Question

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

Short answer

Answer: The Taylor series for the function \(e^x\) at the point \(x_0 = 0\) is given by: $$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ The radius of convergence for the Taylor series is infinite, meaning the Taylor series converges for all values of \(x\).

Step-by-step solution

Find the derivatives of the function at the point \(x_0\)

For the given function \(e^x\), the nth derivative of the function is itself: $$ f^{(n)}(x) = e^x $$ Now, we need to evaluate the nth derivative at the point \(x_0 = 0\): $$ f^{(n)}(0) = e^0 = 1 $$

Determine the Taylor series terms

Now that we have evaluated the nth derivative at the point \(x_0\), we can use the Taylor series formula to write down the terms for the Taylor series. Since \(f^{(n)}(0) = 1\) for all \(n\), our Taylor series terms will be: $$ \frac{f^{(n)}(0)}{n!}(x-0)^n = \frac{1}{n!}x^n $$

Write down the Taylor series for the function

Using the terms we calculated in Step 2, we can now write down the Taylor series for the function \(e^x\) at the point \(x_0 = 0\): $$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$

Determine the radius of convergence

In order to determine the radius of convergence for the Taylor series, we can use the ratio test. For positive series, the ratio test states that the radius of convergence is given by the limit: $$R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}}$$ where \(a_n\) denotes the nth term of the series. In our case, we have: $$ a_n = \frac{x^n}{n!} \quad\text{and}\quad a_{n+1} = \frac{x^{n+1}}{(n+1)!} $$ Now, we calculate the limit: $$ R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{(n+1)!}{n!} \cdot \frac{x^n}{x^{n+1}} = \lim_{n \to \infty} \frac{(n+1)!x^n}{n!x^{n+1}} = \lim_{n \to \infty} \frac{(n+1)x^n}{x^{n+1}} = \lim_{n \to \infty} \frac{n+1}{x} $$ Since the limit is infinity, the radius of convergence \(R\) is infinite. This means that the Taylor series converges for all values of \(x\). To summarise, the Taylor series for the function \(e^x\) at the point \(x_0 = 0\) is given by: $$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ And the radius of convergence for the Taylor series is infinite, which means that the Taylor series converges for all values of \(x\).

Key concepts

Radius of Convergence

The radius of convergence is an important concept when dealing with power series like Taylor series.
The radius of convergence tells us the set of values where the series converges to a function.
To find the radius of convergence, we can often use the ratio test, which analyzes the behavior of the series' terms as they progress to infinity.

• For a series formed by terms \(a_n\), the ratio test indicates convergence if \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\).
• If this limit equals 1, the test is inconclusive; if it's greater than 1, the series diverges.

In the case of the exponential function \(e^x\), the Taylor series is \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\).
By applying the ratio test, it turns out that the radius of convergence is infinite:

• We take \(a_n = \frac{x^n}{n!}\) and \(a_{n+1} = \frac{x^{n+1}}{(n+1)!}\).
• Calculated limit \(\lim_{n \to \infty} \frac{n+1}{1} = \infty\), indicates the series converges for all real numbers of \(x\).

Thus, the Taylor series for \(e^x\) converges for every value of \(x\), allowing us to use it across all real numbers without restriction.

Exponential Functions

Exponential functions are a fundamental concept in mathematics and have applications across many fields.
They are typically expressed in the form \(f(x) = e^x\), where \(e\) is a mathematical constant approximately equal to 2.71828.
Exponential functions are unique because they have the same derivative as themselves:

• For the function \(e^x\), any derivative of \(e^x\) with respect to \(x\) is \(e^x\) itself.
• This property makes them particularly simple for expansion into Taylor series, as the series terms are always building off the same initial function.

Exponential functions grow rapidly, characterized by their increasing slope, which means they model processes like population growth or radioactive decay effectively.
The Taylor series for \(e^x\) at \(x=0\) is a perfect representation of this function:

• It leverages the fact that \(f^{(n)}(0) = 1\) for all derivatives \(n\), which simplifies the Taylor expansion.
• This results in the series form \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\), which is quite easy to calculate.

Understanding exponential functions through their Taylor series helps grasp why they are so powerful and versatile in practical applications.

Power Series

Power series are infinite series that can represent functions, especially around specific points of expansion, known as centers.
A typical power series looks like \(\sum_{n=0}^{\infty} a_n(x-c)^n\), where \(c\) is the center.
Power series are flexible and can approximate many functions when expanded around suitable points:

• Each term of the series incorporates a power of \(x\), multiplied by coefficients \(a_n\), allowing the function to closely align with its value near \(c\).
• Depending on the function, the series might converge for all values of \(x\) or only for values within a specific range — dictated by the radius of convergence.

In our exponential function example, \(e^x\) is approximated about the center \(x=0\), using the Taylor series formula:

• Each coefficient \(a_n\) represents the nth derivative of \(f\) at \(x=0\), divided by \(n!\).
• For \(e^x\), this simplifies heavily since each derivative at \(x=0\) equals 1, meaning \(a_n = \frac{1}{n!}\).

The ease of working with the Taylor series of exponential functions showcases the power of power series in mathematical analysis, providing precise approximations for functions and solving complex equations efficiently.