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Do continued fractions converge?
If the sequence of convergents {xn} approaches a limit the continued fraction is convergent and has a definite value. It may diverge by oscillation (for example, the odd and even convergents may approach two different limits), or it may produce an infinite number of zero denominators Bn.
How do you do continued fraction expansion?
To calculate a continued fraction representation of a number r, write down the integer part (technically the floor) of r. Subtract this integer part from r. If the difference is 0, stop; otherwise find the reciprocal of the difference and repeat.
What is the point of continued fractions?
All continued fraction convergents are best approximations of the first kind, but they satisfy a property even stronger than that. The basic idea is that if you make the denominator larger, you’re paying a price in some way.
Which is the proof of continued fractions-convergence?
Proof: Recall successive convergents get closer to x and alternatively overshoot and undershoot x. Also recall p k q k − 1 − q k p k − 1 = ( − 1) k. The result follows from the next lemma.∎ and a b ′ − a ′ b = − 1. Then d > b and d > b ′. As c b − a d > 0, we find b ′ < d. The proof for b is similar.∎
How to prove the theorem of continued fractions?
Define x k + 1 by the real that satisfies a = [ a 1; a 2,…, a k, x k + 1]. We will show x k + 1 ≥ 1, because this implies we could expand x k + 1 into a continued fraction with a nonzero first term and obtain a continued fraction expansion for a, proving the theorem.
How are successive convergents get closer to X?
Proof: Recall successive convergents get closer to x and alternatively overshoot and undershoot x. Also recall p k q k − 1 − q k p k − 1 = ( − 1) k.
Why are continued fractions better than the mediant?
Roughly speaking, continued fractions are better because they scale up the numerator and denominator of the other convergent by the best possible amount before computing the mediant. Here’s a curiosity.