text:
Definitions

# Lie bracket of vector fields

''See Lie algebra for more on the definition of the Lie bracket and Lie derivative for the derivation

In the mathematical field of differential topology, the Lie bracket of vector fields or Jacobi–Lie bracket is a bilinear differential operator which assigns, to any two vector fields X and Y on a smooth manifold M, a third vector field denoted [X, Y]. It is closely related to, and sometimes also known as, the Lie derivative. In particular, the bracket $\left[X,Y\right]$ equals the Lie derivative $mathcal\left\{L\right\}_X Y$.

It plays an important role in differential geometry and differential topology, and is also fundamental in the geometric theory for nonlinear control systems.

## Definition

Let X and Y be smooth vector fields on a smooth $n$-manifold M. The Jacobi-Lie bracket or simply Lie bracket of X and Y, denoted $\left[X,Y\right]$ is the unique vector field such that

$mathcal\left\{L\right\}_\left\{\left[X,Y\right]\right\} = mathcal\left\{L\right\}_X circ mathcal\left\{L\right\}_Y - mathcal\left\{L\right\}_Y circ mathcal\left\{L\right\}_X$

where $mathcal\left\{L\right\}_X$ is the Lie derivative with respect to the vector field X. For a finite-dimensional manifold M we can define the Jacobi-Lie bracket in local coordinates as

$\left[X,Y\right]^i= sum_\left\{j=1\right\}^n left \left(X^j frac \left\{partial Y^i\right\}\left\{partial x^j\right\} right \right) - left \left(Y^j frac \left\{partial X^i\right\}\left\{partial x^j\right\} right \right)$

where n is the dimension of M.

The Lie bracket of vector fields equips the real vector space $V=Gamma^\left\{infty\right\}\left(TM\right)$ (i.e., smooth sections of the tangent bundle of $M$) with the structure of a Lie algebra, i.e., [.,.] is a map from V$times$V to V with the following properties

• [.,.] is R-bilinear
• $\left[X,Y\right]=-\left[Y,X\right],$
• $\left[X,\left[Y,Z\right]\right]+\left[Z,\left[X,Y\right]\right]+\left[Y,\left[Z,X\right]\right]=0.,$ This is the Jacobi identity.

An immediate consequence of these properties is that $\left[X,X\right]=0$ for any $X$.

## Examples

For a matrix Lie group, smooth vector fields can be locally represented in the corresponding Lie algebra. Since the Lie algebra associated with a Lie group is isomorphic to the group's tangent space at the identity, elements of the Lie algebra of a matrix Lie group are also matrices. Hence the Jacobi-Lie bracket corresponds to the usual commutator for a matrix group:

$\left[X,Y\right] = XY - YX$

where juxtaposition indicates matrix multiplication.

## Applications

The Jacobi–Lie bracket is essential to proving small-time local controllability (STLC) for driftless affine control systems.

## References

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# Calling all Cornell (Johnson) Fall 2010 applicants!

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Director
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Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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21 May 2009, 07:16
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BOOKMARKED
I will be the cheerleader for incoming Fall 2010 applicants! Cornell Johnson has one of the biggest matriculant groups for GMATClub Class of 2011 (so far there are 15 of us) so I assume Class of 2012 will be even bigger!

 Important note for International Applicants:financial-aid-discussion-for-internationals-61406.htmlfinancing-your-mba-international-students-95916.html

Johnson Fall 2010 Applicants:

Total GMATClub Stats:
Total Applicants 0
Johnson Class of 2012 (and 2011 AMBA): 0
Waitlist: 6
Dings: 2
Unknown: 0

==============================================

Applicants: 101 Total

avt106 (AMBA) - Accepted
Ambanii
anotherindian
brass20y
considerator
hserfoj
kidd38s
mjjking
ripr
ubuntu
yuskay

[b]Round 1 GMATClub Stats:
Total Applicants 0
Johnson Class of 2012 (and 2011 AMBA): 0
Waitlist:6
Dings: 2
Unknown: 13

hjf6 - Accepted 12/15 Parks Finalist
Architect - Accepted 12/16
Batman83 - Accepted 12/16
bpr81 (AMBA)- Accepted + $$12/16 Fabian - Accepted 12/16 infanteed - Accepted 12/16 Lexi25 - Accepted 12/16 megachaoren - Accepted 12/16 orgninja - Accepted 12/16 radicalchange - Accepted 12/16 redcarpet - Accepted 12/16 xpiano1978 (AMBA) - Accepted 12/17 adude -WL 12/17 jtk25 - Waitlist 12/17 mbacheckmate - Waitlist 12/17 MeowKat - Waitlist 12/17 sjed - Waitlist Skye10 - Waitlist Reeve015 - Ding w/Int 12/17 rohitmakkad - Ding 12/17 cul3s - Int Invite 10/28 mbahopeful2010 - Int Invite ramang - Int Invite 11/7 achintmba09 alnasl corporateraider fongvs2 get2bscool lulala mhe2 TobiasFuke vishwa vivi72 Round 2 GMATClub Stats: Total Applicants 0 Johnson Class of 2012 (and 2011 AMBA): 0 Admits: 0 Waitlist: 0 Dings: 0 Unknown: 0 gtb2009 - Accepted emickho - Accepted mbaseek12 - Accepted w/$$

mbachallenge - Ding after WL

aabbccddee - Int Invite 1/14
blueblanket - Int Invite
conmisdosmanos - Int Invite 1/4
emickho - Int Invite 1/22
holister - Int Invite 1/13
jessparkle934 - Int Invite 1/4
JinJJa (AMBA) - Int Invite 1/4
jzd (AMBA) - Int Invite 12/17
quantumquark - Int Invite 1/6
sid325 - Int Invite
tangzhijian - Int Invite 1/18
tarunmalik - Int Invite 12/22
Ari123
bandit
bmat
dhanur
elvispresley
Freshair
jackets85
jimmyfunk
Kavsin
optiquezt
paulymba
rajinikanth
RSam79
Sharmar
Silvers
usmba2010 (AMBA)

[b]Round 3 GMATClub Stats:

Total Applicants 0
Johnson Class of 2012 (and 2011 AMBA): 0
Waitlist: 4
Dings: 6
Unknown: 0

jg268 - Admitted with

alpha56789 - Waitlisted
cavorter - Waitlisted
KSmith - Waitlisted
sharmar - Waitlisted - withdrawn for Ross

rainostr - Ding
IndianGuardian - Ding
bschools2010 - Ding
TheDream - Ding
flycolors - Ding after WL
niraja - Ding

tabro - Int Invite

bschools2010
chmny3
ezekiel2517 - withdrawn
fugo
hohoo
ROSS2012
serenityprayer
sj0509
usmbaquest
VictoryMBA
vivekjon
williamthesituation

Round 4 GMATClub Stats:
Total Applicants 8
Johnson Class of 2012 (and 2011 AMBA): 0
Waitlist:
Dings:1
Unknown: 4

ammin - Ding

kekmurph
moiuday
tzv123
superdkny

==============================================

Class of 2007 Employment Report (will update soon): http://www.johnson.cornell.edu/recruiter/resources/Employment%20Report2007.pdf

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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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28 Apr 2010, 01:02
3
KUDOS
There's been some discussion about rankings and I wanted to weigh in.

There was a post a while back wondering if there was a correlation between Cornell's rankings and the overall economy. I plotted the Johnson rankings against the DJIA prices (I tried to find the DJIA price around February because I believe rankings are released around that time).

The first chart is the DJIA on the left axis and Johnson School rankings on the right axis.
The second chart is the % change from both the DJIA and Johnson School rankings.
Does it correlate? I'm no statistician and my methods aren't exactly scientific (seriously, did it in 5 minutes), but some of you may be able to see into the mists and come up with an interpretation for us.

Our rankings seem to go increase (BAD) when the DJIA goes up (GOOD). Hmmm...

Also, this is for anybody considering an international career (I am an expat working in Beijing)

Whenever I tell people where I am going to, they may not know the Johnson School, but they for sure know Cornell University. And then they think Ivy League. When I told someone I was applying to MIT (dinged) they were surprised it even had a business school. My colleagues are from Europe, and they had similar reactions.

So in the US, if people pay attention to rankings, we may not be looked upon as favorably as Harvard, Stanford, etc. BUT, internationally, people will link you to the Cornell name, and that goes a looooong way.
Attachments

File comment: DJIA % change vs Johnson School Rankings % change

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File comment: DJIA vs Johnson School Rankings

Chart1.jpg [ 81.65 KiB | Viewed 1901 times ]

Current Student
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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02 Jun 2010, 08:04
2
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Just got an admit call as well. My feedback session was last week and it went exactly the way tangerinegate described it above. The questions revolved around progress since they put me on the WL, career goals, and in my case improvements in my quantitative aptitude. They also allude to the areas in your profile that you should address after you've updated them on what you've been doing. It was fairly conversational and the person who conducted my session (adcom member) was very nice and definitely put me at ease. I find all of JGSM staff very friendly and outgoing, so that helps.

IMHO getting a feedback session is a factor of a couple of things:

1) self-awareness: have you addressed the areas in your profile they were concerned about (without them telling you what those are)? Know thyself. Also importantly, if you know that Johnson is your top choice - don't hesitate to tell them that! They want to know that one phone call to a WL person is going to fill a spot in the upcoming class. Since the WL is used to manage yield, they want to be sure you will enroll. If you know that you will, tell them that.

2) luck: by luck I really mean things that are out of your control like demographics, profile of admitted students, most recent yield, etc.

We'll most likely see more movement on the WL this week and next as 2nd deposits were due yesterday. If you haven't updated the adcom on your progress in a while and have something worth mentioning, I'd try to get it in as soon as possible.

I hope this helps someone out there. It's really just my perspective, so take it with a grain of salt. Congratulations to everyone admitted and best of luck to everyone else!
Director
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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21 May 2009, 13:12
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*chirp* *chirp* *chirp*
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"George is getting upset!"

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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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22 Oct 2009, 15:12
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hjf6 wrote:
My application is still just "submitted"! Getting anxious now! However, I waited until 5 pm on October 6th to submit my application - I wonder if that has something to do with it?

I wouldn't worry - my application still reads "submitted" and I applied on Sept 28

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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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15 Dec 2009, 20:20
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hjf6 wrote:
Admitted at around 7 tonight! also a Park finalist!

CONGRATS!!!

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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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16 Dec 2009, 12:32
1
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Batman83 wrote:
Architect wrote:
Just got the call! I am in! What a relief/surprise.

Congrats! I was so excited that I missed how we're supposed to get our login info, did you happen to catch that?

are all admitted applicants are updated by email / change in online status or a phone call
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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17 Dec 2009, 07:29
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*crickets*- would be nice to just know one way or another...
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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29 Dec 2009, 16:57
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We had Intel, Chevron, Infosys, Qualcomm, Amgen, HP, and Amazon come to campus this year from out West for various functions. Maybe some others that I cant remember.

serenityprayer wrote:
Hi does anyone know how well Johnson places on the West coast, particularly the bay area into tech? Product management, marketing, corporate strategy, etc.

thanks!

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Current Student
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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03 Jan 2010, 17:47
1
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serenityprayer wrote:
One big draw about cornell is the rest of the university is fair game for taking classes, does a rep here know what are some popular cross enrolled classes / major / specializations and for what future career? ie consulting focus for energy, may take some energy courses or public policy etc?

There is a whole Excel file of the courses on the Johnson website here: http://www.johnson.cornell.edu/currents ... urses.html

See the third link on the bottom of the page.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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08 Jan 2010, 07:41
1
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Just came back from Ithaca for my R2 interview

Questions I can remember:

1) why did you go to your UG?
2) Walk through resume
3) Tell me about a difficult decision
6) Why MBA now? Why Johnson
7) How do you get other people to buy into your view points
8) Questions about a hobby I put on my resume
9) What is your plan for the rest of the day?

There are a couple questions I wish I could have answered a little differently but it is what it is. Pretty standard questions and a lot of behaviorial questions trying to judge the fit with Johnson. Interview was 45 minutes and I asked several questions for the next 15 minutes or so.

Two things of note:

1. I don't recommend against driving to Ithaca in January. But please don't get there like I did AT NIGHT. I am a decent winter driver with snow tires on my 4wd car and I was intimidated driving on small, dark, wooded, hilly sideroads in Ithaca on a snowy night.

2. It seems like if you sign up for on-campus interview they automatically sign your up for a lunch info session with a current student. I did not get email about it. Fortunately I showed up for my interview appointment way early and ended up just having an informal chat with a 1st year who was already waiting for me in the altrium. I was "late" for the info session so hopefully that does not reflect badly on me.

I don't know about why people keep mentioning unhappy/unenthusiastic students, I have visited Johnson twice and EVERY student I met was as about enthusiastic and happy about their experience as they can be. The best advice the current student gives me about the interview was simply be yourself and be sincere.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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01 Feb 2010, 07:21
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I hope the interview went well... Really nervous. The interviewer said it was wonderful talking to me, I hope it is a good sign haha
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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09 Feb 2010, 13:56
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egy wrote:

Egy, you do not need to worry at all. My application had gone under review a full six weeks after I submitted. This is very normal for Cornell.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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24 Feb 2010, 16:15
1
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xpiano1978 wrote:
parore26 wrote:
When did you get the fed-ex notice? I haven't received anything yet.

They probably haven't sent out the packets (your packet) yet? I don't know what the plan for R2 is but somewhere they said there were a few R2 apps they hadn't gotten to yet.

I got my R1 Fedex notice the day after the decision notification via email.

hey xpiano...are you planning to attend the DJ weekend? have you started looking into housing? wat about loan options?
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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01 Mar 2010, 09:54
1
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Hey everyone. I know that the application process is tiring, frustrating, annoying ... use whatever words you want ... but I think the smartest thing to do at this point is to take a few deep breaths, relax, and do something to keep your mind occupied. Go to the gym, take your girlfriend/wife out to dinner, play Rock Band ... whatever !

The decision is out of our hands, and there's nothing we can do. I see a lot of folks adding in unecessary stress to their days, and its just not healthy. So step away from the computer and go do something you love !

Good luck !
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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03 Mar 2010, 09:29
1
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chmny3 wrote:
for those who have visited cornell from nyc.. any recommendations on the best way to get there? i haven't driven a car in about 8 years, so that's out... amtrak to syracuse, then bus?

i have my interview on monday..!

There is supposed to be a good bus that runs up there... I had my visit/interview this past Monday and overheard someone talking about a nice bus with wifi.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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03 Mar 2010, 10:51
1
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chmny3 wrote:
for those who have visited cornell from nyc.. any recommendations on the best way to get there? i haven't driven a car in about 8 years, so that's out... amtrak to syracuse, then bus?

i have my interview on monday..!

cornell runs a bus called "cornell campus to campus" i think it's got 3 stops in nyc. costs \$75 one way and the ride is sweet! check the exact times n pickup stops with the school.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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09 Mar 2010, 15:11
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I got an invitation to interview... at last !!!!!

It will be with alumni in my country......I can not travel to US-Ithaca.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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22 Mar 2010, 11:24
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thanks guys! I interviewed last week.
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Re: Calling all Cornell (Johnson) Fall 2010 applicants! [#permalink]

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22 Mar 2010, 11:25
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I'm in!!
Re: Calling all Cornell (Johnson) Fall 2010 applicants!   [#permalink] 22 Mar 2010, 11:25

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text:
## Basic Analysis: Uniform Convergence

Once again, let $D\subseteq \mathbf{R}$ be a subset. Suppose we now have a sequence of functions

$f_n : D\to\mathbf{R}$, where n = 1, 2, 3, … ,

such that for each x in D, the sequence $(f_n(x))$ converges to some real value. We’ll denote this value by f(x), thus defining a function fD → R. We then say that the sequence of functions fn(x) converges to f(xpointwise.

The big question is: if each fn(x) is continuous at x=a, is it true that f(x) is also continuous at x=a?

It may surprise you that the answer is no. [ Historical note: Cauchy actually believed the answer is yes and provided an erroneous proof. This shouldn’t be held against Cauchy since analysis was still in its infancy in the 19th century. On the contrary, one should use this as an indication of how counter-intuitive results in analysis can be. ]

Example

Consider the sequence of functions

$f_n : [0,1]\to\mathbf{R}, \quad f_n(x) = x^n.$

If 0≤x<1, then clearly fn(x) → 0 as n approaches infinity. On the other hand, for x=1, the sequence is a constant sequence (1, 1, 1, …). Thus fn(x) converges to f(x) pointwise, where f takes any element in [0,1) to 0, and 1 to 1.

(Graphs provided by wolframalpha)

## Uniform Convergence

In order to ensure that f is continuous, we need a further condition.

Definition. The sequence of functions fn(x) is said to converge uniformly to f(x) if:

• for each ε>0, there exists N such that whenever n>N and x is in D, we have |fn(x)-f(x)|<ε.

Pictorially, we have:

[ For large n, the function fn(x) lies between f(x)-ε and f(x)+ε. ]

Let’s consider the difference between pointwise convergence and uniform convergence. In the former case, we have:

• For each x in D and ε>0, there exists N such that for all n>N, we have |fn(x)-f(x)|<ε.

On the other hand, the latter says:

• For each ε>0, there exists N such that for all x in D and n>N, we have |fn(x)-f(x)|<ε.

Using logical quantifiers, the two statements become:

$\forall x,\forall\epsilon, \exists N, P(N,x,\epsilon)$ and $\forall \epsilon, \exists N, \forall x, P(N,x,\epsilon)$,

where P(Nx, ε) says that whenever n>N, we have |fn(x)-f(x)|<ε. Written in this manner, it’s clear that uniform convergence implies pointwise convergence.

Main Therorem. If fn(x) converges to f(x) uniformly, and each fn(x) is continuous at x=a, then so is f(x).

Proof. Let ε>0.

• Since fn(x) → f(x) uniformly, there exists N such that for all x in D and n>N, we have |fn(x)-f(x)|<ε/3. Fix any n>N.
• Since fn(x) is continuous at x=a, there exists δ>0 such that for all x in D satisfying |xa|<δ, we have |fn(x)-fn(a)|<ε/3, which gives:

$|f(x)-f(a)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|<\frac\epsilon 3+\frac\epsilon 3+\frac\epsilon 3=\epsilon.$ ♦

## Alternate Definition

Before going through some examples, we consider an alternate definition of uniform convergence which is somewhat conceptually clearer.

Consider the set V of all functions D → R. This forms a real vector space via pointwise addition and scalar multiplication: i.e.

$(f+g)(x) := f(x) + g(x), \quad (c\cdot f)(x) = c\cdot f(x), \ c\in\mathbf{R}.$

Now we define the norm of f to be $||f|| := \sup\{|f(x)| : x\in D\}$ which is either a non-negative real number or +∞.  It’s not hard to show that $||f+g|| \le ||f||+||g||$, $||fg|| \le ||f||\cdot||g||$ and $||c\cdot f|| = |c|\cdot ||f||$.

[ Indeed, to prove the first inequality, just note that for any x in D, we have |(f+g)(x)| = |f(x)+g(x)| ≤ |f(x)|+|g(x)| ≤ RHS. The same proof holds for the second inequality. ]

Alternate Definition. The sequence $f_n(x) \to f(x)$ uniformly if:

• for every ε>0, there exists N such that when n>N, we have $||f_n - f||<\epsilon.$

Note that this definition is quite similar to that of convergent sequences, which makes it convenient to work with. [ Side note: the acute reader may suspect that one can study analysis on the above function space V and he’d be right! Indeed, this is the primary motivation behind functional analysis. ]

Examples

1. Let’s go back to the example $f_n(x) = x^n$ on D=[0,1]. Since the functions converge to a discontinuous function, the above theorem tells us the convergence is not uniform, but let’s verify this directly. Indeed, the function $f_n-f$ takes xn on [0,1) and 0 at x=1. Thus, $||f_n-f||=1$ for all n, which shows that the convergence is not uniform.
2. On the other hand, if we take $f_n(x) = x^n$ on D=[0, 1/2], then they converge to f(x) = 0 uniformly. Indeed, we have $||f_n - f|| = ||f_n|| = 1/2^n$ for each n, so $\lim_{n\to\infty} ||f_n - f|| = 0$. As expected, since each fn(x) is continuous, so is the resulting converged function f(x).
3. If we now take $f_n(x) = x^n$ on D=[0, 1), then they converge to f(x) = 0, albeit not uniformly. Indeed, it’s easy to see that $||f_n - f|| = 1$ for each nIn other words, one can have pointwise convergence of a sequence of continuous functions to a continuous function, even if the convergence is not uniform.
4. Consider the Riemann zeta function $f(x)=\zeta(x) = 1^{-x} + 2^{-x} + 3^{-x} + \ldots$ on (1, ∞). We know this converges at every x>1, but is the function continuous? Indeed, it is: via the following useful trick. First, write $f_n(x) = 1^{-x} + 2^{-x} + \ldots + n^{-x}$, so that $\lim_{n\to\infty} f_n(x) \to \zeta(x)$ pointwise. Suppose we wish to show that ζ(x) is continuous at x=a>1.
• Pick any b, 1<b<a.
• Let gn (resp. g) be the restriction of fn (resp. f) to [b, ∞).
• Now $||g_n - g||=(n+1)^{-x} + (n+2)^{-x} + \ldots \le (n+1)^{-b} + (n+2)^{-b} + \ldots$ which is a convergent series since b>1.
• Hence $||g_n-g|| \to 0$, which shows that gn converges uniformly to g.
• Thus means g is continuous, in particular f is continuous at a. Hence, f is continuous at any a>1.

## Basic Properties of Uniform Convergence

Let’s figure out some basic properties, specifically suppose $f_n : D\to \mathbf{R}$ and $g_n : D\to\mathbf{R}$ are sequences of functions which converge uniformly to $f, g:D\to\mathbf{R}$ respectively. What about the sum $f_n+g_n$, product $f_n g_n$ and reciprocal $1/g_n$?

1. The case of sum is easy.

Indeed, $f_n + g_n \to f+g$ uniformly. To see why, write:

$||(f_n+g_n)-(f+g)|| = ||(f_n-f)+(g_n-g)|| \le ||f_n-f|| + ||g_n-g||.$

Now, for any $\epsilon>0$, $||f_n - f|| < \frac\epsilon 2$ and $||g_n-g|| < \frac\epsilon 2$ for large n. Thus, we also have $||(f_n+g_n)-(f+g)|| < \epsilon$ for large n.

2. The case of product is true with some additional conditions.

Indeed, let’s write:

$||f_n g_n - fg || = ||(f_n-f)g_n + f(g_n - g)|| \le ||f_n - f||\cdot ||g_n|| + ||f||\cdot ||g_n - g||.$

Now, if ||f|| and ||g|| are both finite, then in particular, we have $||g_n|| \le ||g_n - g|| + ||g|| < ||g||+1$ for large n. It then follows that we can write $||f||, ||g|| < B$ for some positive bound B and $||f_n g_n - fg|| \le B(||f_n - f|| + ||g_n - g||)$. Hence, it also follows that $f_n g_n \to fg$ converges uniformly.

What if either ||f|| or ||g|| is infinite? A counter-example is easy:

$f_n, g_n:\mathbf{R} \to\mathbf{R}, \ f_n(x) = x, g_n(x) = \frac 1 n.$

Then $f_n(x) \to x$ and $g_n(x) \to 0$ uniformly. However, $||f_n g_n - 0|| = \sup\{\frac {|x|} n: x\in\mathbf{R}\} = \infty$.

3. The case of reciprocal is true with some additional conditions.

Let’s consider when $1/g_n \to 1/g$ uniformly, assuming that g(x) and gn(x) are never 0. Taking the difference gives:

$||\frac{1}{g_n} - \frac{1}{g}|| = ||\frac{g_n - g}{g_n g}|| \le ||g_n -g|| \cdot ||\frac{1}{g_n}||\cdot ||\frac 1 g||.$

Hence, suppose there is a positive B such that $||1/g_n|| < B$ for all n. This means $|g_n(x)| > 1/B$ for all n, and $x\in D$. For some N, we have $||g_n - g|| < 1/(2B)$, i.e. $|g_n(x) - g(x)| < 1/(2B)$ for all n>N and $x\in D$. Thus:

$|g(x)| \ge |g_n(x)| - |g_n(x)-g(x)| > 1/B - 1/(2B) = 1/(2B),$ when $x\in D$, i.e. $||1/g||\le 2B$.

Thus, $||\frac 1 {g_n}-\frac 1 g|| \le ||g_n - g||\cdot B\cdot(2B)$ for all n and it follows that $1/g_n \to 1/g$ converges uniformly.

What if such a B doesn’t exist, i.e. the infimum of all $||1/g_n||$ is zero? Consider the example of $g_n: (0, 1]\to\mathbf{R}, g_n(x)=(1+\frac 1 n)x$ $\$. Then $g_n(x)$ converges to g(x)=x uniformly since $||g_n-g|| = ||\frac x n||=\frac 1 n$. On the other hand,

$||\frac 1 {g_n}-\frac 1 g|| = ||\frac n {(n+1)x}-\frac 1 x|| = ||\frac 1 {(n+1)x}||=+\infty,$

so $1/g_n \to 1/g$ pointwise, but not uniformly.

Summary. Suppose $f_n \to f$ and $g_n \to g$ uniformly, where each $f_n, g_n, f, g:D\to\mathbf{R}$. Then:

• $f_n+g_n\to f+g$ uniformly as well;
• if $||f||, ||g||<\infty$, then $f_n g_n \to fg$ uniformly;
• if there is a positive B such that $||g_n||>B$ for all n, then $1/g_n \to 1/g$ uniformly.
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### 1 Response to Basic Analysis: Uniform Convergence

1. l3spam says:

Simple and clean explanation. THANKS!


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# Backbonding in phosphorous pentoxide

I read this today in a book that $$\ce{P=O}$$ in $$\ce{P4O10}$$ consists of a coordinate bond and pπ-dπ backbonding, but why does this happen?

Can't phosphorus share its lone pair with one of the lone pairs available in oxygen to form the double bond?

• Get a better book. Aug 11 '20 at 17:34
• Phosphorus can share its lone pair with oxygen to form a bond $\ce{P-O}$. The same thing happens in a lot of compounds of phosphorus at oxidation number $+5$. This bond may be considered as double or as simple, according to the book you are reading. Aug 11 '20 at 19:30
• There are two different phosphorus oxygen bonds in this molecule. Please clarify which one you are talking about. Please also cite the source for that claim. Aug 12 '20 at 8:56

Coordinate bonds are covalent bonds where both bond electrons stem from only one of the two bond partners. They are formed when a Lewis base donates two electrons into accepting orbitals of the Lewis acid. That's not what you're looking at in this case, where both $$\ce{P}$$ and $$\ce{O}$$ contribute one electron to the common $$\sigma$$ bond. The interesting part is about everything that is to be described beyond the bond order of $$1$$. For example, everything that contributes to PO bonding beyond that single $$\sigma$$ bond.

I will further regard to phosphane oxides $$\ce{P(=O)R3}$$, where the question was originally asking about $$\ce{P=O}$$ bonds in $$\ce{P4O10}$$. The problem might be reduced to PO bonding in phosphoric acid, as $$\ce{P4O10}$$ is nothing more than its anhydride. The general observations regarding the $$\ce{PO}$$ bonding situation are unaffected. Kutzelnigg noticed in $$1977$$ already:

[The PO] bond is somewhere between single and triple and [...] only by chance it may happen to be a double bond.

What we know about the '$$\ce{P=O}$$ bond':

• Its length is remarkably shorter than the P-O bond ($$\pu{1.43 Å}$$ compared to $$\pu{1.60 Å}$$ for $$\ce{P4O10}$$ according to Wikipedia).

• It is highly polar.

• Its bond strength of about $$\pu{540 kJ/mol}$$ is higher than for a corresponding single bond.

A critical review regarding the bonding situation in phosphane oxides was given by Gilheany. If it comes to Lewis valence formula, one might find three reasonable structures:

None of those is suitable to represent all theoretical and experimental findings regarding the characteristics of the $$\ce{PO}$$ bond!

Some scientists prefer to use the zwitterionic representation A, where the predominant $$\sigma$$ character of the $$\ce{PO}$$ bond is highlighted, while its dipolar character is also taken into account. If one chooses to write a Lewis structure as shown in B, one implies (besides one $$\sigma$$ bond) the contribution of a $$\pi$$ bond between oxygen p orbitals and phosphorus d orbitals. However, the $$3$$d orbitals of phosphorus are no suitable $$\pi$$ accepting orbitals for $$2$$p orbitals of oxygen, as they are quite high in energy. Besides compatible symmetry, atomic orbitals (AOs) of the atoms in question require to be somewhat similar in energy to allow for a combination to form molecular orbitals (MOs).

If one allows $$\pi$$ back bonding from oxygen into $$\sigma ^*$$($$\ce{P-R}$$) (anti-bonding orbitals between P and its substituents R), one might come up with Lewis valence formula C. However, it should be noted that the $$\pi$$ bond order for $$\ce{Me3PO}$$ is only $$0.7$$ - and not $$2$$, what is implied by structure C!


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# Associations to the word «Exponent»

## Wiktionary

EXPONENT, noun. One who expounds, represents or advocates.
EXPONENT, noun. (mathematics) The power to which a number, symbol or expression is to be raised. For example, the $$3$$ in $$b^3=p$$.
EXPONENT, noun. (mathematics) The result of a logarithm, between a base and an antilogarithm. For example the $$4$$ in $$\log_b(a)=4$$.
EXPONENT, noun. (linguistics) A manifestation of a morphosyntactic property.
EXPONENT, noun. (mathematics) (rare) The degree to which the root of a radicand is found. For example the $$2$$ in $$\sqrt[2]r=b$$.

## Dictionary definition

EXPONENT, noun. A person who pleads for a cause or propounds an idea.
EXPONENT, noun. Someone who expounds and interprets or explains.
EXPONENT, noun. A mathematical notation indicating the number of times a quantity is multiplied by itself.

## Wise words

Better than a thousand hollow words, is one word that brings peace.
Buddha


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# Which one I should choose for random forest?

I fit a dataset with a binary target class by the random forest. In python, I can do it either by randomforestclassifier or randomforestregressor.

I can get the classification directly from randomforestclassifier or I could run randomforestregressor first and get back a set of estimated probabilities. Then I can find a cutoff value to derive the predicted classes out of the set of probabilities. Both methods can achieve the same goal (i.e. predict the classes for the test data).

Also I can observe that

randomforestclassifier.predict_proba(X_test)[:,1])

is different from

randomforestregressor.predict(X_test)

So I just wanna confirm that both methods are valid and then which one is better in random forest application?

• Binary target class already says it all: classification. Therefore: Random Forest Classifier – Nikolas Rieble Jan 5 '17 at 8:44

If you are doing a classification task, use random forest classifier. If you want probabilities use predict_proba or you can use predict directly to get classes. You might be getting correct results but understand that the random forest regressor works on a different cost function, and is not constrained to give outputs between 0 and 1, so what you are getting out of the regressor are not really probabilities. If you run it on enough datasets it might start giving you outputs greater than 1. Hence, bottom line stick to classifier. Also I will recommend reading up a bit on the differences of regression and classification tasks.

• Thanks for that. In the scope of my dataset, I didn't see the output larger than 1. So you mean the underlying regression was not the logistic regression but rather a linear regression? If so, how can it take the y_train of 0,1 to fit the linear regression? – LUSAQX Jan 5 '17 at 8:01
• You can fit a linear regression on an output of 1,0. Try it. Also, I am not saying its linear regression, random forests are used for fitting non-linear hypotheses. The only thing is that the regressor's cost function works for continuous output and not discrete values. Hence, it is ideal to use the classifier because it is actually churning out probabilities. – Himanshu Rai Jan 5 '17 at 8:06
• Thanks. Yes I also think the regressor was non-linear one. But you mentioned a key concept here, i.e. cost function. I will do some research. On the other hand, even though the output of the regressor was not probability (0-1 metric), I can also find a cutoff value to get the predictive classes and the output can be treated as a general test score rather than probability. So this is not the key, and I will examine the cost function. Thanks for help. – LUSAQX Jan 5 '17 at 8:14
• But how about the methodology to use logistic regression for the classification, i.e. do the classification out of an inner regression model? In that way, it is also to get the discrete class out of a continuous value. – LUSAQX Jan 5 '17 at 8:20
• Cheers. Yes Random Forest Classifier uses cross-entropy function. I dont really know, what regressor uses . Look into it. – Himanshu Rai Jan 5 '17 at 8:20

On the one hand, as Himanshu Rai has mentioned, if you are using RandomForestRegressor for classification tasks, the "probability" you get might greater than 1, or less than 0. That's not what we expected.

On the other hand, RandomForestClassifier uses accuracy_score as loss function, while RandomForestRegressor uses r2_score. That's a big difference.


text:
Question

# For given vectors, and , find the unit vector in the direction of the vector

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Solution

## The given vectors are a → =2 i ^ − j ^ +2 k ^ and b → =− i ^ + j ^ − k ^ . Sum of vector a → and b → is, ( a → + b → )=( 2−1 ) i ^ +( −1+1 ) j ^ +( 2−1 ) k ^ =1 i ^ +0 j ^ +1 k ^ = i ^ + k ^ Consider, c → = i ^ + k ^ . Magnitude of vector c → is, | c → |= 1 2 + 0 2 + 1 2 = 2 The unit vector in direction of c → is, c ^ = c → | c → | = i ^ + k ^ 2 = 1 2 i ^ + 1 2 k ^ Thus, the unit vector in direction of sum of vector a → and b → is 1 2 i ^ + 1 2 j ^ .

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