How can I solve the limit by mathematica?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












I use the Limit and DiscreteLimit to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$










share|improve this question





















  • If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
    – Ulrich Neumann
    Sep 10 at 7:29






  • 1




    @UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
    – LLlAMnYP
    Sep 10 at 7:32










  • @LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
    – Ulrich Neumann
    Sep 10 at 7:39







  • 1




    @King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
    – LLlAMnYP
    Sep 10 at 7:45






  • 1




    Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
    – Mariusz Iwaniuk
    Sep 10 at 9:13














up vote
3
down vote

favorite












I use the Limit and DiscreteLimit to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$










share|improve this question





















  • If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
    – Ulrich Neumann
    Sep 10 at 7:29






  • 1




    @UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
    – LLlAMnYP
    Sep 10 at 7:32










  • @LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
    – Ulrich Neumann
    Sep 10 at 7:39







  • 1




    @King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
    – LLlAMnYP
    Sep 10 at 7:45






  • 1




    Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
    – Mariusz Iwaniuk
    Sep 10 at 9:13












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I use the Limit and DiscreteLimit to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$










share|improve this question













I use the Limit and DiscreteLimit to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$







calculus-and-analysis






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Sep 10 at 7:22









King.Max

161




161











  • If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
    – Ulrich Neumann
    Sep 10 at 7:29






  • 1




    @UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
    – LLlAMnYP
    Sep 10 at 7:32










  • @LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
    – Ulrich Neumann
    Sep 10 at 7:39







  • 1




    @King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
    – LLlAMnYP
    Sep 10 at 7:45






  • 1




    Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
    – Mariusz Iwaniuk
    Sep 10 at 9:13
















  • If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
    – Ulrich Neumann
    Sep 10 at 7:29






  • 1




    @UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
    – LLlAMnYP
    Sep 10 at 7:32










  • @LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
    – Ulrich Neumann
    Sep 10 at 7:39







  • 1




    @King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
    – LLlAMnYP
    Sep 10 at 7:45






  • 1




    Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
    – Mariusz Iwaniuk
    Sep 10 at 9:13















If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
– Ulrich Neumann
Sep 10 at 7:29




If you try Integrate[Tan[x/n]^n, x, 0, Pi/4] MMA evaluates a ConditionalExpression which constrains 0<n<1/2. Might be the limit doesn't exist?
– Ulrich Neumann
Sep 10 at 7:29




1




1




@UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
– LLlAMnYP
Sep 10 at 7:32




@UlrichNeumann I'm quite sure, it does exist. At n->Infinity we can probably replace Tan[x/n] with x/n, then the limit is solvable and gives Pi/4
– LLlAMnYP
Sep 10 at 7:32












@LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
– Ulrich Neumann
Sep 10 at 7:39





@LLlAMnYP The limit would be 1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi]) (not Pi/4). I only tried to indicate, that MMA can't integrate if n>1/2
– Ulrich Neumann
Sep 10 at 7:39





1




1




@King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
– LLlAMnYP
Sep 10 at 7:45




@King.Max as you found, as it is Mathematica doesn't solve the limit. It needs some help (i.e. manual interference from the user). If my suggestion isn't acceptable to you, what kind of help would be acceptable? Maybe one could pass some options to Limit or to Integrate, but I can't say for sure.
– LLlAMnYP
Sep 10 at 7:45




1




1




Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
– Mariusz Iwaniuk
Sep 10 at 9:13




Try: Limit[n*Integrate[Tan[x/n]^n, x, 0, Pi/4, Assumptions -> n > 1]^(1/n) // PowerExpand, n -> Infinity]
– Mariusz Iwaniuk
Sep 10 at 9:13










1 Answer
1






active

oldest

votes

















up vote
6
down vote













Integrate returns an anti-derivative which seems reasonable for large n:



antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, π/4, PlotRange -> All]




And then



Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]



π/4






share|improve this answer




















  • how simple a tricky solution can be...
    – Ulrich Neumann
    Sep 10 at 7:55










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f181593%2fhow-can-i-solve-the-limit-by-mathematica%23new-answer', 'question_page');

);

Post as a guest






























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote













Integrate returns an anti-derivative which seems reasonable for large n:



antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, π/4, PlotRange -> All]




And then



Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]



π/4






share|improve this answer




















  • how simple a tricky solution can be...
    – Ulrich Neumann
    Sep 10 at 7:55














up vote
6
down vote













Integrate returns an anti-derivative which seems reasonable for large n:



antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, π/4, PlotRange -> All]




And then



Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]



π/4






share|improve this answer




















  • how simple a tricky solution can be...
    – Ulrich Neumann
    Sep 10 at 7:55












up vote
6
down vote










up vote
6
down vote









Integrate returns an anti-derivative which seems reasonable for large n:



antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, π/4, PlotRange -> All]




And then



Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]



π/4






share|improve this answer












Integrate returns an anti-derivative which seems reasonable for large n:



antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, π/4, PlotRange -> All]




And then



Limit[n Power[(antiDeri /. x -> π/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]



π/4







share|improve this answer












share|improve this answer



share|improve this answer










answered Sep 10 at 7:45









Coolwater

13.4k32150




13.4k32150











  • how simple a tricky solution can be...
    – Ulrich Neumann
    Sep 10 at 7:55
















  • how simple a tricky solution can be...
    – Ulrich Neumann
    Sep 10 at 7:55















how simple a tricky solution can be...
– Ulrich Neumann
Sep 10 at 7:55




how simple a tricky solution can be...
– Ulrich Neumann
Sep 10 at 7:55

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f181593%2fhow-can-i-solve-the-limit-by-mathematica%23new-answer', 'question_page');

);

Post as a guest













































































這個網誌中的熱門文章

Why am i infinitely getting the same tweet with the Twitter Search API?

Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

Strongly p-embedded subgroups and p-Sylow subgroups.