Question

Prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=\sinh x $$

Short 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 \(x\).

Step-by-step solution

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 \(x = 0\).

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 \(x = 0\).

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 \(x\).

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 \(x\).