Question

Use the Maclaurin series $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1

Short answer

Question: Find the Maclaurin series representation for the function \(\frac{1}{(1 - 4x)^2}\) using the given Maclaurin series \((1+x)^{-2}=1-2x+3x^2-4x^3+\cdots, \text { for } -1<x<1\). Answer: The Maclaurin series representation for the function \(\frac{1}{(1 - 4x)^2}\) is \(1+8x+48x^2+256x^3+\cdots\), and it is valid for \(-0.25<x<0.25\).

Step-by-step solution

Substitute -4x for x in the given Maclaurin series

Substitute \(-4x\) for \(x\) in the Maclaurin series of \((1+x)^{-2}\): $$\left(1+(-4x)\right)^{-2}=1-2(-4x)+3(-4x)^2-4(-4x)^3+\cdots$$

Simplify the series

Now, we want to simplify this series to get the Maclaurin series for the function \(\frac{1}{(1-4x)^2}\). Perform the arithmetic operations for the substitution: $$\left(1-4x\right)^{-2}=1+8x+48x^2+256x^3+\cdots$$

Write down the function and the final Maclaurin series

Now we have the Maclaurin series for the function we were trying to find: $$\frac{1}{(1-4x)^2}=1+8x+48x^2+256x^3+\cdots$$ This is the Maclaurin series representation for the function \(\frac{1}{(1-4x)^2}\) with the terms up to the x^3, and it is valid for \(-1<4x<1\) or \(-0.25<x<0.25\).

Key concepts

Calculus

Calculus, the branch of mathematics that deals with continuous change, is the foundation upon which series and expansion concepts are built. It consists of two complementary areas: differential calculus, which studies the rate of change of functions, and integral calculus, which studies the accumulation of quantities. In the context of Maclaurin and Taylor series, calculus is used to find the derivatives of functions, which are necessary to construct the coefficients of the series. These series serve as tools to approximate complicated functions with polynomials, making them easier to analyze and compute.

Series Expansion

A series expansion is the expression of a function as an infinite sum of terms that are calculated from the values of that function at a certain point. The Maclaurin series is a special type of series expansion, where the function is expanded into an infinite sum about the point zero. The capability to represent complex functions as a series is a powerful application of calculus, enabling functions to be described in simpler forms that can be easily manipulated for tasks such as approximation, integration, and solving differential equations. Maclaurin series is specifically useful in approximating functions with a sum of polynomial terms, particularly when the value of input is close to zero.

Power Series

A power series is an infinite sum of terms in the form of \(a_n x^n\), where \(a_n\) represents the coefficients and \(x^n\) represents the terms of the series. The Maclaurin series is an example of a power series centered at zero. Calculus allows us to determine the coefficients \(a_n\) of these series through differentiation. For example, to obtain the coefficients of the Maclaurin series for a function \(f(x)\), one would take the derivatives of \(f\) at \(x=0\) and divide by the factorial of the order of the derivative. This approach simplifies many complex functions, making them much more tractable in both theoretical and practical applications.

Convergence Interval

The convergence interval of a power series, such as the Maclaurin series, is the range of \(x\) values for which the series converges to the function it represents. In other words, within this interval, the sum of the series will approach the actual value of the function as more terms are added to the series. For the Maclaurin series of \(\frac{1}{(1-4x)^2}\), the convergence interval is \( -0.25 < x < 0.25\), as derived from the original series' requirement that \( -1 < x < 1\). It is important to find this interval to ensure the series will provide accurate approximations within its range. Outside of this interval, the series may diverge, meaning it won’t provide a good representation of the function.