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Mathematics Wolfram Language

[?] Find the limit efficiently containing ArcTan?

Ox Clouding

Posted 6 years ago

Hi everyone, I am trying to evaluate this limit (actually the limit of a complicated expression containing this actan function): Limit[ArcTan[(s + r Cos[\[Phi]])/Sqrt[ a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]]] , s -> (-r Cos[\[Phi]] + Sqrt[ a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), Direction -> "FromBelow"] which is related to some integrals inside a circle of radius a. Anyway, when I set these assumptions: assumptions = {0 < a < r}; Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], s -> (-r Cos[\[Phi]] + Sqrt[ a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), Direction -> "FromBelow", Assumptions -> assumptions] gives the result: DirectedInfinity[Sign[2 a^2 - r^2 + r^2 Cos[2 \[Phi]]]^(1/4)] The region I am interested is where the part inside Sign function is positive, so this will just equals to positive infinity. So I thought this code will work: assumptions = {0 < a < r, 2 a^2 - r^2 + r^2 Cos[2 \[Phi]] > 0}; Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], s -> (-r Cos[\[Phi]] + Sqrt[ a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), Direction -> "FromBelow", Assumptions -> assumptions] On my computer, this code runs for half an hour and did not give any output. I am wondering if there is any simple way to tell Mathematica to ignore the directed infinity thing and just treat it like positive infinity? Thanks in advance!

Hi everyone,

I am trying to evaluate this limit (actually the limit of a complicated expression containing this actan function):

Limit[ArcTan[(s + r Cos[\[Phi]])/Sqrt[
  a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]]] , 
 s -> (-r Cos[\[Phi]] + Sqrt[
    a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), 
 Direction -> "FromBelow"]

which is related to some integrals inside a circle of radius a. Anyway, when I set these assumptions:

assumptions = {0 < a < r};
Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], 
     s -> (-r Cos[\[Phi]] + Sqrt[
        a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), 
     Direction -> "FromBelow", Assumptions -> assumptions]

gives the result:

DirectedInfinity[Sign[2 a^2 - r^2 + r^2 Cos[2 \[Phi]]]^(1/4)]

The region I am interested is where the part inside Sign function is positive, so this will just equals to positive infinity. So I thought this code will work:

assumptions = {0 < a < r, 2 a^2 - r^2 + r^2 Cos[2 \[Phi]] > 0};
Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], 
     s -> (-r Cos[\[Phi]] + Sqrt[
        a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), 
     Direction -> "FromBelow", Assumptions -> assumptions]

On my computer, this code runs for half an hour and did not give any output. I am wondering if there is any simple way to tell Mathematica to ignore the directed infinity thing and just treat it like positive infinity?

Thanks in advance!

POSTED BY: Ox Clouding

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Ox Clouding

Posted 6 years ago

Dear Henrik, Thanks for the help! I tried your code and it worked like magic. Although I don't really get the logic behind this trick. I tried to change the Refine to Simplify: Simplify[Reduce[2 a^2 - r^2 + r^2 Cos[2 \[Phi]] > 0, a, Reals], {\[Phi] \[Element] Reals, 0 < a < r}] which gives: a > r/Abs[Csc[\[Phi]]] With this modification the code run slowly again, and ends with this result: DirectedInfinity[Sign[2 a^2 - r^2 + r^2 Cos[2 \[Phi]]]^(1/4)] It seems only Refine will work. But the original problem is solved, so Thank You regardless!

POSTED BY: Ox Clouding

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

If I rewrite your assumption like so: Refine[Reduce[2 a^2 - r^2 + r^2 Cos[2 \[Phi]] > 0, a, Reals], Assumptions -> {\[Phi] \[Element] Reals, 0 < a < r}] I get: So if I execute using this assumption: assumptions = {0 < r, a > Sqrt[r^2 - r^2 Cos[2 \[Phi]]]/Sqrt[2]}; Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], s -> (-r Cos[\[Phi]] + Sqrt[a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), Direction -> "FromBelow", Assumptions -> assumptions] I get immediately as result `Infinity`. Does that make sense?

If I rewrite your assumption like so:

Refine[Reduce[2 a^2 - r^2 + r^2 Cos[2 \[Phi]] > 0, a, Reals], 
 Assumptions -> {\[Phi] \[Element] Reals, 0 < a < r}]

I get:

enter image description here

So if I execute using this assumption:

assumptions = {0 < r, a > Sqrt[r^2 - r^2 Cos[2 \[Phi]]]/Sqrt[2]};

Limit[(s + r Cos[\[Phi]])/Sqrt[a^2 - r^2 - s^2 - 2 r s Cos[\[Phi]]], 
 s -> (-r Cos[\[Phi]] + 
    Sqrt[a^2 Cos[\[Phi]]^2 + (a - r) (a + r) Sin[\[Phi]]^2]), 
 Direction -> "FromBelow", Assumptions -> assumptions]

I get immediately as result Infinity. Does that make sense?

POSTED BY: Henrik Schachner

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