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Magazine Chapter 7: Problem 1
By computing derivatives, find the Maclaurin series for each function and state where it is valid. (a) \(\sinh z\), (b) \(\cosh z\). (c) \(\log (1+z)\).
Short Answer
Expert verified
(a) Series for \( \sinh z \) is valid for all \( z \); (b) \( \cosh z \) is valid for all \( z \); (c) \( \log(1+z) \) is valid for \( |z|<1 \).
Step by step solution
01
Understanding the Maclaurin Series
The Maclaurin series is a type of Taylor series centered at zero. The general form of a Maclaurin series for a function \( f(z) \) is \( f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \ldots \), where \( f^{(n)}(0) \) is the \( n \)-th derivative of \( f \) evaluated at \( z = 0 \).
02
Finding the Maclaurin Series for \( \sinh z \)
The function \( \sinh z = \frac{e^z - e^{-z}}{2} \). Compute the derivatives at \( z = 0 \):- 1st derivative: \( \sinh'(z) = \cosh z \), \( \cosh(0) = 1 \).- 2nd derivative: \( \sinh''(z) = \sinh z \), \( \sinh(0) = 0 \).- 3rd derivative: \( \sinh'''(z) = \cosh z \), \( \cosh(0) = 1 \).- The patterns repeat. Hence, the series is \( z + \frac{z^3}{3!} + \frac{z^5}{5!} + \ldots \), or in general terms, \( \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} \). This series is valid for all \( z \).
03
Finding the Maclaurin Series for \( \cosh z \)
The function \( \cosh z = \frac{e^z + e^{-z}}{2} \). Compute the derivatives at \( z = 0 \):- 1st derivative: \( \cosh'(z) = \sinh z \), \( \sinh(0) = 0 \).- 2nd derivative: \( \cosh''(z) = \cosh z \), \( \cosh(0) = 1 \).- 3rd derivative: \( \cosh'''(z) = \sinh z \), \( \sinh(0) = 0 \).- The patterns repeat. Hence, the series is \( 1 + \frac{z^2}{2!} + \frac{z^4}{4!} + \ldots \), or \( \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} \). This series is valid for all \( z \).
04
Finding the Maclaurin Series for \( \log(1+z) \)
The series for \( \log(1+z) \) can be derived using derivatives:- \( \log(1+z)' = \frac{1}{1+z} \), evaluated at \( z=0 \) gives \( 1 \).- \( \log(1+z)'' = -\frac{1}{(1+z)^2} \), evaluated at \( z=0 \) gives \( -1 \).- \( \log(1+z)''' = \frac{2}{(1+z)^3} \), evaluated at \( z=0 \) gives \( 2 \).- The general pattern is \( \frac{(-1)^{n+1}}{n}z^n \) for \( n \geq 1 \). Thus, the series is \( z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots \). This series is valid for \( |z|<1 \).
05
Conclusion and Validity
To summarize, the Maclaurin series expansions and their valid regions are:(a) \( \sinh z \): \( \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!} \), valid for all \( z \).(b) \( \cosh z \): \( \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!} \), valid for all \( z \).(c) \( \log(1+z) \): \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}z^n \), valid for \( |z|<1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
The Taylor series is a mathematical concept used to represent functions as an infinite sum of terms calculated from the values of its derivatives at a single point. This is particularly useful for approximating complex functions with infinite series. To derive a Taylor series, you need a function, say \( f(x) \), and a point \( a \) around which the approximation is centered. The series is expressed as:
\[f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \,\ldots\\]where \( f^{(n)}(a) \) denotes the \( n \)-th derivative of \( f \) at \( x = a \). When \( a = 0 \), this series is known as the Maclaurin series, a simplified form of Taylor series centered at zero.
\[f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \,\ldots\\]where \( f^{(n)}(a) \) denotes the \( n \)-th derivative of \( f \) at \( x = a \). When \( a = 0 \), this series is known as the Maclaurin series, a simplified form of Taylor series centered at zero.
- Maclaurin series is particularly handy for functions like exponential, trigonometric, and hyperbolic functions.
- The coefficients of each term are determined by the function's derivatives at the point \( a \).
- Provides a powerful method to approximate analytical functions in calculus.
Derivatives
Derivatives are at the core of calculus and are used to find the rate at which a function is changing at any given point. In the context of the Taylor series, derivatives tell us how the function behaves around a particular point, helping to construct each term of the series.
To compute a derivative, you apply the rules of differentiation to a function. For example, if you have a function \( f(x) \), its first derivative \( f'(x) \) gives the slope or rate of change of \( f(x) \) at any point \( x \). To find higher derivatives, you simply differentiate the derivative itself successively.
To compute a derivative, you apply the rules of differentiation to a function. For example, if you have a function \( f(x) \), its first derivative \( f'(x) \) gives the slope or rate of change of \( f(x) \) at any point \( x \). To find higher derivatives, you simply differentiate the derivative itself successively.
- The first derivative \( f'(x) \) measures the instant rate of change of the function.
- Second derivative \( f''(x) \) describes the curvature, showing how the rate of change is changing.
- Higher derivatives are used when Taylor or Maclaurin series expansions require more terms for accuracy.
- Accurate computation of derivatives allows series to better approximate the function around a point.
Convergence of Series
Convergence of a series is a crucial consideration when working with Taylor or Maclaurin series. It determines whether the sum of an infinite series approaches a definite value as more terms are added. For a Taylor series to be useful, it must converge to the function it represents within a particular range.
For convergence, a series must:
For convergence, a series must:
- Behave in such a way that as you take the limit as the number of terms grows, the series sum approaches a particular value.
- Depend on the region of convergence, affected by the function's behavior and singularities.
- In the case of Maclaurin series like \( \log(1+z) \), convergence is valid for \( |z| < 1 \) because the logarithmic function experiences issues near \( z = -1 \).
