Question

Differentiating Maclaurin Polynomials (a) Differentiate the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) and compare the result with the Maclaurin polynomial of degree 4 for \(g(x)=\cos x\). (b) Differentiate the Maclaurin polynomial of degree 6 for \(f(x)=\cos x\) and compare the result with the Maclaurin polynomial of degree 5 for \(g(x)=\sin x\). (c) Differentiate the Maclaurin polynomial of degree 4 for \(f(x)=e^{x}\). Describe the relationship between the two series.

Short answer

a) The derivative of the Maclaurin series of \(f(x)=\sin x\) with degree 5 matches with the Maclaurin series of \(g(x)=\cos x\) with degree 4. b) The derivative of the Maclaurin series of \(f(x)=\cos x\) with degree 6 corresponds to the negative of the Maclaurin series of \(g(x)=\sin x\) with degree 5. c) The derivative of the Maclaurin series of \(f(x)=e^{x}\) with degree 4 equals the original series minus the last term which is a characteristic property of \(e^{x}\).

Step-by-step solution

Differentiate the Maclaurin polynomial for \(f(x)=\sin x\)

The Maclaurin series for \(f(x)=\sin x\) is given by the expansion:\(f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)Differentiate this series to get a polynomial of degree 4: \(f'(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\)

Compare with Maclaurin polynomial for \(g(x)=\cos x\)

The Maclaurin series for \(g(x)=\cos x\) is given by the expansion:\(g(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\) It can be observed that the differentiated series for \(f(x)=\sin x\) is equal to the Maclaurin series for \(g(x)=\cos x\)

Differentiate the Maclaurin polynomial for \(f(x)=\cos x\)

The Maclaurin series for \(f(x)=\cos x\) is \(f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}\). Differentiate this series:\(f'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!}\)

Compare with Maclaurin polynomial for \(g(x)=\sin x\)

The Maclaurin series for \(g(x)=\sin x\) is: \(g(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\). The differentiated series for \(f(x)=\cos x\) is the negative of the Maclaurin series for \(g(x)=\sin x\)

Differentiate the Maclaurin polynomial for \(f(x)=e^{x}\)

The Maclaurin series for \(f(x)=e^{x}\) is: \(f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\). Differentiate this series: \(f'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}\)

Describe the relationship

The derivative of the Maclaurin series for \(f(x)=e^{x}\) is equal to the original Maclaurin series for \(f(x)=e^{x}\) minus the last term, reflecting the properties of exponentiation and differentiation of \(e^{x}\).

Key concepts

Maclaurin series

Understanding the Maclaurin series is pivotal when delving into the world of calculus. It represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point—in this case, at zero.

The Maclaurin series for some basic functions like sine, cosine, and the exponential function, are standard textbook examples, showing how power series can express elementary functions. With these series, students can see a pattern that repeats, allowing us to approximate these functions for small values of x. When differentiating these Maclaurin polynomials, fascinating relationships emerge, such as the derivative of the Maclaurin series of sine being equal to the Maclaurin series of cosine.

Calculus

Calculus, at its very core, is about understanding change. It comes in two flavors: differential calculus, which studies instantaneous rates of change, and integral calculus, focusing on the accumulation of quantities.

When we look at series such as those in the Maclaurin polynomials, we're engaging with differential calculus. By differentiating term by term, we obtain another series or polynomial that reveals underlying properties of the function we’re examining. In the case of trigonometric functions or the natural exponential function, observing how these series change upon differentiation helps us grasp the inherent relationships between these fundamental functions.

Series expansion

The term 'series expansion' refers to expressing a function as a sum of terms created from the function's values and derivatives at a certain point, often zero.

The Maclaurin series is a special type of series expansion where that point is zero. When we differentiate a series expansion, each term's power decreases by one, and for certain functions like exponentials, we find that the series remains remarkably similar to its original form. This characteristic makes series expansions a powerful tool for approximating and understanding functions.

The improvements provided by differentiating Maclaurin polynomials give insight into the function's behavior, showing the practical importance of series expansions in calculus.

Trigonometric functions

Trigonometric functions like sine and cosine are fundamental in mathematics, defining relationships within a circle. They are periodic, and their Maclaurin series also reflect this periodicity.

When we differentiate the Maclaurin series for \( \sin x \) and \( \cos x \) as in the exercise provided, we observe that the resulting series match up—highlighting a unique trait in trigonometric functions where the derivative of sine is cosine and the derivative of cosine is the negative sine. This interplay is an elegant demonstration of how differentiation affects these functions and helps solidify the understanding of their relationship.