Problem 63
Question
Let \(g(x)=p(x)+x^{n+1} f(x)\), where \(p(x)\) is a polynomial of degree at most \(n\) and \(f\) has derivatives through order \(n\) Show that \(p(x)\) is the Maclaurin polynomial of order \(n\) for \(g\).
Step-by-Step Solution
Verified Answer
\(p(x)\) is the Maclaurin polynomial of order \(n\) for \(g(x)\).
1Step 1: Understand the Problem
We need to show that the polynomial \(p(x)\), which is part of the function \(g(x) = p(x) + x^{n+1} f(x)\), is the Maclaurin polynomial of order \(n\) for \(g(x)\). This means that \(p(x)\) must contain the same terms as the Maclaurin series expansion of \(g(x)\) up to \(x^n\).
2Step 2: Define Maclaurin Polynomial
The Maclaurin polynomial of order \(n\) for a function \(g(x)\) is given by: \[ p_n(x) = g(0) + \frac{g'(0)}{1!} x + \frac{g''(0)}{2!} x^2 + \cdots + \frac{g^{(n)}(0)}{n!} x^n \]. We need \(p(x)\) to satisfy this form.
3Step 3: Determine Derivatives of \(g(x)\)
Consider the derivatives of \(g(x) = p(x) + x^{n+1} f(x)\). Since \(x^{n+1} f(x)\) has all derivatives up to order \(n\) that equal zero at \(x=0\), the contributing terms for the Maclaurin polynomial are solely from \(p(x)\), a polynomial of degree at most \(n\).
4Step 4: Conclude with Properties of Maclaurin Polynomial
Each derivative of \(x^{n+1} f(x)\) at 0 through the \(n ext{-th}\) is zero because the smallest power is \(x^{n+1}\). Therefore, the terms of the Maclaurin polynomial of \(g(x)\) up to degree \(n\) match with those of \(p(x)\). Thus, \(p(x)\) is indeed the Maclaurin polynomial of order \(n\) for \(g(x)\).
Key Concepts
Polynomial functionDerivativeSeries expansionFunction approximation
Polynomial function
A polynomial function is a mathematical expression consisting of variables, coefficients, and the operation of addition, subtraction, and multiplication. The variables in a polynomial have non-negative integer exponents.
For example, a polynomial function of degree at most \( n \) can be expressed as \( p(x) = a_0 + a_1x + a_2x^2 + \, \ldots \, + a_nx^n \).
This means the highest power of \( x \) is \( n \) in the polynomial, and that is what determines its degree. Understanding polynomials is crucial because they appear in numerous applications, making them one of the most fundamental tools in algebra.
For example, a polynomial function of degree at most \( n \) can be expressed as \( p(x) = a_0 + a_1x + a_2x^2 + \, \ldots \, + a_nx^n \).
This means the highest power of \( x \) is \( n \) in the polynomial, and that is what determines its degree. Understanding polynomials is crucial because they appear in numerous applications, making them one of the most fundamental tools in algebra.
Derivative
A derivative represents how a function changes as its input changes. It is the cornerstone of calculus, reflecting the rate of change or the slope of the function's graph at any point.
In the context of polynomials, derivatives are simpler as each term \( a_kx^k \) becomes \( ka_kx^{k-1} \) upon differentiation. Calculating derivatives of polynomial functions like \( p(x) \) is essential to derive Maclaurin polynomials efficiently.
- For a function \( f(x) \), the derivative \( f'(x) \) is calculated as the limit: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
In the context of polynomials, derivatives are simpler as each term \( a_kx^k \) becomes \( ka_kx^{k-1} \) upon differentiation. Calculating derivatives of polynomial functions like \( p(x) \) is essential to derive Maclaurin polynomials efficiently.
Series expansion
Series expansion is a method of expressing a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is a special case of the Taylor series, where we expand the function around \( x=0 \).
The Maclaurin series for a function \( g(x) \) is represented by:
\[ g(x) = g(0) + \frac{g'(0)}{1!} x + \frac{g''(0)}{2!} x^2 + \cdots + \frac{g^{(n)}(0)}{n!} x^n + \cdots \]
This series provides a polynomial approximation of \( g(x) \), allowing us to express complex functions in a simpler polynomial form. Understanding how to form such expansions is crucial for function approximations in calculus.
The Maclaurin series for a function \( g(x) \) is represented by:
\[ g(x) = g(0) + \frac{g'(0)}{1!} x + \frac{g''(0)}{2!} x^2 + \cdots + \frac{g^{(n)}(0)}{n!} x^n + \cdots \]
This series provides a polynomial approximation of \( g(x) \), allowing us to express complex functions in a simpler polynomial form. Understanding how to form such expansions is crucial for function approximations in calculus.
Function approximation
Function approximation is a technique used to estimate a function by using simpler, more easily calculable forms, such as polynomials.
When dealing with complex functions, approximations like the Maclaurin polynomial are invaluable because they simplify analysis and computation.
When dealing with complex functions, approximations like the Maclaurin polynomial are invaluable because they simplify analysis and computation.
- The Maclaurin polynomial, derived from the Maclaurin series, provides a polynomial approximation of a function around \( x=0 \).
- This approximation involves using the function's derivative values at 0.
Other exercises in this chapter
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