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= k sinh J + (1 - k) cosh J = HI - 1(2) sinh J + HI + 1 (2) cosh J ieJ(1 + 0102e-2J) = ie J (1 + 1 2 tanhJ*) where we have defined J* by tanh J* == e- 2J , with a view to re-expressing the

above expression as an exponential:

h (cosh J* + 1 2 sinh J*) cos J* [2 sinh (2J) 1/2 exp (J*OI ( 2 ) 2

One instance where the Fibonacci retracement is especially useful is after a stock has made a major move, either up or down. After a large movement with a lot of volatility, traders are extremely uncertain as to where the next support or resistance level is going to be. Take yourself back to the autumn of 2002. All the U.S. stock indexes, particularly the NASDAQ, had suffered a brutal bear market. The NASDAQ 100 ETF, better known as the QQQ (or cubes ) had lost about 80 percent of its value. Suppose you believed the selling was overdone and you took a major long position in this security. (Take note that even though the nancial instrument is known as the QQQ, the symbol is QQQQ.) QQQ begins to move higher, producing a pro t for you. But how long do you hold on Unless your intent is to hold the stock forever, you should probably try to optimize your holding period so that you get out when the recovery seems to have been fully realized. In this instance, the retracement is once again invaluable. Figure 13.15 shows the retracement levels placed on top of the QQQQ graph. The stock bottomed out at about $20 and the rst retracement level is at about $43, a more

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exp ( J*

(ij)

We formulate the two-dimensional Ising model in terms of matrices as J preliminary step toward an exact solution of the model. Consider a square lattice of N = n 2 spins consisting of n rows and n columns, as shown in Fig. 15.1. Let us imagine the lattice to be enlarged by one row and one column with the requirement that the configuration of the (n + l)th row and column be identical with that of the first row and column, respectively. This boundary condition endows the lattice with the topology of a torus, as depicted in Fig. 15.2. Let }l" (a = 1, ... , n) denote the collection of all the spin coordinates of the ath row'

(15.11

{15.21

A configuration of the entire lattice is then specified by {ILl' ' ILn}. B~ assumption, the ath row interacts only with the (a - 1)th and the (a + 1)th row Let E(ILa' ILa+l) be the interaction energy between the ath and the (a + l)th row. Let E(ILa) be the interaction energy of the spins within the ath row plm their interaction energy with an external magnetic field. We can write

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E{IL, IL'}

100 stocks, virtually no study except the Fibonacci retracement could have provided a good prediction as to how high QQQ would be able to climb back.

E{IL} =

L skSk+l -

{15.31

3 n

+ 1= 1

+ 1 = 1 0----<::>---()---0

neighboring rows:

(15.4)

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