Square Root Formula

[visualAd]

I am using a square formula in a programming assignment the formula is as follows in pseudo-code:

INPUT number
INPUT guess

LOOP
  oldguess = newguess

  newguess = 1/2 * (number / oldguess + oldguess)
WHILE newguess - oldguess > 0.00005

The problem I'm having is writing this method as a mathematical forumla. I can vaugely remember using square brackets to represent repitition but I honestly can't remember. So I'm wondering if any of you guys could help???

[sql_lall]

you could always do this:

U₁ = 1
U_n = (K/U_n-1 + U_n-1) /2
sqrt(K) = lim(f->inf) U_f

[NoteMe]

VB is like pseudo code for me...

Public Function SquareRoot(thenumber As Double) as Double
Dim theVals() As Double
ReDim theVals(1 To thenumber)
theVals(1) = (1 / 2) + (thenumber / 2)
For i = 2 To thenumber
    theVals(i) = (theVals(i - 1) / 2) + (thenumber / (2 * theVals(i - 1)))
    If theVals(i) = theVals(i - 1) Then
        SquareRoot = theVals(i)
        Exit Function
    End If
Next i
End Function

[NoteMe]

BTW...keep in mind that this is not a perfect solution........found it on this site a while back....and never have had time to make it perfect....

[NoteMe]

Ahha...in VB it will work better like this...

Public Function SquareRoot(thenumber As Double) As Double
Dim theVals() As Double
ReDim theVals(1 To thenumber) As Double
theVals(1) = (1# / 2#) + (thenumber / 2#)
 
For i = 2 To thenumber
    theVals(i) = (theVals(i - 1) / 2#) + (thenumber / (2# * theVals(i - 1)))
    
        SquareRoot = theVals(i)
    
Next i
End Function

But for pseudo code the other code is better....

[NoteMe]

Or actually this is how I would have done it....

Public Function SquareRoot(thenumber As Double) As Double
 
    Dim Last As Double
    Dim This As Double
    
    Last = (1# / 2#) + (thenumber / 2#)
    This = (Last / 2#) + (thenumber / (2# * Last))
    
    Do While (Last <> This)
        Last = This
        This = (Last / 2#) + (thenumber / (2# * Last))
    Loop
    
    SquareRoot = This
 
End Function

[visualAd]

Originally posted by sql_lall
you could always do this:

U₁ = 1
U_n = (K/U_n-1 + U_n-1) /2
sqrt(K) = lim(f->inf) U_f

NoteMe - unfortunatly pseudo code is not mathematical notation.

sql_lall: What does that formula mean.

[sql_lall]

Ok, a brief explanation:

1) Have you heard of recursion? Stuff like this:
A₀=0, A₁=1,
A_n = A_n-1 + A_n+1

This is the Fibonacci series, as you can figure out that:
A₂ = 1, A₃ = 2, A₄=3...

2) U₁ = 1: U_n = (K/U_n-1 + U_n-1) /2 <--- this is also recursion

Basically, this is the part which works out the next term from the last one. The square root of the number K is the 'last' term of this series. However, as there is no 'last' term, it is simply the asymptote of the values, which is written as:
lim(f->inf) U_f

[NoteMe]

Originally posted by visualAd
NoteMe - unfortunatly pseudo code is not mathematical notation.

What did you not understand with it....? ALl of it?

[kedaman]

whats wrong with sqrt(x)?

[NoteMe]

Originally posted by kedaman
whats wrong with sqrt(x)?

It is a programming assignment....

[kedaman]

well he is asking for the mathematical notation.. not the code

I think this is newton raphson approximation btw so you can write
x= (x0 - n/x0)/2 or without x0 as divisor x= 3/2 x0 (1- x0^2/n)

[NoteMe]

I am not going to argue with you on this too...but that is not pseudo code either....

[kedaman]

I don't think that was what he was looking for either

[visualAd]

I want to impress the tutor by putting it in mathematical notation.

I think I knew all this once but - after a year with no math(s) lessons I have gone all rusty