Bharat Rakshak

All the talk about Fermat's theorem being "disproved" reminded me of this little curiosity - had quite a bit of trouble when I first saw this in college.

Given the sequence:



It's possible to prove that this converges, but assuming that a[100] is close to this value, what is a[100]?

Cute! (It was 61st birthday of one of my family member .. not kidding it is actually on april 1.. and was mentioning that it is 5^2+6^2)..

I just got 3rd, 4th term .. (341/61, 1921/341) and saw the pattern!

So obviously nth term is :

(5^(n+1)+6^(n+1))/(5^n+6^n) which is easy to check...
(Yes, 111*25-1130*5+3000=5^3 and 111*36-1130*6+3000=6^3))

Added later: the answer (for a(100) ) is

3919911780443470697630731873534775187362173746498050984709582036730929382067381
/
653318631388679958306808321275343473365006007205019222633302014072286448118001

.... The term, obviously tend to 6. (for a(100) approx = 5.999999988)

Aditya - Can you tell me the source, of your problem (I may use the concept for problem given in math contest .. so I would like to get the source of the origin of the problem).. it is more interesting than I first thought.

For example, if one "tries" to solve this by writing a computer program.. the answer for a(100) will most likely come about 100 !! (which is, of course not correct - correct answer is near 6)
In fact, if one just changes the initial conditions as a(0)=5.5 and a(1)=5.54545455 (which is nearly same as 61/11 .. the answer will come out as 100)

In fact the answer, except for a very few initial conditions will converge to 100.

Amber G, exactly!

Something like 140 digits of precision are required to avoid error propagation when this computation is repeated 100 times or so using floating-point numbers. Doing it by hand, or using rational numbers avoids this problem, of course.

This is a sequence discovered by Jean-Michel Muller, cited here and various other places, though obviously I didn't know it at that time

Guido Maharaj ki Jai!



a[100] is 3919911780443470697630731873534775187362173746498050984709582036730929382067381/653318631388679958306808321275343473365006007205019222633302014072286448118001 or approx. 5.99999998793

ArmenT, wouldn't using



give the wrong results, even when, as you write, rational numbers are available? I think this problem was designed to show what happens if values are plugged in blindly to a formula. economic collapse, anyone?

Aditya, Armen -
Thanks for Jean-Michel Muller ref.. let me google and do some more checking.. (I tired this problem on one of the brighter students and he did got the solution correctly)
To be honest, too much "jai" to Python may not be completely unique ...(Disclaimer: I don't know python or have nothing against it )..The math version of Fortran (formath?) I routinely used for numbers (for math purposes) in late 1960's, and IBM's.. Rexx, not to mention "UBaisc" (program I loaded from shareware 20 years ago, will do the math correctly).. (So basically recognizing that fact that "blind" iteration may accumulate lot of errors due to rounding is key to get the right result) (BTW I used maple to get numeric answer - (I knew the answer was (5^101+6^101)/(5^100+6^100) so it was easy to plug)

One needs to solve,
x^3-111x^2+1130x-3000=(x-5)(x-6)(x-100)=0, the roots are 5,6,100 so answer would be (ratios of) the form ==> a*5^n+b*6^n+c*100^n
(a,b,c determined by initial conditions.. and unless c=0, the answer would converge to 100, and if both b and c are 0, it will converge to 5..and in our case (a=b=!0; c=0), it converges to 6.
(Fun problem! I recognized the pattern 2,11,61,341 as 5^n+6^n almost right away but when I saw there is third root, and it is 100, it occurred to me if c is not zero, the terms would tend to 100) ...

Also .. any answers to my previous problem ( first 100 digits after decimal of (7+5sqrt(2))^100.. IMO it is not that hard (or uninteresting )... how about that python.. does it give the right result... just curious.

And here I thought the jai to GvR was something like "the BDFL is retiring. Long live the BDFL!" for April 1 Historically, extended-precision reals do come from fortran IIRC, so you're right.

Your analysis of the problem, is of course, complete and correct in all respects.

Let me try not to cheat to give you the answer to your question...

Edit: okay, I haven't solved it, but it will be of the form (sqrt(a)+sqrt(a+1))=a+b*sqrt(2), where one of the two is a perfect square. it can be expressed as a power of 1+sqrt(2), only to find you've put this in an earlier post The series is also on OEIS, which gives me a nice recursion for the answer, but that still doesn't help in an "easy" way to solve this.

Nice problem - if I remember correctly that is the use of Laplace transforms which was also used to give the closed form solution for the Fibonacci sequence (one of the many methods that was used).

I am leaning towards the use of continued fractions for the [7+5Sqrt(2)] problem, since I can say this is 14+ 1/(14+1/(14+1/(14+.....) and successive convergents will get you closer to the actual solution; Am I on the right path?

Yes!! wrt to continued fraction / Fibonacci type recurring relationship is there for (7+5sqrt(2))^n
(similar to Aditya's problem - Interesting how different problems have something in common)

if (7+5sqrt(2))^n = a[n] + b[n]* sqrt(2)

we have a[n] = 14* a[n-1] + a[n-2]
(same for b[n] = 14 b[n-1] + b [n-2])
with initial conditions a[0]=1, a[1]=7 and b[0]=0 and b[1]=5

(Amazing thing is: Brhamgupta knew these types of techniques centuries before Fibonacci)
(This may or may not help to get answer to the original question)
******
Wrt to Aditya's problem, If I am a little "meaner" I might change say, a[0]=3 and a[1]=1, and keeping the same recursive relation. (so the initial values are integers - still the convergence would be to 6 (and not 100) and the answer would be a little harder to guess... or how about make the recursive relation deceptive simpler (coefficients as -101,99,-100) so one of the root is 100 but the other roots are not rational and the correct answer would be something equal to golden ratio.
******

My Jais to Guido were because he introduced the fractions module and bignum features into python by default, which made the code pretty much resemble the statement of the problem, even without knowing that 61 = 5^2 + 6^2 . That way, poor dimbulb like me could bang out a solution

AmberG: A possible solution hit me in the shower. Unfortunately I had to head to work right afterwards though and it bugged me all day. Here's my attempt after coming back from work:

Going by Brahmagupta's revelation as retold by AmberG (Jai Brahmagupta and Jai AmberG ):
7 + 5*sqrt(2) = (1 + sqrt(2))^3

Therefore instead of working out (7 + 5sqrt(2))^100, we can work out (1 + sqrt(2))^300 instead.

Now we know from the binomial theorem (Jai Google Devata & All hail Professor Moriarty):
1^300 + (300C1 * 1^299 * (sqrt(2))^1) + (300C2 * 1^298 * (sqrt(2))^2)) + (300C3 * 1^297 * (sqrt(2))^3) + ....
+ (300C298 * 1^2 * sqrt(2)^298) + (300C299 * 1^1 * sqrt(2)^299) + (300C300 * 1^0 * sqrt(2)^300)

The first term (1^300) has nothing to do with the digits after the decimal point, so we can ignore it .

Also, even powers of sqrt(2) don't contribute anything to the digits after the decimal point, since sqrt(2)^2 is a whole number. Thus we can ignore those from the calculation as well.

Clearly, the only terms that contribute to the digits after the decimal are the ones where the sqrt(2) term has an odd power (eg.) (300C1 * 1^299 * (sqrt(2))^1), (300C3 * 1^297 * (sqrt(2))^3) (300C5 * 1^295 * (sqrt(2))^5)...)

Now we see that powers of 1 don't contribute anything to the product, so we can ignore those from the calculation as well. Then the above reduces to
(300C1 * sqrt(2)^1) + (300C3 * sqrt(2)^3) + (300C5 * sqrt(2)^5) + ... + (300C299 * sqrt(2)^299)

So all we have to determine is the value of those terms alone, to determine the digits after the decimal point. Taking sqrt(2) as common, we have:
sqrt(2) * ( 300C1 + (300C3 * sqrt(2)^2) + (300C5 * sqrt(2)^4) + (300C7 * sqrt(2)^6) + ... + (300C299 * sqrt(2)^298))

or more simply:
sqrt(2) * ((300C1 * 2^0) + (300C3 * 2^1) + (300C5 * 2^2) + (300C7 + 2^3) + ....(300C299 * 2^149)

We can evaluate the above in a quick script, or we can go for some more fun and games. We know that 300C1 = 300C299 and 300C3 = 300C297 and so on. Therefore we can merge some of the terms above.
sqrt(2) * (300C1(2^0 + 2^149)) + 300C3(2^1 + 2^148) + ...)

Either way, we can write a quick python script to determine the answer (Jai Guido Maharaj for including fractions, bignums and arbitrary precision math by default into his language):

Code:

#!/usr/bin/env python
# Need python 2.6 for the fractions module. Get yours from http://www.python.org/ today!
from fractions import Fraction
from decimal import Decimal, getcontext

def nCr(n, r):
    num = n
    den = 1
    for i in range(1, r):
        num *= (n - i)
        den *= i
    den *= r
    return Fraction(num, den)

sum = 0   
for i in range(1, 300, 2):
    cmb = nCr(300, i)
    # print "300C%d = %s" % (i, str(cmb))
    pwr = (i - 1) / 2
    power_of_two = 2**pwr
    # print "Raising two to the power of", pwr, " is", power_of_two
    term = cmb * power_of_two
    sum += term

sum = int(sum)
print "sum = ", sum
# Shove our precision up to 200 places to be sure :)
getcontext().prec = 200
sqrt_of_two = Decimal(2) ** Decimal('0.5')
prod = Decimal(sum) * sqrt_of_two
print "prod = ", prod
str_prod = str(prod)
decimal_position = str_prod.find('.')
print "Digits after the decimal: ", str_prod[decimal_position:]

and we get this output:

Code:

sum =  2405252129711623484710495558522357173667400588427864372185101521045397503910768684922934668936725440491728003546500
prod =  3401540182764948741830406364060242471221242172434177281122811716713140384942391994469530780004006403241546412250001.0000000000000000000000000000000000000000000000000000000000000000000000000000000000001
Digits after the decimal:  .0000000000000000000000000000000000000000000000000000000000000000000000000000000000001

i.e. a bunch of zeros, which was pretty much my guess in the first place . You can tweak the precision in the above code by editing the one line where I set .prec. For higher precisions I got a bunch of .999999....

Whether this answer is correct or not, I'd like to dedicate this solution to Motörhead for providing me so many years of earsplitting pleasure.

Nice one, ArmenT!


The powers of 1 + sqrt(2):

sqrt(1) + sqrt(2)
sqrt(8) + sqrt(9)
sqrt(49) + sqrt(50)
sqrt(288) + sqrt(289)
sqrt(1681) + sqrt(1682)
...

the sequence gives quite a few recurrence relations to calculate the lesser term (I didn't build a large enough table of differences to spot the recurrence) , which was where I was stuck.

The square roots of the perfect squares terms are related to the CF expansion for sqrt(2) . btw, OEIS has gotten the sequence heading wrong, I think those are denominators.


I try to explain your answer (after the fact, of course):

The 300th power of (1+root2) is basically sqrt(d^2-1)+sqrt(d^2), where n/d is an accurate approximation to root2.

Ignoring the integral part, we need to look at
sqrt(d^2-1)
= d * sqrt(1 - 1/d^2)
~ d * (1 -1/2d^2)
~ d
and therefore d-floor(d) is zero. (nothing special about the value of d, except that it grows much faster than the power - so with 300, d is approximately 3.4e114, and so the first 100 digits after decimal point of 1/d are zero. Doesn't that mean that the d-1/2d has all 9s in its fractional part?)


Completely off-topic - nice free textbook on generating functions, if anyone hasn't seen it already.

Yes, that is what I had in mind, and this in simpler terms.

Maha Hint 1: 50 is a bigger number than 49.
So Maha Hint 2: , of course, 5*sqrt(2)= sqrt(50) > 7
So our number, let us call it x, x=7+ sqrt(50) > 14

Also (sqrt(50) - 7) < .1 (and it is >0)
(You don't need a calculator to check that, above is nothing but
1/x so it is less than 1/14 so obviously less than .1)

Rest is easy,
x^(-100) = (7-5sqrt(2))^100 < 10^(-100)
but the sum (ie (7+5 sqrt(2)^100 + (7-5sqrt(2))^100 is obviously an integer!
so first 100 digits (after decimal) have to be .9999 .......
(actually first n digits of x^n (after the decimal point) is either 0 (if n is odd) or 9 (if n is even.
*******

Okay, little quiet here … so…

First, a simple problem from the same contest (for high school students)..( Inspired to be placed here because of comments made in previous posts)

P1:

There are 15 chairs placed in a straight line, occupied by 15 mathematicians. Now mathematicians get up and sit down again. Each person sits back either in the same chair (same chair as before) or in an adjacent chair.
(Note, each chair, except the first and the last have two “adjacent” – first (and the last) has only one “adjacent” chair)

In how many ways this seating can take place?

P2:

This relates to the observation made by AdityaS in previous post: namely:


Note here, that the bolde part (eg 1, 9, 49, 288) squares are perfect squares, while others (2, 8, 50, 288 ) are twice a perfect squares. (and they are also +-1 of the previous numbers).

Now instead of 2, I take a case of 61..…
So here is a problem, Can you find a sequence, (a_1, a_2, a_3 …) such that a is a perfect square, and (a+1)/61 is also a perfect square?

P1: Would it be 13!/10!=1716. It's 13! since the first and last chair are part of the second and second-to-last chair's permutation.

I'm not sure that answer is correct, because Amber says the first and last chairs have only one neighbour, rather than the chairs being in a circle.


The 'how to solve it' is, I guess, to place 7 markers that divide the chairs into 7 groups of two (who can either sit in their own chairs, or swap), and 1 single (who always sits in her own chair). 2^7 arrangements you can have there.

Number of ways to place markers is the permutations of 7 identical and one different object, 8!/7!=8, so the answer is 2^10?



P2 can be restated to find all 61*n^2 - 1 that are a perfect square.

Thanks for the ref. to book for generating functions...
AdityaS, Restatement of P2 is of course correct.
For P1, I don't think answer is correct. I think there is an over-count. (For example just take the case of 3 chairs, where answer is 3 ( ie 123,132,213) - don't think your method will give the same answer)..

For Pell2, I have 29718 and 3805 as minimal solutions; Have the continued fraction for sqrt(61) with usual notation as [7; 1,4,3,1,2,2,1,3,4,1,14] using till the 1 we get, 29718/3805

For P1, I think Since each additional chair involves both the previous value i.e. F(n-1) and the one preceding F(n-2), this is a Fibonacci with initial conditions of a1= 1, a2 = 2 - Am I correct?

Najunamar - Of course! Your comments a few posts above were the one reminded me of the problems..Interesting how Fibonacci comes into play in P1. (and similar recursive relation applies for P2 too ... by starting with your solution)

Solutions are, of course, correct (for both)

Just to change tack a bit now (too much Number Theory! although I love the subject) - I heard this from a good Prof.,
Find the value of Sin(1)*Sin(2)......*Sin(90). (i.e. the product of all except Sin (0) which would make it trivial).

Nice problem, (It is a little harder than (sin 1) (sin3)(sin 5) ....sin(179) )- which is a little easier).



As you know, (The problem is very popular (This thread also has one eg cos 20*cos 40* cos 60* cos 80 -- which is same as sin 10*sin 30* sin 50* sin 70 .... eg products of degrees in AP) and in similar form has appeared in many contests)

The answer for the problem is, if some one wants to check out, (and if I am not making a silly mistake - as I am doing it while typing this) is (sqrt(90)/(2^89))
(There are some nice easy and fun ways to solve this or similar problems//)
Added: Typo in above correct.

Bah, my mistake The overcounting should have been obvious.


Shouldn't that be 89, btw?

Some comments: (Sorry for long message - will tiny the fonts so if you like to ignore it..

For P1: For n chairs/mathematicians the answer is the integer closest to:
(phi)^(n+1)/sqrt(5)

Where phi = Golden ratio (Najunamar’s Fibbonachi) = 1.618034 (about).
Interesting How that relates to the previous problem(s).. also the value (phi^n/sqrt(5)) will be very close to integer for large n.. (lots and lots of zeros or 9’s..if you calculate with high precession…)

The Method, as Najunamar pointed out: (Just expanding it to make, perhaps, a little clearer)
Suppose T[n] is the answer for n ..
Then in First chair Either:
1 is sitting - Now there are T[n-1] ways for the rest to get arranged.
Or Mathematician 2 is sitting in First chair, this means, Mathematician 1 has to sit in the second chair, and now (n-2) chairs remains = = > T[n-2] ways
So T[n] = T[n-1]+T[n-2]
We know T[1]=1, T[2]=2, so T[3]=2+1=3, T[4]=3+2=5, T[5]=5+3=8 …
We get the Fibonacci = 1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,…

For Pell2: Pell is not really the person who worked on “Pell’s equations” According to Wiki


Of course, Lord Brouncker was familiar with Brhamgupta’s method (as Brahmgupta’s work was tranlated in Europe by then)
What, I found really brilliant about Brahmgupta’s method, is this. (I am translating his method in modern math language – some thing I have not seen done that clearly)
You notice that 61 is between 7^2 and 8^2 and start with:
(8+sqrt(61))(8-sqrt(61))=3 (eq 1)
(7+sqrt(61))(7-sqrt(61))=-12 (eq 2)
Multiply eq 1 and 2 and you get
(117+15 sqrt(61))(117-15 sqrt(61) = -36 = -6^2
Divide each factor by 6 and you get:
(39/2+(5/2)sqrt(61))(39/2-(5/2)sqrt(61)=-1
(You are almost there, except that you have “half” instead of “full” integer – Brhagupta says – just cube it to get rid of “half”) so
(39/2+(5/2) sqrt(61))^3 = 29718+3805 sqrt(61)
And you have the answer (29718,3805) as given earlier by Najunamar ..The method is simpler.. And this was done in 7th century!
All other such pairs, one can get by calculating:
(29718+3805 sqrt(61)^n …
(For odd n, difference would be -1, for even n it will be +1)

Okay, If you like to give it a try, this is the 1st problem of the second day of this year's USA's national Math Olympiad:

For n>1
let a_1, a_2, ..., a_n be positive real numbers such that:

(a_1 + a_2 + ....+ a_n) (1/a_1 + 1/a_2+ ....1/a_n) <= (n+1/2)^2
Prove that
max(a_1, a_2, .... a_n) <= 4* min(a_1, a_2, ...., a_n)

If you want to see the other 2 problems from the same day here is the link:
<link>

FYI: International Math Olympiad (IMO 2009) is in Germany this year and IMO celebrates its 50th anniversary.

Here is the web page of the IMO 2009 in Bremen.

Some fun facts:

Brunei has a girls dominated team
Peru has an 11 year old: Raúl Arturo Chávez Sarmiento
Pakistan is sending a team.
Here is the Indian team


1.Akashnil Dutta (Contestant 1)
2.Akshay Mittal (Contestant 2)
3.Gaurav Digambar Patil (Contestant 3)
4.Ananth Shankar (Contestant 4)
5.Sameer Wagh (Contestant 5)
6.Subhadip Chowdhury (Contestant 6)


It is fairly fresh this time

( Akashnil Dutta is in 10th grade and supposed to be very good - Predict that he will win a gold - but others are also very good)

Here is the first problem (rather easy) of IMO 2009: If you like, try it:


Problem 3 is not that hard either (relatively speaking as IMO problems go)


Problem 2 is geometry: (Again IMO rather easy for IMO)

Think I've solved the first one.
The trivial case (a_1 = 1) cannot exist, because then the next term a_2 will have to be of the form (c*n + 1) for the condition (a_i) * (a_{i+1} - 1) to be a multiple of n to be satisfied (where c is some integer) and we're given that a_1...a_k are numbers in the set of (1...n) and here we have a_2 > n, which is not allowed.

Also from the above, we can infer a_k cannot be 1, since a_i * (a_{i+1} - 1) cannot be equal to n for i = k - 1 (since (a_k - 1) will be 0 then)

For the rest of the cases, where (a_i) * (a_{i+1} - 1) is a multiple of n, let's say we have two numbers (d, e) which are divisors of n. This means that a_i has to be a multiple of d and a_{i+1} - 1 has to be a multiple of e. For instance a_i could be p * d and a_{i+1} - 1 could be q * e (where p and q are some integers) and (a_i) * (a_{i+1} - 1) = c * n (where c = p * q and is a positive integer) and of course p * d and q * e both have to be < n since we are given that that all of a_1...a_k are in the range (1 ... n), but (p * d) * (q * e) has to be >= n

Also for sequences of k > 2, a_{i+1} - 1 and a_{i+1} both have to be multiples of divisors of n. The only exceptions to this rule are a_1 and a_k, where a_1 has to be of the form (p * divisor of n) and a_k - 1 has be be of the form (q * divisor of n)
Hence, (a_1 - 1) will be ((p * divisor of n) - 1) and a_k = (q * divisor of n) + 1. As above, let the divisors of n be (d, e). Therefore (a_1 - 1) = ((p * d) - 1) and a_k = (q * e) + 1.

Therefore (a_1 - 1) * a_k = (pd - 1) * (qe + 1) = pdqe + pd - qe - 1. Since we know that pdqe = c * n, the only way that pdqe + pd - qe - 1 can be a multiple of n (say r * n, where r is another integer) is if pd - qe - 1 is a multiple of n as well (say s * n). Now this is impossible, since we proved above that pd and qe both have to be < n, hence pd - qe - 1 can never be a multiple of n.

Therefore (a_1 - 1) * a_k cannot be divisible by n

Q.E.D.

For Whatever its worth:




Actually a_k can (always) be 1, as 0 is considered divisible (by normal definition) by any n.



May be you meant something more specific, but this is not true for any divisors, (the confusing part is "a_i" has to be a multiple of d, (for example, 4(=say d) and 25 (say =e) are divisors of 100, and 26*(51-1) is a multiple of 100, yet 26 is not a multiple of 4.

In any case, as you have guessed it, the problem is not too hard.

Problem 3 is harder than what I thought at first blush. (Rumor has it that no indian student got it) . this problem was proposed by an American examiner, I know. (actually a MIT grad student).

Next day's problem 4, 5, 6 (guess, could be gotten from IMO's website) - 4 was not hard but prob 6 was harder (Rumor - 5 out of 6 chinese (considered one of the strongest team( did not get it)..

Results would be out next week.

One can try, if one is curious Prob 4 (Geometry - and at least looks easy):

Math Olympiads - China gets 6 Gold, USA 2 Gold 4 Silver, India 3 Silver, 2 Bronze, 1 Honorable Mention.

For India:

Sameer Wagh Silver
Akashnil Dutta Silver
Ananth Shankar Silver
Subhadip Chowdhury Bronze
Gaurav Digambar Patil Bronze
Akshay Mittal Honorable Mention

Problem 1 (which I mention above) , everyone (India) got it, Prob 3 None (some got partial credit)

USA Rank 6
John Berman - Gold
Eric Larson - Gold

So I need to set up and solve matrices for something I am working on. What is a good math software that can be used for this?

Matlab ?

^^ or if you want something free try octave.

Thanks for the suggestions. I will look these up.

cmlib is a stable public domain library with BLAS as subset. Coded usually in fortran, and available in compiled callable library - with c ports available. Best part is that the code is also available and almost anything you may want to do with matrices (including complex) are available.

Gentlemen,

I often engage in writing mails myself on issues supporting India. But here is something that I need Mathematicians/Maths fans to do.

Prof.Marcus du Sautoy is doing some great service in the United Kingdom popularizing Maths. He comes often on radio and TV. And in general a very open person. He recently aired a programme called Story of Maths where he traces the origins of Mathematics.
His home page is here: http://people.maths.ox.ac.uk/~dusautoy/


As far as I know, he is the first person to have acknowledged the works of Kerala School of Mathematicians in public domain in the West
See here:
http://www.telegraph.co.uk/education/31 ... nswer.html




It is important that we use this guy and pass on more details to him.
So someone eqipped with a good idea of Indian Maths/Astronomy books like Surya Siddhanta, Panchange etc should contact him and let him know the details.

There is a book written by Gheverghese George from a UK university in the early 90's which mentions all this explicitly. I think more Indians are probably unaware of this origin than the westerners. Kerala had strong links via trade and missionaries to the west a few hundred years ago and knowledge was transferred.

FWIW: even in BRF there are dozens of places where Madhava and other Kerala School of Mathematicians are discussed.
(Just search Madhava - you will get many hits)

For example not too long ago on Pi day we have


In West, many (most standard math history books and articles give credit to Madhava ) (at least now) give (though not as many times as it should) credit and calls art tan series as Madhava-Gregory series. (all my math students know that.. -

But point is well taken, contribution to math from many of such mathematicians are still not given the due credit .

Look for SciLab - Very similar to Matlab in basic functionality but free
http://www.scilab.org/

um i was just wondering if a mathematical function, such as sin, cos etc, can be written as a continued fraction?

Yes, very much. Look at Lambert's derivation. Better Gauss's continued fraction expansion for Hypergeometric function. for example
$\tan(z)=\frac{z}{1-\frac{z^2}{3-\frac{z^2}{5-\ldots}}}$
paste this in any LaTeX processor. The numbers are odd integers.