5 ms·
This seems slightly silly if one is allowed to use, as Ed Pegg suggested, log and square root. If log and square root are allowed, then the obvious solution is
by kdavis 14y ago
This seems slightly silly if one is allowed to use, as Ed Pegg suggested, log and square root.
If log and square root are allowed, then the obvious solution is log(-1)/(sqrt(-1)*log(e)) which is accurate to an infinite number of digits.
- scott_s 14y agoe is not rational.
- kdavis 14y agoln(-1)/sqrt(-1)
- scott_s 14y agoThe natural log - ln - is not rational, as it is the logarithm with base e. That is, ln(x) answers the question, to what power would we have to raise e in order for it to equal x?
- kdavis 14y agoTwo points: 1. As soon as you allow square roots, you allow irrational numbers. (This sqrt(2).) 2. "The natural log - ln - is not rational" is a different statement than "e is irrational". A rational function is one that can be written as the ratio of two polynomials.
- scott_s 14y agoRegarding point 1, you had many things wrong. I picked what was the most obvious to me at the moment - in order for it not to apply, there only needs to be one thing wrong with it. And my point with the natural logarithm is that once you introduce it, you have introduced an irrational number. Overall, I'm not sure what your point has been.
- ColinWright 14y agoWhich ln are you using? Or for that matter, which sqrt(-1) are you using?
- mhartl 14y agoMaybe you're just trolling, but rational numbers are numbers that can be expressed as p/q, where p and q are integers. Neither i e nor i is an integer, so your proposed quotient has no bearing on the (ir)rationality of e.
- kdavis 14y agoI'll assume you are not trolling. My point is that log(-1)/(sqrt(-1)*log(e)) = ln(-1)/sqrt(-1) I am just writing the same equation using a log with another base so I don't have e in the equation explicitly.
- ColinWright 14y agoI'm confused - he's addressing the point made in the article by Ed Pegg that things are interseting if you allow log and sqrt. Surely that means he's allowed to use log and sqrt, and in particular, to use them in rational expressions. So I don't really understand what your point is.
- scott_s 14y agoThe point of the exercise is to find a short approximation of pi, no? If you allow the use of e, then you can define pi. What he wrote is not an approximation, it is pi. Surely at that point you've defeated the point of the exercise.
- kdavis 14y agoThis is more-or-less my point. Once you allow sqrt and ln (or sqrt, log, and e) the problem is silly. He explicitly allows, see the bottom of the article, sqrt, log, and irrational numbers.
- scott_s 14y agoI think it's reasonable to assume he did not introduce imaginary and transcendental numbers.
- kdavis 14y agoNot really. Once he introduces the square root he introduces imaginary numbers, sqrt(-1), and transcendental numbers, for example the Gelfond–Schneider constant 2^sqrt(2).
- scott_s 14y agoNow we're really down the rabbit hole of someone else's intent, but personally, I assumed he was still trying to maintain some restriction. So, no imaginary numbers, no trascendental numbers. It's easy to restrict what we take the root of, and what we do with the potential irrational result of such roots, to ensure that. As you pointed out, to not do so defeats the purpose of the exercise, and I think my assumption is both reasonable and charitable.