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b p/b q = b(p q) Let s call this the subtraction-of-exponents (SOE) rule Now we ll work out an example Suppose we have b = 10, p = 5, and q = 3 Then 105/103 = 100,000/1,000 = 100 and 10(5 3) = 102 = 100

1 where Km+1 = Pm+1 Hm+1 Rm+1 and eqn (539) was used to obtain the second to the last line of the derivation Although a linear estimator with the form of eqn (533) was initially stated as an objective, such a linear relationship was never imposed as a constraint The linear update relationship of eqn (540) is a natural consequence of the problem formulation Eqns (537) and (540) provide recursive formulas for the estimation of the vector x With proper initialization, the estimate is exactly the same as that attained by use of a batch approach for m+1 measurements using eqns (527) and (528) Table 51 shows that, for large m, the memory and computational requirements of the batch algorithm are O(n2 m) and O(nm), respectively These requirements are necessary even if the estimate is known for (m 1) measurements prior to the m-th measurement The computational and memory requirements for incorporating a single additional scalar measurement using the recursive algorithm of eqns (537) and (540) are evaluated in Table 52 For the recursive algorithm, the memory and computational requirements for incorporating each measurement are determined only by the dimension of the estimated vector Eqn (537) computes P 1 where Pm+1 is required for the calculation m+1 of K; therefore, the algorithm as written requires matrix inversion which is not desirable Table 52 shows that the matrix inversion plays a dominant

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Are you confused (yet)

101 121 131 151 181 221 331 471 561 681 751 821

Computation Hm+1 r = ym+1 Hm+1 xm d1 = Fm+1 = Fm + d1 Hm+1 Pm+1 = F 1 m+1 K = Pm+1 d1 xm+1 = xm + Kr Total

Was that too easy for you Let s try a slightly tougher example Let b = 2, p = 3, and q = 4 Then ( 2)3/( 2)4 = 8/16 = 1/2 and ( 2)(3 4) = ( 2) 1 = 1/( 2) = 1/2

5 2n

The AOE and SOE rules work not only when the exponents are integers, but for any rational numbers You might call these facts the generalized addition-of-exponents (GAOE) rule and the generalized subtractionof-exponents (GSOE) rule

047 05 056 068 1 2 22 33 47 56 1 2

Here s a challenge!

1 2 2n

Using the SOE rule, provide a demonstration of why any nonzero quantity to the 0th power is equal to 1 Also show why 00 is not defined

1 2 2n

Look again at the formula that translates division of quantities into subtraction of exponents That formula is b p/b q = b( p q) where b is the base and p and q are integers Now let s think of the formula in reverse We can transpose the left-hand and right-hand sides of the equation to get b( p q) = b p/b q There s nothing in the rule book that says we can t have p and q be the same Let s do that, and call them both p Then we have b( p p) = b p/b p The left-hand side of this equation is b raised to the (p p)th power, which must be b raised to the 0th power because p p is always 0 The right-hand side is b p divided by itself, which has to equal 1 as long as b 0 Now we get to the 00 situation Let s violate the rule book and let b = 0 in the above equation Then we get 0( p p) = 0 p/0 p No matter what nonzero value we choose for p, we get 00 on the left-hand side of this equation, and 0/0 on the right So 00 = 0/0 The quantity 0/0 is not defined, so 00 can t be, either

1 2 3 4 8 7 6 5

Table 52: Computational and memory requirements for the recursive least squares algorithm of eqns (537) and (540) with x Rn , Hm+1 R1 n , y R, and Rm+1 R The matrix Fm+1 Rn n represents the information matrix

Numbers can be raised to powers more than once In this section we ll see what happens when you raise a quantity to a power, and then raise the result to another power

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