Question

Suppose that \(\sum_{l=1}^{n} a_{i}=\sum_{j=1}^{m} b_{j}=1\) where \(0 \leq a_{i}, b_{j} \leq 1 .\) Use the Product of Sums Principle to prove that \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i} \cdot b_{j}=1 .\) Does the result hold if some of the \(a_{i}\) and \(b_{j}\) can be less than zero and greater than one?

Short answer

The result is true; it doesn't hold if some \(a_i\) or \(b_j\) aren't between 0 and 1.

Step-by-step solution

Understand the Problem

We need to show that the double sum \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i} \cdot b_{j} = 1\), given that \(\sum_{l=1}^{n} a_{l} = 1\) and \(\sum_{j=1}^{m} b_{j} = 1\). Each \(a_i\) and \(b_j\) are between 0 and 1.

Apply the Product of Sums Principle

The Product of Sums Principle states that multiplying two sums is equivalent to summing all products of terms. Here, we first write the product of sums as \(\left( \sum_{i=1}^{n} a_i \right) \left( \sum_{j=1}^{m} b_j \right)\).

Expand the Product of Sums

This expansion is equivalent to computing \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_i \cdot b_j\). Since each sum is 1, this becomes \(1 \times 1 = 1\).

Conclude the Proof

By expanding the product of sums, we have shown \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_i \cdot b_j = 1\). This verifies the initial claim.

Consider Variations in \(a_i\) and \(b_j\) Conditions

If \(a_i\) or \(b_j\) can be less than 0 or greater than 1, then the sums \(\sum_{i=1}^{n} a_i\) and \(\sum_{j=1}^{m} b_j\) may not equal 1. Therefore, the result does not hold under these conditions.

Key concepts

Double Sum

In mathematics, a double sum is an elegant method for adding up a series of terms that depend on two different indices. In this context, the double sum notation \( \sum_{i=1}^{n} \sum_{j=1}^{m} a_i \cdot b_j \) is used to express the summation of products over pairs of indices \( i \) and \( j \). Here, you first do the inner sum by multiplying each \( a_i \) by every \( b_j \), and then the outer sum, which adds up all these products.

This approach is instrumental when evaluating expressions resulting from multiplying series or sums, particularly when both series are finite, as in this problem. The double sum simplifies the representation by stacking the operations, ensuring that each combination of indices is considered.

Proof by Expansion

Proof by expansion is a clear way to validate equations involving sums by directly calculating and comparing both sides. In this exercise, the Product of Sums Principle is applied. This principle, when expanded, allows us to take separate sums and combine them into a comprehensive expression as seen in \( \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{j=1}^{m} b_j \right) \).

Expanding this, we compute every possible product \( a_i \cdot b_j \) for all \( i \) and \( j \), equivalent to our original double sum \( \sum_{i=1}^{n} \sum_{j=1}^{m} a_i \cdot b_j \). By completing the expansion, each term’s contribution is explicitly laid out and summed, proving the statement algebraically. In this example, both sums equal to 1, simplifying the equation to \( 1 \times 1 = 1 \).

• By expanding, each \( a_i \) gets multiplied by every \( b_j \).
• The outer and inner summations simplify calculations and highlight the correlation between elements.

Conditions of Summation

The conditions of summation significantly affect the outcome of sums computed through principles like the Product of Sums. For the original problem, each \( a_i \) and \( b_j \) must lie between 0 and 1, which ensures that \( \sum_{i=1}^{n} a_i = 1 \) and \( \sum_{j=1}^{m} b_j = 1 \). These constraints are crucial, as they maintain balance and result in the statement being true.

Failing to meet these conditions, such as allowing elements outside of the 0 to 1 range, disrupts this balance, and the sums might not equate to 1 anymore. When elements are not confined to this range, it leads to different results due to potential over or under-contribution from terms multiplying to values outside the normalized range.

Thus, adhering to summation conditions ensures product consistency, maintaining proofs like these valid and integral.

Negative or Greater-than-One Values

Using negative or greater-than-one values in the sums can disrupt the results significantly. When \( a_i \) or \( b_j \) exceed this range, the originally balanced sum \( \sum_{i=1}^{n} a_i \) or \( \sum_{j=1}^{m} b_j = 1 \) no longer holds, leading to a mismatch when trying to prove \( \sum_{i=1}^{n} \sum_{j=1}^{m} a_i \cdot b_j = 1 \).

Negative values can result in subtractive terms rather than additive contributions, effectively reducing the overall sum. Similarly, values greater than 1 can inflate the sum, leading to products that exceed the expected unity.

• Negative values can decrease the sum total, deviating from 1.
• Greater-than-one values might inflate the summation, also leading away from 1.

Therefore, understanding the effect of values outside the designated range is critical in safeguarding mathematical logic and ensuring accurate results across varied calculations.