Christian Frétigny
Some applications of the Remainders theorem
E. Estenave / C. Fretigny
A bucket load of formulae !
I can not put them all down, there's too
many of them ! You will discover them as you read this page.
Nevertheless here are a few examples:
but also :
...
Around
The first serie at the top of this page for k=0
seems to be due to E. Estenave, Docteur ès Sciences, and was
published by its author with two quite confusing proofs in a little
book "Sur une série servant à définir
approximativement le nombre Pi" ("On
a serie allowing to define approximatly the number Pi") in 1901.
Christian Frétigny,
researcher and lecture at ESPCI in
the field of quantum physics, found this book in a bookshop in Paris.
He also told me that an exigeant and anonymous hand barred in pencil
the word "define" and replaced it with "determine" with the annotation:
"Speak french" ! (This makes more sense in French)
Anyway it stands that C. Frétigny
offered this serie and one of its generalisation so to complete my
collection, by showing me an other proof which used the remainders
theorem. After reading upon the subject (I've never studied the
holomorphes function, unfindable at uni!), we started a little
correspondance in october as more and more series appeared using the
same method. This page tries to synthestise the result of this
exploration, of course very incomplete, seing the richness of the
method. Several trails are offered, with at least one example for each
of them, when the time or intuition failed to explore more deeply this
idea.
We don't know if this is a complete study or a classification exist on
this subject, or even if certain of those series, they have a poor
performance, appear simply through rearrangement, or even if they are
known (well, here we should get back to reality and Maple knows a fair
few of them!) The calculating interes is not great, obviously, and this
is why maybe I have not found those formulae, but the theory, the
generalisation and the esthics of those series are very interesting.
Without counting that we can reprove a good part of Euler's
formulae very quickly.
To my knowledge, and with a bit of research, nothing exist on the web,
so if you have some info, don't hesitate!
The Remaiders' theorem
First of all, lets us remember some
definations before the famous remainders' theorem which is used for
almost all calculation starting point:
The holormophes functions are complex variable functions on an open E of C in C, and differentiables. Let us
place ourself in the more general case of meromophes functions, ratio
of two holomorphes functions f(z)=g(z)/h(z), where the main
problem is that h can be cancelled in certain z_{i}
of course. So let those famous z_{i} with iI be the "poles" of f (that is to said
rigourously that for all i of I, there exists a minimal
integer n_{i} , called the pole order, such that (zz_{i})^{ni}f(z)
is differentialble in z_{i}).
The general developpement of Laurent of a complex function, on a ring
(space in between two circles, for convergence problem of the sum in + and ), is given by :
For the meromorphes functions, the summation
is simply finite on the negative numbers, basicly the a_{i}
are zero for i<(N), N>0, N being the
pole order of z_{0}.
We define the remainder of f in z_{i} as the
coefficient a_{1} in the developpement of Laurent of f
with z_{i} in the place of z_{0}. When z_{i}
is a simple pole, this comes down to calculating :
For a pole of order m, we have the
heavier formula :
Good, up till here, it's just been
definitions, but what can we do with our meromorphe function that is
getting bored?
Well we then consider a closed curve G of classe C^{1}
contained in E and which includes the poles z_{i},
iJ, JI.
We defined the indice of G in z_{i}
(winding number) by :
which correspond in practical terms to the
number of passage of the curve G (counted in the trigonometric
sense) around the point z_{i} hence it's easy to
calculate for nice contours.
Hence, finaly, the remainders' theorem gives us :
Intuitevely, this tells us that we can know
the integral of f on a closed
curve if we know the behaviour of f
at it's singular points that are his poles. This surprising theorem is
very important in complex analysis, and, as you are going to see,
equally in the researche of series tending to !
Application of the theorem to a class of
functions
In all the sequences, the curves used are
simple squared curves, hence the indices of those curves at any point
inside will always be 1 after the definition (useful !). Intuitevely,
we have now on one side the sum of remainders, which will give us the
series, and the other an integral, that we hope can be easily
calculated. For certain functions, this integrals is null when we make
the curve tends to infinity, it's this kind of functions we introduce
as a class here :
Let . We consider a closed curve G(N), on
which we are going to integrate, the square of the sumits : . We then avoid goingthrough the poles of cotan
(denoted cot), that is the zeroes of tan(z), even more specific the nZ (phew..).
Let us now show that integral of g on the curve G(N)
tends towards 0 when N tends towards infinity.
On the right hand vertical arm, z=N+1/2+i.y, y
varying from N1/2 to N+1/2, we have :
because
and on the top horizontal arm, z=x+i(N+1/2), x
varying from N1/2 to N+1/2, we have :
because exp((2N+1))1>2 for N0 and
hence, on the two arms, cot(z)2, and since cot
is odd, we can deduce we have this result on the whole curve. Since we
are in R, we can put an upper limit on the integral with the
length of the curve * the maximum of the function.
The square having for perimeter 4*(2N+1), we obtain :
But, for z on the square, for
example of the right vertical arm, we have z=(N+1/2)+iy and
z(N+1/2) (similarly for all the other arms). We can even
more affine :
Let us now tend the curve towards infinity,
that is let us tend N to infinity or even z towards infinity for z
on the squared curve. It remains to choose a function f such
that and the integral will tend towards 0.
This form reminds us a bit of Riemann's criteria for real series, and
even, all functions f of the form of a rational fraction having
a denominator degree greater than the numerator degree will do. In
particular, all f of the form (z^{a}+c)^{1}
with a>1 will do.
We used tan in g(z) but we can also look on the side of
sin and cos. For cos, no problem if we
put 1/cos in it's exponential form, since we have an upper
limit exponential sum on the denominator in the same way as for cot.
To put an upper limit on 1/sin, we just need to use the
relation , since we've put an upper bound on
cot(z) in modulo by 2. We
can still use the equivalent hyperbolic functions (cotanh, 1/sinh,
1/cosh) but this seems to be all.
We are now left with an integral tending to 0 if the curve
tends to infinity. But this also means that all those poles z_{n}
of our function g for n in I will be contained
in this infinite curve.
The great advantage is that cot(z), as it's
equivalents in 1/cos and 1/sin, admits an infinite
number of poles, which here are the z_{n}=n for n
in Z, and the sum of remainders at those point will hence
produce an infinte serie.
After taking care of the integral which is null, and as we have since
that the infinite indices of the curve at the poles z_{i}
will be equal to 1, our dear remainder theorem comes down to
choose f correctly :
Our objectives, since the series are going
to have a slow convergence, is that at least they will be rational. Let
us now examine a first application so to observe a simple method before
explaining in more details why this works and then orientate the leads
of research:
Preliminary example
*** Case m=4
and p=4
It's really the one that works the most, hence I will detail the
calculations, that I will definitely ignore afterwards! It is first of
all clear that this function is contained in the previous studied case,
it's integral on it's infinite curve is hence nul.
Let us now look at the poles of g and their remainders :
* 0 is a pole (for z in the denominatori), it's
remainder is obtained by calculating
* The (2n+1) for nZ are the poles
corresponding to the squares of cos, their respective
remainders are :
Excellent, on one side, we have in the first
remainder, and the other the sum of remainders in 2n+1 give a
serie for nZ that is quite nice !
Yes, but the problem, there is still some remainders in g at
the mth roots of p which are also the poles of g
because they cancel z^{m}+p. It would be better if
their sum was nice, most of them zero, will be perfect. In the present
case the 4th roots of 4 are 1+i,1i, 1+i,
1i, are simple luckuly, and that's what will limit the
posibilities for other values of m.
* Let us calculate the remainders in those roots. Notice first, that in
the general case, that when we manipulate a mth root of p,
do not forget that z^{m}=p. But, with z(z^{m}+p)
in the denominator of g, in the calculation of the remainder of
g a mth root of p, denote it a_{i},
we need to take the value in a_{i} of
Hence this term does not depend on a_{i} and only the cos
will depend on it, the remainder of g in an mth root
of p is :
in our case m=4, p=4
But this, is a really good new because the sum of those remainders is
nul. In fact, each remainder of a root is cancelled with the root of
it's conjugate
We just now need to apply the remainder's theorem by writting that the
sum of the remainder of g is nul :
and finaly (phew !)
Generalisation of the case m=4
Consider the function
(g1 for first type of generalisation) with k in N
(or in Z but here, (2k+1)^{4} take the same
value since k is negative because of the power 4)
Here and for the rest, I will use a little table to represent the poles
and the remainders, it is still more clear than if the results are
drowned in the middle of the text. Furthermore, only the sum of the
remainders are important for the calculations, not the remainder of a
root, except in particular case, since we know it's die to the sin
that the sum cancel down. I will hence often odmit the expression of
the only remainder of a root, especially during the generalisation, the
result is quite annoying !
Poles

Remainders

0


(2n+1)


+/(2k+1)+/i(2k+1)


sum of the remainders
of roots

0

Which allows us similarly to previous work
to construct the following serie :
Note that if we make k tend to
infinity, the ratio of the terms in front of the sum and the
denominator tends toward 4, and we refind the sequence of
Leibniz ! Ok, rigour is slightly overlooked since we did not justify
the passage to the limit, but the note was valid especially for the
intuitivity of the result.
Increasing the degree of the
denominator
Harder !
The degree of the denominator, ignoring the z, is of 4
with the z^{4}+4. Well, we can increase it so to make
the sequence converge faster.
Let us consider the function :
Let us calculate the remainders :
Poles

Remainders

0


(2n+1)


sum of the roots'
remainders

0

We then obtain the serie :
We can be interested in the limite case when j tends to
infinity, this comes down to calculating the infinite product :
We can start for this from the well know expression (see for example
the encyclopédie
by Eric Weisstein) :
This is completly intuitive, since the
infinte product is cancel as soon as z takes an odd value,
hence it's a (2n+1)
perodic function, this must be the cosinus... We have then :
and finally :
Of course, this is not really interested to
calculate the decimals and here again rigour has been a bit forgotten,
but the formula is quite esthetic, no?
Note that I think we can bring it back to the identities on the Riemann
function foun by Simon Plouffe
and the Borwein in 1997/1998 (see his
personal page
). An identity used is in fact the sum
The proof is given on one of the pages if
this site dedicated to Plouffe and other
formulae of the same kind are there.
So, how does it works and how to continue
?
The major point to study is the one about the form of the mth
roots of p. We have seen that the cancellation of the sum of
remainders for those roots depends on the transformation of cos into sin and that sin is an odd function. Since we
start from a cos, we need a translation of since cos(x)=+/sin(x).
In the previous example, since we start from cos(z/2), for the calculation of the remainder, the mth
roots of p needs to have a real part of the form (2k+1),
hence integer and odd. This was the case previously for m=4 and
p=4 since the roots were the +/1+/i.
This property essentially depends on the degree of the power of m
through the form of the mth untiy roots. To find the roots of z^{m}+p,
we just need simply to multiply the mth roots of unity by p^{1/m}.
If we denote a_{k} the real part of the kth mth
root of unity, a sufficient condition for this to work is that there
exist integer p such that
a_{k}.p^{1/m}=(2k+1)
and that this is valid for all the roots
with the same p.
We will see later on that only m=2, m=3, m=4
and m=6 seems to work, because above degree 6, for
example for degree 8 which seems favourable, we find one p for 4
roots and an other p for the 4 other roots, and nothing
more work, we have a constant gap of 2^{1/2} ! Note also that,
to take advantage of integer solution,we will sometime inpractise be
more interested in z^{m}+/q^{m}.
And the generalisation 
why does it work ?
When the degree m works and when we have found p,
nothing stop us from multiplying p by (2k+1)^{m},
owe know we can multiply then the mth roots of p by (2k+1),
but since their real part is odd and that the product of two odd number
is odd, the new roots have still the good property on their real part.
This exaplain the generalisation on the previous example works.
It is not forbidden to put on the denominator as many polynomial of the
form z^{m}+p(2k+1)^{m} as we want. This creates
some product that increase the degree of the denominator and leads to
some limiting cases as we have seen
So here was a brief overview of the
principle of exploration, it is not impossible that some particular
cases that do not fall in the abover hypotheses appears, but so far
none has appeared. All the following example are hence varients on this
principle. Here are the various leads that were considered :
Lead 1 : uses of cos, of z
and an other polynomial of degree m : z^{m}+p, m>1
let
*** Case m=4
and p=4
This one, we have seen it as a preliminary example
*** Case m=2
* p=4(2k+1)^{2}, p=4(2k)^{2},
p=4(6k+/1)^{2}/3
m=2 is a bit of a particular case, the simplicity of the
roots allows us to study the general case while initially leaving p
as a parameter in g. We will hence naturally find the values of
p chosen above :
Here is the usual table :
Poles

Remainders

0


(2n+1)


sum of the roots
remainders


The i in the cos is very
annoying if we want a nice relation and not land on some cosh
with nastier effects. To cancel it down we just need to take negative p,
and another i
will come out of the roots.
Good, next we need to make sure we don't cancel down the cos on
the denominator, we justneed a factor of 4 under the roots so
to land on the cos that are multiples of which we know the value.
Finally, we can take a value p of either 4(2k+1)^{2}
or 4(2k)^{2}. And this choice is far from being
random, see or yourself:
With p=4(2k+1)^{2} , we obtain for integer k ,
still by cancelling down the sum of remainders above :
But with p=4(2k)^{2} ,
surprise, we get :
because the remainder in 0 cancel
downs the one with the roots since cos(2k)=1.
Hence wether we chose the odd 2k+1 or even 2k, we have
a sum equal to or nul, this is an interesting
phenomene !
In fact, more generally, if we choose p=4a^{2}, we
obtain according to the table of remainders the following very
beautiful result, which explains our curiosity :
The only other rational value for the cos
concern the value a=(6k+/1)/3, which gives the new serie :
Increasing the degree of the
denominator
this gives :
and for the limit case when making j tend towards infinity :
This last serie is already cited according
to C. Frétigny in
Gradshteyn, I.S. and Ryzhik, I.M. " Tables of Integrals, Series,
and Products " San Diego, CA: Academic Press, 1996 (in CD version)
I unfortunatly don't have either the book nor the CD, but I think it
will please me !
There is no limiting case for the second nul serie, we land on the sum
of (1)^{n} whose result, if normally undeterminable,
correspond intuitevely to the value 0 of the initial serie.
We can increase the degree of the denominator in more exotic ways.
Since we have found a serie whoes sum is null. we can substract it from
the serie giving Pi to find a new serie. Hence, we have :
* p=(2k+1)^{2}
There is this other interesting case for m=2. It correspond to
the case where we know the denominator will cancel for the value n=k
since we will get (2n+1)^{2}(2k+1)^{2} in the
serie. The order of the poles in z=+/(2k+1) is then just 1.
But we just need to calculate the remainders in those values with the
differential formula given in the secion "Remainders' Theorem", and the
remainders of the other values 2n+1 using classic methods, this
gives :
Poles

Remainders

0


(2n+1) (n#k)


+/(2k+1)


By cancelling the sum of those remainders, we can then write for all
integer k :
The generalisation by increasing the degree
of the denominator seems more complex simply because it force us to
remove other values of n of the serie. We can see in all case
that there exists no limiting case (because there would remain no
values of n in the serie !!)
*** Case m=3 and p=2^{3}
A very classical case, looks a lot like the case m=4, and
easily generalised, it's in fact from here we are going to start since
the case p=8 correspond to k=0 in the generalisation :
Remember the function used for this generalisation :
Here is the table of poles/remainder :
Poles

Remainders

0


(2n+1)


24k (1st
root)


1+2k+/I(2k+1)3^{1/2
}(the sum of the roots are cancelled)


We can see, the only perticularity of this
case is to have a first root independant of the others (due to the
degree 3), but its remainder is perfectly usable, and we can
form the sum
(which was quite predictable !) :
A little varient anyway, the sum is this
time down on Z, still due to the degree 3, the sum on N
is not equal to the sum on Z^{}. I think that it is
for this reason that we don't need to just make k
tend toward infinity to refind the sequence by Leibniz
like before with the case m=2.
However, if we combine the sums on Z^{} and Z^{+},
we still manage to obtain a serie on N which is exactly the
same as the one presented in the following section "Case m=6
and p=2^{6}"
Increassing the degree on the
denominator
The principle is known now, let us avoid charging straight in this page
is already long enough. We can with no problem apply the same process
as for m=2 or m=4, we obtain for all integer j :
The limiting case does not seem easy to
calculate due to the infinite product (is it even calculatable?), if
someone has a result... All that we can say is that its a (2n+1)
periodic function on R^{+} and not on R^{}.
This remind us a bit of the problem concerning the inverse of odd
powers, that is the zêta
function for the odd value.
*** Case m=6 and p=2^{6}
Last value that is apparently possible, this case does not give any
theoritical difficulties. The roots of degree 6 are :
+/(2+4k) which gives like in degree 3 a
multiple of Pi
+/(2+4k){j, j^{2}} which has an odd real part,
hence the cos is transformed as always into sin and the
remainders of the conjugate roots cancel down.
We hence quickly move on, to fill in the table of poles/remainder and
on the results :
Poles

Remainders

0


(2n+1)


sum of the remainder of
roots

0

This give :
Increasing the
degree on the denominator
No problem here either, we easily get :
For the limiting case, we use the same
procedure as for m=4, know that the infinite product of degree 2
is equal cos(z/2). The calculation is very
hard and the size (already quite big) of this page significant, we are
going to avoid to detail the calculations. To obtain the infinite
product of degree 6, we are going to multiply three time this
infinte product of degree 2 is worth cos, with z=(n+1/2),
z=j.(n+1/2) et z=j^{2}(n+1/2). We hence find :
The I will gives some developped cosh , (but it's not very
nice, better leave this under it's compact form) which will assure a
fast convegence.
Lead 2 : using sin, and an other
polynomial of degree m : z^{m}+p, m>1 :
Since we have obtained odd series with the cos, that is with a (2n+1)
in the denominator of the general term of the serie, we can expect to
obtain some even series with the sin. But unfortunatly, not
everything work out nicely...
En fact, contrary to the case (2n+1), 0 is a particular
case of 2n, and the remainder in 0
is hence harder to calculate, it is in fact nul, does not appear anymore...
We need to find a trick, and shift the z on the denominator to z1,
that is consider the function
The 1 is chosen so to obtain a nice value of sin, but
we could have an other shift, as long as its not in the form of 2n.
All of this is nice, but with habitude of this method, we can easily
see that this shift will ruin our hope to find even series with this
lead since we will have on the denominator something of the form q.na.
Very bad...
*** Case m=4 and p=4
A big problem of this case, the shift stops the sum of remainders of
the roots from completly cancelling down and we can see some cosh
with harsher effects...
So let us see the pole/remainder table :
Pole

Remainder

1


2n


sum of the remainder of
roots +/1+/i


Which gives the serie :
We really need to get rid of this cosh
term, otherwise the serie has not much use...
One solution is to write this term under the form of a serie as the
same method of remainder, then bring back the two serie together.
Hence, we consider :
With the usual method, the remainder in 2n and the sum of those
for the roots give the serie :
Side remarque, note the extraordinary purity
of the serie on the right, of exceptional simplicity, and the horror of
the left term! Does the form of the member on the left has a nice
properties so to be equal to this kind of serie?
Still if we regroup the two previous equations, we get :
Yet another nice serie, but can we find
another more satisfying result ?
Generalisation
How to generalise it ? As with the usual way with the (2k+1),
we are going to try and be a bit more precise by introducing the
function :
We are going to look for the condition on a, b, and c
such that the sum of the remainders of roots of z^{4}+4
cancell down.
For this, let us see the calculation of remainders for z=1+/i
(while not forgetting that z^{2}=+/2i and z^{4}=4)
Ok, we know that z^{2}=+/2i hence, for the sum
of those two remainders cancell down, we just need to use this property
and that the terms after the z on the numerator is hence
proportional to z, that is :
(+/2a.i+b)(+/2ic)=4ab.c+/2i(a.c+b)=1+/i
The serie previously considered correspond to a=1, b=6
and c=1
The sum of the two other remainders cancel down because the square of
the roots are real (+/2).
Let us interest ourself now to the other poles, and sum using the
remainder's theorem as always. For 2n, clasical calculation
that I won't show.
Since the sum of the remainders for the roots of z^{2}c
give , we have in all interest to choose
c=(2k+1)^{2} so that the sin be equal to
1, which gives after numerous simplification (phew !):
The original formula is the case k=0.
As long as it works, this method is not really made for the
generalisation and depends mostly on the properties on the function.
Here, the degree 4 is ideal for the roots 1+/i have as
a square +/2i, which allows the cancellation of the remainder
of the conjugated roots of z^{4}+4.
The degree 2, 3 and 8 did not seem to be
favourable, but there mst be a lot more to explore on this side !
Lead 3 : uses of tan and another
polynomial of degree m : z^{m}+p, m>1
Intuitevely, the derivative of tan
: (1+tan^{2}) make us things that we can obtain some
nonalternating series, unlike to the sin and to the cos.
Because the (1)^{n} came from the expression of the
derivative of cos and sin in (2n+1) or (2n).
The use of the tan is very similar to the use of the sin,
because their behavious are equivalent originally (noteworthy the poles
are the same). Hence, we will come across the same problem as for the sin,
and we can shift z to z1/2 this time (for tan(/4)=1) so to avoid having a nul remainder in 0.
*** Case m=4 and p=4
Similarly as for the sin, it's the degree 4 which seems to work
the most for the same reason. So let us bypass this quickly and run
towards the results!
By considering :
we obtain the following pole/remainder table :
Pôles

Résidus

1/2


2n


somme des
résidus des racines +/1+/i


and the serie :
Then, studying the same function g_{3}
without the z1/2 leads to a nice expression :
which, injected in the previous serie, gives
:
I have not yet managed to generalised the
result... and I have not yet tried with other degree than 4. If
amateurs want to try !
Note that compare to the usual serie, we have this time 4n1
instead of 2n+/1. This research of other a*n+/1 will
be the object of lead 6.
Lead 4 : using z^{q} and
an other polynomial of degree m : z^{m}+p, m>1
:
*** With no
polynomial z^{m}+p
We consider first of all the functions with no polynomial of degree m,
that is of type : or even with the cos
or the tan.
Those functions are used to reprove the numerous formulae of Euler/Bernoulli. We
will calculate the remainder in 0 with the differential
formulae in the section "Remainders' theorem" for a pole of order m.
One thing to simply notice is that the sum on Z
(without the 0 when we don't need it, of course) of the inverse
of powers is null for certain powers due to the parity. Hence, this
explain why for the sin and q=2, the remainder in 0
is not null and is equal to although it is for q=3. In fact, we
have . And hence  now as we we are used
to, we can even ignore the pole/remainder table  q=2 give :
More generally, for the sin, we can use the
powers q=2k to give some series of the same kind as previously.
The tan uses also q=2k and gives the non alternating series.
For the cos, it's obviously the same phenomene shifted since 0
is not the pole of the cos. We therefore need to use q=2k+1 so
to find the sum of the inverse of odd powers. So, q=3 gives :
Which is really annoying, it's that the
phenomene is the reason for which we can not
calculate with this method the value of (3), Apéry's constant. And this is valid for
all the odd powers, except for alternating series like the one above,
unfortunatly, we can go from an alternative serie to a non alternative
with the odd compared with the evens.
*** With polynomial z^{m}+p
Unfortunatly, things works a lot less here when q=1
(z only). The sum of remainders of the roots do not cancel down for
example in the nice case of m=4 and p=4 even with the
cos. We obtain a hyperbolic varient like cosh/sinh and we need
again to use the trick like for the sin, that is to use the
expression of cosh/sinh so to replace it in the serie.
But an other problem arrise : this expression, for example does not converge as fast as the one we
obtain in the serie since we are varying the power of (2n+1)^{q},
expression missing in the serie with the cosh. And hence
impossible to sort it our so to increasse the speed due to (2n+1)^{q}.
For example, by considering q=2, m=4 and p=4
and the sin :
we obtain the expression :
and the expression cosh above allows
to obtain the serie :
But this is not really useful, and seems
hard to generalise.
For certain powers q, we manage to cancel the sum of the
remainders of the roots of z^{4}+4 (for the cos
and q=5 for example) but since however it does not want to
work, the remainder in 0 is 1/192*(5^{4}96) and we can not
isolate in this expression because the corresponding
polynomial 1/192*x(5x^{4}96)=c does not have roots
that can be express in radical form...
For the degree q=2, we have the same problem in the sense where
we obtain a serie in the polynomial of second degree of the variable . But if we isolate , some square roots
will appear but, our original objectif was to stay in the
rational series, hence with no roots. Having a serie of sum ^{2} like as above does not give too much
trouble, but if we start to introduce some roots, this page will have a
length tending seriously towards infinity !
So I only give one example of what we can get :
q=2, m=2, p=(2k+1)^{2} give :
Hence, we will stop there on this lead that
seems to be a bit blocked if we stay on the original goal.
Lead 5 : using shifted poles for the
polynomials
This method is very rich to obtain nice
generalised formulae.It consist of taking a z we take
systematically a za , of course understanding that we can then
add other polynomial to the denominator. For the loading speed of this
page (but will someone even get this far??), I will odmit some calculus
details, when it is very classical.
Example 1 : two poles simply shifted
The study of gives :
And by making a tends to b, then b towards
0, we refind the serie by Euler (the very least we expect !)
The tan and cos (taken in z/2) gives exactly the same type of formula with the
each of the charecteristic : tan instead of the sin and
the non alternating serie for the tan, and cos instead
of the sin and 2n+1 instead of n for the cos
The result is easily generalised for j+1 distinct on the
denominator and this gives :
Unfortunatly, since the only rational values
for the sin are taken for the rationals a_{k}
in /3 and /2, in this study, we can not have more than two
terms in the product and hence go further than j=1.
Example 2 : a
shifted pole of order q
We will take q=3 here, because there does not seems to be any
real general case that exist, the calculation depends on the used
power. Hence consider the function . The application of the raminder method gives
:
if I'm not wrong. The case m=1/3
or m=1/2 seems more interesting (and seems to be the only ones)
if we want to obtain an expression with no irrational terms (not
counting of course !).
We can also choose tan instead of the sin. We obtain :
Lead 6 : using a polynomial z^{m}+p,
and seatch of coefficients a.n+/1 in the denominiator of the
serie
This lead was touch upon at the end of the
third lead. Two methods can be used to achieve it :
* either we use a general shift (z+/b) and then if we look on
what values of b (of the form 1/r) allows to change the
2n+/b into 2r.n+/1.
* or we shart from a shift inside the cos/sin that is c..z+d, such that the poles are
directly in the form a.n+1.
The two cases are clearly equivalent because they both correspond to a
shift, the second being more practical to find what will work and to
form series.
Hence, if we look more at the second method, and if we base ourself
still on the cancellation of the remainders of the poles of z^{m}+p,
let us fix notation, and find a.n+1 on the denominator with cos(b.z+c.) that is the function .
Let us see the constrains :
1. Since we have z on the denominator, 0 is a pole and
hence we need that the remainder in this pole, i.e. we want cos(c.) to be rational
(no problems for the rationality of p !). This gives the
condition (1) :
(1)
We exclud the case c=1/2 which gives
one pole of cos (not practical !)
2. a.n+1 must be a pole, hence b..(a.n+1)+c.=(2n+1)/2 which is equivalent to condition
(2) :
a.b=1 (2a)
b+c=1/2 (2b)
3. Finally, so that there is a
transformation of cos in sin, the real part z_{1}
of z must satisfy the condition (3) valid for the
generalisations :
(3)
Let us solve this system by using different
possible value of c :
Note that c=0 is the case we have studied in the previous leads
this gives
c 
c=0 
c=1/3 
c=1/3 
c=2/3 
c=2/3 
b 
b=1/2 
b=1/6 
b=5/6 
b=1/6 
b=7/6 
a 
a=2 
a=6 
a=6/5 
a=6 
a=6/7 
z_{1} 
2k+1 
6k+1 
(6k+1)/5 
16k 
(6k+1)/7 
Possible value of a.n+1

2n+1 
6n+1 
6n+5~6n1 
16n~6n1 
6n+7~6n+1 
All is clear now ! Yet another remarque
still, the form of z_{1} forces us to consider only a
degree 2, maybe 3. In fact, 4 roots can not
have the same real part. We will see that there still exist a
particular case for the degree 4.
But let us stay for the moment in the application of table, after
clearing up, it's quite easy.
Hence consider the example of degree 3 with c=1/3 and b=1/6
(3rd columns of the table) :
gives the serie :
Unfortunatly, no increase of the denominator
due to the degree 3, as we predicted in the 1st lead...
Otherwise there exist a particular case, the degree 4. The
obligatory form of z_{1} seen above stops the remainder
of the conjugated 4th roots of z^{4}+4 cancel two by
two. But the sum of the four remainder is nul! I don't think this
phenomene is found at any greater degree, this is due to the perfect
symetries of roots for the degree 4, it's unfortunate...
Still by considering p=4*(6k+3)^{4}, or even c=1/3
et b=1/6, we obtain the following serie :
And by increasing the degree on the
denominator, we find :
And in the limiting case, we have :
The study of the sin is equivalent of the
one of cos with a translation of /2, hence, we
will not find any other new series there.
With the tan, the only possibility of non nul rational tang is
for :
tan(/4)=1. By following the same
reasoning as for the cos, we naturally find a serie in 4n+1,
here for the degree 2, which seems the only possibility :
We can increase the degree on the
denominator and obtain :
and the limiting case :
Other leads ?
With this method, we can see that we can
reproof a fair number of formulae, and of course they are not all there
!
We can look for other functions to put on the denominators some
functions g. For example, we could put the complex exponential
instead of the cos/sin
etc... Only problem, exponential does not cancel down, so we
need to couple it with a cos.
More generally, we can use functions which are not properly defined on
the whole complex domain. For example, we can use cos(*log) so to obtain some series like the Simon Plouffe one because the logarithm is
only defined in the complex domain if we cut part of it
(supressing the right half from the origin) in the definition of the
complex domain. Otherwise, we could turn around and have two different
value for z and z*exp(2i) (log(z)
and log(z)+2i). Which would be the least
annoying !
A part from the polynomial, I have no other application ideas for the
instant, but tell me if you do !
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