A claim regarding Fourier Series

[Raihan amin]

TL;DR

Claim : "A periodic function $f(u)$ satisfying $\int_{0}^{1}f(u)du=0$ can generally expanded into a Fourier Series: $f(u)=\sum_{m=1}^{\infty} [a_m \sin{(2\pi mu)}+b_m\cos{(2\pi mu)}]$ "

This is written on Greiner's Classical Mechanics when solving a Tautochrone problem.
Firstly,I don’t understand why we didn’t use the term $m=0$
and Sencondly, how the integrand helps us to fulfill the Dirichlet conditions. That means,how do we know that the period is 1?Thanks

 

[BvU]

Hi,

why we didn’t use the term $m=0$

$\int_{0}^{1}f(u)du=0$ makes $b_0=0$. The coefficient $a_0$ is arbitrary, so $m=0$ is better left out.

Periodicity is built in through $\sin\left(2\pi m (u+1) \right )$ $= \sin\left(2\pi m u \right )$ and idem $\cos$.

 

[soloenergy]

for any help!Hi there,

I'm not sure what specific Tautochrone problem you are referring to, but I can try to help clarify some things.

To answer your first question, the term $m=0$ is typically used in physics to represent a massless particle or object. In classical mechanics, we often assume that particles have some nonzero mass, so it may seem strange to use $m=0$ in a problem. However, in certain situations, such as when dealing with a pendulum of negligible mass or a massless string, it can be useful to simplify the equations by setting $m=0$.

As for your second question, the integrand in a Tautochrone problem typically represents the kinetic energy of the system. By integrating this expression over time, we can determine the total energy of the system, which is constant due to the conservation of energy. This allows us to set up the Dirichlet conditions, which state that the period of the system must be the same regardless of the initial conditions. In other words, the system will always take the same amount of time to complete one cycle, regardless of how it is initially set up. By solving for the period, we can then determine that it is equal to 1.

I hope this helps! Let me know if you have any further questions.