# Chaitanya's Random Pages

## November 20, 2010

### A collection of algebraic identities

Filed under: mathematics — ckrao @ 5:35 am

In this post I will mainly focus on proving the following nice result that a friend gave to me. Then I will list some other nice identities I have collected from books and the internet.

Let $z_1, z_2, \ldots, z_n$ be n distinct numbers (they could even be complex-valued). Then

$\displaystyle \sum_{i=1}^n \prod_{j\neq i} \frac{1}{z_i-z_j} = 0.\quad \quad ... (1)$

For example, for the case n = 3 we have

$\displaystyle \frac{1}{(z_1 - z_2)(z_1 - z_3)} + \frac{1}{(z_2 - z_3)(z_2 - z_1)} + \frac{1}{(z_3 - z_1)(z_3 - z_2)} = 0.$

To prove (1), consider the following partial fraction expansion:

$\displaystyle \frac{1}{\prod_{i=1}^n (z-z_i)} = \sum_{i=1}^n \frac{A_i}{z-z_i}.\quad \quad ... (2)$

To find the $A_i$, we multiply both sides of (2) by $(z-z_i)$ and then set $z$ to $z_i$, for each $i$. All terms on the right become zero except the one with $A_i$:

$\displaystyle \prod_{j\neq i} \frac{1}{z_i-z_j} = A_i.\quad \quad ... (3)$

Next, multiply both sides of (2) by $\prod_{i=1}^n (z-z_i)$ to obtain

$\displaystyle 1= \sum_{i=1}^n A_i\prod_{j \neq i} (z-z_i).\quad \quad ... (4)$

Taking the coefficient of $z^{n-1}$ of both sides of (4) (treated as a polynomial in z) gives

$\displaystyle 0 = \sum_{i=1}^n A_i,$

which after comparing (1) and (3) is what we wished to show. Note that taking coefficients of other powers of $z$ in both sides of (4) reveals more identities.

I found this result quite cute and it made me dig up other identities I have seen over the years. I will prove one more and the rest can be left as exercises for the interested reader.

If $x + y + z = 0$,

$\displaystyle \left(\frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y}\right)\left(\frac{z}{x-y} + \frac{x}{y-z} + \frac{y}{z-x}\right) = 9.$

An easy way of doing this eludes me. Adding the first and second terms inside the first pair of brackets gives

$\begin{array}{lcl} \frac{x-y}{z} + \frac{y-z}{x} &=& \frac{x^2 - xy + zy - z^2}{zx}\\&=& \frac{(x-z)(x+z-y)}{zx}\\&=&\frac{(x-z)(-2y)}{zx},\end{array}$

where in the last equality we used the fact that $x + y + z = 0$. Multiplying this by $\frac{y}{z-x}$ in the right bracket gives

$\begin{array}{lcl} \frac{-2y^2(x-z)}{(z-x)zx} &=& \frac{2y^2}{zx}.\end{array}$

Summing this cyclically over the other possible products gives

$\begin{array} {lcl} \left(\frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y}\right)\left(\frac{z}{x-y} + \frac{x}{y-z} + \frac{y}{z-x}\right) &=& 1 + 1 + 1 + \frac{2y^2}{zx} + \frac{2z^2}{xy} + \frac{2x^2}{yz}\\&=& 3 + 2 \frac{x^3 + y^3 + z^3}{xyz}\\& = & 3 + 2.3\\ &=& 9,\end{array}$

where the penultimate equality follows from one of the identities shown below.

Finally here is a list of interesting identities, with potential to grow. We assume they are defined for complex numbers for which any denominator shown is non-zero.

### Algebraic Identities

• $\displaystyle x^2 - y^2 = (x+y)(x-y)$ (elementary but very useful!)
• More generally, $\displaystyle x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$
• If $n$ is odd, $\displaystyle x^n + y^n = (x+y)(x^{n-1} - x^{n-2}y + \ldots - xy^{n-2} + y^{n-1}).$
• Binomial theorem: $\displaystyle (x+y)^n = \sum_{k=0}^n \binom{n}{k}x^ky^{n-k}$
• In particular, $\displaystyle (x+y)^2 = x^2 + 2xy + y^2, \ (x+y)^3 = x^3 + 3xy(x+y) + y^3.$
• $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$
• Multinomial theorem: $\displaystyle \left(x_1 + x_2 + \ldots + x_m\right)^n = \sum_{k_1+k_2 + \ldots k_m=n}\binom{n}{k_1, k_2, \ldots, k_m}x_1^{k_1}x_2^{k_2}\ldots x_m^{k_m},$
where $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m} = \frac{n!}{k_1! k_2! \ldots k_m!}.$
• $\begin{array}{lcl} x^3 + y^3 + z^3 - 3xyz &=& (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\\ &=& \frac{1}{2}(x + y + z)\left((x-y)^2 + (y-z)^2 + (z-x)^2\right) \end{array}$
(this implies $x^3 + y^3 + z^3 = 3xyz$ if $x + y + z = 0)$
• $\displaystyle (x-y)^3 + (y-z)^3 + (z-x)^3 = 3(x-y)(y-z)(z-x)$ (follows from the previous result)
• $\displaystyle x^4 + 4y^4 = x^4 + 4x^2y^2 + 4y^4 - 4x^2 y^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$
• $\displaystyle x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$
• $\displaystyle \left(\sum_i x_i^2 \right)\left(\sum_i y_i^2 \right) = \left(\sum_i x_i y_i\right)^2 + \sum_{i < j} (x_i y_j - x_j y_i)^2$ (Lagrange’s identity)
Complex case: $\displaystyle \left(\sum_i |x_i|^2 \right)\left(\sum_i |y_i|^2 \right) = \left|\sum_i x_i y_i\right|^2 + \sum_{i < j} \left|\overline{x_i} y_j - x_j \overline{y_i}\right|^2$
• Special case: $\displaystyle (a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad + bc)^2 = (ac + bd)^2 +(ad - bc)^2.$ (Brahmagupta-Fibonacci identity, or complex number multiplication)
• $\displaystyle (a^2 + nb^2)(c^2 + nd^2) = (ac - nbd)^2 + n(ad+bc)^2 = (ac + nbd)^2 + n(ad - bc)^2.$ (proving that numbers of the form $x^2 +ny^2$ are closed under multiplication – see the Wikipedia entry on the Brahmagupta-Fibonacci identity)

$\begin{array}{l}(x_1^2 + x_2^2 + x_3^2 + x_4^2)(y_1^2 + y_2^2 + y_3^2 + y_4^2) \ = \\ \quad (x_1y_1 - x_2y_2 - x_3y_3 - x_4y_4)^2 \ + \\ \quad (x_1y_2 + x_2y_1 + x_3y_4 - x_4y_3)^2 \ + \\ \quad (x_1y_3 - x_2y_4 +x_3y_1 + x_4y_2)^2 \ + \\ \quad (x_1y_4 + x_2y_3 - x_3y_2 + x_4y_1)^2\end{array}$

• $\displaystyle (x + y + z)^3 - (x^3 + y^3 + z^3) = 3(x+y)(y+z)(z+x)$
• $\displaystyle (x + y + z)^5 - (x^5 + y^5 + z^5) = 5(x+y)(y+z)(z+x)(x^2 + y^2 + z^2 + xy + yz + zx)$
• $\displaystyle (x + y + z)(xy + yz + zx) - xyz = (x + y)(y+z)(z+x)$
• $\displaystyle x^2(y+z) + y^2(z+x) + z^2(x+y) + 2xyz = (x + y)(y+z)(z+x)$
• $\displaystyle xyz(x^3 + y^3 + z^3) - (x^3y^3 + y^3z^3 + z^3 x^3) = (x^2-yz)(y^2 - zx)(z^2 - xy)$
• $\displaystyle (x^2 + y^2 + z^2)^2 - 2(x^4 + y^4 + z^4) = (x + y + z)(-x + y + z)(x - y + z)(x + y -z)$
(used in Heron’s formula for the area of a triangle given side lengths $x, y, z$)
• If $x + y + z = xyz$, $\displaystyle x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2) = 4xyz.$
• $\displaystyle \sum_{\text{cyc}}(zx + 1)(yz+1)(x-y) = \sum_{\text{cyc}}x^2(z-y)$
• $(x+y)^5 - x^5- y^5 = 5xy(x+y)(x^2 + xy + y^2)$
• $(x+y)^7 - x^7 - y^7 = 7xy(x+y)(x^2 + xy + y^2)^2$
• $(x+y)^9 - x^9 - y^9 = 3xy(x+y)\left[3(x^2 + xy + y^2)^3 + x^2y^2(x+y)^2\right]$
• $\displaystyle \sum_{\text{cyc}} \frac{x^3}{(x-y)(x-z)} = x + y + z$
• $\displaystyle \sum_{\text{cyc}} \frac{x-y}{z} + \prod_{\text{cyc}} \frac{x-y}{z} = 0$

Here the $\displaystyle \sum_{\text{cyc}}$ or $\displaystyle \prod_{\text{cyc}}$ notation is used to denote a cyclic sum or product through the indices $x,y,z$, so that the three permutations (x,y,z), (y,z,x) and (z,x,y) are used. It is also worth knowing that any symmetric polynomial can be written as a polynomial of elementary symmetric polynomials (which are sums of all possible products of a fixed number of the variables, for example $x + y + z, xy + yz + zx$, and $xyz$).

## 3 Comments »

1. Yo CK man!

Comment by Radhika — November 23, 2010 @ 11:24 pm

2. excellent master !!!

Comment by fernando — September 16, 2011 @ 2:42 pm

3. fantastic altogether

Comment by babloo — November 24, 2012 @ 6:04 am

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