Given the sequence:
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a[0] = 11/2
a[1] = 61/11
a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
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]
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:Code: Select all
a[0] = 11/2
a[1] = 61/11
a[n+1] = 111 - (1130 - (3000 / a[n-1])) / a[n]
It's possible to prove that this converges, but assuming that a[100] is close to this value, what is a[100]?
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.
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#!/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])
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a[0]=11/2
a[1]=61/11
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#!/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:]
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sum = 2405252129711623484710495558522357173667400588427864372185101521045397503910768684922934668936725440491728003546500
prod = 3401540182764948741830406364060242471221242172434177281122811716713140384942391994469530780004006403241546412250001.0000000000000000000000000000000000000000000000000000000000000000000000000000000000001
Digits after the decimal: .0000000000000000000000000000000000000000000000000000000000000000000000000000000000001
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.
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,
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)
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
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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)
.... 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 .
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.
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
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
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 - .....
Anujan wrote: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?
Matlab ?
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