Problem 35 - A Graphical Approach to Precalculus with Limits

Question

Write the binomial expansion for each expression. $$\left(\frac{m}{2}-1\right)^{6}$$

Step-by-Step Solution

Answer

Expand: \(\frac{m^6}{64} - \frac{3m^5}{8} + \frac{15m^4}{16} - 5m^3 + \frac{15m^2}{4} - 3m + 1\).

Step 1: Understand Binomial Expansion

The binomial theorem states that \((a+b)^n\) can be expanded into a series involving terms of the form \(C(n, k) \cdot a^{n-k} \cdot b^k\). Here, \(C(n, k)\) is the binomial coefficient given by \(\frac{n!}{k!(n-k)!}\).

Step 2: Identify Terms

For the expression \(\left(\frac{m}{2}-1\right)^{6}\), let \(a = \frac{m}{2}\) and \(b = -1\), with \(n = 6\). Identify these terms to apply the binomial theorem accordingly.

Step 3: Write General Term

The general term in the expansion is \(T_k = C(6, k) \cdot \left(\frac{m}{2}\right)^{6-k} \cdot (-1)^k\). You will calculate each term for \(k = 0\) to \(6\).

Step 4: Calculate Binomial Coefficients

For each \(k\), calculate the binomial coefficient \(C(6, k) = \frac{6!}{k!(6-k)!}\): - \(C(6, 0) = 1\)- \(C(6, 1) = 6\)- \(C(6, 2) = 15\)- \(C(6, 3) = 20\)- \(C(6, 4) = 15\)- \(C(6, 5) = 6\)- \(C(6, 6) = 1\)

Step 5: Calculate Each Term

Using \(T_k = C(6, k) \cdot \left(\frac{m}{2}\right)^{6-k} \cdot (-1)^k\):- \(T_0 = 1 \cdot \left(\frac{m}{2}\right)^6 = \frac{m^6}{64}\)- \(T_1 = 6 \cdot \left(\frac{m}{2}\right)^5 \cdot (-1) = -\frac{3m^5}{8}\)- \(T_2 = 15 \cdot \left(\frac{m}{2}\right)^4 = \frac{15m^4}{16}\)- \(T_3 = 20 \cdot \left(\frac{m}{2}\right)^3 \cdot (-1) = -5m^3\)- \(T_4 = 15 \cdot \left(\frac{m}{2}\right)^2 = \frac{15m^2}{4}\)- \(T_5 = 6 \cdot \left(\frac{m}{2}\right) \cdot (-1) = -3m\)- \(T_6 = 1 \cdot (-1)^6 = 1\)

Step 6: Write the Final Expansion

Combine all the terms calculated in Step 5:\[\left(\frac{m}{2} - 1\right)^{6} = \frac{m^6}{64} - \frac{3m^5}{8} + \frac{15m^4}{16} - 5m^3 + \frac{15m^2}{4} - 3m + 1\]

Key Concepts

Binomial Expansion

The concept of a binomial expansion stems from the Binomial Theorem, which allows us to expand expressions of the form \((a + b)^n\). This theorem is incredibly useful, especially when working with polynomials and algebraic expressions. The expansion involves summing up terms that contain binomial coefficients, powers of the first term \(a\), and powers of the second term \(b\).

In practice, each term of the expansion is represented as \(C(n, k) \cdot a^{n-k} \cdot b^k\), where:

• \(n\) is the exponent or the total number of terms minus one.
• \(k\) is the term number starting from zero.
• \(C(n, k)\) is the binomial coefficient, dictating the weight of each term.

The expansion does not only simplify complex exponential expressions but also prepares them for further operations like integration and differentiation.

Binomial Coefficient

The binomial coefficient is a key component of the binomial expansion. It provides the numerical value assigned to each term in the expansion. Mathematically, it is represented as \(C(n, k)\) and calculated using factorials: \[C(n, k) = \frac{n!}{k!(n-k)!}\]

Here, \(n!\) (read as \(n\) factorial) is the product of all positive integers up to \(n\). Factorials allow us to compute combinations or different ways to choose \(k\) elements from \(n\) without regard to order.

For example, in the provided exercise where \((\frac{m}{2} - 1)^6\) is expanded, the coefficients result from calculating each \(C(6, k)\), giving values like 1, 6, 15, and so forth. These coefficients determine the contribution of each term to the entire polynomial.

Polynomial Expansion

Polynomial expansion extends the binomial concept by taking it to expressions containing more than just two terms. It involves representing polynomials as a sum of multiple terms, each term being a product of coefficients and variables raised to powers.

For the exercise in question, after applying the binomial expansion, the result is a polynomial: \[\frac{m^6}{64} - \frac{3m^5}{8} + \frac{15m^4}{16} - 5m^3 + \frac{15m^2}{4} - 3m + 1\]

Each term stands as a part of a larger polynomial, emphasizing how complex expressions can be broken down into simpler, manageable parts. This facilitates understanding and manipulation of algebraic expressions in various mathematical contexts.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of elements within sets. It's particularly important in understanding binomial coefficients, as these coefficients represent combinations of sets.

When we calculate terms like \(C(6, 2)\) or \(C(6, 3)\), we are essentially determining how many ways we can choose 2 or 3 elements from a set of 6. This illustrates the link between combinatorics and the coefficients found in binomial expansions.

Combinatorial principles help categorize complex processes into more logical segments, making them easier to solve. It's fascinating to see how these principles manifest in various applications ranging from algebra to probability theory, emphasizing a universal mathematical language.