I want do exponentiation of floats, so i had a look at Raymonds FPU tutorial and found sample code. This is what i use for testing:
include <windows.inc>
includelib kernel32.lib
.data
r1 real10 2.0
r2 real10 -2.4
r3 real10 -100.123
i4 dword 4
qf qword ?
qs qword ?
qe qword ?
.code
MAIN PROC
;*************************************************************************************
; main proc
;*************************************************************************************
invoke QueryPerformanceCounter, addr qs
FINIT
xor esi, esi
.Repeat
FILD i4
; FLD r1
FLD r2
; FLD r3
FYL2X ;ST(0)=y*log2(x), ST(1)=zzz
FLD st(0) ;make a second copy, ST(0)=y*log2(x), ST(1)=y*log2(x), ST(2)=zzz
FRNDINT ;round it to an integer, ST(0)=int[y*log2(x)], ST(1)=y*log2(x), ST(2)=zzz
FSUB st(1), st(0) ;this will leave only a fractional portion in ST(1), ST(0)=int[y*log2(x)], ST(1)=y*log2(x)-int[y*log2(x)], ST(2)=zzz
FXCH ;ST(0)=y*log2(x)-int[y*log2(x)], ST(1)=int[y*log2(x)], ST(2)=zzz
F2XM1 ;get the fractional power of 2 (minus 1), ST(0)=2ST(0)-1, ST(1)=int[y*log2(x)], ST(2)=zzz
FLD1 ;ST(0)=1, ST(1)=2ST(0)-1, ST(2)=int[y*log2(x)], ST(3)=zzz
FADD ;add the 1 to ST(1) and POP ST(0), ST(0)=2ST(0), ST(1)=int[y*log2(x)], ST(2)=zzz
FSCALE ;add the integer in ST(1) to the exponent of ST(0), effectively multiplying the content of ST(0) by 2int, and yielding the final result of x^y, ST(0)=x^y, ST(1)=int[
FSTP st(1) ;the content of ST(1) has become useless
;overwrite the content of ST(1) with the result and POP ST(0)
;ST(0)=x^y, ST(1)=zzz
inc esi
.Until esi>=1000000
invoke QueryPerformanceCounter, addr qe
invoke QueryPerformanceFrequency, addr qf
FILD qe
FILD qs
FSUBP
FILD qf
FDIVP
int 3
RET
MAIN ENDP
;*************************************************************************************
END MAIN
The code delivers correct results. Running it 1 mio. times in a loop takes about 1 second on my (old) laptop. Running the same code (x = -2.4^4) in a high level language takes about 0.1 seconds, that is 1/10 of the above code. Obviously there is method of doing it 10 times faster - how?
JK
FILD i4
; FLD r1
FLD r2
; FLD r3
FYL2X ;ST(0)=y*log2(x), ST(1)=zzz
At this point, you have a NAN on your FPU.
Intel(R) Core(TM) i5-2450M CPU @ 2.50GHz
493 ms for JK's code yielding 33.178
423 ms for JJ's code yielding 33.178
490 ms for JK's code yielding 33.178
422 ms for JJ's code yielding 33.178
491 ms for JK's code yielding 33.178
427 ms for JJ's code yielding 33.178
494 ms for JK's code yielding 33.178
432 ms for JJ's code yielding 33.178
489 ms for JK's code yielding 33.178
425 ms for JJ's code yielding 33.178
This is 10 Million times, so it's a bit faster than the HLL. Even your (slightly modified) version is a factor 2 faster than the HLL, but of course, it's difficult to compare without having the HLL executable.
ExpXY (https://www.jj2007.eu/MasmBasicQuickReference.htm#Mb1483)(2.4, 4, result) ; calculates 2.4^4 and saves it to a REAL8 variable called result
That's what I use :cool:
What hll?
Cpp for example use only float=real4 and double =real8
There is precision settings in fpu control word you can test
Newer math library can take advantage of SSE which is easier to perform several math ops simultaneously, instead of your scalar fpu
Quote from: daydreamer on July 31, 2022, 06:47:50 AMNewer math library can take advantage of SSE which is easier to perform several math ops simultaneously, instead of your scalar fpu
Interesting. Can you link to one that performs -2.4^4 using SIMD instructions? I'm curious.
Thanks jj,
:sad: my bad, my code is buggy and a logarithm of a negative number results in a nan (as i should have learned in school). So the speed of code is ok, but my implementation definitely needs improvement.
What must i do, if x is negative in x^y (which basically is not forbidden)? But my implementation leads to calculating the logarithm of a negative number (which is forbidden).
JK
Quote from: JK on July 31, 2022, 07:43:24 AMmy code is buggy
Your code was almost perfect, but you forgot to pop the result, see
fstp result.
QuoteWhat must i do, if x is negative in x^y (which basically is not forbidden)?
It is ok if your exponent is an integer. But launch the Windows calculator and try -2.4^3.999 :rolleyes:
Quote from: JK on July 31, 2022, 06:18:09 AM
Obviously there is method of doing it 10 times faster - how?
-2.4 * -2.4 * -2.4 *-2.4
Yes, i forgot to pop the FPU. If you google for -2.4^3.999 you get a calculator and as a result -33.1485667591 (which implies -(2.4^3.999, but (-2.4)^3.999 yields the same as a positve result), my old pocket calculator can do it too. Why the Windows calculator rejects it as "improper entry" - i don´t know.
Mathematically -2.4^3.999 is not forbidden, so what must i do, if x is negative in x^y? Flip sign if negative, look at the exponent, if the exponent rounded to an integer mod 2 = 0 (that is, an even number) the result is positve, negative otherwise - is this correct?
@HSE: yes , i already thought of this, but what if the exponent is not an integer but a float?
JK
Quote from: JK on July 31, 2022, 08:32:35 AMMathematically -2.4^3.999 is not forbidden, so what must i do, if x is negative in x^y?
Windows calculator:
-2.4^3.999999999999= -33.1776
-2.4^4= +33.1776That an even integer exponent produces a positive number is a convention, not math. If your number is negative, flip the sign, perform the calc as usual, then flip the sign for the result.
Windows calculator doesn't handle complex numbers, when you try to raise -2.4 ^ 3.999 it's the same as (-2.4) ^ 3.999 which give a complex number
-2.4 ^ 3.999 can be written as -(2.4 ^ 3.999) which is totally different from (-2.4) ^ 3.999
Quote from: JK on July 31, 2022, 08:32:35 AMBut what if the exponent is not an integer but a float?
Make another tests :biggrin:
Don't get it. How is -2.4 different from (-2.4)? All those parens do is form a group, which so far as I can tell does nothing here. Or am I missing something?
And since when does -2.4 ^ 3.999 yield a complex number?
BTW, Windows calculator doesn't allow fractional exponents, at least so far as I could tell. (Gives me "invalid input".)
hi NoCforMe
the expression -2.4 is interpreted as the negative of 2.4, the - operator (sign) is not bound to the number 2.4
the expression -2.4 ^ 3.999 means the negative of 2.4 ^ 3.999 which is not complex but (-2.4) ^ 3.999 is complex
when you enter 2.4 in the calculator and then negate the number and then try to raise to some power it's as if you had entered (-2.4) ^ some power
I still think you're wrong. Perfectly OK to take the power of a negative number: the sign of the result flips between positive and negative:
-2^2 = 4
-2^3 = -8
-2^4 = 16
-2^5 = -32
and so on.
So far as complex numbers go, I think you're thinking of roots, not powers, as in i = sqrt(-1).
Amiright?
This might been here earlier, in year 2019 ?
#include <stdio.h>
#include <math.h>
int __cdecl main(void)
{
for (double d = 1.7; d < 2.3; d += 0.1)
printf("%f\n", pow(-1.23, d));
return 0;
}
nan
nan
nan
1.512900
nan
nan
So where are all those "nans" coming from?
Quote from: NoCforMe on July 31, 2022, 01:33:54 PM
I still think you're wrong. Perfectly OK to take the power of a negative number: the sign of the result flips between positive and negative:
-2^2 = 4
-2^3 = -8
-2^4 = 16
-2^5 = -32
and so on.
So far as complex numbers go, I think you're thinking of roots, not powers, as in i = sqrt(-1).
Amiright?
if you did the above in the Windows calculator then like I said, when you enter a number in the calculator and then negate that number then it would be the same as the math expression (-2)^n
try the same with the online calculator https://keisan.casio.com/calculator
this is all I am going to say, you believe what you want, I don't care
Quote from: TimoVJL on July 31, 2022, 02:03:29 PM
This might been here earlier, in year 2019 ?
To make it clearer:
#include <stdio.h>
#include <math.h>
int __cdecl main(void) {
printf("test\n\n\n");
for (double d = 1.7; d < 2.3; d += 0.1)
printf("-1.23^%1.1f=%3.5f\n", d,pow(-1.23, d));
return 0;
}-1.23^1.7=nan
-1.23^1.8=nan
-1.23^1.9=nan
-1.23^2.0=1.51290
-1.23^2.1=nan
-1.23^2.2=nanNow the same without math.h:
-1.23^1.7=0.00000
-1.23^1.8=-92559641157289296446000000000000000000000000000000000000000000.00000
-1.23^1.9=19035993442894622016370000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000.00000
-1.23^2.0=0.00000
-1.23^2.1=-92559641157289296446000000000000000000000000000000000000000000.00000
-1.23^2.2=-0.00000Good ol' C always gives you nice surprises :tongue:
Quote from: NoCforMe on July 31, 2022, 02:42:16 PM
So where are all those "nans" coming from?
Some CforYou:
#include <stdio.h>
#include <math.h>
int __cdecl main(void) {
printf("test\n\n\n");
double result;
for (double d = 1.7; d < 2.3; d += 0.1) {
_asm int 3;
result=pow(-1.23, d);
_asm nop;
_asm nop;
printf("-1.23^%1.1f=%3.5f\n", d,pow(-1.23, d));
}
return 0;
}
Somewhere deep in the guts of
pow the fpu says "baaaaad nummmber!"
Well, -(2^2) is not the same as (-2)^2 !
The former yields -4, the latter results in 4, so what is the result of -2^2 ?
According to math rules there is an order of math operations, parenthesis having the highest order (= must be calculated first), next is exponentiation and so on (https://en.wikipedia.org/wiki/Order_of_operations). The minus sign used for negation (unary minus) comes after exponentiation. But there are implementations not following this rule by giving negation a higher precedence than exponentiation.
Back to my question: how to get the result of (-2.4)^3.999 (the number "minus 2 point 4" to the power of "3 point 999"). We don´t need complex numbers here btw.
I refine my previous question: Mathematically (-x)^y is not forbidden, so what must i do in my code, if x is negative? Flip sign, if (x) negative, look at the exponent (y). If the exponent rounded to an integer mod 2 = 0 (that is, an even number), the result is positive, negative otherwise (flip sign afterwards) - is this correct?
JK
Quote from: JK on July 31, 2022, 07:11:01 PM
Well, -(2^2) is not the same as (-2)^2 !
Quote from: jj2007 on July 31, 2022, 04:54:23 PMWolfram Alpha, to my knowledge the highest authority for online math, says f*ck the convention, a negative number yields a negative number regardless of the exponent.
-2^2 -4
-(2^2) -4
(-2)^2 +4 :rolleyes:
Sorry jj,
maybe i got this wrong, but to me in x^y, y is the exponent. So "regardless of the exponent" contradicts your sample calculations:
-(2^2) = - 4
(-2)^2 = + 4
More from Wolfram Alpha:
(-2)^3 = -8
-(2^3) = -8
-(-2)^3 = 8
-(-2)^4 = -16
-(-2^4) = 16
-((-2)^4) = -16
I already know now, what i must know about order of precedence of math operators. So, please could someone (more experienced in math) help me with my coding question ...
Thanks
JK
Hi,
Quote from: JK on July 31, 2022, 07:11:01 PM
Back to my question: how to get the result of (-2.4)^3.999 (the number "minus 2 point 4" to the power of "3 point 999"). We don´t need complex numbers here btw.
I disagree, a negative number to a non integer exponent will
normally require complex numbers to be accurate. Anyway,
here is a FORTRAN program to illustrate.
PROGRAM MASMEX
C 31 July 2022, silly question from the MASM Forum.
COMPLEX BASE, EXPO, RESU
DATA BASE / (-2.4, 0.0) /, EXPO / (3.999, 0.0) /
RESU = BASE ** EXPO
WRITE (*,100) BASE, EXPO, RESU
100 FORMAT ( ' BASE = ', 2F8.3, ' EXPO = ', 2F8.3,
1 ' RESU = ', 2F8.3 )
STOP
END
And the output:
BASE = -2.400 0.000 EXPO = 3.999 0.000 RESU = 33.148 -0.104 Gack, first such program in seven years. About the same
number of errors as lines of code.
Cheers,
Steve N.
Quote from: jj2007 on July 31, 2022, 07:11:53 AM
Quote from: daydreamer on July 31, 2022, 06:47:50 AMNewer math library can take advantage of SSE which is easier to perform several math ops simultaneously, instead of your scalar fpu
Interesting. Can you link to one that performs -2.4^4 using SIMD instructions? I'm curious.
I already used x^4 it in cosine taylor series code
.data
align 16
numbers real4 -2.4,-2.3,-2.1,-2.5
result real4 0.0,0.0,0.0,0.0
.code
mov ecx,1000000/4 ;1/4 for packed loop,1/1 for scalar loop
@@L1:
movaps xmm0,numbers
movaps xmm1,xmm0
mulps xmm0,xmm1;x^2
mulps xmm0,xmm1;x^3
mulps xmm0,xmm1;x^4
movaps result,xmm0
sub ecx,1
jne @@L1
Quote from: JK on July 31, 2022, 07:11:01 PM
Well, -(2^2) is not the same as (-2)^2 !
The former yields -4, the latter results in 4, so what is the result of -2^2 ?
According to math rules there is an order of math operations, parenthesis having the highest order (= must be calculated first), next is exponentiation
Hi Jürgen,
You are right, and I was wrong. The Windows calculator is somewhat confusing to work with, but I know that's not an excuse ;-)
So: -2.4^3.9999 = -33.1747 because that's -(2.4^3.9999) and does not require complex numbers :cool:
If I'm not wrong
(-) * (-) = (+)
(-) * (+) = (-)
(+) * (-) = (-)
(+) * (+) = (+)
Even or odd is only for integer numbers
For any number a, a * a *... * a an even number of times is always positive
If we have a^b = a * a * ... *a (b times)
- If "a" is positive, a^b will ve always positive
- If "a" is negative, a^b will be positive only if b is even, negative if b is odd. Hence, if a is negative and b is not integer, we cannot say if a^b will be positive or negative, hence the error message, I guess.
I don't understand why a^b should be understood as complex. Real numbers are actually a subset of complex numbers, where the imaginary part is 0. But a^b will always be real.
c = a + bi, where i = sqrt(-1).
Here we are always dealing with b = 0.
Quote from: caballero on August 01, 2022, 02:07:03 AMa^b will always be real
Even with a=-2.4, as in JK's original code, and b=3.99?
I have found this one:
The expression a^(m/n) with fractional exponent is defined for when the nth root of a is a real number.
a^(m/n) = root(n)(a^m)
(-2)^(3/2) is not defined because sqrt(-2) is not a real number.
Only roots with odd index have a result when the radicand is negative
2^3.999 = 2^(4-0.001) = 2^4/(2^0.001), 2^0.001 = 2^(1/1000), 1000 is even. (understand (-2))
https://ekuatio.com/exponente-fraccionario/
> Even with a=-2.4, as in JK's original code, and b=3.99?
I'm really not sure here, sqrt(-2) is not real, so, if we have to go to complex numbers to get the solution, then I guess the solution cannot be expresed as a real number.
if JK wants to raise both positive and negative reals to arbitrary power then he certainly needs complex math
here is a complex library in cpp as a reference https://github.com/RobTillaart/Arduino/tree/master/libraries/Complex
Hey jj,
it is not so easy as it seems at first glance, i remember having to deal with such things in school (and i wasn´t that bad in math), but this is decades ago and my knowledge became rusty over the years.
@all,
thanks for your contributions, i think, i´m going to repeat a few math lessons ... and come back later
JK
Hi,
Just for fun, I output the values produced by the base of -2.4
and the range of the exponent from 3.0 to 4.0 by 0.01.
-2.400 0.000 3.000 0.000 -13.824 0.000
-2.400 0.000 3.010 0.000 -13.939 -0.438
-2.400 0.000 3.020 0.000 -14.040 -0.883
-2.400 0.000 3.030 0.000 -14.129 -1.336
-2.400 0.000 3.040 0.000 -14.204 -1.794
-2.400 0.000 3.050 0.000 -14.265 -2.259
-2.400 0.000 3.060 0.000 -14.311 -2.730
-2.400 0.000 3.070 0.000 -14.344 -3.206
-2.400 0.000 3.080 0.000 -14.361 -3.687
-2.400 0.000 3.090 0.000 -14.363 -4.173
-2.400 0.000 3.100 0.000 -14.350 -4.663
-2.400 0.000 3.110 0.000 -14.322 -5.156
-2.400 0.000 3.120 0.000 -14.277 -5.653
-2.400 0.000 3.130 0.000 -14.216 -6.152
-2.400 0.000 3.140 0.000 -14.139 -6.653
-2.400 0.000 3.150 0.000 -14.046 -7.157
-2.400 0.000 3.160 0.000 -13.936 -7.661
-2.400 0.000 3.170 0.000 -13.808 -8.166
-2.400 0.000 3.180 0.000 -13.664 -8.672
-2.400 0.000 3.190 0.000 -13.503 -9.176
-2.400 0.000 3.200 0.000 -13.324 -9.680
-2.400 0.000 3.210 0.000 -13.128 -10.183
-2.400 0.000 3.220 0.000 -12.914 -10.683
-2.400 0.000 3.230 0.000 -12.683 -11.181
-2.400 0.000 3.240 0.000 -12.433 -11.676
-2.400 0.000 3.250 0.000 -12.167 -12.167
-2.400 0.000 3.260 0.000 -11.882 -12.653
-2.400 0.000 3.270 0.000 -11.580 -13.135
-2.400 0.000 3.280 0.000 -11.260 -13.610
-2.400 0.000 3.290 0.000 -10.922 -14.080
-2.400 0.000 3.300 0.000 -10.566 -14.543
-2.400 0.000 3.310 0.000 -10.193 -14.998
-2.400 0.000 3.320 0.000 -9.802 -15.446
-2.400 0.000 3.330 0.000 -9.394 -15.885
-2.400 0.000 3.340 0.000 -8.969 -16.314
-2.400 0.000 3.350 0.000 -8.526 -16.734
-2.400 0.000 3.360 0.000 -8.067 -17.143
-2.400 0.000 3.370 0.000 -7.590 -17.540
-2.400 0.000 3.380 0.000 -7.098 -17.926
-2.400 0.000 3.390 0.000 -6.588 -18.300
-2.400 0.000 3.400 0.000 -6.063 -18.661
-2.400 0.000 3.410 0.000 -5.522 -19.007
-2.400 0.000 3.420 0.000 -4.966 -19.340
-2.400 0.000 3.430 0.000 -4.394 -19.658
-2.400 0.000 3.440 0.000 -3.808 -19.960
-2.400 0.000 3.450 0.000 -3.207 -20.246
-2.400 0.000 3.460 0.000 -2.592 -20.516
-2.400 0.000 3.470 0.000 -1.963 -20.768
-2.400 0.000 3.480 0.000 -1.321 -21.003
-2.400 0.000 3.490 0.000 -0.667 -21.219
-2.400 0.000 3.500 0.000 0.000 -21.416
-2.400 0.000 3.510 0.000 0.679 -21.594
-2.400 0.000 3.520 0.000 1.368 -21.751
-2.400 0.000 3.530 0.000 2.069 -21.888
-2.400 0.000 3.540 0.000 2.780 -22.004
-2.400 0.000 3.550 0.000 3.500 -22.099
-2.400 0.000 3.560 0.000 4.229 -22.171
-2.400 0.000 3.570 0.000 4.967 -22.221
-2.400 0.000 3.580 0.000 5.712 -22.248
-2.400 0.000 3.590 0.000 6.465 -22.252
-2.400 0.000 3.600 0.000 7.223 -22.231
-2.400 0.000 3.610 0.000 7.988 -22.187
-2.400 0.000 3.620 0.000 8.757 -22.118
-2.400 0.000 3.630 0.000 9.531 -22.024
-2.400 0.000 3.640 0.000 10.308 -21.905
-2.400 0.000 3.650 0.000 11.087 -21.760
-2.400 0.000 3.660 0.000 11.869 -21.589
-2.400 0.000 3.670 0.000 12.651 -21.392
-2.400 0.000 3.680 0.000 13.434 -21.168
-2.400 0.000 3.690 0.000 14.216 -20.918
-2.400 0.000 3.700 0.000 14.997 -20.641
-2.400 0.000 3.710 0.000 15.775 -20.337
-2.400 0.000 3.720 0.000 16.551 -20.006
-2.400 0.000 3.730 0.000 17.322 -19.648
-2.400 0.000 3.740 0.000 18.088 -19.262
-2.400 0.000 3.750 0.000 18.849 -18.849
-2.400 0.000 3.760 0.000 19.602 -18.408
-2.400 0.000 3.770 0.000 20.348 -17.939
-2.400 0.000 3.780 0.000 21.085 -17.443
-2.400 0.000 3.790 0.000 21.813 -16.920
-2.400 0.000 3.800 0.000 22.530 -16.369
-2.400 0.000 3.810 0.000 23.236 -15.791
-2.400 0.000 3.820 0.000 23.929 -15.186
-2.400 0.000 3.830 0.000 24.608 -14.553
-2.400 0.000 3.840 0.000 25.274 -13.894
-2.400 0.000 3.850 0.000 25.924 -13.209
-2.400 0.000 3.860 0.000 26.557 -12.497
-2.400 0.000 3.870 0.000 27.173 -11.759
-2.400 0.000 3.880 0.000 27.771 -10.995
-2.400 0.000 3.890 0.000 28.350 -10.207
-2.400 0.000 3.900 0.000 28.909 -9.393
-2.400 0.000 3.910 0.000 29.446 -8.555
-2.400 0.000 3.920 0.000 29.962 -7.693
-2.400 0.000 3.930 0.000 30.454 -6.807
-2.400 0.000 3.940 0.000 30.922 -5.899
-2.400 0.000 3.950 0.000 31.366 -4.968
-2.400 0.000 3.960 0.000 31.783 -4.015
-2.400 0.000 3.970 0.000 32.174 -3.041
-2.400 0.000 3.980 0.000 32.537 -2.047
-2.400 0.000 3.990 0.000 32.872 -1.033
-2.400 0.000 4.000 0.000 33.178 0.000
Regards,
Steve N.
Quote from: daydreamer on August 01, 2022, 01:22:34 AM
.data
align 16
numbers real4 -2.4,-2.3,-2.1,-2.5
result real4 0.0,0.0,0.0,0.0
.code
mov ecx,1000000/4 ;1/4 for packed loop,1/1 for scalar loop
@@L1:
movaps xmm0,numbers
movaps xmm1,xmm0
mulps xmm0,xmm1;x^2
mulps xmm0,xmm1;x^3
mulps xmm0,xmm1;x^4
movaps result,xmm0
sub ecx,1
jne @@L1
33.18 27.98 19.45 39.06
Press any key to continue...
:thumbsup:
Now -2.4^3.99 :biggrin:
Quote from: HSE on August 01, 2022, 04:31:47 AM
Quote from: daydreamer on August 01, 2022, 01:22:34 AM
.data
align 16
numbers real4 -2.4,-2.3,-2.1,-2.5
result real4 0.0,0.0,0.0,0.0
.code
mov ecx,1000000/4 ;1/4 for packed loop,1/1 for scalar loop
@@L1:
movaps xmm0,numbers
movaps xmm1,xmm0
mulps xmm0,xmm1;x^2
mulps xmm0,xmm1;x^3
mulps xmm0,xmm1;x^4
movaps result,xmm0
sub ecx,1
jne @@L1
33.18 27.98 19.45 39.06
Press any key to continue...
:thumbsup:
Now -2.4^3.99 :biggrin:
thanks Hector :thumbsup:
something similar to taylor series e^x maybe?
e^x=1+x+x^2/2!+x^3/3!+x^4/4!...
64bit might be worth it to check caps for avx 256bit SIMD
Sorry, JK, to take this discussion away from the practical problem you asked about. But the issues raised here demand discussion and clarification.
There's a lot of what seems to be outright misinformation floating around here, concerning the sign of the result of exponentiation and the issue of complex numbers.
"FORTRANS" said:
QuoteI disagree, a negative number to a non integer exponent will normally require complex numbers to be accurate.
This goes against everything I ever learned about math. Whether the exponent is an integer or not
makes no difference to the result. Why would it? Exponentiation operates on any real numbers, integer or not. And it certainly does
not introduce complex numbers to the equation.
To review, a complex number is defined by
i = sqrt (-1). Since we're not taking the square root of negative numbers, no complex numbers.
And I have to take extreme issue with something JJ mentioned up there, which was that someone in the Intertubes had apparently "redefined" mathematics to declare that multiplying a negative number by a negative number yielded ... a negative number! (If I'm stating the case correctly: please correct me if I'm wrong.) This is just
wrong, wrong, wrong, wrong, wrong! A negative number multiplied by a negative number always yields a positive number. And you can't just "declare" a "new paradigm" or whatever.
Quote from: NoCforMe on August 01, 2022, 05:07:16 AMa complex number is defined by i = sqrt (-1). Since we're not taking the square root of negative numbers, no complex numbers.
Sure?
i = (-1)^0.5 :cool:
QuoteAnd I have to take extreme issue with something JJ mentioned up there, which was that someone in the Intertubes had apparently "redefined" mathematics to declare that multiplying a negative number by a negative number yielded ... a negative number!
Hmmmm... so what is
(-2.4)^1.8? Is that multiplying a negative number by a negative number, or is it "something else"?
I'd say that
(-2.4)^3.999 = (-2.4)^(4-0.001) = (-2.4)^4/(-2.4)^0.001 = (-2.4)^4/((-2.4)^(1/1000)) = (-2.4)^4 / (sqr(1000)(-2.4))
Since 1000 is an even number, sqr(1000)(-2.4) is not real, but complex, hence (-2.4)^3.999 cannot be represented by a real number.
Bear in mind that sqr(3)(-1) = -1 because (-1)*(-1)*(-1) = -1, but sqr(2)(-1) = i, imaginary number.
In the case of (2.4)^3.999 = (2.4)^4 / (sqr(1000)(2.4)) = 33,1776 / 1,0008758520719666033451730665834 = 33,148566759121300335443929072659
Ie, a^(b/c) can only be calculated (as a real number) when sqr(c)(a) is a real number <=> c is an odd number, which is not the case of 1000
BTW: (-1)^0.5 = (-1)^(1/2) = sqrt(-1) = i
Quote from: jj2007 on August 01, 2022, 05:36:32 AM
Quote from: NoCforMe on August 01, 2022, 05:07:16 AMa complex number is defined by i = sqrt (-1). Since we're not taking the square root of negative numbers, no complex numbers.
Sure? i = (-1)^0.5 :cool:
Yes I'm sure; that's just a restatement of "square root" (the 1/2 power).
Quote
QuoteAnd I have to take extreme issue with something JJ mentioned up there, which was that someone in the Intertubes had apparently "redefined" mathematics to declare that multiplying a negative number by a negative number yielded ... a negative number!
Hmmmm... so what is (-2.4)^1.8? Is that multiplying a negative number by a negative number, or is it "something else"?
I would say that this function "breaks" at integer values:
-x^1: negative
-x^2: positive
-x^3: negative
-x^4: positive, etc.
So for exponent values between 1 and 2, the result would be negative. For values between 2 and 3, positive, etc.
I think ...
Quote from: caballero on August 01, 2022, 05:51:58 AM
I'd say that
(-2.4)^3.999 = (-2.4)^(4-0.001) = (-2.4)^4/(-2.4)^0.001 = (-2.4)^4/((-2.4)^(1/1000)) = (-2.4)^4 / (sqr(1000)(-2.4))
Since 1000 is an even number, sqr(1000)(-2.4) is not real, but complex, hence (-2.4)^3.999 cannot be represented by a real number.
Again, no. How do you get complex numbers out of this? We're not taking (even) roots of negative numbers, so there are no complex numbers.
Please explain how you arrive at this conclusion, that
goes against everything I know about math.
Please stop arguing, it´s me who is to blame for forgetting some basics of exponentiation
After some research, i found this one (https://www.rhetos.de/html/lex/gebrochener_exponent.htm), which puts it short and nice and easy to understand. Unfortunately it´s in German, but the diagram on the right side is all you need. Translation of the text in this diagram: (a must be a real number)
Quotea > 0
m must be an integer
n must be a natural number (integer > 0)
a < 0
m must be an integer
n must be a natural uneven number (integer > 0, e.g. 1,3,5,7,...)
the fraction m/n must be shortened as far as possible
So, this means for positve numbers there is no restriction, for negative numbers, it depends...
It depends on the exponent, for some exponents we will get a valid real result, for some other exponents the result is undefined in the range of real number. In fact you need complex numbers. E.g (-2)^0.5 (radix of minus 2), which is undefined in the range of real numbers.
There is no problem for negative numbers and integer exponents, fractional or real exponents might work or might not work (as described above) - this isn´t going to make coding easier :sad:
JK
Quote from: jj2007 on July 31, 2022, 08:58:38 AM
Windows calculator:
-2.4^3.999999999999= -33.1776
-2.4^4= +33.1776
Entering -2.4^ 3.999999999999 in Windows calculator result in "Entrada no válida"
You have to enter 2.4 and make it negative with "+/-" :thumbsup:
> How do you get complex numbers out of this?
Are nth roots of unity familiar to you?
https://en.wikipedia.org/wiki/Root_of_unity
The complex plane is basically R2. The real roots are those that rest on the abscissa axis, where the real values are. The one on the ordinates is for the imaginary part.
Obviously, sqr(1000)(-2.4) is not a real number. Therefore, we are dealing with a complex number and, therefore, not representable on the real line.
sqrt(-1) = i, sqrt(-2) = sqrt(2*(-1)) = sqrt(2)*sqrt(-1) = sqrt(2)*i
Quote from: NoCforMe on August 01, 2022, 05:07:16 AM
"FORTRANS" said:
QuoteI disagree, a negative number to a non integer exponent will normally require complex numbers to be accurate.
This goes against everything I ever learned about math. Whether the exponent is an integer or not makes no difference to the result. Why would it? Exponentiation operates on any real numbers, integer or not. And it certainly does not introduce complex numbers to the equation.
To review, a complex number is defined by i = sqrt (-1). Since we're not taking the square root of negative numbers, no complex numbers.
If you look at the numbers in Reply #31 that I posted and look
for the line with -2.400^3.500. -2.4 can be factored to (2.4)*(-1).
The exponent is 3.500, which can be factored to (7)*(1/2). Take
the square root of -1 and raise it to the seventh power. You get i^7,
which is -i. Take the square root of 2.4 and raise it to the seventh
power and you get 21.41604+. Multiply those and you get -21.416i
as shown.
caballero in Reply #27 pointed out that "a^(m/n)"..."is defined for
when the nth root of a is real number." Arbitrary fractions as an
exponent rarely works that way with negative numbers.
Steve N.
P.S.
Further posts while I was composing this make this simplistic maybe?
SRN
> Arbitrary fractions as an exponent rarely works that way with negative numbers
Maybe not so, there are infinite numerable odd numbers.
a^(b/(2n+1)) where n is a natural number, 2n+1 is always odd.
Quote from: JK on August 01, 2022, 06:07:42 AM
So, this means for positve numbers there is no restriction, for negative numbers, it depends...
It depends on the exponent, for some exponents we will get a valid real result, for some other exponents the result is undefined in the range of real number. In fact you need complex numbers. E.g (-2)^0.5 (radix of minus 2), which is undefined in the range of real numbers.
OK, now we seem to be getting somewhere.
So what is the test of which exponents yield a valid (i.e., real) result and which don't? Or is it not so simple?
z = a^(b/c)
If a < 0, z is a real number <=> c is an odd number. Hence (-2.4)^3.999 = 33.148 must not be true since we have already seen that this number is not real (sqr(1000)(-2.4) is not real), so it must be something like z = a + bi, where b <> 0
Does
this page (https://www.cuemath.com/algebra/exponent-rules/) help?
One thing from there:
Quote
If a fraction has a negative exponent, then we take the reciprocal of the fraction to make the exponent positive, i.e., (a/b)^-m = (b/a)^m
Quote from: caballero on August 01, 2022, 06:30:10 AM
> Arbitrary fractions as an exponent rarely works that way with negative numbers
Maybe not so, there are infinite numerable odd numbers.
A countable infinity. Irrational numbers are a larger infinity.
Rational numbers are therefore "rare".
Quote
a^(b/(2n+1)) where n is a natural number, 2n+1 is always odd.
Not sure what you are trying to say here. For n = 1, you get 3.
But (-5)^(1/3) is complex.
Actually (-5)^I/3) for integer I is
complex.Regards,
Steve N.
Edit.
That last part is wrong. Sorry.
SRN
QuoteSo what is the test of which exponents yield a valid (i.e., real) result and which don't? Or is it not so simple?
Yes, that´s the problem, i have no idea how to do that (make a fraction out of a real number, shorten it as far as possible and then look, if the denominator is even [=result undefined] or not even [=result in real range]).
Well, i could make my life easier and round exponents to an integer, if the number is negative. This guarantees a valid result in real range (avoiding cases where complex numbers come into play). But the result might not always be, what you would expect. Or could go the "just don´t do that" route and declare exponentiation of negative numbers (regardless of the exponent) as "forbidden" (invalid input/whatever)...
JK
Quote from: NoCforMe on August 01, 2022, 05:54:53 AM
Quote from: jj2007 on August 01, 2022, 05:36:32 AM
Quote from: NoCforMe on August 01, 2022, 05:07:16 AMa complex number is defined by i = sqrt (-1). Since we're not taking the square root of negative numbers, no complex numbers.
Sure? i = (-1)^0.5 :cool:
Yes I'm sure; that's just a restatement of "square root" (the 1/2 power).
Indeed :biggrin:
But the whole thread talks about (-a)^b, and that includes (-1)^0.5 alias
i = sqrt (-1)...
> (-5)^(1/3) is complex
Let's say (-8)^(1/3) = -2 (real), because (-2)*(-2)*(-2) = -8
I use -8 because is a more comfortable number than -5
sorry caballero but you are wrong, the answer is 1 + sqrt(3) * I
Hi,
Actually as a qube root, there are three roots. -2, 1+(3^0.5)i,
and 1-(3^0.5)i. Equally spaced around the origin in the complex
plane.
Cheers,
Steve N.
yes you are right, sorry caballero :smiley:
Hi,
Quote from: FORTRANS on August 01, 2022, 06:57:40 AM
Quote from: caballero on August 01, 2022, 06:30:10 AM
a^(b/(2n+1)) where n is a natural number, 2n+1 is always odd.
Not sure what you are trying to say here. For n = 1, you get 3.
But (-5)^(1/3) is complex.
Okay, if I look at Reply #50 and Reply #52 I can now see what
you were saying. It can take awhile to figure things out. So,
thanks to caballero, it kind of makes sense now.
Regards,
Steve
Quote from: HSE on August 01, 2022, 04:31:47 AM
Quote from: daydreamer on August 01, 2022, 01:22:34 AM
.data
align 16
numbers real4 -2.4,-2.3,-2.1,-2.5
result real4 0.0,0.0,0.0,0.0
.code
mov ecx,1000000/4 ;1/4 for packed loop,1/1 for scalar loop
@@L1:
movaps xmm0,numbers
movaps xmm1,xmm0
mulps xmm0,xmm1;x^2
mulps xmm0,xmm1;x^3
mulps xmm0,xmm1;x^4
movaps result,xmm0
sub ecx,1
jne @@L1
33.18 27.98 19.45 39.06
Press any key to continue...
:thumbsup:
Now -2.4^3.99 :biggrin:
like this SSE exp2 proc?:
exp2 :2.71828 (^1.0)
exp2 2.0 : 4(^2.0) 8(^3.0) 16(^4.0) 15.9889(^3.999)
Quote from: daydreamer on August 06, 2022, 04:16:33 AM
like this SSE exp2 proc?:
exp2 :2.71828 (^1.0)
exp2 2.0 : 4(^2.0) 8(^3.0) 16(^4.0) 15.9889(^3.999)
:thumbsup: I think that it's the idea: to use a Taylor serie (your notation is a mistery, I assume you used a serie accord to previous post).
To check I'm using the Complex Number Zlib (http://www.ray.masmcode.com/complex.html) by Raymond Filiatreault.
[ 2.0000e+000 0.0000e+000i] ^ [ 3.9900e+000 0.0000e+000i] = [ 1.5889e+001 0.0000e+000i]
[ -2.4000e+000 0.0000e+000i] ^ [ 3.9900e+000 0.0000e+000i] = [ 3.2872e+001 -1.0330e+000i]
this is my SSE2 natural e^x exponent code,together with fakultet x! and reciprocal 1/x! calculation
:thumbsup: Look good!!
Quote from: HSE on August 06, 2022, 08:51:14 AM
:thumbsup: Look good!!
thanks Hector :thumbsup: