Correctness of Terminating Continued Fraction Computations (GeneralizedContinuedFraction.of
) #
Summary #
We show the correctness of the
algorithm computing continued fractions (GeneralizedContinuedFraction.of
) in case of termination
in the following sense:
At every step n : ℕ
, we can obtain the value v
by adding a specific residual term to the last
denominator of the fraction described by (GeneralizedContinuedFraction.of v).convergents' n
.
The residual term will be zero exactly when the continued fraction terminated; otherwise, the
residual term will be given by the fractional part stored in
GeneralizedContinuedFraction.IntFractPair.stream v n
.
For an example, refer to
GeneralizedContinuedFraction.compExactValue_correctness_of_stream_eq_some
and for more
information about the computation process, refer to Algebra.ContinuedFractions.Computation.Basic
.
Main definitions #
GeneralizedContinuedFraction.compExactValue
can be used to compute the exact value approximated by the continued fractionGeneralizedContinuedFraction.of v
by adding a residual term as described in the summary.
Main Theorems #
GeneralizedContinuedFraction.compExactValue_correctness_of_stream_eq_some
shows thatGeneralizedContinuedFraction.compExactValue
indeed returns the valuev
when given the convergent and fractional part as described in the summary.GeneralizedContinuedFraction.of_correctness_of_terminatedAt
shows the equalityv = (GeneralizedContinuedFraction.of v).convergents n
ifGeneralizedContinuedFraction.of v
terminated at positionn
.
Given two continuants pconts
and conts
and a value fr
, this function returns
conts.a / conts.b
iffr = 0
exact_conts.a / exact_conts.b
whereexact_conts = nextContinuants 1 fr⁻¹ pconts conts
otherwise.
This function can be used to compute the exact value approximated by a continued fraction
GeneralizedContinuedFraction.of v
as described in lemma
compExactValue_correctness_of_stream_eq_some
.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Just a computational lemma we need for the next main proof.
Shows the correctness of compExactValue
in case the continued fraction
GeneralizedContinuedFraction.of v
did not terminate at position n
. That is, we obtain the
value v
if we pass the two successive (auxiliary) continuants at positions n
and n + 1
as well
as the fractional part at IntFractPair.stream n
to compExactValue
.
The correctness might be seen more readily if one uses convergents'
to evaluate the continued
fraction. Here is an example to illustrate the idea:
Let (v : ℚ) := 3.4
. We have
GeneralizedContinuedFraction.IntFractPair.stream v 0 = some ⟨3, 0.4⟩
, andGeneralizedContinuedFraction.IntFractPair.stream v 1 = some ⟨2, 0.5⟩
. Now(GeneralizedContinuedFraction.of v).convergents' 1 = 3 + 1/2
, and our fractional term at position2
is0.5
. We hence havev = 3 + 1/(2 + 0.5) = 3 + 1/2.5 = 3.4
. This computation corresponds exactly to the one using the recurrence equation incompExactValue
.
The convergent of GeneralizedContinuedFraction.of v
at step n - 1
is exactly v
if the
IntFractPair.stream
of the corresponding continued fraction terminated at step n
.
If GeneralizedContinuedFraction.of v
terminated at step n
, then the n
th convergent is
exactly v
.
If GeneralizedContinuedFraction.of v
terminates, then there is n : ℕ
such that the n
th
convergent is exactly v
.
If GeneralizedContinuedFraction.of v
terminates, then its convergents will eventually always
be v
.