text:
# Hamiltonian of linear harmonic oscilator

1. May 27, 2013

### LagrangeEuler

Could hamiltonian of linear harmonic oscilator be written in the form?
$\hat{H}=\sum^{\infty}_{n=0}(n+\frac{1}{2})\hbar\omega |n\rangle \langle n|$

2. May 27, 2013

### lugita15

Yes, it can. The ket vectors |n> form a basis for the Hilbert space, so two operators on the Hilbert space are equal if the yield the same answer for each element of the basis. So you just need to check that when the left-hand side acts on an arbitrary |n>, it yields the same answer as when the right hand side acts on it.

3. May 27, 2013

### dextercioby

What you wrote there is called the spectral decomposition of the Hamiltonian operator. According to the spectral theorem, every self-adjoint operator has one.

4. May 27, 2013

### LagrangeEuler

$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$
Its hermitian. What is his spectral form?

5. May 27, 2013

### lugita15

That's the position space representation of the Hamiltonian of a free particle. Well, the Hamiltonian of a free particle can be written in terms of momentum eigenstates as $\hat{H}=\int^{\infty}_{-\infty}\frac{p^{2}}{2m}|p\rangle \langle p| dp$. Or did you want to express the Hamiltonian in terms of position eigenstates instead?

Last edited: May 27, 2013
6. May 27, 2013

### LagrangeEuler

I want in terms of positional eigenstates.

7. May 27, 2013

### dextercioby

It's more complicated than that. I'll take the free spinless particle in R.

If you Trans-Fourier it to momentum space (ignore hbar and m/2), it's the unit operator times p^2, so that the spectral decomposition would be

$\mathcal{F}\left(-\frac{d}{dx^2}\right) = \int_{k\in\mathbb{R}} k^2 \delta (k-p) dk$

8. May 27, 2013

### lugita15

You can only write an operator in terms of outer products formed from eigenstates of that operator. So you can't write the Hamiltonian of a free particle in terms of position states.

EDIT: Of course, you can always say $\hat{H}=\int^{\infty}_{-\infty}\hat{H}|x\rangle \langle x| dx$, but that's trivial. You can't write it the same way you can write it in terms of momentum eigenstates, because momentum eigenstates are eigenstates of the Hamiltonian of a free particle, while position eigenstates are not.

Last edited: May 27, 2013
9. May 27, 2013

### lugita15

Was that addressed to me? Did I say something wrong?

10. May 28, 2013

### LagrangeEuler

$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}$
$\psi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}$
Could you write Hamiltonian from here?

11. May 28, 2013

### Jorriss

What do you think about it?

12. May 28, 2013

### LagrangeEuler

I think it could be done, but I really don't know how?

Last edited: May 28, 2013
13. Jun 1, 2013

### lugita15

Yes, it's just $\hat{H}=\sum^{\infty}_{n=1} E_n |n\rangle \langle n|$, where $\hat{H}|n\rangle= E_n|n\rangle$.


text:
SCIENCE CHINA Information Sciences, Volume 60 , Issue 10 : 108101(2017) https://doi.org/10.1007/s11432-016-9080-3

## Relative influence maximization in competitive social networks

• AcceptedApr 24, 2017
• PublishedAug 9, 2017
Share
Rating

### Acknowledgment

This work was supported by National Natural Science Foundation of China (Grant Nos. 61472400, 61300105), Research Fund for Doctoral Program of Higher Education of China (Grant No. 2012351410010), Key Project of Science and Technology of Fujian (Grant No. 2013H6012), Key Laboratory of Network Data Science & Technology, and Chinese Science and Technology Foundation (Grant No. CASNDST20140X).

### References

[1] Chen W, Collins A, Cummings R, et al. Influence maximization in social networks when negative opinions may emerge and propagate. In: Proceedings of 2012 SIAM International Conference on Data Mining, Mesa, 2011. 379--390. Google Scholar

[2] Budak C, Agrawal D, El Abbadi A. Limiting the spread of misinformation in social networks. In: Proceedings of the 20th International Conference on World Wide Web, Hyderabad, 2011. 665--674. Google Scholar

[3] Kempe D, Kleinberg J, Tardos É. Maximizing the spread of influence through a social network. In: Proceedings of the 9th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, Washington, 2003. 137--146. Google Scholar

[4] Cheng S Q, Shen H W, Huang J M, et al. Staticgreedy: solving the scalability-accuracy dilemma in influence maximization. In: Proceedings of the 22nd ACM International Conference on Information & Knowledge Management, San Francisco, 2013. 509--518. Google Scholar

[5] Cheng S Q, Shen H W, Huang J M, et al. IMRank: influence maximization via finding self-consistent ranking. In: Proceedings of the 37th International ACM SIGIR Conference on Research & Development in Information Retrieval, Gold Coast, 2014. 475--484. Google Scholar

[6] Chen W, Wang Y J, Yang S Y. Efficient influence maximization in social networks. In: Proceedings of the 15th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining, Paris, 2009. 199--208. Google Scholar

• Figure 1

(Color online) Experimental results. (a) Relative influence of greedy algorithms with different $k$ in NetHEPT; (b) relative influence of greedy algorithms with different $k$ in Geom; (c) relative influence of heuristics with different $k$ in NetHEPT; (d) relative influence of heuristics with different $k$ in Geom; (e) running times with $k$ =100 in NetHEPT and Geom; (f) running times with $k$ =100 in synthetic networks.

Citations

Altmetric

format: plain


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Feb 2019, 17:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Algebra Webinar

February 17, 2019

February 17, 2019

07:00 AM PST

09:00 AM PST

Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
• ### Free GMAT Strategy Webinar

February 16, 2019

February 16, 2019

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# If the sum of all positive factors of an integer n is 2n, n is a perfe

Author Message
TAGS:

### Hide Tags

Senior DS Moderator
Joined: 27 Oct 2017
Posts: 1205
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

Updated on: 07 Oct 2018, 04:48
2
00:00

Difficulty:

65% (hard)

Question Stats:

45% (02:08) correct 55% (02:24) wrong based on 64 sessions

### HideShow timer Statistics

If the sum of all positive factors of an integer n is 2n, n is a perfect number. For example, the factors of 6 are 1, 2, 3, and 6, and from the sum 1+2+3+6=12=2*6, the sum of the factors of 6 becomes 12=2*6, thus 6 is the first perfect number. Then, what is the number of factors of the second perfect number?

A. 4
B. 5
C. 6
D. 8
E. 12

Weekly Quant Quiz #3 Question No 5

_________________

Originally posted by gmatbusters on 06 Oct 2018, 09:21.
Last edited by gmatbusters on 07 Oct 2018, 04:48, edited 2 times in total.
Renamed the topic and edited the question.
Senior DS Moderator
Joined: 27 Oct 2017
Posts: 1205
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 09:23

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.
_________________
Manager
Joined: 16 Sep 2011
Posts: 96
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

Updated on: 06 Oct 2018, 11:31
12 factors = 1,2, 3, 6, 12
sp 1+2+3+6+12=24 which is 2*12

Originally posted by pk123 on 06 Oct 2018, 09:26.
Last edited by pk123 on 06 Oct 2018, 11:31, edited 1 time in total.
Senior PS Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 722
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 09:31
Second perfect number comes out to be 28 after a little trial and error. The factors of 28 are 1,2,4,7,14,28 and hence it has 6 factors.

So Option (C) is correct.

Best,
G

Posted from my mobile device
_________________

Regards,

“Do. Or do not. There is no try.” - Yoda (The Empire Strikes Back)

BSchool Forum Moderator
Joined: 23 May 2018
Posts: 428
Location: Pakistan
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 09:40
E because factors of 12 are 12, 6, 3, 2, and 1.

$$12+6+3+2+1=24=12*2$$
_________________

If you can dream it, you can do it.

Practice makes you perfect.

Kudos are appreciated.

Manager
Joined: 26 Sep 2018
Posts: 59
Location: Sweden
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 09:42
Cheers

Using the options.
Manager
Joined: 21 Jul 2017
Posts: 192
Location: India
GMAT 1: 660 Q47 V34
GPA: 4
WE: Project Management (Education)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 09:46
28 is the other such number, factors of 28 are 6. Answer C
Intern
Joined: 06 Feb 2018
Posts: 16
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 10:48
6 2*3 = 1 + 2 + 3

The key here is that 2+1 = 3

A few trials with other primes shows 5 = 2 + 2 + 1
but we can't have two repeat has a factor
next option is 4
Therefore 1 + 2 + 4 = 7
i.e 2^2*7 = 1 + 2 + 4 + 7 + 14
number is 28
Factors 3*2 = 6

C
RC Moderator
Joined: 24 Aug 2016
Posts: 673
Concentration: Entrepreneurship, Operations
GMAT 1: 630 Q48 V28
GMAT 2: 540 Q49 V16
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

06 Oct 2018, 11:14
Really not sure if there is a mathematical way of deduction.......
..................but by iteration got the next perfect number as 28.
Factors of 28= 1,2,4,7,14,28
There are 6 factors
Ans C

_________________

Please let me know if I am going in wrong direction.
Thanks in appreciation.

Manager
Joined: 17 Jun 2018
Posts: 50
Location: India
Schools: IMD '20
GPA: 2.84
WE: Engineering (Consulting)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

30 Jan 2019, 10:06
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

how did you arrive at the formulation (2^(n-1))(2^n-1)?
Senior DS Moderator
Joined: 27 Oct 2017
Posts: 1205
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe  [#permalink]

### Show Tags

30 Jan 2019, 19:17
This is just another approach.

We can solve the question by iteration as shown by gladiator.

arpitkansal wrote:
gmatbusters wrote:

Official Explanation

In general, perfect number is (2^(n-1))(2^n-1) when n=prime number.
In other words, when n=2, (2^(2-1))(2^2-1)=(2)(3)=6 is the first perfect number.
The 2nd perfect number is when n=3, and with substitution, you get (2^(3-1))(2^3-1)=(4)(7)=28.
In other words, the sum of all factors of 28=1+2+4+7+14+28=2(28), hence it is a perfect number.
If $$N = a^p*b^q...$$, No of factors =$$(p+1)(q+1)...$$
From 28 = $$(2^2)(7^1)$$, the number of factors is (2+1)(1+1)=6. The answer is C.

how did you arrive at the formulation (2^(n-1))(2^n-1)?

Posted from my mobile device
_________________
Re: If the sum of all positive factors of an integer n is 2n, n is a perfe   [#permalink] 30 Jan 2019, 19:17
Display posts from previous: Sort by


text:
How do you solve the system 8x+2y=2 and -4x=y+8?

Aug 1, 2018

$\text{no solution}$

Explanation:

$8 x + 2 y = 2 \to \left(1\right)$

$- 4 x = y + 8 \to \left(2\right)$

$\text{from } \left(2\right) \textcolor{w h i t e}{x} y = - 4 x - 8$

$\text{substitute "y=-4x-8" into equation } \left(1\right)$

$8 x + 2 \left(- 4 x - 8\right) = 2$

$8 x - 8 x - 16 = 2$

$- 16 = 2 \leftarrow \text{not possible}$

$\text{There is no solution to this system}$


text:
# How to understand effective interaction in non-interacting quantum gas models (Fermi-Dirac, Bose-Einstein statistics)?

It's known that Fermi-Dirac and Bose-Einstein statistics describe non-interacting indistinguishable particles at thermodynamic equilibrium. However, if we solve some problems wtih using such statistics, we get curious results, which are interpreted as an effective interaction presence. Let's consider few examples:

### 1. Fermi gas free expansion

$$N$$ fermions with spin $$\frac{1}{2}$$ are concluded in a small box with volume $$V_1$$ and have energy $$E_1=\frac{3}{5}N\varepsilon_F$$, where $$\varepsilon_F=(3\pi^2)^{\frac{2}{3}}\frac{\hbar^2}{2m}(\frac{N}{V_1})^{\frac{2}{3}}$$ is the Fermi energy. Their temperature $$T_1=0$$ ($$T_1\ll \varepsilon_F$$, the gas is degenerate). The small box is separated from a big one with volume $$V_2\gg V_1$$ by a small partition. After removing the partition fermions occupy the big box. The problem is to find their final temperature $$T_2$$ knowing that $$E_2=E_1$$ (an isolated system).

If we suppose that the gas is still degenerate and described by Fermi-Dirac statistics we will obtain $$T_2\propto \tilde{\varepsilon}_F\left(\frac{V_2}{V_1}\right)^{\frac{1}{3}}\gg \tilde{\varepsilon}_F$$, where $$\tilde{\varepsilon}_F=\varepsilon_F\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}$$ is the new Fermi energy. It means the assumption is wrong.

Then we can use classical expressions for gas values in the final state, which are derived from Boltzmann statistics: $$E_2=\frac{3}{2}NT_2$$. From energy conservation we obtain $$T_2=\frac{2}{5}\varepsilon_F\propto \left(\frac{N}{V_1}\right)^{\frac{2}{3}}$$, which is obviously much greater than $$T_1$$ whilst we consider macroscopical concentration.

So, Fermi gas is heated during free expansion high enough to be non-degenerate. At the same time Boltzmann ideal gas is neither heated or cooled during free expansion. To explain this some people claim that there is a kind of repulsion between fermions, which is related to the Pauli exclusion principle.

### 2. Bose gas condensation in a harmonic trap

$$N$$ bosons sit in a 3D harmonic trap with frequencies $$\omega_1, \omega_2, \omega_3 \ll \frac{T}{\hbar}$$. It can be shown that density of states $$g(\varepsilon)=\frac{\varepsilon^2}{2\hbar^3\tilde{\omega}^3}$$, where $$\tilde{\omega}=(\omega_1\omega_2\omega_3)^{\frac{1}{3}}$$ is a characteristic frequency. After some calculations it's possible to obtain the following relations: $$\Omega(T,V,\mu)\equiv F(T,V,N)-\mu N=-\frac{1}{3}E(S,V,N)$$ $$\frac{\partial E}{\partial \mu}=-3\frac{\partial \Omega}{\partial \mu}=3N$$ $$E=E_0+3\mu N$$ $$\mu=\begin{cases} 0 & ,\quad T $$C_V=\begin{cases} \frac{2\pi^4T^3}{15\hbar^3\tilde{\omega}^3} & ,\quad T where $$T_0=\left(\frac{N}{\zeta(3)}\right)^{\frac{1}{3}}\hbar\tilde{\omega}$$ is the critical temperature.

As we can see there is a negative jump of $$C_V$$ at $$T=T_0$$, which means that for some reason we spend more energy to heat up the condensate. A similar behaviour of $$C_V$$ is observed in superconductors, but it's clear why it happens there: we need some energy to break Cooper pairs. But Bose ideal gas does't have any bonds between particles and remains Bose ideal gas after "evaporation".

Nevertheless, some people claim that there is an effective attraction between bosons in the condensate.

### 3. Quantum gas at high temperature

In "Statistical Physics, Part 1: Volume 5" L. D. Landau demonstrates how to obtain the following expression of the equation of state for ideal quantum gases at high temperature: $$PV=NT\left ( 1\pm \frac{\pi^{\frac{3}{2}}N\hbar^3}{2V(mT)^{\frac{3}{2}}} \right )$$ where the sign of the second term depends on statistics. In the limit $$T\rightarrow \infty$$ it turns into the equation of state for a classical ideal gas.

Then Landau gives a comment:

Thus we see that the deviations of an ideal gas from classical properties, occuring when the temperature is lowered at constant density (the gas then being said to become degenerate), cause in Fermi statistics an increase in pressure as compared with its value in an ordinary gas; we may say that in this case the quantum exchange effects lead to the occurrence of an additional effective repulsion between the particles.

In Bose statistics, on the other hand, the value of the gas pressure changes in the opposite direction, becoming less than the classical value; we may say that here there is an effectife attraction between the particles.

Unfortunately, he doesn't explain why there is such opportunity to consider initially non-interacting particles as interacting.

### 4. Quantum gas at zero temperature

It's the last and brief example.

For fermions at zero temperature their pressure isn't zero: $$P=\frac{(3\pi^2)^{\frac{2}{3}}\hbar^2}{5m}\left(\frac{N}{V}\right)^{\frac{5}{3}}$$ (according to Landau)

For bosons it is, because all particles sit on the ground level with energy $$\varepsilon=0$$.

The difference is often explained by the same effective interaction.

Now, let me remind my question. Unfortunately, I'm not able to understand how it happens when we consider non-interacting particles and obtain many consequences, which apparently need an interction in the system. Those qualitative explanations with help of the Pauli principle and bonds in the condensate don't satisfy me. They seem contrary to the initital problem statement.

Is it possible to show mathematically on what step of statistics deriving we take into account interactions after all?

I would recommend you take a look at this paper on the subject, which is referred to in the "Introduction to Quantum Mechanics" book by D.J. Griffiths on the subject. There they show how all the phenomena you go through can be understood without reference to any effective interaction.

The short answer is that the "effective interaction" is another way to express the requirement that the relative fermionic wavefunction must be zero for all configurations which leave two fermions in the same location. This means that squeezing two fermions together requires a larger investment of momentum than in the bosonic case. This might look like an interaction, but the paper gives a number of reasons for why such an interpretation is, if not wrong then somewhat dangerous.

A compound and strongly interacting system at low temperatures may often be represented as an ensebmle of nearly non interacting quasi-particles - normal modes of the system. It is the strongly interacting system that dictates the quasi-particle properties. We should not take the quasi-particles for "literally particles". In QM there are always some "boundary conditions" created by a strongly interacting system, let us not forget it.

According to my understanding, those interactions are due to 'exchange force', which happens due to symmetry properties of wave functions. Fermions should always have anti symmetric wave function and bosons should always have symmetric wave function. It will lead to a kind of repulsive/attractive interactions which can be shown mathematically. See derivations in Griffiths Quantum Mechanics, Identical Particles>exchange force. I tried to explain about identical particles here: https://youtu.be/aq5tMYlzFjo

Term non-interacting particles has different meaning in classical/quantum mechanics and in statistical physics. In the former case it literally means the absence of interactios. In statistical physics it means that the interactions are weak and can be neglected in the equations; nevertheless the interactions are present and essential for establishing tehrmal equilibrium (e.g., in classical statistical emchanics of an ideal gas, the collisions between the atoms and the atoms and the walls of the container play teh role of such interactions.) The more detailed reasoning can be found in any good statistical physics textbook, although this is easily overlooked in the first reading, when one focuses on the equations.


text:
Question

# (i) How is ammonia gas prepared in laboratory starting from NH4Cl? State the conditions and balanced equation for the preparation. (ii) How is gas collected in a gas jar? (iii) Name the substance used for drying ammonia gas. Why cannot substances such as conc. H2SO4, anhydrous calcium chloride and phosphorus pentoxide be used for drying ammonia gas?

Open in App
Solution

## (i) Preparation: Ammonium chloride (NH2Cl) and dry slaked lime [Ca(OH)2] are used in the preparation of ammonia in the laboratory. Both the compounds are taken in the ratio 2:3 by weight and ground thoroughly before being placed in a flask and heated. The apparatus used for the preparation of ammonia is shown in the following figure: Conditions: The reactants must be dry and grounded. A slightly higher ratio of slaked lime is used to counteract any loss of ammonium chloride due to sublimation. Reaction: $\underset{\mathrm{Ammonium}\mathrm{chloride}}{2{\mathrm{NH}}_{4}\mathrm{Cl}}+\underset{\mathrm{Slaked}\mathrm{lime}}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_{2}}\stackrel{△}{\to }\underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_{3}}+{\mathrm{CaCl}}_{2}+{\mathrm{H}}_{2}\mathrm{O}$ (ii) Ammonia is lighter than air; therefore, it is collected in the gas jar by downward displacement of air. (iii) Quick lime, also known as calcium oxide (CaO), is used for drying ammonia. Sulphuric acid, calcium chloride and phosphorus pentoxide cannot be used for drying ammonia.This is due to the reason that ammonia being basic reacts with them.

Suggest Corrections
0


text:
## Help finishing connectedness proof.

Been having trouble with connectedness. Tried out a problem to practice and not sure how to finish it.

The problem is this:

Let A and B be closed subsets of the space X. If A U B and A n B are connected, so also are A and B. Show by counter-example that both A and B must be closed, i.e., if one of A and B is not closed then in general the thm fails.

My proof went like this:

Let C = A ᴜ B. Suppose that U, V give a separation of C, where U and V are open in C.
Since A is connected, we must have A ᴜ U or A ᴜ V (otherwise, A ∩ U and A ∩ V give a separation of A).
Without loss of generality, assume A ᴜ U. Since A ∩ B = Ø and B is connected,
we must have B ᴜ U. This implies that C ᴜ U, and therefore, V = Ø.
This contradicts that U, V is a separation of C. Hence the sets A or B must be closed,

To prove your result, assume (to get a contradiction) that one of the closed sets, A or B is not connected. WOLOG, we may assume it is A. Then there are open sets V and W in X s.t. A= (A ∩V) U (A∩ W) and = (A ∩V) ∩ (A∩ W)=ᴓ.
Now, you should be able to use V and W to build a separation of either AUB or A∩B. (Try “unioning“ B to both sides of the above equations, and it may give you an idea of what to use for a separation.)

I thought I was doing it right, but by my professor's response, I'm not sure. Could anyone tell me what I'm doing wrong and what he means by "unioning" B to both sides?

Any feedback would be greatly appreciated.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

 Quote by Mister Eric Been having trouble with connectedness. Tried out a problem to practice and not sure how to finish it. The problem is this: Let A and B be closed subsets of the space X. If A U B and A n B are connected, so also are A and B. Show by counter-example that both A and B must be closed, i.e., if one of A and B is not closed then in general the thm fails. My proof went like this: Let C = A ᴜ B. Suppose that U, V give a separation of C, where U and V are open in C. Since A is connected, we must have A ᴜ U or A ᴜ V (otherwise, A ∩ U and A ∩ V give a separation of A). Without loss of generality, assume A ᴜ U.

You already repeated this several times to be a mere mistake: you write "we must have $\,A\cup U\,\,or\,\,A\cup V\,$", and then again " assume $\,A\cup U\,$"

WHAT are we assuming, anyway?? My guess is that you actually mean "assume $\,A\cup U=A\,$" , but then what do you expect from this?

I think you first must be clear above in order to meet your instructor's criticism.

 Since A ∩ B = Ø and B is connected, we must have B ᴜ U. This implies that C ᴜ U, and therefore, V = Ø. This contradicts that U, V is a separation of C. Hence the sets A or B must be closed, The response I got back was this: To prove your result, assume (to get a contradiction) that one of the closed sets, A or B is not connected. WOLOG, we may assume it is A. Then there are open sets V and W in X s.t. A= (A ∩V) U (A∩ W) and = (A ∩V) ∩ (A∩ W)=ᴓ. Now, you should be able to use V and W to build a separation of either AUB or A∩B. (Try “unioning“ B to both sides of the above equations, and it may give you an idea of what to use for a separation.) I thought I was doing it right, but by my professor's response, I'm not sure. Could anyone tell me what I'm doing wrong and what he means by "unioning" B to both sides? Any feedback would be greatly appreciated.

DonAntonio


text:
# the conditions vec(r)* vec(s) =0, and vec(r) * vec(x) =c, and vec(r) xx vec(x) = vec(s) . Find x in each of the three mutually orthogonal directions, vec(r), vec(s) , and vec(r) xx vec(s).

The conditions $\stackrel{\to }{r}\cdot \stackrel{\to }{s}=0$, and $\stackrel{\to }{r}\cdot \stackrel{\to }{x}=c$, and $\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}$ . Find x in each of the three mutually orthogonal directions, $\stackrel{\to }{r}$ , $\stackrel{\to }{s}$ , and $\stackrel{\to }{r}×\stackrel{\to }{s}$
So far $\stackrel{\to }{{x}_{r}}=\frac{\stackrel{\to }{x}\cdot \stackrel{\to }{r}}{|r|}\frac{\stackrel{\to }{r}}{|r|}=\frac{c}{|r{|}^{2}}\stackrel{\to }{r}$ , and $\stackrel{\to }{{x}_{s}}=\frac{\stackrel{\to }{x}\cdot \stackrel{\to }{s}}{|s|}\frac{\stackrel{\to }{s}}{|s|}=\frac{\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)}{|s|}\frac{\stackrel{\to }{s}}{|s|}=0$ Since $\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)$ is the volumn of a degenerate parallelepiped. Where I'm having most difficulty is in...
$\stackrel{\to }{{x}_{\stackrel{\to }{r}×\stackrel{\to }{s}}}=\frac{\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{s}\right)}{|\stackrel{\to }{r}×\stackrel{\to }{s}{|}^{2}}\stackrel{\to }{r}×\stackrel{\to }{s}$ . Since r and s are orthogonal does that mean $|\stackrel{\to }{r}×\stackrel{\to }{s}|=|r||s|$. And also can I calculate $\stackrel{\to }{x}\cdot \left(\stackrel{\to }{r}×\stackrel{\to }{s}\right)$using the triple product to be $|r||s||\stackrel{\to }{{x}_{s\perp }}|=|r||s||\stackrel{\to }{x}-\stackrel{\to }{{x}_{s}}|=|r||s||\stackrel{\to }{x}|$. Is there a simpler simplification of this?
So does $\stackrel{\to }{{x}_{\stackrel{\to }{r}×\stackrel{\to }{s}}}=\frac{|\stackrel{\to }{x}|}{|\stackrel{\to }{r}||\stackrel{\to }{s}|}\stackrel{\to }{x}$?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Lorenzo Acosta
Just to simplify the notation a bit, let $\stackrel{\to }{t}=\stackrel{\to }{r}×\stackrel{\to }{s}$
Because ${\stackrel{\to }{x}}_{\stackrel{\to }{s}}=0$, we know $\stackrel{\to }{x}={\stackrel{\to }{x}}_{\stackrel{\to }{r}}+{\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ . Therefore
$\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{r}×\left({\stackrel{\to }{x}}_{\stackrel{\to }{r}}+{\stackrel{\to }{x}}_{\stackrel{\to }{t}}\right)=\stackrel{\to }{s}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{r}}+\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}.$
But $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{r}}=0$, so we conclude that $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}$ . Moreover, $\stackrel{\to }{r}$ and ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ are orthogonal, so $|\stackrel{\to }{r}||{\stackrel{\to }{x}}_{\stackrel{\to }{t}}|=|\stackrel{\to }{s}|$
This tells us $|{\stackrel{\to }{x}}_{\stackrel{\to }{t}}|=\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}|}$, so ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}=±\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}|}\frac{\stackrel{\to }{t}}{|\stackrel{\to }{t}|}=±\frac{|\stackrel{\to }{s}|}{|\stackrel{\to }{r}||\stackrel{\to }{t}|}\stackrel{\to }{t}$. We also know $|\stackrel{\to }{t}|=|\stackrel{\to }{r}||\stackrel{\to }{s}|$, further simplifying ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}$ to $±\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{t}$
Which sign is it? Well, we want $\stackrel{\to }{r}×{\stackrel{\to }{x}}_{\stackrel{\to }{t}}=\stackrel{\to }{s}$ , but $\stackrel{\to }{r}×\stackrel{\to }{t}$ is proportional to $-\stackrel{\to }{s}$. Therefore ${\stackrel{\to }{x}}_{\stackrel{\to }{t}}=-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{t}$
###### Did you like this example?
Marcus Bass
consider $\stackrel{\to }{r}×\stackrel{\to }{x}=\stackrel{\to }{s}$
taking cross product from both sides
$\stackrel{\to }{r}×\left(\stackrel{\to }{r}×\stackrel{\to }{x}\right)=\stackrel{\to }{r}×\stackrel{\to }{s}$
you can use vector triple cross product formula
then
$\left(\stackrel{\to }{r}.\stackrel{\to }{x}\right)\stackrel{\to }{r}-\left(\stackrel{\to }{r}.\stackrel{\to }{r}\right)\stackrel{\to }{x}=\stackrel{\to }{r}×\stackrel{\to }{s}$
$\left(\stackrel{\to }{r}.\stackrel{\to }{x}\right)\stackrel{\to }{r}-|\stackrel{\to }{r}{|}^{2}\stackrel{\to }{x}=\stackrel{\to }{r}×\stackrel{\to }{s}$
rearranging
$\stackrel{\to }{x}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}×\stackrel{\to }{s}$
since $\stackrel{\to }{r}$ ,$\stackrel{\to }{s}$ and $\stackrel{\to }{r}×\stackrel{\to }{s}$ are mutually orthogonal they are linearly independent therefore you can get $\stackrel{\to }{x}$ , as a linear combination of these $\stackrel{\to }{r}$ ,$\stackrel{\to }{s}$ and $\stackrel{\to }{r}×\stackrel{\to }{s}$ vectors
$\stackrel{\to }{x}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{r}-\frac{1}{|\stackrel{\to }{r}{|}^{2}}\stackrel{\to }{\left(}r×\stackrel{\to }{s}\right)-0\stackrel{\to }{s}$
so you can take component as
${\stackrel{\to }{x}}_{\stackrel{\to }{r}}=\frac{c}{|\stackrel{\to }{r}{|}^{2}}$
${\stackrel{\to }{x}}_{\stackrel{\to }{r}×\stackrel{\to }{s}}=\frac{-1}{|\stackrel{\to }{r}{|}^{2}}$
and
${\stackrel{\to }{x}}_{\stackrel{\to }{s}}=0$


text:
# Factoring by Grouping

## Use associative and commutative properties with factoring

Estimated13 minsto complete
%
Progress
Practice Factoring by Grouping

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated13 minsto complete
%
Factoring by Grouping

What if you had a polynomial expression like \begin{align*}3x^2 - 6x + 2x - 4\end{align*} in which some of the terms shared a common factor but not all of them? How could you factor this expression? After completing this Concept, you'll be able to factor polynomials like this one by grouping.

Fisch Video: Factoring Polynomials by Grouping

### Guidance

Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping.

The next example illustrates how this process works.

#### Example A

Factor \begin{align*}2x+2y+ax+ay\end{align*}.

Solution

There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of \begin{align*}a\end{align*}. Factor 2 from the first two terms and factor \begin{align*}a\end{align*} from the last two terms:

\begin{align*}2x + 2y + ax + ay = 2(x + y) + a(x + y)\end{align*}

Now we notice that the binomial \begin{align*}(x + y)\end{align*} is common to both terms. We factor the common binomial and get:

\begin{align*}(x + y)(2 + a)\end{align*}

#### Example B

Factor \begin{align*}3x^2+6x+4x+8\end{align*}.

Solution

We factor 3x from the first two terms and factor 4 from the last two terms:

\begin{align*}3x(x+2)+4(x+2)\end{align*}

Now factor \begin{align*}(x+2)\end{align*} from both terms: \begin{align*}(x+2)(3x+4)\end{align*}.

Now the polynomial is factored completely.

### Practice

Factor by grouping.

1. \begin{align*}6x^2-9x+10x-15\end{align*}
2. \begin{align*}5x^2-35x+x-7\end{align*}
3. \begin{align*}9x^2-9x-x+1\end{align*}
4. \begin{align*}4x^2+32x-5x-40\end{align*}
5. \begin{align*}2a^2-6ab+3ab-9b^2\end{align*}
6. \begin{align*}5x^2+15x-2xy-6y\end{align*}

### Additional Practice Opportunities (including self-check)

1. Braingenie: Factoring Polynomials by Grouping Terms

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes


text:
# Need image of absorption and emission spectra for sodium

Stanley Tran
Hi,

I need images of absorption and emission spectra for sodium to be used in a physics student book.


text:
# Complex polynomial

1. Dec 4, 2015

### cdummie

1. The problem statement, all variables and given/known data
It is known that roots of complex polynomial:

$P_n (z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0$

are the following complex numbers:

$\alpha_1, \alpha_2, \cdots, \alpha_n$

Find the product:

$\prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)$
2. Relevant equations

3. The attempt at a solution
I am pretty sure that i have to use Vieta formulas somehow, and since roots are all complex then it means that there are n/2 pairs of complex-complex conjugate roots. If i multiply all of this i would end up with product of all roots $\alpha_1* \alpha_2* \cdots* \alpha_n$ and by Vieta formulas i can easily determine that one, but there are more elements of this product after multiplying, what can i do with them?

2. Dec 4, 2015

### geoffrey159

Ok so it has $n$ roots without counting multiplicity.
So your polynomial has the form $P_n(z) = A(z) (z-\alpha_1)...(z-\alpha_n)$.
Explain why $A = 1$

3. Dec 4, 2015

### HallsofIvy

This is only true if all coefficients are real numbers but you said "complex polynomial" which means that is not necessarily true.

4. Dec 4, 2015

### Ray Vickson

The requirement of having complex-conjugate root pairs is true only if all the coefficients are real, and can fail if some of the coefficients are non-real complex numbers. For example, the equation $x^2 + (2+i)x - 3 i = 0$ has roots $\alpha_1 \doteq 0.522 +0.788\, i$ and $\alpha_2 \doteq -2.552 - 1.788\, i$.

5. Dec 4, 2015

### cdummie

I am not really sure, is it maybe because i have coefficient 1 with zn?

6. Dec 4, 2015

### geoffrey159

yes, but if you want to convince yourself, you can justify that $\text{deg}{(A)} = 0$, so that $A$ is a non-zero constant, and with your argument above, $A = 1$.

Ok, so now you have the answer, what is $P_n(-1)$ ?

7. Dec 4, 2015

### cdummie

Well, it's $P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$
But how this gets me any closer to the solution?

8. Dec 4, 2015

### geoffrey159

You must have worked on this problem for too long :-)
Think about it 10 more minutes

EDIT: or don't think about it the next 10 minutes :-)

EDIT (+10 minutes) : What did we just say ? $P_n(z) = (z-\alpha_1)...(z-\alpha_n)$. So what is $P_n(-1)$?

Last edited: Dec 4, 2015
9. Dec 4, 2015

### cdummie

I am sorry, i really don't know anything than this $P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$ which is obvious and it still isn't clear to me how this is going to help me solving this.

10. Dec 4, 2015

### Ray Vickson

Does this formula (up to a possible sign) express your desired product in terms of the coefficients $a_0, a_1, \ldots, a_{n-1}$? Did the problem ask you to do more than that?

11. Dec 4, 2015

### cdummie

I really don't understand what are you trying to say, do you mean that $a_0 + a_1 + \ldots + a_{n-1} = \prod = (\alpha_1 + 1)(\alpha_2 + 1)\cdots(\alpha_n +1)$?

12. Dec 4, 2015

### Samy_A

You know that $P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$.

You also know another way to find $P_n(-1)$:
Now put the two together.

13. Dec 4, 2015

### Ray Vickson

No, absolutely not. YOU said that $P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$ (not $\sum a_j$). My question amounts to asking: why do you want to compute $P_n(-1)$? What do you gain by doing that? How is that related to the product you started with?

14. Dec 5, 2015

### cdummie

Let's see: I have $P_n(-1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$ and $P_n(-1) =(-1-\alpha_1) \cdots(-1-\alpha_n)$

Now, taking all -1's out i have $P_n(-1) =(-1)^n(\alpha_1+1) \cdots(\alpha_n+1) = (-1)^n + a_{n-1}(-1)^{n-1} + \cdots + a_1(-1) + a_0$

Multiplying both sides by $(-1)^{-n}$ i have:

$(\alpha_1+1) \cdots(\alpha_n+1) = 1 + a_{n-1}(-1)^{-1} + \cdots + a_1(-1)^{1-n} + a_0(-1)^{-n}$

Now, if this is correct i have another question, my idea was to use Vieta's formulas, i mean i have a given product which i have to find:

$\prod=(\alpha_1 + 1)\cdots(\alpha_n+1)$ i am sure that there would be $\alpha_1\alpha_2\cdots\alpha_n$ but what is with rest of it , i mean i would have 1 at the end, but in order to use Vieta's formulas i should get something like $\alpha_1\alpha_2 + \alpha_1\alpha_3 + \cdots + \alpha_1\alpha_n + + \alpha_2\alpha_3 + \alpha_2\alpha_4 +\cdots + \alpha_{n-1}\alpha_n$ and then sum of products of three and so on all the way to the product of all of them. Is this correct?

15. Dec 5, 2015

### Samy_A

Correct.

Using Vieta's formulas should also work, as you indeed get all possible combinations of products of roots when you expand $\prod(\alpha_1 + 1)\cdots(\alpha_n+1)$.


text:
# Do you have a better chance of getting hit by lightning or winning the lottery?

The actual problem is stated as follows:

The chance of getting hit by lightning is 1 in 600,000. In the lottery you will play, you'll chose 6 numbers out of the first 30. Do you have a better chance of getting hit by lightning or winning the lottery?

To me, I feel like this problem is ambiguous and has left out some needed things. For example, wouldn't we need to know the total population when trying to find the chance of getting hit by lightning? And for the lottery wouldn't we need some more information as well?

My first thought was that this is an unanswerable question. If this is true, could I just make up some arbitrary numbers? And if I can, what would that look like in this case?

• Ignore the lightning part for now. Can you compute the probability of winning the lottery? Jun 14, 2016 at 16:20
• No, making up arbitrary numbers is not feasible. What if you decide that the population is $1,000,000,000,000,000$? But, making estimates might be allowable, if they can be realistically justified. Maybe one of those events is so much less likely than the other that realistic estimates will demonstrate it. Jun 14, 2016 at 16:21

The problem isn't ambiguous, just not worded terribly well. The given probability of being struck by lightning assumes some information about frequency of lightning strikes, population, exposure to risk, and so on. You are to assume the probability that you will, at some point, be struck by lightning is $\frac{1}{600000}$.
Is the probability of choosing a particular set of six integers between $1$ and $30$ greater than, equal to, or less than $\frac{1}{600000}?$


text:
Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# Choose the correct answer in area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is

$(a)\;\pi\qquad(b)\;2 \pi\qquad(c)\;3 \pi\qquad(d)\;4 \pi$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• The area bounded by the curve f(x),x-axis and the ordinate x=a,x=b is given by$A=\int_a^by\;dx=\int_a^{-b}f(x)dx.$
Here the area of the region bounded by the line x=0 and x=2 and the circle $x^2+y^2=4$ is the shaded portion as shown in the fig:
Hence $A=\int_0^2y dx$.
Here y=$\sqrt{4-x^2}dx$
$A=\int_0^2\sqrt {4-x^2}dx.$
On integrating we get,
$A=\begin{bmatrix}\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\big(\frac{x}{2}\big)\end{bmatrix}_0^2$
On applying limits we get,
$A=\begin{bmatrix}\frac{2}{2}\sqrt{4-4}+\frac{4}{2}\sin^{-1}\big(\frac{2}{2}\big)\end{bmatrix}$
$\;\;\;=0+2.\frac{\pi}{2}$
$\;\;\;=\pi\; sq.units.$
Hence A is the correct answer.
answered Dec 20, 2013 by

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer


text:
# Theoretically, Which d orbital participates in sp3d and sp3d2?

This question is regarding the old way (inaccurate) $\ce{PCl5}$ and $\ce{SF6}$ are taught in elementary chemistry.

In the crystal field splitting diagram for triagonal bypyramidal geometry, $\mathrm{d}_{z^2}$ has the highest energy. Now, since $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$ of same shell hybridise, then the $\mathrm{d}$ orbital closest in energy to $\mathrm{s}$ and $\mathrm{p}$ should take part, which should be $\mathrm{d}_{xz}$ or $\mathrm{d}_{yz}$. Instead, the hybridisation is said to be $\mathrm{sp^3d}_{z^2}$.

A similar problem arises for molecules with an octahedral shape. It has $\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$ which are higher in energy than the rest and have higher energy difference with $\mathrm{s}$ and $\mathrm{p}$ orbitals of the same shell. There is no such problem for $\mathrm{d^2sp^3}$, $\mathrm{d^3s}$, $\mathrm{dsp^2}$.

EDIT: Let me change the question to ask the same thing. In outer shell octahedral complexes ($\mathrm{s,p,d}$ of same principal quantum numbers take part) why are the $\mathrm{e_g}$ orbitals taking part in sigma bond formation when they are furthest away from $n\mathrm{s}$ and $n\mathrm{p}$ orbitals?

There is no such problem for inner shell complexes.

• Since the assignment of a coordinate system is completely arbitrary, this question cannot be answered. – Martin - マーチン Jan 20 '17 at 9:03
• Let's assume in sp3d, the terminal atoms/ligands approach from ±z, +x and y=x√3,z=0 line and y=-x√3,z=0 line. My main question was, was I correct in assuming that the orbitals closest in energy hybridise or there is some symmetry issue too? Secondly, how the people came up with the fact that dz2 supposedly takes part, not the rest? – Mrigank Jan 20 '17 at 9:42
• I just want to point out that this concept of hybridization is ancient and if someone has an answer, it doesn't really has any meaning. It's a helpful tool to teach chemistry to students for sure, but that's about it. – AMT Jan 20 '17 at 15:46
• None, because neither the $\mathrm{sp^3d}$ of $\ce{PCl5}$ nor the $\mathrm{sp^3d^2}$ depiction of $\ce{SF6}$ are close to the truth. – Jan Jan 20 '17 at 20:13
• The new version of the question is in no way the same thing. It is very different. And the answer is symmetry. – Martin - マーチン Jan 21 '17 at 7:52

The answer to this lies in the symmetry of the orbitals involved.

Naturally, an octahedral complex will have the point group $O_\mathrm{h}$ — the octahedral point group. In molecular orbital theory, we will first need to work out how the ligand group orbitals transform; then we can find out which metal orbitals transform identically, i.e. belong to the same irreducible representation; only these will form bonds with each other. To arrive at the group orbitals’ irreducible representations, we need to see how they transform under the given symmetry operations of the group; thankfully, a copy of the character table can be found on meta (Attention: MathJax heavy link! Give the page about 30 seconds time to load!) Since your diagram does not include π interactions, I will ignore them, too.

\begin{array}{lcccccccccc}\hline \text{Irrep} & E & 8 C_3 & 6 C_2 & 6 C_4 & 3 C_4^2 & i & 6 S_4 & 8 S_6 & 3 \sigma_\mathrm{h} & 6 \sigma_\mathrm{d} \\ \hline 6 \sigma_{\ce{L\bond{->}M}} & 6 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 4 & 2 \\ \hline\end{array}

We now need to use the reduction formula to determine how many times each irreducible representation is part of this reducible representation.

$$n = \frac 1h \sum_{\chi} N \chi_\mathrm{r} \chi_\mathrm{i}\tag{1}$$

Wherein $n$ is the number of times an irreducible representation is included, $N$ is the coefficient of a symmetry operation group, $h$ is the sum of all coefficients ($48$ in $O_\mathrm{h}$) and $\chi$ is the character of the irreducible or reducible representation, respectively. Thankfully, a large number of reducible group characters end up as $0$, easing our calculations. This should be the result:

\begin{align}n(\mathrm{a_{1g}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\ n(\mathrm{a_{2g}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\\ n(\mathrm{e_g}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times 2 + 2 \times 6 \times 0) &&= 1\\ n(\mathrm{t_{1g}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\ n(\mathrm{t_{2g}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\ n(\mathrm{a_{1u}}) &= \frac 1{48} (6 + 2 \times 6 \times 1 + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times (-1)) &&= 0\\ n(\mathrm{a_{2u}}) &= \frac 1{48} (6 + 2 \times 6 \times (-1) + 2 \times 3 \times 1 + 4 \times 3 \times (-1) + 2 \times 6 \times 1) &&= 0\\ n(\mathrm{e_u}) &= \frac 1{48} (12 + 2 \times 6 \times 0 + 2 \times 3 \times 2 + 4 \times 3 \times (-2) + 2 \times 6 \times 0) &&= 0\\ n(\mathrm{t_{1u}}) &= \frac 1{48} (18 + 2 \times 6 \times 1 + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times 1) &&= 1\\ n(\mathrm{t_{2u}}) &= \frac 1{48} (18 + 2 \times 6 \times (-1) + 2 \times 3 \times (-1) + 4 \times 3 \times 1 + 2 \times 6 \times (-1)) &&= 0\end{align}

We see, that the six σ-symmetric ligand orbitals transform as $\mathrm{a_{1g} + e_g + t_{1u}}$ — this is also shown in your picture. Now that we have this settled, we need to ask ourselves which metal orbitals these irreducibles correspond to. $\mathrm{a_{1g}}$, the totally symmetric irreducible is easy; it must be the $\mathrm{s}$ orbital. Furthermore, the character table tells us that $(x,y,z)$ transforms as $\mathrm{t_{1u}}$ which means that the $\mathrm{p}$ orbitals must transform as that. (You can verify this by performing the same exercise on the three central $\mathrm{p}$ orbitals.)

This leaves the $\mathrm{d}$ orbitals which — for symmetry reasons as there is no pentadegenerate irreducible — have to split up into at least a group of three and two ($\mathrm{t}_x$ and $\mathrm{e}_x$). Again, glancing at the character table tells us that $\mathrm{d}_{xy}, \mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ transform as $\mathrm{t_{2g}}$ and $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ as $\mathrm{e_g}$. The first can easily be verified by the same exercise as for the $\mathrm{p}$ orbitals; the second not so easily due to the ‘unnatural’ shape of the $\mathrm{d}_{z^2}$ orbital. However, do trust the calculations that they are equivalent.

Therefore, it is only $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$ that can interact with the ligand group orbitals and which therefore form the σ bonds as your scheme shows. And this also makes sense from a purely qualitative point of view: The $\mathrm{d}_{z^2}$ orbital is pointing towards the $z$-axis ligands (and the ring is somewhat interacting with the $x,y$-axis ligands) while each of the four lobes of $\mathrm{d}_{x^2-y^2}$ is pointing towards a ligand on the $x$ or $y$ axis. The $\mathrm{t_{2g}}$ orbitals are pointing in between the coordinate axis directions and therefore between the ligands and are thus nonbonding with respect to σ bonds.


text:
# Kibble Analytics

## Mar 30, 2018 · 2488 words · 12 minutes read R statistics

From late April to early July last year, I was cat sitting for two lap cats. Originally alley cats from Montreal, they’re now living highly sheltered lives in a one-bedroom apartment in Toronto. Breakfast and dinner consists of wet food, and they have a near continuous supply of kibble to snack on throughout the day.

When their owner went off to Rhode Island for three months, I got the honour of looking after both the cats and the apartment. At the time, I was wrapping up my master’s degree, working part time on MPEDS, and looking for full time jobs. Without a lot of other fun things going on in my life, my dedication to the cats quickly spiralled out of control.

What started out as feeding them twice a day soon turned into an elaborate effort to catalog their food consumption. I kept two Google Sheets with the weight of their food bowls at various points throughout the day, and visualized the data through a few custom pipeline scripts in R. Or, as I told my friends, I was pioneering the field of kibble analytics! At the risk of seeming less sane than I am, I’ve done a full write-up of the results in this post.

## Data Pipeline

With the possible exception of primary school projects, this project was my first foray into manual data collection. I’m used to having to deal with other people’s messy data, not generating it myself.

To minimize the effort involved, I kept track of the weight of the bowls themselves rather than the food. Every time I refilled food, I would first do an initial measurement to see how much food was left from the last meal. After refilling, I would measure the weight of the bowls again before making the food available to the cats. As part of the R pipeline, I would subtract the weight of the empty bowl to get the weight of the food.

Jenny Bryan’s googlesheets package has worked its way into most areas of my life over the past year and a half, and it seemed like the obvious choice for this project. I kept all of the raw data on Google Drive, and wrote a few helper functions (prepare.kibble() and prepare.wet.food()) to load the data into R.

In total I ended up with 663 measurements. 398 of these were of the wet food bowls, and the remaining 265 were kibble bowl measurements. Predictably, I was the most enthusiastic about the project in the beginning. At my best, I made 28 measurements in a single day! After about two weeks I got lazier, but still managed to keep up the data collection at a rate of at least four measurements a day until I moved out in early July.

The cats were similarly consistent in eating the food I gave them. Aside from two days in late May when we ran out of kibble, they ate around 50 grams of kibble and one to two cans of wet food every day. Interestingly, the three-day kibble deprivation appears to have tipped food consumption in favour of kibble for about a week after it was put back on the menu – the first two days had the highest kibble consumption of the entire period.

## Analysis

With the data and processing pipelines in place, it was time to start the analysis. At first, I didn’t have any particular questions I wanted to investigate. Like many modern businesses, I had data but no clear ideas on how it could help me. My first few weeks were spent making basic plots. A kibble dashboard, if you will.

I regained some sense of purpose after a friend of mine introduced me to CatterPlots. The package allows you to make R plots with cat-shaped points. It’s admittedly a visualization technique for the few, but it seemed perfect for this analysis. Armed with kitty-shaped plots, I embarked on some actual statistics.

### Food Preference

I started by looking at the difference between wet food and kibble consumption. Kibble was generally available to the cats at all times, while wet food was distributed at most twice a day. It seemed pretty clear that the cats preferred wet food as long as it was available and fresh, and used kibble more as a last resort. To delve deeper into this, I tried modelling the food consumed as a function of the time since wet food was last distributed.

The wet food consumption is clearly non-linear in time, forcing me to look beyond linear regression.

In the beginning, I didn’t have enough data for good non-parametric estimates. Standard scatterplot smoothers like LOESS model the response based on data in a neighbourhood, and only imposes a smoothness restriction on the global function. In my case that resulted in local dips in the function, which didn’t make sense. On average the cats should have consumed at least as much food after five hours as after four, and I wanted a regression method that would accommodate this constraint.

Isotonic regression is a basic approach to non-decreasing, non-parametric regression. For a set of response variables $y_i$ ordered by a covariate $x_i$, isotonic regression seeks to find a set of predicted values $\hat{y}_i$ that minimize

$$\sum_{i=1}^n w_i \left( y_i - \hat{y}_i \right)^2$$

subject to the constraint $\hat{y}_1 \leq \hat{y}_2 \leq … \leq \hat{y}_n$. If we ignore the weights $w_i$, this amounts to finding the step function that minimizes the least squares criterion for the response $y$.

The R function isoreg() implements isotonic regression. It takes vectors of $x$ and $y$ values as input and returns an object that can be converted into a step function. The step function can then be called to predict the response $y$ for any new value $x$.

While obtaining the regression estimates was easy, superimposing them on the plot required some acrobatics. Internally, CatterPlots works with data on a $[0, 1]$ scale. To add the regression curves to the plot, I had to first scale the $x$ and $y$-values with the non-exported CatterPlots:::scaleData() function1.

# limits on plot
hack.args <- list(
xlim = c(0, 24),
ylim = c(0, 150)
);

kibble.regression <- isoreg(
y = kibble$food_consumed, x = kibble$time_elapsed
);

x <- seq(0, 24, by = 0.01);
scaled.kibble.regression <- CatterPlots:::scaleData(
x,
as.stepfun(kibble.regression)(x),
args = hack.args
);

lines(
scaled.kibble.regression$xscale, scaled.kibble.regression$yscale,
lwd = 3,
col = kibble.colour
);

The result is a rather uninviting step function overlaid on the cats.

Frustrated with the decreased aesthetic appeal, I set out in search for non-parametric regression methods that would give me smooth, monotone increasing functions. The R package monreg implements an approach based on a monotone transformations. Given an initial non-parametric regression function $\hat{m}(x)$, it estimates the inverse of the monotonized version through the formula

$$\hat{m}_I^{-1} = \frac{1}{Nh_d} \sum^N_{i=1} \int_{-\infty}^t K_d \left( \frac{\hat{m}\left(\frac{1}{N}\right) - u}{h_d} \right) du$$

where $K_d$ is a kernel density estimator with bandwidth $h_d$. Afterwards, a monotone non-parametric function $\hat{m}_I$ is obtained by taking the inverse of $\hat{m}_I^{-1}$. The result is a much more pleasing, smooth function that looks similar to what you would obtain using standard standard scatterplot smoothers.

The smooth functions also make it easier to see patterns. Kibble consumption is roughly linear, while wet food is eaten more quickly before consumption levels off. A can contains around 150 grams of wet food, and on average the cats ate 127 grams in the first 12 hours. In most cases they would have been served a new meal at that point, but on days that they weren’t, wet food consumption continued to increase very slowly – presumably because food had either run out or gone stale. By contrast, the average kibble consumption by 12 hours was 27 grams, and this continued to grow linearly to an average of 56 grams eaten by the 24 hour mark.

The differences are smaller when we consider the amount of nutrients rather than the weight. Like the name suggests, wet food contains a considerable amount of water. Repeating the analysis for protein consumption rather than total food consumption reveals that kibble overtakes wet food as the main source of protein after roughly 18 hours.

Nutrients can also explain some of the cats’ eagerness to eat kibble after the three-day deprivation of late May. When only eating wet food, they would get less than half of their normal intake of fibre.

Nutrient Kibble (%) Wet food (%)
Protein 30 10
Fat 14 5
Fibre 10.5 1
Moisture 10 78
Ash - 3
Taurine - 0.5

### Bowl Preference

After concluding on the kibble vs wet food question, I started digging deeper into the wet food dynamics. The cats had two wet food bowls with different shapes. When refilling food, I would try to split it evenly between the two bowls.

Initially, the different shapes of the bowls were simply an easy way to keep track of the food consumption. I created separate columns in my spreadsheet asymmetric_bowl_weight and symmetric_bowl_weight, and only calculated the total food remaining as part of my R pipeline. This setup made it easy to extend the analysis to consider the two bowls separately.

At first sight, it could seem the cats start by eating more from from the asymmetric bowl, although it’s not clear that the difference is statistically significant. There’s also a confounding factor in the spatial arrangement of the bowls. I lined the bowls up along the side of the kitchen, with one bowl closer to the living room (and the rest of the apartment) than the other. It seemed entirely plausible that the cats could exhibit a bias towards the first bowl they encountered on their quest for food.

To disentangle the effect of the bowl shape from the effect of the spatial arrangement, I launched a new experiment. Every time I refilled the bowls, I would randomize their relative position and record it in a new column in my spreadsheet.

Analyzing this data turned out to be more difficult than expected. The non-parametric monotone regression framework does not lend itself easily to significance testing, and I reverted to a standard linear regression model. To make those assumptions slightly more realistic, I focused on the first twelve hours after wet food had been served.

I also wanted to restrict the analysis to time points when the cats had a choice. If one of the bowls was empty, I imagined they would go for the other one, regardless of any personal preferences for the shape of the bowl. To avoid biasing the analysis, I only included time points where each bowl had at least five grams of food left.

There’s one further complication. On some days, I would take multiple measurements within a short time frame. These observations are typically highly correlated, and in a standard linear regression model they would be given equal weight to observations from days when I only took a single measurement. To account for the correlation in the data, I grouped all of the observations according to the refill time stamp and included this group as a random effect.

Fixed effects: food_consumed ~ bowl + closest_livingroom * time_since_refill +      bowl * time_since_refill
Value Std.Error DF   t-value p-value
(Intercept)                              24.416994  4.126351 71  5.917333  0.0000
bowlsymmetric                            -3.189013  5.306992 71 -0.600908  0.5498
closest_livingroomTRUE                   14.078871  5.238359 71  2.687649  0.0090
time_since_refill                         3.841742  0.593709 71  6.470746  0.0000
closest_livingroomTRUE:time_since_refill -0.586184  0.720174 71 -0.813947  0.4184
bowlsymmetric:time_since_refill          -0.410452  0.721227 71 -0.569101  0.5711

Contrary to what I expected, there’s no statistically significant difference in the rate of consumption over time. However, the bowl closest to the living room has a significantly higher baseline consumption. When served a new meal, the cats would immediately eat an average of 25 grams from the kitchen side bowl and 39 grams from the bowl on the living room side. By contrast, there’s no evidence of any preference based on the shape of the bowl.

I don’t have a good explanation for this. One unexplored possibility is that it could relate to the personal preferences of each cat. As everyone who visited me that summer would remark, one of the cats was considerably bigger than the other. It’s possible that the cat that would eat the most also tended to position himself on the living room side. Without a more invasive follow-up study that records the food consumption of the individual cat, there’s no way to validate this hypothesis.

### Association with Temperature

Finally, I looked at whether there was any association between the cats’ food consumption and the weather. I downloaded daily temperature data for Toronto Pearson International airport from the Government of Canada website and correlated it with the total food consumption (kibble and wet food) of the cats.

Spring in Canada is typically miserable. For my first few weeks of cat-sitting, the temperatures hovered around 10 degrees Celsius. Towards the end of my stay they reached the 20-25 degree range, and at first glance it seems the average food consumption was slightly lower.

This effect gets far more pronounced when we look at the data in a CatterPlot. The Spearman’s correlation is a staggering -0.62! Translating this into a regression framework, we expect the cats to eat seven grams less a day for every degree increase in average daily temperature.

## Conclusion

Since moving to London, I have spent considerably less time on cats. The only four-legged creatures in my current apartment are mice, and the food we give them would turn similar projects into survival analyses. But even if I won’t be doing a follow-up study any time soon, I’ve thought about some limitations of this project.

For one thing, I never randomized the measurement times. To avoid data collection feeling too much like a chore, I only weighed the food bowls when I felt like it. Of course, this opens up the possibility that I could have been influenced by the cats’ behaviour. If I noticed they were eating at unusual times, I might have been more inclined to get off the couch and take out my coffee scale.

Another problem related to the cats themselves. Not a friend of science, Cat 1 was caught splashing water into their kibble bowl on several occasions. While I tried to let the disaster area dry out before weighing the bowl, I’m not sure I succeeded.

If you think you can do a better study yourself, all of the code is available on GitHub.

## Acknowledgements

None of this would have been possible without the love and support of the experimental subjects themselves, Smoke and Brindi. I also have to thank VK for supplying the cats, Vinny for introducing me to CatterPlots, and Paul for making sure I normalized my data.

## References

Dette, Holger and Regine Scheder. “Strictly Monotone and smooth nonparametric regression for two or more variables”. (2005)

Mair, Patrick, Kurt Hornik, and Jan de Leeuw. “Isotone optimization in R: pool-adjacent-violators algorithm (PAVA) and active set methods.” Journal of statistical software 32.5 (2009): 1-24.

1. For all of the code, see the GitHub repository [return]


text:
% Copyright (C) 2003 David Roundy % % This program is free software; you can redistribute it and/or modify % it under the terms of the GNU General Public License as published by % the Free Software Foundation; either version 2, or (at your option) % any later version. % % This program is distributed in the hope that it will be useful, % but WITHOUT ANY WARRANTY; without even the implied warranty of % MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the % GNU General Public License for more details. % % You should have received a copy of the GNU General Public License % along with this program; see the file COPYING. If not, write to % the Free Software Foundation, Inc., 51 Franklin Street, Fifth Floor, % Boston, MA 02110-1301, USA. \begin{code}
module Darcs.ArgumentDefaults ( get_default_flags ) where
import Data.Maybe ( catMaybes, listToMaybe, mapMaybe )

import Darcs.Arguments ( DarcsFlag,
DarcsOption( DarcsArgOption, DarcsNoArgOption, DarcsMultipleChoiceOption ),
arein, isin )
import Darcs.Commands ( CommandControl( CommandData ),
commandAlloptions )
import Darcs.Commands.Help ( commandControlList )
import Darcs.Repository.Prefs ( getGlobal, getPreflist )

\end{code} \paragraph{defaults}\label{defaults} Default values for darcs commands can be configured on a per-repository basis by editing (and possibly creating) the \verb!_darcs/prefs/defaults! file. Each line of this file has the following form: \begin{verbatim} COMMAND FLAG VALUE \end{verbatim} where \verb!COMMAND! is either the name of the command to which the default applies, or \verb!ALL! to indicate that the default applies to all commands accepting that flag. The \verb!FLAG! term is the name of the long argument option without the \verb!--!'', i.e.\ \verb!verbose! rather than \verb!--verbose!. Finally, the \verb!VALUE! option can be omitted if the flag is one such as \verb!verbose! that doesn't involve a value. If the value has spaces in it, use single quotes, not double quotes, to surround it. Each line only takes one flag. To set multiple defaults for the same command (or for \verb!ALL! commands), use multiple lines. Note that the use of \verb|ALL| easily can have unpredicted consequences, especially if commands in newer versions of darcs accepts flags that they didn't in previous versions. A command like \verb|obliterate| could be devastating with the wrong'' flags (for example --all). Only use safe flags with \verb|ALL|. \begin{tabular}{ll} {\tt \verb!~/.darcs/defaults!} & provides defaults for this user account, on MS Windows~\ref{ms_win} \\ {\tt \verb!repo/_darcs/prefs/defaults!} & provides defaults for one project,\\ & overrules changes per user \\ \end{tabular} For example, if your system clock is bizarre, you could instruct darcs to always ignore the file modification times by adding the following line to your \verb!_darcs/prefs/defaults! file. (Note that this would have to be done for each repository!) \begin{verbatim} ALL ignore-times \end{verbatim} If you never want to run a test when recording to a particular repository (but still want to do so when running \verb'check' on that repository), and like to name all your patches Stupid patch'', you could use the following: \begin{verbatim} record no-test record patch-name Stupid patch \end{verbatim} If you would like a command to be run every time patches are recorded in a particular repository (for example if you have one central repository, that all developers contribute to), then you can set apply to always run a command when apply is successful. For example, if you need to make sure that the files in the repository have the correct access rights you might use the following. There are two things to note about using darcs this way: \begin{itemize} \item Without the second line you will get errors, because the sub process that runs apply cannot prompt interactively. \item Whatever script is run by the post apply command should not be added to the repository with \verb!darcs add!; doing so would allow people to modify that file and then run arbitrary scripts on your main repository, possibly damaging or violating security. \end{itemize} \begin{verbatim} apply posthook chmod -R a+r * apply run-posthook \end{verbatim} Similarly, if you need a command to run automatically before darcs performs an action you can use a prehook. Using prehooks it could be possible to canonicalize line endings before recording patches. There are some options which are meant specifically for use in \verb!_darcs/prefs/defaults!. One of them is \verb!--disable!. As the name suggests, this option will disable every command that got it as argument. So, if you are afraid that you could damage your repositories by inadvertent use of a command like amend-record, add the following line to \verb!_darcs/prefs/defaults!: \begin{verbatim} amend-record disable \end{verbatim} Also, a global preferences file can be created with the name \verb!.darcs/defaults! in your home directory, on MS Windows~\ref{ms_win}. Options present there will be added to the repository-specific preferences. If they conflict with repository-specific options, the repository-specific ones will take precedence. \begin{code}
get_default_flags :: String -> [DarcsOption] -> [DarcsFlag] -> IO [DarcsFlag]
get_default_flags com com_opts already = do
repo_defs   <- default_content $getPreflist "defaults" global_defs <- default_content$ getGlobal   "defaults"
let repo_flags = get_flags_from com com_opts already repo_defs
global_flags = get_flags_from com com_opts
return $repo_flags ++ global_flags get_flags_from :: String -> [DarcsOption] -> [DarcsFlag] -> [(String,String,String)] -> [DarcsFlag] get_flags_from com com_opts already defs = options_for com_defs com_opts com_opts ++ options_for all_defs com_opts all_opts where com_defs = filter (\ (c,_,_) -> c == com) defs all_defs = filter (\ (c,_,_) -> c == "ALL") defs options_for d o ao = concatMap (find_option o ao already) d all_opts = concatMap get_opts commandControlList get_opts (CommandData c) = let (o1, o2) = commandAlloptions c in o1 ++ o2 get_opts _ = [] find_option :: [DarcsOption] -> [DarcsOption] -> [DarcsFlag] -> (String,String,String) -> [DarcsFlag] find_option opts all_opts already (c, f, d) = if null$ mapMaybe choose_option all_opts
then error $"Bad default option: command '"++c++"' has no option '"++f++"'." else concat$ mapMaybe choose_option opts
where choose_option (DarcsNoArgOption _ fls o _)
| o elem already = Just []
| f elem fls = if null d
then Just [o]
else error $"Bad default option: '"++f ++"' takes no argument, but '"++d ++"' argument given." choose_option (DarcsArgOption _ fls o _ _) | o isin already = Just [] | f elem fls = if null d then error$ "Bad default option: '"++f
++"' requires an argument, but no "
++"argument given."
else Just [o d]
choose_option (DarcsMultipleChoiceOption os)
| os arein already = Just []
| otherwise = listToMaybe \$ mapMaybe choose_option os
choose_option _ = Nothing

default_content :: IO [String] -> IO [(String,String,String)]
default_content = fmap (catMaybes . map (doline.words))
where doline (c:a:r) = Just (c, drop_dashdash a, unwords r)
doline _ = Nothing
drop_dashdash ('-':'-':a) = a
drop_dashdash a = a

\end{code}


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Aug 2018, 11:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Applying AM-GM inequality into finding extreme/absolute value

Author Message
TAGS:

### Hide Tags

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1394
Location: Viet Nam
Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

27 Nov 2016, 21:11
5
9
1. AM-GM inequality

First, I would like to introduce standard AM-GM inequality:$x + y \geq 2\sqrt{xy} \quad \forall x,y \geq 0$
Sign "=" occurs $$\iff x=y$$

Other version of AM-GM inequality is: $$(x+y)^2 \geq 4xy \quad \forall x,y \in R$$

The extended version of AM-GM inequality:
$x_1+x_2+...+x_n \geq n\sqrt[n]{x_1x_2...x_n} \quad \forall x_k \geq 0$
Sign "=" occurs $$\iff x_1=x_2=...=x_n$$

In GMAT, for $$n=2$$ and $$n=3$$ are enough to solve every questions.

Note that sign $$\forall$$ means for all
Sign $$\implies$$ means imply. For example, $$a \implies b$$ means $$a$$ implies $$b$$ but $$b$$ doesnot imply $$a$$
Sign $$\iff$$ means if and only if. For example, $$a \iff b$$ means $$a$$ occurs if and only if $$b$$ occurs. This could also be expressed like this: $$a$$ implies $$b$$ and also $$b$$ implies $$a$$, or $$a \implies b$$ and $$b \implies a$$.

2. Proving AM-GM inequality

Since $$x,y \geq 0$$, we have $$x + y \geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy \iff x^2 +2xy +y^2 \geq 4xy \iff (x-y)^2 \geq 0$$
The inequality $$(x-y)^2 \geq 0$$ is true for all $$(x,y)$$, so AM-GM inequality is true.

3. Applying AM-GM inequality into finding extreme value
Here are some examples to illustrate method of applying AM-GM inequality.

Example 1. For x and y are positive and x+y=10. Find the maximum value of xy.
Solve:
Since $$x,y>0$$, $$x,y$$ are satisfied to apply AM_GM inequality.

$$x+y \geq 2\sqrt{xy} \iff (x+y)^2 \geq 4xy \iff xy \leq \frac{(x+y)^2}{4} =\frac{10^2}{4}=25$$
$$max(xy)=25 \iff x=y=5$$.

Example 2. For x and y are positive and xy=36. Find the minimum value of x+y.
http://gmatclub.com/forum/which-one-of- ... l#p1768552
Solve:
Since $$x,y>0$$, $$x,y$$ are satisfied to apply AM_GM inequality.
$$x+y \geq 2\sqrt{xy} = 2\sqrt{36} =12$$
$$min(x+y)=12 \iff x=y=6$$.

Exmaple 3. For x and y are positive and 2x+3y=12. Find the maximum value of xy.
Solve:
Since $$x,y>0$$, $$2x,3y$$ are satisfied to apply AM_GM inequality.
$$2x+3y \geq 2\sqrt{2x \times 3y} = 2\sqrt{6xy} \iff xy \leq \frac{(2x+3y)^2}{24}=\frac{12^2}{24}=6$$
$$max(xy)=12 \iff 2x=3y=6 \iff x=3 \quad y=2$$.

Example 4. Triangle ABC has perimeter 30. What is the lagest area of triangle ABC?
http://gmatclub.com/forum/patricia-buil ... l#p1766993

Solve:
Triangle ABC has 3 side lengths are $$a,b,c$$.

Heron's formula to calculate triangle's area based on the lengths of 3 sides:
$$S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)} \quad$$ with $$p=\frac{a+b+c}{2}$$

We have $$p=30/2=15$$.

Since $$p-a,p-b,p-c>0$$, so $$p-a,p-b,p-c$$ are satisfied to apply AM_GM inequality for 3 positive numbers.

The AM-GM inequality for 3 positive numbers: $$x+y+z \geq 3\sqrt[3]{xyz} \iff xyz \leq \frac{(x+y+z)^3}{27}$$

$$S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)} \leq \sqrt{p \times \frac{(p-a+p-b+p-c)^3}{27}}=\sqrt{p \times \frac{p^3}{27}}=$$ $$\sqrt{\frac{p^4}{27}}=\frac{p^2}{3\sqrt{3}}=25\sqrt{3}$$

$$max(S_{ABC})=25\sqrt{3} \iff a=b=c=10$$ or ABC is equilateral triangle

4. Applying AM-GM inequality into solving equality with absolute
(in)equations with absolute are generally quite complicated to solve since the absolute value is based on specific value of expression. To solve these (in)equation, the simpliest way is to get rid of absolute site by specifying each cases. However, there are some kinds of (in)equation that could be solve quickly by using AM-GM inequality.

Example 5. Proving that $$|a|+|b| \geq |a+b| \quad \forall a,b \in R$$.
Solve:
$$|a|+|b| \geq |a+b| \iff (|a|+|b|)^2 \geq (a+b)^2 \iff a^2 + 2|a||b| + b^2 \geq a^2 + 2ab + b^2 \iff |a||b| \geq ab$$.
Since $$|a||b| \geq ab$$ is true for all $$(a,b)$$, so the inquality is proved.

$$|a|+|b| = |a+b| \iff ab \geq 0$$

Example 6. Find the minium value of $$|x+3| + |x-7|$$.
Solve:
$$|x+3| + |x-7| = |x+3| + |7-x| \geq |x+3+7-x|=10$$
$$min(|x+3| + |x-7|)=10 \implies (x+3)(7-x) \geq 0 \iff (x+3)(x-7) \leq 0 \implies -3 \leq x \leq 7$$.

To solve inequilitys like $$(x-a)(a-b)\leq 0$$, $$(x-a)(a-b)\geq 0$$, I'll discuss in new post later.

Example 7. (Bunuel's questions) http://gmatclub.com/forum/inequality-an ... ml#p653853
Is $$|x+y|>|x-y|$$?
(1) |x| > |y|
(2) |x-y| < |x|

Solve:
$$|x+y|>|x-y| \iff |x+y|^2>|x-y|^2 \iff x^2+2xy+y^2>x^2-2xy+y^2 \iff 4xy>0 \iff xy>0$$
So solving this example is simply consider $$xy$$ is positive or not.

Example 8. (Bunuel's questions) http://gmatclub.com/forum/inequality-an ... ml#p653870
Is $$r=s$$?
(1) $$-s<=r<=s$$
(2) $$|r|>=s$$

$$(1) \implies s>0 \quad |r| \leq s$$
$$(2) \implies |r| \geq s$$

Combine (1) and (2) $$\implies s>0 \quad and \, s \leq |r| \leq |s| \implies |r|=s \implies r=s \: or \: r=-s$$. Answer is E.

Example 9. (Bunuel's questions) http://gmatclub.com/forum/inequality-an ... ml#p653783
$$|x+2|=|y+2|$$ what is the value of $$x+y$$?
(1) $$xy<0$$
(2) $$x>2 \:\;\; y<2$$

Solve:
$$|x+2|=|y+2| \iff (x+2)^2=(y+2)^2 \iff x^2+4x+4=y^2+4y+4$$ $$\iff x^2 +4x - y^2-4y=0 \iff (x-y)(x+y-4)=0$$

(1) If $$x=y \implies xy=x^2 \geq 0$$. Hence we must have $$x+y=4$$. Sufficient
(2) Clearly $$x \neq y \implies x+y=4$$. Sufficient.

The AM-GM inequality is so powerful to solve some kinds of inequalities as well as some kinds of equations. This tool could assist us in finding the solution immediately and saving a lot of time. I know that this tool is quite relevant to advanced mathematics and is somewhat hard to apply into actual GMAT questions. However, I believe that this post is useful for you in finding new (and maybe quick) ways to solve quant questions in GMAT.

Last but not least, the AM-GM inequality and its other versions are nothing but just tools to help us solve math quicker. The most important is that we need to specify what kind of each question is and dig for its solutions.
_________________
Intern
Joined: 17 May 2017
Posts: 20
GMAT 1: 690 Q50 V33
Re: Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

17 May 2017, 10:56
Does GMAT asks AM and GM concepts i have seen them in only CAT (entrance exam for MBA in India)
Never seen them in GMAT
_________________

Mention why you eliminated each option

Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1394
Location: Viet Nam
Re: Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

17 May 2017, 17:59
1
shwetachauhan wrote:
Does GMAT asks AM and GM concepts i have seen them in only CAT (entrance exam for MBA in India)
Never seen them in GMAT

Actually this is just a tool for solving question. I mean that there could be many ways to solve a math problem. Sometime using AM-GM concept you lead you to the result faster.

Generally, GMAT doesn't require AM-GM concept. Also, you could solve equation without using this tool. I just offer another useful tool that I often use in solving math Use it or not, mainly based on your decision
_________________
SVP
Joined: 26 Mar 2013
Posts: 1783
Re: Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

27 May 2017, 09:17
nguyendinhtuong wrote:

Example 9. (Bunuel's questions) http://gmatclub.com/forum/inequality-an ... ml#p653783
$$|x+2|=|y+2|$$ what is the value of $$x+y$$?
(1) $$xy<0$$
(2) $$x>2 \: y<2$$

Solve:
$$|x+2|=|y+2| \iff (x+2)^2=(y+2)^2 \iff x^2+4x+4=y^2+4y+4$$ $$\iff x^2 +4x - y^2-4y=0 \iff (x-y)(x+y-4)=0$$
If $$x \neq y \implies x+y=4$$, so we simply consider $$x=y$$ or not.

(1) Insufficient. No thing indicates that $$x=y$$ or not.
(2) Clearly $$x \neq y \implies x+y=4$$. Sufficient.

The answer posted by Bunuel is D niot B
Senior CR Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1394
Location: Viet Nam
Re: Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

27 May 2017, 09:55
Mo2men wrote:
nguyendinhtuong wrote:

Example 9. (Bunuel's questions) http://gmatclub.com/forum/inequality-an ... ml#p653783
$$|x+2|=|y+2|$$ what is the value of $$x+y$$?
(1) $$xy<0$$
(2) $$x>2 \: y<2$$

Solve:
$$|x+2|=|y+2| \iff (x+2)^2=(y+2)^2 \iff x^2+4x+4=y^2+4y+4$$ $$\iff x^2 +4x - y^2-4y=0 \iff (x-y)(x+y-4)=0$$
If $$x \neq y \implies x+y=4$$, so we simply consider $$x=y$$ or not.

(1) Insufficient. No thing indicates that $$x=y$$ or not.
(2) Clearly $$x \neq y \implies x+y=4$$. Sufficient.

The answer posted by Bunuel is D niot B

Oh, my carelessness Thank you, I've edited my post
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 7764
Re: Applying AM-GM inequality into finding extreme/absolute value  [#permalink]

### Show Tags

31 May 2018, 10:38
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Applying AM-GM inequality into finding extreme/absolute value &nbs [#permalink] 31 May 2018, 10:38
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


text:
# E. T. Whittaker

Edmund Taylor Whittaker
Born 24 October 1873
Southport, Lancashire
Died 24 March 1956 (aged 82)
Citizenship United Kingdom
Nationality English
Fields mathematics
Institutions
Alma mater Trinity College, Cambridge
Doctoral students
Other notable students W. V. D. Hodge
Marion Cameron Gray
Known for
Notable awards

Edmund Taylor Whittaker FRS[2] FRSE (24 October 1873 – 24 March 1956)[2][3][4] was an English mathematician who contributed widely to applied mathematics, mathematical physics and the theory of special functions. He had a particular interest in numerical analysis, but also worked on celestial mechanics and the history of physics. Near the end of his career he received the Copley Medal, the most prestigious honorary award in British science. The School of Mathematics of the University of Edinburgh holds The Whittaker Colloquium, a yearly lecture in his honour.[1][5]

## Biography

Whittaker was born in Southport, in Lancashire. He was educated at Manchester Grammar School and Trinity College, Cambridge from 1892.[6] He graduated as Second Wrangler in the examination in 1895 and also received the Tyson Medal for Mathematics and Astronomy. In 1896, Whittaker was elected as a fellow of Trinity College, Cambridge, and remained at Cambridge as a teacher until 1906. Between 1906 and 1911 he was the Royal Astronomer of Ireland and professor of astronomy at Trinity College Dublin where he taught mathematical physics. In 1911 Whittaker became professor at the University of Edinburgh and remained there for the rest of his career.

Whittaker was a Christian and became a convert to the Roman Catholic Church (1930). In relation to that he was a member of the Pontifical Academy of Sciences from 1936 onward and was president of a Newman Society. Earlier at Cambridge in 1901 he married the daughter of a Presbyterian minister. They had five children, including the mathematician John Macnaghten Whittaker, (1905-1984) and his elder daughter, Beatrice, married E.T. Copson, who would later become Professor of Mathematics at St. Andrew's University.[7]

Whittaker wrote the biography of a famous Italian mathematician, Vito Volterra for the Royal Society in 1941. In 1954, he was awarded the Copley Medal by the Royal Society, their highest award, "for his distinguished contributions to both pure and applied mathematics and to theoretical physics". Back in 1931 Whittaker had received the Royal Society's Sylvester Medal "for his original contributions to both pure and applied mathematics". Whittaker died in Edinburgh, Scotland.

## Whittaker & Watson

Whittaker is remembered as the author of A Course of Modern Analysis (1902), which in its 1915 second edition in collaboration with George Neville Watson became Whittaker and Watson, one of the handful of mathematics texts of its era that was considered indispensable. This work has remained in print continuously for over a century.[7]

## Special functions

Whittaker is the eponym of the Whittaker function or Whittaker integral, in the theory of confluent hypergeometric functions. This makes him also the eponym of the Whittaker model in the local theory of automorphic representations. He published also on algebraic functions and automorphic functions. He gave expressions for the Bessel functions as integrals involving Legendre functions.

## Partial differential equations

In the theory of partial differential equations, Whittaker developed a general solution of the Laplace equation in three dimensions and the solution of the wave equation. He developed the electrical potential field as a bi-directional flow of energy (sometimes referred to as alternating currents). Whittaker's pair of papers in 1903 and 1904 indicated that any potential can be analysed by a Fourier-like series of waves, such as a planet's gravitational field point-charge. The superpositions of inward and outward wave pairs produce the "static" fields (or scalar potential). These were harmonically-related. By this conception, the structure of electric potential is created from two opposite, though balanced, parts. Whittaker suggested that gravity possessed a wavelike "undulatory" character.

## History of science

In 1910, Whittaker wrote "A History of the Theories of Aether and Electricity", which gave a very detailed account of the aether theories from René Descartes to Hendrik Lorentz and Albert Einstein, including the contributions of Hermann Minkowski, and which made Whittaker a respected historian of science.

In 1951 (Vol. 1) and 1953 (Vol. 2), he published an extended and revised edition of his book in two volumes. The second volume contains some interesting historical remarks. For example, it contains a chapter named "The Relativity Theory of Poincaré and Lorentz", where Whittaker credited Henri Poincaré and Lorentz for developing special relativity, and he attributed to Albert Einstein's relativity paper only little importance. He also attributed the formula $E=mc^2$ to Poincaré. In 1984 Clifford Truesdell wrote that Whittaker "aroused colossal antagonism by trying to set the record straight on the basis of print and record rather than recollection and folklore and professional propaganda,..."[8] On the other hand Abraham Pais wrote that "Whittaker's treatment of special relativity shows how well the author's lack of physical insight matches his ignorance of the literature".[9] According to Roberto Torretti,[10] "Whittaker's views on the origin of special relativity have been rejected by the great majority of scholars", and he cites Max Born (1956), Gerald Holton (1960,1964), Schribner (1964), Goldberg (1967), Zahar (1973), Hirosige (1976), Schaffner (1976), and Arthur I. Miller (1981).

## Applied mathematics and mathematical physics

Whittaker wrote The Calculus of Observations: A Treatise on Numerical Mathematics (1924)[11] and Treatise on the Analytical Dynamics of Particles and Rigid Bodies: With an Introduction to the Problem of Three Bodies (1937).[12][13] He was the editor of Eddington's Fundamental Theory (1946), and wrote From Euclid to Eddington, A Study of Conceptions of the External World (1949), including a first scholarly account of some of the research between 1900 to 1925.

## References

1. ^ a b c
2. ^ a b c Temple, G. (1956). "Edmund Taylor Whittaker. 1873-1956". Biographical Memoirs of Fellows of the Royal Society 2: 299–226. doi:10.1098/rsbm.1956.0021.
3. ^ Dingle, H. (1956). "Edmund T. Whittaker, Mathematician and Historian". Science 124 (3214): 208–209. doi:10.1126/science.124.3214.208. PMID 17838072.
4. ^ Aitken, A. C. (2009). "The Contributions of E. T. Whittaker to Algebra and Numerical Analysis". Proceedings of the Edinburgh Mathematical Society 11: 31. doi:10.1017/S001309150001436X.
5. ^
6. ^ "Whittaker, Edmund Taylor (WHTR892ET)". A Cambridge Alumni Database. University of Cambridge.
7. ^ a b http://www.lms.ac.uk/newsletter/322/322_03.html
8. ^ Clifford Truesdell, An Idiot's Fugitive Essays on Science, page 432, Springer ISBN 0-387-90703-3
9. ^ Pais, Abraham, "Subtle is the Lord", 1982
10. ^ Roberto Torretti, "Relativity and Geometry", 1983
11. ^ Lehmer, D. N. (1925). "Review: The calculus of observations: a treatise on numerical mathematics, by E. T. Whittaker and G. Robinson". Bull. Amer. Math. Soc. 31 (3): 181–182. doi:10.1090/s0002-9904-1925-04014-8.
12. ^ Buchanan, Herbert Earle (1938). "Review: Treatise on the Analytical Dynamics of Particles and Rigid Bodies, by E. T. Whittaker". Bull. Amer. Math. Soc. 44 (5): 316. doi:10.1090/s0002-9904-1938-06728-6.
13. ^ Wilson, E. B. (1906). "Treatise on the Analytical Dynamics of Particles and Rigid Bodies, by E. T. Whittaker". Bull. Amer. Math. Soc. 12 (9): 451–458. doi:10.1090/s0002-9904-1906-01372-6.