## Tag Archives: Algebra for RMO

### Basic Algebra for IITJEE Main and RMO

More basic algebra for you guys who are thirsting for more…The following is a nice problem indicating some basic concepts or tricks in problems involving logarithms/powers.

Solve for x: $4^{x}-3^{x-(1/2)}=3^{x+(1/2)}-2^{2x-1}$ (IITJEE 1978)

Solution:

Writing $4^{x}$ as $2^{2x}$ and bringing powers of the same number on the same side we get,

$2^{2x}+2^{2x-1}=3^{x+(1/2)}+3^{x-(1/2)}$

The first term on the LHS can be written as $2^{2x-1} \times 2$, and hence, a common factor of

$2^{2x-1}$ comes out from the terms on the LHS. As for RHS, we can rewrite the first term as

$3^{x-(1/2)} \times 3$ and then the factor $3^{x-(1/2)}$ comes out as common. So, we get

$2^{2x-1}(2+1)=3^{x-(1/2)}(3+1)$, that is, $3 \times 2^{2x-1}=4 \times 3^{x-(1/2)}$

Bringing all powers of 2 to  the left and all powers of 3 to the right, we get

$2^{2x-3}=3^{x-(3/2)}$

By inspection, $x=3/2$ is a solution. But, how do we arrive at it systematically? Also, how do we know that there is no other solution? It is tempting to try to do this by saying that a power of 2 can equal a power of 3 only when are both are equal to 1. (Such a reasoning is indeed useful in solving equations in Number Theory where we mostly deal with positive integers and their factorization into integers). But, here it is inapplicable because we do not  know that the exponents are integers. Instead, let us express both the sides as the power of the same number. One way to do this  is to write 3 as $2^{\log_{2}3}$ in the RHS. Then, we can get

$2^{2x-3}=2^{(\log_{2}3)(x-(3/2)}$

As the bases are the same, the equality of powers implies that of the exponents. So, we have

$2x-3=(\log_{2}3)(x-(3/2))$

This can be solved easily to give $x=\frac{3-(3/2)(\log_{2}3)}{2-\log_{2}3}$

which simply equals $3/2$.

Hence, $x=3/2$ is the only solution.

### Basic Algebra for IITJEE Main

Main Problem: For an integer $n \geq 5$, let $S_{n}$ denote the sum of the products of the integers from 1 to n taken three at a time. Then, what is the value of $S_{10}$?

First Hint: Consider the terms in the expansion of $(1+2+3+ \ldots +10)^{3}$.

Second Hint: Classify these terms according to the number of times they occur.

Solution: There are totally 1000 terms in the expansion of

$(1+2+3+ \ldots +10)^{3}$. Terms of the form $ijk$ with i,j,k all distinct appear 6 times each. We are interested in the sum of these terms each such term considered only once. Then, there are terms of the form $i^{2}jk$ with

$i \neq j$. Each such product appears three times. Finally, there are terms of  the form $i^{3}$. Each such term appears once only. Hence,

$A=6S_{10}+3B+C$ equation I

where $A=(\sum_{i=1}^{10})^{3}$ and $C=\sum_{i=1}^{i=10}i^{3}$ and

$\sum_{i \neq j}^{10}i^{2}j$.

Using the well-known formulae given below:

$1+2+3+ \ldots +n = \frac {n(n+1)}{2}$ equation II

$1^{2}+2^{2}+3^{2}+ \ldots + n^{3}= \frac {n^{2}(n+1)^{2}}{4}$ equation III

A and C come out to be respectively $(55)^{3}$ and $55^{2}$.

To calculate B, note that if we add to B, products of the form $i^{2}i$, then we get all the terms of the product

$(1^{2}+2^{2}+3^{2}+ \ldots + n^{2})(1+2+3+\ldots +n)$.

The first factor equals $55 \times 7$ using the formula

$\sum n^{2}=\frac {n(n+1)(2n+1)}{6}$ equation IV

Hence, $B=6 \times (55)^{2}$. Putting these values in equation I, we get

$S_{10}=(1/6)(A-3B-C)=18150$.

More later

Nalin Pithwa

### A variant of mathematical induction — for IITJEE Math and RMO Math

As is generally known, the principal of mathematical induction is used in the following form: We are supposed to prove a proposition for natural numbers. We check whether it is true for the natural number 1, and then assume it to be true for another natural number m. Then, we got to  show  that the proposition also holds for the natural number $m+1$.

Here is a variation of the principle of mathematical induction:

Assume the proposition to be true for a first natural number k (you have to  find the particular value of k under the conditions of the given problem). Then, assume it to be true for the natural number m. Finally, we gotta show that the proposition is true for the natural number $m+1$.

Example. Show that $n! > 3^{n}$ if n is large enough.

Solution. How large is large? We have to find that “first” value for which the above proposition is true. We actually need to check whether the inequality holds for $n=1, 2, 3, 4, 5, 6, 7, 8 \ldots$

So, if we tabulate the values of LHS and RHS, we find that the first natural number for which this holds is $n=7$..

Hence, the proposition $n! > 3^{n}$ is true for n=7.

Assume the proposition to be true for some natural number m where $m > 7$.

That is, $m! > 3^{m}$ holds true. Multiplying both sides by $m+1$, we get

$(m+1)m! > 3^{m} (m+1)$ that is,

$(m+1)! > m \times 3^{m} + 3^{m}$, but $m>7$. Hence, we get

$(m+1)! > 7 \times 3^{m} + 3^{m}$ which equals $8 \times 3^{m}$.

Hence, obviously, $(m+1)! > 3^{m} \times 3$ that is,

$(m+1)! > 3^{m+1}$. That is, the proposition holds for all natural numbers. Thus, the proof.

More later,

Nalin Pithwa

### Reciprocal equation for IITJEE and RMO/INMO

In this set of little exercises, you will get a grip on reciprocal equations.

reciprocal polynomial has the form

$ax^{n}+bx^{n-1}+cx^{n-2}+...+cx^{2}+bx+a$

in which $a \neq 0$ and the coefficients are symmetric about the middle one. A reciprocal equation is of the form $p(t)=0$ with $p(t)$ a reciprocal polynomial.

1(a) Verify that each of the following polynomials is a reciprocal polynomial:

$x^{3}+4x^{2}+4x+1$

$3x^{6}-7x^{5}+5x^{4}+2x^{3}+5x^{2}-7x+3$

1(b) Show that 0 is not a zero of any reciprocal polynomial.

1(c) Show that -1 is a zero of any reciprocal polynomial of odd degree, and deduce that any reciprocal polynomial of odd degree can be written in the form $(x+1)q(x)$, with $q(x)$ a reciprocal polynomial of even degree.

1(d) Show that, if r is a root of a reciprocal equation, then so also is $1/r$.

2(a) Let $ax^{2k}+bx^{2k-1}+...+rx^{k}+...+bx+a$ be a reciprocal equation of even degree $2k$. Show that this equation can be rewritten as

$a(x^{k}+x^{-k})+b(x^{k-1}+x^{-k+1})+...+r=0$

2(b) Let $t=x+x^{-1}$. Verify that $x^{2}+x^{-2}=t^{2}-2$ and that $x^{3}+x^{-3}=t^{3}-3t$. Prove that, in general, $x^{m}+x^{-m}$ is a polynomial of degree m in t.

2(c) Use the substitution in 2b to show that the reciprocal equation in 2a can be rewritten as an equation of degree k in the variable t. Deduce that the solution of a reciprocal equation of degree $2k$ can in general be reduced to solving one polynomial equation of degree k as well as at most k quadratic equations.

3(a) Show that a product of reciprocal polynomials is a reciprocal polynomial.

3(b) Show that, if f, g,  h are polynomials with $f=gh$ and f and h are both reciprocal polynomials, then g is also a reciprocal polynomial.

More later…

Nalin Pithwa

### Polynomials — quartics

Let us continue our exploration of polynomials. Just as in the previous blog, let me present to you an outline of some method(s) to solve Quartics. As I said earlier, “filling up the gaps” will kindle your intellect. Above all, the main aim of all teaching is teaching “to think on one’s own”.

I) The Quartic Equation. Descartes’s Method (1637).

(a) Argue that any quartic equation can be solved once one has a method to handle quartic equations of  the form:

$t^{4}+pt^{2}+qt+r=0$

(b) Show that the quartic polynomial in (a) can be written as the product of two factors

$(t^{2}+ut+v)(t^{2}-ut+w)$

where u, v, w satisfy the simultaneous system

$v + w - u^{2}=p$

$u(w-v)=q$

$vw=r$

Eliminate v and w to obtain a cubic equation in $u^{2}$.

(c) Show how any solution u obtained in (b) can be used to find all the roots of the quartic equation.

(d) Use Descartes’s Method to solve the following:

$t^{4}+t^{2}+4t-3=0$

$t^{4}-2t^{2}+8t-3=0$

II) The Quartic Equation. Ferrari’s Method:

(a) Let a quartic equation be presented in the form:

$t^{4}+2pt^{3}+qt^{2}+2rt+s=0$

The strategy is to complete the square on the left side in such a way as to incorporate the cubic term. Show that the equation can be rewritten in the form

$(t^{2}+pt+u)^{2}=(p^{2}-q+2u)t^{2}+2(pu-r)t+(u^{2}-s)$

where u is indeterminate.

(b) Show that the right side of the transformed equation in (a) is the square of  a linear polynomial if u satisfies a certain cubic equation. Explain how such a value of u can be used to completely solve the quartic.

(c) Use Ferrari’s Method to solve the following:

$t^{4}+t^{2}+4t-3=0$

$t^{4}-2t^{3}-5t^{2}+10t-3=0$.

Happy problem-solving 🙂

More later…

Nalin

### Polynomials — commuting and cubics

Let us start delving deeper in Algebra. But, I will be providing only an outline to you in the present article. I encourage you to fill in the  details. This is a well-known way to develop mathematical aptitude/thinking. (This method of  learning works even in Physics and esoteric/hardcore programming).

Definition. Commuting polynomials. Two polynomials are said to commute under composition if and only if $(p \circ q)(t)=(q \circ p)(t)$

(i.e., $p(q(t)=q(p(t)))$). We define the composition powers of a polynomial as follows

$p^{(2)}(t)=p(p(t))$

$p^{(3)}(t)=p(p(p(t)))$

and in general, $p^{(k)}(t)=p(p^{(k-1)}(t)$ for $k=2,3 \ldots$

Show that any two composition powers of the same polynomial commute with each other.

One might ask whether two commuting polynomials must be composition powers of the same polynomial. The  answer is no. Show that any pair of  polynomials in the  following  two sets commute

I. ${t^{n}: n=1,2 \ldots}$

II. ${T^{n}(t):n=1,2 \ldots}$

Let a and b be any constants with a not equal to zero. Show that, if p and q are two polynomials which commute under composition, then the polynomials

$(t/a-b/a) \circ p \circ (at+b)$ and $(t/a-b/a) \circ q \circ (at+b)$ also  commute under the composition. Use this fact to find from sets I and II other  families which commute under composition.

Can you find pairs of polynomials not  comprised in the foregoing discussion which commute under composition? Find families of polynomials which commute under composition and within which  there is exactly one polynomial of each positive degree.

The Cubic Equation. Cardan’s Method. An elegant way to solve the general cubic is due to Cardan. The strategy is to replace an equation in one variable by one in two variables. This provides an extra degree of  freedom by which we can impose a convenient second constraint, allowing us to  reduce the problem to that of solving a quadratic.

(a) Suppose  the given equation is $t^{3}+pt+q=0$. Set $t=u+v$ andn obtain the equation $u^{3}+v^{3}+(3uv+p)(u+v)+q=0$.

Impose the second condition $3uv+p=0$ (why do we do  this?) and argue that we can obtain solutions for  the cubic by solving the system

$u^{3}+v^{3}=-q$

$uv = -p/3$

(b) Show that $u^{3}$ and $v^{3}$ are roots of  the quadratic equation

$x^{2}+qx-p^{3}/27=0$

(c) Let $D=27q^{2}+4p^{3}$. Suppose that p and q are both real and that $D>0$. Show that the quadratic in (b) has real solutions, and that if

$u_{0}$ and $v_{0}$ are the real cubic roots of  these solutions, then the system in (a) is satisfied by

$(u,v)=(u_{0},v_{0}), (u_{0}\omega, v_{0}\omega^{2}), (u_{0}\omega^{2}, v_{0}\omega)$

where $\omega$ is the imaginary cube root $(0.5)(-1+\sqrt{-3})$ of  unity. Deduce that the cubic polynomial $t^{3}+pt+q$ has one real and two nonreal zeros.

(d) Suppose that p and q are both real and that $D=0$. Let $u_{0}$ be the real cube root of the solution of  the quadratic in (b). Show  that, in this case, the cubic has all its zeros real, and in fact can be written in the form

$sy^{2}$ where $y=(t+u_{0})$ and $s=t-2u_{0}$

(e) Suppose that p and q are both real and that $D<0$. Show that the solutions  of  the quadratic equation in (b) are nonreal complex conjugates, and that it is possible to choose cube roots u and v of  these solutions which are complex conjugates and satisfy the system in (a). If

$u=r(cos \theta + isin \theta)$ and $v=r(cos \theta - isin \theta)$, show that the three roots of  the cubic equation are the reals

$2r cos \theta$

$2r cos (\theta + (2/3)\pi)$

$2r cos(\theta + (4/3)\pi)$.

(f) Prove that every cubic equation with real coefficients has at least one real root.

Use Cardan’s Method to solve the cubic equation.

(a) $x^{3}-6x+9=0$

(b) $x^{3}-7x+6=0$.

Part (b) above will require  the use of a pocket calculator and some trigonometry. You will also need De Moivre’s Theorem and give a solution to an accuracy of 3 decimal places.

More later…

-Nalin

### Some non trivial factorization examples

I hope to give you a flavour of some non-trivial factorization examples using the following identity:

$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab -bc-ca)$ Example 1. Let n be a positive integer. Factorize

$3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1$

Solution. Observe that

$3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1 = a^{3}+b^{3}+c^{3}-3abc$ where $a=3^{3^{n-1}}$, $b=9^{3^{n-1}}$, and $c=-1$.

Thus, using the above factorization identity, we get the following factorization:

$(3^{3^{n-1}}+9^{3^{n-1}}-1)(9^{3^{n-1}}+81^{3^{n-1}}+1-27^{3^{n-1}}+3^{3^{n-1}}+9^{3^{n-1}})$

Example 2. Let a, b, c be distinct positive integers and let k be a positive integer such that $ab+bc+ca \geq 3k^{2}-1$.

Prove that $(1/3)(a^{3}+b^{3}+c^{3})-abc \geq 3k$.

Solution. The desired inequality is equivalent to

$a^{3}+b^{3}+c^{3}-3abc \geq 9k$.

Suppose without loss of generality, that $a>b>c$.

Then, since a, b, and c are distinct positive integers, we $a-b \geq 1$, and

$(b-c) \geq 1$ and $a-c \geq 2$.

It follows that $a^{2}+b^{2}+c^{2}-ab-bc-ca = (1/2)((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \geq (1/2)(1+1+4)=3$

We obtain

$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \geq 3(a+b+c)$ so it suffices to prove that $3(a+b+c) \geq 9k$ or $(a+b+c) \geq 3k$

But, $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca = a^{2}+b^{2}+c^{2} - ab -bc- ca +3(ab+bc+ca) \geq 3+3(3k^{2}-1) = 9k^{2}$,

and the conclusion follows.

More later… Nalin

### Factorization uses and tricks

One of the first concepts of elementary algebra is factorization. Apart from its intrinsic use in Math itself, it is applied in several areas like pole-placement method in Control System Design and design of digital filters for use in digital communications, the latter being the foundation stone of IT. Another use is in error control coding which is used in wireless/mobile/cellular communications. One of the trickiest uses of factorization is in Trigonometry, which of course, has countless applications. Let me present to you some tricks, which are not so obvious in factorizing complicated algebraic expressions.

1) Factorization by guessing! Solve the equation $x^{3}-3x^{2}-6x+16=0$.

Solution: The product of the coefficient of highest degree term and constant is 16. So, think of the factors of 16. You will see that the LHS vanishes when $x=2$. (We have used the factor theorem here).

Hence, $x=2$ is one root of the equation and corresponding to this root, we have a factor $x-2$.

The equation may now be written $x^{2}(x-2)-x(x-2)-8(x-2)=0$ or $(x^{2}-x-8)(x-2)=0$

Removing the factor $x-2$, we have $x^{2}-x-8=0$

hence, the three roots of the above cubic $x = 0.5(1+\sqrt{33}$ or $x = 0.5(1-\sqrt{33})$ or $x=2$.

Moral of the story: Guessing the factor reduces the degree of the equation. There will be some trials and errors though.

2) The Method of Undetermined Coefficients:

Resolve into two factors: $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy$

Solution: This equation might seem not factorizable, at first sight, as the *middle term* involves no *c*. (Note: a, b, and c are parameters, and the variables are x and y). Also, the brute force technique for using the formula method for quadratics is way too cumbersome. So, let’s give a shot to it using the method of undetermined coefficients.

Let $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy=(Px+Qy)(Mx+Ny)$

Hence, equating like coefficients, $PM=2a^{2}$, $NQ= -2(3b-4c)(b-c)$, $MQ+NP=ab$.

Again, you have to guess!

Let’s try: let $N=2(b-c)$, $Q=-(3b-4c)$, $M=a$, $P=2a$

We need to check: $MQ+NP=-a(3b-4c)+2(b-c)(2a)=ab$ Bingo:-)

So, the factors are $(ax+2(b-c)y)(2ax-y(3b-4c))$

3) An important factorization identity is:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Would you like to try proving this?(It is not so easy!)

More later, — Nalin