Question

the following periodic function with period (2pi) is defined by specifying its values on the interval (0,2pi). find the fourier coefficients (a_0,a_k), and (b_k) for the trigonometric polynomial (p_n) of degree (n) for the given function. plot (p_n) for various values of (n) over the domain (-2pi,4pi).

f(x)=\begin{cases}9e^{x},&0leq x < 2pi\\9,&x = 2piend{cases}

find the fourier coefficient (a_0).

(a_0=square) (type an exact answer, using (pi) as needed.)

Explanation:

Step1: Recall the formula for \(a_0\)

The formula for \(a_0\) in the Fourier series of a function \(f(x)\) with period \(2\pi\) is \(a_0=\frac{1}{\pi}\int_{0}^{2\pi}f(x)dx\). Since \(f(x) = 9e^{x}\) for \(0\leq x<2\pi\) and \(f(2\pi)=9\), but the integral of a function over an interval is not affected by a single - point value (because \(\int_{a}^{b}f(x)dx=\lim_{\epsilon
ightarrow0}\int_{a}^{b - \epsilon}f(x)dx\) when \(f(x)\) has a single - point discontinuity at \(x = b\)). So \(a_0=\frac{1}{\pi}\int_{0}^{2\pi}9e^{x}dx\).

Step2: Integrate \(9e^{x}\)

We know that \(\int e^{x}dx=e^{x}+C\). Then \(\frac{1}{\pi}\int_{0}^{2\pi}9e^{x}dx=\frac{9}{\pi}[e^{x}]_{0}^{2\pi}\).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus \(F(b)-F(a)\) (where \(F(x)\) is an antiderivative of \(f(x)\)), we have \(\frac{9}{\pi}(e^{2\pi}-e^{0})\).

Answer:

\(\frac{9(e^{2\pi}- 1)}{\pi}\)