## Tag Archives: algebra for IIT

### 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

### 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