Question

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$\frac{1}{(3+4 x)^{2}}$$

Short answer

Question: Find the first four nonzero terms of the Taylor series for the function \(f(x)=\frac{1}{(3+4x)^2}\) centered at \(x=0\). Answer: The first four nonzero terms of the Taylor series for the function \(f(x)=\frac{1}{(3+4x)^2}\) centered at \(x=0\) are \(\frac{1}{9} - \frac{8}{27}x + \frac{16}{27}x^2 - \frac{256}{243}x^3\).

Step-by-step solution

Rewrite the function using substitution

First, we rewrite the function \(\frac{1}{(3+4x)^2}\) in a form similar to \((1+x)^{-2}\): $$\frac{1}{(3+4x)^2}=\frac{1}{9}\cdot\frac{1}{\left(1+\frac{4}{3}x\right)^2} = \frac{1}{9}\cdot \frac{1}{\left(1 + u\right)^2},$$ where \(u=\frac{4}{3}x\).

Substitute the Taylor series for \((1+u)^{-2}\)

Now, we substitute the given Taylor series \((1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \cdots\) and use the substitution \(u=\frac{4}{3}x\): $$\frac{1}{9}\cdot \frac{1}{(1+u)^2} = \frac{1}{9}\left(1 - 2u + 3u^2 - 4u^3 + \cdots \right).$$

Replace \(u\) with \(\frac{4}{3}x\) and simplify

Now, we replace \(u\) with the initial substitution \(\frac{4}{3}x\) and simplify the expression: $$\frac{1}{9}\left(1 - 2\left(\frac{4}{3}x\right) + 3\left(\frac{4}{3}x\right)^2 - 4\left(\frac{4}{3}x\right)^3 + \cdots \right).$$ Expand each term: $$\frac{1}{9}\left(1 - \frac{8}{3}x + \frac{16}{3}x^2 - \frac{256}{27}x^3 + \cdots \right).$$

Multiply each term by the constant factor and find the first four nonzero terms

Finally, we multiply each term by the constant factor \(\frac{1}{9}\) and find the first four nonzero terms of the Taylor series: $$\frac{1}{9}\left(1-\frac{8}{3}x+\frac{16}{3}x^{2}-\frac{256}{27}x^{3}\right)=\frac{1}{9}-\frac{8}{27}x+\frac{16}{27}x^{2}-\frac{256}{243}x^{3}.$$ Thus, the first four nonzero terms of the Taylor series for the function \(\frac{1}{(3+4x)^2}\) centered at \(0\) are: $$\frac{1}{9} - \frac{8}{27}x + \frac{16}{27}x^2 - \frac{256}{243}x^3.$$

Key concepts

Power Series

A power series is an infinite series that is used to represent functions as a sum of terms with increasing powers of a variable. This is especially useful in calculus and mathematical analysis. In the context of finding the Taylor series, power series help us break down complex functions into simpler, piecewise components that are easier to work with and manipulate.

Power series generally look like this:

• Summation form: \(\sum_{n=0}^{\infty} a_n(x-c)^n\), where \(a_n\) are the coefficients of the series, \(x\) is the variable, and \(c\) is the center of the series.
• Essentially, a power series is like an expanded polynomial that can extend indefinitely.
• Power series converge within a certain range of \(x\), known as the radius of convergence.

By expressing functions as power series, mathematicians and students are able to approximate function values and conduct calculus operations like differentiation and integration more conveniently.

Substitution Method

The substitution method involves replacing a variable within a mathematical expression with another expression to simplify the mathematics involved. This approach is particularly useful when working with series like Taylor series, where the goal is to replace complex functions with more manageable series.

Here’s how substitution makes the Taylor series problem more straightforward:

• Identify the part of the original function that matches the known series structure—like the series for \((1+x)^{-2}\) in this example.
• Substitute part of the expression to align with this known form; replace \(x\) with another term, say \(u\), to make substitution clearer and manageable.
• The simplified series representation is then used to expand and re-substitute the original terms back, ultimately simplifying the full expression.

In the provided problem, by substituting \(u=\frac{4}{3}x\), the function \(\frac{1}{(3+4x)^2}\) becomes comparable to \((1+u)^{-2}\), allowing for the direct application of a known Taylor series.

Factoring Constants

Factoring constants out of an expression is a useful algebraic technique that can simplify calculations, especially when dealing with series. By factoring out constants, you align the expression closer to well-known series forms that can be further manipulated.

Here’s why factoring constants is important:

• It reduces the problem to simpler terms, so you can apply known series expansions more easily.
• Algebraic manipulation is more straightforward when constants are outside the main series operation.
• This technique avoids unnecessary compounding of terms during substitution steps or expansion processes.

In our exercise, the constant \(\frac{1}{9}\) was factored out from the expression \(\frac{1}{(3+4x)^2}\), simplifying the application of the Taylor series and making the resulting computation of terms much clearer.

Calculus

Calculus, the mathematical study of continuous change, underpins much of the work we do with power series and Taylor series. Concepts from calculus allow us to describe and work with rates of change and the behavior of functions more deeply.

Key relationships in calculus help when using Taylor series:

• Derivatives tell us about the slope or rate of change in functions, crucial when establishing Taylor series terms, which rely on function differentiation.
• Taylor series themselves are derived through repeated differentiation of a function and use calculus to sum these differentiated terms centered around a point.
• Understanding convergence and behavior of functions via derivatives enables more accurate usage and creation of series to approximate functions effectively.

In our exercise, calculus principles are less explicitly mentioned but are inherently part of devising the Taylor series itself, showcasing the fundamental role calculus plays in the broader understanding and application of series in mathematics.