Question

Write the Taylor series for \(f(x)=\ln (1+x)\) about 0 and find the interval of convergence. Evaluate \(f\left(-\frac{1}{2}\right)\) to find the value of \(\sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}.\)

Short answer

Answer: The value of the function \(f\left(-\frac{1}{2}\right)\) is given by the infinite series \(\sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}\).

Step-by-step solution

Find the Taylor series for \(f(x)=\ln(1+x)\)

To find the Taylor series, we will first find the derivatives of the function, and then substitute \(x=0\) in them. Let's start by finding the series. 1. \(f(x) = \ln(1+x)\) 2. \(f'(x) = \frac{1}{1+x}\) 3. \(f''(x) = -\frac{1}{(1+x)^2}\) 4. \(f'''(x) = \frac{2}{(1+x)^3}\) 5. And so on... Notice a pattern in the higher derivatives: they consist of a factorial in the numerator (starting with 0-factorial), and a power of x in the denominator (starting with power 0). Now let's find the Taylor series.

Substituting \(x=0\) in derivatives and finding the Taylor series

1. \(f(0) = \ln(1+0) = 0\) 2. \(f'(0) = \frac{1}{1+0} = 1\) 3. \(f''(0) = -\frac{1}{(1+0)^2} = -1\) 4. \(f'''(0) = \frac{2}{(1+0)^3} = 2\) 5. And so on... Using these values, we can write the Taylor series of the function as \(f(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} x^k\)

Find the interval of convergence

To find the interval of convergence, we will use the ratio test. Find the absolute value of \(\frac{a_{n+1}}{a_n}\), and set this expression less than 1. Let's start with the ratio test: \(R = \lim_{n\to\infty} \left|\frac{(-1)^{n+2} x^{n+1}}{(n+1)} \cdot \frac{n}{(-1)^{n+1} x^n}\right|\) Simplify the ratio: \(R = \lim_{n\to\infty} \frac{n}{n+1} |x| = |x|\) For the series to converge, we need \(R<1\), which means: \(|x| < 1\) This gives us the interval of convergence: \(-1 < x < 1\).

Evaluate \(f\left(-\frac{1}{2}\right)\)

Now, let's substitute \(x=-\frac{1}{2}\) into the Taylor series and find the value: \(f\left(-\frac{1}{2}\right) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\left(-\frac{1}{2}\right)^k = \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}\) Now we found the value of the infinite series, which is equal to the Taylor series evaluated at \(x=-\frac{1}{2}\): \(\sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}\)

Key concepts

Interval of Convergence

When working with power series like the Taylor series, understanding the interval of convergence is key. This interval indicates the set of values for which the series converges to a finite value. For the function given, \( f(x) = \ln(1+x) \), we used the ratio test to determine its interval of convergence.

The ratio test is a method that helps us find the range of \( x \) values where the series converges. Convergence is crucial because, outside this interval, the series may diverge, leading to incorrect or undefined values. To apply this test, calculate the limit of the absolute value of the ratio of subsequent terms in the series. If this limit, \( R \), is less than 1, the series converges.

For our series, the calculation leads to \( |x| < 1 \). This means the interval of convergence for the series \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} x^k \) is \(-1 < x < 1\). Understanding this ensures you know where the series accurately represents the function \( \ln(1+x) \).

Ratio Test

The Ratio Test is a powerful tool in calculus, particularly for figuring out where a series converges. The basic idea behind it is to compare the size of consecutive terms as \( n \) approaches infinity. For a series \( \sum a_n \), the test involves taking the limit \( R = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| \).

If \( R < 1 \), the series converges absolutely. If \( R > 1 \), or if \( R \) is infinite, the series diverges. If \( R = 1 \), the test is inconclusive, and you would need other methods to determine convergence.

In our example, the series is \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} x^k \), and applying the ratio test, we find \( R = |x| \). Therefore, for the series to converge, \( |x| < 1 \). This process provides a clear mechanism to ascertain the values of \( x \) where the series behaves well, linking directly back to our interval of convergence.

Logarithmic Functions

Logarithmic functions, such as \( \ln(x) \), are essential in calculus and mathematical analysis due to their unique properties, such as converting multiplicative processes into additive ones. Consider \( f(x)=\ln(1+x) \): this is a specific form of logarithmic function that allows us to model growth, decay, and understand various limits.

When expanding \( \ln(1+x) \) into a Taylor series, we're approximating the function by an infinite sum of terms derived from its derivatives. The series begins with \( \ln(1) = 0\). The derivatives alternate in sign, reflecting the alternating nature of the logarithm as you move further from 0.

• The first derivative \( f'(x) = \frac{1}{1+x} \)
• The second derivative \( f''(x) = -\frac{1}{(1+x)^2} \)

Ultimately, this series is constructed by taking these derivatives at 0 and scaling by \( x^k/k! \). This approach efficiently captures the behavior of the logarithmic function near \( x = 0 \) and explains why logarithms are so crucial in expanding and simplifying complex exponential calculations in both pure and applied math.