To find the Maclaurin series for $\cosh(x)$, we can use the given definition:

$\cosh(x) = \frac{1}{2} \left( e^x + e^{-x} \right)$

We already know the Maclaurin series for $e^x$:

$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

The Maclaurin series for $e^{-x}$ can be obtained by replacing $x$ with $-x$ in the series for $e^x$:

$e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots$

Now, add the series for $e^x$ and $e^{-x}$:

$e^x + e^{-x} = \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\right) + \left(1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots\right)$

This simplifies to:

$e^x + e^{-x} = 2 + \frac{2x^2}{2!} + \frac{2x^4}{4!} + \cdots$

Finally, divide by 2 to get $\cosh(x)$:

$\cosh(x) = \frac{1}{2} \left(e^x + e^{-x}\right) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$

Thus, the Maclaurin series for $\cosh(x)$ is:

$\cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

Do you want more details or have any questions?

Here are some related questions:

1. What is the Maclaurin series for $\sinh(x)$?
2. How can you use the Maclaurin series to approximate $\cosh(x)$ for small values of $x$?
3. What are the derivatives of $\cosh(x)$ at $x = 0$?
4. How does the series for $\cosh(x)$ compare to the series for $\cos(x)$?
5. What is the radius of convergence for the Maclaurin series of $\cosh(x)$?

Tip: For even functions like $\cosh(x)$, only even powers of $x$ appear in the series expansion, making it easier to compute specific values.