partial quotient造句
例句與造句
- The badly approximable numbers are precisely those with bounded partial quotients.
- Any Liouville number must have unbounded partial quotients in its continued fraction expansion.
- Historically, mathematicians studied periodic continued fractions before considering the more general concept of restricted partial quotients.
- An irrational number is badly approximable if and only if the partial quotients of its continued fraction are bounded.
- Using a counting argument one can show that there exist transcendental numbers which have bounded partial quotients and hence are not Liouville numbers.
- It's difficult to find partial quotient in a sentence. 用partial quotient造句挺難的
- Serret's theorem implies that the regular continued fraction expansions of equivalent quadratic irrationalities are eventually the same, that is, their sequences of partial quotients have the same tail.
- It breaks down a Chunking ( also known as the partial quotients method or the hangman method ) is a less-efficient form of long division which may be easier to understand.
- Using the explicit continued fraction expansion of " e ", one can show that " e " is not a Liouville number ( although the partial quotients in its continued fraction expansion are unbounded ).
- In two dimensions, with the cone generated by \ textstyle \ { ( 1, \ alpha ), ( 1, 0 ) \ }, they are just the partial quotients of the continued fraction of \ textstyle \ alpha.
- Clearly any regular periodic continued fraction consists of restricted partial quotients, since none of the partial denominators can be greater than the largest of " a " 0 through " a " " k " + " m ".
- For example, truncating before the 4th partial quotient, we obtain the partial sum 10 / 81 = 0 . \ overline { 123456790 }, which approximates Champernowne's constant with an error of about, while truncating just before the 18th partial quotient, we get
- For example, truncating before the 4th partial quotient, we obtain the partial sum 10 / 81 = 0 . \ overline { 123456790 }, which approximates Champernowne's constant with an error of about, while truncating just before the 18th partial quotient, we get
- If " ? " is the golden ratio, then all the partial quotients " a " " n " are equal to 1, the denominators " q " " n " are the Fibonacci numbers and we recover Zeckendorf's theorem on the Fibonacci representation of positive integers as a sum of distinct non-consecutive Fibonacci numbers.