A function \(f\) has Maclaurin series given by \(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...+ \frac{x^{2n}}{(2n)!} + ... \). Which of the following is an expression for \(f(x)\) ?

Summary Submit Skip Question

The expansion of \(e^x\) is:

$$ 1+x+\frac{x^2}{2!}+\frac{x^3}{3!} + ...$$

If we want only the even powers, we can add the expansion of \(e^{-x}\):

$$ 1-x+\frac{x^2}{2!}-\frac{x^3}{3!} + ...$$

Adding these two expansions result in:

$$ 2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+... $$

Dividing this by \(2\) results in the requested series.

$$ \frac{e^x-e^{-x}}{2} $$