Answer:
Show that the Fourier series of \( f \) can be term-by-term differentiated.
To show this let's first recall the expression for the Fourier series of a function \( f \) with period \( 2 p \) given by:
\[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \left(\frac{n\pi x}{p}\right) + b_n \sin \left(\frac{n\pi x}{p}\right) \right \]
where the Fourier coefficients are given by:
\[ a_0 = \frac{1}{p} \int_{-p}^{p} f(x) \ dx \]
\[ a_n = \frac{1}{p} \int_{-p}^{p} f(x) \cos \left(\frac{n\pi x}{p}\right) \ dx \]
\[ b_n = \frac{1}{p} \int_{-p}^{p} f(x) \sin \left(\frac{n\pi x}{p}\right) \ dx. \]
Now let's consider the function \( g(x) = f'(x) \).
The Fourier series representation of \( g(x) \) is given by:
\[ g(x) = \frac{a_0'}{2} + \sum_{n=1}^{\infty} \left( a_n' \cos \left(\frac{n\pi x}{p}\right) + b_n' \sin \left(\frac{n\pi x}{p}\right) \right \]
where the Fourier coefficients \( a_n' \) and \( b_n' \) are obtained by differentiating the previous expressions for \( a_n \) and \( b_n \) with respect to \( x \).
Taking the derivatives term by term we have:
\[ a_n' = \frac{1}{p} \frac{d}{dx} \int_{-p}^{p} f(x) \cos \left(\frac{n\pi x}{p}\right) \ dx \]
\[ b_n' = \frac{1}{p} \frac{d}{dx} \int_{-p}^{p} f(x) \sin \left(\frac{n\pi x}{p}\right) \ dx. \]
By interchanging the order of differentiation and integration we can differentiate under the integral sign. Hence we obtain:
\[ a_n' = \frac{1}{p} \int_{-p}^{p} \frac{dfrac