Power Series & Taylor Series: Videos & Practice Problems

In recent lessons, we explored the concept of infinite series, and now we will focus on a specific type known as a power series. A power series is an infinite series that includes a variable, typically denoted as \( x \), and can be viewed as an infinite polynomial. This series is expressed in the form:

\[\sum_{n=0}^{\infty} c_n (x - a)^n\]

where \( c_n \) represents the coefficients, and \( a \) is the center of the power series. To illustrate this, consider the example where \( c_n = \frac{1}{n!} \). The first four terms of the corresponding infinite series can be calculated by substituting the first four values of \( n \) (from 0 to 3) into the series:

\[\sum_{n=0}^{3} \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} = \frac{11}{6}\]

Next, to derive the first four terms of the power series centered at \( a = 2 \), we substitute \( c_n \) and the center into the power series formula:

\[\sum_{n=0}^{\infty} \frac{1}{n!} (x - 2)^n\]

Calculating the first four terms involves substituting \( n = 0, 1, 2, 3 \):

\[\begin{align*}\text{For } n=0: & \quad \frac{1}{0!} (x - 2)^0 = 1 \\\text{For } n=1: & \quad \frac{1}{1!} (x - 2)^1 = (x - 2) \\\text{For } n=2: & \quad \frac{1}{2!} (x - 2)^2 = \frac{1}{2} (x - 2)^2 \\\text{For } n=3: & \quad \frac{1}{3!} (x - 2)^3 = \frac{1}{6} (x - 2)^3 \\\end{align*}\]

Thus, the first four terms of the power series are:

\[1 + (x - 2) + \frac{1}{2} (x - 2)^2 + \frac{1}{6} (x - 2)^3\]

This illustrates how power series can be thought of as infinite polynomials, where the coefficients \( c_n \) dictate the behavior of the series. As we progress, we will explore various operations involving power series, including convergence analysis, addition, subtraction, multiplication, division, differentiation, and integration. Additionally, we will learn how to represent functions using power series, which is one of their most significant applications.

To analyze the given power series, we start with the expression:

$$\sum_{n=0}^{\infty} \frac{(x + 1)^n}{(n + 1) \cdot 3^n}$$

First, we identify the center of the power series. Power series are typically expressed in the form $$\sum_{n=0}^{\infty} a_n (x - a)^n$$, where $$a$$ is the center. In this case, we have $$x + 1$$ raised to the power of $$n$$, which can be rewritten as $$x - (-1)$$. Thus, the center of this power series is at $$a = -1$$.

Next, we will find the first four terms of the series by substituting values of $$n$$ from 0 to 3:

1. For $$n = 0$$:

$$\frac{(x + 1)^0}{(0 + 1) \cdot 3^0} = \frac{1}{1 \cdot 1} = 1$$

2. For $$n = 1$$:

$$\frac{(x + 1)^1}{(1 + 1) \cdot 3^1} = \frac{x + 1}{2 \cdot 3} = \frac{x + 1}{6}$$

3. For $$n = 2$$:

$$\frac{(x + 1)^2}{(2 + 1) \cdot 3^2} = \frac{(x + 1)^2}{3 \cdot 9} = \frac{(x + 1)^2}{27}$$

4. For $$n = 3$$:

$$\frac{(x + 1)^3}{(3 + 1) \cdot 3^3} = \frac{(x + 1)^3}{4 \cdot 27} = \frac{(x + 1)^3}{108}$$

Combining these results, the first four terms of the power series are:

$$1, \quad \frac{x + 1}{6}, \quad \frac{(x + 1)^2}{27}, \quad \frac{(x + 1)^3}{108}$$

In summary, the power series is centered at $$-1$$, and the first four terms are:

$$1, \quad \frac{x + 1}{6}, \quad \frac{(x + 1)^2}{27}, \quad \frac{(x + 1)^3}{108}$$

In this problem, we analyze a power series defined as the sum from \( n = 0 \) to infinity of \( (-1)^n n! x^n \). The first step is to determine the center of the power series. Power series typically take the form \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where \( a \) is the center. In this case, since the series is expressed as \( x^n \) without any subtraction from \( x \), it can be rewritten as \( (x - 0)^n \). Therefore, the center \( a \) is \( 0 \).

Next, we need to find the first four terms of the series. We do this by substituting \( n = 0, 1, 2, \) and \( 3 \) into the series:

1. For \( n = 0 \): \[ (-1)^0 \cdot 0! \cdot x^0 = 1 \] 2. For \( n = 1 \): \[ (-1)^1 \cdot 1! \cdot x^1 = -x \] 3. For \( n = 2 \): \[ (-1)^2 \cdot 2! \cdot x^2 = 2x^2 \] 4. For \( n = 3 \): \[ (-1)^3 \cdot 3! \cdot x^3 = -6x^3 \]

Now, we can combine these terms to express the series up to the fourth term:

Thus, the first four terms of the series are:

1 - x + 2x² - 6x³

This series demonstrates the alternating nature of the terms due to the factor \( (-1)^n \), which causes the signs to alternate between positive and negative. The factorial growth in the coefficients also indicates that the series diverges for all \( x \neq 0 \). Overall, we have successfully identified the center of the power series and derived the first four terms.

In the study of power series, a crucial aspect is determining the convergence of the series, which depends on the values of \( x \). A power series converges only for specific values of \( x \), creating an interval defined by the inequality \( |x - a| < r \), where \( a \) is the center of the series and \( r \) is the radius of convergence. This radius indicates how far the series converges from its center.

To find the radius of convergence, we typically apply convergence tests such as the ratio test, root test, or geometric series test. The ratio test is particularly common for power series. It involves evaluating the limit of the absolute value of the ratio of successive terms in the series. For a series expressed as \( \sum_{n=0}^{\infty} n \cdot \frac{(x - 2)^n}{3^n} \), we can apply the ratio test by examining the limit:

$$ \lim_{n \to \infty} \left| \frac{(n + 1)(x - 2)^{n + 1}/3^{n + 1}}{n(x - 2)^n/3^n} \right| $$

This simplifies to:

$$ \lim_{n \to \infty} \left| \frac{(n + 1)(x - 2)}{3n} \right| $$

As \( n \) approaches infinity, the limit evaluates to \( \frac{1}{3} |x - 2| \). For convergence, we require:

$$ \frac{1}{3} |x - 2| < 1 $$

Multiplying both sides by 3 gives:

$$ |x - 2| < 3 $$

This indicates that the radius of convergence \( r \) is 3, meaning the power series converges for values of \( x \) within 3 units of the center \( a = 2 \). Thus, the series converges for \( x \) in the interval \( (2 - 3, 2 + 3) \) or \( (-1, 5) \).

It is important to note that the radius of convergence can vary. It may be zero if the limit diverges to infinity, or it may be infinite if the limit equals zero. Understanding these concepts is essential for analyzing the behavior of power series and their convergence properties.

To determine the radius of convergence for the series given by:

\[\sum_{n=0}^{\infty} \frac{2^n (x - 1)^n}{n!}\]

we will apply the ratio test, a common method for analyzing the convergence of series. The ratio test involves examining the limit of the absolute value of the ratio of consecutive terms in the series as \( n \) approaches infinity.

We start by setting up the limit:

\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]

where \( a_n = \frac{2^n (x - 1)^n}{n!} \). Thus, we have:

\[a_{n+1} = \frac{2^{n+1} (x - 1)^{n+1}}{(n+1)!}\]

Now, substituting \( a_{n+1} \) and \( a_n \) into the limit gives us:

\[L = \lim_{n \to \infty} \left| \frac{2^{n+1} (x - 1)^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n (x - 1)^n} \right|\]

By simplifying this expression, we can rewrite it as:

\[L = \lim_{n \to \infty} \left| \frac{2^{n+1}}{2^n} \cdot \frac{(x - 1)^{n+1}}{(x - 1)^n} \cdot \frac{n!}{(n+1)!} \right| = \lim_{n \to \infty} \left| \frac{2 (x - 1)}{n + 1} \right|\]

As \( n \) approaches infinity, the term \( n + 1 \) in the denominator grows without bound, causing the entire limit to approach zero:

\[L = \lim_{n \to \infty} \frac{2 (x - 1)}{n + 1} = 0\]

Since \( 0 < 1 \), the ratio test indicates that the series converges for all values of \( x \). Therefore, the radius of convergence \( R \) is infinite:

\[R = \infty\]

This means that the series converges for every real number \( x \). In conclusion, the radius of convergence for the given series is infinite, indicating that it converges everywhere.

To determine the radius of convergence for the series given by the sum from \( n = 0 \) to infinity of \( n^n x^n \), we can effectively utilize the Root Test. This approach is particularly suitable here due to the presence of both \( n \) and \( x \) raised to the power of \( n \).

We start by applying the Root Test, which involves calculating the limit:

This expression can be simplified as follows:

Recognizing that \( \sqrt[n]{|x^n|} = |x| \), we can rewrite the limit as:

As \( n \) approaches infinity, the term \( n \cdot |x| \) will also approach infinity unless \( |x| = 0 \). This leads us to the inequality:

However, since \( n \) tends to infinity, this inequality can never hold true for any non-zero value of \( |x| \). Therefore, the only scenario where the limit remains bounded is when \( |x| = 0 \).

Consequently, we conclude that the radius of convergence \( r \) for this series is:

This indicates that the series converges only at \( x = 0 \). Further practice with convergence tests will enhance understanding of these concepts.

To determine the radius of convergence for the series given by the expression from \( n = 1 \) to infinity of \( \frac{(x + 3)^n}{n \cdot 4^n} \), we will utilize the ratio test, a common method for analyzing the convergence of series.

We start by setting up the limit as \( n \) approaches infinity of the absolute value of the ratio of the terms of the series with \( n + 1 \) and \( n \). This can be expressed as:

\[L = \lim_{n \to \infty} \left| \frac{(x + 3)^{n + 1}}{(n + 1) \cdot 4^{n + 1}} \cdot \frac{n \cdot 4^n}{(x + 3)^n} \right|\]

By simplifying this expression, we can rewrite \( (x + 3)^{n + 1} \) as \( (x + 3)^n \cdot (x + 3) \) and \( 4^{n + 1} \) as \( 4^n \cdot 4 \). This leads to:

\[L = \lim_{n \to \infty} \left| \frac{(x + 3) \cdot n}{(n + 1) \cdot 4} \right|\]

Next, we can cancel \( (x + 3)^n \) and \( 4^n \) from the numerator and denominator, resulting in:

\[L = \lim_{n \to \infty} \left| \frac{n}{4(n + 1)} \cdot (x + 3) \right|\]

As \( n \) approaches infinity, the terms \( n \) and \( n + 1 \) both grow large, and since they are of the same degree, the limit simplifies to the ratio of their coefficients:

\[L = \left| \frac{1}{4} (x + 3) \right|\]

For the series to converge, we require that \( L < 1 \). Thus, we set up the inequality:

\[\left| \frac{1}{4} (x + 3) \right| < 1\]

Multiplying both sides by 4 gives:

\[|x + 3| < 4\]

This inequality can be interpreted in the standard form \( |x - a| < r \), where \( a = -3 \) and \( r = 4 \). Therefore, the radius of convergence \( R \) is:

\[R = 4\]

In conclusion, the radius of convergence for the given series is 4, indicating that the series converges for values of \( x \) within 4 units of -3.

The radius of convergence, denoted as \( r \), indicates how far a power series converges from its center \( a \). To determine the interval of convergence, we start by solving the inequality \( |x - a| < r \). This inequality helps us identify the range of \( x \) values for which the power series converges.

When analyzing the radius of convergence, three scenarios can arise:

1. **Zero Radius of Convergence**: If the limit leads to an inequality like \( \infty < 1 \), this is false, indicating \( r = 0 \). Consequently, the power series converges only at the center \( x = a \) and diverges elsewhere.

2. **Infinite Radius of Convergence**: If the limit results in \( 0 < 1 \), this is always true, leading to \( r = \infty \). Thus, the power series converges for all \( x \) values, from \( -\infty \) to \( +\infty \).

3. **Finite Radius of Convergence**: If the limit gives a finite number, such as \( |x + 3| < 4 \), we can solve this to find a specific range of \( x \). The power series converges within the interval \( (a - r, a + r) \) and diverges outside this range.

To illustrate, consider the power series \( \sum_{n=1}^{\infty} \frac{(x + 3)^n}{n \cdot 4^n} \). Previously, we determined the radius of convergence \( r = 4 \) from the inequality \( |x + 3| < 4 \). The center \( a \) is \( -3 \), leading to the interval of convergence \( (-7, 1) \) after solving the inequality algebraically or using a number line.

However, to finalize the interval of convergence, we must check the endpoints \( x = -7 \) and \( x = 1 \) for convergence. For \( x = -7 \), substituting into the series yields \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), an alternating series that converges by the alternating series test. For \( x = 1 \), the series simplifies to \( \sum_{n=1}^{\infty} \frac{1}{n} \), a harmonic series that diverges.

Thus, the interval of convergence is \( [-7, 1) \), indicating convergence at \( x = -7 \) and divergence at \( x = 1 \). This process of determining the interval of convergence through solving inequalities and testing endpoints is essential for understanding the behavior of power series.

To determine the interval of convergence for the series given by the sum from \( k = 0 \) to infinity of \( k! \cdot x^{2k} / 3^k \), we first need to find the radius of convergence using the ratio test. The ratio test is a common method for assessing the convergence of series.

We start by setting up the limit as \( k \) approaches infinity of the absolute value of the ratio of the terms of the series with \( k + 1 \) and \( k \). This can be expressed as:

\[\lim_{k \to \infty} \left| \frac{(k + 1)! \cdot x^{2(k + 1)} / 3^{k + 1}}{k! \cdot x^{2k} / 3^k} \right|\]

Rearranging this expression involves multiplying by the reciprocal of the denominator:

\[= \lim_{k \to \infty} \left| \frac{(k + 1)! \cdot x^{2(k + 1)} \cdot 3^k}{k! \cdot x^{2k} \cdot 3^{k + 1}} \right|\]

Next, we simplify the expression. Notably, \( (k + 1)! = (k + 1) \cdot k! \) and \( x^{2(k + 1)} = x^{2k} \cdot x^2 \). This leads to:

\[= \lim_{k \to \infty} \left| \frac{(k + 1) \cdot x^{2k} \cdot x^2 \cdot 3^k}{k! \cdot x^{2k} \cdot 3^{k + 1}} \right|\]

After canceling common terms, we are left with:

\[= \lim_{k \to \infty} \left| \frac{(k + 1) \cdot x^2}{3} \right|\]

As \( k \) approaches infinity, the term \( (k + 1) \) grows without bound, leading to:

\[\lim_{k \to \infty} \left| \frac{(k + 1) \cdot x^2}{3} \right| = \infty\]

This result indicates that the series diverges for any \( x \neq 0 \). Therefore, the radius of convergence \( r \) is zero, which implies that the series converges only at its center.

In this case, since the series is expressed in the form \( x^{2k} \) without any subtraction from a center \( a \), we identify the center as \( 0 \). Thus, the interval of convergence is simply the single point where the series converges:

\[\text{Interval of convergence: } \{0\}\]

In summary, the series converges only at \( x = 0 \) and diverges for all other values of \( x \).

To determine the interval of convergence for the series given by the sum from \( n = 1 \) to infinity of \( \frac{n!}{2 \cdot 4 \cdot 6 \cdots (2n)} x^n \), we first need to simplify the denominator. The product \( 2 \cdot 4 \cdot 6 \cdots (2n) \) can be rewritten as \( 2^n \cdot n! \). This is because we can factor out \( 2 \) from each term, resulting in \( 2^n \) multiplied by the factorial of \( n \). Thus, the series can be rewritten as:

\[\sum_{n=1}^{\infty} \frac{n!}{2^n \cdot n!} x^n = \sum_{n=1}^{\infty} \frac{x^n}{2^n}\]

This simplifies further to:

\[\sum_{n=1}^{\infty} \left(\frac{x}{2}\right)^n\]

This series is a geometric series with a common ratio \( r = \frac{x}{2} \). For a geometric series to converge, the absolute value of the common ratio must be less than 1:

\[\left|\frac{x}{2}\right| < 1\]

Multiplying both sides by 2 gives:

\[|x| < 2\]

This indicates that the radius of convergence is 2, centered at 0, leading to the interval of convergence being \( (-2, 2) \).

Next, we need to check the endpoints \( x = -2 \) and \( x = 2 \) to determine if they should be included in the interval. For \( x = -2 \), substituting into the series gives:

\[\sum_{n=1}^{\infty} \left(-1\right)^n\]

This is an alternating series that diverges because it does not approach zero. For \( x = 2 \), substituting gives:

\[\sum_{n=1}^{\infty} 1\]

This series also diverges since it sums to infinity. Therefore, both endpoints do not contribute to convergence.

In conclusion, the interval of convergence for the series is:

\[(-2, 2)\]

Power series are essential tools in mathematics, particularly for representing complex functions in a more manageable form. By expressing functions as power series, we can approximate and manipulate them more easily, which is crucial in various applications across calculus and analysis.

To represent a function as a power series, we often start with the geometric series, which is defined as:

$$ S = \frac{a}{1 - r} $$

where \( a \) is the first term and \( r \) is the common ratio. A convergent geometric series has a sum that converges when the absolute value of \( r \) is less than one.

For example, consider the function \( f(x) = \frac{1}{x} \). To express this function as a power series centered at \( a = 1 \), we need to rewrite it in the form of a geometric series. We start by manipulating the denominator:

$$ f(x) = \frac{1}{x} = \frac{1}{1 - (1 - x)} $$

Here, we can identify \( a = 1 \) and \( r = 1 - x \). Thus, the power series representation becomes:

$$ \sum_{n=0}^{\infty} (1 - x)^n $$

Next, we determine the interval of convergence for this power series using the geometric series test, which requires that the absolute value of \( r \) be less than one:

$$ |1 - x| < 1 $$

This inequality can be solved as follows:

$$ -1 < 1 - x < 1 $$

Rearranging gives:

$$ 0 < x < 2 $$

Thus, the interval of convergence is \( (0, 2) \).

However, we must also check the endpoints \( x = 0 \) and \( x = 2 \) to confirm whether they are included in the interval. Plugging in \( x = 0 \) into the power series results in:

$$ \sum_{n=0}^{\infty} 1^n $$

This series diverges since the common ratio \( r = 1 \) does not satisfy the convergence condition.

For \( x = 2 \), we have:

$$ \sum_{n=0}^{\infty} (-1)^n $$

This series also diverges, as the absolute value of the common ratio \( r = -1 \) is equal to one.

Consequently, the final interval of convergence for the power series representation of \( f(x) = \frac{1}{x} \) is \( (0, 2) \), excluding the endpoints. This process illustrates how to derive a power series from a function and determine its convergence, which is a fundamental skill in calculus.

To find the power series representation centered at \( x = 0 \) for the function \( f(x) = \frac{2}{2 + x} \), we start by rewriting the function in a form suitable for a geometric series. The geometric series formula is given by \( \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. We can manipulate the function as follows:

First, factor out a 2 from the denominator:

\[ f(x) = \frac{2}{2 + x} = \frac{2}{2(1 + \frac{x}{2})} = \frac{1}{1 + \frac{x}{2}} \]

Next, we rewrite this in the form of a geometric series:

\[ f(x) = \frac{1}{1 - (-\frac{x}{2})} \]

Here, we identify \( a = 1 \) and \( r = -\frac{x}{2} \). The power series representation can now be expressed as:

\[ f(x) = \sum_{n=0}^{\infty} a r^n = \sum_{n=0}^{\infty} \left(-\frac{x}{2}\right)^n = \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{2}\right)^n \]

This series is an alternating series, as indicated by the term \( (-1)^n \). Now, we need to determine the interval of convergence for this series. The series converges when the absolute value of the common ratio \( r \) is less than 1:

\[ \left| -\frac{x}{2} \right| < 1 \]

Removing the negative sign, we have:

\[ \frac{|x|}{2} < 1 \]

Multiplying both sides by 2 gives:

\[ |x| < 2 \]

This indicates that the interval of convergence is \( -2 < x < 2 \), or in interval notation, \( (-2, 2) \).

Next, we must check the endpoints \( x = -2 \) and \( x = 2 \) to determine if they are included in the interval of convergence. For \( x = -2 \):

\[ f(-2) = \sum_{n=0}^{\infty} (-1)^n \left(-1\right)^n = \sum_{n=0}^{\infty} 1 \]

This series diverges. Now for \( x = 2 \):

\[ f(2) = \sum_{n=0}^{\infty} (-1)^n \left(1\right)^n = \sum_{n=0}^{\infty} (-1)^n \]

This series also diverges. Since both endpoints diverge, they are not included in the interval of convergence.

In conclusion, the power series representation of the function \( f(x) = \frac{2}{2 + x} \) is:

\[ \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{2}\right)^n \]

And the interval of convergence is \( (-2, 2) \).

A power series is essentially a polynomial with an infinite number of terms, represented by the sum from \( n = 0 \) to infinity of \( c_n (x - a)^n \). A specific type of power series is the Taylor series, which utilizes the derivatives of a function to determine its coefficients. The Taylor series for a function \( f(x) \) centered at \( a \) is expressed as:

\[\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n\]

In this formula, \( f^{(n)}(a) \) denotes the \( n \)-th derivative of the function evaluated at the point \( a \), and \( n! \) (n factorial) is the product of all positive integers up to \( n \). The first few terms of the Taylor series can be derived by evaluating the function and its derivatives at the center \( a \).

For example, to find the Taylor series for \( f(x) = \ln(x) \) centered at \( a = 1 \), we start by calculating the derivatives:

• \( f(1) = \ln(1) = 0 \)
• \( f'(x) = \frac{1}{x} \) so \( f'(1) = 1 \)
• \( f''(x) = -\frac{1}{x^2} \) so \( f''(1) = -1 \)
• \( f'''(x) = \frac{2}{x^3} \) so \( f'''(1) = 2 \)

Using these derivatives, we can construct the first few terms of the Taylor series:

\[\begin{align*}\text{For } n = 0: & \quad \frac{0}{0!} (x - 1)^0 = 0 \\\text{For } n = 1: & \quad \frac{1}{1!} (x - 1)^1 = (x - 1) \\\text{For } n = 2: & \quad \frac{-1}{2!} (x - 1)^2 = -\frac{1}{2} (x - 1)^2 \\\text{For } n = 3: & \quad \frac{2}{3!} (x - 1)^3 = \frac{1}{3} (x - 1)^3 \\\end{align*}\]

Combining these terms, we can express the Taylor series as:

\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x - 1)^n\]

This series is an alternating series, as indicated by the alternating signs in the coefficients. Additionally, a special case of the Taylor series is the Maclaurin series, which is simply a Taylor series centered at zero. The Maclaurin series for a function is given by:

\[\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]

Understanding Taylor and Maclaurin series is crucial for approximating functions and analyzing their behavior near specific points. As you continue to practice, you'll become more adept at deriving these series for various functions.

The Maclaurin series for the function \( f(x) = e^x \) can be derived using the Taylor series centered at zero. The Maclaurin series is expressed as:

For the function \( e^x \), all derivatives evaluated at zero yield \( f^{(n)}(0) = 1 \). This leads to the series:

Calculating the first few terms, we find:

• For \( n=0 \): \( \frac{1}{0!} x^0 = 1 \)
• For \( n=1 \): \( \frac{1}{1!} x^1 = x \)
• For \( n=2 \): \( \frac{1}{2!} x^2 = \frac{x^2}{2} \)
• For \( n=3 \): \( \frac{1}{3!} x^3 = \frac{x^3}{6} \)
• For \( n=4 \): \( \frac{1}{4!} x^4 = \frac{x^4}{24} \)

Thus, the Maclaurin series for \( e^x \) can be succinctly written in summation notation as:

Next, to find the Maclaurin series for \( f(x) = e^{2x} \), we can utilize the previously established series for \( e^x \) by substituting \( 2x \) for \( x \). This results in:

Rearranging this expression gives:

Therefore, the Maclaurin series for \( f(x) = e^{2x} \) is:

This series effectively captures the behavior of the function \( e^{2x} \) around the point \( x = 0 \).

Understanding the convergence of Taylor and Maclaurin series is essential for analyzing their behavior. The process for determining the interval of convergence mirrors that of basic power series, utilizing convergence tests to establish an inequality that leads to the radius of convergence. This radius is then used to find the interval of convergence, which is further refined by testing the endpoints.

To illustrate this, consider the Taylor series for the function \( f(x) = \ln(x) \) centered at \( x = 1 \). The general term of the Taylor series can be expressed as:

$$ \sum_{n=0}^{\infty} \frac{(-1)^n (x - 1)^{n + 1}}{n + 1} $$

Next, we apply the ratio test to determine convergence. The ratio test involves calculating the limit as \( n \) approaches infinity of the absolute value of the ratio of consecutive terms. For our series, this leads to:

$$ \lim_{n \to \infty} \left| \frac{(x - 1)^{n + 2}}{n + 2} \cdot \frac{n + 1}{(x - 1)^{n + 1}} \right| < 1 $$

After simplification, we find:

$$ \lim_{n \to \infty} \frac{n + 1}{n + 2} \cdot |x - 1| = |x - 1| < 1 $$

This indicates that the radius of convergence is 1, leading to an interval of convergence from 0 to 2. To finalize this, we must test the endpoints \( x = 0 \) and \( x = 2 \).

At \( x = 0 \), substituting into the series yields:

$$ \sum_{n=0}^{\infty} \frac{(-1)^{2n + 1}}{n + 1} $$

This series diverges, as it resembles a harmonic series. Conversely, at \( x = 2 \), the series becomes:

$$ \sum_{n=0}^{\infty} \frac{(-1)^n}{n + 1} $$

This is an alternating series, and applying the alternating series test confirms convergence since the limit of the non-alternating part approaches zero and the terms decrease.

Thus, the interval of convergence for the Taylor series of \( \ln(x) \) is \( (0, 2] \), indicating divergence at 0 and convergence at 2. This process exemplifies the systematic approach to determining convergence for Taylor and Maclaurin series, which will be further explored in subsequent discussions.

In recent lessons, we have explored Taylor series and their application through Taylor polynomials, which serve as partial sums of a Taylor series. Taylor polynomials are particularly useful for approximating functions. A Taylor polynomial of degree \( n \), denoted as \( p_n(x) \), is constructed by summing the first \( n \) terms of the Taylor series centered at a specific point, often denoted as \( a \).

When the Taylor series is centered at zero, it is referred to as a Maclaurin series, and similarly, the corresponding Taylor polynomials are called Maclaurin polynomials. For example, to find the Maclaurin polynomials for the function \( f(x) = e^x \), we calculate \( p_0, p_1, p_2, \) and \( p_3 \).

The zeroth degree Maclaurin polynomial, \( p_0(x) \), is simply the value of the function at zero, which is \( f(0) = e^0 = 1 \). The first degree polynomial, \( p_1(x) \), includes the zeroth term and the first derivative evaluated at zero, yielding:

\[ p_1(x) = 1 + \frac{e^0}{1!}(x - 0) = 1 + x. \]

For the second degree polynomial, \( p_2(x) \), we add the second term, which involves the second derivative:

\[ p_2(x) = 1 + x + \frac{e^0}{2!}(x - 0)^2 = 1 + x + \frac{1}{2}x^2. \]

Finally, the third degree polynomial, \( p_3(x) \), incorporates the third derivative:

\[ p_3(x) = 1 + x + \frac{1}{2}x^2 + \frac{e^0}{3!}(x - 0)^3 = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3. \]

To approximate \( e^{0.2} \) using the third degree Maclaurin polynomial, we substitute \( x = 0.2 \) into \( p_3(x) \):

\[ p_3(0.2) = 1 + 0.2 + \frac{1}{2}(0.2)^2 + \frac{1}{6}(0.2)^3. \]

Calculating this gives an approximation of \( 1.2213 \). In contrast, directly calculating \( e^{0.2} \) yields \( 1.2214 \), demonstrating the effectiveness of the Taylor polynomial for approximation. Higher degree Taylor polynomials will yield even more accurate results, making them a powerful tool for function approximation.