1. The problem is to find the Maclaurin series (Taylor series at $x=0$) expansion of the function $f(x) = e^{\cos x}$. 2. Recall the Maclaurin series formula for a function $f(x)$: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$ where $f^{(n)}(0)$ is the $n$-th derivative of $f$ evaluated at 0. 3. We start by expanding $\cos x$ as its Maclaurin series: $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ 4. Substitute this into the exponential function: $$e^{\cos x} = e^{1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots} = e \cdot e^{-\frac{x^2}{2} + \frac{x^4}{24} - \cdots}$$ 5. Use the exponential series expansion for $e^u$ where $u = -\frac{x^2}{2} + \frac{x^4}{24} - \cdots$: $$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$$ 6. Up to the $x^4$ term, approximate $u$ and $u^2$: $$u = -\frac{x^2}{2} + \frac{x^4}{24}$$ $$u^2 = \left(-\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$$ 7. Substitute back: $$e^u \approx 1 + \left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{1}{2} \cdot \frac{x^4}{4} = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^4}{8}$$ 8. Combine like terms for $x^4$: $$\frac{1}{24} + \frac{1}{8} = \frac{1}{24} + \frac{3}{24} = \frac{4}{24} = \frac{1}{6}$$ 9. So, $$e^u \approx 1 - \frac{x^2}{2} + \frac{x^4}{6}$$ 10. Multiply by $e$: $$e^{\cos x} \approx e \left(1 - \frac{x^2}{2} + \frac{x^4}{6}\right) = e - \frac{e x^2}{2} + \frac{e x^4}{6}$$ This is the Maclaurin series expansion of $e^{\cos x}$ up to the $x^4$ term. Final answer: $$e^{\cos x} = e - \frac{e x^2}{2} + \frac{e x^4}{6} + \cdots$$