Question

Determine the Taylor series about the point \(x_{0}\) for the given function. Also determine the radius of convergence of the series. \(\frac{1}{1-x}, \quad x_{0}=2\)

Short answer

Question: Find the Taylor series of the function \(\frac{1}{1-x}\) centered at the point \(x_0 = 2\), and determine its radius of convergence. Answer: The Taylor series of the function \(\frac{1}{1-x}\) centered at \(x_0 = 2\) is given by: $$\sum_{n=0}^{\infty} (-1)^n(x-2)^n$$ And the radius of convergence is \(1\).

Step-by-step solution

Find the derivatives of the function

First, we need to find the first few derivatives of the function: \(f(x) = \frac{1}{1-x}\) Find the first derivative: \(f'(x) = -\frac{1}{(1-x)^2}\) Find the second derivative: \(f''(x) = \frac{2}{(1-x)^3}\) Find the third derivative: \(f'''(x) = -\frac{6}{(1-x)^4}\) The pattern is clear, and we can write the general form for the nth derivative as: \(f^{(n)}(x) = (-1)^n\frac{n!}{(1-x)^{n+1}}\)

Evaluate the derivatives at \(x_0 = 2\)

Now, we need to evaluate these derivatives at \(x_0 = 2\): \(f(2) = \frac{1}{1 - 2} = -1\) \(f'(2) = -\frac{1}{(1 - 2)^2} = 1\) \(f''(2) = \frac{2}{(1 - 2)^3} = -2\) \(f'''(2) = -\frac{6}{(1 - 2)^4} = 6\) In general, we get: \(f^{(n)}(2) = (-1)^n n!\)

Plug the values into the Taylor series formula

Using the Taylor series formula, we can write the series as: $$\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} (-1)^n(x-2)^n$$

Determine the radius of convergence

To find the radius of convergence, we'll use the Ratio Test: $$\lim_{n\to\infty}\left|\frac{(-1)^{n+1}(x-2)^{n+1}}{(-1)^n(x-2)^n}\right|=|x-2|$$ For the series to converge, this limit must be less than 1: $$|x-2|<1$$ Therefore, the radius of convergence is \(1\). The interval of convergence is \((1, 3)\). In conclusion, the Taylor series of the function \(\frac{1}{1-x}\) centered at \(x_0 = 2\) is given by: $$\sum_{n=0}^{\infty} (-1)^n(x-2)^n$$ And the radius of convergence is \(1\).

Key concepts

Radius of Convergence

The radius of convergence is a term we use to describe the extent of the interval around the center point of a power series where the series is guaranteed to converge. In simpler terms, it tells us how far we can stray from a central point before our series stops making sense and starts diverging (going to infinity). Understanding the radius of convergence is crucial for functions represented by power series, like the Taylor series, because it dictates the region where the series accurately represents the function. For the Taylor series of the function \(\frac{1}{1-x}\), centered at \(x_0 = 2\), by using the Ratio Test for convergence, we find that the series will converge (i.e., stay well-behaved) when \(x\) is within 1 unit away from 2 – that is, between 1 and 3 – making the radius of convergence equal to 1.

Derivative of a Function

A derivative represents the rate at which a function is changing at any point. For those of you who drive, imagine your speedometer: it shows how fast you're going at any moment. That's what a derivative measures, but for functions instead of cars. When we say we 'take a derivative,' we are looking for this exact rate of change, which depends on where we are in the function. The process of finding the derivatives of the function \(\frac{1}{1-x}\) involved calculating how the function's values change as we move along the \(x\)-axis. As we continue taking higher-order derivatives (meaning derivatives of derivatives, and so on), patterns emerge, which helps in constructing the corresponding Taylor series. In our exercise, after finding several derivatives of our function, a clear pattern allowed us to express the \(n\)th derivative as a formula involving factorial \(n!\).

Power Series

A power series is like a polynomial with potentially infinite terms that follows a specific structure: coefficients multiplied by increasing powers of \(x\), typically centered around a point \(x_0\). Each term looks a bit like \(a_n(x-x_0)^n\), where \(a_n\) is the coefficient for the \(n\)th term. It's an incredibly flexible way to represent a wide range of functions, and, importantly, it gives us a way to do calculations with functions that would be otherwise difficult to handle. For the given function \(\frac{1}{1-x}\), we expressed it as a Taylor series, which is a kind of power series with coefficients derived from the function's derivatives at the point \(x_0\). When our series is centered at \(x_0 = 2\), each coefficient is determined by evaluating the derivatives at that specific point, leading to our final power series representing the function in a certain interval.

Ratio Test for Convergence

Determining whether a series will converge or diverge can be tricky, but the Ratio Test is here to help us out. The test checks the ratio of consecutive terms in a series. As we head towards infinitely many terms (which is what makes up a series after all), if this ratio is less than 1, the series converges; if it's more than 1 or oscillates without settling down, the series diverges. Applying this test to a power series gives us not just a yes-or-no answer to convergence but also tells us something about where the series converges in relation to the center point, leading us back to our radius of convergence. This is exactly what we did for the Taylor series of \(\frac{1}{1-x}\) to determine that the series converges when the absolute value of \(x-2\) is less than 1.