Re: BR Maths Corner-1
Posted: 18 Aug 2008 22:34
AmberG is la pata after that post?
Host cant quit without notice. See first post of this thread.
Host cant quit without notice. See first post of this thread.
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Not noticing the bolded part could lead to confusionWiki wrote: Any two closed curves in space can be moved into exactly one of the following standard positions. This determines the linking number.
Each curve may pass through itself during this motion, but the two curves must remain separated throughout.
Seems too low-funda for this thread.SK Mody wrote:A number of men and women stand in a line holding hands. There are an unspecified number of men and women, the only condition being that the there is a woman at the beginning of the line and a man at the end. If two members of the opposite sex hold hands, they are obliged to have a baby. Show that there will be an odd number of babies.
Hmm.. Is that because it is a derivative of a potential function and not path dependent and hence the line integral around any random closed loop is the same as that of a circle kind of thing like you would use to prove Green's theorem ? After all, if the complex function is differentiable, it is infinitely differentiable arent' they?vsunder wrote:Take a circle or any loop(it needs a lemma that it does not matter what curve you take) around z=0.
Saar. Question here. If I remember my complex analysis from close to a decade and a half ago, one of the transformation under conformal mapping of z^2 is a straight line isn't it? (not sure..maybe it was a different function , but a circle transforms to a straight line is one I remember from the depths hidden away somewhere under layers of grime and disuse), so, if in such a transform you go around either clock wise or anti clockwise in a circle in the z plane, in the w plane (the mapped plane), since you come back to the starting point (I guess in the transformed plane, you would go up and come back down to the starting point right?) , so if as you state below that the winding numbers are additiveLook at the image of this loop in the w-plane. Now walk say counterclockwise around the loop in the z-plane.
How many times have you walked around the loop in the w-plane. Answer=2 times COUNTERCLOCKWISE. So the winding
number=H(f,0)=2. Now change the situation a little bit. Take w=g(z)=(bar z)^2, where bar z is the complex conjugate.
Do the same construction. Now as you travel counterclockwise in the z-plane you will now travel CLOCKWISE twice
in the w plane so the winding number H(g,0)=-2.
So wouldn't the winding number in that case be zero ?Now take w=1. Take w=z^2 again. Then the pre-images are z=1 and z=-1. Now take little circles around each point z=1 and z=-1
and see what happens as you walk around once counterclockwise in the z plane. You will find that it will be once counterclockwise in the w plane for both circles around z=1 and z=-1. The winding number is then the SUM
of the winding numbers you get from z=1 and z=-1 and hence H(f,w=1)=2. So one adds the winding numbers
from each local contribution for each and every pre-image.
Ah. Thanks, now I understand where you are coming from. I got thrown when you started using "maps" and brought in complex analysis notation and I immediately jumped to conformal mapping and Jacobian and all that transformation rubbish.vsunder wrote: I just brought in complex analysis simply because it is easy to write a formula for f using complex notation or else I would have to write for w=f=z^2.
f:(x,y)------> ( x^2-y^2, 2xy) --------(1)
What a horrible way to write something!! If you wish you can use the notation (1). I find it easier to use
the formula w=z^2.
In the case of w=z^2. Take a circle around z=0. The circle can be parametrized as z=exp(it) with 0<t<2 pi
or as (cos t, sin t) if you prefer. Now what is w(t)=z^2 it is w=exp(2it). This is the image curve. If you wish you can write this as w(t)=( cos 2t, sin 2t). Now note you will go around twice in the image curve. Maybe I am misreading what you write. If you use the formula (1) you will get for x=cos t, y=sin t
(cos^2 t-sin^2 t, 2 sin t cos t). Since all of you think passing JEE is some sort of mathematical greatness
then you will tell me that every JEE topper knows immediately that cos 2t= cos^2 t-sin ^2 t, 2 sin t cos t=sin 2t.
and so w(t)=(cos 2t, sin 2t). But I prefer using complex analysis to do the trivial calculation.
Neat. In that case try this one.Nandu wrote:Seems too low-funda for this thread.SK Mody wrote:A number of men and women stand in a line holding hands. There are an unspecified number of men and women, the only condition being that the there is a woman at the beginning of the line and a man at the end. If two members of the opposite sex hold hands, they are obliged to have a baby. Show that there will be an odd number of babies.
Start tracking gender starting from one end of the thread.
Gender changes each time there is a baby and an odd number of gender changes are needed for the other end to be of opposite gender.
Neat. In that case try this one.
You are given a triangle ABC. Call this triangle the main triangle M. This M is further subdivided into a finite number of sub-triangles in such a way that every side of every interior sub-triangle is common to exactly two triangles (a side of a triangle is defined as the line between two adjacent vertexes). The sides that lie on the boundary of ABC are of course not shared. Now each vertex of each sub-triangle is given any one of the labels A, B or C at whim with the condition that vertices on the side AB of M are only given A or B labels, vertices on the side AC of M are given only A or C labels and vertices on the side BC of M are given only B or C labels.
A sub-triangle is called complete if it has labels ABC.
Show that there are an odd number of complete sub-triangles.
(Note: Sub-triangle means "elementary" sub-triangles, that is there are no other triangles inside the sub-triangle).
(Note2: If you have seen this before - apologies)
Najunamar wrote:Vinaji,
I am sure you meant exp(it)=cos t + i sin t and not sin t + i cost t as every JEE person can vouchsafe! (and quite a few non JEE folks too....).
This need not be the case. There can be any number of interior triangles.sugriva wrote:
The only way a vertex of a sub-triangle will have a label at all is if it lies on one of the original edges AB, BC, CA.
Sorry, I have been out for a while. Let me just touch a few points which I think are necessary.AmberG is la pata after that post?
Host cant quit without notice. See first post of this thread.
Ramana – I have re-read my post, and honestly do not find that I was insulting anyone. The problem made no sense to me as it was not clear(to me at least), and I in fact asked specific questions about the ambiguity (you get c=3 for n=1 and c=2 for n=2), and asked it to be explained. There was no personal attack, only asking if problem can be made unambiguous.AmberG, Vsunder is respected mathematician. his work was appreciated by the greats like S Chandrasekhar himself. I understand your strong point is physics. If one didn't understand the problem one can ask for clarification. No need to consider it humorous etc. I am sure you didnt mean to insult. So please edit your posts. Thanks, ramana
Vsunder, please don't go off. We do need you. Ramana
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(this is after:
Another link which tells Here is one which eventually appeared in the American Journal of Mathematics one of the leading journals of the world in 2004. points toThe lemma 2 that makes no sense appears here. If you feel it does not make sense write to the editorial board.
Vsunder does not know me, how does he know that I have no mathematical research to my credit? Isn’t that rather rude and arrogant? (and inaccurate, to put it mildly).Amber you need to be polite to people, you are very arrogant and have no mathematical research to your credit.
Shivji- I have warning in my in box, and can not reply as PM hence the message is here.shiv wrote:Sagun, I admit that the post by AmberG (quoted below) was unnecessarily aggressive and egotistic. Since I know you personally I would like to apologize for this idiocy. Postor AmberG has not spared even a few microseconds to think that anyone else might know any math, which is sad for a person who wanted to test the math ability of forum members on the nuclear discussion thread a few weeks ago.vsunder wrote: Forum admins: I must say the forum has degenerated into a cesspool of name calling arrogance by petty foggers
and sundry others. I have decided to stop coming here. It was nice in the old days and I am happy I learnt many things from people.
Sagun
Amber G. wrote: It's not April 1, but I will bite. If this is not a joke, could one please write it in a way so it makes sense mathematically by being precise in definition.
For example: What exactly the sentence mean "Take a square it works for a cube in any dimension." ?? If rest of the part is particular example in D2, then state so, if not, then were exactly, say 4 in 4^n comes from?
Also what is "N" is is different than "n"? for example, when n=1, we have only one choice (3 squares) making C=3 (Assuming N and n are same) while for n=2, one can easily get an example where you need only 2 squares to cover.
In short it makes no sense.
If this was a joke, can the spam be removed.
(I am not even commenting on the rest of the post.. as it makes even less sense to me)
Thanks in advance.
Asking if n is same as N, or asking clarify the problem, or asking if the problem is a serious one ,
Postor AmberG has not spared even a few microseconds to think that anyone else might know any math, which is sad for a person who wanted to test the math ability of forum members on the nuclear discussion thread a few weeks ago.
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(How do you or the postor know if I have “no mathematical research to my credit?”Amber you need to be polite to people, you are very arrogant and have no mathematical research to your credit.
One statement was probably okay, but piling on five at a time is just enough to press vsunders buttons. So yes amberg, maybe you need to rephrase your statements better, think twice and see the posiblity that the other person is serious and not joking/being silly/spamming the board.It's not April 1, but I will bite.
If this is not a joke
In short it makes no sense.
If this was a joke, can the spam be removed.
(I am not even commenting on the rest of the post.. as it makes even less sense to me)
OKAmber you need to be polite to people
Sort of OK again in light of over criticism by Amber, you are very arrogant
here he loses it - how does vsunder know who what AmberG does for a living?have no mathematical research to your credit.
The above seems to be a favorite refrain of anyone who cant take heat.I have decided to stop coming here
Nandu - Big hint is SK mody's first problem, Take line A B Call A man, B woman then the number of 'AB' type pairs (change in parity) on line AB has to be odd.AFAICS, the two unsolved problems are the Ankan one from Amber G. and the triangle division one from SK Mody, yes?
It would be interesting to know if this has any implications for cryptography.Amber G. wrote:Rumor is that a new "largest known prime" (mersenne prime number) number is found..
One possible Solution:given any number n , there is a multiple of n which can be written as 11111...000...
the above obviously as ankan =2 (you add 0's if ankan =1 and you want to make it 2)
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You get [b]n [/b] pinjra (or pigen-holes) number 0,1,2,3, ... (n-1)
Now get kabootars...
1
11
111
1111
etc...
in each case, you divide (say 11111) by n, what ever the remainder you get (it has to be between 0 and n-1) send that kabootar to the pinjra which has the remainder number.
(eg suppose n=4, send first kabborar (1) to 1 , second kabootar (11) to (11/4 = 2*4+3) plus remainder 3, third kabootar (111) to 3 (If you divide 111 by 4 you get remainder 3) etc...)
Now you must get one hole where there are two kabootars..(In our example 111 and 11) which have the same remainders .. subtract smaller from the other... the difference is obviously divisible by n and, of course it will look like (1111.....) -(111...) = 111....000
which will have ankan=2
The Mersenne prime search project has been going on for years, and other than eating up CPU cycles that could otherwise be used for code breaking, I don't think there are any implications for cryptography.SK Mody wrote:It would be interesting to know if this has any implications for cryptography.Amber G. wrote:Rumor is that a new "largest known prime" (mersenne prime number) number is found..
Yeah, I did it the same way as Vriksh (i.e.) basic 8th grade arithmetic principles. Actually the longest part of the one minute was not thinking about it, but getting the syntax straight (I'd missed one of the parenthesis in my first tryKaran Dixit wrote: ArmenT,
I do not know about Ruby or Python. However, it is easy to do it in C# using .Net Framework function. But he is not allowed to use the .Net Framework functions. Were you able to do it in one minute without using .Net Framework functions?
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#!/usr/bin/env python
array = [1, 4, 2, 3, 1]
# Here's my one line solution
answer = sum(array) - ((len(array) - 1) * (len(array)))/2
print answer
Well, one nice thing about Mersenne primes is its relationship to perfect numbers. Basically there is a proof by Euclid that states that if 2^p -1 is prime (this is the definition of a mersenne prime), then (2^p-1) * (2^(p-1)) is a perfect number. For those who don't know what a perfect number is, any number that is equal to the sum of its factors is a perfect number. 6 is one example (6's factors are 1, 2 and 3 and 1+2+3 is 6) and 28 is another (28's factors are 1, 2, 4, 7, 14 and 1 + 2 + 4 + 7 + 14 = 28)Nandu wrote:In fact, other than being able to claim "largest known prime" and laying another possible claim to "largest co-operative computing effort on the internet", I am not sure the project has any theoretical significance. The Lucas-Lehmer test used to check for primality is from 1930.SK Mody wrote: It would be interesting to know if this has any implications for cryptography.
An anonymous reader writes "The Great Internet Mersenne Prime Search (GIMPS) has apparently discovered a new world-record prime number. A GIMPS client computer reported the number on August 23rd, and verification is currently under way. The verification could take up to two weeks to complete. The last Mersenne prime discovered was over 9.8 million digits long, strongly suggesting that the new value may break the 10 million digit barrier — qualifying for the EFF's $100000 prize!"
Yeah, but I am not going to post or even claim it because all I did is type in factor _number_ on my linux box.Amber G. wrote:
And speaking of numbers like (2^n +-1)has any one found factors of 2^58+1 (other than 5) ...