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There is so-called Power Transform technique in signal processing where the unresolved signal is sharpened by raising each data to a constant positive power. The example shown below is a sum of two Gaussians, $G_1$ and $G_2$ plotted against time. In order to resolve the peaks one can apply a power of 5 to each data point. The result is the resolved pair as shown in the bottom figure.

The above operating is equivalent to raising the sum of Gaussians $G_1$ and $G_2$ to the power 5, i.e., $(G_1 + G_2)^5$ then according to the binomial theorem, we should get 6 peaks, because not only we have the first and the second peak raised to power 5, we will have a "mixture" of the two peaks as well- the other terms of the binomial theorem. No matter how well we zoom, we cannot see the missing $G_1^4$$G_2$, $G_1^3$$G_2^2$ peaks? If power transform of 5 is applied to two peaks, we are left with two peaks not six.

Where do the peaks corresponding to the other binomial terms go?

Thanks.

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The binomial theorem is not necessarily involved: the top waveform is simply pointwise raised to the 5-th power, as others have noted, and there are no missing peaks that should be seen. To illustrate, consider the following crude hand-drawn figure:

Raw peaks

I drew this in my simulation software, so it was automatically "digitized" as I drew it. Obviously, I am poor at freehand drawing with a computer mouse, but I was only trying to make two peaks, with no underlying peak shape in mind.

Then the second figure shows what happens if the digitized values from the first figure are raised to the 5-th power and plotted:

Peaks after nonlinear gain

Still ugly, but better resolved.

Now taking into account a more recent answer by @MBaz (https://dsp.stackexchange.com/a/60750/41790), there is still one loose end. To see this, consider two Gaussians, $G_1(x)$ and $G_2(x)$, and adding them. Their sum function, i.e., $G_1(x)$ + $G_2(x)$, does not look like either function. Of course, we know the sum function's two constituent summand functions. But the sum function does not show them: it only shows their sum.

Now square the sum function. Then the pointwise squaring works as expected. Alternatively, if we use the expansion, then we get three constituent functions: $G_1^2(x)$, $G_2^2(x)$ and $2G_1(x)G_2(x)$. So the squared sum function is just the sum of these three constituent functions and none of the three are 'visible' to us: we only see the sum of the three constituent functions. So this is why we can never see the 'extra' peak functions. Indeed, we cannot see any of the constituent functions. We can only see the their sum and should not have expected that sum to display its constituent functions.

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  • $\begingroup$ I think I am missing something fundamental: Why element-wise powering of the sum of two Gaussians is not the $same$ as $(G_1 + G_2)^5$? $\endgroup$ – M. Farooq Sep 16 at 14:46
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    $\begingroup$ Each point, whether from the hand-drawn plot or the plot of 2 overlapping Gaussian functions, is essentially independent of its neighbors: it is just a numerical value. So raising it to the fifth power just does that numerical operation. The power operation is a nonlinear scaling point by point. Indeed, you could just select subsets of points to raise to the fifth power. It would be rather silly except that it illustrates that the individual points are all that matter. $\endgroup$ – Ed V Sep 16 at 15:04
  • $\begingroup$ How would we go about implementing a discrete version of $(G_1+G_2)^5$? I thought they would yield identical results? Thanks. $\endgroup$ – M. Farooq Sep 16 at 15:18
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    $\begingroup$ I would just numerically evaluate the two Gaussians, add them pointwise, then each resulting point would be raised to the fifth power. I will take a closer look at this, to see if I am missing something obvious. If I find anything pertinent, I will update the answer. $\endgroup$ – Ed V Sep 16 at 15:23
  • $\begingroup$ This is exactly what I did for the second figure. It is a point by point power (5) of the sums of Gaussians. Thats why I am wondering how are they different. $\endgroup$ – M. Farooq Sep 16 at 16:30
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then according to the binomial theorem, we should get 6 peaks

Where does that statement come from and why do you think it applies here?

Applying the binomial theorem we get terms that consists of Gaussian raised to different powers. However, when raising the power on a Gaussian, the peak stays where it is so there is no mechanism that would move it someplace else.

You may be confusing this with raising the sum of sine waves to a certain power.

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