Correctness of Terminating Continued Fraction Computations (GenContFract.of
) #
Summary #
We show the correctness of the algorithm computing continued fractions (GenContFract.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 (GenContFract.of v).convs' 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 GenContFract.IntFractPair.stream v n
.
For an example, refer to
GenContFract.compExactValue_correctness_of_stream_eq_some
and for more
information about the computation process, refer to Algebra.ContinuedFractions.Computation.Basic
.
Main definitions #
GenContFract.compExactValue
can be used to compute the exact value approximated by the continued fractionGenContFract.of v
by adding a residual term as described in the summary.
Main Theorems #
GenContFract.compExactValue_correctness_of_stream_eq_some
shows thatGenContFract.compExactValue
indeed returns the valuev
when given the convergent and fractional part as described in the summary.GenContFract.of_correctness_of_terminatedAt
shows the equalityv = (GenContFract.of v).convs n
ifGenContFract.of v
terminated at positionn
.
Given two continuants pconts
and conts
and a value fr
, this function returns
conts.a / conts.b
iffr = 0
exactConts.a / exactConts.b
whereexactConts = nextConts 1 fr⁻¹ pconts conts
otherwise.
This function can be used to compute the exact value approximated by a continued fraction
GenContFract.of v
as described in lemma compExactValue_correctness_of_stream_eq_some
.
Equations
- GenContFract.compExactValue pconts conts fr = if fr = 0 then conts.a / conts.b else let exactConts := GenContFract.nextConts 1 fr⁻¹ pconts conts; exactConts.a / exactConts.b
Instances For
Just a computational lemma we need for the next main proof.
Shows the correctness of compExactValue
in case the continued fraction
GenContFract.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 convs'
to evaluate the continued
fraction. Here is an example to illustrate the idea:
Let (v : ℚ) := 3.4
. We have
GenContFract.IntFractPair.stream v 0 = some ⟨3, 0.4⟩
, andGenContFract.IntFractPair.stream v 1 = some ⟨2, 0.5⟩
. Now(GenContFract.of v).convs' 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 GenContFract.of v
at step n - 1
is exactly v
if the
IntFractPair.stream
of the corresponding continued fraction terminated at step n
.
If GenContFract.of v
terminated at step n
, then the n
th convergent is exactly v
.
If GenContFract.of v
terminates, then there is n : ℕ
such that the n
th convergent is
exactly v
.
If GenContFract.of v
terminates, then its convergents will eventually always be v
.