consider the following equation:

NF = 1.6;
df[dn_] = NF - dn;
DeltaF[mis_, theta_, 
   dn_] = (dn Tan[theta] Cot[ArcSin[mis Sin[theta]]] + df[dn])/NF;

If, for example, mis is fixed to 0.875 and dn to 0.1, I'm expecting this function to tend to 1 as theta approaches 0, with a singularity in theta=1. However, the limit gives this result:

Limit[DeltaF[0.875, theta, 0.1], theta -> 0]]

If the precision is increased with SetPrecision, I get a bunch of number, but still far from what I'd expect. Is it mathematically correct or there something wrong with the precision here? Basically, I'd like to plot this equation in a way I can see it going to 1 as theta goes to 0, with obviously the singularity in 0.Thanks.

  • $\begingroup$ Substituting exact rational numbers for your approximate ones, I get a value of exactly 113/112 for theta=0. So, I think your expectation is wrong. $\endgroup$
    – John Doty
    Dec 6, 2017 at 20:29
  • $\begingroup$ Yes, which is the value I get. If you try to do it now by using the raw equation (0.1 Tan[0] Cot[ArcSin[0.875 Sin[0]]] + 1.5)/1.6, instead of calling DeltaF[0.875, 0, 0.1], you'll get Indeterminete, as said. This makes sense, since it goes to 0*Infinite. Why that? $\endgroup$
    – MicheleG
    Dec 6, 2017 at 20:41
  • 1
    $\begingroup$ It's because you used = to define DeltaF, thereby symbolically evaluating the formula in the definition.This reduces Tan[theta] Cot[ArcSin[mis Sin[theta]]] to (Sec[theta] Sqrt[1 - mis^2 Sin[theta]^2])/mis, thus removing the removable singularity. If you substitute numbers before evaluation, it doesn't do that. $\endgroup$
    – John Doty
    Dec 6, 2017 at 20:58
  • $\begingroup$ All right, I didn't notice that. Thanks for spotting! Effectively, simplifying the equation it is possible to get something without singularity. I though that, since the equation could be reduced to something similar to Tan[x] Cot[x] as x->0, the whole equation was going to 1, but actually it's not. $\endgroup$
    – MicheleG
    Dec 7, 2017 at 20:21


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