text:
Search for supersymmetry with jets, missing transverse momentum and at least one hadronically decaying tau lepton in proton-proton collisions at sqrt(s) = 7 TeV with the ATLAS detector

5 Laboratoire de Physique Corpusculaire
LPC - Laboratoire de Physique Corpusculaire - Clermont-Ferrand
Abstract : A search for production of supersymmetric particles in final states containing jets, missing transverse momentum, and at least one hadronically decaying tau lepton is presented. The data were recorded by the ATLAS experiment in sqrt(s) = 7 TeV proton-proton collisions at the Large Hadron Collider. No excess above the Standard Model background expectation was observed in 2.05 fb-1 of data. The results are interpreted in the context of gauge mediated supersymmetry breaking models with Mmess = 250 TeV, N5 = 3, mu > 0, and Cgrav = 1. The production of supersymmetric particles is excluded at 95% C.L. up to a supersymmetry breaking scale Lambda = 30 Tev, independent of tan(beta), and up to Lambda = 43 TeV for large tan(beta).
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-00688572
Contributor : Emmanuelle Vernay <>
Submitted on : Wednesday, April 18, 2012 - 8:51:45 AM
Last modification on : Thursday, June 17, 2021 - 3:47:47 AM

Citation

G. Aad, S. Albrand, M.L. Andrieux, Q. Buat, B. Clement, et al.. Search for supersymmetry with jets, missing transverse momentum and at least one hadronically decaying tau lepton in proton-proton collisions at sqrt(s) = 7 TeV with the ATLAS detector. Physics Letters B, Elsevier, 2012, 714, pp.197-214. ⟨10.1016/j.physletb.2012.06.061⟩. ⟨in2p3-00688572⟩

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format: plain


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# zone reflect command

Syntax

zone reflect keyword ... <range>

Primary keywords:

All zones within the given range are reflected across the plane. Note that group information is copied from the original to the new zones.

The following keywords are available to control how this reflection is accomplished.

dip-direction f (3D ONLY)

Specify the dip direction f [degrees] of the plane measured in the global $$xy$$-plane clockwise from the positive $$y$$-axis. (The default is f = 0.)

dip f

Specify (in FLAC3D) the dip angle f [degrees] of the plane, measured in the negative $$z$$-direction from the global $$xy$$-plane. (The default is f = 0.)

In FLAC2D, this specifies the angle f [degrees] of the plane measured clockwise from the global $$x$$-axis. (The default is f = 0.)

normal v

Specify a normal vector to the plane, with the components specified as v. If normal is specified, dip and dip-direction are not required.

origin v

Specify the location of one point through which the plane crosses. The coordinates of the point are v.

merge b

If b is off, then boundary gridpoints of the newly created zones are not automatically merged with previous existing gridpoints. By default merge is on.


text:
Q

# Log2,log(2^n-1),log(2^n+3) in AP, then n=?

a 5÷2   b log5base2   c log5base3   d 3÷2

9 Views

As we learned that in AP

$a_{2}-a_{1}=a_{3}-a_{2}$

$\log (2^{n}-1) -\log(2)=\log(2^{n}+3)-\log(2^{n}-1)$

$\log(2^{n}-1/2)= \log(2^{n}+3/2^{n}-1)$

$(2^{n}-1/2) = (2^{n}+3/2^{n}-1)$

Let $2^{n} = x$

$(x-1)^2 = 2x+6\Rightarrow x^2 -2x +1 = 2x +6$

$x^2 -4x-5=0\Rightarrow here$ X = 5 & X=-1 (X = -1 not possible)

So X = 5
$2^{n}= 5$ (Taking log both sides we get)

$n\log2 = \log5 \Rightarrow n = {log_{2}}^{5}$

Hence option (B) is correct

As we learned that in AP

$a_{2}-a_{1}=a_{3}-a_{2}$

$\log (2^{n}-1) -\log(2)=\log(2^{n}+3)-\log(2^{n}-1)$

$\log(2^{n}-1/2)= \log(2^{n}+3/2^{n}-1)$

$(2^{n}-1/2) = (2^{n}+3/2^{n}-1)$

Let $2^{n} = x$

$(x-1)^2 = 2x+6\Rightarrow x^2 -2x +1 = 2x +6$

$x^2 -4x-5=0\Rightarrow here$ X = 5 & X=-1 (X = -1 not possible)

So X = 5
$2^{n}= 5$ (Taking log both sides we get)

$n\log2 = \log5 \Rightarrow n = {log_{2}}^{5}$

Hence option (B) is correct

Exams
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# Tag Info

9

QRP only means low power, often 1W or less. While you could technically have a 1W SSB transceiver and call it QRP, the practical range of such a radio would be limited. It is much easier to hear a single tone over the noise than it is to hear a voice which has its energy spread over a wide range of frequencies. Thus we say CW is a more sensitive mode: it ...

8

My understanding of "QRP" is simply the use of as little power as possible to make contacts over interesting distances. There isn't a specific power level that equates to "QRP" - it is more a function of what's less than expected. 1W on 2m/UHF for 10 mile simplex (or local/regional repeater communications) isn't QRP. 0.1W on 2m/UHF for 250 mile simplex ...

8

My personal opinion is that QRP means using significantly less power than is customary for a given mode of operation. In that sense, a 5W HT would not be QRP because pretty much everybody else with an HT is also operating at (or near) 5W. 0.5W might be considered QRP in this case.

7

If you look up "link dipole" or "linked dipole" you will find several designs that do just that. It works well for QRP or even for 100W. The ones I've seen have used several different connectors: alligator clips, quick disconnect spade terminals, Anderson PowerPoles, and banana plugs, to name a few. Of those, alligator clips are probably ...

6

20 meters. This assumes you're wanting to work back to the States or into Europe. If your primary interest is local in South America, then I'd go with 40 meters. An advantage is the ease of dealing with a smaller antenna. While not a single-band radio, I'd seriously consider one of the KD1JV rigs. Occasionally available as kits, but now commercially ...

5

Doppler spreading is the change in received frequency from a distance transmitter due to the rise and fall of the ionosphere along the signal path. When the effective height of the ionosphere rises, this lengthens the path and causes the received frequency to drop; when it falls the path decreases and the frequency rises. You can measure this frequency ...

5

Without getting into specific product recommendations, I believe the simplest way is as follows: Obtain a license. An amateur license is required for using amateur bands. I am not aware of any non-amateur bands using FT8. A technician-level license is sufficient in USA for VHF and UHF but it looks like a General level license is needed for the HF bands. ...

5

A QRP transceiver is just a transceiver that operates on low power (usually less than 5 watts). If you look at the international Q Code, you will see that QRP simply means 'low power', and QRO means 'high power'. However, what you are talking about is a QRP transceiver kit. Such kits are usually by necessity a fairly simple design, and the simplest ...

5

Generally, making an antenna shorter makes it less efficient. This means some fraction of your transmitter's power is used to make the antenna and/or surrounding soil warmer instead of radiating where someone might receive it. The fraction of power lost is the same whether the transmit power is 3W or 300W. So the detrimental effects of a loading coil are not ...

4

CW is preferred over SSB for low power (QRP) operation because CW delivers better signal-to-noise ratio (S/N or SNR) than SSB. Because a CW signal occupies about an order of magnitude less bandwidth than an SSB signal, a narrower receiving filter can be used, which admits all of the signal and a much smaller amount of noise energy than a wider SSB filter. ...

4

Q1 forms a Colpitts crystal oscillator. It uses the reverse biased D2 as a varicap (variable capacitor) to pull the crystal frequency by 700 to 800 Hz, This frequency beats with the received frequency to produce the audible CW tone. The adjustability of the pull through W1 also allows a simple, limited range RIT function during receive.

4

It makes a difference of: $$10 \times \log_{10}\left(3 \over 5\right) = 2.2 \:\mathrm{dB}$$ How big is 2.2 dB? See How big is a decibel? It's also relevant to note: if the 12V battery is lead acid, the charged voltage can be as high as 13.7. Furthermore, while a boost converter might further increase your transmit power, it's likely to make noise which ...

4

I see what your problem is. That's a 1.4 meter long, based-loaded, telescoping, end-fed vertical antenna. It has significant losses on 40 meters. Furthermore, all end-fed antennas require a good RF return, such as a resonant elevated radial or counterpoise. Since you don't mention that, that indicates to me that very little of your signal is actually ...

4

The output impedance isn't especially important: in fact I believe it uses a nonlinear amplifier so the concept doesn't really apply. What does matter is the intended load impedance, which for any amateur radio application you can assume to be 50 ohms unless otherwise specified. To verify, I modelled the low-pass filter part of the circuit from the manual: ...

3

Although it's highly opinionated, it may be defined as operating at 5% or less than standard 'barefoot' operating power on your band, by convention. So, for most HF, 100W is the standard power, 5W is considered QRP. For VHF, most handhelds (which are the most frequently used on that band) operate at 5W, so QRP would be 250mW. Of course, that doesn't ...

3

If a transmitter is moving toward or away from you, the received frequency will get shifted up or down, depending on direction and rate of movement. Even if the transmitter and receiver are not moving relative to each other, but a reflector, reflecting the signal between them, is moving, the you can get the same Doppler effect. It's well known (at minimum, ...

3

When you ask if a QRP transmitter could be used to solve your problem, I presume that you mean a low-powered transmitter on HF, meaning roughly 3 – 30 MHz. That would solve the problem of limited range associated with the VHF and higher-frequency bands, but a large antenna would be necessary. The problem is that radio spectrum, especially in the HF ...

3

If you haven't done much QRP operating, QRP is a funny thing. Sometimes propagation is great, and you wonder why anyone would bother with 100 W or more. I'll never forget my first DX QSO, from Oregon to Estonia, running 11 W to a vertical, by gray-line propagation at the bottom of the sunspot cycle in 2006. Other times it seems like a complete waste of ...

2

Clandestine can be very effective. A thin wire hanging out of your window is hard to see, especially if you only have it outside while operating. Commercially built tuners aren't expensive either. Another option to consider is MCW (modulated CW) where you send the Morse tones over audio via VHF/UHF FM. If you have enough CW fans there you may be able ...

2

It happens slowly and organically. Yes, most of these folks have probably mentioned on bulletin boards, forums, or mailing lists that they have a new project they're working on. They aren't running around putting those notes in all the lists though. They're putting it on lists in which they're active members and in ways they think that community might like, ...

2

As the saying goes, if you build a better mousetrap, then the world will beat a path to your door; but in the modern era, you do need to get the word out first somehow. A website with complete information about your awesome kit is essential. That way, interested hams can find your site either directly from a URL, or from a search engine. Don't neglect the ...

2

There's at least one video on YouTube concerning a Pixie (normally a 40m QRP CW-only transceiver, with rather questionable control of harmonics) converted to use the output audio amplifier IC as a modulator for AM transmission. Another Pixie could be used as a receiver, as long as they have the same frequency crystal. This is about as simple and cheap as ...

2

ITU Recommendation F.1487-0 defines methods for testing HF ionospheric paths for bandwidths up to 12 kHz. While ionospheric propagation can be complex, this document provides a starting point for the widely applicable basics. It characterizes an HF channel with two parameters: multipath differential time delay, and Doppler spread. The multipath ...

2

If you read contest rules, the QRP section is for those using 5W or less (sometimes 10W or less) no matter what type of transmitter you use. So you can enter the FM QRP section of a VHF or UHF contest when using a handheld on full (5w) power.

2

A tuned antenna should result in a real load impedance for your transceiver. This impedance can deviate from 50 Ohm, resulting in an SWR deviating from 1:1. The EFHW antenna impedance can be as high as 1500 Ohm and can only be connected to the TX when there is a transformer at the feedpoint of the antenna; the base I assume. Without a transformer (common 1:5 ...

1

Any transmissive data channel will hit a lower power limit for readability when the signal no longer exceeds the noise floor of the transmission mode. This is another way of saying that the minimum-required signal strength for establishing data exchange is dependent on the position of the noise floor. So understanding the physics and math behind S/N ratios ...

1

If you are just transmitting GPS coordinates and at a slow rate and need a bandwidth of about 50kbps or less then have a look at LoRa technology. It is low power, low data rate and spread spectrum but it is very reliable with a range over 10km (some as far as 50km) on licence free VHF and UHF bands. It is designed for IoT uses in a noisy environment it could ...

1

QRP simply refers to low output power. What modulation you use is up to you. Voice, CW, digital mode like JS8Call .... Using 5 Watts to contact a DX station in SSB is quite a thrill.

1

As others pointed out, the shorter the antenna, the lower its efficiency. Trying to use short wire in combination with a QRP transmitter is very likely to be a very frustrating experience (I know this from my own experience). If I simplify, it is the current in the antenna that causes RF transmission. Therefore, if your antenna is a short wire, it will ...

1

One of my favorite things to do is QRP field operation with compromised antennas. The easiest is just a piece of wire with one end thrown over a tree branch and a counterpoise wire laying on the ground. I use a sling shot to launch 20 lb fishing line up and over a branch (up to about 15 to 20 feet high). My antenna is a single wire about 40 to 65 feet ...

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


text:
Notes on Circle | Grade 11 > Mathematics > Coordinate Geometry straight line | KULLABS.COM

• Note
• Things to remember
• Exercise

### Introduction:

Circle is a closed shape plane figure whose boundary (the circumference) consists

of points equidistant from a fixed point (the centre).

Circle may also be defined as a locus of a point which moves so that its distance from a fixed point is constant. The fixed point is called the centre and the constant distance is called the radius of the circle.

The circle has been known since before the beginning of recorded history. Natural circles would have been observed, such as the Moon, Sun, and a short plant stalk blowing in the wind on sand, which forms a circle shape in the sand. The circle is the basis for the wheel, which, with related inventions such asgears, makes much of modern machinery possible. In mathematics, the study of the circle has helped inspire the development of geometry, astronomy, and calculus.

Some of the elements in the circle are given in alongside figure.

### Equations of the circle.

#### (i)Center at origin(standard form):-

If r be the radius of the circle from the origin and (x,y) be any coordinates on circumference of the circle then.

$$x^2+y^2=0$$

#### (ii)Center at any point.

If (h,k) be the center of circle of radius r and (x,y) be any point on the circumference, then.

$$(x-h)^2+(y-k)^2=0$$

#### (iii) General equation of circle:

Consider an equation

$$x^2+y^2+2gx+2fy+c+0.......(i)$$

$$or,(x^2+2gx+g^2)+(y^2+2fy+f^2)=g^2+f^2+c$$

$$or,(x+g)^2+(y+f)^2=g^2+f^2+c$$

$$or,[x-(-g)]^2+[y-(-f)]^2=\sqrt{(g^2+f^2-c)^2}$$

Comparing the above equation with

$$(x-h)^2+(y-k)^2=r^2$$

Now we get

$$h=-g$$

$$k=-f$$

$$r=\sqrt{g^2+f^2-c}$$

Hence eqution (i) represent an quation of circle with:-

$$centre=(-g,-f)$$

$$radius=\sqrt{g^2+f^2-c}$$

#### Equation of circle at particular cases:

As usual let (h,k) be center of circle and r be radius of it in all cases . Now

(i) Equation of circle touching x-axis

$$(x-h)^2+(y-k)^2=k^2$$

This is because if the circle touches x-axis then the radius of circle(r)=k

(ii) Equation of circle touching y-axis

$$(x-h)^2+(y-k)^2=h^2$$

This is because if the circle touches y-axis then the radius of circle(r)=h

(iii) Equation of circle touching both axes

$$(x-h)^2+(y-h)^2=h^2$$

This is because if the circle touches both axes then the radius of circle(r)=h=k

(iv) Equation of circle in diameter form

Let the co-ordinates of diameter is as shown in figure .

Also, construct a straight line from both A and B meeting at the same point P on the circumference .

Now we can see right angle triangle APB where slopes of

$$(i)AP=\frac{y-y_1}{x-x_1}$$

$$(ii)BP=\frac{y-y_2}{x-x_2}$$

Being AP and BP perpendicular so their product should be -1.

$$or,\frac{y-y_1}{x-x_1}.\frac{y-y_2}{x-x_2}=-1$$

$$(x-x_1)(x-x_2)=(y-y_1)(y-y_2)$$

It is the required equation of circle in diameter form.

#### Some important Questions

1.Find the equation of circle which touches the axes at (2,0) and (0,2)?

2.Find the equation of circle whose center is at (3,4) and touches the x-axis

3.Find the center and radius of the circle x2+y2+4x-6y+4=0?

4.Proove that the circles x+y+2ax+c=0 and x+y+2by+c=0 touch each other if

$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$$

5.Find the equation of circle which touches the coordinate axes at (a,0) and (0,a)?

6.Find the equation of tangent to the circle x2+y2=a2the point (x1,y1) ?

7.Find the equation of circle with two of the diameter are x+y=6 and x+2y+8 and radius=10?

8.If one end of the diameter of the circle x2+y2-2x+6y-15=0 is (4,1), find the coordinate of the end?

Some formulae or keypoints are:

General equation of circle

$$x^2+y^2+2gx+2fy+c=0$$

$$\sqrt{g^2+f^2-c}$$

And centre

$$(-g,-f)$$

Standard form of circle is

$$x^2+y^2=r^2$$

Standard form of circle with center at any point

$$(x-h)^2+(y-k)^2=r^2$$

Equation of circle in diameter form

$$(x-x_1)(x-x_2)=(y-y_1)(y-y_2)$$

.

#### Click on the questions below to reveal the answers

Solution:

Let (h,k) be the centre of the circle.

Now in the circle touching both the axes we have the equation as:

$$(x-h)^2+(y-h)^2=h^2$$

Where h=r=k so,

$$(x-2)^2+(y-2)^2=2^2$$

$$x^2+y^2-12=0$$Is the required equation

Solution:

We can assume the equation of circle as :

$$x^2+y^2+2gx+2fy+c=0$$

Since it passes through (1,0), (2,-2), (3,1) so it must satisfy the above quation.Hence substituting the equation in corresponding coordinates we get

$$2g+c=1$$

$$-4f+4g+c=8$$

$$2f+6g+c=-10$$

Solving this we get,

$$g=\frac{-5}{2}$$

$$f=\frac{1}{2}$$

$$c=4$$

Hence equation of the circle is

$$x^2+y^2-5x+y+4=0$$

Solution:

Let O(h,k) be centre of the circle.Since the circle touches x-axis so radius of circle(r)=k and also from given coordinate h=3

Now equation of circle is:

$$(x-3)^2+(y-k)^2=k^2...(i)$$

Since the circle (i) passes through the point (1,2), so

$$4+(2-k)^2=k^2$$

$$or, 4k=8$$

$$Therefore,k=2$$

Now required equation of circle is

$$(x-3)^2(y-2)^2=2^2$$

$$Therefore,x^2+y^2-6x-4y+9=0$$

is the required expression

0%

## ASK ANY QUESTION ON Circle

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text:
# Does the prefix Ortho have some significance? [duplicate]

Are orthoboric acid and boric acid one and the same thing or we can not write H3BO3 as B(OH)3?

The reference given by Andselisk does not bring an answer to the question about the origin of the prefix "ortho" in inorganic nomenclature.

The prefix "ortho" is added before the name of an acid $$H_aX_bO_c$$ if first $$b = 1$$ and if it corresponds to the highest possible number of $$OH$$ groups attached to the central X atom. For example $$H_4SiO_4$$ is the orthosilicic acid, $$H_3BO_3$$ is the orthoboric acid. $$H_4CO_4$$ is the orthocarbonic acid (which only exists in esters). $$H_3PO_4$$ is the orthophosphoric acid (where the prefix is often skipped). $$H_3AsO_4$$ is the orthoarsenic acid. The use of the prefix does not depend on the oxidation number of $$X$$. For example, if $$P$$ and $$As$$ are taken with a lower oxidation number, one obtains $$H_3PO_3$$ as the orthophosphorous acid, and $$H_3AsO_3$$ as the orthoarsenous acid.

The prefix "meta" is added to the name of an acid, if its formula is formed by removing one molecule $$H_2O$$ from the formula of the ortho-acid. $$H_3BO_3$$ - $$H_2O$$ = $$HBO_2$$ and this is the metaboric acid. $$H_3PO_4$$ - $$H_2O$$ = $$HPO_3$$ which is the metaphosphoric acid. Repeating this process, one gets :$$H_2CO_3$$ is the metacarbonic acid (often written without the prefix meta). $$H_2SiO_3$$ is the metasilicic acid. As with ortho-acids, the use of "meta-" does not depend of the oxidation number. $$HPO_2$$ is the metaphosphorous acid. $$HAsO_2$$ is the metaarsenous acid.

It should also be mentioned that removing one $$H_2O$$ out of two molecules of an ortho- acid gives a pyro- acid. Example : $$2 H_3PO_4 - H_2O = H_4P_2O_7$$ and this is the pyrophosphoric acid. $$H_4As_2O_7$$ is the pyroarsenic acid. This prefix pyro- is even used for acids existing only under the ortho- definition. For example, a pyro derivate from the sulfuric and sulfurous acids $$H_2SO_4$$ and $$H_2SO_3$$ exist. They are the pyrosulfuric acid $$H_2S_2O_7$$. and the pyrosulfurous acid $$H_2S_2O_5$$

It should also be mentioned that the prefixes ortho-, meta-, and pyro- are never used with acids having already a prefix like hypo- and per-.

All these rules have been edited in the middle of the 19th century.

• I admit the link I provided wasn't the most relevant one, thank you for pointing this out. I corrected my comment and included the link that IMO fits better. Either way, the question appears to a be a duplicate. – andselisk Jan 8 at 13:54


text:
Home

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
 By selling an article for Rs 110 a man loses 12 %. For how much should he sell it to gain 8 %. [ Calculator, Method and examples ]Solution:Your problem -> By selling an article for Rs 110 a man loses 12 %. For how much should he sell it to gain 8 %.Here "S.P." = "Rs. "110," Loss" = 12%"C.P." = "S.P." × 100/(100 - "Loss"%) = 110 × 100/(100 - 12) = 125" Rs."Now, "C.P." = "Rs. "125," Gain" = 8%"S.P." = "C.P." × (100 + "Gain"%)/100 = 125 × (100 + 8)/100 = 135" Rs."

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
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text:
2 edited body

to Leonard-

An important example of homology pull-back is given by the intersection product. Let $M$ be a closed oriented manifold then we define this product as: $$H_i(M)\otimes H_j(M)\stackrel{\times}{\rightarrow} H_{i+j}(M\times M)\stackrel{\Delta^{*}}{\rightarrow} H_{i+j-d}(M)$$ where $\Delta^*$ is the homology pull-back of the diagonal $\Delta:M\rightarrow M\times M$. The intersection product was first introduced by means of geometric transverse intersection of cycles. Bredon "Geometry and Topology" explains it and makes the relationship with the cup product which is its Poincaré dual. He does not explain how to do when the cycle is not representable by a manifold but this is a good starter.

Fore general cycles, as I explained in my other answer either you play with simplexes viewed as manifolds with corners or you play with manifolds with singularities. M. Kreck in his book "Differential algebraic topology" did it for a particular type of manifold manifolds with singularities and he proves all the tranversality results you want in order to define homology pull-backs. M. Goresky in his PhD thesis studied it in the case of stratified pseudomanifolds and also build a geometric homology and a geometric cohomology thanks to what he called Whitney chains and cochainshere . Here you need transversality for stratified pseudomanifolds (it works inductively strata by strata). This is in the smooth world, if you are more a PL-guy then you should have a look at:

S. Buoncristiano, C. P. Rourke and B. J. Sanderson "A geometric approach to homology theory", Cambridge Univ. Press, Cambridge, Mass., 1976

Another good reference is: Jakob, Martin "Bivariant theories for smooth manifolds." Papers in honour of the seventieth birthday of Professor Heinrich Kleisli (Fribourg, 2000). Appl. Categ. Structures 10 (2002), no. 3, 279–290.

It is another geometric approach to homology theories where homology pull-backs are geometric pull-backs. In fact this all goes back to

Quillen, D. "Elementary proofs of some results of cobordism theory using Steenrod operations." Advances in Math. 7 1971 29–56 (1971).

I like and recommand Jakob's paper because it is related to the framework of

Fulton, W.; MacPherson, R. "Categorical framework for the study of singular spaces." Mem. Amer. Math. Soc. 31 (1981), no. 243

which explains how to mix homology covariant and contravariant morphism morphisms in order to have a good framework to have get Riemann-Roch types theorem.

To conclude this geometric interpretation of homology pull-backs certainly goes back to the first days of homological techniques homology in algebraic topology before cohomology was invented.

1

to Leonard-

An important example of homology pull-back is given by the intersection product. Let $M$ a closed oriented manifold then we define this product as: $$H_i(M)\otimes H_j(M)\stackrel{\times}{\rightarrow} H_{i+j}(M\times M)\stackrel{\Delta^{*}}{\rightarrow} H_{i+j-d}(M)$$ where $\Delta^*$ is the homology pull-back of the diagonal $\Delta:M\rightarrow M\times M$. The intersection product was first introduced by means of geometric transverse intersection of cycles. Bredon "Geometry and Topology" explains it and makes the relationship with the cup product which is its Poincaré dual. He does not explain how to do when the cycle is not representable by a manifold but this is a good starter.

Fore general cycles, as I explained in my other answer either you play with simplexes viewed as manifolds with corners or you play with manifolds with singularities. M. Kreck in his book "Differential algebraic topology" did it for a particular type of manifold with singularities and he proves all the tranversality results you want to define homology pull-backs. M. Goresky in his PhD thesis studied it in the case of stratified pseudomanifolds and also build a geometric homology and a geometric cohomology thanks to what he called Whitney chains and cochains here you need transversality for stratified pseudomanifolds (it works inductively strata by strata). This is in the smooth world, if you are more a PL-guy then you should have a look at:

S. Buoncristiano, C. P. Rourke and B. J. Sanderson "A geometric approach to homology theory", Cambridge Univ. Press, Cambridge, Mass., 1976

Another good reference is: Jakob, Martin "Bivariant theories for smooth manifolds." Papers in honour of the seventieth birthday of Professor Heinrich Kleisli (Fribourg, 2000). Appl. Categ. Structures 10 (2002), no. 3, 279–290.

It is another geometric approach to homology theories where homology pull-backs are geometric pull-backs. In fact this all goes back to

Quillen, D. "Elementary proofs of some results of cobordism theory using Steenrod operations." Advances in Math. 7 1971 29–56 (1971).

I like and recommand Jakob's paper because it is related to the framework of

Fulton, W.; MacPherson, R. "Categorical framework for the study of singular spaces." Mem. Amer. Math. Soc. 31 (1981), no. 243

which explains how to mix homology covariant and contravariant morphism in order to have a good framework to have Riemann-Roch types theorem.

To conclude this geometric interpretation of homology pull-backs certainly goes back to the first days of homological techniques in algebraic topology before cohomology was invented.


text:
# Thread: Finding Probability of Bayesian Network

1. ## Finding Probability of Bayesian Network

I know I can do:
P(~s|~y, ~h) = P(~s|~y)*P(~h|~y)

problem is P(~s|~y) and P(~h|~y) are not given and i'm not sure how to find their probabilities.

2. ## Re: Finding Probability of Bayesian Network

We have to assume the decisions of Yale and Harvard are independent of one another.

Let's first look at the event $Y^c \wedge H^c$

$P[Y^c \wedge H^c] = P[Y^c \wedge H^c |S]P[S] + P[Y^c \wedge H^c |S^c]P[S^c]$

$P[Y^c \wedge H^c |S] = P[Y^c|S]P[H^c|S] = (1-P[Y|S])(1-P[H|S]) = (1-0.94)(1-0.87)=0.0078$

Similarly

$P[Y^c \wedge H^c | S^c] = (1-0.15)(1-0.12) = 0.748$

$P[Y^c \wedge H^c] = (0.0078)(0.005) + (0.748)(0.995) = 0.744299$

Now

$P[S^c | Y^c \wedge H^c] = \dfrac{P[ Y^c \wedge H^c | S^c]P[S^c]}{P[Y^c \wedge H^c]}$

$P[S^c | Y^c \wedge H^c] = \dfrac{(0.748)(0.995)}{0.744299} = 0.999948$


text:
# Typeset Like Jackson's E&M [closed]

I was wondering if there's a package or style to match that of Classical Electrodynamics by J.D. Jackson (there is a multiple pages preview). His integrals are almost vertical and everything related to E&M just looks proper in his style.

-

## closed as too localized by yo', Joseph Wright♦Jan 5 '13 at 21:10

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Welcome to TeX.sx! Can you sample a page or somehow describe what you mean? Otherwise we can't do much. – percusse Dec 15 '12 at 0:02
I am not sure whether we can include here a small image under Fair use. At Amazon there is a free preview showing the typesetting of equations: Jackson on Amazon. It does not look too special to me; vectors are bold and the typeface needs to be identified to emulate the layout. – Alexander Dec 15 '12 at 11:18
Have you tried writing Prof Jackson and asking him if he has a LaTeX style, or at least knows the fonts? – Jim Hefferon Dec 15 '12 at 11:29
the typeface looks like times, – barbara beeton Dec 15 '12 at 20:32
I found a one I like! hea-www.harvard.edu/~alexey/emulateapj/apjfonts.sty – user23332 Dec 15 '12 at 21:32


text:
# Is there an easy proof of this equation related to simultaneous Pell equations?

Working with the famous Baker-Davenport system of simultaneous Pell equations \begin{align} 3x^2-2 &= y^2, & 8x^2-7 &= z^2, \qquad(\star) \end{align} I am left, after a series of substitutions and reductions, with the equation $$u(u+1)\left(96u(u+1)+11\right) = v(v+1). \qquad(\dagger)$$ The point is now to prove that the only solutions with $u \ge 0$ are $u=0$ and $u=5$, yielding, respectively, the two known (and only) solutions $x=1$ and $x=11$ in ($\star$).

Are there any ways of attacking the formulation ($\dagger$), different from those which would naturally be applied to ($\star$), which might lead to a relatively straight-forward proof?

Thanks,
Kieren.

EDIT: For example, by unique factorization of integers, we can write $(v,v+1)=(abc,rst)$ for postive integers $a,b,c,r,s,t$ with $gcd(abc,rst)=gcd(v,v+1)=1$. Now, with appropriate permutation, we can write \begin{align} u &= ar, & u+1 &= bs, & 96u(u+1)+11=ct. \end{align} But all methods and manipulations I've tried from that point seem to lead to dead ends.

-
Noting for documentation sake that OP posted, and then deleted, another question about these equation recently, mathoverflow.net/questions/143301/… – Gerry Myerson Sep 29 '13 at 5:36
@GerryMyerson: I found soon after I posted that the result was trivial, so I didn't think it worthy of remaining on MO. I did not know (1) that deleted posts are still visible, nor (2) that such documentation was required or expected. As you can tell from my reputation points and/or profile, I'm fairly new to MO and MSE — are these rules and expectations listed somewhere? Sorry, Kieren. – Kieren MacMillan Sep 29 '13 at 11:22
Deleted posts are visible to those with enough points. I don't know that documentation is required, but speaking for myself, I like people posting here to put all their cards on the table. – Gerry Myerson Sep 29 '13 at 12:18
@GerryMyerson: I'm certainly not trying to hide anything. I just didn't think that deleted observation was worth MO bandwidth, once I saw how trivial it was. – Kieren MacMillan Sep 29 '13 at 14:17

Sorry, it's probably no easier: both are looking for integer points on a genus-$1$ curve. In fact it's not hard to get from Baker-Davenport to your equation: multiplying the two equations in $(\star)$ yields $(3x^2-2)(8x^2-7) = (yz)^2$, and both $x$ and $yz$ must be odd, so we can write $(x,yz) = (2u+1,2v+1)$, and then subtracting $1$ and dividing by $4$ yields $(\dagger)$.
Yes, that's my "series of substitutions and reductions". I was (am) hoping that there would be (e.g.) a factorization approach applicable to ($\dagger$) that isn't applicable to ($\star$) directly. – Kieren MacMillan Sep 29 '13 at 11:25


text:
# INFRARED SPECTROSCOPY OF LARGE $CO_{2}$ CLUSTERS

Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/18030

Files Size Format View
1989-WF-08.jpg 55.83Kb JPEG image

 Title: INFRARED SPECTROSCOPY OF LARGE $CO_{2}$ CLUSTERS Creators: Disselkamp, R.; Ewing, George E. Issue Date: 1989 Publisher: Ohio State University Abstract: Carbon dioxide clusters were formed by admitting monomeric $CO_{2}$, diluted in an inert gas, into a large cell cooled to $77K^{1}$. The resulting suspension of clusters, each containing of the order of $10^{2}$ molecules, were spectroscopically studied through FTIR techniques. The cluster size is such that the number of molecules in the interior is comparable to the number on the exterior (faces, edges and corners). Spectroscopic features associated with vibrational motions of these interior and exterior molecules are observed. Quantum mechanical and electromagnetic classical descriptions of the experimental results will be presented. Description: $^{1}$ G. Ewing and De T. Sheng, J. Phys. Chem. 92 (1988) 4063. Author Institution: Department of Chemistry, Indiana University URI: http://hdl.handle.net/1811/18030 Other Identifiers: 1989-WF-8


text:
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# Open Physics

### formerly Central European Journal of Physics

Editor-in-Chief: Seidel, Sally

Managing Editor: Lesna-Szreter, Paulina

IMPACT FACTOR 2018: 1.005

CiteScore 2018: 1.01

SCImago Journal Rank (SJR) 2018: 0.237
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Volume 16, Issue 1

# Experimental investigation on optical vortex tweezers for microbubble trapping

Xiaoming Zhou
• College of Information Science and Engineering, Fujian Key Laboratory of Light Propagation and Transformation, Huaqiao University, Xiamen, Fujian 361021, China
• Other articles by this author:
/ Ziyang Chen
• College of Information Science and Engineering, Fujian Key Laboratory of Light Propagation and Transformation, Huaqiao University, Xiamen, Fujian 361021, China
• Other articles by this author:
/ Zetian Liu
• College of Information Science and Engineering, Fujian Key Laboratory of Light Propagation and Transformation, Huaqiao University, Xiamen, Fujian 361021, China
• Other articles by this author:
/ Jixiong Pu
• Corresponding author
• College of Information Science and Engineering, Fujian Key Laboratory of Light Propagation and Transformation, Huaqiao University, Xiamen, Fujian 361021, China
• Email
• Other articles by this author:
Published Online: 2018-07-09 | DOI: https://doi.org/10.1515/phys-2018-0052

## Abstract

In this paper, we investigated the microbubble trapping using optical vortex tweezers. It is shown that the microbubble can be trapped by the vortex optical tweezers, in which the trapping light beam is of vortex beam. We studied a relationship between the transverse capture gradient force and the topological charges of the illuminating vortex beam. For objective lenses with numerical aperture of 1.25 (100×), the force measurement was performed by the power spectral density (PSD) roll-off method. It was shown that transverse trapping forces of vortex optical tweezers increase with the increment of the laser power for fixed topological charge. Whereas, the increase in the topological charges of vortex beam for the same laser power results in the decrease of the transverse trapping forces. The experimental results demonstrated that the laser mode (topological charge) has significant influence on the lateral trapping force.

## 1 Introduction

Ashkin et al. demonstrated that micro-particles can be trapped in three dimensions by laser beams, indicating that laser beams are a very powerful tool in particle trapping and manipulation [1]. Since this pioneer work, optical trapping, which is known as optical tweezers, has attracted much attention. Studies showed that the trapping force depends on particle diameter, polarization direction, depth of the optical trap and numerical aperture [2, 3, 4] In the traditional Gaussian beam-based optical tweezers, some special beams have been introduced to trap particles as well. Allen et al. found that beams with a helical phase structure, i.e. vortex beams, carry orbital angular momentum (OAM) [5]. Jixiong Pu et al. used the structured beams, generated from Devil’s vortex-lens (DVL), to trap microspheres [6]. The OAM of vortex beams can be transferred to particles and make them rotate. The influence of amplitude profile, polarization and laser power on this rotation has been investigated [7, 8, 9, 10, 11]. Owing to the phase singularity, the intensity center of vortex beams is a central dark core. It has been indicated that particles with refractive index lower than the ambient can be trapped by beams of dark core. Therefore, some studies have focused on trapping the particles with refractive index lower than the ambient by optical vortex tweezers [12, 13, 14, 15]. For vortex optical tweezers trapping the particles with refractive index lower than the ambient, to the best of our knowledge, there have been no experimental studies on the relationship between the transverse gradient force and the topological charges of vortex beams. The aim of the study is to experimentally investigate this relationship. In this paper, we used the power spectral density (PSD) method [16, 17, 18, 19, 20] to demonstrate the relationship between the transverse gradient force and the topological charge of the vortex beam.

## 2 Theory of force calibration

There have been several papers studying the optical tweezers for trapping particles using vortex beams. Ng et al. theoretically studied optical trapping by an optical vortex beam. They found that the trap stiffness of cylindrically symmetric optical vortex has real and imaginary numbers [21]. In our experiment, no rotation was observed in our experiment. Thus, the trap stiffness is always real. Therefore, the optical tweezers can be modeled by Hooke’s law [22].

$F=−kixi,$(1)

where ki is the spring constant (trap stiffness) of the trap and xi is the displacement from the centre of the trap. The trap stiffness can be measured using the PSD method based on the OTKBFM-CAL (Thorlabs) software. According to PSD method, the equation of Brownian motion for particles in the optical trap can be described by [20, 23]:

$md2xi/dt2+βdxi/dt+kixi(t)=F(t),$(2)

where m is the mass of the particle, β=6 π η a is the drag coefficient, ηis the fluid viscosity, and a is the radius of the particle. F(t) is a random force, xi(t) is the displacement at time t, dxi/dt is the velocity, and d2xi/dt2 is the acceleration in the xi direction. When the inertial force can be neglected compared with the viscous force, Eq. (2) can be described by a simplified equation as follows [16, 20, 23]:

$βdxi/dt+kixi(t)=F(t).$(3)

By solving the Fourier transform of the above equation, the power spectrum of the captured microsphere is expressed as [16]

$X2(f)=kbT/π2β(f2+fc2),$(4)

where f is the frequency, kb is Boltzmann’s constant, T is the temperature, and the roll-off frequency fc is

$fc=ki/2πβ.$(5)

In the experiment, the dependence of the power spectrum X2(f) on frequency f can be measured, and the roll-off frequency fc can be determined by Eq. (4). The transverse trap stiffness ki can be obtained by substituting the calculated roll-off frequency into Eq. (5) [16, 20].

## 3 Experimental results

The experimental setup of optical vortex tweezers was shown in Figure 1. The laser source was a single-mode laser diode with a maximum output power of P = 340 mW of a wavelength of 975 nm. Spiral phase plates (SPP) were used to transform the incident Gaussian beam into a vortex one. The topological charge of the vortex beam was determined by the phase structure of the SPP. The expanded vortex beam was focused by an objective with NA of 1.25. The particles could be trapped near the focus of the objective if the gradient force was larger than that of scattering and absorption force.

Figure 1

Schematic of the experimental setup. L-lens, M-reflection mirror, OB-objective lens, DM-dichroic mirror, LDC-laser diode and controller, BE-beam expander, F1(IR)-IR filters, F2(ND)-neutral density filters, CCD-charge coupled device, QPD-quadrant position detector, spp-Spiral phase plates, LED-Light-Emitting Diode.

In the PSD method, the quadrant position detector (QPD) needed a high bandwidth to record the particle Brownian motion information. To do the measurements with broad bandwidth and high-resolution, a QPD was placed in the conjugate back-focal plane of the condenser lens (Nikon 10X Air Condenser, numerical aperture of 0.25). The signal generated by the QPD was sensitive to the relative displacement of the trapping particles from the centre of the optical trap, which could be used to determine the position, stiffness, and force of the optical tweezers. The experimental data (voltage signal) from the QPD was loaded into the mechanical acquisition module OTKBFM-CAL (Thorlabs Products). The position calibration and the PSD could be obtained by the software. The work principle of the software is described as follows:

The voltage signal measured in the experiment is V(t) = ξxi(t), ξ is constant. The displacement signal V(t) was measured by QDP. Then we got V(t)-t curve from the software. According to V2(f) = ξ2X2(f)=ξ2kbT/π2β(f2+f2c) and Eq. (4), PSD function can be obtained by performing calculation with software, which is shown in Figure 2. According to Figure 2, we could get the roll-off frequency fc (corner frequency) by Lorentzian fitting PSD function, which is the abscissa value of the left blue line. Finally, the trap stiffness could be obtained by substituting fc into the Eq. (5).

Figure 2

PSD as a function of frequency, for finding the roll-off frequency fc, being 2.3923 × 10 Hz

In the experiment, we focused on the transverse forces of vortex optical tweezers acting on microbubbles (F-30VS, Matsumoto Yushi-Seiyaku). The microbubble was an alkane sphere (n3 = 1.4) surrounded by a transparent dielectric spherical shell (n2 = 1.6). The radius of the microbubbles ranges between 3 to 7 μm, and that of alkane sphere ranges between 2 to 5 μm. Compared with other types of microbubbles, this type of microbubble was found to be more stable. The influence of the temperature on the shape of the microbubble could be neglected as the non-recoverable thermal expansion occurs only when the sample was heated to 80°C. In the experiment, the power of the laser was chosen to be 100 mW, which could heat the sample by approximately 1° C.

We placed the sample between two plastic slides with built-in channels. To avoid excessively concentrating the sample, we mixed it with deionized water to a dilution ratio of 1:10. The sample was placed near the focus of the objective, and the plastic slides were placed on a displacement stage with a differential regulator (MAX300 NanoMax 3-Axis Flexure Stage, Thorlabs). The displacement stage provided a 4 mm (0.16 inch) coarse stroke and 300 μm fine adjustment stroke, with coarse adjuster 10 μm graduated scale, and the micro adjuster 1 μm graduated scale. To trap particles, the light beam should be tightly focused, and the tightly focused pattern was appropriately adjusted with a NanoMax 3-Axis flexure stage. We showed that the particle in the sample could be trapped and moved by the optical tweezers, as shown in Figure 3. The trapped particle was indicated by the arrow of Figure 3.

Figure 3

(a) A free particle; (b), (c), and (d) a captured particle (the arrow indicates the captured particle; the others are uncaptured microbubbles as a reference)

We wanted to study the influence of the topological charge of the incident light beams on the optical trapping force. The incident vortex beams with topological charge of l = 1, 2 and 3 were generated for trapping microbubbles. The transverse stiffness of optical vortex tweezers was calculated by PSD. For a vortex beam with fixed topological charge, we performed measurements ten times. Owing to that the experimental data are uncertain, we use Box-plot to express the discrete distribution of the experimental data, which is given in Figure 4. It was shown that from the experimental measurements that the transverse stiffness decreases with the increasing topological charge of the vortex beams. This was consistent with theoretical simulations [24, 25]. This result could be understood by the intensity distribution of the vortex beams. The intensity pattern of a tightly focused vortex beam was a dark core, surrounding by a bright annulus. The size of the dark core increases with the increasing topological charge of the vortex beam. Therefore, the magnitude of the intensity of the bright annulus decreased, which led to a lower stiffness.

Figure 4

Relationship between the transverse stiffness and the topological charge. The power of the laser is 6.073 mW

In a Gaussian beam-based optical tweezers, the trapping was influenced by the power of the incident laser. In general, the trapping force increased with a higher laser power [2]. We also investigated the influence of the laser power on the optical vortex tweezers. We took the vortex beam with topological charge of l = 3 as an example. The influence of the laser power on the transverse stiffness was presented in Figure 5. It was found that the increasing laser power results in larger transverse stiffness. This result was similar to that of Gaussian beam-based optical tweezers.

Figure 5

The transverse stiffness as a function of the incident laser power. The topological charge of the vortex beam was chosen as l = 3.

## 4 Conclusions

We had shown that the particles could be trapped and moved in an optical vortex tweezers, which was achieved by introducing a vortex beam as the incident beam. The effect of the topological charge and the power of the incident beam on the transverse stiffness was investigated. A high stiffness could be obtained by vortex beam with lower topological charge and stronger power. This finding was particularly significant for research on transverse trapping forces based on vortex beam.

## Acknowledgement

This work was supported by the National Natural Science Foundation of China (Grant Nos. 61575070, 11674111, and 61505059) and the Promotion Program for Young and Middle-Aged Teachers in Science and Technology Research of Huaqiao University (ZQN-PY209)

## References

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Ashkin A., Dziedzic J.M., Bjorkholm J.E., Chu S., Observation of a single-beam gradient force optical trap for dielectric particles, Opt. Lett., 1986, 11(5), 288.

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Wright W.H., Sonek G.J., Berns M.W., Parametric study of the forces on microspheres held by optical tweezers, Appl. Opt., 1994, 33(9), 1735-48.

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Zhong M., Zhou J., Wu J., Li Y., Determination of the axial stiffness of an optical trap with information entropy signals, Proceedings of SPIE, Int. Soc. Opt. Eng., 2009, 7507, 75070J-75070J-7. Google Scholar

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Liang Y.Z., Ou Q.R., Bo L., Liang R.Q., Improvement of transverse trapping efficiency of optical tweezers, Chinese Phys. Lett., 2008, 25(6), 2300-2302.

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Allen L., Beijersbergen M.W., Spreeuw R.J., Woerdman J.P., Orbital angular momentum of light and the transformation of laguerre-gaussian laser modes, Phys. Rev. A Atom. Mol. Opt. Phys., 1992, 45(11), 8185.

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Pu J., Jones P.H., Devil’s lens optical tweezers, Opt. Expr., 2015, 23(7), 8190-9.

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Volkesepúlveda K., Chávezcerda S., Garcéschávez V., Dholakia K., Three-dimensional optical forces and transfer of orbital angular momentum from multiringed light beams to spherical microparticles, J. Opt. Soc. Amer. B, 2004, 21(10), 1749-1757.

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Kovalev A.A., Kotlyar V.V., Porfirev A.P., Optical trapping and moving of microparticles by using asymmetrical laguerre-gaussian beams, Opt. Lett., 2016, 41(11), 2426.

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Li M., Yan S., Yao B., Liang Y., Zhang P., Spinning and orbiting motion of particles in vortex beams with circular or radial polarizations, Opt. Expr., 2016, 24(18), 20604-20612.

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Gao H., Rotation dynamics of yeast cell in vortex optical tweezers, Chin. J. Lasers, 2011, 38(4), 0404002, (in Chinese). Google Scholar

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O’Neil A.T., Padgett M.J., Axial and lateral trapping efficiency of laguerre–gaussian modes in inverted optical tweezers, Opt. Comm., 2001, 193(1), 45-50.

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Prentice P.A., Macdonald M.P., Frank T.G., Cuschieri A., Spalding G.C., Sibbett W. et al., Manipulation and filtration of low index particles with holographic laguerre-gaussian optical trap arrays, Opt. Expr., 2004, 12(4), 593-600.

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Swartzlander G.A., Gahagan K.T., Trapping of low-index microparticles in an optical vortex, J. Opt. Soc. Amer. B, 1998, 15(15), 524-534.

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Ahluwalia B.S., Løvhaugen P., Hellesø O.G., Waveguide trapping of hollow glass spheres, Opt. Lett., 2011, 36(17), 3347-9.

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Fury C., Harfield C., Memoli G., Jones P.H., Stride E., Experimental characterisation of holographic optical traps for microbubbles, Proceedings of SPIE - Int. Soc. Opt. Eng., 2014, 9126, 91263L-10, 1117/12.2055889. Google Scholar

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Sarshar M., Wong W.T., Anvari B., Comparative study of methods to calibrate the stiffness of a single-beam gradient-force optical tweezers over various laser trapping powers, J. Biomed. Opt., 2014, 19(11), 115001.

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Fei P., Yao B.L., Ming L., Yan S.H., Measurement of optical trapping force and stiffness of micro-particles with the drag-force method, J. Optoel. Laser, 2010, 21(1), 78-82, (in Chinese). Google Scholar

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Berg-Sørensen K., Flyvbjerg H., Power spectrum analysis for optical tweezers, Rev. Sci. Instr., 2004, 75(3), 594-612.

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Ng J., Lin Z., Chan C.T., Theory of optical trapping by an optical vortex beam, Phys. Rev. Lett., 2010, 104(10), 103601.

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Mclaren M., Sidderashaddad E., Forbes A., Accurate measurement of microscopic forces and torques using optical tweezers, S. Afr. J. Sci., 2011, 107(9/10), 67-74.

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Mazur P., On the theory of brownian motion, Physica, 1959, 25(1-6), 149-162.

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Wang J., Ren H., Trapping forces of core-shell particles using laguerre-gaussian beams, Chinese J. Lasers, 2015, 42(6), 0608001, (in Chinese). Google Scholar

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Accepted: 2018-05-15

Published Online: 2018-07-09

Citation Information: Open Physics, Volume 16, Issue 1, Pages 383–386, ISSN (Online) 2391-5471,

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# Convert from Nested Square Roots to Sum of Square Roots

I am looking for a way to easily discover how to go from a nested root to a sum of roots. For example,

$$\sqrt{10-2\sqrt{21}}=\sqrt{3}-\sqrt{7}$$

I know that if I set $\alpha=\sqrt{10-2\sqrt{21}}$, square both sides, I get

$$\alpha^2=10-2\sqrt{21}$$

Now I recognize that we have a situation where $10=3+7$ and $21=7\cdot 3$, so I can immediately see that we have

$$\alpha^2=10-2\sqrt{21}=3-2\sqrt{21}+7=\sqrt{3}^2-2\sqrt{3}\sqrt{7}+\sqrt{7}^2=(\sqrt{3}-\sqrt{7})^2$$

My question is, it this the only way to approach this problem? This approach mirrors basic algebra 1 methods of factoring quadratics, but I was curious to know if thre are other techniques that can be used to quickly deduce that a nested radical can be simplified to the sum of two radicals. Mathematically, suppose $a,b,c,m,n,r,s\in\mathbb{N}$. Is there a way to quickly determine $m,n,r,s$ in the equation

$$\sqrt{a\pm b\sqrt{c}}=m\sqrt{r}\pm n\sqrt{s}$$

## 2 Answers

Suppose $a+b\sqrt{c}=(m\sqrt{r}+n\sqrt{s})^2$, where $a,b,c,m,n,r,s\in\mathbb{Z}$ and $r,s,c>0$. Then you have the following: $$a=m^2r+n^2s\\ b=2mn\\ c=rs\\ a+b\sqrt{c}>0$$

This shows that $b$ must be even, and the gcd's of $m,n$ and $r,s$ each divide $a$. I see no reason why this set of diophantine equations has a particularly clean solution. As an example of a case where you cannot get the desired form we have $\sqrt{64+35\sqrt{3}}$.

• What I thought. Rearranging with a formula just doesn't work. It's like finding the formula of a number's HCF. – Bradman175 May 8 '16 at 3:37
• Yeah, if I had a formula for that, I'd be rich. – J. David Taylor May 8 '16 at 4:27
• you could make a program to solve it lol. – Bradman175 May 8 '16 at 4:28
• Oh, yeah. I suppose the Euclidean algorithm does the trick. I somehow read your comment as implying an efficient algorithm for factoring integers. hehe – J. David Taylor May 8 '16 at 4:29

Let $a$ and $b$ be non-negative rational numbers such that $$\sqrt{10 - 2\sqrt{21}} = \sqrt{a} - \sqrt{b}$$ Squaring both sides of the equation yields $$10 - 2\sqrt{21} = a - 2\sqrt{ab} + b$$ Matching rational and irrational parts yields the system of equations \begin{align*} a + b & = 10 \tag{1}\\ -2\sqrt{ab} & = -2\sqrt{21} \tag{2} \end{align*} Dividing both sides of equation 2 by $-2$ and squaring yields $$ab = 21$$ Solving for $b$ yields $$b = \frac{21}{a}$$ Substituting this expression in equation 1 yields $$a + \frac{21}{a} = 10$$ Multiplying both sides of the equation by $a$ and solving the resulting quadratic equation yields \begin{align*} a^2 + 21 & = 10a\\ a^2 - 10a + 21 & = 0\\ (a - 3)(a - 7) & = 0 \end{align*} The roots are $a = 3$ and $a = 7$. If $a = 3$, then $b = 21/a = 21/3 = 7$. However, $$\sqrt{3} - \sqrt{7} < 0 \implies \sqrt{10 - 2\sqrt{21}} \neq \sqrt{3} - \sqrt{7}$$ since the principal square root of a number cannot be negative.

If $a = 7$, then $b = 21/a = 21/7 = 3$. Hence, $$\sqrt{10 - 2\sqrt{21}} = \sqrt{7} - \sqrt{3}$$

If you instead had to evaluate $\sqrt{7 + 4\sqrt{3}}$, set $$\sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b}$$


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Mathematical Modelling: A case studies approach
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Mathematical modelling is a subject without boundaries. It is the means by which mathematics becomes useful to virtually any subject. Moreover, modelling has been and continues to be a driving force for the development of mathematics itself. This book explains the process of modelling real situations to obtain mathematical problems that can be analyzed, thus solving the original problem.

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text:
# How can I get Latin Modern/Computer Modern serif double-struck/blackboard-bold characters which copy/paste as unicode when using unicode-math?

I want my characters to copy-paste as Unicode when I copy them from the PDF. I also want them to look like the traditional Latin Modern font. For unrelated reasons, I need to use LuaLaTeX rather than pdflatex. The unicode-math package works fine in most cases, but produces strange sans-serif double-struck characters. How do I fix this?

Example:

\documentclass{article}
\usepackage{amssymb}
\begin{document}
$\alpha\beta\gamma\pi\sum\prod1+2+3+4+5+6+7+8+9+0/1!$
$abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$
$\mathbb{abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ}$
\end{document}

in lualatex gives

whereas

\documentclass{article}
\usepackage{amssymb}
\usepackage{unicode-math}
\begin{document}
$\alpha\beta\gamma\pi\sum\prod1+2+3+4+5+6+7+8+9+0/1!$
$abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$
$\mathbb{abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ}$
\end{document}

in lualatex gives

While most of the characters are identical are nearly identical, the capital double-struck characters are clearly not. They don't even have serifs, unlike the rest of the characters. It's cool that I can now get lowercase double-struck characters, but this doesn't make up for the jarring and stylistically very-different uppercase double-struck characters. The answer at https://tex.stackexchange.com/a/360609/2066 gives the correct characters, but then the characters don't copy-paste as double-struck. How do I get the Latin Modern / Computer Modern double-struck characters back without losing the unicode-copy-paste feature of unicode-math?

(I'm also curious about the history here; why are these characters so different? https://tex.stackexchange.com/a/525235/2066 says "Latin Modern Math is a clone of Knuth’s Computer Modern, and unicode-math produces nearly-identical output under the default settings (with a few caveats: the \mathcal alphabet is based on Euler Calligraphic instead of traditional LaTeX).")

• You say "traditional latin modern" but your first code block doesn't use latin modern at all, the first two rows use computer modern fonts and the last \mathbb uses AMS msbm10 font, the AMS fonts don't really blend that well with computer modern at all, and are based on designs aimed to fit with the in house AMS publishing fonts. Conversely with unicode-math all three rows are set from the same font, latin modern Math. Jan 5, 2021 at 0:43
• You might prefer an alternative computer modern based OTF math font \setmathfont{New Computer Modern Math} Jan 5, 2021 at 0:46
• You can also select any OpenType or TrueType font as \mathbb with \setmathfont[range=bb]. Jan 5, 2021 at 4:00

The classic TeX double struck font you show in the first code block is not derived from Computer Modern at all it is the AMS symbol font msbm10 (and the first two rows are set in Computer Modern, not Latin Modern)

The second block is using a consistent font for all three rows, Latin Modern Math Opentype.

There are other Computer Modern inspired OpenType Math fonts, notably NewComputerModernMath. You may prefer the double struck math in this font:

\documentclass{article}
\usepackage{amssymb}
\usepackage{unicode-math}
\setmathfont{New Computer Modern Math}
\begin{document}
$\alpha\beta\gamma\pi\sum\prod1+2+3+4+5+6+7+8+9+0/1!$
$abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ$
$\mathbb{abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ}$
\end{document}

• The uppercase double struck letters do not have serifs on my system (updated TeX Live 2021) for some reason. (with the newcomputermodern fonts) Aug 30, 2021 at 13:59
• @ApoorvPotnis the above document makes this output in a texlive 2021 updated two minutes ago. If you get something different make a test document and post as a question Aug 30, 2021 at 14:11


text:
# Importance Sampling Simulation Method

FIND A SOLUTION AT Academic Writers Bay

The principle is also illustrated in Figure 11.4. Fxj (xi)
z ‘ci Zji
Simulated sample
Figure 11.4 Principle for simulation of a random variable.
This process is the continued until all components of the vector fi have been generated.
Importance Sampling Simulation Method As already mentioned the problem in using Equation (11.45) is that the sampling function fa(x) typically is located in a region far away from the region where the indicator function /[g(x) 5 0] attains contributions. The success rate in the performed simulations is thus low. In practical reliability assessment problems where typical failure probabilities are in the order of 10 3 — le this in turn leads to the effect that the variance of the estimate of failure probability will be rather large unless a substantial amount of simulations are performed. To overcome this problem different variance reduction techniques have been proposed aiming at, with the same number of simulations to reduce the variance of the probability estimate. In the following we shall briefly consider one of the most commonly applied techniques for variance reduction in structural reliability applications, namely the importance sampling method.
The importance sampling method takes basis in the utilisation of prior information about the domain contribution to the probability integral, i.e. the region that contributes to the indicator function. Let us first assume that we know which point in the sample space x. contributes the most to the failure probability. Then by centring the simulations on this point, the important point, we would obtain a higher success rate in the simulations and the variance of the estimated failure probability would be reduced. Sampling centred on an important point may be accomplished by rewriting Equation (11.42) in the following way: () Pf = lig(x) Olfx(x)clx = l[g(x) 5 0 jfxx f,(x)dx (11.48)fs (x)

READ ALSO...   Poverty has a strong influence

QUALITY: 100% ORIGINAL PAPERNO PLAGIARISM – CUSTOM PAPER


text:
Next: Subspace Iteration Up: Single- and Multiple-Vector Iterations Previous: Power Method   Contents   Index

## Inverse Iteration

Inverse iteration, described in Algorithm 4.2, can also be used to solve the NHEP without any apparent change.

As in the Hermitian case, assume that and are an eigenvalue and eigenvector pair of so that is the largest eigenvalue of in magnitude. The inverse power method converges if the starting vector is not perpendicular to . The convergence rate is , where is an eigenvalue of such that is the second largest eigenvalue of in magnitude.

In general, inverse iteration tends to have much more rapid convergence than the power method if is chosen to be very close to a desired eigenvalue. However, inverse iteration does require a factorization of the matrix , making it less attractive when this factorization is expensive.

Susan Blackford 2000-11-20


text:
# 8.3: Worldwide Voyage

To Prepare for next activity:

1. Read this description of the daily life on Hōkūle`a: http://pvs.kcc.hawaii.edu/ike/canoe_living/daily_life.html.
2. Watch the video about the Worldwide Voyage:

From the webpage above, you learned:

The quartermaster is responsible for provisioning the canoe — loading food, water and all needed supplies, and for maintaining Hōkūle`a’s inventory. While this is not an on board job, it is critical to the safe and efficient sailing of the canoe.

Problem 3

Imagine that you are part of the crew for the Worldwide Voyage, and you are going to help the quartermaster and the captain with provisioning the canoe for one leg of the voyage. You need to write a preliminary report for the quartermaster, documenting:

1. Which leg of the trip are you focused on? (See the map below.)
2. How long will that leg of the trip take? Explain how you figured that out.
3. How much food and water will you need for the voyage? Explain how you figured that out.

The rest of this section contains pointers to information that may or may not be helpful to you as you make your plans and create your report. Your job is to do the relevant research and then write your report. You should include enough detail about how you came to your conclusions that the quartermaster can understand your reasoning.

## Pick a leg of the route:

Here’s a picture of the route planned for the Worldwide Voyage, which you can find at the Worldwide Voyage website: http://www.hokulea.com/worldwide-voyage/ and a full-sized map here: https://tinyurl.com/WWVmap. On the map, the different colors correspond to different years of the voyage. A “leg” means a dot-to-dot route on the map.

After you pick a leg of the voyage, you’ll need to figure out the total distance of that leg. This tool might help (or you can find another way): http://www.acscdg.com/.

Here is some relevant information to help you figure out how long it will take Hōkūle`a to complete your chosen leg:

• The first trip from Hawai`i to Tahiti in 1976 took a total of 34 days. (You probably want to use the tool above to compute the number of nautical miles.)

## Plan the provisions:

Here is some information about provisions.

• Hōkūle`a can carry about 11,000 pounds, including the weight of the crew, provisions, supplies, and personal gear.
• The supplies (sails, cooking equipment, safety equipment, communications equipment, etc.) account for about 3,500 pounds.
• The crew eats three meals per day and each crew member gets 0.8 gallons of water per day.
• For a trip that is expected to take 30 days, the quartermaster plans for 40 days’ worth of supplies, in case of bad weather and other delays.


text:
How would you determine the oxidation number of Sulpher in SO_3^-2?

The oxidation state of sulfur in $S {O}_{3}^{2 -}$ is $+ I V$.
The sum of the oxidation numbers/states of the individual elements in the species must equal the charge on the ion. $O$ normally takes a $- I I$ oxidation state and it does here.
Thus $- 2 = 3 \times \left(- 2\right) + O {N}_{\text{sulfur}}$
And so $O {N}_{\text{sulfur}} = - 2 + 6 = + 4 ,$ i.e. $+ I V$.


text:
# SPSS mixed models - using classroom as an effect in the model

I'm looking at data from a health intervention study done in one middle school (16 classes/"clusters" at that school). Half assigned to control, half to intervention.

Is this an appropriate way to include the classes/clusters in the mixed model syntax in SPSS? I am unsure how to include this in the model. I have read that it would be a random effect, but I have also seen "schools" used as a fixed effect in others' models.

DV1 = my main dependent variable
Time = 4 time points
RandomizedGrp = 0 (intervention), 1 (control)
Class = coded 1-16 individual classes
(x) used in some places to denote multiplication

MIXED DV1 BY Time RandomizedGrp Class

/CRITERIA=CIN(95) MXITER(100) MXSTEP(10) SCORING(1) SINGULAR(0.000000000001) HCONVERGE(0,
ABSOLUTE) LCONVERGE(0, ABSOLUTE) PCONVERGE(0.000001, ABSOLUTE)

/FIXED=Time RandomizedGrp Class Time(x)Randomized Time(x)Randomized(x)Class  | SSTYPE(3)

/METHOD=ML

/PRINT=SOLUTION TESTCOV

/REPEATED=Time | SUBJECT(ID) COVTYPE(UN)

/EMMEANS=TABLES(Time)

/EMMEANS=TABLES(RandomizedGrp)

/EMMEANS=TABLES(Time*RandomizedGrp)

/EMMEANS=TABLES(Time*RandomizedGrp*Class).


text:
# 9.1 Multi-Stage Sampling: Two Stages with S.R.S at Each Stage

Printer-friendly version
 Unit Summary multi-stage design two stages with simple random sampling at each stage unbiased estimator ratio estimator

We have learned about cluster sampling where one selects the primary units and then all of the cases from the secondary units. With multi-stage sampling we will only select some of the units from the secondary stages.

For example, in two-stage sampling:

• 1st stage samples n primary units
• 2nd stage, for the ith primary unit, selects mi (not all) secondary units

Multistage designs are used in many practical cases. These are just a few:

1. Large surveys involving the sampling of housing units - The U.S. Census Bureau selects geographical areas within each state and then select housing units within each selected geographical area.
2. Practical quality control problems often involve two (or more) stages of sampling. For example, Ford wants to inspect the quality of a supplier of air filters. They first sample some cartons and then inspect some air filters inside these selected cartons.
3. Gallop poll samples approximately 300 election districts. At the second stage, they select 5 households per district.

Notation:

• N : number of primary units in the population
• Mi : number of secondary units in the ith primary unit
• $y_i=\sum\limits_{j=1}^{M_i}y_{ij}$
• population total : $\tau=\sum\limits_{i=1}^N \sum\limits_{j=1}^{M_i}y_{ij}$
• $\mu=\dfrac{\tau}{M}$ where $M=\sum\limits_{i=1}^N M_i$
• n : number of primary units selected in the first stage
• mi : number of secondary units selected in the second stage

Two-stage sampling includes both one-stage cluster sampling and stratified random sampling as special cases.  When does two-stage sampling reduce to cluster sampling?  When does two-stage sampling reduce to stratified random sampling?

[Come up with an answer to this question and then click on the icon to reveal the solution.]

#### Multistage Design

This is something that arises in practice quite often. As a result, we need to be able to figure out how this type of sampling design is implemented. Most of the time this deals with two stages of sample with simple random sampling at each stage.

Let's take a look at this graph as a means of understanding how this type of sampling design plays out.

N = 50 for both graphs

(i) . Two-stage sample of 10 primary units
and four secondary units per primary unit.

Here is another graph for another example of two-stage sample

(ii)  Two-stage sample of 20 primary units
and two secondary units per primary unit.

#### Two-stage cluster sampling with simple random sampling at each stage

We will discuss two possible estimators for this sampling design: unbiased estimator and ratio estimator.

A. Unbiased Estimator

Since simple random sampling is used in the second stage, an unbiased estimator of the total y-value for the ith primary unit is:

$\hat{y}_i=M_i \dfrac{\sum\limits_{j=1}^{m_i}y_{ij}}{m_i}=M_i \bar{y}_i$   where  $\bar{y}_i=\dfrac{\sum\limits_{j=1}^{m_i}y_{ij}}{m_i}$

The first part of this formula is also known as the expansion estimator.

Also, since simple random sampling is used in the first stage, an unbiased estimator for the population total is:

$\hat{\tau}=N\cdot \dfrac{\sum\limits_{i=1}^n\hat{y}_i}{n}=N \cdot \dfrac{\sum\limits_{i=1}^n M_i \bar{y}_i}{n}$

Now we have the expansion estimators from each stage. The next thing we need is the variance.

The estimated variance of $\hat{\tau}$ is:

$\hat{V}ar(\hat{\tau})=N(N-n)\dfrac{s^2_u}{n}+\dfrac{N}{n}\sum\limits_{i=1}^n M_i (M_i-m_i) \dfrac{s^2_i}{m_i}$

su2 is the sample variance among the primary unit totals,
si2 is the sample variance within the ith primary unit, here

$s^2_u=\dfrac{1}{n-1}\sum\limits_{i=1}^n \left(\hat{y}_i-\dfrac{\sum\limits_{i=1}^n \hat{y}_i}{n}\right)^2$, and $s^2_i=\dfrac{1}{m_i-1}\sum\limits_{j=1}^{m_i}(y_{ij}-\bar{y}_i)^2$

To estimate the population mean μ = τ / M, the estimators and the estimated variance are:

$\hat{\mu}=\dfrac{N}{M}\cdot \dfrac{\sum\limits_{i=1}^n \hat{y}_i}{n}$, and $\hat{V}ar(\hat{\mu})=\dfrac{1}{M^2}\hat{V}ar(\hat{\tau})$

Let's take a look at an example where we can compute both the estimates and their variances.

#### Example - Restaurant Employee Satisfaction

A restaurant chain wants to estimate the average employee satisfaction with their job (the scale is from 1 to 7). They have 120 restaurants the total number of employees in the chain is 6860. They use simple random sampling to sample 10 restaurants. They then use simple random sampling to sample and interview about 20% of the employees in those restaurants,. The data are given as follows.

 Restaurant Mi mi Employee Satisfaction $\bar{y}_i$ si 1 54 10 5, 7, 6, 5, 4, 7, 6, 6, 4, 5 5.50 1.08 2 48 10 7, 7, 7, 6, 5, 4, 7, 7, 6, 6 6.20 1.03 3 68 14 5, 6, 5, 6, 4, 5, 6, 5, 4, 5, 4, 6, 5, 6 5.14 0.77 4 70 14 6, 5, 7, 6, 7, 6, 5, 7, 5, 7, 6, 5, 7, 6 6.07 0.83 5 52 10 4, 5, 4, 5, 5, 6, 5, 4, 4, 4 4.60 0.70 6 62 12 5, 7, 6, 7, 4, 3, 1, 5, 4, 6, 4, 5 4.75 1.71 7 41 8 7, 6, 7, 7, 6, 6, 5, 7 6.38 0.74 8 53 11 6, 6, 5, 4, 6, 7, 5, 5, 7, 6, 5 5.64 0.92 9 64 12 7, 6, 5, 4, 6, 5, 7, 4, 3, 6, 5, 7 5.42 1.31 10 43 9 7, 6, 6, 5, 7, 3, 5, 4, 5 5.33 1.32

Minitab output:

Here we have output from Minitab that provides the descriptive statistics that you will need to compute the estimators and variance.

#### Application Exercise

Find the unbiased estimator for the mean employee satisfaction score.

[Come up with an answer to this question and then click on the icon to reveal the solution.]

The estimated variance of the unbiased estimator is then:

$\hat{V}ar(\hat{\tau})=N(N-n)\dfrac{s^2_u}{n}+\dfrac{N}{n}\sum\limits_{i=1}^n M_i (M_i-m_i) \dfrac{s^2_i}{m_i}$

su2 is the sample variance of $\hat{y}_1,\ \hat{y}_2,\cdots,\ \hat{y}_{10}$. From the Minitab output, su2 = (58.1)2 = 3375.61

si2 is the sample variance within the primary unit.

$s^2_i=\dfrac{1}{m_i-1}\sum\limits_{j=1}^{m_i}(y_{ij}-\bar{y}_i)^2$

si has been computed and given in the table.

#### Application Exercise

Find the estimated variance of the unbiased estimator for the mean employee satisfaction score.

[Come up with an answer to this question and then click on the icon to reveal the solution.]

Remark: If M is unknown, we cannot use the unbiased estimator $\hat{\mu}$.

If the cluster total is proportional to the cluster size, then the ratio estimate is appropriate.  We will discuss the ratio estimator in the following:

B. Ratio estimator

For the population total, the ratio estimator and its estimated variance are:

$\hat{\tau}_r=\dfrac{\sum\limits_{i=1}^n \hat{y}_i}{\sum\limits_{i=1}^n M_i}\cdot M=\hat{r}M$

$\hat{V}ar(\hat{\tau}_r)=\dfrac{N(N-n)}{n}\cdot \dfrac{1}{n-1}\sum\limits_{i=1}^n(\hat{y}_i-M_i\hat{r})^2+\dfrac{N}{n}\sum\limits_{i=1}^n M_i(M_i-m_i)\dfrac{s^2_i}{m_i}$

A similar question can be asked of the population mean. Therefore, for the population mean, the ratio estimator and its estimated variance are:

$\hat{\mu}_r=\hat{r}$

$\hat{V}ar(\hat{\mu}_r)=\dfrac{1}{M^2}\hat{V}ar(\hat{\tau}_r)$

#### Application Exercise

For the example using the Restaurant Employee's Satisfaction data above, find the ratio estimator for the population mean and it estimated variance.

[Come up with an answer to this question and then click on the icon to reveal the solution.]

Remark: If M is unknown, one can use $\hat{\mu}_r$ and estimate M by:

$\dfrac{\sum\limits_{i=1}^n M_i}{n}\times N$

Recall: $M=\sum\limits_{i=1}^N M_i$


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# A grocer is storing soap boxes in cartons that measure 25

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A grocer is storing soap boxes in cartons that measure 25 [#permalink]  03 Apr 2013, 06:33
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A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 5 inches, then what is the maximum number of soap boxes that can be placed in each carton?

A. 210
B. 252
C. 280
D. 300
E. 420
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Apr 2013, 07:04, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Soap boxes in a carton [#permalink]  03 Apr 2013, 06:43
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deepri0812 wrote:
A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 5 inches, then what is the maximum number of soap boxes that can be placed in each carton?

A -210
B - 252
C - 280
D -300
E-420

Area of the box = $$25*42*60=63000$$
Area of one soap = $$7*6*5=210$$
Tot number = $$\frac{63000}{210}=300$$

You have to calculate the area of the box, then divide it by the area of one soap, and you ll find the number of soaps that will fit into the box
D
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Re: Soap boxes in a carton [#permalink]  03 Apr 2013, 06:52
deepri0812 wrote:
A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 5 inches, then what is the maximum number of soap boxes that can be placed in each carton?

A -210
B - 252
C - 280
D -300
E-420

Please suggest the method to solve such problems.

Maximum number of soap boxes that can be placed in each carton = Volume of each carton/ volume of each box
= (25*42*60)/(7*6*5)= 300(option D)
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Re: Soap boxes in a carton [#permalink]  03 Apr 2013, 07:28
Zarrolou wrote:
deepri0812 wrote:
A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 5 inches, then what is the maximum number of soap boxes that can be placed in each carton?

A -210
B - 252
C - 280
D -300
E-420

Area of the box = $$25*42*60=63000$$
Area of one soap = $$7*6*5=210$$
Tot number = $$\frac{63000}{210}=300$$

You have to calculate the area of the box, then divide it by the area of one soap, and you ll find the number of soaps that will fit into the box
D

I am getting confused between volume approach as done by you and my thought process:

Think about lining up the dimensions of the small boxes along the dimensions of the carton so that you can fill the carton with no wasted space. Since one edge of the small box is 7 inches, that edge of the small boxes should be placed along the side of the carton that is 42 inches long, since 42 is the only dimension that can be divided by 7 evenly. Thus you can put six small boxes along that side, but if you do that 6*6 soap boxes will have width 36 which would be greater than 25 & will exceed the carton width. so how can we just divide the volume of the carton & the soap's volume.

Or am I thinking too much ??
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Re: Soap boxes in a carton [#permalink]  03 Apr 2013, 08:20
I guess you are thinking too muck
For example you say:
"Since one edge of the small box is 7 inches, that edge of the small boxes should be placed along the side of the carton that is 42 inches long, since 42 is the only dimension that can be divided by 7 evenly."
Are you sure? I mean, what you say isn't wrong, but is not correct either (if we want to answer this particular question). On a side of 42 you can place 6 soaps on their "7-side"(6*7=42) or 2 soaps on their "6-side" and 6 soaps on their "5 side"(6*2+6*5=42). You can also rotate those soaps...
The correct approach is also the simplest in this case.

Hope it's clear
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  16 Apr 2013, 08:24
The answer works in this case... what if the dimension of the soap box was 6 7 8... This approach would fail then?
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  16 Apr 2013, 08:50
I'm not convinced... need some experts to help here....

I think the final answer should also include some sort of surface area calculations as well.

Hear my logic out

you have a box which is 42 60 25.. The best way to fit the soap box is 7 6 8

So you are going to have an array of 42/7=6 by 60/6=10 = 60 boxes

Now in 25 8 can only go 3 times

So total no of boxes that can be filled is 60x3=180
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  15 Nov 2014, 07:44
Much easier if we use fraction:

25*42*60/5*7*6=5*6*10=300

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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  23 Nov 2014, 17:59

$$\frac{25*42*60}{5*7*6} = 5*6*10 = 300$$
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  23 Feb 2015, 06:23
Hi all,

Cartoon LBH = 25 * 42 * 60

Box LBH = 5 * 7 * 6

One layer of boxes in cartoon = (5*2)*(4*2) = 80
For H of 60, we can have 10 such layers. Hence # of boxes = 800

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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  23 Feb 2015, 12:56
Expert's post
Hi thorinoakenshield,

To maximize the number of boxes that we can put in this carton, we have to use as much of (if not all of) the space in the carton. Thus, we have to "orient" the boxes in such a way that we use up all of the space in the carton.

The carton's dimensions are 25 x 42 x 60 and each box has dimensions of 5 x 6 x 7

To maximize the number of boxes, I'm going to "line up" the 5 with the 25 and the 7 with the 42.

25/5 = 5 and 42/7 = 6, so the "bottom layer" of boxes will be 30 boxes.

Now the 6 will "line up" with the 60. 60/6 = 10, so we'll have 10 layers of 30 boxes...and all of carton will be completely full.

(10)(30) = 300 boxes.

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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  23 Feb 2015, 17:53
thorinoakenshield wrote:
Hi all,

Cartoon LBH = 25 * 42 * 60

Box LBH = 5 * 7 * 6

One layer of boxes in cartoon [color=#ff0000]= (5*2)*(4*2) = 80

For H of 60, we can have 10 such layers. Hence # of boxes = 800

TO[/color]

As shown in my earlier post, you just have to divide the Carton LBH by box LBH

$$= \frac{25}{5} * \frac{42}{7} * \frac{60}{6} = 5*6*10 = 300$$
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  24 Feb 2015, 08:41
however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers
Dimensions of the box =25*42*60
Dimensions of the soap = 5*6*7
placing the 7 inch side along 42 inch side we get 6 soaps in a line and in a similar way 5 along 25 and 6 along 60
we get = 5x6x10 = 300
so the question is why this particular arrangement, in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number
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Re: A grocer is storing soap boxes in cartons that measure 25 [#permalink]  24 Feb 2015, 11:17
Zarrolou wrote:
deepri0812 wrote:
A grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each soap box is 7 inches by 6 inches by 5 inches, then what is the maximum number of soap boxes that can be placed in each carton?

A -210
B - 252
C - 280
D -300
E-420

Area of the box = $$25*42*60=63000$$
Area of one soap = $$7*6*5=210$$
Tot number = $$\frac{63000}{210}=300$$

You have to calculate the area of the box, then divide it by the area of one soap, and you ll find the number of soaps that will fit into the box
D

I would just like to comment that in this case this appproach works fine, but in the case that the boxes weren't proportional to the container it would not be as simple as dividing the total volume of the container by the volume of each box. One would have to maximize the boxes that can fit in every direction and then multiply those.
Re: A grocer is storing soap boxes in cartons that measure 25   [#permalink] 24 Feb 2015, 11:17
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text:
Generates data that can be used to plot a critical differences plot. Computes the critical differences according to either the "Bonferroni-Dunn" test or the "Nemenyi" test.
"Bonferroni-Dunn" usually yields higher power as it does not compare all algorithms to each other, but all algorithms to a baseline instead.
Learners are drawn on the y-axis according to their average rank.
For test = "nemenyi" a bar is drawn, connecting all groups of not significantly different learners.
For test = "bd" an interval is drawn arround the algorithm selected as a baseline. All learners within this interval are not signifcantly different from the baseline.
Calculation: $$CD = q_{\alpha} \sqrt{\left(\frac{k(k+1)}{6N}\right)}$$
Where $$q_\alpha$$ is based on the studentized range statistic. See references for details.

generateCritDifferencesData(
bmr,
measure = NULL,
p.value = 0.05,
baseline = NULL,
test = "bd"
)

## Arguments

bmr (BenchmarkResult) Benchmark result. (Measure) Performance measure. Default is the first measure used in the benchmark experiment. (numeric(1)) P-value for the critical difference. Default: 0.05 (character(1)): (learner.id) Select a learner.id as baseline for the test = "bd" ("Bonferroni-Dunn") critical differences diagram. The critical difference interval will then be positioned arround this learner. Defaults to best performing algorithm. For test = "nemenyi", no baseline is needed as it performs all pairwise comparisons. (character(1)) Test for which the critical differences are computed. “bd” for the Bonferroni-Dunn Test, which is comparing all classifiers to a baseline, thus performing a comparison of one classifier to all others. Algorithms not connected by a single line are statistically different from the baseline. “nemenyi” for the PMCMRplus::frdAllPairsNemenyiTest which is comparing all classifiers to each other. The null hypothesis that there is a difference between the classifiers can not be rejected for all classifiers that have a single grey bar connecting them.

## Value

(critDifferencesData). List containing:

data

(data.frame) containing the info for the descriptive part of the plot

friedman.nemenyi.test

(list) of class pairwise.htest
contains the calculated PMCMRplus::frdAllPairsNemenyiTest

cd.info

(list) containing info on the critical difference and its positioning

baseline

baseline chosen for plotting

p.value

p.value used for the PMCMRplus::frdAllPairsNemenyiTest and for computation of the critical difference

Other generate_plot_data: generateCalibrationData(), generateFeatureImportanceData(), generateFilterValuesData(), generateLearningCurveData(), generatePartialDependenceData(), generateThreshVsPerfData(), plotFilterValues()
Other benchmark: BenchmarkResult, batchmark(), benchmark(), convertBMRToRankMatrix(), friedmanPostHocTestBMR(), friedmanTestBMR(), getBMRAggrPerformances(), getBMRFeatSelResults(), getBMRFilteredFeatures(), getBMRLearnerIds(), getBMRLearnerShortNames(), getBMRLearners(), getBMRMeasureIds(), getBMRMeasures(), getBMRModels(), getBMRPerformances(), getBMRPredictions(), getBMRTaskDescs(), getBMRTaskIds(), getBMRTuneResults(), plotBMRBoxplots(), plotBMRRanksAsBarChart(), plotBMRSummary(), plotCritDifferences(), reduceBatchmarkResults()


text:
### Studies of Critical Phenomena in Causal Dynamical Triangulations on a Torus

This is a Ph.D. thesis that presents the author's findings in the area of Causal Dynamical Triangulations. In compliance with Jagiellonian University of Krak\'ow regulations, the document consists of six publications and a general summary, which serves as a guide to assist readers in navigating through the publications. Although the six publications that constitute the main content of the thesis are not included in this version of the text on arXiv, they are referred to frequently throughout the document. The original document is available at the following link: https://fais.uj.edu.pl/documents/41628/150115897/Thesis_DN-skompresowany.pdf

### Staggered bosons

A model with a half boson degree of freedom per lattice site in one dimension is developed. The boson is protected from developing a gap by translation symmetry: while the left movers are at zero quasi-momentum, the associated right movers are at the midpoint of the quasi-momentum period. The model has different properties depending on if a periodic lattice has an even or an odd number of sites and similar features are found for open boundary conditions. A special case of the non-linear half boson model where even and odd lattice sites contribute differently to the Hamiltonian gives rise to the Toda chain and a more symmetric generalization of the Toda chain is found. Upon periodic identifications of the half bosons degrees of freedom under a shift, the total Hilbert space has a finite dimension and can be encoded in finitely many qubits per unit length. This way one finds interesting critical spin chains, examples of which include the critical Ising model in a transverse magnetic field and the 3-state Potts model at criticality. Extensions to higher dimensions are considered. Models obtained this way automatically produce dynamical systems of gapless fractons.

### Light double-gluon hybrid states with the exotic quantum numbers $J^{PC} = 1^{-+}$ and $3^{-+}$

We apply the QCD sum rule method to study the double-gluon hybrid states with the quark-gluon contents $\bar q q gg$ ($q=u/d$) and $\bar s s gg$. We construct twenty-eight double-gluon hybrid currents, eleven of which are found to be zero due to some internal symmetries between the two gluons fields. We concentrate on the non-vanishing currents with the exotic quantum numbers $J^{PC} = 1^{-+}$ and $3^{-+}$. Their masses are calculated to be $M_{|\bar q q gg;1^{-+}\rangle} = 4.35^{+0.26}_{-0.30}$ GeV, $M_{|\bar s s gg;1^{-+}\rangle} = 4.49^{+0.25}_{-0.30}$ GeV, $M_{|\bar q q gg;3^{-+}\rangle} = 3.02^{+0.24}_{-0.31}$ GeV, and $M_{|\bar s s gg;3^{-+}\rangle} = 3.16^{+0.22}_{-0.28}$ GeV. The decay behaviors of the $J^{PC} = 3^{-+}$ states are studied, and we propose to search for them in the $\pi a_1(1260)/\rho \omega/\phi \phi$ channels in future particle experiments.

### The isospin-3 three-particle $K$-matrix at NLO in ChPT

The three-particle $K$-matrix, $\mathcal{K}_{\mathrm{df},3}$, is a scheme-dependent quantity that parametrizes short-range three-particle interactions in the relativistic-field-theory three-particle finite-volume formalism. In this work, we compute its value for systems of three pions at maximal isospin through next-to-leading order (NLO) in Chiral Perturbation Theory (ChPT). We compare the values to existing lattice QCD results and find that the agreement between lattice QCD data and ChPT in the first two coefficients of the threshold expansion of $\mathcal{K}_{\mathrm{df},3}$ is significantly improved with respect to leading order once NLO effects are incorporated.