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However, we give a practical ‘cookbook’ recipe for determining the coefficients. (a) To determine the coefficient Ak , imagine the denominator h(x) written as the product (x − α1 )(x − α2 ) · · · (x − αn ), with any m-fold repeated root giving rise to m factors in parentheses. (b) Now set x equal to αk and evaluate the expression obtained after omitting the factor that reads αk − αk . (c) Divide the value so obtained into g(αk ); the result is the required coefficient Ak . e. 1. e in agreement with our other evaluations.
E. all mk = 1 and r = n, and this we will do. Sometimes it is desirable to be able to reverse this process, in particular when one exact zero has been found by some method and the remaining zeros are to be investigated. 11) PRELIMINARY ALGEBRA where f1 (x) is a polynomial of degree n−1. How can we find f1 (x)? The procedure is much more complicated to describe in a general form than to carry out for an equation with given numerical coefficients ai . If such manipulations are too complicated to be carried out mentally, they could be laid out along the lines of an algebraic ‘long division’ sum.
Only the most elementary ones, given earlier, are worth committing to memory but, as illustrations, we now derive two results involving sums of binomial coefficients. 27 PRELIMINARY ALGEBRA The first is a further application of the method of induction. Consider the proposal that, for any n ≥ 1 and k ≥ 0, n−1 k+s Ck = n+k Ck+1 . 53) s=0 Notice that here n, the number of terms in the sum, is the parameter that varies, k is a fixed parameter, whilst s is a summation index and does not appear on the RHS of the equation.