Problem 21 - Calculus Single Variable

Question

In each of Exercises \(21-26,\) a function \(f\) and a point \(c\) are given. Use the equation $$ \frac{1}{1-(t-c)}=\sum_{n=0}^{\infty}(t-c)^{n}, \quad|t-c|<1 $$ together with some algebra, to express \(f(x)\) as a power series with base point \(c\). State the radius of convergence \(R\). \(f(x)=\frac{1}{6-x} \quad c=5\)

Step-by-Step Solution

Answer

The power series is \(\sum_{n=0}^{\infty}(-1)^n(x-5)^n\), with radius of convergence \(R=1\).

Step 1: Identify the Function Transformation

The given function is \( f(x) = \frac{1}{6-x} \). We need to express it in a form that matches the power series formula \( \frac{1}{1-(t-c)} \). To do this, start by rewriting \( f(x) \) as \( \frac{1}{1-(x-6)} \).

Step 2: Adjust Function for Base Point

We need the function to be in the form \( \frac{1}{1-(t-c)} \) where \( t-c = x - 5 \). Rewrite it as: \[ f(x) = \frac{1}{1-((x-5) - 1)} = \frac{1}{1 - ((x-5) -1)}.\] This makes \( t-c = (x-5) - 1 \).

Step 3: Apply Power Series Formula

Using the formula, substitute \( t-c = (x-5) - 1 \) into the series expansion: \[ f(x) = \sum_{n=0}^{\infty}((x-5) - 1)^{n}.\] Simplify this to: \[ \sum_{n=0}^{\infty}(-1)^n(x-5)^n.\]

Step 4: Determine Radius of Convergence

The series converges when \(|t-c| < 1\). In this context, \(|x-6| < 1\) must be satisfied. Since \(t-c = x-6\), the absolute value condition becomes \(|x-5|<1\), which indicates that the radius of convergence \(R\) is 1.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when dealing with power series. It indicates the interval around the base point where the series converges to a finite value. In practical terms, it means how far you can move from the center (base point) and still have the series sum up to the function it's meant to represent.

To find the radius of convergence, one usually analyzes the condition \(|t-c|
Ultimately, \(R=1\) here tells us that as long as x is within 1 unit from 5, our power series correctly represents the function. This radius is pivotal because beyond this interval, our series diverges, failing to match the function.

Series Expansion

Series expansion is like breaking a complex function into an infinite sum of simpler terms, which are easier to handle and compute. It lets us express functions in an elegant and often side-manageable form. With a good understanding of series expansion, you can tackle otherwise challenging problems with more ease.

The formula \(\frac{1}{1-(t-c)}=\sum_{n=0}^{\infty}(t-c)^{n}\) is the foundation for these expansions. This formula leverages the simple geometric series concept where each term builds upon the previous, but with powers of \(t-c\). Such expansions often involve rewriting the original function to fit the desired series form, like turning \(f(x) = \frac{1}{6-x}\) into a more relatable form \((1/(1-(x-6)))\).

• Rewriting the function helps us fit it into the power series equation.
• This sets the stage for applying the power series formula and computing terms.

In essence, series expansion allows mathematicians and scientists to approximate functions up to any desired accuracy by including enough terms from the series.

Algebraic Manipulation

Algebraic manipulation plays a key role in power series problems. It essentially enables the transformation of functions into forms that fit nicely with series formulas. Mastering this skill is about learning to rewrite expressions using algebra so that it becomes possible to use power series in practical situations.

For functions like \(f(x) = \frac{1}{6-x}\), algebraic manipulation helped us craft a structure that made using the power series formula possible. Adjusting it involved recognizing we could simplify and rewrite this fraction into a base form \(\frac{1}{1-(x-6)}\). From here, further algebraic tinkering aligned it perfectly with \(\frac{1}{1-(t-c)}\).

• Simplify and substitute parts of the function to fit the necessary form.
• Skillful rewriting can expose the series potential hidden within complex expressions.

Therefore, algebraic manipulation is about seeing and restructuring functions into a series-ready form, thus transforming a complex mathematical obstacle into merely another series expansion challenge.