The delta Dirac function can be presented in the form of Fourier series expansion as $$ \delta(x)=\frac{1}{2\pi} +\frac{1}{\pi} \sum_{n\ge 1} \cos (nx) \, . $$ The proof is straightforward and can be found e.g. in this Mathworld Wolfram article.
In a mathematical physics problem, I would like to make use of the Fourier series technique in order to solve a system of differential equations involving a delta Dirac function, such that the solution has the form $$ f(x) = \sum_{n\ge 1} f_n \cos \left( (2n-1)x \right) \, . $$ That is, only the odd indices have to be need to be included in the Fourier series representation of the delta Dirac function.
What I tries is: $$ \delta(x) = \frac{2}{\pi} \sum_{n\ge 1} \cos \left( (2n-1)x \right) \, , $$ which seems to lead to correct final results. I am wondering whether this can be proved in a rigorous way and/or whether this is true in the first place. Any help or comment would be very helpful.
Mickhausen
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that: $$ \delta(x) - \delta(x-\pi) = \frac{2}{\pi} \sum_{n ~\text{odd}} \cos( n x) $$ Note that the series expansion implies that you consider functions with a $2\pi$-periodicity, and that one also could have written it in a more symmetric form as $$ \delta(x) - \frac{1}{2} \left[ \delta(x-\pi) + \delta(x+\pi)\right] $$ that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series: $$ f(x) = \sum_n A_n e^{\imath n x} $$ with $$ A_n = \frac{1}{2 \pi} \int_{-\pi/2}^{3 \pi/2} f(x) e^{-\imath n x} \text{d} x $$ which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.