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 power series representation of the function \(\frac{1}{(3+4x)^2}\) given the Maclaurin series for \((1+x)^{-2}\) is given by \(1 - 2x + 3x^2 - 4x^3 + \cdots\). Answer: The power series representation of the function \(\frac{1}{(3+4x)^2}\) is \[1 - \frac{8x}{3} + \frac{16x^2}{3} - \frac{256x^3}{27} + \cdots\]

Step-by-step solution

Rewrite the given function in the form \((1+x)^{-2}\)

Observe that we can rewrite the given function as follows: $$\frac{1}{(3+4x)^2} = \frac{1}{3^2(1+\frac{4x}{3})^2} = \frac{1}{9}\left(1+\frac{4x}{3}\right)^{-2}$$ Now, we have the function in the form \((1+x)^{-2}\) with \(x\) replaced by \(\frac{4x}{3}\).

Substitute the Maclaurin series

Replace \((1+x)^{-2}\) in the given Maclaurin series with our rewritten function: $$\frac{1}{9}\left(1+\frac{4x}{3}\right)^{-2} = \frac{1}{9}\left[\left(1+\frac{4x}{3}\right)^{-2}\right]$$ Then we have: $$\frac{1}{9}\left[1 - 2\left(\frac{4x}{3}\right) + 3\left(\frac{4x}{3}\right)^2 - 4\left(\frac{4x}{3}\right)^3 + \cdots\right]$$

Simplify the expression

Now, simplify the expression inside the brackets to obtain the power series representation of the given function: $$\frac{1}{9}\left[1 - \frac{8x}{3} + 3\left(\frac{16x^2}{9}\right) - 4\left(\frac{64x^3}{27}\right) + \cdots\right]=1 - \frac{8x}{3} + \frac{16x^2}{3} - \frac{256x^3}{27} + \cdots$$ The final power series representation of the function \(\frac{1}{(3+4x)^2}\) is: $$\frac{1}{(3+4x)^2} = 1 - \frac{8x}{3} + \frac{16x^2}{3} - \frac{256x^3}{27} + \cdots$$

Key concepts

Power Series Representation

A power series representation is a way to express a function as an infinite sum of terms, each involving increasing powers of a variable. This is particularly useful in calculus for solving problems and approximating complex functions.

In the original exercise, we want to find a power series representation for the function \( \frac{1}{(3+4x)^{2}} \), and we do this by using an existing known series: the Maclaurin series of \((1+x)^{-2}\).

The general form of a power series is \( a_0 + a_1x + a_2x^2 + a_3x^3 +\cdots \), where \( a_n \) are coefficients. By utilizing known series expansions and substituting them appropriately, we can transform complex functions into simpler series forms. This way, calculations become more manageable.

Understanding this process of representing a function as a power series is fundamental, as it helps in analyzing how functions behave across intervals. It enables mathematicians to use series for precise calculations, especially when exact answers are difficult to obtain.

Function Transformation

Function transformation involves changing the appearance or form of a mathematical function to achieve a different perspective or another desired result. This concept often entails modifying variables or parameters to express the function in a more usable form.

In the calculus problem presented, the function \( \frac{1}{(3+4x)^2} \) is transformed by expressing it in a form similar to the known Maclaurin series for \((1+x)^{-2}\). This transformation occurs by dividing both the numerator and denominator by \(3^2\) and substituting \(x\) with \(\frac{4x}{3}\).

Such transformations are essential in mathematics because they simplify complex expressions and allow us to apply known series, theorems, or rules. In calculus, being able to transform functions cleverly helps in solving integrals, derivatives, and understanding function behavior through approximations.

Calculus Problem-Solving

Calculus problem-solving involves various strategies and techniques to work through and find solutions to mathematical problems involving change and motion. The use of power series and function transformations forms an integral part of calculus problem-solving strategies.

In working through the problem provided, understanding the relation between the function \( \frac{1}{(3+4x)^2} \) and its power series representation allows us to handle the function more easily in a series form. This is extremely helpful when exact arithmetic solutions aren't feasible, providing approximations that can solve calculus problems effectively.

Problem-solving in calculus requires understanding the core principles like differentiation and integration, as well as the clever use of series expansions for approximations. By manipulating functions into more recognizable forms, students can tackle a wide range of problems, breaking them down into manageable pieces. This is crucial not just for solving equations, but also for visualizing how functions behave and predicting future outcomes.

In summary, power series offer a useful toolset within the broader, exciting world of calculus problem-solving.