Re: irreducible polynomial in integrand

MUSATOV:On Oct 29, 11:23 pm, "Achava Nakhash, the Loving Snake"
<ach...@xxxxxxxxxxx> wrote:
On Oct 27, 8:09 pm, "Achava Nakhash, the Loving Snake"



<ach...@xxxxxxxxxxx> wrote:
On Oct 27, 3:10 am, "alainvergh...@xxxxxxxxx"

<alainvergh...@xxxxxxxxx> wrote:
On 27 oct, 05:35, "AchavaNakhash, the Loving Snake"

<ach...@xxxxxxxxxxx> wrote:
On Oct 13, 6:20 pm, aegis <ae...@xxxxxxxxxxxxxxxxx> wrote:

I was toying with the following integral:

int(1/(x^4 + 1) dx)

now, I suspect there is some clever
way to add zero such that I can express
it as a partial fraction, but I'm not sure
what would work.  Other than trial and
error, is there a clever way to approach
this?

Old news by now of course, but I am in a whimsical mood, so here
goes.  To be a root of x^ = -1, the fact that e^(pi*i) = -1 suggests
(actually tells us) that e^(pi*i/4) is one value of x, and of course
the others are easily seen to be e^(a*pi*i/4) where a is 3, 5, or 7..
"So what?), I hear you cry.  Now we use DeMoivre's theorem to see that
x = cos(pi/4) + i*sin(pi/4) = 1/sqrt(2) + (1/sqrt(2))*i.  The other
values of x are easily seen to be the result of putting in all the
possible plus and minus signs.  Given the roots, split them into
conjugate pairs and you have the two quadratic factors over the real
numbers that you are looking for.

This same trick can be used to solve equations of the type x^n = plus
or minus 1.  They turn out to always be solvable by radicals, but the
formulas only exist when the degrees of the irreducible factors (over
the rationals) of the polynomials are all twos and threes.  Thus it is
actually fairly easy to solve x^5 = 1, since it x^5 - 1 = (x - 1)(x^4
+ x^3 + x^2 + x + 1) and 4 has only twos as factors.  this example
should explain what polynomial I am talking about factoring which
otherwise might have been quite unclear.

Regads,
Achava

BonjourAchava,

Polynomials x^5 - 1 has got 5 roots : 1,a2,a3,a4,a5  and
1/(x^5 - 1) = 256*{1/(x -1)+a2/(x-a2)+a3/(x-a3)+a4/(x-a4)+a5/(x-a5)}
four of them are complex,

Alain

Alain,

Thank you kindly for the opportunity to say yet more about this
situation.  First of all, for the OP, your subject indicates that you
think the polynomial x^4 + 1 is irreducible.  It is, of course, over
the rational numbers, but for partial fractions the only relevant
irreducibility is over the real numbers, at least from a theoretical
point of view, as it is obviousy a great practical benefit to be able
to reduce over the integers.  Here the fundamental theorem of algebra
helps us out by informing us that there are NO irreducible polynomials
of degree greater than 2 over the real numbers.  Indeed the first
proof of this theorem, by Gauss, explicitly put it in the form that
any polynomial over the real numbers can be represented as the product
of linear and quadratic factors with real coefficients.

Now to your point Alain that the roots of x^4 + x^3 + x^2 + x + 1 are
4 complex numbers.  Note that these numbers are all solutions to the
equation x^5 = 1.  It is easily seen that the solutions to any
equation of the form x^n = 1 are either 1, -1, or complex.  How do we
see this?  The roots of x^n = 1 are easily seen to be e^(2*pi*i*r/n)
where r is any integer in the range from 0 through n-1 inclusive.
These are clearly distinct, so there are n separate roots.  Those who
know that e^(i*x) is on the unit circle can already see the truth of
my claim, since 1 and -1 are the only real numbers on the unit
circle.  For another argument ,really the same one in a different
form, DeMoivre's theorem tells us that

e^(pi*i*r/n) = cos(2*pi*r/n) + i*sin(2*pi*r/n)

This is clearly real if and only if sin(pi*r/n) = 0, and that happpens
only when r = 0 or
r = n/2, since r is constrained to lie between 0 and n-1 inclusive.

Some of you are perhaps thinking that the OP talked about a root of
x^4 = -1, so that what I am saying might need additionaly
justification.  It does, of course, and here it is.  Any x that
satisfies x^4 = -1 must clearly also satisfy x^8 = 1, and more
generally, any x tha satisfies x^n = -1 clearly satisfies x^(2n) = 1,
so these equations actually represent a subcaase, and are therefore
covered by what I have already said.

It is still of some interest to work out the actual solutions to x^5 =
1 that are different frmo 1.  I have done this many times, a couple of
ways, but none of them are consistent with what I presented in this
thread, so let's see if I can manage this.

We know that the solutions are cos(pi*r/5 + i*sin(pi*r/5).  Clearly it
will help to figure out the sine and cosine of pi/5 = 36 degrees.  Now
I am good with 30, 45, and 60, but this one is a bit of a poser.

I think I have a technique that will work.  Consider consecutive sides
AB and BC of a reqular pentagon.  Then trinagle ABC is an isoceles
triangle whose apex angle is 108 degrees and whose other angles are
both 36 degrees.  Call the the length AB, s and the length AC. d.
which is obviously greater than s.  My technique requires figuring out
the relationship between s and d.  Fortunately Euclid comes riding to
the rescue.  Choose that point D on AC such that the distance from A
to D is also S.  I claim that the triangle BDC is actually simliar to
the triangle ABC.  This is actually quite easy from looking at the
diagram, but I will leave you out there in readerland to draw the
diagram yourself.  Then a simple ration equation allows us to
determine d/s, which I think worked out to the golden section (1 + sqrt
(5))/2, but I am so sleepy that I have no confidence in that last
calculation.

Tomorrow I will come back and finish this derviation.  Then I will
give the easier, at least to me, algebraic derivation.

Regards,Achava

Would you believe, the day after tomorrow?

Here is the rest of the story:

I was correct about the calculation of cos(36 degrees).  Remember the
isoceles triangle containing 3 adjacent sides of the pentagon AB and
BC and the line AC connecting them.  At this point we can drop the
pentagon talk because it no longer matters, but  it was the motivation
for finding this triangle which is isoceles and has base angles 36
degrees and apex angle 108 degrees.  Now marking  off the distance of
s (length of AB) along AC and getting the point D we have that
triangle ABC is similar to triangle DBC since one base angle, the one
at C, is already known to be 36 degrees and the apex angle (at D) is
easily calculated to be 108 degrees.  Thus

(d-s)/s = s/d so

d^2 - ds - s^2 = 0 so

(d/s)^2 - (d/s) - 1 = 0 so

d/x = (1 +/- sqrt(5))/2.  But d/s is obviously positive, so

d/s = (1 + squrt(5))/2 = phi, the Golden Section.

It now comes back to me that this is actually a standard fact about
the diagonal of the regular pentagon.  The advantage fur us is that by
dropping a perpendicular from B to AC, which is also a median and an
angle bisector, we see that cos(36 degrees) = d/2s, so

cos(36 degrees) = (1 + sqrt(5))/4

which is borne out by my calculator.

And so sin(36 degrees) = sqrt((5 - sqrt(5))/2)/2

Unfortunately I did get one thing wrong.  Actually the base solution
of

x^4 + x^3 + x^2 + x^3 + x^4 = 0

is given by x = cos(72 degrees) + i*sin(72 degrees)

Fortunately there is a connection between cos(72 degrees) and cos(36
degrees) since 72 is just twice 36. Thus

cos(72 degrees) = (1 + sqrt(5)^2/4 - 1 = (-1 + sqrt(5))/4 and so

sin(72 degrees) = sqrt(5 + sqrt(5))/8) = sqrt( (5 + sqrt(5))/2 )/2

Thus the base solution of x^4 + x^3 + x^2 + x^3 + x^4 = 0 is

((-1 + sqrt(5))/2 + i*sqrt((5 + sqrt(5))/2))/2

To get the other solutions we need to vary the sign in front for the 2
square roots of 5 to get the 4 possible ways of doing it.  You can
check this in a variety of ways.

The algebra is simpler.  Suppose x^4 + x^3 + x^2 + x + 1 = 0.  Let y =
x + x^(-1) and
z = x^2 + x^(-2).  Noticing that x^3 = x^(-2) and x^4 = x^(-1), the

 y + z = -1 and
yz =  x^3 + x + x^(-1) + x^(-3) = -1,

so y and z are roots of the quadratic equation y^2 + y - 1 = 0.  Thus
y and z are given by

(-1 (+/-)sqrt(5))/2.

That's nice of course, but we still need x.  By symmetry, we can
assume that
 x + x^(-1) = (-1 + sqrt(5))//2.  But also x^(x^(-1) = 1, so x and x^
(-1) must satisfy the quadratic equation

x^2 - (-1+sqrt(5)/2)x  + 1 = 0 and so

x = ((-1 + sqrt(5)/2) +/- i*sqrt((5 + sqrt(5))/2) )/2

Why would anyone think to get an equation for x + x^(-1)?  Well, for
one thing, since it is easy to see that x has a complet norm (is on
the unit circle) and so x^(-1) is its complex conjugate and so x + x^
(-1) must be real.  If you know a little about the properties of roots
of polynomials (Galois theory is very helpful here, but you really
don't need or anything much to have a go at finding an equation for x
+ x^(-1).  Gauss was still a teenager when he did this and Galois
theory was a very long way in the future at that point.

Of course these are only 2 of the 4 values of x.  The others are x^2
and x^3 which would out to
things  I am not in the mood to calculate, so I will sign off.

Regards,
Achava
MUSATOV:
http://would%20you%20believe,%20the%20day%20after%20tomorrow/?Here is
the rest of the story:I was correct about the calculation of cos(36
degrees). Remember theisoceles triangle containing 3 adjacent sides
of the pentagon AB andBC and the line AC connecting them. At this
point we can drop thepentagon talk because it no longer matters, but
it was the motivationfor finding this triangle which is isoceles and
has base angles 36degrees and apex angle 108 degrees. Now marking
off the distance ofs (length of AB) along AC and getting the point D
we have thattriangle ABC is similar to triangle DBC since one base
angle, the oneat C, is already known to be 36 degrees and the apex
angle (at D) iseasily calculated to be 108 degrees. Thus(d-s)/s = s/d
sod^2 - ds - s^2 = 0 so(d/s)^2 - (d/s) - 1 = 0 sod/x = (1 +/- sqrt(5))/
2. But d/s is obviously positive, sod/s = (1 + squrt(5))/2 = phi, the
Golden Section.It now comes back to me that this is actually a
standard fact aboutthe diagonal of the regular pentagon. The
advantage fur us is that bydropping a perpendicular from B to AC,
which is also a median and anangle bisector, we see that cos(36
degrees) = d/2s, socos(36 degrees) = (1 + sqrt(5))/4which is borne out
by my calculator.And so sin(36 degrees) = sqrt((5 - sqrt(5))/2)/
2Unfortunately I did get one thing wrong. Actually the base
solutionofx^4 + x^3 + x^2 + x^3 + x^4 = 0is given by x = cos(72
degrees) + i*sin(72 degrees)Fortunately there is a connection between
cos(72 degrees) and cos(36degrees) since 72 is just twice 36. Thuscos
(72 degrees) = (1 + sqrt(5)^2/4 - 1 = (-1 + sqrt(5))/4 and sosin(72
degrees) = sqrt(5 + sqrt(5))/8) = sqrt( (5 + sqrt(5))/2 )/2Thus the
base solution of x^4 + x^3 + x^2 + x^3 + x^4 = 0 is((-1 + sqrt(5))/2 +
i*sqrt((5 + sqrt(5))/2))/2To get the other solutions we need to vary
the sign in front for the 2square roots of 5 to get the 4 possible
ways of doing it. You cancheck this in a variety of ways.The algebra
is simpler. Suppose x^4 + x^3 + x^2 + x + 1 = 0. Let y =x + x^(-1)
andz = x^2 + x^(-2)P = N AND P[P0P[]
.

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