Using Binomial Expansion to Evaluate [2 + sqrt(3)]^50
Date: 11/29/2006 at 10:45:55 From: Matt Subject: Why is (2 + root3)^50 so close to an integer? When you work out (2 + root3)^50, why is it so close to an integer? I have worked out (2 + root3)^50 on a computer and got the answer 39571031999226139563162735373.999999999999999999999999999999999974728 ... I have tried expanding it using all the expansions I know (starting with the binomial expansion) but have had no luck seeing why it comes out so close to an integer. Do you have any ideas?
Date: 11/29/2006 at 11:31:07 From: Doctor Douglas Subject: Re: Why is (2 + root3)^50 so close to an integer? Hi Matt. Are you sure that you're not simply running into roundoff error from the finite precision of the computer arithmetic? Calculations such as these on a computer or calculator require considerable care because they usually don't keep enough significant digits in memory to be resistant against roundoff errors. This might not matter if you're only looking for a result to a few significant figures, but if the question is evaluating how close the original number is to an integer, one can encounter problems. Your idea of writing out the original number X using the binomial expansion is an excellent one: X = [2 + sqrt(3)]^50 = C(50,0) * 2^50 * sqrt(3)^0 + C(51,1) * 2^49 * sqrt(3)^1 + C(52,2) * 2^48 * sqrt(3)^2 + ... + + C(50,50) * 2^0 * sqrt(3)^50 Notice how many of these terms (in particular those with sqrt(3)^[something even] are integers, and do not affect the fractional part of X. So we can safely ignore them, as far as the integer part of X is concerned. Then, notice that everything else is of the form D[k] * sqrt(3)^k, D[k] is an integer, k is odd, where k is odd, and if we subsume all of the even powers of sqrt(3), we can condense this down to D[k] * (n) * sqrt(3)^1 k=2n+1, n=0,...,48. Thus we see that the only terms that affect the fractional part of X are of the form E[k] * sqrt(3), E[k]=D[k]*n is an integer, and the only remaining step is to evaluate the sum E = E + E + E + ... + E. This should be a lot easier for your computer, since this is guaranteed to be an integer, and you should be protected somewhat from roundoff errors. But note that the numbers in the middle of this sum, say E, will have 27 or 28 decimal digits or so, and you will still have to retain at least this many digits in these integer computations. The leading digits of E give you information about how close the original number X is to an integer. If the leading digits are close to 1/sqrt(3) = .577350269..., then when finally multiplied by sqrt(3)^1, the product is near an integer. If E is zero, then X is an integer (ask yourself, can that happen?). - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
Date: 10/22/2011 at 09:46:34 From: Mike Subject: re:69763.html : Why is (2 + root3)^50 close to an integer I read the cool question from Matt. I wanted a theoretical answer, and figured something out. Please check it out: Let x = 2 + sqrt(3). Clearly, 1/x = 2 - sqrt(3). By binomial expansion, we get that ... x^n = a + b sqrt(3) ... and ... (1/x)^n = a - b sqrt(3), ... where a, b are integers. Therefore, this is also an integer: x^n + (1/x)^n Since lim(1/x)^n = 0, the fractional part of x^n tends to 1. That's why (2 + root3)^50 is so close to an integer. No need to use computers! The most difficult is to conjecture that x^n + (1/x)^n is an integer. Feel free to use my solution.
Date: 10/24/2011 at 21:08:50 From: Doctor Douglas Subject: re:69763.html : Why is (2 + root3)^50 close to an integer Hi Mike, I think that this is a very nice approach. Can you generalize it to other constants, such as 2 + sqrt(5)? - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
Date: 10/25/2011 at 13:03:06 From: Mike Subject: Thank you (re:69763.html : Why is (2 + root3)^50 close to an integer) Dear Doctor Douglas, It is a great idea to consider (2 + sqrt(5))^n and generalize. This, too, is "close to an integer" since i) (2 + sqrt(5))^n + (2 - sqrt(5))^n is an integer ii) lim (2 - sqrt(5))^n = 0. The same holds for ... a + sqrt(b), ... where a, b are integers and |a - sqrt(b)| < 1. Mike
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