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Suppose there is a function $f(\theta,\phi)$ defined on the surface of a sphere, and $\theta$ and $\phi$ are the polar and azimuthal angles respectively. Similar with the fact that a function defined on a line or a circle can be expanded with respect to the Fourier bases, this function $f$ can be expanded in terms of the spherical harmonics $Y_l^m(\theta, \phi)$: $$f(\theta,\phi)=\sum_{l=0}^{\infty} \sum_{m=-l}^l F_l^m Y_l^m(\theta,\phi),$$ where $l$ and $m$ are the degree and order of the harmonics. In order for the expansion to achieve a perfect fitting to the original function, $l$ needs to go from 0 to $\infty$.

In our experiment design, however, we cannot detect directly the value of the original function but can only detect the expansion coefficients $F_l^m$, and unfortunately, the orders of $F_l^m$ we are able to detect is limited, ie. $l \leq l_0$, so in fact we are using a truncated summation $$\sum_{l=0}^{l_0} \sum_{m=-l}^l F_l^m Y_l^m(\theta,\phi)$$ to simulate the original function. When $l_0$ is small, even though the truncated function captures the overall variation trends of the original function well, the amplitudes are significantly different from the originals.

So my question is: is there a way to mathematically scale the truncated function value to a direct comparable level with the original function? If so, how to do it please? Any input would be greatly appreciated, thanks a lot!

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    $\begingroup$ That doesn't make sense. If the amplitude differs significantly from the original, then the $L^2$ difference between your reconstructed signal and the original is large, then there is a lot of higher frequency information missing. So I cannot believe that you have the overall variation trends captured. $\endgroup$ Jun 13, 2014 at 15:44
  • $\begingroup$ If you are working with crystalline fields, the expansion is usually truncated by selection rules. $\endgroup$ Jun 13, 2014 at 23:35
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    $\begingroup$ @WillieWong, Yes there is a lot of high frequency information missing, you're right. But the amplitudes of these high frequency terms are also smaller compared to the lower degree ones, so it can still show the overall map. It like drawing a portrait, we cannot capture the details like the eyes, nose, or the mouth, but we can have a rough idea of where the face is, the belly is, the legs are, and to us this rough information in this experiment is also important. $\endgroup$
    – Lampard
    Jun 14, 2014 at 9:18
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    $\begingroup$ @Lampard: to your final question in your previous comment: no. The whole point of the spherical harmonic expansion is that the different harmonics are independent of each other, and so given an arbitrary signal information up through the 8th order will give you absolutely zero information about the 9th through 16th order. An arbitrary signal can easily have the first 8 orders being identically zero and the 10th order being rather large. On the other hand, if your data is coming from a physical experiment, the physical laws often dictate that there can be some correlation between modes. $\endgroup$ Jun 16, 2014 at 8:02
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    $\begingroup$ (So in other words your signal is not truly "arbitrary".) But you will need to tell us what the set-up is and preferably what are the known correlation for the question to be answerable. On the other hand, I suspect as @FelixMarin said, much of these is classical and are already captured in the truncation "by selection rules". And quite likely you cannot beat that (in the sense that if you truncate even more you start to lose information that cannot be recovered). $\endgroup$ Jun 16, 2014 at 8:05

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