I'll do one of these terms: the others (apart from the 3/8) should be similar.
Let z = exp(i x), dz = i z dx. As x goes from 0 to 2 pi, z goes once around unit circle C counterclockwise. We have cos(2 sin(3 x)) = cos(-i z^3 + i/z^3) = cosh(z^3-1/z^3) and cos(2 x) = (z^2 + 1/z^2)/2. So
When expanded in a Laurent series about z=0, cosh(z^3 - 1/z^3) will have only terms involving z to powers divisible by 3. Thus the residue of (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=0 (which is the only singularity inside the circle) is 0, so this integral is 0. -- Robert Israel isr...@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
> I'll do one of these terms: the others (apart from the 3/8) > should be similar.
> Let z = exp(i x), dz = i z dx. > As x goes from 0 to 2 pi, z goes once around unit circle C > counterclockwise. We have > cos(2 sin(3 x)) = cos(-i z^3 + i/z^3) = cosh(z^3-1/z^3) > and cos(2 x) = (z^2 + 1/z^2)/2. So
> When expanded in a Laurent series about z=0, cosh(z^3 - 1/z^3) will have > only > terms involving z to powers divisible by 3. Thus the residue of > (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=0 (which is the only singularity > inside > the circle) is 0, so this integral is 0. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada
Thank you very much for your idea. :))
But this problem taken from highschool math contest.
Student don't know Laurent series or anything of hyperbolic sine.
I will try to follow your idea.
But if anyone can solve this problem in elementary ways.
Bra...@legend.org wrote: > On 23-Nov-2009, William Elliot <ma...@rdrop.remove.com> wrote: >> On Sun, 22 Nov 2009 Bra...@legend.org wrote: >>>>> integration of sin^4(x + sin 3x) dx
>>>>> from x = 0 to x = pi.
>>>>> any idea.? >>>> By symmetry, integrate from 0 to 2 pi and divide by 2.
>> I'll do one of these terms: the others (apart from the 3/8) >> should be similar.
>> Let z = exp(i x), dz = i z dx. >> As x goes from 0 to 2 pi, z goes once around unit circle C >> counterclockwise. We have >> cos(2 sin(3 x)) = cos(-i z^3 + i/z^3) = cosh(z^3-1/z^3) >> and cos(2 x) = (z^2 + 1/z^2)/2. So
>> When expanded in a Laurent series about z=0, cosh(z^3 - 1/z^3) will have >> only >> terms involving z to powers divisible by 3. Thus the residue of >> (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=0 (which is the only singularity >> inside >> the circle) is 0, so this integral is 0. >> -- >> Robert Israel isr...@math.MyUniversitysInitials.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada
>Thank you very much for your idea. :))
>But this problem taken from highschool math contest.
>Student don't know Laurent series or anything of hyperbolic sine.
>I will try to follow your idea.
>But if anyone can solve this problem in elementary ways.
>Please tell me.
That solution suggests the usefulness of a threefold symmetry.
Now for f(x) = sin^4(x + sin 3x) one has
f(x + 2pi/3) = sin^4(x + sin 3x + 2pi/3)). It is easily seen that
cos t + cos (t + 2pi/3) + cos (t + 4pi/3) = 0 for all t and combined
with sin^4 t = 3/8 - 1/2 cos 2t + 1/8 cos 4t it follows that
f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8. Time to say etcetera.
> Bra...@legend.org wrote: > > On 23-Nov-2009, William Elliot <ma...@rdrop.remove.com> wrote: > >> On Sun, 22 Nov 2009 Bra...@legend.org wrote: > >>>>> integration of sin^4(x + sin 3x) dx
> >>>>> from x = 0 to x = pi.
> >>>>> any idea.? > >>>> By symmetry, integrate from 0 to 2 pi and divide by 2.
> >> I'll do one of these terms: the others (apart from the 3/8) > >> should be similar.
> >> Let z = exp(i x), dz = i z dx. > >> As x goes from 0 to 2 pi, z goes once around unit circle C > >> counterclockwise. We have > >> cos(2 sin(3 x)) = cos(-i z^3 + i/z^3) = cosh(z^3-1/z^3) > >> and cos(2 x) = (z^2 + 1/z^2)/2. So
> >> When expanded in a Laurent series about z=0, cosh(z^3 - 1/z^3) will > >> have > >> only > >> terms involving z to powers divisible by 3. Thus the residue of > >> (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=0 (which is the only singularity > >> inside > >> the circle) is 0, so this integral is 0. > >> -- > >> Robert Israel isr...@math.MyUniversitysInitials.ca > >> Department of Mathematics http://www.math.ubc.ca/~israel > >> University of British Columbia Vancouver, BC, Canada
> >Thank you very much for your idea. :))
> >But this problem taken from highschool math contest.
> >Student don't know Laurent series or anything of hyperbolic sine.
> >I will try to follow your idea.
> >But if anyone can solve this problem in elementary ways.
> >Please tell me.
> That solution suggests the usefulness of a threefold symmetry.
> Now for f(x) = sin^4(x + sin 3x) one has
> f(x + 2pi/3) = sin^4(x + sin 3x + 2pi/3)). It is easily seen that
> cos t + cos (t + 2pi/3) + cos (t + 4pi/3) = 0 for all t and combined
> with sin^4 t = 3/8 - 1/2 cos 2t + 1/8 cos 4t it follows that
> f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8. Time to say etcetera.
> Leon
Hello Leon
Your idea look great :))
But I don't understand all.
What does it mean when f(x + 2pi/3) = sin^4(x + sin 3x + 2pi/3)) .?
You mean that int f(x) = int (x + 2pi/3) = int(x + 4pi/3) ? [x = 0 to pi]
> I'll do one of these terms: the others (apart from the 3/8) > should be similar.
> Let z = exp(i x), dz = i z dx. > As x goes from 0 to 2 pi, z goes once around unit circle C > counterclockwise. We have > cos(2 sin(3 x)) = cos(-i z^3 + i/z^3) = cosh(z^3-1/z^3) > and cos(2 x) = (z^2 + 1/z^2)/2. So
> When expanded in a Laurent series about z=0, cosh(z^3 - 1/z^3) will have only > terms involving z to powers divisible by 3. Thus the residue of > (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=0 (which is the only singularity inside > the circle) is 0, so this integral is 0. > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada
I hate to be fussy, but how can the integral of the positive quantity $ \sin^4(x+\sin 3x))$ be zero?
Tim Norfolk <timsn...@aol.com> wrote: > On Nov 22, 3:17=EF=BF=BDpm, Robert Israel > <isr...@math.MyUniversitysInitials.ca> wrote: > > Bra...@lenend.org writes: > > > integration of sin^4(x + sin 3x) dx
> > > from x =3D 0 to x =3D pi.
> > > any idea.?
> > By symmetry, integrate from 0 to 2 pi and divide by 2.
> > I'll do one of these terms: the others (apart from the 3/8) > > should be similar.
> > Let z =3D exp(i x), dz =3D i z dx. =EF=BF=BD > > As x goes from 0 to 2 pi, z goes once around unit circle C > > counterclockwise. =EF=BF=BD =EF=BF=BDWe have > > cos(2 sin(3 x)) =3D cos(-i z^3 + i/z^3) =3D cosh(z^3-1/z^3) > > and cos(2 x) =3D (z^2 + 1/z^2)/2. =EF=BF=BDSo
> > When expanded in a Laurent series about z=3D0, cosh(z^3 - 1/z^3) will > > hav= > e only > > terms involving z to powers divisible by 3. =EF=BF=BDThus the residue > > of (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=3D0 (which is the only > > singularity= > inside > > the circle) is 0, so this integral is 0. > > -- > > Robert Israel =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD > > =EF=BF=BD= > =EF=BF=BDisr...@math.MyUniversitysInitials.ca > > Department of Mathematics =EF=BF=BD =EF=BF=BD =EF=BF=BD > > =EF=BF=BDhttp://w= > ww.math.ubc.ca/~israel > > University of British Columbia =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD > > = > =EF=BF=BD =EF=BF=BDVancouver, BC, Canada
> I hate to be fussy,
If you actually were fussy, you would have read Robert's post more carefully.
> but how can the integral of the positive quantity $ > \sin^4(x+\sin 3x))$ be zero?
It can't be. But clearly, when he said "so this integral is 0", he was referring to int_0^{2 pi} cos(2 x) cos(2 sin(3 x)) dx, rather than to the originally posed integral (the value of which is 3/8 * pi).
Leon Aigret wrote: > On Tue, 24 Nov 2009 16:01:30 GMT, Bra...@legend.org wrote: >> On 23-Nov-2009, snipthis.aig...@myrealbox.com (Leon Aigret) wrote: >>> On Sun, 22 Nov 2009 21:36:11 GMT, Bra...@legend.org wrote: >>>> On 23-Nov-2009, Robert Israel <isr...@math.MyUniversitysInitials.ca> >>>> wrote: >>>>> Bra...@lenend.org writes: >>>>>> integration of sin^4(x + sin 3x) dx
>>>>>> from x = 0 to x = pi.
>>>>>> any idea.? >>>>> By symmetry, integrate from 0 to 2 pi and divide by 2.
> [skipping remainder of solution]
>>>> Thank you very much for your idea. :))
>>>> But this problem taken from highschool math contest.
>>>> Student don't know Laurent series or anything of hyperbolic sine.
<snipped>
> As in the previous post, the final step from > f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8 to > Int f(x) dx over x = 0 ... 2pi equals 3pi/4 > is left as an exercise for the reader :))
> Hope this helps,
> Leon
But (at least for me) it seems unclear who to show that without residue calculus and I do not see why it is easier than
> Tim Norfolk <timsn...@aol.com> wrote: > > On Nov 22, 3:17=EF=BF=BDpm, Robert Israel > > <isr...@math.MyUniversitysInitials.ca> wrote: > > > Bra...@lenend.org writes: > > > > integration of sin^4(x + sin 3x) dx
> > > > from x =3D 0 to x =3D pi.
> > > > any idea.?
> > > By symmetry, integrate from 0 to 2 pi and divide by 2.
> > > I'll do one of these terms: the others (apart from the 3/8) > > > should be similar.
> > > Let z =3D exp(i x), dz =3D i z dx. =EF=BF=BD > > > As x goes from 0 to 2 pi, z goes once around unit circle C > > > counterclockwise. =EF=BF=BD =EF=BF=BDWe have > > > cos(2 sin(3 x)) =3D cos(-i z^3 + i/z^3) =3D cosh(z^3-1/z^3) > > > and cos(2 x) =3D (z^2 + 1/z^2)/2. =EF=BF=BDSo
> > > When expanded in a Laurent series about z=3D0, cosh(z^3 - 1/z^3) will > > > hav= > > e only > > > terms involving z to powers divisible by 3. =EF=BF=BDThus the residue > > > of (z^2 + 1/z^2)/2 cosh(z^3 - 1/z^3) at z=3D0 (which is the only > > > singularity= > > inside > > > the circle) is 0, so this integral is 0. > > > -- > > > Robert Israel =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD > > > =EF=BF=BD= > > =EF=BF=BDisr...@math.MyUniversitysInitials.ca > > > Department of Mathematics =EF=BF=BD =EF=BF=BD =EF=BF=BD > > > =EF=BF=BDhttp://w= > > ww.math.ubc.ca/~israel > > > University of British Columbia =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD > > > = > > =EF=BF=BD =EF=BF=BDVancouver, BC, Canada
> > I hate to be fussy,
> If you actually were fussy, you would have read Robert's post more > carefully.
> > but how can the integral of the positive quantity $ > > \sin^4(x+\sin 3x))$ be zero?
> It can't be. But clearly, when he said "so this integral is 0", he was > referring to int_0^{2 pi} cos(2 x) cos(2 sin(3 x)) dx, rather than to the > originally posed integral (the value of which is 3/8 * pi).
> David- Hide quoted text -
> - Show quoted text -
My apologies to everyone for posting late at night.
On Wed, 25 Nov 2009 15:09:33 +0100, Axel Vogt <&nore...@axelvogt.de> wrote:
>Leon Aigret wrote: >> As in the previous post, the final step from >> f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8 to >> Int f(x) dx over x = 0 ... 2pi equals 3pi/4 >> is left as an exercise for the reader :)) >But (at least for me) it seems unclear who to show >that without residue calculus
Well, one way would be to write the integral of f(x) over [0, 2pi] as a sum of three integrals over [0, 2pi/3], [2pi/3, 4pi/3] and [4pi/3, 2pi] respectively, transform the last two integrals into integrals over [0, 2pi/3] through the obvious change of integration parameter and recombine them.
I must have missed the part of this thread where that type of integral was computeded without using a Laurent series or a hyperbolic function, as the OP requested.
Leon Aigret wrote: > On Wed, 25 Nov 2009 15:09:33 +0100, Axel Vogt <&nore...@axelvogt.de> > wrote:
>> Leon Aigret wrote:
>>> As in the previous post, the final step from >>> f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8 to >>> Int f(x) dx over x = 0 ... 2pi equals 3pi/4 >>> is left as an exercise for the reader :))
>> But (at least for me) it seems unclear who to show >> that without residue calculus
> Well, one way would be to write the integral of f(x) over [0, 2pi] as > a sum of three integrals over [0, 2pi/3], [2pi/3, 4pi/3] and > [4pi/3, 2pi] respectively, transform the last two integrals into > integrals over [0, 2pi/3] through the obvious change of integration > parameter and recombine them.
> I must have missed the part of this thread where that type of integral > was computeded without using a Laurent series or a hyperbolic > function, as the OP requested.
> On Wed, 25 Nov 2009 15:09:33 +0100, Axel Vogt <&nore...@axelvogt.de> > wrote:
> >Leon Aigret wrote:
> >> As in the previous post, the final step from > >> f(x) + f(x + 2pi/3) + f(x + 4pi/3) = 9/8 to > >> Int f(x) dx over x = 0 ... 2pi equals 3pi/4 > >> is left as an exercise for the reader :))
> >But (at least for me) it seems unclear who to show > >that without residue calculus
> Well, one way would be to write the integral of f(x) over [0, 2pi] as > a sum of three integrals over [0, 2pi/3], [2pi/3, 4pi/3] and > [4pi/3, 2pi] respectively, transform the last two integrals into > integrals over [0, 2pi/3] through the obvious change of integration > parameter and recombine them.
> I must have missed the part of this thread where that type of integral > was computeded without using a Laurent series or a hyperbolic > function, as the OP requested.
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