inconsistent with the observed dark matter density in our galaxy. You can get around this with theoretical models that incorporate new kinds of long-range forces. However, the observed antiproton flux, according to Sarkar, is consistent with background. Therefore dark matter would have to be able to annihilate into leptons (electrons and positrons, muons, neutrinos, and so on) but not quarks. Such models exist, but now we’re starting to severely restrict our model space. Moreover, dark matter annihilating in the early universe near the time of recombination should leave visible imprints in the Cosmic Microwave Background (CMB), which have not yet been seen. CMB experiments such as Planck therefore disfavor a dark matter explanation for the observed peak in positron fraction.

Sarkar then goes on to present an alternate model where secondary cosmic ray particles such as positrons are accelerated by the same mechanisms (magnetic shocks from supernovae, pulsars, and other cosmic accelerators) that accelerate primary cosmic rays. Then, if there are invisible accelerators in our nearby galactic neighborhood, as seems likely because electrons and positrons can’t propagate very far without losing energy due to interactions with starlight and the CMB, it could be possible to get very large fluctuations in the cosmic ray flux due purely to the randomness of how these accelerators are distributed around us.

Regardless of whether or not the AMS has actually seen a dark matter signal, the data are finally beginning to be precise enough that we can start really pinning down how cosmic rays backgrounds are created and propagated. I encourage you to check out some of the webcasts at “AMS Days at CERN” for yourself. Although the event is over the webcasts are still available in the CERN document archive here.

### New beamline at Fermilab Test Beam Facility delivers beam to LArIAT, a liquid-argon detector for neutrino research

Tuesday, May 5th, 2015

Technicians John Cornele, Pat Healey and Skyler Sherwin have been crucial in preparing the LArIAT detector for beam. The liquid-argon-filled detector saw first beam on Thursday. Photo: Jen Raaf

Fermilab’s Test Beam Facility (FTBF) now runs a second beamline to provide particles for R&D experiments. The MCenter beamline came back to life last year after an eight-year slumber to join the facility’s other beamline, MTest.

On Thursday, April 30, accelerator operators began using the revived beamline to send particles to its first major experiment, Liquid Argon TPC in a Test Beam (LArIAT), which will help advance particle detector technologies for neutrino experiments.

The FTBF provides experiments with different types of particle beams with a range of energies. Its main purpose is the research and development of particle detectors. It is one of only two sites in the world that provides this service with high-energy hadrons, which are particles made of quarks. Since 2005, the FTBF, with its distinctive orange and blue corrugated-steel roof, has staged more than 50 experiments, conducted by scientists from more than 170 institutions in 30 countries.

“We’re very busy and fully subscribed,” said JJ Schmidt, deputy facility manager at FTBF. “The existence of two beams allows us to serve a broader class of experiments.”

Not only does the new beamline allow FTBF to serve a larger number of users, it also provides room for a greater diversity of experiments. While MTest is aimed at experiments with a turnover of about one to four weeks, MCenter caters to more long-term experiments like LArIAT that will last for months, or even years.

Beautiful tracks at first try
LArIAT is a liquid-argon time projection chamber. Charged particles traveling through the sea of liquid argon ionize the argon atoms, and an electric field causes liberated electrons to drift toward the detector readout. Different particles cause different amounts of ionization, allowing researchers to distinguish between particles such as pions, kaons and protons.

This plot shows LArIAT’s first tracks: two views of a charged particle interacting inside the LArIAT detector, which is filled with liquid argon.

The first spill of particles delivered to LArIAT led to immediate success. The detector recorded picture-perfect tracks of charged particles.

Like the test beam, LArIAT will act as a research and development vehicle for future projects. Because neutrinos can be studied only through the particles produced when they interact with material inside a particle detector, being able to reliably characterize these other particles is of great importance.

“This is going to be fantastic not only for LArIAT but all the neutrino experiments that will use its results,” said Jen Raaf, co-spokesperson for LArIAT.

LArIAT will run the test beam for 24 hours a day while experimenters take data. The first run will last about three months, after which the detector’s cryogenic system will undergo upgrades to prepare for longer follow-up runs.

“It’s great that we have a facility where a small experiment can take beam over a long term,” said Brian Rebel, a scientist involved in LArIAT.

About 75 people from 22 institutions from the United States, Europe and Japan work on this experiment.

“Most are young postdocs and Ph.D. students that are enthusiastically doing a great job,” said Flavio Cavanna, LArIAT co-spokesperson.

“It’s an exciting combination of many years of work by the Accelerator, Particle Physics, Neutrino and Scientific Computing divisions to have the capability to do research that is important for making this the premier neutrino laboratory in the world,” Schmidt said.

Diana Kwon

### Final piece of large magnet built at Fermilab arrives at Jefferson Lab

Friday, May 1st, 2015

Fermilab Director Nigel Lockyer shakes hands with Jefferson Lab Director Hugh Montgomery by a superconducting coil and its development and fabrication team at Fermilab. Six coils have been made and shipped to Jefferson Lab for use in the CLAS12 experiment. Photo: Reidar Hahn

A group of Fermilab physicists and engineers was faced with a unique challenge when Jefferson Lab asked them to make the superconducting coils for an upgrade to their CEBAF Large Acceptance Spectrometer experiments. These are some of the largest coils Fermilab has ever built.

Despite obstacles, the sixth coil was completed, packed on a truck and sent to Jefferson Lab to become the last piece of the torus magnet in the lab’s CLAS detector. It arrived on Thursday.

The CLAS detector’s upgrade (CLAS12) will allow it to accept electron beams of up to 11 GeV, matching the beam energy of the Virginia laboratory’s CEBAF electron accelerator after five passes. These improvements will allow Jefferson Lab to more accurately study the properties of atomic nuclei.

A major component of the enhanced detector is the torus magnet, which will be made from the six superconducting coils created at Fermilab. Aside from cleaning, insulating and winding the coils, one of the most important parts of the process is vacuum epoxy impregnation. During this step, air and water vapor are removed from the coils and replaced with an epoxy.

This process is particularly difficult when you’re working on magnets as big as the CLAS12 coils, which are 14 feet long and seven feet wide. Fermilab’s Magnet Systems Department fabrication team, the group responsible for making these massive coils, encountered a major obstacle at the end of March 2014 after finishing the first practice coil.

What they found were dry areas within the coil where the epoxy couldn’t penetrate. These were places where the coils weren’t fixed into place, meaning they could move and generate heat and resistance. This can lead to magnet quench, the transition from superconducting to a normal state — a highly undesirable consequence.

The Fermilab group and Jefferson Lab staff collaborated to come up with a solution. By trying new materials, new temperature profiles and adjusting the time that the epoxy was left to sit and be adsorbed, the team was able to prevent the dry areas from forming.

Fred Nobrega, the lead engineer at Fermilab for the CLAS12 coil project, joined the effort last August.

“It was rewarding for me to join the project near its low point, be able to help get through the hurdle and see this completed,” he said.

Production has been steady since December, with Fermilab sending roughly one coil a month to Jefferson Lab. Although the sixth coil will become the last piece of the torus magnet, the project isn’t complete just yet — the ultimate goal is to make eight identical coils, the six for the magnet and two spares.

“We’re succeeding because we have great people and a productive collaboration with Jefferson Lab, who helped us at difficult moments,” said George Velev, head of the Magnet Systems Department. “We worked together on a tough problem and now we see the results.”

Diana Kwon

### Italian neutrino experiment to move to the US

Wednesday, April 22nd, 2015

The world’s largest liquid-argon neutrino detector will help with the search for sterile neutrinos at Fermilab. Photo: INFN

Mysterious particles called neutrinos seem to come in three varieties. However, peculiar findings in experiments over the past two decades make scientists wonder if a fourth is lurking just out of sight.

To help solve this mystery, a group of scientists spearheaded by Nobel laureate Carlo Rubbia plans to bring ICARUS, the world’s largest liquid-argon neutrino detector, across the Atlantic Ocean to the United States. The detector is currently being refurbished at CERN, where it is the first beneficiary of a new test facility for neutrino detectors.

Neutrinos are some of the most abundant and yet also most mysterious particles in the universe. They have tiny masses, but no one is sure why—or where those masses come from. They interact so rarely that they can pass through the entire Earth as if it weren’t there. They oscillate from one type to another, so that even if you start out with one kind of neutrino, it might change to another kind by the time you detect it.

Many theories in particle physics predict the existence of a sterile neutrino, which would behave differently from the three known types of neutrino.

“Finding a fourth type of neutrinos would change the whole picture we’re trying to address with current and future experiments,” says Peter Wilson, a scientist at Fermi National Accelerator Laboratory.

The Program Advisory Committee at Fermilab recently endorsed a plan, managed by Wilson, to place a suite of three detectors in a neutrino beam at the laboratory to study neutrinos—and determine whether sterile neutrinos exist.

Over the last 20 years, experiments have seen clues pointing to the possible existence of sterile neutrinos. Their influence may have caused two different types of unexpected neutrino behavior seen at the Liquid Scintillator Neutrino Detector experiment at Los Alamos National Laboratory in New Mexico and the MiniBooNE experiment at Fermilab.

Both experiments saw indications that a surprisingly large number of neutrinos may be morphing from one kind to another a short distance from a neutrino source. The existence of a fourth type of neutrino could encourage this fast transition.

The new three-detector formation at Fermilab could provide the answer to this mystery.

In the suite of experiments, a 260-ton detector called Short Baseline Neutrino Detector will sit closest to the source of the beam, so close that it will be able to detect the neutrinos before they’ve had a chance to change from one type into another. This will give scientists a baseline to compare with results from the other two detectors. SBND is under construction by a team of scientists and engineers from universities in the United Kingdom, the United States and Switzerland, working with several national laboratories in Europe and the US.

The SBND detector will be filled with liquid argon, which gives off flashes of light when other particles pass through it.

“Liquid argon is an extremely exciting technology to make precision measurements with neutrinos,” says University of Manchester physicist Stefan Soldner-Rembold, who leads the UK project building a large section of the detector. “It’s the technology we’ll be using for the next 20 to 30 years of neutrino research.”

Farther from the beam will be the existing 170-ton MicroBooNE detector, which is complete and will begin operation at Fermilab this year. The MicroBooNE detector was designed to find out whether the excess of particles seen by MiniBooNE was caused by a new type of neutrino or a new type of background. Identifying either would have major implications for future neutrino experiments.

Finally, farthest from the beam would be a liquid-argon detector more than four times the size of MicroBooNE. The 760-ton detector was used in the ICARUS experiment, which studied neutrino oscillations at Gran Sasso Laboratory in Italy using a beam of neutrinos produced at CERN from 2010 to 2014.

Its original beam at CERN is not optimized for the next stage of the sterile neutrino search. “The Fermilab beamline is the only game in town for this type of experiment,” says physicist Steve Brice, deputy head of Fermilab’s Neutrino Division.

And the ICARUS detector “is the best detector in the world to detect this kind of particle,” says Alberto Scaramelli, the former technical director of Gran Sasso National Laboratory. “We should use it.”

Rubbia, who initiated construction of ICARUS and leads the ICARUS collaboration, proposed bringing the detector to Fermilab in August 2013. Since then, the ICARUS, MicroBooNE and SBND groups have banded together to create the current proposal. The updated plan received approval from the Fermilab Program Advisory Committee in February.

“The end product was really great because it went through the full scrutiny of three different collaborations,” says MicroBooNE co-leader Sam Zeller. “The detectors all have complementary strengths.”

In December, scientists shipped the ICARUS detector from the Gran Sasso laboratory to CERN, where it is currently undergoing upgrades. The three-detector short-baseline neutrino program at Fermilab is scheduled to begin operation in 2018.

Kathryn Jepsen

### Mu2e breaks ground on experiment seeking new physics

Tuesday, April 21st, 2015

Fermilab’s Mu2e groundbreaking ceremony took place on Saturday, April 18. From left: Alan Stone (DOE Office of High Energy Physics), Nigel Lockyer (Fermilab director), Jim Siegrist (DOE Office of High Energy Physics director), Ron Ray (Mu2e project manager), Paul Philp (Mu2e federal project director at the Fermi Site Office), Jim Miller (Mu2e co-spokesperson), Doug Glenzinski (Mu2e co-spokesperson), Martha Michels (Fermilab ESH&Q head), Mike Shrader (Middough architecture firm), Julie Whitmore (Mu2e deputy project manager), Jason Whittaker (Whittaker Construction), Tom Lackowski (FESS). Photo: Reidar Hahn

This weekend, members of the Mu2e collaboration dug their shovels into the ground of Fermilab’s Muon Campus for the experiment that will search for the direct conversion of a muon into an electron in the hunt for new physics.

For decades, the Standard Model has stood as the best explanation of the subatomic world, describing the properties of the basic building blocks of matter and the forces that govern them. However, challenges remain, including that of unifying gravity with the other fundamental forces or explaining the matter-antimatter asymmetry that allows our universe to exist. Physicists have since developed new models, and detecting the direct conversion of a muon to an electron would provide evidence for many of these alternative theories.

“There’s a real possibility that we’ll see a signal because so many theories beyond the Standard Model naturally allow muon-to-electron conversion,” said Jim Miller, a co-spokesperson for Mu2e. “It’ll also be exciting if we don’t see anything, since it will greatly constrain the parameters of these models.”

Muons and electrons are two different flavors in the charged-lepton family. Muons are 200 times more massive than electrons and decay quickly into lighter particles, while electrons are stable and live forever. Most of the time, a muon decays into an electron and two neutrinos, but physicists have reason to believe that once in a blue moon, muons will convert directly into an electron without releasing any neutrinos. This is physics beyond the Standard Model.

Under the Standard Model, the muon-to-electron direct conversion happens too rarely to ever observe. In more sophisticated models, however, this occurs just frequently enough for an extremely sensitive machine to detect.

The Mu2e detector, when complete, will be the instrument to do this. The 92-foot-long apparatus will have three sections, each with its own superconducting magnet. Its unique S-shape was designed to capture as many slow muons as possible with an aluminum target. The direct conversion of a muon to an electron in an aluminum nucleus would release exactly 105 million electronvolts of energy, which means that if it occurs, the signal in the detector will be unmistakable. Scientists expect Mu2e to be 10,000 times more sensitive than previous attempts to see this process.

Construction will now begin on a new experimental hall for Mu2e. This hall will eventually house the detector and the infrastructure needed to conduct the experiment, such as the cryogenic systems to cool the superconducting magnets and the power systems to keep the machine running.

“What’s nice about the groundbreaking is that it becomes a real thing. It’s a long haul, but we’ll get there eventually, and this is a start,” said Julie Whitmore, deputy project manager for Mu2e.

The detector hall will be complete in late 2016. The experiment, funded mainly by the Department of Energy Office of Science, is expected to begin in 2020 and run for three years until peak sensitivity is reached.

“This is a project that will be moving along for many years. It won’t just be one shot,” said Stefano Miscetti, the leader of the Italian INFN group, Mu2e’s largest international collaborator. “If we observe something, we will want to measure it better. If we don’t, we will want to increase the sensitivity.”

Physicists around the world are working to extend the frontiers of the Standard Model. One hundred seventy-eight people from 31 institutions are coming together for Mu2e to make a significant impact on this venture.

“We’re sensitive to the same new physics that scientists are searching for at the Large Hadron Collider, we just look for it in a complementary way,” said Ron Ray, Mu2e project manager. “Even if the LHC doesn’t see new physics, we could see new physics here.”

Diana Kwon

See a two-minute video on the ceremony

### In Defense of Scientism and the Joys of Self-Publishing.

Friday, April 17th, 2015

As long-time readers of Quantum Diaries know I have been publishing here for a number of years and this is my 85th and last post[1]. A couple of years ago, I collected the then current collection, titled it “In Defense of Scientism,” after the title of one of the essays, and sent it off to a commercial publisher. Six months later, I got an e-mail from the editor complaining that he had lost the file and only found it by accident, and he somehow inferred that it was my fault. After that experience, it was no surprise he did not publish it.

With all the talk of self-publishing these days, I thought I would give it a try. It is easy, at least compared to finding the Higgs boson! There are a variety of options that give different levels of control, so one can pick and choose preferences – like off an á la carte menu. The simplest form of self-publishing is to go to a large commercial publisher.  The one I found would, for $50.00 USD up front and$12.00 a year, supply print on demand and e-books to a number of suppliers. Not sure that I could recover the costs from the revenue – and being a cheapskate – I decided not to go that route. There are also commissioned alternatives with no upfront costs, but I decided to interact directly with three (maybe four, if I can jump over the humps the fourth has put up) companies.  One of the companies treated their print-on-demand and digital distribution arms as distinct, even to the point of requiring different reimbursement methods. That is the disadvantage of doing it yourself, sorting it all out. The advantage of working directly with the suppliers is more control over the detailed formatting and distribution.

From then on things got fiddly[2], for example, reimbursement. Some companies would only allow payment by electronic fund transfer, others only by check. The weirdest example was one company that did electronic fund transfers unless the book was sold in Brazil or Mexico. In those cases, it is by check but only after \$100.00 has been accumulated. One company verified, during account setup, that the fund transfer worked by transferring a small amount, in my case 16 cents. And then of course there are special rules if you earn any money in the USA. For USA earnings there is a 30% withholding tax unless you can document that there is a tax treaty that allows you to get around it. The USA is the only country that requires this. Fine, being an academic, I am used to jumping through hoops.

Next was the question of an International Standard Book Number (ISBN). They are not required but are recommended. That is fine since in Canada you can get them for free. Just as well since each version of the book needs a different number. The paperback needs a different number from the electronic and each different electronic format requires its own number. As I said, it is a good thing it is free. Along with the ISBN, I got a reminder that the Library of Canada requires one copy of each book that sells more than four copies and two copies if it goes over a hundred and of course a separate electronic copy if you publish electronically. Fun, fun, fun[3]. There are other implications of getting you own ISBN number. Some of the publishers would supply an ISBN free of charge but then would put the book out under their own imprint and, in some cases, give wider distribution to those books. But again, getting your own number ultimately gives you more control.

With all this research in hand, it was time to create and format the content. I had the content from the four years’ worth of Quantum Diary posts and all I had to do was put it together and edit for consistency. Actually, Microsoft Word worked quite well with various formatting features to help. I then gave it to my wife to proofread. That was a mistake; she is still laughing at some of the typos. At least there is now an order of magnitude fewer errors. I should also acknowledge the many editorial comments from successive members of the TRIUMF communications team.

The next step was to design the book cover. There comes a point in every researcher’s career when they need support and talent outside of themselves. Originally, I had wanted to superimpose a picture of a model boat on a blackboard of equations. With that vision in mind, I set about the hallways to seek and enroll the talent of a few staff members who could make it happen. After normal working hours, of course. A co-op communication student suggested that the boat be drawn on the blackboard rather than a picture superimposed. The equations were already on a blackboard and are legitimate. The boat was hand drawn by a talented lady in accounting, drawing it first onto an overhead transparency[4] and then projecting it onto a blackboard. A co-op student in the communications team produced the final cover layout according to the various colour codes and margin bleeds dictated by each publisher. For both my own and your sanity, I won’t go into all the details. In the end, I rather like how the cover turned out.

Now, several months after I started the publishing process, I have jumped through all the hoops! All I have to do is lay back and let the money roll in so I can take early retirement. Well, at my age, early retirement is no longer a priori possible but at least I hope to get enough money to buy the people who helped me prepare the book a coffee. So everyone, please rush out and buy a copy. Come on, at least one of you.

As a final point, you may wonder why there is a drawing of a boat on the cover of a book about the scientific method. Sorry, to find out you will have to read the book. But I will give you a hint. It is not that I like to go sailing. I get seasick.

[1] I know, I have promised this before, but his time trust me. I am not like Lucy in the Charlie Brown cartoons pulling the football away.

[2] Epicurus, who made the lack of hassle the greatest good, would not have approved.

[3] Reminds me of an old Beach Boys song.

[4] An old overhead projector was found in a closet.

[5] Hey! We got through an entire conversation about formatting and word processing software without mentioning LaTeX despite me having been the TRIUMF LaTeX guru before I went over to the dark side and administration.

### Life Underground: Anything Anyone Would Teach Me

Friday, April 17th, 2015

Going underground most days for work is probably the weirdest-sounding this about this job. At Laboratori Nazionali del Gran Sasso, we use the lab to be underground because of the protection it affords us from cosmic rays, weather, and other disruptions, and with it we get a shorthand description of all the weirdness of lab life. It’s all just “underground.”

The last kilometer of road before reaching the above-ground labs of LNGS

Some labs for low background physics are in mines, like SURF where fellow Quantum Diariest Sally Shaw works. One of the great things about LNGS is that we’re located off a highway tunnel, so it’s relatively easy to reach the lab: we just drive in. There’s a regular shuttle schedule every day, even weekends. When there are snowstorms that close parts of the highway, the shuttle still goes, it just takes a longer route all the way to the next easy exit. The ride is a particularly good time to start drafting blog posts. On days when the shuttle schedule is inconvenient or our work is unpredictable, we can drive individual cars, provided they’ve passed emissions standards.

The guards underground keep a running list of all the people underground at any time, just like in a mine. So, each time I enter or leave, I give my name to the guards. This leads to some fun interactions where Italian speakers try to pronounce names from all over. I didn’t think too much of it before I got here, but in retrospect I had expected that any name of European etymology would be easy, and others somewhat more difficult. In fact, the difficult names are those that don’t end in vowels: “GladStone” become “Glad-eh-Stone-eh”. But longer vowel-filled names are fine, and easy to pronounce, even though they’re sometimes just waved off as “the long one” with a gesture.

There’s constantly water dripping in the tunnel. Every experiment has to be housed in something waterproof, and gutters line all the hallways, usually with algae growing in them. The walls are coated with waterproofing, more to keep any potential chemical spill from us from getting into the local groundwater than to keep the water off our experiments. When we walk from the tunnel entrance to the experimental halls, the cue for me to don a hardhat is the first drip on my head from the ceiling. Somehow, it’s always right next to the shuttle stop, no matter where the shuttle parks.

And, because this is Italy, the side room for emergencies has a bathroom and a coffee machine. There’s probably emergency air tanks too, but the important thing is the coffee machine, to stave off epic caffeine withdrawal headaches. And of course, “coffee” means “espresso” unless otherwise stated– but that’s another whole post right there.

When I meet people in the neighboring villages, at the gym or buying groceries or whatever, they always ask what an “American girl” is doing so far away from the cities, and “lavoro a Laboratorio Gran Sasso” is immediately understood. The lab is even the economic engine that’s kept the nearest village alive: it has restaurants, hotels, and rental apartments all catering to people from the lab (and the local ski lift), but no grocery stores, ATMs, gyms, or post offices that would make life more convenient for long-term residents.

Every once in a while, when someone mentions going underground, I can’t help thinking back to the song “Underground” from the movie Labyrinth that I saw too many times growing up. Labyrinth and The Princess Bride were the “Frozen” of my childhood (despite not passing the Bechtel test).

Just like Sarah, my adventures underground are alternately shocking and exactly what I expected from the stories, and filled with logic puzzles and funny characters. Even my first night here, when I was delirious with jetlag, I saw a black cat scamper across a deserted medieval street, and heard the clock tower strike 13 times. And just like Wesley, “it was a fine time for me, I was learning to fence, to fight–anything anyone would teach me–” (except that in my case it’s more soldering, cryogenics plumbing, and ping-pong, and less fighting). The day hasn’t arrived where the Dread Pirate Roberts calls me to his office and gives me a professorship.

And now the shuttle has arrived back to the office, so we’re done. Ciao, a dopo.

(ps the clock striking 13 times was because it has separate tones for the hour and the 15-minute chunks. The 13 was really 11+2 for 11:30.)

### Building a Neutrino Detector

Thursday, April 16th, 2015

Ever wanted to see all the steps necessary for building a neutrino detector? Well now you can, check out this awesome video of constructing the near detector for the Double Chooz reactor neutrino experiment in France.

This is the second of two identical detectors near the Chooz nuclear power station in northern France. The experiment, along with competing experiments, already showed that the neutrino mixing angle, Theta_13, was non-zero. A second detector measuring the same flux of neutrinos from the two reactor cores will drastically reduce the final measurement uncertainty.


text:
AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic Unit Exercise Textbook Questions and Answers.

## AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic Unit Exercise

Question 1.
Draw a four sided closed figure and divide it into some number of equal parts. Shade the figure with any colour so that the ratio of shaded parts to unshaded parts is 1 : 3.

Given ratio of shaded parts: unshaded parts = 1 : 3
Total parts of a closed figure = 1 + 3 = 4
Consider a four sided closed figure square.
Divide it into 4 parts and shade one part.
Now, from the figure, shaded parts : unshaded parts = 1 : 3

Question 2.
Ramu spent $$\frac{2}{5}$$th of his money on a story book. Find the ratio of money spent to the money with him at present.
Given money spent on story book = $$\frac{2}{5}$$ th of his money.
That means from total 5 parts of money he spent 2 parts of money on story book. Then, out of 5 parts remaining parts of money with him is 3 parts.
So, ratio of money spent to the money with him at present
= parts of money spent : parts of money at him = 2 : 3

Question 3.
Divide Rs. 72,000 between Kesav and David in the ratio of 5 : 4.
Amount to be divided = Rs. 72,000
Ratio of Kesav and David’s Shares = 5 : 4
Total shares = 5 + 4 = 9
Kesav’s share = $$\frac{5}{9}$$ of 72000
$$\frac{5}{9}$$ × 72,000 = 40,000/-
David’s share = $$\frac{4}{9}$$ of 72,000
$$\frac{4}{9}$$ × 72,000 = 32,000/-

Question 4.
The income of Kumar for 3 months is Rs. 15,000. If he earns the same amount for a month,
a) How much will he earn in 5 months?
b) In how many months, will he earn Rs. 95,000?
Given Kumar’s 3 months income = Rs. 15,000
Kumar’s one month income = 15,000 ÷ 3
= $$\frac{15000}{3}$$ = Rs. 5000
a) Kumar’s one month income = Rs. 5000
Kumar’s 5 months income = 5 × 5000 = Rs. 25,000
b) Time taken by Kumar to earn Rs. 5000 = 1 month
Time taken by Kumar to earn Rs. 95,000 = 95,000 ÷ 5000
= $$\frac{15000}{3}$$ = 19 months = 1 year 7 months.

Question 5.
The cost of 16 chairs is Rs. 4,800. Find the number of chairs that can be purchased for Rs. 6,600.
Cost of 16 chairs = Rs. 4800
Number of chairs that can be pruchased for Rs. 6600 = x (say)
∴ 16 : 4800 : : x : 6600
So, 16 x 6600 = 4800 x x
x = $$\frac{16 \times 6600}{4800}$$ = 22 chairs
(OR)
Given cost of 16 chairs = Rs. 4800
Cost of one chair = 4800 ÷ 16
= $$\frac{4800}{3}$$ = Rs. 300
Number of chairs can purchased for Rs. 6600 = 6600 ÷ cost of one chair
= 6600 ÷ 300
= $$\frac{6600}{300}$$ = 22 chairs

Question 6.
What percentage of numbers from 1 to 30 has 1 or 9 in the unit’s digit?
Total number from 1 to 30 = 30
Numbers ending with 1 or 9 are 1, 9,11, 19, 21, 29 = 6 numbers
Ratio = 6 : 30 = $$\frac{6}{30}$$ = $$\frac{1}{5}$$
Percentage = $$\frac{1}{5}$$ × 100% = 20%

Question 7.
In a college, 63% of students are less than 20 years of age. The number of students more than 20 years of age is — of number of students of 20 years of age which is 42. What is the total number of students in the college?
Then, 37% of x = 42 + $$\frac{2}{3}$$ of 48
$$\frac{37}{100}$$ × x = 74
∴ x = 74 × $$\frac{100}{37}$$ = 200


text:
## Installation

by Alex Jordan -
Number of replies: 5

I've installed MariaDB (apt install mariadb-server) and that seems to have been successful.

Now I am trying to install DBD::MariaDB using MCPAN. I get:

cpan[1]> install DBD::MariaDBReading '/root/.cpan/Metadata'  Database was generated on Wed, 16 Jun 2021 01:29:02 GMTRunning install for module 'DBD::MariaDB'Checksum for /root/.cpan/sources/authors/id/P/PA/PALI/DBD-MariaDB-1.21.tar.gz okScanning cache /root/.cpan/build for sizes............................................................................DONEConfiguring P/PA/PALI/DBD-MariaDB-1.21.tar.gz with Makefile.PLCannot find the file 'mariadb_config' nor 'mysql_config'! Your execution PATH doesn't seemnot contain the path to mariadb_config or mysql_config. Resorting to guessed values!PLEASE NOTE:For 'make test' to run properly, you must ensure that thedatabase user 'root' can connect to your MariaDB or MySQL serverand has the proper privileges that these tests require suchas 'drop table', 'create table', 'drop procedure', 'create procedure'as well as others.mysql> grant all privileges on test.* to 'root'@'localhost' identified by 's3kr1t';You can also optionally set the user to run 'make test' with:perl Makefile.PL --testuser=usernameFailed to determine directory of mysql.h. Use  perl Makefile.PL --cflags=-I<dir>to set this directory. For details see DBD::MariaDB::INSTALL,section "C Compiler flags" or type  perl Makefile.PL --helpWarning: No success on command[/usr/bin/perl Makefile.PL INSTALLDIRS=site]  PALI/DBD-MariaDB-1.21.tar.gz  /usr/bin/perl Makefile.PL INSTALLDIRS=site -- NOT OKFailed during this command: PALI/DBD-MariaDB-1.21.tar.gz                 : writemakefile NO '/usr/bin/perl Makefile.PL INSTALLDIRS=site' returned status 512

This doesn't seem right. How can I get it to find those config files? "Resorting to guessed values" is not working out. If I try to go to the admin course for example, I get:

error instantiating DB driver WeBWorK::DB::Driver::SQL for table achievement_user: install_driver(MariaDB) failed: Can't locate DBD/MariaDB.pm in @INC (you may need to install the DBD::MariaDB module) (@INC contains: /opt/webwork/pg/lib /opt/webwork/webwork2/lib /etc/perl /usr/local/lib/x86_64-linux-gnu/perl/5.26.1 /usr/local/share/perl/5.26.1 /usr/lib/x86_64-linux-gnu/perl5/5.26 /usr/share/perl5 /usr/lib/x86_64-linux-gnu/perl/5.26 /usr/share/perl/5.26 /usr/local/lib/site_perl /etc/apache2) at (eval 996) line 3. Perhaps the DBD::MariaDB perl module hasn't been fully installed, or perhaps the capitalisation of 'MariaDB' isn't right. Available drivers: DBM, ExampleP, File, Gofer, Mem, Proxy, Sponge, mysql. at /opt/webwork/webwork2/lib/WeBWorK/DB/Driver/SQL.pm line 79. at /opt/webwork/webwork2/lib/WeBWorK.pm line 317.

by Alex Jordan -

After going down too many roads to recount here, I was able to return my server to using mysql 5.7 and I will just leave it at that for now. Shout out to Danny who helped me out.

In the end, I completely reinstalled mysql. Then I placed a backup of my /var/lib/mysql folder back in place at /var/lib/mysql. It appears that with active courses and new courses, things are working.

However now when I unarchive an old course, it has database structure errors and won't initially load. I can use the admin course to update the course I unarchived. But the course has lost its users and problem sets. I had to use the command line wwsh to add myself back into the course, where I can get in to see that these things are missing. Files form the course are still there, it's just the database elements that are gone.

A less concerning issue is that I seem to not be able to use extended characters. I tried creating a course where the title had an em dash — in its title. There was no error, but then that course's title appeared with some accented letter a in place of the em dash. So character encoding is not behaving as hoped for.

by Danny Glin -

I can't offer too much help since my system is still running 2.14, and I haven't braved the migration to UTF8.

It sounds like there is something going on with the character sets between your existing courses and the new version of WeBWorK.

There are two pieces to this:

1. Changing the default character set for the database
2. Changing the tables themselves to use the new character set (utf8mb4)

There are two ways to accomplish this:

1. By manually following the instructions at https://webwork.maa.org/wiki/Converting_the_webwork_database_from_the_latin1_to_the_utf8mb4_character_set
2. Using the upgrade-database-to-utf8mb4.pl script, which can be found in the /opt/webwork/webwork2/bin directory

I recommend doing a mysqldump of your whole database before trying any of these changes, as it is possible that some of these adjustments to the database may be hard to undo if need-be.

by Alex Jordan -
This more serious issue I have (about archiving/unarchiving courses) seems now to be separate from my recent databse woes. I am experiencing the same thing on two other 2.16 servers. Either something is not right with version 2.16, or I have failed to update database configuration settings in the right way.

On my 2.16 servers, if I attempt to archive a course, I get:

Warning messages

Use of uninitialized value \$var in hash element at /opt/webwork/webwork2/lib/WeBWorK/DB/Schema/NewSQL/Std.pm line 285.

The same error message comes up if I attempt to unarchive a course. (Which is the main issue that led me to start this thread, but at the time I thought that was a symptom of my attempt to move to MariaDB.)
I found my configuration error. I had missed a few lines of the new site.conf.dist.


text:
LCM of two numbers is 2079 and their HCF is 27. If one of the number is 189, the other number is :
[A]189
[B]216
[C]297
[D]584

297
$\because Second\ Number = \frac{HCF\times LCM}{First\ Number}$
$= \frac{27\times 2079}{189} = 297$
Hence option [C] is correct answer.


text:
# Beatles -If I Needed Someone chords version 1

Chords:
Transpose
```#-------------------------------PLEASE NOTE-------------------------------------#
# This file is the author's own work and represents their interpretation of the #
# song. You may only use this file for private study, scholarship, or research. #
#-------------------------------------------------------------------------------#
If I Needed Someone chords
The Beatles *

A

AIf I needed someone to love
GYou're the one that I'd be thinking of
AIf I needed someone

AIf I had some more time to spend
GThen I guess I'd be with you my friend
AIf I needed someone

Em                F#Had you come some other day
BmThen it might not have been like this
Em              F#              Bm   EBut you see now I'm too much in love

ACarve your number on my wall
GAnd maybe you will get a call from me
AIf I needed someone

A       G   AAh, ah, ah, ah

AIf I had some more time to spend
GThen I guess I'd be with you my friend
AIf I needed someone

Em                F#Had you come some other day
BmThen it might not have been like this
Em              F#              Bm   EBut you see now I'm too much in love

ACarve your number on my wall
GAnd maybe you will get a call from me
AIf I needed someone

A       G   AAh, ah, ah, ah

* Alternate:

Capo II

A  = G
G  = F
Em = Dm
F# = E
Bm = Am
E  = D

Set8


text:
# if a2 + b2 =1 , then find the value of 1+b+ia/1+b-ia

the given expression is: $\frac{1+b+ia}{1+b-ia}$
multiplying by (b+ia) to the numerator and denominator:

$=b+ia$

hope this helps you

• 95
What are you looking for?


text:
Creating a live web page with your publications is now easy to do and there are many ways to do so. I will show one that uses Zotero and GitHub.

## Prerequisites

1. Github page with gh-pages set up
2. Zotero database, synced with zotero.org

## Optional

1. A web redirection so that you don’t have to use the default web address: https://username.github.io . I won’d go into this here, but, it may be desireable to use a web address that is not the default.

## Steps

1. Go to bibBase
2. Click on the link to the right of Zotero part way down the page (Generate BibBase page …)
 <script src="https://bibbase.org/show?bib=[URL-OF-BIBTEX-FILE]&jsonp=1"></script>

where you replace [URL-OF-BIBTEX-FILE] with the URL you just copied.


text:
# Serving RLlib Models¶

In this guide, we will train and deploy a simple Ray RLlib model. In particular, we show:

• How to train and store an RLlib model.

• How to load this model from a checkpoint.

• How to parse the JSON request and evaluate the payload in RLlib.

We will train and checkpoint a simple PPO model with the CartPole-v0 environment from gym. In this tutorial we simply write to local disk, but in production you might want to consider using a cloud storage solution like S3 or a shared file system.

Let’s get started by defining a PPOTrainer instance, training it for one iteration and then creating a checkpoint:

import ray
import ray.rllib.agents.ppo as ppo
from ray import serve

def train_ppo_model():
trainer = ppo.PPOTrainer(
config={"framework": "torch", "num_workers": 0},
env="CartPole-v0",
)
# Train for one iteration
trainer.train()
trainer.save("/tmp/rllib_checkpoint")
return "/tmp/rllib_checkpoint/checkpoint_000001/checkpoint-1"

checkpoint_path = train_ppo_model()

You create deployments with Ray Serve by using the @serve.deployment on a class that implements two methods:

• The __init__ call creates the deployment instance and loads your data once. In the below example we restore our PPOTrainer from the checkpoint we just created.

• The __call__ method will be invoked every request. For each incoming request, this method has access to a request object, which is a Starlette Request.

We can load the request body as a JSON object and, assuming there is a key called observation, in your deployment you can use request.json()["observation"] to retrieve observations (obs) and pass them into the restored trainer using the compute_single_action method.

from starlette.requests import Request

@serve.deployment(route_prefix="/cartpole-ppo")
class ServePPOModel:
def __init__(self, checkpoint_path) -> None:
self.trainer = ppo.PPOTrainer(
config={
"framework": "torch",
"num_workers": 0,
},
env="CartPole-v0",
)
self.trainer.restore(checkpoint_path)

async def __call__(self, request: Request):
json_input = await request.json()
obs = json_input["observation"]

action = self.trainer.compute_single_action(obs)
return {"action": int(action)}

Tip

Although we used a single input and trainer.compute_single_action(...) here, you can process a batch of input using Ray Serve’s batching feature and use trainer.compute_actions(...) to process a batch of inputs.

Now that we’ve defined our ServePPOModel service, let’s deploy it to Ray Serve. The deployment will be exposed through the /cartpole-ppo route.

serve.start()
ServePPOModel.deploy(checkpoint_path)

Note that the checkpoint_path that we passed to the deploy() method will be passed to the __init__ method of the ServePPOModel class that we defined above.

Now that the model is deployed, let’s query it!

import gym
import requests

for _ in range(5):
env = gym.make("CartPole-v0")
obs = env.reset()

print(f"-> Sending observation {obs}")
resp = requests.get(
"http://localhost:8000/cartpole-ppo", json={"observation": obs.tolist()}
)

You should see output like this (observation values will differ):

<- Received response {'action': 1}
-> Sending observation [0.04228249 0.02289503 0.00690076 0.03095441]

In this example the client used the requests library to send a request to the server. We defined a json object with an observation key and a Python list of observations (obs.tolist()). Since obs = env.reset() is a numpy.ndarray, we used tolist() for conversion. On the server side, we used obs = json_input["observation"] to retrieve the observations again, which has list type. In the simple case of an RLlib trainer with a simple observation space, it’s possible to pass this obs list to the trainer.compute_single_action(...) method. We could also have created a numpy array from it first and then passed it into the trainer.
In more complex cases with tuple or dict observation spaces, you will have to do some preprocessing of your json_input before passing it to your trainer instance. The exact way to process your input depends on how you serialize your observations on the client.


text:
# The position x of a particle varies with time t as x = at2 - bt3, where a and b are constants. What is velocity of particle at t = 1 sec?

1. a - 3b
2. 2a - b
3. 2a - 3b
4. None of the above

Option 3 : 2a - 3b

## Detailed Solution

Concept:

Position: The position of the body defines its location based on the given frame and with respect to time.

Displacement is defined as the change in position.

The rate of change of position with respect to time is called velocity

$$v = \frac{dx}{dt}$$

Calculation:

Given Position is x =  at2 - bt3

So, velocity is given as

$$\Rightarrow v = \frac{d(at^2\ -\ bt^3)}{dt}$$

$$\Rightarrow v = \frac{d(at^2\ -\ bt^3)}{dt}$$

v = 2 at - 3 bt2

So, we have to find the speed at t = 1

So, v = 2 a (1) - 3 b (1) 2 = 2a - 3b

So, the correct option is 2a - 3b


text:
# Tag Info

## New answers tagged computational-complexity

1

That you don't create new matrices $D^{(n)}$ but just modify the values in the one matrix. Well, not in one matrix but you have two where one is the previous one ($D^{(n - 1)}$) and the second is the one you compute. When a new $D^{(n)}$ is computed, it becomes $D^{(n - 1)}$ in the next iteration and the previous $D^{(n - 1)}$ (now $D^{(n - 2)}$) becomes a ...

1

We set $h=O(f+g)$. So it suffices to prove that $h=O(\max\{f,g\})$, i.e. that $\exists c>0, n_0 \in \mathbb{N}$ such that $\forall n \geq n_0$: $h \leq c \max\{ f,g \}$ We know that $h=O(f+g)$. By definition that means that $\exists c_1>0, n_1 \in \mathbb{N}$ such that $\forall n \geq n_1:$ $$h \leq c_1(f+g)$$ So, we have $\forall n \geq n_1:$ ...

1

$$h(x)=\mathcal O(f(x)+g(x))\iff \exists \delta>0,\exists C>0:\forall |x|<\delta, \left|h(x)\right|<C|f(x)+g(x)|$$$$<2C\max\{|f(x)|,|g(x)|\}$$ therefore $$h(x)=\mathcal O(\max\{f(x),g(x)\})$$ I let you do the other one, the proof goes in the same way ;-)

1

To show that $O(f+g)\subseteq O(\max(f,g))$, say, let $r$ be an element of $O(f+g)$. Then $r$ is a function of $n$ with the property that there exist positive numbers $C$ and $N$ such that $n \ge N$ implies $|r(n)| \le C(f(n)+g(n))$. Now, note that $$a + b = \min(a,b) + \max(a,b) \le \max(a, b) + \max(a, b) = 2\max(a,b)$$ for any numbers $a$ and $b$. Can you ...

1

IP problems are generally solved by using a relaxation to the (N)LP formulation. Branch and bound algorithms and cutting plane algorithms utilize a relaxation of some sort. Non-Linear Programming is much harder than LP, especially when the feasible region is non-convex, as it can be hard to determine which points are actually optimal. So if you have a ...

2

Quicksort boils down to: Selecting a pivot Separate big list into a "smaller than pivot" list $S_{1}$ and "larger than pivot" list $S_{2}$ Sort $S_{1}$ and $S_{2}$ separately, call result $S_{1}'$ and $S_{2}'$. Combine $S_{1}'$ and $S_{2}'$ and pivot into final, sorted list $S'$. Selecting a pivot randomly is constant time. Hence, it follows that $T(n) ... 0 Note that$f(n) = O(g(n))$iff$f(x) \leq c * g(x)$for some constant$c$and$x \geq x_{0}$. Big-Omega switches the inequality. So if$f(n) = O(g(n))$and$f(n) = \Omega(g(n))$, then we have$f(n) = c * |g(n)|$, which satisfies the definition of$\Theta(n)$. To test this, consider$L = \lim_{n \to \infty} \frac{f(n)}{g(n)}$. If$0 < L < \infty$, ... 2 Run the algorithm repeatedly (and independently) until you don't get '?'. 1 This just adds some information about things you've already mentioned -- I'm quite interested as well in any methods. I note: To the authors’ knowledge, the only known prime generation algorithms for which the statistical distance to the uniform distribution can be bounded are the one proposed by Maurer [19,20] on the one hand, and the trivial ... 0 This looks like the knapsack problem, which is$\operatorname{NP}$-complete. You can easily verify the problems are similar if you consider a related problem where each company offers only one mode of transportation. Then you would have: $$\left[\begin{matrix} C_1\\C_2\\...\\C_i\end{matrix}\right] = \left[\begin{matrix}(s_1, d_1)\\(s_2, d_2)\\...\\(s_i, ... 0 As for asymptotic bounds, note \log n! \in \Theta(n \log n). This is immediate from n^n \geq n! \geq (n/2)^{n/2}. 1 Note that n! = \prod_{i=1}^{n} i \leq \prod_{i=1}^{n} n = n^{n}. So log(n!) \leq log(n^{n}) = n log(n). And so log(n!) = O(n log(n)). 0 Probably you need Stirling's theorem. 0 If G has n vertices, you just want to compare to the path P_{n}. So you reduce HP \leq L, where HP is Hamiltonian Path. So let G be a graph with a Hamiltonian Path. We construct an instance of L with parameters (G, P_{n}). So suppose G has a Hamiltonian Path. As the Hamiltonian path is a path on n vertices, so P_{n} \subset G and L ... 2 A Hamiltonian path is a spanning tree isomorphic to a linear graph. 0 In a certain sense made rigorous elsewhere in a mathematical logic conjecture whose origin I can't remember, every mathematician's proof is expressible in terms of basic semantics that a computer can understand given basic axioms and deduction rules. If you accept that, then no, there is no computer undecidable problem that a mathematician can figure out a ... 0 It is common to find that undecidable problems often have a solution in certain special cases even though there is no general solution. The halting problem is an excellent example; certainly it is possible to prove that (most) programs written by humans intended for a practical purpose do eventually halt. 0 Use the definitions. Say f and g are positive. f is o(g) means: for every constant k>0, for large enough n we have f(n) \le k g(n). f is \omega(g) means: for every constant m>0, for large enough n we have f(n) \ge m g(n). Now try k=1/2, m=2, for example. For large enough n we have$$ 2 g(n) \le f(n) \le (1/2) g(n) $$... 0 No. Little-o and Little-omega are strict inequalities. Big-O and Big-Omega are weak inequalities. So if f(n) \in O(g(n)) and f(n) \in \Omega(g(n)), then f(n) \in \Theta(g(n)). Little-o states that |f(n)| < c * |g(n)| (for some sufficiently large constant c, and all n > n_{0}). Little-omega is comparable, just use a strict greater-than ... 0 Ok, I fear I have now to give it a try. The problem is that solving this requires some understanding of the meaning of the Landau symbols, but I try to make it as elementary as possible avoiding all order manipulation of the terms. First, it does not matter if one uses the \ln or the \log_{10} for asymptotic estimates here, since changing from one to ... 0 In order to prove the asymptotic relationship f \in O(g) you have to choose both constant c and x_0, such that the inequality f(x) \le c \cdot g(x) becomes true for all x > x_0. In your case you can choose c = 8 and x_0 = 4. I'll leave all the rest to you. Just a hint - you can replace x + 2 by 2 \cdot x in this case (for simplicity, ... 0 If you run an elimination tournament among the elements, you can find the largest in n-1 comparisons. The second largest could be any one of the elements that lost to he final winner. How many is that? 1 Since I cannot post a comment I make this into an answer. This answer provides the desired lower bound for finding only the second largest element and provides an algorithm which finds both the largest and the second largest element in desired number of comparisons. So it solves your question. 1 The classes \mathcal{P}, \mathcal{NP} are defined for decision problems. The class \mathcal{NP}-HARD does not contain only decision problems. So \mathcal{NP}-Complete problems are the hardest problems in \mathcal{NP}, but the optimization problems of such \mathcal{NP}-Complete problems are \mathcal{NP}-Hard. Consider for example the ... 0 It's actually co-NP-Complete. Here's the proof. First we have to show that it's in co-NP. This isn't hard, since it's quick to verify a counterexample showing that the two Boolean formulas are not equivalent. Then we have to show a problem known to be co-NP-complete reduces to your problem, which we'll call Boolean Equivalence. We'll reduce the ... 2 No, it is not tractable in the general case. Simply specialize the result to b=0 and A a polytope, and you have maximization of 1-norm in polytope, which is known to be intractable Mangasarian, O., & Shiau, T. H. (1986). A variable-complexity norm maximization problem. SIAM Journal on Algebraic and Discrete Methods, 7(3), 455–461. 0 I don't believe it is. I have a deleted answer sitting here as a scratchpad for me to try some various reformulations to a tractable form, without success. It resembles a saddle-point problem, but the joint convexity of the objective is an obstacle. The inner minimization produces a convex function of a; e.g.,$$\begin{array}{ll} \text{maximize} & ... 1 For each variable y, build two vertices, one for y and one for ~y. Connect both y and ~y to X, the unique global node that is connected in a triangle with T(rue) and F(alse). These connections will ensure that one of {y,~y} gets a color of T and the other gets a color of F. This is what gives you the connection between a coloring and a truth assignment. ... 1 Actually, there is no 3-coloring if the 3 literals are colored as F. This is because the node after the "merge" will be colored as F, while the blank below left of T will be as X. The remaining node (upper left of T) can have no valid color in this case. 3$\let\mr\mathrm \mr{BPP}$is in the second level of polynomial hierarchy by the Gács–Sipser–Lautemann theorem, and it is an easy exercise to show that $$\mr{P^{ZPP}}=\mr{ZPP}\subseteq\mr{P^{RP}}\subseteq\mr{P^{BPP}}=\mr{BPP}.$$ The least trivial of these is the inclusion$\mr{P^{BPP}}\subseteq\mr{BPP}$: let$L_0$be a language decidable by a machine$M_0$... 1 Well you can do this with$2*k$multiplications i.e with$C=2$. The algorithm is outlined below: Initialize$temp = 1$for i=$1$to k do$b_i$= temp temp = temp*$a_i$end temp =$1$for i=k to$1$do$b_i$= temp temp = temp*$a_i$end Example ,$k=4$First Pass$b_1$= 1$b_2$=$a_1$... 0 This is a case of the weapons-target assignment problem: http://en.wikipedia.org/wiki/Weapon_target_assignment_problem 1 Here is a (stronger) result (with an easier proof) which implies yours: Assume that$\liminf |f|\gt0$, then, for every finite$c$,$f+c\in O(f)$. To apply this to your setting, note that the hypothesis that$\lim f=+\infty$implies that$\liminf |f|\gt0$, and that if$f+c\in O(f)$and$f\in O(g)$then$f+c\in O(g)$. Now can you prove the result above? ... 1 First ask "Is 1 occupied?", then, "Is 2 occupied?", "Is 4 occupied?", "Is 8 occupied?", and so on, until you get the answer, "no". This will take$\log_2n$questions (give or take a fraction). You now know$k$such that$n$is between$2^k$and$2^{k+1}$, and binary search inside that interval will get you the answer in another$\log_2n$questions. 1 You're basically looking for a series of questions that resemble a binary search. Binary search runs average O($\log n$) and worst O($\log n$). Questions like, "is position n occupied", should suffice. Make sure each subsequent question tries to cut the problem size in half. Note: Try asking for an extremely large value, if yes, double the value size, if ... 1 Let$A$be$m\times n$and$B$be$n\times p$(assuming both sparse) and let$e_i$be the$i$th column of the identity matrix of an appropriate size. If the matrices are stored by rows, the product$C:=AB$should be computed by rows as well. We have $$\tag{1} e_i^TC=\sum_j\underbrace{(e_i^TAe_j)}_{a_{ij}}e_j^TB$$ so the$i$th row of$C$is given by linear ... 1 The bound$nnz(AB)=nnz(A)nnz(B)$is tight: Take $$A = \pmatrix{1 & 0 & \dots & 0 \\ \vdots & \vdots & & \vdots \\1 & 0 & \dots & 0 }\in \mathbb R^{m_a,n_a}, \quad B = \pmatrix{ 1 & \dots & 1 \\0 & \dots & 0 \\\vdots & &\vdots \\0 & \dots & 0 \\}\in \mathbb R^{m_b,n_b}.$$ Then$AB$is a ... 1 First I imagine that$n$is an integer and that you want to study this problem when$n$goes to$+\infty$. That's probably obvious but better to be sure! Then please write the definition of$2^n = O(3^n)$. That should be something like there exists$C>0$real and$N$integer such that for$n > N$you have$|2^n| \le C|3^n|$. Now what do you think of ... 1 Apply the same techniques I've shown you in your other threads: $$L = \lim_{n \to \infty} \frac{2^{n}}{3^{n}} = \lim_{n \to \infty} (\frac{2}{3})^{n} = 0$$ So$3^{n}$strictly dominates$2^{n}$, and$L = 0$implies your result. Check out this tutorial on complexity for more on the limit test and your calculus techniques: ... 0 To prove$15n^{2} \in \Omega(3 * 2^{n})$, we consider$L = \lim_{n \to \infty} \frac{15n^{2}}{3 * 2^{n}}$. If$L = 0$, then we conclude$15n^{2} \in \omega(3 * 2^{n})$. Little-omega is a strict inequality, and so it implies Big-Omega. Applying L'Hospital's rule, we get: $$L = \lim_{n \to \infty} \frac{30}{3 * (ln(2))^{2} * 2^{n}} = 0$$ I applied ... 2$T(n)=2T(\frac{n}{2} )+nT(n)= 2[2T(\frac{n}{4})+\color{red}{\frac n2}] +n = 4T(\frac{n}{4})+\color{red}2nT(n)=8T(\frac{n}{8})+\color{red}3n...T(n)= 2^kT(\frac{n}{2^k})+\color{red}kn$Finally$\frac{n}{2^k}=1$and$\therefore n=2^k0 According to Maple, \eqalign{m^e &\equiv 14178120117339266261904109624890227407523673482786024455164108238811626728989472\cr &1928886528279400665340976615948053755946789302972467196231829204061441862114262 \cr &11704026304276898946101664519229545142128929712389990917298990673103791915511466 \cr ... 1 An alternative is to use the limit test. Consider f(n) = 8n^{3} + \sqrt{n} and:L = lim_{n \to \infty} \frac{f(n)}{n^{3}}$$Note f(n) \in o(n^{3}) iff L = 0 (and little-o is the strict inequality, which implies Big-O). Similarly, 0 < L < \infty \implies f(n) \in \Theta(n^{3}). And finally, L = \infty \implies f(n) \in \omega(n^{3}) ... 1 Your proof is correct, however all you have to observe is that the degree of the LHS equals the degree of the RHS. :) 2 An example of a correct proof would be:$$\log n!=\sum_{i=1}^n{\log i}\leq \sum_{i=1}^n{\log n}=n\log n$so$\log n!\in O(n\log n)$since$\log n!\leq n\log n$for all$n$. 1 The second algorithm uses less space, but the first does fewer comparisons in general. In both algorithms you compare every node with alphabetically with the target until you find one that’s greater. In the second you also test each of those nodes except the last to see whether it has the desired property; in the first, however, you test only those back to ... 1 The "known" subset-sum problem takes as input a list L of n integers and a target sum A. Partition L into n lists (each containing one of the n integers). Append a zero to each of the n lists. Now, each list contains an integer from L and a$0$. Now, feed the n lists and the target A into an algorithm that solves your problem. If your algorithm solves ... 1 It is just simpler to describe the situation with a single input tape -- in particular, if you want to compare the power or efficiency of machines with differing number of tapes, it is convenient that the format of the inputs to all the machines are the same. But it is not a particularly deep choice -- if you have input on one tape and want it spread across ... 0 Too long for a comment: When multiplying in bases other than$2$you have two different sorts of carries. In base$2$, each individual bit multiplication cannot carry. The only carries you get are when you add the bit multiplications together. In other bases you can get carries in the multiplication step. In base$10$, for example, when multiplying$79 ...

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# Experimental patterns in LHC quarkonium production from pp to AA

## by Pietro Faccioli

Tuesday, 5 June 2018 from to (Europe/Vienna)
 Description Mid-rapidity LHC measurements of quarkonium production show unexpectedly simple scaling patterns: 1) the J/ψ, ψ(2S), χc1, χc2 and ϒ(nS) pT-differential cross sections are compatible with one universal shape function, an observation complementing the observed lack of polarization in all measured cases; 2) cross-section normalizations and binding energies of the S-wave quarkonia show a remarkable correlation, identical for 7 and 13 TeV data. These results provide a strong experimental evidence supporting the validity of the factorization ansatz, a cornerstone of non-relativistic QCD, still lacking theoretical demonstration. Moreover, they are crucial information for quantitative interpretations of quarkonium production data, including studies of the suppression patterns measured in nucleus-nucleus collisions.