Question

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\ln \left(\frac{3}{2}\right)$$

Short answer

Answer: The first four nonzero terms of the Taylor series expansion of ln(3/2) are approximately (1/2) - (1/8) + (1/24) - (1/64).

Step-by-step solution

Identify the appropriate Taylor series

The function provided is \(\ln(\frac{3}{2})\). To expand it using Taylor series, we need to rewrite it in the form of \(\ln(1+x)\). Observe that \(\ln(\frac{3}{2}) = \ln(1+\frac{1}{2})\); thus \(x = \frac{1}{2}\). The Taylor series expansion for \(\ln(1+x)\) is given as: $$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n$$ Now we can plug in \(x=\frac{1}{2}\) and find the first four nonzero terms.

Substitute the value of x and expand the series

Plugging in \(x=\frac{1}{2}\), we get: $$\ln \left(\frac{3}{2}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \left(\frac{1}{2}\right)^n$$ Now, let's expand the series up to the fourth nonzero term: $$\ln \left(\frac{3}{2}\right) \approx \frac{1}{1} \left(\frac{1}{2}\right)^1 - \frac{1}{2} \left(\frac{1}{2}\right)^2 + \frac{1}{3} \left(\frac{1}{2}\right)^3 - \frac{1}{4} \left(\frac{1}{2}\right)^4$$

Simplify the expression

Simplify the expression: $$\ln \left(\frac{3}{2}\right) \approx \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64}$$

Final result

Thus, the first four nonzero terms of the Taylor series expansion of \(\ln(\frac{3}{2})\) are: $$\ln \left(\frac{3}{2}\right) \approx \frac{1}{2} - \frac{1}{8} + \frac{1}{24} - \frac{1}{64}$$

Key concepts

Logarithmic Functions

Logarithmic functions are an important concept in mathematics, primarily used to describe the inverse process of exponentiation. A logarithm tells us what power we need to raise a specific base number to produce another number. For example, if we say that \ \( \log_b(a) = c \ \), it means that \ \(b^c = a\ \).

The natural logarithm, denoted as \ \( \ln(x)\ \), is a specific logarithmic function with a base of "e", where "e" is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is extensively used in mathematics due to its unique properties and appearance in many natural processes.

• Inverse Property: Natural logs undo exponentials: \ \( \ln(e^x) = x \ \).
• Product Property: \ \( \ln(ab) = \ln(a) + \ln(b) \ \).
• Quotient Property: \ \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \ \).

Many problems in calculus and advanced mathematics, such as finding the value of \ \(\ln\left(\frac{3}{2}\right)\ \), involve manipulating and expanding logarithmic functions into series to approximate their values.

Series Expansion

Series expansion is a mathematical method used to represent functions as sums of simpler terms. This is particularly useful when evaluating functions that are too complex to deal with directly. One of the most popular series expansions is the Taylor series.

In the context of logarithmic functions, Taylor series takes the form of an expansion that can approximate a function, like \ \(\ln(1+x)\ \), by summing a sequence of polynomial terms. Each term depends on powers of \ \(x\ \) and the derivatives of the function at a single point.

The general formula for the Taylor series of \ \(\ln(1+x)\ \) is:
\[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \]

Where:

• "n" represents the term number.
• "x" is the variable input to the logarithmic function.
• The coefficients \ \( \frac{(-1)^{n+1}}{n} \ \) determine the sign and size of each term.

Using this series, one can approximate values for \ \(\ln\ \) when given an x-value, as with \ \(\ln\left(\frac{3}{2}\right)\ \). You simply insert \ \(x=\frac{1}{2}\ \) and sum the terms until a desired accuracy is achieved.

Calculus

Calculus is a branch of mathematics focused on change and motion, using derivatives and integrals to solve problems. It plays a crucial role in understanding concepts such as logarithmic functions and their series expansions.

The fundamental idea behind derivatives is finding the rate at which a function is changing at any point. This comes into play in Taylor series, where a function is expressed in terms of its derivatives. For a function like \ \(f(x)\ \), the Taylor series:

• Includes terms that are derived from the derivatives of \ \(f(x)\ \).
• Each term scales with a power of \ \(x\ \) to provide an approximation in a neighborhood close to a specified point.

With \ \(\ln(1+x)\ \), calculus helps us find each derivative necessary for building up the series from simpler components. Integrals in calculus sometimes work complementarily, calculating the whole from rates of change.

When approximating using series:

• Understanding the derivative relationships in \ \(\ln(1+x)\ \) allows calculation of successive terms.
• These contributions add precision to the approximated value, guiding us to solutions like the series expansion of \ \(\ln\left(\frac{3}{2}\right)\ \).

This iterative building block approach is fundamental in calculus, providing a solid foundation for more complex mathematical modeling and solution-finding.