An inequality for the christmas eve !!

I usually categorize inequalities into 4 types (which is a rather weird kind of classification) keeping in mind the problem solving strategies available:

TYPE 1 : <summation> $sign$ <summation>
TYPE 2 : <summation> $sign$ <product>
TYPE 3 : <product> $sign$ <summation>
TYPE 4 : <product> $sign$ <product>

 "<summation>" denotes expressions majorly involving sums and differences which may be symmetric or asymmetric like $ \sum ab$ or $\sum \frac{ab}{c-1}$ or $a+b^2+c^3$ or $\sqrt{\frac{a^2+bc}{b+c}}$ or more complicated stuff.
"<product>" denotes symmetric or asymmetric expressions involving products like $\prod (a-1)$ or $abcd$ or $(\sum ab)^2 (\sum a)(abc)$ etc ...
"$sign$" may be $\ge$or$\le$or$<$or$>$.

I'll basically be talking of algebraic inequalities in this article while some of the rules are inherited by geometric inequalities too but they need a seperate discussion (topic of the next post).


TYPE 1, 2 engulf a very large portion of the whole subject and some of the techniques and theorems are easily applicable to them.The statements of most of the theorems in inequalities belong to these categories (like Weighted means, Cauchy-Schwarz, Tchebycheff's, Jensen's, Muirhead's, ...)
TYPE 4 can be converted to TYPE 1 by suitable transformations and substitutions (simplest of the methods being applying log on both sides) or solved using Muirhead's inequality.
TYPE 3 is a difficult type. My experience shows me that Schur's inequality is a strong technique to prove problems of this category (Muirhead's is also sometimes helpful). Most of the time we have to rely on some ingenuine and innovative methods of substitution(s) and transformations which differ from problem to problem which either solve the problem directly or reduce it to that of TYPE 1 or 2.

Here I will present a small new technique to prove certain symmetric (sometimes asymmetric also) inequalities of TYPE 4 based on Lagrange's identity which states :

$(a^2 +nb^2)(c^2+nd^2)=(ac-nbd)^2+n(ad+bc)^2=(ac+nbd)^2+n(ad-bc)^2$

Let me demonstrate with an example :

Prove the following :

$(a^2+2)(b^2+2)(c^2+2) \ge 9 \sum ab$

Proof:

$(a^2+2)(b^2+2)(c^2+2) = ((\sqrt{2}a+\sqrt{2}b)^2+(ab-2)^2)(c^2+2)$

                      $= (2a+2b+2c-abc)^2+(\sqrt{2}ab+\sqrt{2}bc+\sqrt{2}ca-2\sqrt{2})^2 $

                      $=(2\sum a-abc)^2+2(\sum ab -2)^2 $

(we have basically used the identity to put the expression as a combination of elementary symmetric functions)


Let $ \sum a =p,\; \sum ab=q,\,  abc=r $. Then we have the following basic inequalities relating elementary symmetric functions of $a, \; b, \; c$ :

$p \ge \sqrt{3q} \ge (27r)^\frac{1}{3}\\ q^2 \ge 3pr $

What we need to prove is :

$4p^2+r^2-4pr+2q^2+8 \ge 17q$

We now have to carefully break down the terms on the left side to obtain the result using the above basic inequalities. Looking at the equality case always helps! Equality clearly holds when $a=b=c=1$ i.e. $p=3, \; q=3, \; r=1$. The terms which are creating a 'problem' are $r^2$ and $8$. Notice how we split the terms with an aim to apply AM-GM inequality :

$4p^2+r^2-4pr+2q^2+8 = (\frac{p^2}{9}+r^2)+\frac{35p^2}{9}-4pr+(\frac{8q^2}{9}+8)+\frac{10q^2}{9}$

                   $\ge \frac{2pr}{3}+\frac{35p^2}{9} -4pr+\frac{16q}{3}+\frac{10q^2}{9}$

                   $=\frac{35p^2}{9}+\frac{16q}{3}+\frac{10q^2}{9}-\frac{10pr}{3}$

                   $\ge \frac{35p^2}{9}+\frac{16q}{3}$

                   $\ge \frac{35q}{3}+\frac{16q}{3}$

                   $=17q$
Q.E.D.

Any other solutions for the above inequality are welcome :-)

COMING UP NEXT : On Geometric Inequalities....

A Merry Christmas to all !!

AAKANSH

Mathematician = Poetician ?

Of lately I have been reading many poetic blogs so couldn't help this seemingly "absurd" comparison emerge into my mind......

Karl Weierstrass who is often cited as the "father of modern analysis" once said:

"A mathematician who is not also something of a poet will never be a complete mathematician."

Some will condemn this to be rubbish but I say to such people they don't have that insight into a mathematicians brain nor a poets mind. What kinda stuff are we talking.....finding truth in the above quote......?

A person who has studied mathematics deeply enough will know the delight at an elegant proof, the same can be found on the face of a poet enchanted by the grace of a poem. Indeed a mathematical proof has a rhythmic structure which if probed reveals its musical and poetic beauty.

Ezra Pound who is known for his work in developing imagism has been quoted as saying:

"Poetry is a sort of inspired mathematics, which gives us equations, not for abstract figures, triangles, squares, and the like, but for the human emotions. If one has a mind which inclines to magic rather than science, one will prefer to speak of these equations as spells or incantations; it sounds more arcane, mysterious, recondite."

A mathematician's work has an artistic blend, and deep thoughts and mysticism of accurate guesses involved makes him stand on the same platform of a poet. Just as that of a poet his style is unique in expressing his words and thoughts in what is known as a 'proof'. Both live in an entirely different world of their creations.

Mathematics and art have a long historical relationship. 
Many think those who learn mathematics stand aloof from art and literature. Mathematics being too rigorous a subject is incapable of estimating the essence of nature and art. Yet this is not true at all. Mathematical ability easily transforms to artistic and poetic abilities and there have been great poets and mathematicians: Omar Khayyam, Raymond Queneau, Sarah Glaz and MarionCohen to name a few.

As for what is mathematics to nature, I say mathematics is fully embedded in nature. Let me tell you the role Fractals, Golden rectangles, spirals and Fibonacci numbers play in nature's creations. They occur mysteriously and naturally!!( In fact Fibonacci numbers have been called the nature's numbering system). 

Romanesco broccoli showing a

naturally occuring fractal.
For more on fractals click here.

Golden ratio($\phi$) [ defined by $\phi = 1+ \frac{1}{\phi}$] is approximately 1.6180339887.

The two divisions shown are in the golden ratio.
Click here for animation to see more illustrations of the Golden ratio.

A Golden spiral is a logarithmic spiral which gets wider by a factor of $\phi$, the Golden ratio for every quarter turn it makes.

 

Nautilus Shell cutaway showing the pattern to be the same as a Golden spiral

Even the finger prints have a Golden spiral  hidden in them! 

                                                 
Parethon, this ancient temple in Athens fits almost exactly in a Golden rectangle 

 The number of petals in commonly occuring flowers are terms of Fibonacci sequence:

3   petals  : Lily, Iris
5   petals  : Wildrose, Buttercup
8   petals  : Delphiniums
13 petals  : Ragwort, Cornmarigold
21 petals  : Chicory, Aster


What more? Golden ratio, rectangles, spirals are themselves related to Fibonacci numbers!! 


Isn't all this Amazing!! 

Let me end up with the poem Geometry (written by Rita Dove) that captures the ecstasy of discovery:

I prove a theorem and the house expands:
the windows jerk free to hover near the ceiling, 

the ceiling floats away with a sigh.

. 

. 

. 

NOTE:

Some interesting links and articles relating to poetry and mathematics can be found at the following blog :  http://poetrywithmathematics.blogspot.com

AAKANSH 

Decimal representation of $\frac{1}{p^k}$

Here I talk about the decimal representations of numbers of the form $\frac{1}{p^k}$ where $p$ is a prime.

If $p= 2 $ or $5$ the decimal expansion is a finite one otherwise it is recurring. The aim is to study the length of recurring cycle of the decimal expansion of $\frac{1}{p^k}$ $(p \ne 2,5)$. 

The expansion consists of some initial number say $M$ of $m$ digits after which some number say $N$ (which has $n$ digits) starts repeating as shown:

$ \frac{1}{p^k} = 0.MNNNN... =\frac{M}{10^m}+ \frac{N}{10^{m+n}}+ \frac{N}{10^{m+2n}}+..... = \frac{M}{10^m} + \frac{N}{10^{m}(10^{n}-1)}$

$\Rightarrow $ $Np^k + M (10^{n}-1)p^k = 10^m(10^n-1) $

Thus

$10^n \equiv 1 (mod \; p^k) $ ($p \neq 2,5$) where $n$ is the recurring cycle of the decimal expansion and is thus the least non zero solution of the above congruence i.e. the order of $10$ modulo $p^k$.

CLAIM: If $r $ is the order of $10$ modulo $p$ then the order of $10$ modulo $p^k$ is of the form $rp^{j}$ where $j$ is an integer. 

PROOF: 

Since $ 10^r \equiv 1(mod \; p) $ we can write $10^r = pu+1$ where $u$ is an integer.

$\Rightarrow$ $10^{rp^{k-1}}\equiv {(pu+1)}^{p^{k-1}} \equiv 1+p^ku+..... \equiv 1 (mod \; p^k)

Thus if $t$ is the order of $10$ modulo $p^k$ then $ t \mid rp^{k-1}$. Also $10^t \equiv 1 ( mod \; p) $ implies $ r \mid  t$.

Thus $t$ must be of the form $ rp^j$ where $0\leq j\leq k-1 $.

We enhance the result further.....

Let $s_1$ be the minimum value of $i$ for which $r$ is the order of $10$ modulo $p^i$. Then what is the order of $10$ modulo $p^{s_1 +1}$ ?....... It is $rp$ .

Call the order modulo $p^i$ as $r_i$. ( $r_1 = r_2 = ... r_{s_1} = r $). Let $10^r = p^{s_1}u_1 +1$

We know $r_{s_1+1}$ is of the form $rp^j$, $j \ne 0$. We show that $j=1$ works.

$10^{rp} \equiv {(p^{s_1}u_1 +1)}^p \equiv 1+p^{s_1+1}u_1 +... \equiv 1 (mod \; p^{s_1 +1})$

Thus rp is the order.

Now if $s_2$ is the minimum value of $i$ for which $rp$ is the order of $10$ modulo $p^i$, then using similar arguments as above we can show that $rp^2$ is the order modulo $p^{s_2 +1}$ and so on.......

We thus obtain that:

Order of $10$ modulo $p,p^2,...,p^{s_1}$ is $r$

Order of $10$ modulo $p^{s_1+1},...,p^{s_2}$ is $rp$

Order of $10$ modulo $p^{s_2 +1},..,p^{s_3}$ is $rp^2$

.

.

.

.

and the length of recurrence cycle is order itself.

NOTE:

1. By Fermats theorem $r$ is clearly a divisor of $p-1$.
2. There is nothing special about 10 in the whole discussion except of course that $(10,p)=1$.So the discussion  can be extended to representation in any general base $b$ and p chosen such that $(b,p)=1$

Three mirrors and a ray !!

 You may sense this to have some physical flavour but I am talking something of mathematical interest ( and perhaps of physical too! )

Consider a system of three mutually perpendicular plane mirrors P, Q, R as shown :

Theorem :
A ray of light incident on the above system of mirrors gets reflected at most 3 times before escaping !!

Proof:
We employ the fact that an incident ray (in above system) if reflected exactly 3 times then the emergent ray is parallel and in a direction opposite to that of the incident ray. The proof employs an easy use of vectors and is left to the reader as an exercise. ( Use Ê = Î - 2 (Î . ñ)ñ  where Î is the unit vector along the incident ray, Ê is the unit vector along the reflected ray and ñ is the unit vector along the normal to a plane mirror, during reflection.) [I discovered the above theorem while solving this.]

Suppose there is an incident ray a₁ that gets reflected more than 3 times....we will obtain a contradiction.
Without loss of generality, let incident ray a₁ strikes P and gets reflected (call this ray a₂ ) to strike the mirror Q. The ray obtained after reflection at Q (call it a₃ ) will again get reflected (since more than 3 reflections take place) at R. Call the now emerging ray a₄. Since 3 reflections have taken place until now,  a₄ has to be parallel and opposite to a₁. And since more than three reflections take place in all a₄ will strike some mirror again and hence a₁ being parallel to a₄, when extended backwards (opposite to incident direction) will also intersect some mirrror...... both a₁ and a₄ being parallel will intersect the same pair of mirrors and they are thus P (which a₁ strikes), R (which a₄ emerges from).

The situation looks like this: a₁ strikes P ( and has a direction from R to P), gets reflected, then strikes Q, then R and emerges as a₄ which moves to P to strike it again, thus a₁ and a₄ both are parallel and have same direction ( from R to P)....a contradiction to the fact that they have opposite direction.

 Q.E.D.

Ok! Now lets get a little serious......

There was a blonde, a brunette, and a redhead. They came across this mirror... Written on the mirror it said "Tell the truth and get a million bucks, tell a lie and disappear forever."
So the brunette's like "I think I'm the prettiest girl in the school"
*POOF* she disappeared.
Then the redhead steps up and says "I think I'm the smartest girl in the whole school"
*POOF* she dissappeared.
Then the blonde steps up and says"I think -"
*POOF* she dissappeared.

Allright....... that must have really made u serious!!!!

٩(●̮̮̃•̃)۶

An ingenuine use of Prime Number Theorem

The following problem appeared in Selection Test II of International Maths Olympiad Training Camp 2010, India :

Given an integer k > 1, show that there exists an integer n > 1 and distinct positive integers a₁, a₂ ,..., a_n ( > 1) such that the sums  ∑a_j and ∑φ(a_j) are both k-th powers of some integers. ( Here φ(m) denotes the number of positive integers less than m and relatively prime to m.)

The line of action I took was to select a_j 's which are square free so that φ(a_j) is related to a_j in a simple way [here if a_j = p₁ p₂ ... p_r  is the prime factorisation into distinct primes then φ(a_j) = (p₁-1)(p₂-1) ... (p_r -1) ]  .....
The next thought was to select a_j 's so that ∑φ(a_j) is also related to ∑a_j in a simple way so that "latter is a k-th power" easily implies "former is a k-th power" ....
Let  ∑a_j = λ^k
Then comes the main idea ..... why not take λ to be a number that can be written as a sum of 2 distinct primes in k different ways!! ...... write the above as .... ∑a_j = λ.λ....λ (k times) = (sum of primes in first way)(second way)......(kth way) .... expand the product of brackets and take each term in the expansion to be our a_j 's (each of which is square free and distinct)
And since in φ(a_j)'s prime factorisation is obtained by replacing each prime factor p of a_j by (p-1), we have ∑φ(a_j) = (λ-2)^k .

What remains to prove is a beautiful result that has emerged out :

Given k > 1, there exists a number λ that can be written as a sum of two distinct primes in at least k different ways.

The proof I present is a nice use of the Prime Number Theorem ( which states that lim_x→∞ ( π(x) log x ÷ x ) = 1 ; π(x) denotes the number of primes less than or equal to x). In fact the well known  elementary result π(n) > (n ÷ 6 log n)  [here n is an integer] is sufficient to prove it. The proof is due to Dr. M.A. Prasad who is a regular faculty at the IMOTC :

Proof:
There are  π(n) prime numbers less than or equal to n and thus C₂^π(n) ways of choosing two of them and summing them ... thus there are C₂^π(n) distinct sums less than or equal to 2n .... therefore at least          (C₂^π(n) ÷ 2n) of them are equal say equal to λ. What remains to show is that there is an n such that      (C₂^π(n) ÷ 2n) is at least k. But (C₂^π(n) ÷ 2n)  = [π(n)][π(n) – 1] ÷ 4n  >  (n – 6 log n) ÷ (12 log n)²  [ using π(n) > (n ÷ 6 log n) ] which is an increasing function and therefore (C₂^π(n) ÷ 2n) is an increasing function and thus is greater than k for large n.
We have thus obtained  λ that can be written as a sum of two distinct primes in at least k different ways.

Q.E.D.

٩(●̮̮̃•̃)۶

Welcome to my mathematical fantasy...

The universe is a manifestation of mathematics. But only a part of it is manifested as the physical world.

Complicated  rare creatures who can sit focused for hours without showing any signs of life and suddenly come up with an idea that can change the World......These are mathematicians......

They play with everything and anything..... this is their art and living....

This blog is not a set of problems for the readers to solve but a set of ideas for the readers to ponder. My primary aim will always remain  "creating ideas out of nowhere" ...and that is the real mathematics. I will occasionally be including some philosophy and abstracts providing insight into a mathematicians mind besides putting together new ideas evolving in mathematics, new methods of solving problems, innovative proofs of  some old problems and the thoughts that inspired them. This is going to be a collection of some really weird elementary mathematics!!! :-)

Aakansh