Problem 13
Question
Prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=\sinh x $$
Step-by-Step Solution
Verified Answer
The Maclaurin series for the function \(f(x) = \sinh x\) is \(\sum_{n=0}^{\infty} \frac {x^{2n+1}}{(2n+1)!}\). This series converges to the function for all \(x\).
1Step 1: Understanding the Function and Maclaurin Series
The function in question is the hyperbolic sine function which is defined as \(\sinh x = \frac {e^x - e^{-x}}{2}\). The Maclaurin series for a function \(f(x)\) at \(x = 0\) is defined as \(\sum_{n=0}^{\infty} \frac {f^{(n)}(0)}{n!} x^n \) where \(f^{(n)}(0)\) is the \(n\)-th derivative of the function at \(x = 0\).
2Step 2: Computing the Derivatives of the Function
Calculate the first few derivatives of \(f(x) = \sinh x\) at \(x = 0\). The first derivative of \(f(x)\) is \(f'(x) = \cosh x\), and at \(x = 0\), \(f'(0) = 1\). The second derivative is \(f''(x) = \sinh x\) and \(f''(0) = 0\). The third derivative is \(f'''(x) = \cosh x\) and \(f'''(0) = 1\). The pattern repeats with even derivatives being 0 and odd derivatives being 1 at \(x = 0\).
3Step 3: Forming the Maclaurin Series
Using the definitions from Step 1 and the evaluations from Step 2, the Maclaurin series for \(f(x) = \sinh x\) becomes \(\sum_{n=0}^{\infty} \frac {f^{(n)}(0)}{n!} x^n = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ... = \sum_{n=0}^{\infty} \frac {x^{2n+1}}{(2n+1)!}\). This series includes all odd powers of \(x\).
4Step 4: Convergence of the Maclaurin Series
We notice that the series derived for \(f(x) = \sinh x\) is an example of an alternating series, so the Alternating Series Test can be applied. Since \( a_n = \frac {1}{(2n+1)!}\) is decreasing and the limit \(n \rightarrow \infty\) is 0, then the Alternating Series Test confirms the convergence of the series. Therefore, it can be concluded that the Maclaurin series of \(f(x) = \sinh x\) converges to the function for all \(x\).
Key Concepts
Hyperbolic FunctionsConvergenceSeries Expansion
Hyperbolic Functions
Hyperbolic functions are mathematical functions that are analogs of trigonometric functions. While trigonometric functions relate to circles and oscillatory behavior, hyperbolic functions deal with hyperbolas and exponential growth. The two most common hyperbolic functions are hyperbolic sine (\(\sinh x\)) and hyperbolic cosine (\(\cosh x\)). These functions help us describe the geometry of hyperbolic curves and model hyperbolic angles in mathematics and physics.
- The hyperbolic sine function, \(\sinh x\), is defined as \(\sinh x = \frac{e^x - e^{-x}}{2}\).
- Similarly, the hyperbolic cosine function, \(\cosh x\), is given by \(\cosh x = \frac{e^x + e^{-x}}{2}\).
- The hyperbolic sine function, like its circular counterpart, displays unique symmetry and properties. Importantly, it is odd as \(\sinh(-x) = -\sinh x\), and it describes the measure of hyperbolic angles in hyperbolic geometry.
- The hyperbolic cosine function is even, behaving like \(\cosh(-x) = \cosh x\), and it is always non-negative.
Convergence
Convergence is a critical concept in mathematics that describes the behavior of sequences and series. When dealing with infinite series like the Maclaurin series, we need to know whether they converge to a particular value or function as the number of terms increases to infinity.
The Alternating Series Test states that an alternating series \(\sum (-1)^n a_n\) converges if:
- A series converges if the sum of its terms approaches a finite limit as the number of terms increases.
- If a series does not converge, it diverges, meaning it does not settle on any fixed value.
The Alternating Series Test states that an alternating series \(\sum (-1)^n a_n\) converges if:
- The terms \(a_n\) are positive.
- The sequence \(a_n\) is decreasing, i.e., \(a_{n+1} \le a_n\).
- The limit of \(a_n\) as \(n \to \infty\) is 0.
Series Expansion
Series expansion helps represent complex functions as infinite sums of simpler terms. The power series, a type of series expansion, expresses functions using powers of \(x\). A common form of this is the Maclaurin series, which expands a function into a power series around \(x = 0\).
- The general formula for the Maclaurin series is \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\).
- Here, \(f^{(n)}(0)\) represents the \(n\)-th derivative of function \(f\) evaluated at 0.
- The first derivative is \(\cosh x\).
- The second derivative is \(\sinh x\).
- This alternates, showing zero for even derivatives at \(x = 0\) and non-zero for odd ones.
- \(\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}\),
Other exercises in this chapter
Problem 13
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{2 n}{n^{2}+1} $$
View solution Problem 13
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=x e^{x}, \quad n=4 $$
View solution Problem 13
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \sqrt{n}}{n+2} $$
View solution Problem 13
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ f(x)=\frac{2}{1-x^{2}}, \quad c=0 $$
View solution