Chebyshev's Polynomials, Pi, and the "look
BBP"
Some funny formulae
A first example :
Note that it's a serie with rational terms !
The sequence *U*_{n} and *V*_{n} are none
the other that the value of the Chebyshev's polynomial, respectivaly *T(n,99/100)*
for *U*_{n} and *T(n,99/4780)* for *V*_{n}.
And for the more perceptive of you who have already
notice a certain similarity between the recurence of *U*_{n}
and *V*_{n} coeffiecent and with the formulae of *Arctan*
(think back to Machin !), well done, here is
a more general formula where we use those famous formulae :
If we know a relation of type : then we can construct the sequence *U*_{n}^{k}
and the following sequence :
You want something different to *10* as the denominator? Sure,
here is a even more general formula for *p*^{2}>1,
still starting from the relation with the *Arctan* :
with *T(n,x)* Chebyshev's polynomials.
But where does it originate ?
This algorithm is nearly of the same style
as the series of type Machin, but here we
allow ourself to have a common denominator of powers (*10* in our
first example). Those formulae have unfortunatly from the BBP series only the look. Since even if we have
at last obtained a *10* in the denominator instead of *16*
from the BBP formula, and that the rest of
the sum is rational, it is a function of powers. Indeed, the sequences *U*_{n}
and *V*_{n} are recurrent, therefore we can find the
general term under the form *U*_{n}=a.r_{1}^{n}+b.r_{2}^{n}
where *r*_{1} and *r*_{2} are the roots of *x*^{2}-c.x-d=0
such that *U*_{n}=c.U_{n-1}+d.U_{n-2}.
For Delta non zero of course, but go and find me a repeated root ! And
even in this case, we write *U*_{n}=(a+b.n)r^{n}
where *r* is the repeated root. *a* and *b* are found
by using *U*_{1} and *U*_{2}, but the
expressions are so horrible that I prefer not waste any time to write
them here. Anyways, as always, there's some roots in the manipulation
and we don't even understand how the expression can be rational in the
end.
Here again, I have not found this kind of expression on the net, and
even the *Gradshteyn* does not mention it. And yet, it is from
this book that the idea came to me and we are finnaly going to see in
the proof why I called this page Chebyshev's polynomial and *Pi*.
Preliminary definition of Chebyshev's
polynomials
A little
word on Chebyshev (also written Chebychev or Chebyshov) :
Born in 1821 in Okatovo from a noble and cultivated family, he studied
at the University of Moscow. His family ruined, he refuse the lecture
position that he is offered and lives in poverty until 1847 where he
becomes lecturer at Saint-Petersbourg. He then starts to travel and to
make the most of his engineer talents, he is in fact very skilled.
On the other hand, he was horrible to his students, he did not like
wasting his time, and at the bad habit of stopping his lecture to the
second even if in the middle of a sentence, so it is told by *"Des
mathématiciens de A à Z"*.
During his carrier, he was mostly interested in number theory and did
great progression on the path of proofs for the prime numbers' theorem.
We can define his famous polynomial with degree *n* with the
following relation :
*T(n,cos(x))=cos(n.x)*, *T(0,y)=1*
In other words, it's the polynomial which allows us to express the *cos(n.x)*
as a function of *cos*^{k}(x), *kn*. It is unique if we choose to make it equal *1*
with degree *0* : *T(0,y)=1*.
Indeed, it checks the following recurrence relation (genius) :
*T(n,x)=2x.T(n-1,x)-T(n-2,x)*
It's this that will help us to build the
sequences *U*_{n}^{k}.
Proof
First of all, we need to make a little
preliminary calculation :
How is that going to be any use to us? Well by noting that
we get
(the convergence of the serie is uniform on the whole compact included
in *[0,1[* and *p<1*, which justify the reversal of the
sum and the integral)
This formula which is recounted in the Gradshteyn/Ryzhik (5th ed.
1.448.6). Working out the proof wasn't that hard, plus you must have
noticed that I shamefully did the calculation backward in the first
place, since the "by nothing" above is not easy to see straight away!
Then simply replace *p* by *1/p<1* to find the formula
which will use afterwards :
And at this stage, I told myself "But
the *cos(n.x)*, we know them as function of the powers *cos*^{n}(x)
thanks to Chebyshev's polynomial, which if we remind ourself, verify
that *T(n,cos(x))=cos(n.x)*.
So we consider the relation . Hence we choose *x=x*_{k} such
that . (yes, ok, the term inside *arccos*
has to be less than *1*, but even if this is not the case, we get
a complex *x* and that work as well in the end with *a*_{k}=1...).
Note that if the realtion on the *arctan* is nice, the *a*_{k}
are rational and thanks to the Chebyshev's polynomials, *cos((2k-1)x)=T(2k-1,x)*
is also.
With this expression, it comes down to :
Now, we're nearly there, we use the
recurence of Chebyshev's polynomials' relation to calculate the *T(n,x)*.
Like
and *T(n,x)=2x.T(n-1,x)-T(n-2,x)* we then build the final
algorithm :
Funny, isn't it ?
Trials
In the case of the first sequence, I
had chossen *p=10* and use Machin's
formula. Here are the numerical values :
*n=2* |
*3,141545* |
*n=5* |
*3,1415926535919* |
*n=10* |
*20 decimals* |
*n=20* |
*41 decimals* |
We note that the Chebyshev's polynomials' part decrease very slowly
since the convergence is nearly of *2n*, which is due to the
factor *1/10*^{2} in the denominator. In my view, it's an
interesting alternative solution to the series of type Machin...
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