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Limit superior of a sequence of oscillating functions related to Chebyshev polynomials
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Let $n in mathbb N$ and consider the polynomial function $f_n colon mathbb R to mathbb R$ defined by
$$f_n(x) = sum_k=0^n (-1)^k binom 2n+1 2k+1 (1 - x^2)^n-k x^2k$$
for any $x in mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:
As $n to infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_50$:
Let $f colon D to mathbb R$ be defined by
$$f(x) = limsup_n to infty f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):
$f$ is an even function, since all of the $f_n$'s are even.
$0 notin D$. Indeed, $f_n(0) = 2n + 1 to infty$ as $n to infty$.
$f(pm 1) = 1$, because $f_n(pm 1) = (-1)^n$ for any $n in mathbb N$.
$f left (pm frac sqrt 2 2 right ) = 1$. This is because:
$$f_n left ( pm frac sqrt 2 2 right ) = sum_k=0^n (-1)^k binom 2n+1 2k+1 left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor 2^n left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor le 1$$
In particular, $f_4m left (pm frac sqrt 2 2 right ) = 1$ for any $m in mathbb N$, so $limsup_n to infty f_n left ( pm frac sqrt 2 2 right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.
What else can we say about $f$? Is it possible to find a "simple" expression?
Thank you in advance for any reply.
Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(sin alpha) = frac sin ((2n+1) alpha)sin alpha$$
for any $alpha in mathbb R smallsetminus pi mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n left (sin frac pi 6 right ) = fracsin left ( (2n+1) frac pi 6 right )sin frac pi 6 le frac 1 frac 1 2 = 2$$
In particular,
$$f_6m+1 left (sin frac pi 6 right ) = fracsin left ( (12 m + 3) frac pi 6 right )sin frac pi 6 = fracsin left ( 2 m pi + frac pi 2 right )sin frac pi 6 = frac 1 frac 1 2 = 2$$
for any $m in mathbb N$, and thus $f left (pm frac 1 2 right ) = 2$.
How can we get a simple expression for $f$ using this information?
sequences-and-series functions polynomials limsup-and-liminf chebyshev-polynomials
asked Jul 19 at 20:01
Luca Bressan
3,84021036
add a comment |Â
up vote
3
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Let $n in mathbb N$ and consider the polynomial function $f_n colon mathbb R to mathbb R$ defined by
$$f_n(x) = sum_k=0^n (-1)^k binom 2n+1 2k+1 (1 - x^2)^n-k x^2k$$
for any $x in mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:
As $n to infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_50$:
Let $f colon D to mathbb R$ be defined by
$$f(x) = limsup_n to infty f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):
$f$ is an even function, since all of the $f_n$'s are even.
$0 notin D$. Indeed, $f_n(0) = 2n + 1 to infty$ as $n to infty$.
$f(pm 1) = 1$, because $f_n(pm 1) = (-1)^n$ for any $n in mathbb N$.
$f left (pm frac sqrt 2 2 right ) = 1$. This is because:
$$f_n left ( pm frac sqrt 2 2 right ) = sum_k=0^n (-1)^k binom 2n+1 2k+1 left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor 2^n left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor le 1$$
In particular, $f_4m left (pm frac sqrt 2 2 right ) = 1$ for any $m in mathbb N$, so $limsup_n to infty f_n left ( pm frac sqrt 2 2 right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.
What else can we say about $f$? Is it possible to find a "simple" expression?
Thank you in advance for any reply.
Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(sin alpha) = frac sin ((2n+1) alpha)sin alpha$$
for any $alpha in mathbb R smallsetminus pi mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n left (sin frac pi 6 right ) = fracsin left ( (2n+1) frac pi 6 right )sin frac pi 6 le frac 1 frac 1 2 = 2$$
In particular,
$$f_6m+1 left (sin frac pi 6 right ) = fracsin left ( (12 m + 3) frac pi 6 right )sin frac pi 6 = fracsin left ( 2 m pi + frac pi 2 right )sin frac pi 6 = frac 1 frac 1 2 = 2$$
for any $m in mathbb N$, and thus $f left (pm frac 1 2 right ) = 2$.
How can we get a simple expression for $f$ using this information?
sequences-and-series functions polynomials limsup-and-liminf chebyshev-polynomials
asked Jul 19 at 20:01
Luca Bressan
3,84021036
One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $n in mathbb N$ and consider the polynomial function $f_n colon mathbb R to mathbb R$ defined by
$$f_n(x) = sum_k=0^n (-1)^k binom 2n+1 2k+1 (1 - x^2)^n-k x^2k$$
for any $x in mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:
As $n to infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_50$:
Let $f colon D to mathbb R$ be defined by
$$f(x) = limsup_n to infty f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):
$f$ is an even function, since all of the $f_n$'s are even.
$0 notin D$. Indeed, $f_n(0) = 2n + 1 to infty$ as $n to infty$.
$f(pm 1) = 1$, because $f_n(pm 1) = (-1)^n$ for any $n in mathbb N$.
$f left (pm frac sqrt 2 2 right ) = 1$. This is because:
$$f_n left ( pm frac sqrt 2 2 right ) = sum_k=0^n (-1)^k binom 2n+1 2k+1 left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor 2^n left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor le 1$$
In particular, $f_4m left (pm frac sqrt 2 2 right ) = 1$ for any $m in mathbb N$, so $limsup_n to infty f_n left ( pm frac sqrt 2 2 right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.
What else can we say about $f$? Is it possible to find a "simple" expression?
Thank you in advance for any reply.
Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(sin alpha) = frac sin ((2n+1) alpha)sin alpha$$
for any $alpha in mathbb R smallsetminus pi mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n left (sin frac pi 6 right ) = fracsin left ( (2n+1) frac pi 6 right )sin frac pi 6 le frac 1 frac 1 2 = 2$$
In particular,
$$f_6m+1 left (sin frac pi 6 right ) = fracsin left ( (12 m + 3) frac pi 6 right )sin frac pi 6 = fracsin left ( 2 m pi + frac pi 2 right )sin frac pi 6 = frac 1 frac 1 2 = 2$$
for any $m in mathbb N$, and thus $f left (pm frac 1 2 right ) = 2$.
How can we get a simple expression for $f$ using this information?
sequences-and-series functions polynomials limsup-and-liminf chebyshev-polynomials
asked Jul 19 at 20:01
Luca Bressan
3,84021036
Let $n in mathbb N$ and consider the polynomial function $f_n colon mathbb R to mathbb R$ defined by
$$f_n(x) = sum_k=0^n (-1)^k binom 2n+1 2k+1 (1 - x^2)^n-k x^2k$$
for any $x in mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)
By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:
As $n to infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_50$:
Let $f colon D to mathbb R$ be defined by
$$f(x) = limsup_n to infty f_n(x)$$
whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.
So far, I have only been able to show the following (see the update below):
$f$ is an even function, since all of the $f_n$'s are even.
$0 notin D$. Indeed, $f_n(0) = 2n + 1 to infty$ as $n to infty$.
$f(pm 1) = 1$, because $f_n(pm 1) = (-1)^n$ for any $n in mathbb N$.
$f left (pm frac sqrt 2 2 right ) = 1$. This is because:
$$f_n left ( pm frac sqrt 2 2 right ) = sum_k=0^n (-1)^k binom 2n+1 2k+1 left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor 2^n left ( frac 1 2 right )^n = (-1)^left lfloor frac n 2 right rfloor le 1$$
In particular, $f_4m left (pm frac sqrt 2 2 right ) = 1$ for any $m in mathbb N$, so $limsup_n to infty f_n left ( pm frac sqrt 2 2 right ) = 1$.
By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.
What else can we say about $f$? Is it possible to find a "simple" expression?
Thank you in advance for any reply.
Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that
$$f_n(sin alpha) = frac sin ((2n+1) alpha)sin alpha$$
for any $alpha in mathbb R smallsetminus pi mathbb Z$, which immediately provides us with many other values of $f$. For instance,
$$f_n left (sin frac pi 6 right ) = fracsin left ( (2n+1) frac pi 6 right )sin frac pi 6 le frac 1 frac 1 2 = 2$$
In particular,
$$f_6m+1 left (sin frac pi 6 right ) = fracsin left ( (12 m + 3) frac pi 6 right )sin frac pi 6 = fracsin left ( 2 m pi + frac pi 2 right )sin frac pi 6 = frac 1 frac 1 2 = 2$$
for any $m in mathbb N$, and thus $f left (pm frac 1 2 right ) = 2$.
How can we get a simple expression for $f$ using this information?
sequences-and-series functions polynomials limsup-and-liminf chebyshev-polynomials
asked Jul 19 at 20:01
Luca Bressan
3,84021036
edited Jul 26 at 7:46
asked Jul 19 at 20:01
Luca Bressan
3,84021036
asked Jul 19 at 20:01
Luca Bressan
3,84021036
asked Jul 19 at 20:01
Luca Bressan
3,84021036
3,84021036
One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10
add a comment |Â
One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10
One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $alpha in (0,pi/2]$ where $sin(alpha)>0$ $$f(sin(alpha)) = fracsup_n sin((2n+1)alpha)sin(alpha).$$
If $alpha$ is an irrational multiple of $2pi$ then $(2n+1)alpha pmod2pi$ is dense in $[0, 2pi]$ and therefore $sup_n sin((2n+1)alpha)=1$. So in this case $$f(sin(alpha))=sin^-1(alpha).$$ If $alpha = 2pi fracpq$ for some coprime integers $p, q$ then a case by case examination on $q pmod 8$ shows
$$s(q) = sup_n sin((2n+1)alpha) = begincases
cos(fracpi2q) & textrmIf $q$ is odd\
cos(fracpiq) & textrmIf $q equiv 2, 6 pmod 8$\
cos(frac2piq) & textrmIf $q equiv 0 pmod 8$\
1 & textrmIf $q equiv 4 pmod 8$
endcases$$
so in this case $$f(sin(alpha))=fracs(q)sin(alpha).$$ Since you already noted that $f$ is even this completely describes $f$.
answered Jul 24 at 19:17
WimC
23.7k22860
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
add a comment |Â
up vote
2
down vote
This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 sqrtx^2 left(x^2-1right) qquad textandqquad v=1-2 x^2+2 sqrtx^2 left(x^2-1right)$$ a CAS produced
$$f_n(x)=frac left(u^n+v^nright)2 +frac left(u^n-v^nright)2 ,fracsqrtx^2 left(x^2-1right)
x^2$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_nleft(sin left(fracpi k12right)right)$ and obtained the (may be) interesting values
$$left(
beginarraycc
k & f_nleft(sin left(fracpi k12right)right) \
0 & 2 n+1 \
1 & cos left(fracn pi 6right)+left(2+sqrt3right) sin left(fracn
pi 6right) \
2 & cos left(fracn pi 3right)+sqrt3 sin left(fracn pi 3right)
\
3 & cos left(fracn pi 2right)+sin left(fracn pi 2right) \
4 & cos left(frac2 n pi 3right)+frac1sqrt3sin left(frac2 n pi
3right) \
5 & cos left(frac5 n pi 6right)+left(2-sqrt3right) sin
left(frac5 n pi 6right) \
6 & (-1)^n
endarray
right)$$
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $alpha in (0,pi/2]$ where $sin(alpha)>0$ $$f(sin(alpha)) = fracsup_n sin((2n+1)alpha)sin(alpha).$$
If $alpha$ is an irrational multiple of $2pi$ then $(2n+1)alpha pmod2pi$ is dense in $[0, 2pi]$ and therefore $sup_n sin((2n+1)alpha)=1$. So in this case $$f(sin(alpha))=sin^-1(alpha).$$ If $alpha = 2pi fracpq$ for some coprime integers $p, q$ then a case by case examination on $q pmod 8$ shows
$$s(q) = sup_n sin((2n+1)alpha) = begincases
cos(fracpi2q) & textrmIf $q$ is odd\
cos(fracpiq) & textrmIf $q equiv 2, 6 pmod 8$\
cos(frac2piq) & textrmIf $q equiv 0 pmod 8$\
1 & textrmIf $q equiv 4 pmod 8$
endcases$$
so in this case $$f(sin(alpha))=fracs(q)sin(alpha).$$ Since you already noted that $f$ is even this completely describes $f$.
answered Jul 24 at 19:17
WimC
23.7k22860
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
add a comment |Â
up vote
1
down vote
accepted
Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $alpha in (0,pi/2]$ where $sin(alpha)>0$ $$f(sin(alpha)) = fracsup_n sin((2n+1)alpha)sin(alpha).$$
If $alpha$ is an irrational multiple of $2pi$ then $(2n+1)alpha pmod2pi$ is dense in $[0, 2pi]$ and therefore $sup_n sin((2n+1)alpha)=1$. So in this case $$f(sin(alpha))=sin^-1(alpha).$$ If $alpha = 2pi fracpq$ for some coprime integers $p, q$ then a case by case examination on $q pmod 8$ shows
$$s(q) = sup_n sin((2n+1)alpha) = begincases
cos(fracpi2q) & textrmIf $q$ is odd\
cos(fracpiq) & textrmIf $q equiv 2, 6 pmod 8$\
cos(frac2piq) & textrmIf $q equiv 0 pmod 8$\
1 & textrmIf $q equiv 4 pmod 8$
endcases$$
so in this case $$f(sin(alpha))=fracs(q)sin(alpha).$$ Since you already noted that $f$ is even this completely describes $f$.
answered Jul 24 at 19:17
WimC
23.7k22860
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $alpha in (0,pi/2]$ where $sin(alpha)>0$ $$f(sin(alpha)) = fracsup_n sin((2n+1)alpha)sin(alpha).$$
If $alpha$ is an irrational multiple of $2pi$ then $(2n+1)alpha pmod2pi$ is dense in $[0, 2pi]$ and therefore $sup_n sin((2n+1)alpha)=1$. So in this case $$f(sin(alpha))=sin^-1(alpha).$$ If $alpha = 2pi fracpq$ for some coprime integers $p, q$ then a case by case examination on $q pmod 8$ shows
$$s(q) = sup_n sin((2n+1)alpha) = begincases
cos(fracpi2q) & textrmIf $q$ is odd\
cos(fracpiq) & textrmIf $q equiv 2, 6 pmod 8$\
cos(frac2piq) & textrmIf $q equiv 0 pmod 8$\
1 & textrmIf $q equiv 4 pmod 8$
endcases$$
so in this case $$f(sin(alpha))=fracs(q)sin(alpha).$$ Since you already noted that $f$ is even this completely describes $f$.
answered Jul 24 at 19:17
WimC
23.7k22860
Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.
Otherwise, for $alpha in (0,pi/2]$ where $sin(alpha)>0$ $$f(sin(alpha)) = fracsup_n sin((2n+1)alpha)sin(alpha).$$
If $alpha$ is an irrational multiple of $2pi$ then $(2n+1)alpha pmod2pi$ is dense in $[0, 2pi]$ and therefore $sup_n sin((2n+1)alpha)=1$. So in this case $$f(sin(alpha))=sin^-1(alpha).$$ If $alpha = 2pi fracpq$ for some coprime integers $p, q$ then a case by case examination on $q pmod 8$ shows
$$s(q) = sup_n sin((2n+1)alpha) = begincases
cos(fracpi2q) & textrmIf $q$ is odd\
cos(fracpiq) & textrmIf $q equiv 2, 6 pmod 8$\
cos(frac2piq) & textrmIf $q equiv 0 pmod 8$\
1 & textrmIf $q equiv 4 pmod 8$
endcases$$
so in this case $$f(sin(alpha))=fracs(q)sin(alpha).$$ Since you already noted that $f$ is even this completely describes $f$.
answered Jul 24 at 19:17
WimC
23.7k22860
edited Jul 25 at 9:25
answered Jul 24 at 19:17
WimC
23.7k22860
answered Jul 24 at 19:17
WimC
23.7k22860
answered Jul 24 at 19:17
WimC
23.7k22860
23.7k22860
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
add a comment |Â
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
That's not true, because $f left ( pm frac sqrt 2 2 right ) = 1 neq sqrt 2$. Indeed, if $alpha = frac pi 4$, there is no $n in mathbb N$ such that $sin((2n + 1) alpha) = 1$, because $sin((2n +1) alpha) = pm frac sqrt 2 2$.
– Luca Bressan
Jul 24 at 19:21
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
@LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this.
– WimC
Jul 24 at 19:22
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
If $operatornameasin(x)$ is not a rational multipe of $pi$ then the supremum is actually equal to $lvert x rvert^-1$.
– WimC
Jul 24 at 19:27
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
I agree that $x mapsto lvert x rvert^-1$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions.
– Luca Bressan
Jul 25 at 9:23
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q equiv 4 pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$sin left ( (2n+1) alpha right ) = sin left ( frac (2n+1) p 2t+1 cdot frac pi 2 right ) = sin left ( p^2 frac pi 2 right ) = 1$$ because $p^2 equiv 1 pmod 4$ for any odd $p$. But the other cases seem quite harder to prove.
– Luca Bressan
Jul 25 at 13:11
add a comment |Â
up vote
2
down vote
This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 sqrtx^2 left(x^2-1right) qquad textandqquad v=1-2 x^2+2 sqrtx^2 left(x^2-1right)$$ a CAS produced
$$f_n(x)=frac left(u^n+v^nright)2 +frac left(u^n-v^nright)2 ,fracsqrtx^2 left(x^2-1right)
x^2$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_nleft(sin left(fracpi k12right)right)$ and obtained the (may be) interesting values
$$left(
beginarraycc
k & f_nleft(sin left(fracpi k12right)right) \
0 & 2 n+1 \
1 & cos left(fracn pi 6right)+left(2+sqrt3right) sin left(fracn
pi 6right) \
2 & cos left(fracn pi 3right)+sqrt3 sin left(fracn pi 3right)
\
3 & cos left(fracn pi 2right)+sin left(fracn pi 2right) \
4 & cos left(frac2 n pi 3right)+frac1sqrt3sin left(frac2 n pi
3right) \
5 & cos left(frac5 n pi 6right)+left(2-sqrt3right) sin
left(frac5 n pi 6right) \
6 & (-1)^n
endarray
right)$$
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
add a comment |Â
up vote
2
down vote
This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 sqrtx^2 left(x^2-1right) qquad textandqquad v=1-2 x^2+2 sqrtx^2 left(x^2-1right)$$ a CAS produced
$$f_n(x)=frac left(u^n+v^nright)2 +frac left(u^n-v^nright)2 ,fracsqrtx^2 left(x^2-1right)
x^2$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_nleft(sin left(fracpi k12right)right)$ and obtained the (may be) interesting values
$$left(
beginarraycc
k & f_nleft(sin left(fracpi k12right)right) \
0 & 2 n+1 \
1 & cos left(fracn pi 6right)+left(2+sqrt3right) sin left(fracn
pi 6right) \
2 & cos left(fracn pi 3right)+sqrt3 sin left(fracn pi 3right)
\
3 & cos left(fracn pi 2right)+sin left(fracn pi 2right) \
4 & cos left(frac2 n pi 3right)+frac1sqrt3sin left(frac2 n pi
3right) \
5 & cos left(frac5 n pi 6right)+left(2-sqrt3right) sin
left(frac5 n pi 6right) \
6 & (-1)^n
endarray
right)$$
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 sqrtx^2 left(x^2-1right) qquad textandqquad v=1-2 x^2+2 sqrtx^2 left(x^2-1right)$$ a CAS produced
$$f_n(x)=frac left(u^n+v^nright)2 +frac left(u^n-v^nright)2 ,fracsqrtx^2 left(x^2-1right)
x^2$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_nleft(sin left(fracpi k12right)right)$ and obtained the (may be) interesting values
$$left(
beginarraycc
k & f_nleft(sin left(fracpi k12right)right) \
0 & 2 n+1 \
1 & cos left(fracn pi 6right)+left(2+sqrt3right) sin left(fracn
pi 6right) \
2 & cos left(fracn pi 3right)+sqrt3 sin left(fracn pi 3right)
\
3 & cos left(fracn pi 2right)+sin left(fracn pi 2right) \
4 & cos left(frac2 n pi 3right)+frac1sqrt3sin left(frac2 n pi
3right) \
5 & cos left(frac5 n pi 6right)+left(2-sqrt3right) sin
left(frac5 n pi 6right) \
6 & (-1)^n
endarray
right)$$
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
This is not an answer since it is just the result from a CAS.
Defining $$u=1-2 x^2-2 sqrtx^2 left(x^2-1right) qquad textandqquad v=1-2 x^2+2 sqrtx^2 left(x^2-1right)$$ a CAS produced
$$f_n(x)=frac left(u^n+v^nright)2 +frac left(u^n-v^nright)2 ,fracsqrtx^2 left(x^2-1right)
x^2$$
Edit
This will not help much, I am afraid, but after your edit, I computed
$f_nleft(sin left(fracpi k12right)right)$ and obtained the (may be) interesting values
$$left(
beginarraycc
k & f_nleft(sin left(fracpi k12right)right) \
0 & 2 n+1 \
1 & cos left(fracn pi 6right)+left(2+sqrt3right) sin left(fracn
pi 6right) \
2 & cos left(fracn pi 3right)+sqrt3 sin left(fracn pi 3right)
\
3 & cos left(fracn pi 2right)+sin left(fracn pi 2right) \
4 & cos left(frac2 n pi 3right)+frac1sqrt3sin left(frac2 n pi
3right) \
5 & cos left(frac5 n pi 6right)+left(2-sqrt3right) sin
left(frac5 n pi 6right) \
6 & (-1)^n
endarray
right)$$
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
edited Jul 25 at 4:08
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
answered Jul 20 at 5:05
Claude Leibovici
111k1055126
111k1055126
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
add a comment |Â
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique.
– Hans
Jul 20 at 5:28
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
@Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same.
– Claude Leibovici
Jul 20 at 5:39
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function.
– Hans
Jul 20 at 5:43
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I tried Wolfram with no success.
– Claude Leibovici
Jul 20 at 5:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
I've updated my question, maybe you'd like to look at it again.
– Luca Bressan
Jul 24 at 18:59
add a comment |Â
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One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-frac23 left(2 n^3+3 n^2+nright) x^2+frac215 left(2 n^5+5 n^4-5 n^2-2 nright) x^4+Oleft(x^6right)$$
– Claude Leibovici
Jul 20 at 8:09
Interested by your observations, I added a few things (of no use, I am afraid).
– Claude Leibovici
Jul 25 at 4:10