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All Rational Approximations of Pi Are Useless

blog.wolfram.com

1–10 of 113 posts

Re: All Rational Approximations of Pi Are Useless

#6
Ok, I'm confused, what exactly is the point here? If you don't care, 3.14 is sufficient, if you do care then you use π, and if you really care you use τ[1].

[1](http://tauday.com/tau-manifesto)

(And if you really really really care, you use a different font then the one HN defaults too.)

Re: All Rational Approximations of Pi Are Useless

#7
Agreed

22/7 is the worse. Too much trouble for too little benefit

If you need the value of pi to do a hand calculation, 3.14 is more than enough

And if you need to "produce" pi just remember pi/4 = 1 - 1/3 + 1/5 - 1/7... (there are formulas that are better, sure, but less memorizable)

Re: All Rational Approximations of Pi Are Useless

#8
post #6

Ok, I'm confused, what exactly is the point here? If you don't care, 3.14 is sufficient, if you do care then you use π, and if you really care you use τ[1]. [1]( http://tauday.com/tau-manifesto ) (And if you really really really care, you use a different font then the one HN defaults too.)

The point is that it is easier to remember X digits of PI than a fraction, because the fraction will have more digits in total than X.

Re: All Rational Approximations of Pi Are Useless

#9

Upto 10 digits, pi is sqrt(sqrt(2143 / 22)) .. and that is 9 characters as per his definition or 8 if you allow ()^(1/4) as an elementary operation. 3.141592652 vs pi ~= 3.141592653

He likely bets the person who was taught 22/7 as an approximation will not be able to do the sqrt of a sqrt of a fraction in their head or on paper. If they have a calculator they might as well use the dedicated pi key.

Re: All Rational Approximations of Pi Are Useless

#10
I think blog.wolfram.com is probably the best company blog I've seen, from a marketing perspective. A large number of the entries are basically of the form "Here is an interesting problem, and here's how I solved it with Wolfram products". They generally let the problem have the spotlight rather than focus on the Wolfram products, so it doesn't feel like you are getting pitched.

Here's a related problem, but for e: using each digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 at most once each, and only the operators +, -, x, /, and ^ (exponentiation), and parenthesis for grouping, how close can you get to e? Digits may not be concatenated--for instance, you cannot get a 23 by simply placing the 2 next to the 3.

My best, after about 15 minutes of fiddling, were:

   (3x(4x7+1))/2^5 = 2.71875 ≈ e + 0.000468172
and

    2x(9x6-1)/(3x(8+5)) = 2.717948... ≈ e - 0.000333111
but it turns out you can do FAR better.
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