I would like to solve
nf = 1.9; ns = 1.89; c = 3*^8; H = 1.250*^-6; B = Table[Sqrt[ (A/c)^2 nf^2 - (FindRoot[Tan[k*H/2]==Sqrt[(A/c)^2 (nf^2 -ns^2) - k^2]/k, {k,20}][[1, 2]])^2], {A, 0, 3*^15}] ListPlot[B] Am I missing something really obvious??
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1 Answer
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You have many many syntactic errors:
c = 3*^8 makes no sense. Presumably you mean c = 3 10^8
H = 1.250*^-6 is wrong in several ways. Presumably you mean H = 1.250 10^(-6)
You must assign f.
You should use a large step size (as @MarcoB writes). And so on.
You should simplify by substituting values whenever possible.
You should avoid using upper-case letters to label a variable ALWAYS.
You need semicolon at the end of each functional line.
A start:
f = 7; c = 3 10^8; h = 1.250 10^(-6); b = Table[ Sqrt[(a/c)^2 1.9 f^2 - (FindRoot[ Tan[k h/2] == Sqrt[(a/c)^2 (1.9^2 - 1.89^2) - k^2]/k, {k, 20}][[1, 2]])^2], {a, 0, 3 10^(15), 10^(14)}]; ListPlot[b] 
SystemException["MemoryAllocationFailure"]? My first step would be to look at the equation insideFindRoot. This function looks very weird and doesn't seem to have a root at all. $\endgroup$Table[..., {A, 0, 3*^15}]implies evaluation of all values of $A$ between 0 and $3\times 10^{15}$ with a step of $1$. That seems unreasonable. If each evaluation took only 1 ms, the whole exercise would take on the order of 100,000 years, not to mention immense amounts of memory. Start by setting a reasonable step size, e.g.{A, 0, 3*^15, 1*^14}. You will at least get some results then. $\endgroup$