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)=left{\begin{array}{ll}13 e^{x}, & 0 leq x<2 pi \\ 13, & x = 2 piend{array}
ight. )

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 Fourier series is \(a_0=\frac{1}{\pi}\int_{0}^{2\pi}f(x)dx\).

Step2: Split the integral

Since \(f(x) = 13e^{x}\) for \(0\leq x<2\pi\) and \(f(x) = 13\) at \(x = 2\pi\), and the integral of a function over an interval \([a,b]\) is not affected by the value of the function at a single - point. So \(a_0=\frac{1}{\pi}\int_{0}^{2\pi}13e^{x}dx\).

Step3: Integrate \(13e^{x}\)

We know that \(\int e^{x}dx=e^{x}+C\). So \(\frac{1}{\pi}\int_{0}^{2\pi}13e^{x}dx=\frac{13}{\pi}[e^{x}]_{0}^{2\pi}\).

Step4: Evaluate the definite integral

Using the fundamental theorem of calculus \(F(b)-F(a)\) where \(F(x)=e^{x}\), we have \(\frac{13}{\pi}(e^{2\pi}-e^{0})\).

Answer:

\(\frac{13(e^{2\pi}- 1)}{\pi}\)