Hard trigonometry integration
Bra...@lenend.org
from x = 0 to x = pi.
any idea.?
------------------
Here is I tried
If Let f(x) = sin(x + sin 3x)
I find that f(x) = f(-x)
There are 3 value of x such that f(x) = 0,
f(0) = f(pi) = 0 and f(some value) = 0.
no symetry about x = pi/2
----------------
Can anyone solve this problem.?
TIA
Braver
Robert Israel
> 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.
sin^4(x+t) = 3/8 - 1/2 cos(2 x + 2 t) + 1/8 cos(4 x + 4 t)
= 3/8 - 1/2 cos(2 x) cos(2 t) + 1/2 sin(2 x) sin(2 t)
+ 1/8 cos(4 x) cos(4 t) - 1/8 sin(4 x) sin(4 t)
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
int_0^{2 pi} cos(2 x) cos(2 sin(3 x)) dx
= int_C (z^2 + 1/z^2)/2 cosh(z^3-1/z^3) (iz)^(-1) dz
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
Bra...@legend.org
On 23-Nov-2009, Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:
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> Newsgroups: sci.math
> Subject: Re: Hard trigonometry integration
> Date: Sun, 22 Nov 2009 14:17:36 -0600
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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.
TIA
Braver
William Elliot
>>
>>> 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.
>>
>> sin^4(x+t) = 3/8 - 1/2 cos(2 x + 2 t) + 1/8 cos(4 x + 4 t)
>> = 3/8 - 1/2 cos(2 x) cos(2 t) + 1/2 sin(2 x) sin(2 t)
>> + 1/8 cos(4 x) cos(4 t) - 1/8 sin(4 x) sin(4 t)
>>
>> I'll do one of these terms: the others (apart from the 3/8)
>> should be similar.
>>
>
> 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.
>
integeral (sin^4 x + sin 3x) dx
was made.
----
Bra...@legend.org
On 23-Nov-2009, William Elliot <ma...@rdrop.remove.com> wrote:
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Hello Wiliiam
It's integeral sin^4(x + sin 3x) dx not integeral (sin^4 x + sin 3x) dx
Braver
Axel Vogt
Here is another attempt, incomplete, may be you can make it sound.
It seems that Int( sin(2^k*x+sin(3*x))^4, x = 0 .. Pi) is the same for
all k, k a positive integer.
Asymptotical 2^k*x+sin(3*x) ~ 2^k*x+O(1), (2^k*x+c)/(2^k*x+sin(3*x))=1
for any c.
Now Int( sin(2^k*x+c)^4, x = 0 .. Pi) can be computed and if k is some
positive integer it is 3/8*Pi.
However I used Maple to play with that question ...
Leon Aigret
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
A N Niel
the plain x, so that we get
(1/Pi)*int(int(sin(x+t)^4,x=0..Pi)/sqrt(1-t^2),t=-1..1);
3
- Pi
8
Axel Vogt
Would you mind to say a bit about that? It puzzles me.
Bra...@legend.org
On 23-Nov-2009, Axel Vogt <&nor...@axelvogt.de> wrote:
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> From: Axel Vogt <&nor...@axelvogt.de>
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> Subject: Re: Hard trigonometry integration
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Thanks Alex :))
Braver
Bra...@legend.org
On 23-Nov-2009, snipthi...@myrealbox.com (Leon Aigret) wrote:
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> From: snipthi...@myrealbox.com (Leon Aigret)
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> Subject: Re: Hard trigonometry integration
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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]
Can you explain me more.
TIA
Braver
Tim Norfolk
I hate to be fussy, but how can the integral of the positive quantity $
\sin^4(x+\sin 3x))$ be zero?
David W. Cantrell
> 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
> >
> >
> > > any idea.?
> >
> > By symmetry, integrate from 0 to 2 pi and divide by 2.
> >
> > =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD =EF=BF=BD=3D 3/8 -
> > 1/2 =
> cos(2 x) cos(2 t) + 1/2 sin(2 x) sin(2 t)
> > =EF=
> =BF=BD + 1/8 cos(4 x) cos(4 t) - 1/8 sin(4 x) sin(4 t)
> >
> > I'll do one of these terms: the others (apart from the 3/8)
> > should be similar.
> >
> > As x goes from 0 to 2 pi, z goes once around unit circle C
> > 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
> >
> > int_0^{2 pi} cos(2 x) cos(2 sin(3 x)) dx
> >
> > 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.
> > --
> > =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
Leon Aigret
>On 23-Nov-2009, snipthi...@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.
>> >
>> >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.
[skipping remainder of sketched solution]
>Your idea look great :))
>
>But I don't understand all.
Sorry, that was a very sketchy sketch of a solution. A more formal
version would have been:
Let u = u(x) = x + sin 3x, so f(x) = sin^4(u). and also
f(x + 2pi/3) = sin^4(x + 2pi/3 + sin 3(x + 2pi/3)) =
= sin^4(x + 2pi/3 + sin 3x) = sin^4(u + 2pi/3) and
f(x + 4pi/3) = sin^4(u + 4pi/3).
Using the easily verified identities
sin^4 t = 3/8 - 1/2 cos 2t + 1/8 cos 4t and
cos t + cos (t + 2pi/3) + cos (t + 4pi/3) = 0, one gets
f(x) + f(x + 2pi/3) + f(x + 4pi/3) =
= sin^4(u) + sin^4(u + 2pi/3) + sin^4(u + 4pi/3) =
= 3/8 - 1/2 cos 2u + 1/8 cos 4u +
+ 3/8 - 1/2 cos 2(u + 2pi/3) + 1/8 cos 4(u + 2pi/3) +
+ 3/8 - 1/2 cos 2(u + 4pi/3) + 1/8 cos 4(u + 4pi/3) =
= 3/8 - 1/2 cos 2u + 1/8 cos 4u +
+ 3/8 - 1/2 cos (2u + 4pi/3) + 1/8 cos (4u + 2pi/3) +
+ 3/8 - 1/2 cos (2u + 2pi/3) + 1/8 cos (4u + 4pi/3) =
= 9/8
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
Axel Vogt
> On Tue, 24 Nov 2009 16:01:30 GMT, Bra...@legend.org wrote:
>> On 23-Nov-2009, snipthi...@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.
> 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
Int(1/8*cos(4*x+4*sin(3*x)),x = 0 .. Pi) = 0 and
Int(1/2*cos(2*x+2*sin(3*x)),x = 0 .. Pi) = 0
Tim Norfolk
>
> - Show quoted text -
My apologies to everyone for posting late at night.
Leon Aigret
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.
>and I do not see why
>it is easier than
>
>Int(1/8*cos(4*x+4*sin(3*x)),x = 0 .. Pi) = 0 and
>Int(1/2*cos(2*x+2*sin(3*x)),x = 0 .. Pi) = 0
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
Axel Vogt
> On Wed, 25 Nov 2009 15:09:33 +0100, Axel Vogt <&nor...@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.
A cute way, overall! Thx, now I got it.
Bra...@legend.org
On 25-Nov-2009, snipthi...@myrealbox.com (Leon Aigret) wrote:
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> From: snipthi...@myrealbox.com (Leon Aigret)
> Newsgroups: sci.math
> Subject: Re: Hard trigonometry integration
> Date: Wed, 25 Nov 2009 15:39:34 GMT
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Now I understand all. !!!
Thanksssssssssss LEON :)))
Braver