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Evaluate an expression inside Module[]

Posted 3 years ago

Let

f = 2 n + t

Now I want to evaluate f given values for n and t. So what I've done is

Module[{n = 1, t = 5},
 f]
(*f*)

What I expected to get is 7, of course. But the output is just f.

However,

Module[{n = 1, t = 5},
 2 n + t]
(*7*)

returns a calculation.

Now I moved on to something more complicated. Let

u = a Sqrt[k^2 n1^2 - \[Beta]^2]

and

w = a Sqrt[\[Beta]^2 - k^2 n0^2]

I want to get the value of

 BesselJ[1, u]/(u BesselJ[0, u]) == -(BesselK[1, w]/(w BesselK[0, w]))]

for some values of n1, n2 and a, as a function of Beta and k (and then plot a ContourPlot).

The same trick does no calculation:

Module[{n1 = 1.4475, n2 = 1.4444, a = 5},
 Evaluate[BesselJ[1, u]/(u BesselJ[0, u])] == 
  Evaluate[-(BesselK[1, w]/(w BesselK[0, w]))]]

nor

Module[{n1 = 1.4475, n2 = 1.4444, a = 5},
 Evaluate[BesselJ[1, u]]/Evaluate[u BesselJ[0, u]] == 
  Evaluate[-(BesselK[1, w]/(w BesselK[0, w]))]]

or any other combination of wrapping Evaluate[] around the u and w.

Can I use Module at all for this kind of numerical substitution? Is Module the advised tool?

Of course, I can set n1, n2 and a to these numerical values, but I good bookkeeping in my code.

Thanks very much for your help!

POSTED BY: Ehud Behar
3 Replies

You can also use Block:

f = 2 n + t;
Block[{n = 1, t = 5}, f]

or Replace:

f /. {n -> 1, t -> 5}

or Simplify:

Simplify[f, n == 1 && t == 5]
POSTED BY: Gianluca Gorni
Posted 3 years ago

Not exactly sure what you are trying to do. Anyway, I think n0 should be n2, the right scoping construct for this is With.

u = a Sqrt[k^2 n1^2 - β^2]
w = a Sqrt[β^2 - k^2 n2^2]

With[{n1 = 1.4475, n2 = 1.4444, a = 5}, 
 Evaluate[BesselJ[1, u]/(u BesselJ[0, u]) == -(BesselK[1, w]/(w BesselK[0, w]))]]

BesselJ[1, 5 Sqrt[2.09526 k^2 - β^2]] / (5 Sqrt[2.09526 k^2 - β^2] BesselJ[0, 5 Sqrt[2.09526 k^2 - β^2]]) == -(
  BesselK[1, 5 Sqrt[-2.08629 k^2 + β^2]] / (5 Sqrt[-2.08629 k^2 + β^2] BesselK[0, 5 Sqrt[-2.08629 k^2 + β^2]]))
POSTED BY: Rohit Namjoshi
Posted 3 years ago

With is what I was looking for.

Thanks a lot!

POSTED BY: Ehud Behar
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