... about high frequency ripple whose amplitude is modulated at a slower rate (i.e. sin(ax)*sin(bx)). If you do the analysis this scenario actually has zero frequency content at the slower rate...
If it is a ripple, I understand that it is riding on a DC with a small amplitude. I mean DC+(small percentage of DC)*sin(100*t).
Because the ripple current is proportional to the ripple voltage, the power lost is small (I^2)*R loss
On the other hand, the 70 kHz is really not a ripple because it is not riding on a DC. If it is the output after the filtering, then we can call this a ripple because it will be mostly DC with 70 kHz ripple.
I also think that ESR will be higher at 100 Hz compared to 70 kHz but the proportions are not large (perhaps within a factor of 2). I plead my ignorance.
Why the two effects (losses) cannot be added, even if it is an approximation?
70 KHz ripple:
DC=alpha; ripple=beta (percent of DC); some people specify ripple in absolute volts.
voltage=alpha+alpha*beta*0.01*sin(100*t);
100 Hz ripple; gamma (percent of DC);
voltage=alpha+alpha*beta*gamma*0.001*sin(100*t)*sin(70000*t);
using sin(a)*sin(b)=-0.5*(cos(70100*t)-cos(69900*t))
I guess 100 Hz frequency is not going to have much effect.