10 things to do if your friend forgets to sign off his Gmail account

November 21, 2009 in Humour:) | 5 comments

:P has left the lab computer without signing off his Gmail account. You are excited to avenge all his misdoings including that unforgivable lab incident; but can’t really decide the extent of turmoil. Try doing some or all of the following things:-

  • Delete all mails and clear trash (God forgive you)
  • Send an “I-love-you” message to all :P ’s frequently mailed female contacts.
  • Message :P ’s family and friends saying that somehow suicide seems the only option
  • Start chatting with all online contacts. Past chats may help.
  • Mail all contacts that windows7sux@gmail.com is my ( :P ’s) new email id.
  • Send random waves to all (requires Google Wave invite)
  • Send an offliner to oft-chatters that they are on the verge of getting blocked
  • Open Picasa gallery and comment on others’ pics, how dumb they look in front of a camera
  • Subscribe the email address to porn sites (nasty one)
  • Open Google Web history, print-screen and save the most wicked searches :P ever did and blackmail him if he tries to thrash you for doing the above things!

Help needed to expand this list!

Problem statement for the IITB Maths Olympiad 2010

November 16, 2009 in TeXnical | Leave a comment

What is a good Olympiad problem?

It must be easy to state and understand but difficult to solve. The solution should not require any prerequisites but cleverness. Indeed, once the solution is known, the solver may enjoy its simplicity and/or condemn himself for not being able to solve it! A high school student should not be at a disadvantage as compared to a professional mathematician.

Example:
What is the maximum number of terms in a sequence with the property that every 7-sum (sum of 7 consecutive terms) is negative but every 11-sum is positive? }

It is an earnest request to at least understand (the difficulty of) the problem before looking up at the (Tinkle see-n-smile) solution at the end of the post.

The problem is a gem for another reason. It employs the oft-used trick in Maths to count the same objects in two different ways and come up with interesting results. I have seen such a counting method in a proof of the Quadratic Reciprocity (the recipe of which goes, Gauss’ Lemma \to Eisenstein’s Lemma \to lattice-point counting), another in computing the famous 2-sum \sum_{n=1}^{\infty} \frac{1}{n^2} by evaluating the same integral, (cf. Fubini) and in obtaining different basic combinatorial formulae.

More on this later.

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Solution to the above problem:

a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 < 0
a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 < 0
\vdots
a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} + a_{17} < 0

Answer: 16

For those who did not understand even after staring at this picture for 300 seconds, click here for an explanation.
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Yeh Dilli hai mere yaar…

November 8, 2009 in Travel&Living | 4 comments

Yeh Dilli Hai Mere Yaar, Bas Ishq Mohabbat Pyaar
Basti Hai Mastano Ki Dilli Dilli, Gali Hai Deewano Ki Dilli

Somehow all this ishq, mohabbat and pyaar seemed to evade me as I passed through the gallis of Dilli. All I could see around me were the topi-like helmet cladded men riding their ‘Hamara Bajaj’ chetaks; the moti aunties with mote ear-rings and yet moti bindiyaan as their pillions along with their cantankerous Punjabi kids wailing and howling at the top of their larynges. All my communication to the surrounding people was reciprocated in a heavy North-Indian accent that made me ask them to repeat every word.

I left CSI Airport on Friday morning for the Maths GRE exam at Delhi. At the airport, the low-cost carriers’ low air fares seemed to have made almost no effect on the exponentially increasing food costs. A flight seemed more affordable than supper at the CCD. Nevertheless, I had some coffee there and soon landed in the aircraft. As I took off, I saw the Powai lake and the IIT from the plane. It was a different perspective. There was an unusual feeling that I was going to miss something. An inexpressible and strange but short-lived fear gripped my mind. But I soon recovered. A Course in Arithmetic and some Floyd kept me occupied for the rest of the flight.

I reached Delhi on Friday and had to bear one full day of boredom. I solved a mock paper by evening and called up Abhimanyu. I believed he works at TCS Delhi, but it turns out that his office was at Gurgaon, an hour from New Delhi. Somehow, I found it unreasonable to call him to spend his weekend with me instead of his newly-married bride (of an year or so). By evening, I did some window shopping at Chroma and R-Fresh, some shopping for Pooja, travelled by the famous Delhi Metro and saw the exam center.

DelhiMetro

The day of the exam was very gloomy and depressing. I could, for the first time, see with my naked eyes the “smog” that was seen in the newspapers. Pollution has now taken its toll on the capital. The smog didn’t clear off even by afternoon.

It was only in the dull train journey that I really regretted choosing Delhi instead of Bangalore. The latter is quite a lively place and I could have met some of the VNIT friends there. The food barrage in the Rajdhani was the only thing that kept me away from bouts of boredom during the lonesome 16-hour journey. And lastly, the longer a homeward journey, more blissful is your time as every minute takes you nearer to your sweet home!

(Yet) another proof of the Reciprocity

October 31, 2009 in TeXnical | Leave a comment

In this section, I present a simple proof of Gauss’ Quadratic Reciprocity due to G. Rousseau that uses the Chinese Remainder theorem, simple congruences and counting arguments and evades the Gauss’ Lemma!

Theorem Let {p} and {q} be distinct odd primes. Then
\displaystyle \displaystyle\left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}

Proof:

\displaystyle G:= ( ({\mathbb Z}/p{\mathbb Z})^* \times ({\mathbb Z}/q{\mathbb Z})^* )/ U

where {U= \{ (1,1), (-1,-1) \}} is a normal subgroup of {G}. We count the product {\pi} of cosets of {U} in three ways and equate them. The two products so obtained will be, upto a sign, be equal.
Clearly, {\mathcal S:= \{ (i,j) | i = 1,2,\dots ,(p-1) \text{ and } j=1,2,\dots,(\frac{q-1}{2})\}} is a system of representatives for the cosets of {U}.

[For example, if {j=\frac{q+r}{2}}, then (i,\frac{q+r}{2})U = (i,\frac{q+r}{2}).(-1,-1)U = (-i,q-\frac{q+r}{2})U = (p-i,\frac{q-r}{2})U \in \mathcal S .]

The left product is
\displaystyle\prod_{i=1}^{p-1} \prod_{j=1}^{\frac{q-1}{2}} i = \displaystyle\left(\prod_{i=1}^{p-1}i\right)^{\frac{q-1}{2}} = \displaystyle((p-1)!)^{\frac{q-1}{2}} .
Similarly,
\displaystyle\prod_{i=1}^{p-1} \prod_{j=1}^{\frac{q-1}{2}} j = \displaystyle\left(\prod_{j=1}^{\frac{q-1}{2}}j\right)^{p-1} = \displaystyle\left(\left(\frac{q-1)}{2}\right)!\right)^{p-1} .
Now note that

\displaystyle\left(\left(\frac{q-1)}{2}\right)!\right)^2 = (1.2.\cdots . (\frac{q-1}{2}))^2
= 1.2.\cdots .(\frac{q-1}{2}).(\frac{q-1}{2}).(\frac{q-3}{2}).\cdots .1
\equiv 1.2.\cdots .(\frac{q-1}{2}).(\frac{q-1}{2}-q).(\frac{q-3}{2}-q).\cdots .(1-q) (\mod q)
\equiv 1.2.\cdots .(\frac{q-1}{2}).(-1).(q-\frac{q-1}{2}).(-1).(q-\frac{q-3}{2}).\cdots .(-1)(q-1) (\mod q)
\equiv 1.2.\cdots .(\frac{q-1}{2}).(-1)^{\frac{q-1}{2}} .(\frac{q+1}{2}).(\frac{q+3}{2}).\cdots .(q-1) (\mod q)
\equiv (-1)^{\frac{q-1}{2}} . (q-1)! (\mod q).

Using this,

\boxed{\pi = \displaystyle\left( ((p-1)!)^{\frac{q-1}{2}} , (-1)^{\frac{q-1}{2}\frac{p-1}{2}} . ((q-1)!)^{\frac{p-1}{2}} \right) U}.

We now claim that that

\displaystyle \mathcal T := \{ (k \mod p, k \mod q) | k=1,2,\dots, \textstyle(\frac{pq-1}{2}) \text{ such that } (k,pq)=1 \}

is another set of representatives of cosets of {U}. Here, we invoke the Chinese Remainder Theorem which asserts that

\displaystyle ({\mathbb Z}/p{\mathbb Z})^*\times ({\mathbb Z}/q{\mathbb Z})^* \cong ({\mathbb Z}/pq{\mathbb Z})^* = \{ k \mod pq| \text{ with } (k,pq)=1\}

The product of {k}, modulo {p} is,

\frac{ (\prod_{i=1}^{p-1} i) . (\prod_{i=1}^{p-1} p+i) .\cdots . (\prod_{i=1}^{p-1} (\frac{q-1}{2}-1)p+i)(\prod_{i=1}^{\frac{p-1}{2}} \frac{q-1}{2}p+i)}{q.2q.\cdots.\frac{p-1}{2}q}
\equiv \frac{ ((p-1)!)^{\frac{q-1}{2}} }{ q^{\frac{p-1}{2}} } (\mod p)
\equiv \frac{ ((p-1)!)^{\frac{q-1}{2}} q^{\frac{p-1}{2}} }{ q^{p-1} } (\mod p)
\equiv \frac{ ((p-1)!)^{\frac{q-1}{2}} (\frac{q}{p}) }{1} (\mod p),

where the first step is justified after observing that {\frac{p-1}{2}q} (respectively, {\frac{q-1}{2}p}) is the highest multiple of {q} (resp. {p}) which is less than {\frac{pq-1}{2}} and the last step follows from Euler’s criterion and Fermat’s little theorem. Similarly, we have a similar expression for {k} modulo q and hence, {\pi} now becomes, \boxed{\pi = \Large\left( ((p-1)!)^{\frac{q-1}{2}} \large\left(\frac{q}{p} \right) , ((q-1)!)^{\frac{p-1}{2}} \large\left(\frac{p}{q} \right) \right) U}

Comparing the above two expressions for {\pi} gives,

\displaystyle \large\left(1,(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\right)U = \large\left( \large\left(\frac{q}{p}\right), \large\left(\frac{p}{q}\right) \right) U

and hence, the reciprocity law,
\displaystyle \large\left(\frac{q}{p}\right) = \large\left(\frac{p}{q}\right) (-1)^{\frac{p-1}{2}\frac{q-1}{2}}

\blacksquare

Math Overflow

October 23, 2009 in TolerableMaths | Leave a comment

With many latest mathematician-bloggers turning their heads and posts towards the newly opened Math Overflow website (including my favourite, Terry), I think its time for me to write something about this.

The Math Overflow Q&A website was launched two weeks ago by some UC Berkeley folks and funded by Ravi Vakil. To increase popular interest, they have given reputation points to active members who post questions and answers regularly and accurately. Also, there is no pain of registration as such; an OpenID like Google will do. One glitch is that there is no LaTeX support but users seem to be typing off the TeX code directly. With increase in popularity, we can expect this feature to be added.

Presently, there are many active posts in Algebraic Geometry but hopefully we shall soon see other ‘exciting’ (sic) topics like PDE. Many interesting topics were categorized and posted. Tag lines like, “A gentleman never chooses a basis” give the reader the required kick! I also liked the mathematics jokes post. One cool joke which had me rolling on the floor with laughter was – “A comathematician is a device for turning cotheorems into ffee” in response to the popular quote, “A mathematician is a device for turning coffee into theorems”. I could also find a satisfactory answer to the question What is the “best” proof of the quadratic reciprocity?.

The project is in its nascent stage but has already gripped me. Also, immense contributions by big names like J S Milne, Ben Webster, Charles Siegel and Andy Putman will add to the success of this venture. Hopefully, the process of understanding mathematics will be further simplified by this website, especially at the undergraduate level.

About me

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. An ardent lover of Chess & Counter-Strike.

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Quotable Quotes

ABHISHEK PARAB
“Do not think; let the equation think for you”

PAUL HALMOS
”You cannot be perfect, but if you won’t try, you won’t be good enough”

ALBERT EINSTEIN
“Don’t worry about your maths problems; I assure you, mine are greater”

THE BEST MATH JOKE
"A comathematician is a device for turning cotheorems into ffee"

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