Problem 32
Question
Expand \(g(x)\) as indicated. \(g(x)=(x-1)^{n} \quad\) in powers of \(x\).
Step-by-Step Solution
Verified Answer
To expand the given function \(g(x) = (x-1)^n\) in powers of x using the binomial theorem, we have: \[g(x) = \sum_{k=0}^n {n \choose k} (-1)^k x^{n-k}\] This expression represents \(g(x)\) expanded in powers of \(x\).
1Step 1: Apply the binomial theorem
We are given \(g(x) = (x-1)^n\). Let \(a = x\) and \(b = -1\). We can now apply the binomial theorem to expand this expression:
\[g(x) = \sum_{k=0}^n {n \choose k} x^{n-k}(-1)^k\]
2Step 2: Simplify the expression
Now we need to simplify this expression by incorporating the \((-1)^k\) term into the expansion, and then write \(g(x)\) in powers of \(x\). We have:
\[g(x) = \sum_{k=0}^n {n \choose k} (-1)^k x^{n-k}\]
This expression represents the function \(g(x)\) expanded in powers of \(x\).
Key Concepts
Expansion of expressionsPower functionsCombinatorial coefficients
Expansion of expressions
Expansion of expressions refers to rewriting algebraic expressions in an extended form. Using the Binomial Theorem, we expand expressions like \[(x-1)^n\]into a sum of terms with different powers of \(x\). The formula usually involves combinations and powers. Understanding the basics is crucial before diving into more complex expression expansions.
- Identify components: Find base terms like \(a = x\) and \(b = -1\).
- Apply Binomial Theorem: Use it to form a series expansion.
- Repetitive process: Break down terms step-by-step.
Power functions
Power functions involve expressions with variables raised to a power, such as \(x^n\). When dealing with expansions, understanding the distribution of powers in each term is essential. Each term in an expanded expression has a power of \(x\) determined by the Binomial Theorem. This power decreases successively from the initial exponent \(n\) in the formula \[ {x}^{n-k} \]. Knowing how these power functions behave is important, as they give shape to the polynomial resulting from an expansion. Understanding power functions helps predict the expansion's behavior:- **Exponent Relation**: Each term's power relates to its position in the expansion.- **Decreasing Order**: Powers of \(x\) lessen as you move through the series.Recognizing how to distribute and simplify these powers helps unravel complex polynomials during expansion.
Combinatorial coefficients
Combinatorial coefficients, often expressed as \({n \choose k}\), also known as "n choose k," are a crucial part of expressions involving binomial expansions. They determine the coefficients that appear with each term in the expansion.These coefficients are derived from combinations, which tell us how many ways we can select \(k\) elements from \(n\) total elements, following the formula:\[ {n \choose k} = \frac{n!}{k!(n-k)!} \]The combinatorial coefficients assign weight to each term, indicating its relative contribution to the polynomial. Here’s how they work in practice:
- **Number of Terms**: Given \(n\), the expansion has \(n+1\) terms.
- **Coefficient Calculation**: Apply the formula to compute each coefficient for terms \(k=0\) to \(n\).
- **Symmetry Property**: Coefficients are symmetric, meaning \({n \choose k} = {n \choose n-k}\).
Other exercises in this chapter
Problem 31
Determine whether the series converges or diverse. $$\sum \frac{2+\sin k}{k^{2}}$$
View solution Problem 32
(a) Prove that if \(\sum_{k=0}^{\infty} a_{k}\) is a convergent series with all terms nonzero, then \(\sum_{x=0}^{\infty}\left(1 / a_{k}\right)\) diverges. (b)
View solution Problem 32
Use Taylor polynomials to estimate the following within 0.01. $$\cos 6^{\circ}$$
View solution Problem 32
Determine whether the series converges or diverges. $$\sum \frac{\ln k}{k^{5 / 4}}$$
View solution