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Expand each of the following as a polynomial in powers of \(x-1\) a. \(f(x)=x^{3}-2 x^{2}+x-1\) b. \(f(x)=x^{3}+x+1\) c. \(f(x)=x^{4}\) d. \(f(x)=x^{3}-3 x^{2}+3 x\)

Short Answer

Expert verified
a. \((x-1)^3 + (x-1)^2\) b. \((x-1)^3 + 3(x-1)^2 + 4(x-1) + 3\) c. \((x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1\) d. \((x-1)^3\)

Step by step solution

01

Identifying Binomial Expansion Need

To expand each given polynomial in terms of powers of \(x-1\), we need to transform each function so that \(x\) is re-centered around 1. This is done by substituting \(x = y+1\) where \(y=x-1\).
02

Substitute and Simplify - Part (a)

For \(f(x) = x^3 - 2x^2 + x - 1\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \-2(x)^2 = -2(y+1)^2 = -2(y^2 + 2y + 1) = -2y^2 - 4y - 2 \x = y + 1\)Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 - 2y^2 - 4y - 2 + y + 1 - 1 \)Combine like terms:\(f(y+1) = y^3 + y^2 + 0(y) = (x-1)^3 + (x-1)^2\)
03

Substitute and Simplify - Part (b)

For \(f(x) = x^3 + x + 1\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \x = y + 1\) Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 + y + 1 + 1 \)Combine like terms:\(f(y+1) = y^3 + 3y^2 + 4y + 3 = (x-1)^3 + 3(x-1)^2 + 4(x-1) + 3\)
04

Substitute and Simplify - Part (c)

For \(f(x) = x^4\), substitute \(x = y+1\):\((x)^4 = (y+1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1 \\)Therefore, \(f(y+1) = y^4 + 4y^3 + 6y^2 + 4y + 1 = (x-1)^4 + 4(x-1)^3 + 6(x-1)^2 + 4(x-1) + 1 \)
05

Substitute and Simplify - Part (d)

For \(f(x) = x^3 - 3x^2 + 3x\), substitute \(x = y+1\):\((x)^3 = (y+1)^3 = y^3 + 3y^2 + 3y + 1 \-3(x)^2 = -3(y+1)^2 = -3(y^2 + 2y + 1) = -3y^2 - 6y - 3 \3x = 3(y + 1)= 3y + 3\)Gather terms:\(f(y+1) = y^3 + 3y^2 + 3y + 1 - 3y^2 - 6y - 3 + 3y + 3\)Combine like terms:\(f(y+1) = y^3 = (x-1)^3\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
The Binomial Theorem is a powerful mathematical tool used to expand expressions raised to a power, typically of the form \((a + b)^n\). It provides a formula that makes it straightforward to determine the coefficients of the expanded terms without direct multiplication.
The theorem itself states that:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula involves the binomial coefficients \( \binom{n}{k} \), which are calculated using factorials.
  • Applications: It's extremely useful for polynomial expansions, probability, and series calculations.
  • Coefficients: The coefficients are represented by the binomial coefficients, which specify how many different ways "n" can be chosen "k" times.
The Binomial Theorem can simplify complex polynomial calculations by providing a systematic way to expand binomial expressions.
Substitution Method
The Substitution Method is a straightforward technique utilized in mathematical problems, particularly in polynomial expansion and re-centering processes. In the context of expanding a polynomial in powers of \((x-1)\), substitution is essential to simplify calculations and shift the polynomial's center.
Powers Substitution: In our exercise, we substitute \(x = y + 1\) where \(y = x - 1\). This transformation allows us to express the polynomial terms more conveniently as powers of \(y\). For example, replacing \(x\) with \(y+1\) makes it easier to apply the binomial expansion.
  • Purpose: This changes the polynomial into a series centered at a different point, which in this case is 1.
  • Steps: Each term is modified so that all calculations happen in terms of \(y\), the transformed variable.
This method not only aids in simplification but also offers visual binary tree-like expansion via systematic arrangement in powers of \(y\).
Polynomial Simplification
Polynomial Simplification is a process where a complex polynomial expression is simplified into a more manageable form by combining like terms and reducing the expression's overall complexity.
Key Focus:
  • Combine Like Terms: Terms with the same power of the variable are added or subtracted to provide a simpler expression.
  • Organize Powers: Arrange terms in descending order of their degree, making the polynomial easy to read and understand.
  • Reduce Coefficients: By combining like terms, coefficients often shrink, thus simplifying the overall look and computation.
For example, in our expanded polynomials, after substituting \(x = y + 1\), the combined terms reveal the structure of the polynomial in powers of \(y\), allowing for an expression optimised for further calculations or insights.
Centered Polynomial
A Centered Polynomial refers to a polynomial expression that is restructured to focus around a particular point, typically making further calculations easier. In our context, the polynomials are centered at 1, demonstrated by converting them to powers of \((x-1)\).
Explanation: By substituting \(x = y+1\), where \(y = x-1\), the new polynomial expresses its terms about point 1. This is beneficial because:
  • It simplifies derivatives: Derivatives of centered polynomials are often more straightforward.
  • It provides clarity: Seeing how each term relates to a new central point offers insight, especially in series expansions.
This process is especially useful in cases of power series expansions, providing a different perspective of polynomial structure and aiding in various mathematical computations.

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Most popular questions from this chapter

Show that \(\mathrm{x}=\mathrm{v}\) is the only solution to the equation \(\mathbf{x}+\mathbf{x}=2 \mathbf{v}\) in a vector space \(V\). Cite all axioms used.

Let \(a_{0}, a_{1}, \ldots, a_{n}\) be distinct numbers. If \(f(x)\) and \(g(x)\) in \(\mathbf{P}_{n}\) satisfy \(f\left(a_{i}\right)=g\left(a_{i}\right)\) for all \(i\), show that $f(x)=g(x) .

\(\mathbf{}\) Let \(a\) and \(b\) be two distinct numbers. Assume that \(n \geq 2\) and let $$U_{n}=\left\\{f(x) \text { in } \mathbf{P}_{n} \mid f(a)=0=f(b)\right\\}$$ a. Show that $$U_{n}=\left\\{(x-a)(x-b) p(x) \mid p(x) \text { in } \mathbf{P}_{n-2}\right\\}$$ b. Show that \(\operatorname{dim} U_{n}=n-1\). [Hint: If \(p(x) q(x)=0\) in \(\mathbf{P},\) then either \(p(x)=0\), or \(q(x)=0 .]\) c. Show \(\left\\{(x-a)^{n-1}(x-b),(x-a)^{n-2}(x-b)^{2}\right.\), \(\left.\ldots,(x-a)^{2}(x-b)^{n-2},(x-a)(x-b)^{n-1}\right\\}\) is a ba- sis of $U_{n} .

Let \(U\) and \(W\) be subspaces of \(V\) with bases \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) and \(\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}\right\\}\) respectively. If \(U\) and \(W\) have only the zero vector in common, show that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}, \mathbf{w}_{1}, \mathbf{w}_{2}\right\\}\) is independent.

Let \(V\) be the set of all infinite sequences \(\left(a_{0}, a_{1}, a_{2}, \ldots\right)\) of real numbers. Define addition and scalar multiplication by $$\left(a_{0}, a_{1}, \ldots\right)+\left(b_{0}, b_{1}, \ldots\right)=\left(a_{0}+b_{0}, a_{1}+b_{1}, \ldots\right)$$ and $$r\left(a_{0}, a_{1}, \ldots\right)=\left(r a_{0}, r a_{1}, \ldots\right)$$ a. Show that \(V\) is a vector space. b. Show that \(V\) is not finite dimensional. c. [For those with some calculus.] Show that the set of convergent sequences (that is, \(\lim _{n \rightarrow \infty} a_{n}\) exists) is a subspace, also of infinite dimension.

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