Problem 1 - Calculus Volume 1

Question

State whether the given sums are equal or unequal. $$a\sum_{i=1}^{10} i and \sum_{k=1}^{10} k$$ $$b. \sum_{i=1}^{10} i and \sum_{i=6}^{15}(i-5)$$ $$c. \sum_{i=1}^{10} i(i-1) and \sum_{j=0}^{9}(j+1) j$$ $$d. \sum_{i=1}^{10} i(i-1) \text { and } \sum_{k=1}^{10}\left(k^{2}-k\right)$$

Step-by-Step Solution

Answer

All given pairs of sums are equal.

Step 1: Evaluating Sum for Part a

First, calculate \( \sum_{i=1}^{10} i \). This is the sum of the first 10 natural numbers, which is \( \frac{10 \times 11}{2} = 55 \). Now calculate \( \sum_{k=1}^{10} k \), which is the same series as \( \sum_{i=1}^{10} i \) since the indices are dummy variables. Thus, both sums are equal to 55.

Step 2: Evaluating Sum for Part b

Calculate \( \sum_{i=1}^{10} i \) as done previously, which is 55. Now calculate \( \sum_{i=6}^{15} (i-5) \). Notice that \((i-5)\) translates the series to start from 1 such that \(\sum_{i=6}^{15} (i-5) = \sum_{j=1}^{10} j \) by letting \( j = i-5 \), giving \( \sum_{j=1}^{10} j = 55 \). Both sums are equal to 55.

Step 3: Transform and Compare for Part c

Calculate \( \sum_{i=1}^{10} i(i-1) \) which can be rewritten as \( \sum_{j=1}^{10} j(j-1) \). For \( \sum_{j=0}^{9} (j+1)j \), expand this to \( j^2 + j \). Both sums describe the same series \( \sum_{j=1}^{10} j(j-1) \), hence they are equal.

Step 4: Verifying Sum for Part d

Here, both sums \( \sum_{i=1}^{10} i(i-1) \) and \( \sum_{k=1}^{10} (k^2 - k) \) compute to the same expression because \( k^2 - k = k(k-1) \). Thus, both expressions represent the same sum and are equal.

Key Concepts

Understanding Natural Numbers

Natural numbers are a fundamental concept in mathematics. They are the basic counting numbers that begin from 1 and go onwards like 2, 3, 4, and so forth. Natural numbers are important as they form the building blocks for most mathematical operations involving summation.
In the context of summation, we're often interested in finding the sum of a sequence of natural numbers, which is a straightforward arithmetic series.
For example, the sum of the first 10 natural numbers can be calculated using the formula:\[S = \frac{n(n+1)}{2}\]where \(n\) is the number of terms.
For \(n = 10\), this is \( \frac{10 \times 11}{2} = 55 \). This pattern holds the same way in various series manipulations.

Role of Dummy Variables

When dealing with series notation, the variable used under the summation sign is often a placeholder. Such a placeholder is known as a "dummy variable".
The importance of dummy variables comes from their interchangeable nature. They do not influence the sum's actual values. For example, in \[\sum_{i=1}^{10} i\text{ and } \sum_{k=1}^{10} k\]although different letters \(i\) and \(k\) are used, they calculate the same result of 55.
The choice of letter here is arbitrary and does not change the sum. So, when you see different variables, know that it's the series structure, not the variable name, that's key.

Comparing Series

Series comparison involves examining two sums to determine if they yield the same result. Sometimes this requires deeper exploration beyond just the surface variables.
Consider the example: \[\sum_{i=1}^{10} i(i-1) \text{ and } \sum_{k=1}^{10} (k^2-k)\]Both expressions represent the same computation, because \(k^2 - k\) breaks down to \(k(k-1)\). Here, knowing transformations aids in making such comparisons.
In another instance, shifting indices or reinterpreting terms, like in \[\sum_{i=6}^{15}(i-5)\]and shifting it to start from 1, shows the sums are equivalent to original forms due to index adjustments.

Exploring Index Transformation

Index transformation is a third key concept, dealing with changing the index in a series while maintaining the sum's value. It's useful for aligning series or comparing different formulations.
Take \[\sum_{i=6}^{15}(i-5)\]To transform this, substitute \(j = i-5\), changing the starting point. The series now becomes:\[\sum_{j=1}^{10} j\]This seemingly complex transformation realigns to the straightforward sum of numbers from 1 to 10.
Understanding how to transform indices and recognize equivalent forms of series ensures that you evaluate sums correctly, especially when they don't appear similar at first glance.