f(x)=(a_(0))/(2)+sum _(n=1)^infty (a_(n)cos(npi x)/(L)+b_(n)sin(npi x)/(L)) f(x)=(3)/(2)+sum _(n=1)^infty (0times cos(npi x)/(5)+{ (3(1-cosnpi ))/(npi ))sin(npi x)/(5)) f(x)=(3)/(2)+(6)/(pi )(sin(pi x)/(5)+(1)/(3)sin(3pi x)/(5)+(1)/(5)sin(5pi x)/(5)+... )

The given function is a Fourier series representation of a periodic function. The Fourier series is a way to represent a periodic function as an infinite sum of sines and cosines. In this case, the function is represented as a sum of sines with different frequencies and amplitudes.<br /><br />The coefficients $a_0$, $a_n$, and $b_n$ are constants that determine the shape of the function. The variable $x$ represents the independent variable, and $L$ is the length of the period of the function.<br /><br />In the given function, the coefficient $a_0$ is $\frac{3}{2}$, which represents the average value of the function over one period. The coefficient $a_n$ is 0, which means that the cosine term is not present in the function. The coefficient $b_n$ is $\frac{3(1-\cos n\pi)}{n\pi}$, which represents the amplitude of the sine term at each frequency.<br /><br />The function can be simplified by factoring out the common factor of $\frac{6}{\pi}$ from each term in the sum. This gives us the final expression for the function:<br /><br />$f(x) = \frac{3}{2} + \frac{6}{\pi} \left( \sin \frac{\pi x}{5} + \frac{1}{3} \sin \frac{3\pi x}{5} + \frac{1}{5} \sin \frac{5\pi x}{5} + \cdots \right)$<br /><br />This expression represents the function as an infinite sum of sines with different frequencies and amplitudes. The coefficients of the sines are determined by the Fourier coefficients $b_n$, which are calculated based on the original function.