This note, with the help of strict and merciless mathematics, allows us to prove that the sum of squares and the square of the sum in electrodynamics are equal :) In fact, this approach is true for linear storage electrodynamic systems, an example of which was provided by this paper. In such systems, the accumulation of energy occurs according to the law of the square of the sum, and the cost of each accumulation (iteration) is the square of the current or voltage, which ultimately leads us to the sum of the squares. The energy balance, in this case, is compiled as the ratio of the square of the sum to the sum of the squares. Here we will prove that the formula (17) of the energy balance (or efficiency) will be equal to one for any values of i -- number of iterations. We only recall that we are considering the ideal case, without losses.

Let's write this formula, substituting all the values there: \[\eta_i = {S^2 \left( \sum \limits_{n=1}^i C^{n-1} \right)^2 \over \sum \limits_{m=1}^i \left[ 1 - \left(C - S^2 \sum \limits_{n=1}^{m-1} C^{n- 1} \right)^2 \right] } \tag{1}\] We also recall that here, to simplify the perception and writing of formulas, the following notation is introduced: \[ \sin(\omega_0 \tau) = S, \quad \cos(\omega_0 \tau) = C \tag{2}\] where: \(\omega_0\) is the circular resonant frequency of the circuit, \(\tau\) is the time of one iteration. For this note, such data is not needed, here we will deal exclusively with sines, cosines and their sums.

In the following, we will work with the denominator from formula (1), which we denote as \(D\), and in which we will immediately open the brackets: \[ D = \sum \limits_{m=1}^i \left[ 1 - C^2 + 2 C S^2 \sum \limits_{n=1}^{m-1} C ^{n-1} - S^4 \left( \sum \limits_{n=1}^{m-1} C^{n-1} \right)^2 \right] \tag{3}\] Thanks to the \(S^2 = 1 - C^2\) property, the previous expression becomes: \[ D = S^2 \sum \limits_{m=1}^i \left[ 1 + 2 C \sum \limits_{n=1}^{m-1} C^{n -1} - \left( 1 - C^2 \right) \left( \sum \limits_{n=1}^{m-1} C^{n-1} \right)^ 2 \right] \tag{4}\] Expanding the brackets again, and grouping, we get: \[ D = S^2 \sum \limits_{m=1}^i \left[ \left( 1 + C \sum \limits_{n=1}^{m-1} C ^{n-1} \right)^2 - \left( \sum \limits_{n=1}^{m-1} C^{n-1} \right)^2 \right]\tag{5}\] Now it remains to consider and simplify the following expression: \[ C \sum \limits_{n=1}^{m-1} C^{n-1} = \sum \limits_{n=1}^{m} C^{n-1 } - 1 \tag{6}\] Substituting it into (5) we get \[ D = S^2 \sum \limits_{m=1}^i \left[ \left( \sum \limits_{n=1}^{m} C^{n-1 } \right)^2 - \left( \sum \limits_{n=1}^{m-1} C^{n-1} \right)^2 \right] \tag{ 7}\] where does it come from \[ D = S^2 \left( \sum \limits_{n=1}^i C^{n-1} \right)^2 \tag{8}\] It is now obvious that the numerator and denominator in formula (1) are equal, which means: \[\eta_i = 1 \tag{9}\] This means that the efficiency of a linear storage device in the ideal case (without losses) is equal to unity for any number of iterations. Q.E.D.