How do I make this summation numerical?

Questions
,
1

Hi,

I don’t understand why I can’t make the below numerical?

NSum[NSum[(2^(m + p)), {m, 1, 2}], {p, 1, 2}]

It gives:

Error - NSum::nsnum: Summand (or its derivative) 2^(m+p) is not numerical at point m = 1. >>

2

p is symbolic for the inner NSum

NSum[(2^(m + p)), {m, 1, 2}, {p, 1, 2}]

gives 36. without the error msg.

3

That’s great! Thanks!
Is there a way for this to work? NSum[{m, p}, {m, 1, 3}, {p, 1, 2}]

4 
NSum[#, {m, 1, 3}, {p, 1, 2}] & /@ {m, p}

{12., 9.}
5

Thanks, I wouldn’t have come up with that myself!
The reason is that I’m trying to get the following to plot:

n=2
a=10
R[m_, p_] := {m*a, p*a}
K[m_, p_] := {(2*Pi*m)/a, (2*Pi*p)/a}
A[x_, y_] := Sum[(-(-1)^(m*p + m + p) Exp[(-(K[m, p]*K[m, p])*a^2)/(8*Pi)])*(1 - Cos[K[m, p]*{x, y}]), {m, 1, 5}, {p, 1, 5}]
s[m_, p_] := -R[m, p] ((Sqrt[3]*a^2)/(4*Pi*R[m, p]*R[m, p]) + (0.068*a^4)/(R[m, p]*R[m, p]*R[m, p]*R[m, p]))
w[x_, y_] := A[x, y]/A[x/n, y/n] Product[(A[({x, y} - R[m, p] - s[m, p])/n]/A[({x, y} - R[m, p])/n]), {m, 1, 5}, {p, 1, 5}]
StreamPlot[w[x, y], {x, 1, 10}, {y, 1, 10}]

At the moment it gives a blank plot so I’m attempting to make it numerical in the hope that it may work. However I’m really struggling to get it to be numerical.
Any advice?
Thank you!

6

A powerful debugging trick I often use when some kind of plot doesn’t display anything:
In your latest version of your code change

 StreamPlot[w[x, y], {x, 1, 10}, {y, 1, 10}]

to

Table[w[x,y],{x,1,10},{y,1,10}]

and take a look at the result, you might need to then ask Mathematica to show you more and more output to see enough to be useful.

I see lots of

A[{someconstant,someotherconstant}]

in that and you have no definition for A that involves two items in a list, only an A that accepts two numeric arguments.

So I think the problem lies in several places where you give your A the result of vector operations.

If I make a guess that you might want this to be A[someconstant,someotherconstant] then I can try

A[Sequence@@(yourvectorexpression)]

where i have added an extra layer of () around your vector expression given to A, which should turn your vectors being given to A into a simple pair of arguments.

When I try Table on this I get a variety of warnings about 1/0 and indeterminant and complex infinities, but at least I no longer get unevaluated copies of A[…] and you might be able to tell whether the values in the resulting Table are sensible or not.

When I replace Table with StreamPlot it seems to take longer than I want to wait and this may indicate that I have misunderstood your code and even broken something.
Perhaps something in this will let you diagnose why you are not getting the plot you desire.

EDIT: Wait a minute, well actually perhaps more like ten minutes, and I actually get a StreamPlot from

StreamPlot[w[x,y],{x,1,3},{y,1,3}]

so with time you might also get a result, possibly even for 1 to 10 if you wait long enough

7

Thanks! Seems I should have put:
A[{x_, y_}] :=

8

Using the Table trick after changing to A[{x_,y_}]:=…, it shows you have A[x,y] in some places and A[{x,y}] in other places.
What I should have done was sorted all this out and made all of them consistent.
Instead I cheated and just made two definitions for A,
A[{x_,y_}:=…
A[x_,y_]:=…
with identical right hand sides.
I also separated each of your statements with a semicolon, always a good idea and will keep you out of trouble in the future.
With those changes and after a depressingly long time your stream plot appears.