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November 30, 2015

Lowest highest scores in test matches in recent times

Filed under: cricket,sport — ckrao @ 10:46 am

The past week saw two low scoring test cricket matches between India and South Africa in Nagpur and Australia vs New Zealand in Adelaide. According to this link, only 3 times since the 1930s has a completed test match (i.e. with a result) had no player make 50 in any of the four innings. Here are some other matches since 1980 where the batting average for the match was at most that of the Nagpur test (14.9 runs per wicket). Some of them may ring familiar for cricket fans.

Also only 14 times since 1980 has the top score been less than 66 which was the highest score in Adelaide (one other in Australia when McGrath had match figures of 10/27!).

Here are stats highlights from the two matches:

Lowest top score in an Adelaide Test Cricket – ESPN Cricinfo

Ashwin’s records and lowest top scores Cricket –  ESPN Cricinfo

 

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November 25, 2015

An identity based on three numbers summing to zero

Filed under: mathematics — ckrao @ 11:06 am

Here is a nice identity which according to [1] appeared in a 1957 Chinese mathematics competition.

If x + y + z = 0 then

\displaystyle \left(\frac{x^2 + y^2 + z^2}{2} \right)\left(\frac{x^5 + y^5 + z^5}{5} \right) = \left(\frac{x^7 + y^7 + z^7}{7} \right).\quad \quad (1)

An elegant proof of this avoids any lengthy expansions. Let x, y and z be roots of the cubic polynomial

\begin{aligned} (X-x)(X-y)(X-z) &= X^3 - (x+y+z)X^2 + (xy + yz + zx)X - xyz\\ &:= X^3 + aX + b.\quad \quad (2)\end{aligned}

Then

\begin{aligned} x^2 + y^2 + z^2 &= (x+y+z)^2 - 2(xy + yz + xz)\\ &= 0 -2a\\ &= -2a\quad\quad (3)\end{aligned}

and summing the relation X^3 = -aX - b for each of X=x, X=y and X=z, gives

\begin{aligned}x^3 + y^3 + z^3 &= -a(x + y + z) - 3b\\ &= -3b.\quad\quad (4)\end{aligned}

In a similar manner, X^4 = -aX^2 - bX and so

\begin{aligned} x^4 + y^4 + z^4 &= -a(x^2 + y^2 + z^2) - b(x + y + z)\\ &= -a(-2a)\\ &= 2a^2.\quad \quad (5)\end{aligned}

Next, X^5 = -aX^3 - bX^2 and so

\begin{aligned} x^5 + y^5 + z^5 &= -a(x^3 + y^3 + z^3) - b(x^2 + y^2 + z^2)\\ &= -a(-3b) -b(-2a)\\ &= 5ab.\quad \quad (6)\end{aligned}

Finally, X^7 = -aX^5 - bX^4 and so

\begin{aligned} x^7 + y^7 + z^7 &= -a(x^5 + y^5 + z^5) - b(x^4 + y^4 + z^4)\\ &= -a(5ab) -b(2a^2)\\ &= -7a^2b.\quad \quad (7)\end{aligned}

We then combine (3), (6) and (7) to obtain (1). It seems that x^n + y^n + z^n for higher values of n are more complicated expressions in a and latex $b$, so we don’t get as pretty a relation elsewhere.

Reference

[1] Răzvan Gelca and Titu Andreescu, Putnam and Beyond, Springer, 2007.

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