approximate $\large \frac{2}{7}$ to 7 decimal places. Show all work for a gold medal

Question

kainui · Answer

$$\Large \frac{2}{7} = 2 \frac{1}{1-(-6)}=2(1-6+6^2-6^3+ \cdots)$$

Ok just kidding haha. =P

parthkohli · Answer

long division...?

anonymous · Answer

Calculator ?

parthkohli · Answer

just committed a plebian mistake

anonymous · Answer

oh okay

parthkohli · Answer

did it on paper and it turns out $0.\overline{285714}$

kainui · Answer

I was going to guess something along the lines of this, but thought it would take too much work and didn't really amount to anything $$\Large \frac{2}{7} = \frac{1}{\frac{7}{2}}=\frac{1}{3+\frac{1}{2}}$$

parthkohli · Answer

0.285714285714...
that looks like it'd be 0.2857143 till seven digits.

kainui · Answer

Interesting and relevant http://primes.utm.edu/glossary/xpage/PeriodOfADecimal.html

kainui · Answer

So this says the period of 1/k is the same as the order of 10 mod k. Which means we have $$\Large 10^6 \equiv 1 \mod 7$$ so 6 is the order and also the period. So it suffices that we will only have to compute this up to 6 decimal places instead of 10 I guess haha.

abdullah1995 · Answer

this is some next level stuff lol

xapproachesinfinity · Answer

cool you have you (1+x)^r series expansion @Kainui

xapproachesinfinity · Answer

1-x i meant

kainui · Answer

What do you mean, like the geometric series or something else?

kainui · Answer

Honestly I don't even know why people are throwing medals at me, I didn't get this question right or anything lol

xapproachesinfinity · Answer

i meant  how did you get that 2(1-6+6^2....)
you did you use hehe

xapproachesinfinity · Answer

isn't that Tylor series stuff?

kainui · Answer

Yeah but if you calculate that it is clearly divergent. 1-6=5, 1-6+36=31, 1-6+36-6^3=-185, etc... I didn't and couldn't calculate any digits of the fraction with this.

xapproachesinfinity · Answer

oh i see
i thought that would get you the digits
well it is still cool though

kainui · Answer

Haha yeah whenever you can sum up a divergent series and get a value, this is called its Cesaro sum which is really quite an interesting thing since it's the limit of the average of the terms surprisingly.

anonymous · Answer

2/7 can be written as $$2\div7$$, which you will get 2.0/7 = .2R .6, .60/7 = .08 R .04, .040/7 = .005R .005, .0050/7 = .0007R .0001, .00010/7 = .00001R .00003, .000030/7 = .000004R .000002, and wraps around. This is $$\approx .28571428$$, so it would be rounded (in 7 decimal places) to .2857143