How can I solve the limit by mathematica?

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up vote
3
down vote
favorite
I use the Limit
and DiscreteLimit
to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$
calculus-and-analysis
 |Â
show 2 more comments
up vote
3
down vote
favorite
I use the Limit
and DiscreteLimit
to solve it but failed. How to solve it by Mathematica?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$
calculus-and-analysis
If you tryIntegrate[Tan[x/n]^n, x, 0, Pi/4]
MMA evaluates a ConditionalExpression which constrains0<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. Atn->Infinity
we can probably replaceTan[x/n]
withx/n
, then the limit is solvable and givesPi/4
– LLlAMnYP
Sep 10 at 7:32
@LLlAMnYP The limit would be1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi])
(not Pi/4). I only tried to indicate, that MMA can't integrate ifn>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 toLimit
or toIntegrate
, 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
 |Â
show 2 more comments
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?
$$undersetnto infty textlimn left(int_0^fracpi 4 tan ^nleft(fracxnright) , dxright)^1/n$$
calculus-and-analysis
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
calculus-and-analysis
asked Sep 10 at 7:22
King.Max
161
161
If you tryIntegrate[Tan[x/n]^n, x, 0, Pi/4]
MMA evaluates a ConditionalExpression which constrains0<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. Atn->Infinity
we can probably replaceTan[x/n]
withx/n
, then the limit is solvable and givesPi/4
– LLlAMnYP
Sep 10 at 7:32
@LLlAMnYP The limit would be1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi])
(not Pi/4). I only tried to indicate, that MMA can't integrate ifn>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 toLimit
or toIntegrate
, 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
 |Â
show 2 more comments
If you tryIntegrate[Tan[x/n]^n, x, 0, Pi/4]
MMA evaluates a ConditionalExpression which constrains0<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. Atn->Infinity
we can probably replaceTan[x/n]
withx/n
, then the limit is solvable and givesPi/4
– LLlAMnYP
Sep 10 at 7:32
@LLlAMnYP The limit would be1/4 [Pi]^((4 + [Pi])/[Pi]) (4 + [Pi])^(-4/[Pi])
(not Pi/4). I only tried to indicate, that MMA can't integrate ifn>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 toLimit
or toIntegrate
, 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
 |Â
show 2 more comments
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, À/4, PlotRange -> All]
And then
Limit[n Power[(antiDeri /. x -> À/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
À/4
how simple a tricky solution can be...
– Ulrich Neumann
Sep 10 at 7:55
add a comment |Â
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, À/4, PlotRange -> All]
And then
Limit[n Power[(antiDeri /. x -> À/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
À/4
how simple a tricky solution can be...
– Ulrich Neumann
Sep 10 at 7:55
add a comment |Â
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, À/4, PlotRange -> All]
And then
Limit[n Power[(antiDeri /. x -> À/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
À/4
how simple a tricky solution can be...
– Ulrich Neumann
Sep 10 at 7:55
add a comment |Â
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, À/4, PlotRange -> All]
And then
Limit[n Power[(antiDeri /. x -> À/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
À/4
Integrate
returns an anti-derivative which seems reasonable for large n
:
antiDeri = Integrate[Tan[x/n]^n, x];
Plot[antiDeri /. n -> 18, x, 0, À/4, PlotRange -> All]
And then
Limit[n Power[(antiDeri /. x -> À/4) - (antiDeri /. x -> 0), 1/n], n -> ∞]
À/4
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
add a comment |Â
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
add a comment |Â
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If you try
Integrate[Tan[x/n]^n, x, 0, Pi/4]
MMA evaluates a ConditionalExpression which constrains0<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 replaceTan[x/n]
withx/n
, then the limit is solvable and givesPi/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 ifn>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 toIntegrate
, 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