OK i was in the wrong. Sorry (thought i apologized already?)

RAdlanor you said (You just don't use trigonometry for square roots.)

This is DeMoivres theorom which can be used to find the square, cubic, or any power roots of any real or complex number.

[(any number) ^ (1/n)] = [(cosA +k *360 + sinB +k*360)/n]

That looks like trig to me. The problem is that you still have to find the square root of the radius vector. (i.e. [(any number) ^ (1/n)] )

This thread has got me thinking of trying to find one unique formula for finding roots of members of infinite sequences that uses only squares and normal math operations. I spent most of yesterday working on the sequence ( 2,4,8,16,32.....)

Maybe such formula already exist. I dunno
Gonna work on it anyway.
:alright:
Posted on 2002-06-22 10:50:39 by IwasTitan
I haven't used DeMoivre's theorem since grade 11... but if I remember correctly it's really only good for complex numbers. For real numbers the sin componenet disappears and cos goes to one, because your angle is 0.

So, even if you were being serious about solving it this way... you'd be stuck with your original problem...

However, for complex numbers, it's great.

Personally, I don't see anything wrong with Newton's method for square roots... it should work well enough and you could probably code up an integer version easily

Take care

--Chorus


PS. shouldn't the formula be:

p=x+yi
r^2=x^2+y^2
k=angle (in radians) between abscissa and p
p^(1/n) = r^(1/n)
Posted on 2002-06-22 11:05:24 by chorus
Yep chorus your right about the formula. I get typing and don't watch my brackets and stuff.

Here i have uncovered the formula for finding the roots of the members of the infinite sequence {2,4,8,16....} obviously the easiest sequence to use when basing the formula on the square root of 2.

It Is:
Sqr root of N = ((N^2)/ 2) / (1.4142135)^(1+3*n)


Where N is the number you want the root of.

Where n is position of the number in the infinite sequence. (i.e. 2 = position 0, 4 = position 1, 8 = position 2)

1.4142135 is the root of 2

That was an obviously easy sequence.

I am going to try this one next still based on the the root of 2.

{3,6,9,12...}

:alright:
Posted on 2002-06-22 11:50:39 by IwasTitan
What's wrong with

sqrt (2^n) == 2^(n/2)

which is a lot easier to calculate than (1.414..)^(1+3n)

or if you wanna be fancy, you can use lns

sqrt(2^n) == e^(ln (2)/2)


The sequence {3,6,9,12} is simply 3*{1,2,3,4,...} for which the solution for the square root of the nth number is sqrt(3)*sqrt(n) where sqrt (3) ~= 1.73.

--Chorus
Posted on 2002-06-22 12:50:08 by chorus
Yep that is very nice chorus for the first sequence {2,4,8,...}

sqrt (2^n) == 2^(n/2)

However your solution to finding the roots of the sequence {3,6,9,12...} requires that you find individual roots.

sqrt(3)*sqrt(n)

Here you have a constant sqrt (3) but you then have to calculate the root sqrt(n) for all integer values.

My goal is to find a formula for the series similar to your posted sqrt (2^n) == 2^(n/2) which uses powers. This "is" better than mine.

And of course if the series were {3,9,27,81....} you could use
sqrt(3^n)==3^(n/2)

But the series is { 3,6,9,12 } so you see my point i hope.

Any ideas please reply.
thanx



:alright: :alright:
Posted on 2002-06-22 13:28:57 by IwasTitan
IwasTitan,

my point about {3,6,9,12} is that if there were a nice solution then you would implicitly have a solution for the square root of integers because they are identical problems.

There are a number of solutions, but none as simple as for the specific case of Sqrt (2^n). You can use a Taylor series expansion, for example, which is what Newton's method is based on.

The problem you will find is that most square roots are non-constructable. That is, they, are transcendental and do not boil down to finite algebraic sequences. You pretty much have to define them in terms of other transcendental numbers.

Regards
--Chorus
Posted on 2002-06-22 15:36:20 by chorus
Well that is certainly food for thought chorus. You appear to have a good understanding of number theory.

Maybe you could help me with something. I once had taken a calculas course in which the following equation was stated as being impossible to integrate.


(x^2 + 5) ^ 2 . dx

Any idea why?

:alright:
Posted on 2002-06-22 16:25:20 by IwasTitan
No idea...

f(x)dx == (x^2 + 5) ^ 2 . dx == x^4+ 10x^2+25

Integeral f(x) dx == (x^5)/5 + 10(x^3)/3 + 25x + C1

Seems perfectly integrable (?) to me...

However, there are many equations for which an algebraic solution of an integral is not possible. Every polynomial is though.

--Chorus
Posted on 2002-06-22 18:16:00 by chorus
Hmmmm ...I think what the course may have been implying is that if the X value is at a power greater than 1 ..then you would have to expand just as you did. Expansion isn't necessary for the equation below.

(3x+4)^5.dy/dx = 15 (3x+4)^4

and so

15(3X +4) ^4. dx == 15(3X+4)^5 /(3*5)


Imagine how simple it woul'd become if you didn't have to expand this:

(x^7 + 11) ^ 25

If i remember correctly now their words were:

"There is presently no method" ....which does not mean there is no solution.

With modern computing power would a method that did not require expansion really count for much? i dunno. Maybe they were implying something else.

:alright:
Posted on 2002-06-22 19:18:13 by IwasTitan

something like this would do it

SquareRoot PROC Number:DWORD

mov ecx, 1
@@: inc ecx
mov eax, ecx
mul ecx
cmp eax, Number
jb @b
ja @f
mov eax, ecx
ret
@@: xor eax, eax
ret
SquareRoot ENDP


This just gives 0's if it's not 9 or 4
Posted on 2002-06-22 21:36:04 by andy981
I got it to work, it's pretty cool code!
Posted on 2002-06-22 21:37:25 by andy981
Originally posted by andy981
AnyOne have a FPU version of a SqRoot routine



...

_dq DQ 0

...

push 10
fild DWORD PTR [esp]
fsqrt
fstp _dq
pop eax ;Rebalance stack
_dq contains the squareroot of 10. You can eliminate push 10 ... pop eax if you want to compute directly from a "variable". Then use the FloatToStr from the masm32 library to convert FPU to string. :)
Posted on 2002-06-22 22:16:27 by stryker
Originally posted by stryker



...

_dq DQ 0

...

push 10
fild DWORD PTR [esp]
fsqrt
fstp _dq
pop eax ;Rebalance stack


.data
Number dd ?

.code

mov eax,25
push eax
call SquareRoot

SquareRoot:
mov Number,eax
mov ecx, 1
@@: inc ecx
mov eax, ecx
mul ecx
cmp eax, Number
jb @b
ja @f
mov eax, ecx
ret
@@: xor eax, eax
ret


If you want, can you lay out what you showed me Stryker last night like this
if it can work like that, Thank You for the help so far

Andy981

:stupid:
Posted on 2002-06-23 21:35:01 by andy981
;maybe this one
;but you should add in the start a check if the number is 0,1 or 2 ( i think)
un tested ..
shr eax,2
mov ecx, eax
@@: inc ecx
mov eax, ecx
mul ecx
cmp , eax,; esp if you push eax, number if not
ja @b
sbb ecx,0
@@:
mov eax, ecx ;add esp,4
ret 4 ; or do ret 4


bye
eko
Posted on 2002-06-24 07:01:43 by eko
Originally posted by Andy981
Don't really know how to make this work, but Thank You Stryker
You mean my code? Or the one you posted. If the latter, I'll check that out later, also haven't tested eko's version yet.
.data?


_dq DQ ?
bfr DB 20 DUP(?)

.code
start:
push 10 ;TOS == 10
fild DWORD PTR [esp] ;ST(0) == 10
fsqrt ;ST(0) == 3.162278 == sqrt(10))
fstp _dq ;_dq = ST(0)
pop eax
invoke FloatToStr, _dq, OFFSET bfr
invoke MessageBox, NULL, OFFSET bfr, NULL, MB_OK

;or

fld _dq ;ST(0) == 3.162278
fsqrt ;ST(0) == 1.778279 == sqrt(3.162278)
fstp _dq ;_dq = ST(0)

invoke FloatToStr, _dq, OFFSET bfr
invoke MessageBox, NULL, OFFSET bfr, NULL, MB_OK

invoke ExitProcess,NULL
end start
Just include masm32.inc and masm32.lib for the FloatToStr.
Posted on 2002-06-24 11:21:22 by stryker
Yeah you really come through, Thanks Stryker!

I'd done a little of this floating point several months ago, but never got to this level, as I actually look at it right now,
I remember that fstp is for storing the number.
And by looking up in the Book right now about

DQ is a double, I wondered what
_dq DQ ?

was about



Anyway Thank You Stryker

I made it one complete copy paste for the next person.

And I changed it so 25 is pushed
then Squared to 5
then 5 is Squared

and that last Square is displayed on the Screen









.486
.model flat,stdcall
option casemap:none

includelib \masm32\lib\kernel32.lib
includelib \masm32\lib\user32.lib
includelib \masm32\lib\masm32.lib
include \masm32\include\windows.inc
include \masm32\include\kernel32.inc
include \masm32\include\user32.inc
include \masm32\include\masm32.inc

.data?

_dq DQ ?
bfr DB 20 DUP(?)

.code
start:
push 25 ; You will get 5 ;TOS == 10
fild DWORD PTR ;ST(0) == 10
fsqrt ;ST(0) == 3.162278 == sqrt(10))
fstp _dq ;_dq = ST(0)
pop eax
invoke FloatToStr, _dq, OFFSET bfr
invoke MessageBox, NULL, OFFSET bfr, NULL, MB_OK

;or

fld _dq ;ST(0) == 3.162278
fsqrt ;ST(0) == 1.778279 == sqrt(3.162278)
fstp _dq ;_dq = ST(0)

invoke FloatToStr, _dq, OFFSET bfr
invoke MessageBox, NULL, OFFSET bfr, NULL, MB_OK
invoke StdOut,addr bfr
invoke ExitProcess,NULL


end start
Posted on 2002-06-24 20:45:35 by andy981