swampwiz
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I understand how these expressions are derived, but I don't see how to calculate successive resultant values. Should the continued fraction expression simply be replaced by 1?
swampwiz said:I understand how these expressions are derived, but I don't see how to calculate successive resultant values. Should the continued fraction expression simply be replaced by 1?
swampwiz said:The video says to just presume the fraction is 0, and that presuming it to be 1 gives the wrong result.
All continued fractions in general.Stephen Tashi said:What fraction are you talking about? A particular fraction? All continued fractions in general?
swampwiz said:I think folks are missing the gist of my question. Obviously, if one were to use a continued fraction to compute a result, the part of the expression that is still a continuing fraction needs to be given some value just to plug into the arithmetic calculation.
How about for π?Stephen Tashi said:Your question seems to assume the method for evaluating continued fractions must be done by the "trick" of using a symbol to stand for an infinite expression.
That's the impression I get from your post:
As indicated by @mfb (and the video) the value of a continued fraction is defined as the limit of a sequence, not by whether the trick (also mentioned in the video) works.
It would help if you give a specific example and explain what part of the continued fraction you think must be given a value of 0 or 1. However, evaluating an infinite sequence does not require setting the tail end of the sequence to be 0 or 1. And the definition of how to evaluate a continued fraction does not involve setting the tail of an expression equal to any particular value.
It seems that the first approximation (3) is the result of setting the continuing fraction to 0,Truncating the continued fraction at any point yields a rational approximation for π; the first four of these are 3, 22/7, 333/106, and 355/113.
swampwiz said:How about for π?
https://en.wikipedia.org/wiki/Pi#Continued_fractions
The Wiki article says:
It seems that the first approximation (3) is the result of setting the continuing fraction to 0,