Question

For some function \(f(x)\), the Maclaurin polynomial of degree 4 is \(p_{4}(x)=6+3 x-4 x^{2}+5 x^{3}-7 x^{4}\). What is \(f^{\prime \prime \prime}(0)\) ?

Short answer

The third derivative at zero, \(f^{(3)}(0)\), is 30.

Step-by-step solution

Understand the Maclaurin Polynomial

The Maclaurin polynomial is a Taylor series expansion of a function around 0. For degree 4, the polynomial is given as \(p_{4}(x) = 6 + 3x - 4x^{2} + 5x^{3} - 7x^{4}\). Each term corresponds to the derivatives of the function evaluated at 0.

Identify the Coefficient of Third-Degree Term

In the Maclaurin polynomial \(p_{4}(x)\), the coefficient of the \(x^{3}\) term is 5. This represents the term \(\frac{f^{(3)}(0)}{3!}x^3\) in the series expansion.

Calculate the Third Derivative at Zero

The formula for the coefficient of the \(x^{3}\) term in the Maclaurin series is \(\frac{f^{(3)}(0)}{3!}\). We have \(\frac{f^{(3)}(0)}{6} = 5\). Multiply both sides by 6 to solve for \(f^{(3)}(0)\):\[f^{(3)}(0) = 5 \times 6 = 30\].

Key concepts

Maclaurin Polynomial

A Maclaurin polynomial is a specific type of Taylor series expansion. It's used to approximate a function by summing its derivatives evaluated at 0. This creates a polynomial that represents the function near this point.
The Maclaurin polynomial is expressed as: \[ p_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n \]

• n: Degree of the polynomial
• fᵢ⁽ⁱ⁾(0): Derivatives evaluated at 0
• n! (n factorial): Product of all positive integers up to n

Understanding and identifying the coefficients is crucial for solving related problems, like finding specific derivatives.

Polynomial Expansion

Polynomial expansion is a method of expressing a function as a polynomial. This helps in simplifying and understanding the function's behavior near a certain point.
In the context of Maclaurin and Taylor series:

• Polynomial: A sum of terms with constant coefficients and variable powers.
• Expansion: The process of expressing a complex function in simpler polynomial terms.

For example, a function can be expanded around 0 to make calculations easier, especially when derivatives are used in applications of calculus or approximations. In solving such problems, identifying the degree of each term helps extract information like specific derivative values.

Third Derivative

The third derivative is the result of differentiating a function three times consecutively. It provides information about the function's curvature and rate of curvature change.
When working with Taylor series expansions:

• The coefficient of the third-degree term in the polynomial gives us \(\frac{f^{(3)}(0)}{3!}\).
• For example, if the coefficient is 5, as in the polynomial \(6 + 3x - 4x^2 + 5x^3\), it indicates that \[\frac{f^{(3)}(0)}{6} = 5\]

Solving for \(f^{(3)}(0)\), you multiply both sides by 6, yielding a result of 30. This highlights how derivatives relate directly to polynomial terms, making derivative extraction relatively straightforward with polynomial expressions.

Calculus

Calculus is a branch of mathematics focused on change and motion, involving differentiation and integration. It's a tool for analyzing the behavior of functions.

• Differentiation: Determines how functions change, providing derivatives which measure sensitivity to change.
• Integration: Opposite of differentiation, used to compute areas under curves or the total accumulation of quantities.

Through differentiation, Taylor and Maclaurin series can be created by repeatedly finding derivatives at a specific point (like 0 for Maclaurin polynomials). These tools provide ways to approximate functions, solve complex differential equations, and model natural phenomena efficiently. Understanding calculus can simplify many applied problems, offering insights in physics, engineering, and beyond.