1. **Problem statement:** Find the Fourier coefficients $a_n$ and $b_n$ for the function $f(x) = e^x$ on the interval $-1 < x < 1$ and determine the value of the Fourier series at $x=2$. 2. **Fourier series basics:** For a function with period 2, the Fourier series is given by $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(n \pi x) + b_n \sin(n \pi x)\right)$$ where $$a_n = \int_{-1}^1 e^x \cos(n \pi x) \, dx, \quad b_n = \int_{-1}^1 e^x \sin(n \pi x) \, dx$$ 3. **Cosine coefficients $a_n$:** Start with $$a_n = \frac{2}{2} \int_{-1}^1 e^x \cos(n \pi x) \, dx = \int_{-1}^1 e^x \cos(n \pi x) \, dx$$ Using integration by parts and the given steps, $$a_n = 2(-1)^n \sinh 1 - n^2 \pi^2 a_n$$ Rearranging, $$a_n + n^2 \pi^2 a_n = 2(-1)^n \sinh 1$$ $$a_n(1 + n^2 \pi^2) = 2(-1)^n \sinh 1$$ Therefore, $$a_n = \frac{2(-1)^n \sinh 1}{1 + n^2 \pi^2}$$ 4. **Sine coefficients $b_n$:** Similarly, $$b_n = \int_{-1}^1 e^x \sin(n \pi x) \, dx$$ From the problem, $$b_n = 2(-1)^{n+1} n \pi \sinh 1 - n^2 \pi^2 b_n$$ Rearranging, $$b_n + n^2 \pi^2 b_n = 2(-1)^{n+1} n \pi \sinh 1$$ $$b_n(1 + n^2 \pi^2) = 2(-1)^{n+1} n \pi \sinh 1$$ Therefore, $$b_n = \frac{2(-1)^{n+1} n \pi \sinh 1}{1 + n^2 \pi^2}$$ 5. **Value of the Fourier series at $x=2$:** Since the Fourier series has period 2, the value at $x=2$ equals the value at $x=0$. $$f(0) = e^0 = 1$$ Thus, the Fourier series converges to 1 at $x=2$. **Final answers:** $$a_n = \frac{2(-1)^n \sinh 1}{1 + n^2 \pi^2}, \quad b_n = \frac{2(-1)^{n+1} n \pi \sinh 1}{1 + n^2 \pi^2}, \quad f(2) = 1$$