You are currently browsing the tag archive for the ‘identity’ tag.

The following “polynomial-logarithmic” algebraic identity that one encounters on many occasions turns out to have a rather useful set of applications!

POLYNOMIAL-LOGARITHMIC IDENTITY: If $P(z)$ is a polynomial of degree $n \ge 1$ with roots $a_1, a_2, \ldots, a_n$, then $\displaystyle \frac{P'(z)}{P(z)} = \frac1{z-a_1} + \frac1{z-a_2} + \ldots + \frac1{z-a_n}$.

PROOF: This one is left as a simple exercise. (Hint: Logarithms!)

A nice application of the above identity is found in one of the exercises from the chapter titled Analysis (p120) in Proofs from the Book by Aigner, Ziegler and Hofmann.

EXERCISE: Let $p(x)$ be a non-constant polynomial with only real zeros. Show that $p'(x)^2 \ge p(x) p''(x)$ for all $x \in \mathbb{R}$.

SOLUTION: If $x = a_i$ is a zero of $p(x)$, then the right hand side of the above inequality equals zero, and we are done. So, suppose $x$ is not a root of $p(x)$. Then, differentiating the above identity w.r.t. $x$, we obtain $\displaystyle \frac{p''(x)p(x) - p'(x)^2}{p(x)^2} = - \sum_{k=1}^n \frac1{(x - a_k)^2} < 0$, and we are done.

It turns out that the above identity can also used to prove the well-known Gauss-Lucas theorem.

GAUSS-LUCAS: If $P$ is a non-constant polynomial, then the zeros of $P'$ lie in the convex hull of the roots of $P$.

PROOF: See this

HISTORY: The well-known Russian author V.V. Prasolov in his book Polynomials offers a brief and interesting historical background of the theorem, in which he points out that Gauss’ original proof (in 1836) of a variant of the theorem was motivated by physical concepts, and it was only in 1874 that F. Lucas, a French Engineer, formulated and proved the above theorem. (Note that the Gauss-Lucas theorem can also be thought of as some sort of a generalization (at least, in spirit!) of Rolle’s theorem.)

Even though I knew the aforesaid identity before, it was once again brought to my attention through a nice (and elementary) article, titled On an Algebraic Identity by Roberto Bosch Cabrera, available at Mathematical Reflections. In particular, Cabrera offers a simple solution, based on an application of the given identity, to the following problem (posed in the 2006 4th issue of Mathematical Reflections), the solution to which had either escaped regular problem solvers or required knowledge of some tedious (albeit elementary) technique.

PROBLEM: Evaluate the sum $\displaystyle \sum_{k=0}^{n-1} \frac1{1 + 8\sin^2 (k\pi /n)}$. (proposed by Dorin Andrica and Mihai Piticari.)

There is yet another problem which has a nice solution based again on our beloved identity!

PROBLEM: (Putnam A3/2005) Let $p(z)$ be a polynomial of degree $n$, all of whose zeros have absolute value 1 in the complex plane. Put $g(z) = p(z)/z^{n/2}$. Show that all zeros of $g'(z) = 0$ have absolute value 1.

Those who love (elementary) problem-solving eventually come across the Sophie Germain identity. It has lots of applications in elementary number theory, algebra and so on. The identity states $x^4 + 4y^4 = (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$.

Indeed, note that $x^4 + 4y^4$ $= (x^2)^2 + (2y^2)^2 + 2\cdot x^2 \cdot 2y^2 - 2\cdot x^2 \cdot 2y^2$ $= (x^2 + 2y^2)^2 - (2xy)^2$ $= (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$

More can read about Marie-Sophie Germain, a brilliant mathematician, here and here.

Let us use the above identity in solving a couple of problems. Here is the first one.

Problem 1: Evaluate $\displaystyle \sum_{k=1}^{n}\frac{4k}{4k^4 + 1}$.

Solution: A first glance tells us that the sum should somehow “telescope.” But the denominator looks somewhat nasty! And, it is here that the above identity comes to our rescue. Using the Sophie Germain identity, we first note that $1 + 4k^4 = (1 -2k + 2k^2)(1 + 2k + 2k^2).$

We thus have $\displaystyle \sum_{k=1}^{n} \frac{4k}{4k^4 + 1}$ $\displaystyle = \sum_{k=1}^{n} \left(\frac{(1 + 2k + 2k^2)-(1 -2k + 2k^2)}{(1 -2k + 2k^2)(1 + 2k + 2k^2)}\right)$ $\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 + 2k + 2k^2}\right)$ $\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 - 2(k+1) + 2(k+1)^2} \right)$ $\displaystyle = 1 - \frac1{1 + 2n + 2n^2}$.

Here is another one.

Problem 2: Show that $n^4 + 4^n$ is a prime iff $n=1$, where $n \in \mathbb{N}$.

Solution: Note that if $n$ is even, then the expression is clearly composite. If $n$ is odd, say, $2k+1$ for some $k \in \mathbb{N}$, then we have $n^4 + 4^n$ $= n^4 + 4^{2k+1}$ $= n^4 + 4\cdot (2^k)^4$ $= (n^2 - n\cdot 2^{k+1} + 2^{2k+1})(n^2 + n\cdot 2^{k+1} + 2^{2k+1})$

And, if $n > 1$, then both the factors above are greater than $1$, and hence the expression is composite. Moreover, if $n = 1$, we have $1^4 + 4^n = 5$, which is prime. And, we are done.

### Blog Stats

• 357,713 hits

 Convictions ·… on Continued fraction for e John Favors on Solution to POW-13: Highly… Wayne J. Mann on Solution to POW-12: A graph co… erneststephen on The 54th Carnival of Math… anhtraisg on p^q + q^p is prime prof dr drd horia or… on My first post prof drd horia orasa… on My first post prof dr mircea orasa… on Inequality with log notedscholar on Self-referential Paradoxes, In… prof dr mircea orasa… on Inequality with log prof dr mircea orasa… on Inequality with log prof dr mircea orasa… on 2010 in review kenji on Basic category theory, I prof dr mircea orasa… on Solution to POW-10: Another ha… prof drd horia orasa… on Continued fraction for e