Did I Invent This (REALLY Nerdy Content)

[sthief09]

or is it already invented?

for figuring out decimals for fractions with denominator 7. take the numerator, double it, multiply by 7. then divide by 100. add that to the numerator doubled twice, multiplied by 7, and divided by 100 twice.

take 1/7 for example. 1*2*7/100 = .14. add 1*2*2*7/100/100 = .0028. add 1*2*2*2*7/100/100/100 = .000056. sum that and you get .142845. keep going and it keeps expanding. it works for all 7's. divide all that by 2 and it works for the 14 demoninators. it's pretty useful for quick approximations. you just double ti and append, double it and append, double it and append. once the doubles start getting into triple digits, it starts having a big impact. but for most, it works well for the first 6 digits. for 2/7 it's just .2856 + .000108 so .285708, which you can round to .28571

in series form, it's the sum from n= 1 to fininity of numerator*7*(2^n)/100^n

I'm assuming that the series boils down to numerator/7. does anyone know if someone's already done this?

[partygirluk]

Another cool way to work out 1/7ths is

1/7 is

.142857.....

Remember this order. (If you have trouble, each two digits is roughly twice the previous two)

Now 2/7 must start with a 2, and it is

.285714.....

3/7 starts with a 4 therefore it is....

.428571.....

4/6 starts with a 5 """""""""""""""""

.571428........

etc.

Strange but true

in fact, I think if you show 1/7 in full it just repeats that sequence over and over.

[gazarsgo]

Was that the quickest that an invention has ever been rendered useless? Or is this phenomena perhaps unique to sthief?

[jason_t]

The sum of any series of the form sum from n = 0 to infinity of x^n is 1/(1-x) as long as |x| < 1.

So the sum from n = 1 to infinity of numerator * 7 * (2/100)^n is (numerator) * 7 * (1/(1-2/100) - 1) = numerator * 7 / 49 = numerator / 7.

In general, numerator / d = sum from n = 1 to infinity of numerator * d * [1/(d^2+1)]^k, as you can work out using the above formula.

as for the formula, here is the reason this is true

let s = 1 + x + x^2 + ... + x^n
then x * s = x + x^2 + ... + x^(n+1)

so x * s - s = x^(n+1) - 1

and therefore

(x-1) * s = x^(n+1) - 1

so

s = [ (x^(n+1) - 1) / (x - 1) ]

and if |x| < 1 then letting n->infinty we have x^(n+1)->0 so s->-1/(x-1) = 1/(1-x).

[[censored]]

[ QUOTE ]
The sum of any series of the form sum from n = 0 to infinity of x^n is 1/(1-x) as long as |x| < 1.

So the sum from n = 1 to infinity of numerator * 7 * (2/100)^n is (numerator) * 7 * (1/(1-2/100) - 1) = numerator * 7 / 49 = numerator / 7.

In general, numerator / d = sum from n = 1 to infinity of numerator * d * [1/(d^2+1)]^k, as you can work out using the above formula.

as for the formula, here is the reason this is true

let s = 1 + x + x^2 + ... + x^n
then x * s = x + x^2 + ... + x^(n+1)

so x * s - s = x^(n+1) - 1

and therefore

(x-1) * s = x^(n+1) - 1

so

s = [ (x^(n+1) - 1) / (x - 1) ]

and if |x| < 1 then letting n->infinty we have x^(n+1)->0 so s->-1/(x-1) = 1/(1-x).

[/ QUOTE ]

Yeah, pretty much my thoughts exactly.

[Eurotrash]

[ QUOTE ]
the sum from n= 1 to fininity of numerator*7*(2^n)/100^n

[/ QUOTE ]

if you do this calculation a lot, might as well simplify that (2^n/100^n) to 1/50^n... would take out some of the hassle.

[sthief09]

[ QUOTE ]
Was that the quickest that an invention has ever been rendered useless? Or is this phenomena perhaps unique to sthief?

[/ QUOTE ]


well I would have called it sthief's theory, but I already have one...

(waiting for someone to ask what it is)

[sthief09]

yeah, that's how I came across it I think. it was a while ago. but I told a teacher in middle or high school when I came up with a simpler version of it, and (surprise!) she didn't care

[partygirluk]

[ QUOTE ]
[ QUOTE ]
Was that the quickest that an invention has ever been rendered useless? Or is this phenomena perhaps unique to sthief?

[/ QUOTE ]


well I would have called it sthief's theory, but I already have one...

(waiting for someone to ask what it is)

[/ QUOTE ]

What is it?

[sthief09]

our opponents don't like to value bet in limit hold'em. when an opponent bets into you, and you're deciding whether to call, things are usually better than they appear. this is because your opponent isn't likely to be value betting a mediocre hand like middle pair, and against more inexperienced players, even top pair. this is especially true on a scary board. for example, on a 49Q9 board with a 4 river, it's less likely than normal he has a Q, since there's a good chance he'll check-call or check through with a Q. with showdown-able hands, they don't value bet. that lets you narrow the range of hands down to monster or bluff. since mathematically, there are tons more combinations of bluffs than combinations of monsters, you'll be good a lot more often than it seems. that's why, IMO, party 15/30 isn't as easy as advertised. it's a lot harder to pick off bluffs, because the aggressive players will make good, thin value bets, which balance the bluffs. against a mediocre player who doesn't value bet enough, it's obvious when he's bluffing