Problem 50 - Calculus Single Variable

Question

Which is larger, \(\sum_{i=1}^{N} j^{2}\) or \(\sum_{i=1}^{N^{2}} j ?\) Explain why.

Step-by-Step Solution

Answer

The second sum \( \sum_{j=1}^{N^2} j \) is larger; it grows faster as \( N \) increases.

Step 1: Express each sum mathematically

The first sum is expressed as \( \sum_{j=1}^{N} j^2 \), which represents the sum of squares of the first \( N \) natural numbers. The second sum is \( \sum_{j=1}^{N^2} j \), which represents the sum of the first \( N^2 \) natural numbers.

Step 2: Calculate the value of the first sum

The formula for the sum of squares of the first \( N \) natural numbers is \( \frac{N(N+1)(2N+1)}{6} \). Using this formula, we have:\[ \sum_{j=1}^{N} j^2 = \frac{N(N+1)(2N+1)}{6} \]

Step 3: Calculate the value of the second sum

The formula for the sum of the first \( N^2 \) natural numbers is \( \frac{N^2(N^2 + 1)}{2} \). Thus, we have:\[ \sum_{j=1}^{N^2} j = \frac{N^2 (N^2 + 1)}{2} \]

Step 4: Compare the two expressions

We need to compare:\[ \frac{N(N+1)(2N+1)}{6} \] with\[ \frac{N^2 (N^2 + 1)}{2} \].Let's simplify the expressions:- For the first expression \( \frac{N(N+1)(2N+1)}{6} \) simplifies as is.- For the second expression, after expanding, we have \( \frac{N^4 + N^2}{2} \).Consider their growth:- The first expression grows as \( O(N^3) \).- The second expression grows as \( O(N^4) \), because \( N^4 \) becomes the dominant term as \( N \) becomes large.Thus, as \( N \) grows, the second sum will eventually be greater than the first.

Key Concepts

Sum of Squares

The concept of the sum of squares involves adding up the squares of a sequence of numbers. Specifically, when we talk about the sum of squares of natural numbers, we refer to adding the squares of the numbers starting from 1 up to \( N \).

The formula to find the sum of squares for the first \( N \) natural numbers is given by \[ \sum_{j=1}^{N} j^2 = \frac{N(N+1)(2N+1)}{6} \].

• This formula allows us to quickly calculate the sum without having to square each number and then add them up individually.
• Understanding this formula helps us solve problems involving comparisons of sums, like those in the exercise.

This sum grows proportionally to the cube of \( N \), often written as \( O(N^3) \) in asymptotic notation. This means as \( N \) becomes larger, this sum increases much like a cube function.

Natural Numbers

Natural numbers are the set of positive integers starting from 1. They are the counting numbers that we use when we begin to count objects: 1, 2, 3, and so on.

These numbers play a crucial role in various areas of mathematics because they are the foundation of arithmetic and number theory. When dealing with natural numbers:

• We often look for patterns or formulate rules that allow us to perform calculations efficiently, like summing them up.
• In problems like the one given, natural numbers are used to form sums that we need to evaluate and compare.

The exercise distinguishes between two different sequences of sums based on natural numbers: the sum of the first \( N \) squares and the sum of the first \( N^2 \) numbers.

Mathematical Proof

To solve mathematical problems, especially in discrete mathematics, we must prove which mathematical statement is true by comparing two quantities.

In proving which of two sums is larger, we use:

• Formulas that simplify extensive computations, such as those for the sum of squares and the sum of natural numbers.
• We determine the growth behavior of each expression through asymptotic analysis.

For instance, in the given exercise, the sum of squares grows as \( O(N^3) \), while the sum of the first \( N^2 \) natural numbers grows as \( O(N^4) \). Therefore, by observing their growth patterns, we prove that for large \( N \), the second sum becomes substantially larger.

This shows that mathematical proofs involve not just direct computation but also reasoning about the growth of quantities and their behavior as \( N \) increases.