Something like the SQRT function

[jaenster]

Don't forget the first line of a square root script is always.

If $number < 0 Then

ConsoleWrite("Can't square root a negative number")

else

ConsoleWrite(($number ^ .5) & @CRLF)

endif

The last consolewrite i dont understand.

[1905russell]

The last consolewrite i dont understand.

Which part?

$number^.5

same as ^(1/2)

Square root = (multiply the number with itself two times) = 2/1

Raised to the power (^), always the inverse of the root = 1/2

1/2 = .5

$number^.5 = square root of number. Prove it with logs.

Cubed root = $x*$x*$x = (multiply the number with itself three times) = 3/1 = ^1/3 = cube root calculation

[jaenster]

Which part?

$number^.5

same as ^(1/2)

Square root = (multiply the number with itself two times) = 2/1

Raised to the power (^), always the inverse of the root = 1/2

1/2 = .5

$number^.5 = square root of number. Prove it with logs.

Cubed root = $x*$x*$x = (multiply the number with itself three times) = 3/1 = ^1/3 = cube root calculation

ah.. y i see ^^

[martin]

Cubed root = $x*$x*$x = (multiply the number with itself three times) = 3/1 = ^1/3 = cube root calculation

That's nonsense the way you've written it. I think that what you you were trying to put across was that if $x is the cubed root of $y then

$x*$x*$x = $y

ie

$x^3 = $y

therefore

($x^3)^(1/3) = $y^(1/3)

but since ($x ^n)^p = $x ^(n*p), then ($x^3)^(1/3) = $x^1 = $x

3/1 = three and nothing else

^1/3 is not a cube root calculation

A cube root is $x^(1/3)

$x^1/3 = $x/3 which is only the same as $x^(1/3) when $x = 27^0.5

[1905russell]

That's nonsense the way you've written it. I think that what you you were trying to put across was that if $x is the cubed root of $y then

$x*$x*$x = $y

ie

$x^3 = $y

therefore

($x^3)^(1/3) = $y^(1/3)

but since ($x ^n)^p = $x ^(n*p), then ($x^3)^(1/3) = $x^1 = $x

3/1 = three and nothing else

^1/3 is not a cube root calculation

A cube root is $x^(1/3)

$x^1/3 = $x/3 which is only the same as $x^(1/3) when $x = 27^0.5

I wrote it the way I thought jaenster would understand it and it seems it was understood.

You are wrong saying $x^1/3 = $x/3 (this is truly nonsense)

To cube root a number you ^1/3 or ^(1/3) its the same thing.

To square root a number you ^1/2 or ^(1/2) its the same thing.

It was understood - yours is nonsense not mine.

[martin]

I wrote it the way I thought jaenster would understand it and it seems it was understood.

You are wrong saying $x^1/3 = $x/3 (this is truly nonsense)

To cube root a number you ^1/3 or ^(1/3) its the same thing.

To square root a number you ^1/2 or ^(1/2) its the same thing.

It was understood - yours is nonsense not mine.

OK, then I didn't understand, but you should check what you are saying and I know you haven't.

You think that $x^1/3 = $x/3 is nonsense, but you haven't tried it have you?

^ has higher precedence than / so when the string is parsed the value calculated after ^ will end as soon as an operator of lower precedence is met.

If you try 27^1/3 you will get 9

If you try 27^(1/3) you will get 3

Try this

MsgBox(0,'root 16 is',16^1/2)

Have you tried it?

Who is talking nonsense?

But what I said was nonsense was this

Cubed root = $x*$x*$x = (multiply the number with itself three times) = 3/1 = ^1/3 = cube root calculation

I think you have a hard job in front of you persuading me that isn't nonsense.

[jvanegmond]

; This will calculate the square root of any number..

$Number = 12345

MsgBox(0, "Babylonian Square Root", SqrtBabylonian($Number) )
MsgBox(0, "Bakhshali Square Root", SqrtBakhshali($Number) )

Func SqrtBabylonian($S)
    Dim $X = 3^StringLen($S)
    While 1
        $newX = 0.5 * ($x + ($S/$x) )
        If $newX = $X Then
            ExitLoop
        Else
            $X = $newX
        EndIf
    WEnd
    Return $X
EndFunc

Func SqrtBakhshali($S)
    Dim $N = 0
    While 1
        $N += 1
        If $S - ($N^2) < 1 Then ;bad practice.. but it works
            ExitLoop
        EndIf
    WEnd
    $d = $S - $N^2
    $P = $d / (2*$N)
    $A = $N + $P
    Return $A - ( ($P^2) / ( 2*$A) )
EndFunc

Just playing.

Edit: Full time to make: 12 minutes.

By the way, I'll post Taylor Series for those interested later..

Edited July 23, 2007 by Manadar

[1905russell]

OK, then I didn't understand, but you should check what you are saying and I know you haven't.

You think that $x^1/3 = $x/3 is nonsense, but you haven't tried it have you?

^ has higher precedence than / so when the string is parsed the value calculated after ^ will end as soon as an operator of lower precedence is met.

If you try 27^1/3 you will get 9

If you try 27^(1/3) you will get 3

Try this

MsgBox(0,'root 16 is',16^1/2)

Have you tried it?

Who is talking nonsense?

But what I said was nonsense was this

I think you have a hard job in front of you persuading me that isn't nonsense.

Okay I checked ^(1/3) and ^1/3 and you are right I know () has precedent over ^ (Bedmas) and I see written this way (not raised) makes a difference.

Explaination

Cubed root = $x*$x*$x = (multiply the number with itself three times) = 3/1 = ^1/3 = cube root calculation

This meant if the root is a number timesing itself 3 times (ie $x*$x*$x) then take 3 = 3/1 and inverse it = 1/3 and use ^(1/3) to calculate cube root.

Edited July 23, 2007 by 1905russell

[1905russell]

; This will calculate the square root of any number..

$Number = 12345

MsgBox(0, "Babylonian Square Root", SqrtBabylonian($Number) )
MsgBox(0, "Bakhshali Square Root", SqrtBakhshali($Number) )

Func SqrtBabylonian($S)
    Dim $X = 3^StringLen($S)
    While 1
        $newX = 0.5 * ($x + ($S/$x) )
        If $newX = $X Then
            ExitLoop
        Else
            $X = $newX
        EndIf
    WEnd
    Return $X
EndFunc

Func SqrtBakhshali($S)
    Dim $N = 0
    While 1
        $N += 1
        If $S - ($N^2) < 1 Then ;bad practice.. but it works
            ExitLoop
        EndIf
    WEnd
    $d = $S - $N^2
    $P = $d / (2*$N)
    $A = $N + $P
    Return $A - ( ($P^2) / ( 2*$A) )
EndFunc

Just playing.

Edit: Full time to make: 12 minutes.

By the way, I'll post Taylor Series for those interested later..

Incredible what you did, but why would you do it these ways when simple $Number^(1/2) does the same thing?

It's okay I know the answer.

Edited July 23, 2007 by 1905russell

[lokster]

@1905russell, I think the simple answer to your question is "because some people like to ask the question 'WHY'".

WHY 3^2=9?

WHY Sqrt(9)=3?

Without the "WHY" the life is meaningless...

[jvanegmond]

Incredible what you did, but why would you do it these ways when simple $Number^(1/2) does the same thing?

It's okay I know the answer.

Thanks for that. I'm still going to give you an answer why.

$Number^(1/2) is exactly the same as Sqrt($Number). Even in mathematics that is another way of writing.

The functions I show you are a way of using multiplication and addition to calculate a root. A root could not be calculated the way jaenster wrote it, as it would be too time consuming to calculate the root of a really big number, like 8124746273, and it also does not calculates roots that are more accurate then integers.

@lokster, "why 3^2 = 9"? Because it is an easier form when you have to write 5*5*5*5*5, you just use 5^5.

"Why 9^0.5 = 3"? follows out of notation used here before. This is just convenient when using higher functions.

[martin]

Thanks for that. I'm still going to give you an answer why.

$Number^(1/2) is exactly the same as Sqrt($Number). Even in mathematics that is another way of writing.

The functions I show you are a way of using multiplication and addition to calculate a root. A root could not be calculated the way jaenster wrote it, as it would be too time consuming to calculate the root of a really big number, like 8124746273, and it also does not calculates roots that are more accurate then integers.

@lokster, "why 3^2 = 9"? Because it is an easier form when you have to write 5*5*5*5*5, you just use 5^5.

"Why 9^0.5 = 3"? follows out of notation used here before. This is just convenient when using higher functions.

The Babylonian one is best, and most accurate and fast. The other one isn't even as good as my method. Tell us about the Taylor series.

I don't think lokster meant what you thought, I think he knows about ^ etc, he was just saying (I think) it's good to ask why.

[jaenster]

Can somebody explain how you get the power of .5? How to calulate it?

[tmo]

Can somebody explain how you get the power of .5? How to calulate it?

if the result is a rational number, you just somehow know it due to experience.

if the result is a irrational number, you can use an iterational method like Newton:

f(x) = x^2 - a = 0 <-- the solution is a^0.5

f'(x) = 2x

now u can use the following iteration:

[jvanegmond]

Can somebody explain how you get the power of .5? How to calulate it?

I don't understand.

To the power of .5 is the same as square root. It is just a different notation.

Why would you first provide the answer, and then ask the question?

[jaenster]

I don't understand.

To the power of .5 is the same as square root. It is just a different notation.

Why would you first provide the answer, and then ask the question?

oh ok now understand, So ^ (1/3) is the cube root ?

[RazerM]

Yes,

x^(a/B) = b root(x^a)

[martin]

@Manadar

Can you tell me what the Taylor series is?

[jaenster]

But the topic, Do somebody like my function?

[jvanegmond]

@Manadar

Can you tell me what the Taylor series is?

Wikipedia says this about the Taylor series:

The Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

What it comes down to is, that the formula for the square root is:

Or simpler said:

Edited July 24, 2007 by Manadar