In Exercise 1.15, we proved the result (1.135) for the numbe
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In Exercise 1.15, we proved the result (1.135) for the number of independent parameters in the Mth order term of a D-dim
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In Exercise 1.15, we proved the result (1.135) for the number of independent parameters in the Mth order term of a D-dimensional polynomial. We now find an expression for the total number N(D,M) of independent parameters in all of the terms up to and including the M6th order. First show that N(D,M) satisfies N(D,M) = M m=0 n(D,m) (1.138) where n(D,m) is the number of independent parameters in the term of order m. Now make use of the result (1.137), together with proof by induction, to show that N(d, M) = (D + M)! D! M! . (1.139) This can be done by first proving that the result holds for M = 0 and arbitrary D 1, then assuming that it holds at order M, and hence showing that it holds at order M + 1. Finally, make use of Stirling
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