Earlier quoted context omitted.
It's necessary that one number is absent. The period is 2997 as he mentions. Can't pack 1000 3 digit numbers into that ;)
Chuck Norris could :P
Fun with math: Dividing one by 998001 yields a surprising result
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Re: Fun with math: Dividing one by 998001 yields a surprising result
#72Re: Fun with math: Dividing one by 998001 yields a surprising result
#73Earlier quoted context omitted.
Chuck Norris could :P
No sense of humor. Seriously.
It's not that people here are stuffy, or humorless, or anything; we just want HN to be different, and not decay into a festering cesspool of malfeasance like Slashdot or whatever. There are plenty of places one can go to get that kind of stuff on the 'net. Here at HN, it will usually be downvoted mercilessly.
Re: Fun with math: Dividing one by 998001 yields a surprising result
#74Re: Fun with math: Dividing one by 998001 yields a surprising result
#75Earlier quoted context omitted.
Here is a more detailed explanation 1--------- If x = 0.001 then the sum x^2 + 2 x^3 + 3 x^4…. in its decimal places will have all the three digit numbers except the the second last starting with 000,001 and then till 997,999. As pointed out by someone below,the reason 998 is missing is because after 997998999 the next coefficient is 1000.This overflow 1 will carry over and mess up all the nines to the right until it…
A quicker proof is to just differentiate. For |r| 1 + r + r^2 + ... = 1/(1-r) Differentiate both sides of the equation: 1 + 2r + 3r^2 + ... = 1/(1-r)^2 Here's a bijective combinatorial proof, which I like best of all. It uses the concept of generating functions. As a caveat, it only shows the equality for formal power series, not analytic power series. The series 1 + r + r^2 + ... = 1/(1-r) is the type of tuples with…
Re: Fun with math: Dividing one by 998001 yields a surprising result
#76Earlier quoted context omitted.
For x < 1, 1 + 2x + 3x^2 + 4x^3 + ... converges to 1/(1-x)^2. When x = 0.001, you get 1/.999^2 = 1000000/998001 = 1.002003004005...
Here is a more detailed explanation 1--------- If x = 0.001 then the sum x^2 + 2 x^3 + 3 x^4…. in its decimal places will have all the three digit numbers except the the second last starting with 000,001 and then till 997,999. As pointed out by someone below,the reason 998 is missing is because after 997998999 the next coefficient is 1000.This overflow 1 will carry over and mess up all the nines to the right until it…
x + 2x^2 + 3x^3 + 4x^2..
=
x + x^2 + x^3 + x^4 + ...
+ x^2 + x^3 + x^4 + ...
+ x^3 + x^4 + ...
+ x^4 + ...
= sum i from 0 to infinity (sum j=i to infinity (x^j))
= sum i from 0 to infinity ((x^i)/(1-x))
= 1/(1-x)^2Re: Fun with math: Dividing one by 998001 yields a surprising result
#7712345679 * 9 = 111111111
12345679 * 18 = 222222222
12345679 * 27 = 333333333
12345679 * 36 = 444444444
12345679 * 45 = 555555555
12345679 * 54 = 666666666
12345679 * 63 = 777777777
12345679 * 72 = 888888888
12345679 * 81 = 999999999
12345679 * 999999999 = 12345678987654321
Re: Fun with math: Dividing one by 998001 yields a surprising result
#78http://people.csail.mit.edu/devadas/numerics_demo/chord.html
For an explanation better than I can provide of what they are and how it works, see 6.006 lecture 11 notes!
http://courses.csail.mit.edu/6.006/fall11/lectures/lecture11...
Re: Fun with math: Dividing one by 998001 yields a surprising result
#79Thanks for posting. I'm trying to keep a collection of these type of things so that when she's ready, it'll be another tool to get/keep my daughter excited about math.
Care to share? Would be handy to have for my daughter... :)
Re: Fun with math: Dividing one by 998001 yields a surprising result
#80 0.00
+ 0.0001
+ 0.000002
+ 0.00000003
...
--------------
Each row is equal to x, but shifted over 2x digits. This is the same as dividing by 10^x. This simplifies to the formula: sum k=0 to infinity: k/(10^k)
This is fairly easily calculable, and results in 1/9801. Try it yourself on wolfram alpha: http://www.wolframalpha.com/input/?i=%28sum_%28k%3D1%29%5Ein...