Chapter 12: Q. 2TF (page 987)

Reordering a Double Summation: Explain why
$\sum_{i = 1}^{m} \sum_{j = 1}^{n} f (i, j) = \sum_{j = 1}^{n} \sum_{i = 1}^{m} f (i, j) .$

Short Answer

Expert verified

Since, i and j are independent variables reordering of the summation makes no effect.

Step by step solution

Step 1. Given Information.

Given $\sum_{i = 1}^{m} \sum_{j = 1}^{n} f (i, j) = \sum_{i = 1}^{m} (\sum_{j = 1}^{n} f (i, j)) f r o m p r e v i o u s e x a m p l e .$

Step 2. Proof.

$\sum_{i = 1}^{m} \sum_{j = 1}^{n} f (i, j) = \sum_{i = 1}^{m} (f (i, 1) + f (i, 2) + . . . . . + f (i, n)) = \sum_{i = 1}^{m} (f (i, 1)) + \sum_{i = 1}^{m} (f (i, 2)) + . . . . . + \sum_{i = 1}^{m} (f (i, n)) = (f (1, 1) + f (2, 1) + . . . + f (m, 1)) + (f (1, 2) + f (2, 2) + . . . + f (m, 2)) + (f (1, n) + f (2, n) + . . . + f (m, n)) R e o r d e r i n g w e g e t, = (f (1, 1) + f (1, 2) + . . . + f (1, n)) + (f (2, 1) + f (2, 2) + . . . + f (2, n)) + (f (m, 1) + f (m, 2) + . . . + f (m, n)) = \sum_{j = 1}^{n} (f (1, j)) + \sum_{j = 1}^{n} (f (2, j)) + . . . . . + \sum_{j = 1}^{n} (f (m, j)) = \sum_{j = 1}^{n} (f (1, j) + f (2, j) + . . . + f (m, j)) = \sum_{j = 1}^{n} (\sum_{i = 1}^{m} f (i, j)) = \sum_{j = 1}^{n} \sum_{i = 1}^{m} f (i, j) = R H S . H e n c e, p r o v e d .$

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Reordering a Double Summation: Explain why
$\sum_{i = 1}^{m} \sum_{j = 1}^{n} f (i, j) = \sum_{j = 1}^{n} \sum_{i = 1}^{m} f (i, j) .$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

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Reordering a Double Summation: Explain why∑i=1m∑j=1nf(i,j)=∑j=1n∑i=1mf(i,j).

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Proof.

One App. One Place for Learning.

Most popular questions from this chapter

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Reordering a Double Summation: Explain why
$\sum_{i = 1}^{m} \sum_{j = 1}^{n} f (i, j) = \sum_{j = 1}^{n} \sum_{i = 1}^{m} f (i, j) .$