Given the sequence:
Code: Select all
a[0] = 11/2
a[1] = 61/11
a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
Code: Select all
a[0] = 11/2
a[1] = 61/11
a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
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)..adityaS wrote: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]?Code: Select all
a[0] = 11/2 a[1] = 61/11 a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
Guido Maharaj ki Jai!Amber G. wrote: 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.
Code: Select all
#!/usr/bin/env python
# Use version 2.6 or higher :)
from fractions import Fraction
a = [0] * 101
a[0] = Fraction('11/2')
a[1] = Fraction('61/11')
for n in range(1,100):
a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
print "a[100] is", a[100], " or approx. ", float(a[100])
Code: Select all
a[0]=11/2
a[1]=61/11
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).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.
Of course, Lord Brouncker was familiar with Brhamgupta’s method (as Brahmgupta’s work was tranlated in Europe by then)The name of this equation arose from Leonhard Euler's mistakenly attributing its study to John Pell. Euler was aware of the work of Lord Brouncker, the first European mathematician to find a general solution of the equation, but apparently confused Brouncker with Pell. This equation was first studied extensively in ancient India, starting with Brahmagupta,
Problem 3 is not that hard either (relatively speaking as IMO problems go)Let n be a positive integer and let a_1,a_2,a_3, ....... a_k (k>1) be distinct integers in the set 1,2,...n such that n divides a_i(a_{i + 1} - 1) for i = 1,2,...k - 1 . Prove that n does not divide a_k(a_1 - 1)
Problem 2 is geometry: (Again IMO rather easy for IMO)Suppose that s_1,s_2,s_3, .... is a strictly increasing sequence of positive integers such that the sub-sequences s_{s_1},s_{s_2},s_{s_3},.... and s_{s_1 + 1},s_{s_2 + 1},s_{s_3 + 1}....are both arithmetic progressions. Prove that the sequence s_1,s_2,s_3, ...is itself an arithmetic progression
Let ABC be a triangle with circumcentre O. The points P and Q are interior points of the sides CA and AB respectively. Let K,L and M be the midpoints of the segments BP,CQ and PQ . respectively, and let X be the circle passing through K,L and M . Suppose that the line PQ is tangent to the circle X . Prove that OP = OQ
Code: Select all
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)
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..... 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....
Let ABC be a triangle with AB = AC . The angle bisectors angle CAB of and angle ABC meet the sides BC and CA and at D and E , respectively. Let K be the incentre of triangle ADC . Suppose that angle BEK = 45 degrees . Find all possible values of angle CAB .
Matlab ?vera_k wrote: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?
It therefore came as a huge shock to me to discover recently that a school of Indian mathematicians in Kerala in south India arrived at this formula several centuries earlier. It should, in fact, be called the Madhava formula, in honour of the Hindu scholar who first hit upon it.
π was not the only great mathematical discovery made in India. Negative numbers and zero – concepts that in Europe, as late as the 14th century, were viewed with huge suspicion – were being conjured with on the subcontinent as early as the seventh century.
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.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
FWIW: even in BRF there are dozens of places where Madhava and other Kerala School of Mathematicians are discussed.s 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
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.. -Today is Pi Day. (3/14)! Happy Pi day to all!
A few centuries before Newton, Madhava of Kerala gave us (among other things) interesting relationship of pi ,
(see how pi appears with nothing but numbers like 1,3,5...)
pi/4 = 1-1/3+1/5-1/7+1/9 - .....
Look for SciLab - Very similar to Matlab in basic functionality but freeAnujan wrote:Matlab ?vera_k wrote: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?