An answer of the exercise: Loops and Functions on a tour of Go

answer_exercise_gotour.go

package main

import (
	"fmt"
	"math"
)

const Delta = 0.0001

func isConverged(d float64) bool {
	if d < 0.0 {
		d = -d
	}
	if d < Delta {
		return true
	}
	return false
}

func Sqrt(x float64) float64 {
	z := 1.0
	tmp := 0.0
	for {
		tmp = z - (z * z - x) / 2 * z
		if d := tmp - z; isConverged(d) {
			return tmp
		}
		z = tmp
	}
	return z
}

func main() {
	attempt := Sqrt(2)
	expected := math.Sqrt(2)
	fmt.Printf("attempt = %g (expected = %g) error = %g\n",
		attempt, expected, attempt - expected)
}

Can anyone let me know the different of z -= (z*z - x) / (2 * x) and z -= (z * z - x)/(2 * z) in attached example?
The result is different between these expressions which make me very confuse

You can find the full code here https://go.dev/play/p/ff8LzDMIg8c

package main

import (
	"fmt"
	"math"
)

func Sqrt(x float64) float64 {
	z,p := 1.0, 0.0
	for math.Abs(p-z) >= 1e-8 {
		p,z = z, z - (z*z - x) / (2*z)
	}
	return z
}

func main() {
	fmt.Println(Sqrt(2))
}

package main

import (
	"fmt"
	"math"
)

func Sqrt(x float64) (z float64, iterations int) {
	z = 1.0
	iterations = 0
	for {
		iterations++
		temp := z - (z*z - x) / (2*z)
		if math.Round(temp * 100000) / 100000 == math.Round(z * 100000) / 100000 {
			break
		}
		z = temp
	}
	return
}

func main() {
	fmt.Println(Sqrt(2))
	fmt.Println(math.Sqrt(2))
}

As a total beginner I've done it with rounding

package main

import (
	"fmt"
)

func Sqrt(x float64) float64 {
	z := 1.0
	for i := 0; i <= 10; i++ {
		z -= (z*z - x) / (2 * z)
	}
	return z
}

func main() {
	fmt.Println(Sqrt(23123))
}

package main

import (
"fmt"
"math"
)

func Sqrt(x float64) float64 {
z:=1.0

for n:= 1;n < 10;n++ {
	z = z - ((z*z - x) / (2*z))
}
return z

}

func main() {
y := 169.
fmt.Println(Sqrt(y))
fmt.Println(math.Sqrt(y))
}

func Sqrt(x float64) float64 {
var z, tmp float64 = 1.0, 0.
for ; math.Abs(z - tmp) >= 1e-8 ; z, tmp = z - (zz-x)/(2z), z {}
return z
}