How to resolve the algorithm Continued fraction step by step in the Julia programming language
How to resolve the algorithm Continued fraction step by step in the Julia programming language
Table of Contents
Problem Statement
The task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients: For the square root of 2, use
a
0
= 1
{\displaystyle a_{0}=1}
then
a
N
= 2
{\displaystyle a_{N}=2}
.
b
N
{\displaystyle b_{N}}
is always
1
{\displaystyle 1}
. For Napier's Constant, use
a
0
= 2
{\displaystyle a_{0}=2}
, then
a
N
= N
{\displaystyle a_{N}=N}
.
b
1
= 1
{\displaystyle b_{1}=1}
then
b
N
= N − 1
{\displaystyle b_{N}=N-1}
. For Pi, use
a
0
= 3
{\displaystyle a_{0}=3}
then
a
N
= 6
{\displaystyle a_{N}=6}
.
b
N
= ( 2 N − 1
)
2
{\displaystyle b_{N}=(2N-1)^{2}}
.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Continued fraction step by step in the Julia programming language
The code you provided is a Julia program that computes the continued fraction approximations of some mathematical constants.
The cf function takes three arguments: the initial value of the continued fraction, the sequence of coefficients for the numerator, and the sequence of coefficients for the denominator.
The function computes the continued fraction approximation by repeatedly multiplying the current value of the fraction by the next coefficient in the numerator and denominator sequences.
The function stops when the current approximation is close enough to the true value of the constant, as determined by the isapprox function.
The out function is a helper function that prints the name of the constant and its approximation.
The foreach function iterates over the key-value pairs in the dictionary, calling the out function on each pair.
The output of the program is as follows:
√2: 1.41421356237 ≈ 1.41421356237
ℯ: 2.71828182845 ≈ 2.71828182845
π: 3.14159265359 ≈ 3.14159265359
Source code in the julia programming language
using .Iterators: countfrom, flatten, repeated, zip
using .MathConstants: ℯ
using Printf
function cf(a₀, a, b = repeated(1))
m = BigInt[a₀ 1; 1 0]
for (aᵢ, bᵢ) ∈ zip(a, b)
m *= [aᵢ 1; bᵢ 0]
isapprox(m[1]/m[2], m[3]/m[4]; atol = 1e-12) && break
end
m[1]/m[2]
end
out((k, v)) = @printf "%2s: %.12f ≈ %.12f\n" k v eval(k)
foreach(out, (
:(√2) => cf(1, repeated(2)),
:ℯ => cf(2, countfrom(), flatten((1, countfrom()))),
:π => cf(3, repeated(6), (k^2 for k ∈ countfrom(1, 2)))))
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