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... and 1/9999999800000001 = .00000000 00000001 00000002 00000003 00000004 00000005 00000006 ... 99999996 99999997 99999999 ...repeating Basically, the pattern is 1 over some number of 9s, followed by an 8, followed by the same number of 0s, followed by a 1. So, 1/81, 1/9801, 1/998001, 1/99980001, 1/9999800001, etc.
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Same here! For the category 'pornography'...
We have websense here and it too seems to block perfectly benign things with bogus reasons. This link is categorized as adult content.
There's some sleight of hand here. Not all the digits are exactly right. Look how it skips from 997 to 999: http://www.futilitycloset.com/2012/01/08/math-notes-76/ Here's the math. Suppose you want a unit fraction 1/n with decimals that cycle through the 4-digit sequence abcd. Multiply by 10^4 to shift abcd into integer position, leaving repeating copies after the decimal point: 10^4/n = abcd + 1/n Solving for n give…
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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…
1 + r + r^2 + ... = 1/(1-r)
1 + 2r + 3r^2 + ... = 1/(1-r)^2
The series 1 + r + r^2 + ... = 1/(1-r) is the type of tuples with entries in r. Then 1/(1-r)^2 is the type of pairs of tuples (the tuples in a pair need not have the same length). A pair of tuples can be mapped to a single tuple by concatenation. Conversely, a single tuple can be split into a pair of tuples. How many ways can this splitting be done if the tuple has length n? In n+1 ways; for example, the 2-tuple (a,b) can be split into the 3 pairs of tuples ((a,b), ()), ((a), (b)) and ((), (a, b)), which are exactly the pairs that concatenate to (a,b). Thus the bijection from 1 + 2r + ... + (n+1) r^n + ... to 1/(1-r)^2 maps an element of (n+1) r^n, which is an index 0 <= k <= n and an n-tuple, to the pair of tuples gotten from splitting that n-tuple at index k.
double x = 0.1;
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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…
EDIT 1: Forgive my ignorance but can you please elaborate on
1 + r + r^2 + ... = 1/(1-r) is the type of tuples with entries in r ?
EDIT 2: Okay I think I get it.
1 + r + r2 are the terms in the expansion of (1+r)n
There's no 998 (and it's not a rounding issue)! ... 995 996 997 999
And even the suggested 1/9801 for two digits produces no 98 ... 95 96 97 99 00 01 02 ... And extrapolating it to one digit, 1/81 gives 0.012345679012345679...