Question

Representing functions by power series Identify the functions represented by the following power series. $$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{t}}$$

Short answer

$$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{t}}$$ Answer: The given power series represents the function: $$f(x) = \int \frac{1}{3^{t}}\cdot \frac{k(k+1)\cdot\left(\frac{-x}{3}\right)^{k}}{1+\frac{x}{3}} dx$$

Step-by-step solution

Find the general term of the series

First, write down the given series: $$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{t}}$$ The general term of this series can be represented as: $$a_k = (-1)^{k} \frac{k x^{k+1}}{3^{t}}$$

Decompose the expression

Notice that the given expression can be rewritten as a product of functions, which makes it easier to manipulate: $$a_k = k \cdot \frac{(-x)^{k+1}}{3^{k}} \cdot \frac{1}{3^{t}}$$ Now let's focus on the middle term \(\frac{(-x)^{k+1}}{3^{k}}\). Rewrite it as follows: $$\frac{(-x)^{k+1}}{3^{k}} = \left(\frac{-x}{3}\right)^{k+1}$$

Recognize geometric series

We can now rewrite the general term as: $$a_k = k \cdot \left(\frac{-x}{3}\right)^{k+1} \cdot \frac{1}{3^{t}}$$ Notice that the middle term is now the geometric series with a common ratio \(\frac{-x}{3}\). From this observation, we can continue to find the closed-form function for this series.

Differentiate the geometric series formula

The sum of an infinite geometric series can be represented by the formula: $$S = \frac{a}{1-r}$$ Where \(a\) is the first term, and \(r\) is the common ratio. In our case, the common ratio \(r=\frac{-x}{3}\). Now, let's differentiate the geometric series formula with respect to \(x\): $$\frac{dS}{dx} = \frac{d}{dx} \left(\frac{a}{1-r}\right)$$

Obtain function with differentiation

Before differentiating, let's rewrite the summation using our observation from Step 3: $$\sum_{k=1}^{\infty}(-1)^{k} \frac{k x^{k+1}}{3^{t}} = \frac{1}{3^{t}} \sum_{k=1}^{\infty} k \cdot \left(\frac{-x}{3}\right)^{k+1}$$ Next, differentiate both sides of the equation with respect to \(x\): $$\frac{d}{dx} \left[\frac{1}{3^{t}} \sum_{k=1}^{\infty} k \cdot \left(\frac{-x}{3}\right)^{k+1} \right] = \frac{1}{3^{t}} \sum_{k=1}^{\infty} \frac{d}{dx} \left[ k \cdot \left(\frac{-x}{3}\right)^{k+1} \right]$$ Differentiating the right side with respect to \(x\) gives: $$\frac{1}{3^{t}} \sum_{k=1}^{\infty} k(k+1) \cdot \left(\frac{-x}{3}\right)^{k}$$ Now use the geometric series formula for the differentiated series: $$\frac{dS}{dx} = \frac{1}{3^{t}}\cdot \frac{k(k+1)\cdot\left(\frac{-x}{3}\right)^{k}}{1+\frac{x}{3}}$$

Integrate to find the original function

Finally, integrate both sides with respect to \(x\) to obtain the original function represented by the series: $$S(x) = \int \frac{1}{3^{t}}\cdot \frac{k(k+1)\cdot\left(\frac{-x}{3}\right)^{k}}{1+\frac{x}{3}} dx$$ $$S(x) = \int \frac{1}{3^{t}}\cdot \frac{\sum_{k=1}^{\infty}k(k+1)\cdot\left(\frac{-x}{3}\right)^{k}}{1+\frac{x}{3}} dx$$ The function represented by the given power series is as follows: $$f(x) = \int \frac{1}{3^{t}}\cdot \frac{k(k+1)\cdot\left(\frac{-x}{3}\right)^{k}}{1+\frac{x}{3}} dx$$

Key concepts

Geometric Series

A geometric series is a sequence where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. Consider this simple example: 2, 4, 8, 16. Here, each term is twice the previous one, so the common ratio is 2. In the series presented in the exercise,

• the sequence is defined by a common ratio of \( r = \frac{-x}{3} \).
• The series can be expressed generally as \( a, ar, ar^2, ... \) where \( a \) is the first term.

Understanding the concept of geometric series helps in summing these infinite sequences easily using the formula: \[ S = \frac{a}{1-r} \] provided that \( |r| < 1 \). This fundamental formula bypasses the need to add terms individually.

Differentiation of Series

Differentiation involves finding the rate at which a function changes, which for series like these, helps in transforming and manipulating them. To differentiate a power series:

• We differentiate term by term.
• This operation allows us to turn geometric series into forms that match our observed functions better.

Recall:

• The derivative of \( f(x) = \sum_{k=0}^{\infty} a_k x^k \) is \( f'(x) = \sum_{k=1}^{\infty} k a_k x^{k-1} \).

In the exercise, differentiating helped simplify and progress the series toward finding a function that represents it, showing how differentiation allows navigating through complex series efficiently.

Integration of Series

Integration is the reverse operation of differentiation and is used to find the original function from a derivative. With infinite series:

• Integration can help retrieve the function that the series represents.
• We integrate term by term within the same interval of convergence.

For example, integrating \( f'(x) \) returns us to \( f(x) \), thus revisiting the primary function from its differentiated form. In the given problem:

• The definite integration enabled the finding of a function expressed by the power series.

This process demonstrates the power of integration in bridging gaps between series representation and their corresponding functions.

Infinite Series

An infinite series sums an endless sequence of terms. Representing functions as infinite series allows us to approximate and study their behaviors in new ways. Key points include:

• Infinite series are foundational in calculus and analysis.
• Convergence is crucial - only when an infinite series converges to a certain value, does it provide meaningful results.

In the exercise, the series was infinite, providing a robust lens through which to interpret functions more flexibly. Knowledge of how an infinite series behaves can reveal insights into complex mathematical structures and functions, enabling deeper analytical underpinnings and easier calculation methods.