Question

Which of the following functions is/are expandable using the Maclaurin series?

• A. $\ln(1 + z)$
• B. $\ln z$
• C. $\frac{1}{z^2}$
• D. $\exp(z)$

Hint:

Remember, the Maclaurin series can be used for functions that are analytic (i.e., they have a well-defined derivative at the expansion point, typically \( z = 0 \)). Functions like \( \ln(1 + z) \) and \( \exp(z) \) are good examples of such functions.

Answer 1

The Correct Option is A, D

The Maclaurin series is a special case of the Taylor series, expanded around \( z = 0 \). The general form of the Maclaurin series for a function \( f(z) \) is: \[ f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f^{(3)}(0)}{3!}z^3 + \dots \] Step 1: Understanding each function - Option (A): \(\ln(1 + z)\) - The function \( \ln(1 + z) \) is commonly expanded using a Maclaurin series. The series expansion is valid for \( |z|<1 \), and the Maclaurin series is given by: \[ \ln(1 + z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \dots \] Thus, Option A is correct.
- Option (B): \(\ln z\) - The function \( \ln z \) cannot be expanded using the Maclaurin series around \( z = 0 \) because it is undefined at \( z = 0 \). Therefore, Option B is incorrect.
- Option (C): \( \frac{1}{z^2} \) - The function \( \frac{1}{z^2} \) has a singularity at \( z = 0 \), and therefore cannot be expanded around \( z = 0 \) using a Maclaurin series. Thus, Option C is incorrect.
- Option (D): \( \exp(z) \) - The exponential function \( \exp(z) \) (or \( e^z \)) has a well-known Maclaurin series expansion given by: \[ \exp(z) = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots \] This series is valid for all values of \( z \). Hence, Option D is correct.
Step 2: Conclusion The functions that are expandable using the Maclaurin series are \( \ln(1 + z) \) (Option A) and \( \exp(z) \) (Option D). Therefore, the correct answer is Option A and D.