Problem 6 - Discrete Mathematics with Applications

Question

Using the binomial theorem, expand each. $$(x-y)^{5}$$

Step-by-Step Solution

Answer

Using the binomial theorem, we expand the expression \((x-y)^{5}\) as: \[(x-y)^{5} = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5\]

Step 1: Recall the Binomial Theorem

The binomial theorem states that for any non-negative integer \(n\) and any real numbers \(a\) and \(b\), \[ (a + b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}a^0 b^n \] Here, \(n=5\), \(a=x\), and \(b=-y\). We must substitute these values into the binomial theorem formula.

Step 2: Find the Binomial Coefficients

We will compute the binomial coefficients for our expansion using the following formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] For \(n=5\), we need to find \(\binom{5}{0}\), \(\binom{5}{1}\), \(\binom{5}{2}\), \(\binom{5}{3}\), \(\binom{5}{4}\), and \(\binom{5}{5}\): \[ \binom{5}{0} = 1,\, \binom{5}{1} = 5,\, \binom{5}{2} = 10,\, \binom{5}{3} = 10,\, \binom{5}{4} = 5,\, \binom{5}{5} = 1 \]

Step 3: Apply the Binomial Theorem

Now we will combine the coefficients we found in step 2, with the corresponding powers of \(x\) and \(-y\), using the binomial theorem formula: \[ (x-y)^{5} = \binom{5}{0}x^5(-y)^0 + \binom{5}{1}x^4(-y)^1 + \binom{5}{2}x^3(-y)^2 + \binom{5}{3}x^2(-y)^3 + \binom{5}{4}x^1(-y)^4 + \binom{5}{5}x^0(-y)^5 \] Plug in the values of the binomial coefficients: \[ (x-y)^{5} = 1\cdot x^5(-y)^0 + 5\cdot x^4(-y)^1 + 10\cdot x^3(-y)^2 + 10\cdot x^2(-y)^3 + 5\cdot x^1(-y)^4 + 1\cdot x^0(-y)^5 \]

Step 4: Simplify the Expression

Simplify the expression by evaluating the powers of \(x\) and \(-y\) and multiplying by the coefficients: \[ (x-y)^{5} = x^5 + 5x^4(-y) + 10x^3(y^2) + 10x^2(-y^3) + 5x(y^4) + (-y^5) \] Now, simplify further by combining terms: \[ (x-y)^{5} = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5 \] So, the expanded form of \((x-y)^{5}\) is: \[ (x-y)^{5} = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + 5xy^4 - y^5 \]

Key Concepts

Binomial Coefficients

Binomial coefficients are the numerical factors that appear in the polynomial expansion of a binomial expression. They are represented by the symbol \( \binom{n}{k} \), which is read as "n choose k". This notation arises from combinatorics and is used to find the number of ways to choose \( k \) items from \( n \) items without regard to order. The formula is given by:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, "!" denotes a factorial, which is the product of all positive integers up to a number. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
In the context of expanding a polynomial like \( (x-y)^5 \), calculating the binomial coefficients is crucial because they determine the weight of each term in the expansion. This is why you will often see these coefficients appearing alongside powers of the binomial terms in such expansions.

Polynomial Expansion

Polynomial expansion is the process of expressing a power of a binomial as a sum of multiple terms. Each term in the expansion involves a product of the binomial terms raised to increasingly smaller powers, while the other term raises to increasingly larger powers.
This is systematically outlined by the binomial theorem, which provides a formula to do so:

• \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)

In simpler terms, the formula tells you to sum up all the terms from \( k = 0 \) to \( n \), where each term is formed by multiplying the coefficient \( \binom{n}{k} \) by the powers \( a^{n-k} \) and \( b^k \).
For example, to expand \( (x-y)^5 \), you replace \( a \) with \( x \) and \( b \) with \( -y \), then compute each term using the binomial coefficients found earlier. Each term of the expansion becomes more intuitive if you understand that you are distributing the powers between the two components, \( x \) and \( -y \), in every possible way that adds up to \( n \) (which is 5 in this specific problem).

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of elements within a finite set. One major area of combinatorics involves calculating how many ways we can combine items in a specific order. This is fundamental for understanding binomial expressions, as it directly connects to computing the binomial coefficients.
Using our earlier example, the problem \( (x-y)^5 \) can be thought of in terms of combining \( x \) and \( -y \) in every possible way across 5 positions. Each "choice" impacts the coefficient size in the polynomial expansion.
In this way, combinatorics help us determine the variety and frequency of each type of term in any binomial expansion. Whether you are placing people in seats, arranging books on a shelf, or expanding binomials, combinational mathematics illuminates the path to understanding complex selection processes.