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)

\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).

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3 Comments »

  1. Yo CK man!

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

  2. excellent master !!!

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

  3. fantastic altogether

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


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