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Topic: Trivial Approximation (Read 722 times) 

Sir Col
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Trivial Approximation
« on: Jun 17^{th}, 2008, 12:02pm » 
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A rational approximation, m/n, of an irrational number, a, is defined as trivial if 1/n^{2} < a  m/n < 1/n. If a  m/n < 1/n^{2} then it is defined as a reasonable approximation. (Technically a  m/n < 1/n^{3} is defined as a good approximation, but we shall not concern ourselves with these for this problem.) For example, sqrt(2)  17/12 = 0.002453... < 1/12^{2} = 0.0069444..., and so 17/12 is considered to be a nontrivial approximation. However, if we round sqrt(2) to two decimal places, sqrt(2) = 1.41 (2 d.p.) and sqrt(2)  141/100 = 0.00421356... < 1/100 = 0.01, which is only a trivial approximation. If sqrt(x) is rounded to k > 1 decimal places, does this always leads to a trivial approximation? (Edited to correct m,n mixup pointed out by Hippo.)

« Last Edit: Jun 17^{th}, 2008, 1:04pm by Sir Col » 
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Hippo
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Re: Trivial Approximation
« Reply #1 on: Jun 17^{th}, 2008, 12:48pm » 
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There should be 1/n on place of 1/m and so on in the definition ... otherewise the examples do not correspond to the definition.


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Eigenray
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Re: Trivial Approximation
« Reply #2 on: Jun 18^{th}, 2008, 12:48am » 
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on Jun 17^{th}, 2008, 12:02pm, Sir Col wrote:If sqrt(x) is rounded to k > 1 decimal places, does this always leads to a trivial approximation? 
 If x is a positive integer below 245, then yes. For some larger values, no. I'm not sure about the rest though. E.g., given x and r, can we always find a bound on k for which x*10^{2k}  r is a square?

« Last Edit: Jun 18^{th}, 2008, 1:13am by Eigenray » 
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