Taylor series and interval of convergence

b. Write the power series using summation notation.

f(x) = cosh 3x, a = 0

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Recall the definition of the hyperbolic cosine function: $\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$. This is the Maclaurin series expansion (Taylor series at $a=0$) for $\cosh x$.

Since the function is $f(x) = \cosh 3x$, substitute \$3x$ in place of $x$ in the series expansion. This gives $f(x) = \sum_{n=0}^{\infty} \frac{(3x)^{2n}}{(2n)!}$.

Simplify the power inside the summation: $(3x)^{2n} = 3^{2n} x^{2n}$. So the series becomes $f(x) = \sum_{n=0}^{\infty} \frac{3^{2n} x^{2n}}{(2n)!}$.

Express the power series explicitly in summation notation centered at $a=0$ as $f(x) = \sum_{n=0}^{\infty} \frac{3^{2n}}{(2n)!} (x - 0)^{2n}$.

This is the power series representation of $\cosh 3x$ about $a=0$ in summation notation.

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