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Carl Lange
Carl Lange
10:55
@C.E. Just wanted to say I really like your answer here mathematica.stackexchange.com/questions/239503/… I know SE doesn't encourage comments that just say thanks but I wanted to say thanks! It's very helpful and thorough.
2
Mats Granvik
Mats Granvik
11:07
Can someone run this code on a machine with a newer version of Mathematica than 8.0.1 and tell me the result:
(*start*)
Clear[a, b, c, d, s];
SumConvergence[1/(a*b*c)^s + 1/(a*b*c*d)^s, a]
(*end*)
Takes a few seconds on my machine.
Carl Lange
Carl Lange
11:35
@MatsGranvik Re[s] > 1
(On 12.2)
Mats Granvik
Mats Granvik
12:11
@CarlLange Thanks for the help.
In Mathematica 8 it just returns the expression unevaluated without error messages.
 
1 hour later…
C. E.
C. E.
13:16
@CarlLange I'm glad you like it :)
 
3 hours later…
ThunderBiggi
ThunderBiggi
15:55
Can somebody tell me why Mathematica fails to use Plot inside a Block

Block[{vars, eq, ineqs}, vars = Complement[{x, y, z}, {y, z}];
 eq = 5 x + 3; ineqs = x > 2 && x < 5;
 If[Length[vars] == 1,
  Plot[eq, {vars} \[Element] ImplicitRegion[ineqs, vars]]]]
the above produces errors, whereas

Plot[5 x + 3, {x} \[Element] ImplicitRegion[x > 2 && x < 5, {x}]]

plots fine
I really need to be able to do the first one, as I need to go over a table and the vars, eq and ineqs will change for each element
oh, and the first vars inside Plot, in the first example does not need curly brackets
 
6 hours later…
halirutan
halirutan
21:37
@ThunderBiggi Welcome to the world of how Mathematica evaluates things. Look at this simple example: 
Block[{x, y, z, vars, ineqs},
 vars = Complement[{x, y, z}, {y, z}];
 ineqs = x > 2 && x < 5;
 ImplicitRegion[ineqs, vars]
 ]
Do you see that in the output there is still vars? 
ImplicitRegion[x>2&&x<5,{vars}]
This happens on multiple levels in your short example and the reason it fails has nothing to do with Block. You need to carefully evaluate certain things when you want to plot something like this using variables instead of putting the equations and regions directly into the expression.
Something like this should work: 
Block[{x, y, z, vars, eq, ineqs, region, plotFunc},
 vars = Complement[{x, y, z}, {y, z}];
 eq = 5 x + 3;
 ineqs = x > 2 && x < 5;
 region = ImplicitRegion[ineqs, Evaluate[vars]];
 plotFunc = Function[{equation, spec}, Plot[equation, spec]];
 If[Length[vars] == 1, plotFunc[eq, Element[vars, region]]
  ]
 ]
Where I used Evaluate in ImplicitRegion to say the vars should be expanded to their value. Making the plotFunc will also evaluate the arguments you pass to it.

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