# Solution1 - AE 321 Solution of Homework#1 1 The number of...

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AE 321 – Solution of Homework #1 1. The number of terms is given by the following expression , (1.1) n 3 where n is the number of free indices . Then, Number of free indices n Number of terms n 3 ij A 2 9 kl ij B , 4 81 ji ij C , 0 1 ijk D 3 27 qr pq B A 2 9 2. The number of equations is again obtained from the number of free indices by Equation (1.1). Number of free indices n Number of equations n il kl jk ij D C B A = 2 9 0 , = j ij A 1 3 ijkl kl ij l ijk C B A = + , , 4 81 1

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3.1 ik jk ij C B A = Let’s first work with the dummy index and perform the summation ik k i k i k i C B A B A B A = + + 3 3 2 2 1 1 (1.2) In expression (1.2) we have 2 free indices, therefore it represents 9 equations that are obtained by expanding (1.2) for every combination of i and : i.e. & , & , …. The 9 equations are given in the following table. k 1 = i 1 = k 1 = i 2 = k 1 = k 2 = k 3 = k 1 = i 11 31 13 21 12 11 11 C B A B A B A = + + 12 32 13 22 12 12 11 C B A B A B A = + + 13 33 13 23 12 13 11 C B A B A B A = + + 2 = i 21 31 23 21 22 11 21 C B A B A B A = + + 22 32 23 22 22 12 21 C B A B A B A = + + 23 33 23 23 22 13 21 C B A B A B A = + + 3 = i 31 31 33 21 32 11 31 C B A B A
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## This note was uploaded on 01/18/2010 for the course AE 321 taught by Professor Waller during the Fall '07 term at University of Illinois at Urbana–Champaign.

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Solution1 - AE 321 Solution of Homework#1 1 The number of...

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