e to the i pi

[Lane Straatman]

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <complex.h>
/*
double complex z1, z2, z3;
bool flag;
z1 = .4 + .7I;
z2 = cpow(z1, 2.0);
z3 = z1 * z1;
flag = false;
flag = true;
if (flag)
{
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
printf("%lf %lf\n", creal(z3), cimag(z3));
printf("%d\n", N);
}
*/

int main(int argc, char *argv[])
{
double complex z1, z2, z3, z4, z5;
z1=5 +7I;
z2=cpow(z1, 1I);
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
z5= 0 + I*(3.14159);

z3=2.54 + 0*I;
z4=cpow(z5,z3);
printf("%lf %lf\n", creal(z4), cimag(z4));

system("PAUSE") ;
return 0;
}
Why doesn't e^(i *pi) equal what most folks think it does? LS

[user923005]

On Feb 4, 7:23 am, Richard Heathfield <[e-mail].i nvalidwrote:

Eric Sosman said:

Malcolm McLean wrote:

"Lane Straatman" <inva...@invali d.netwrote in message
Why doesn't e^(i *pi) equal what most folks think it does? LS

Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.

That's plane wrong.

Either you have a couple of axes to grind, or you're misreading what Malcolm
wrote, which rings true.

Actually, it's hilarious (if you consider the deliberate misspelling).
IMO-YMMV.

This is referring to the famous equation e^(i*pi) - 1 = 0
which is certainly true in the complex plane.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Keith Thompson]

jacob navia <ja***@jacob.re mcomp.frwrites:

osmium wrote:

"Fred Kleinschmidt" writes:

>Aren't M_PI and M_E defined in math.h?

Thankfully, no. Only in grotesque extensions to the language made
by certain vendors.
Definitions of constants have no proper place in a programming
language, it just leads to two sets of constants, a private set, for
M_AVOGADRO and the like, and the set that came with the compiler,
M_PI. and friends.

lcc-win32 defines M_PI if not in conforming mode, as gcc does, for
instance.
They are not standard C but are POSIX. I do not see the point
in defining PI again and again...

No, gcc doesn't define M_PI. <math.his part of the C library, not
part of the compiler. gcc uses whatever <math.hfile is provided by
the system (that's glibc on some, but by no means all, systems).
(Minor detail: the "fixinclude s" program creates a modified version of
<math.hwhen gcc is installed, but that doesn't affect M_PI.)

If we're going to discuss specific implementations , we need to know
what we're talking about. gcc is a compiler, not a full
implementation.

The point is that any program that depends on M_PI is non-portable.
A conforming C implementation *may not* define M_PI in <math.h>.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Keith Thompson]

"user923005 " <dc*****@connx. comwrites:

On Feb 4, 7:23 am, Richard Heathfield <[e-mail].i nvalidwrote:

Eric Sosman said:

Malcolm McLean wrote:
>"Lane Straatman" <inva...@invali d.netwrote in message
>>Why doesn't e^(i *pi) equal what most folks think it does? LS

>Most folks would say that if you try to multiply a number by itself an
>imaginary number of times, that is impossible.

That's plane wrong.

Either you have a couple of axes to grind, or you're misreading
what Malcolm wrote, which rings true.

Actually, it's hilarious (if you consider the deliberate misspelling).
IMO-YMMV.

You missed something. Hint: What is the singular of "axes"? (There
are two correct answers.)

This is referring to the famous equation e^(i*pi) - 1 = 0
which is certainly true in the complex plane.

Or perhaps to the even more famous equation e^(i*pi) + 1 = 0
(which I used to have on a t-shirt).

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[user923005]

On Feb 5, 1:10 pm, Keith Thompson <[e-mail] ote:

"user923005 " <dcor...@connx. comwrites:

On Feb 4, 7:23 am, Richard Heathfield <[e-mail].i nvalidwrote:

Eric Sosman said:

Malcolm McLean wrote:
"Lane Straatman" <inva...@invali d.netwrote in message
>Why doesn't e^(i *pi) equal what most folks think it does? LS

Most folks would say that if you try to multiply a number by itself an
imaginary number of times, that is impossible.

That's plane wrong.

Either you have a couple of axes to grind, or you're misreading
what Malcolm wrote, which rings true.

Actually, it's hilarious (if you consider the deliberate misspelling).
IMO-YMMV.

You missed something. Hint: What is the singular of "axes"? (There
are two correct answers.)

Is there anything more ironic than ironically missing the irony?

This is referring to the famous equation e^(i*pi) - 1 = 0
which is certainly true in the complex plane.

Or perhaps to the even more famous equation e^(i*pi) + 1 = 0
(which I used to have on a t-shirt).

Do I have to turn in my "math major" badge now?
:-(

--
Keith Thompson (The_Other_Keit h) [e-mail] <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.- Hide quoted text -

- Show quoted text -

[Lane Straatman]

"Lane Straatman" <in*****@invali d.netwrote in message
news:kY******** *************** *******@comcast .com...

>
"Keith Thompson" <ks***@mib.orgw rote in message

>On the other hand, perhaps Malcolm's subtlety just escaped me, in
which case I'm duly embarrassed to have missed it.

He's got me after complex roots, now, and I've thought about it enough to
want to write it from scratch.

#include <stdio.h>
#include <stdlib.h>
#include <complex.h>
#include <math.h>
int main(void)
{
double complex z1, z3, z4, z5;
double mod, phi, a, b, c, d;
z1 = .4 + .7I;
a = creal(z1);
b = cimag(z1);
mod = sqrt(pow(a, 2) + pow(b, 2));
phi = atan(b/a);
c = ( (sqrt(mod)) * cos( .5 * phi ) );
d = ( (sqrt(mod)) * sin( .5 * phi ) );
z4 = c + d*I;
z5 = cpow(z1, .5);
z3 = cpow(z4, 2.0);
printf("The square root of %lf %lfi\n", creal(z1), cimag(z1));
printf("is : %lf %lfi\n", creal(z4), cimag(z4));
printf("using cpow to get the root: %lf %lfi\n", creal(z5), cimag(z5));
printf("using cpow to check the square: %lf %lfi\n", creal(z3),
cimag(z3));
system("PAUSE") ;
return 0;
}
Am I going to get in trouble with atan as I've got it? LS

[Lane Straatman]

"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:jd******** *************@b t.com...

Lane Straatman said:

>>
"Tim Prince" <tp*****@nospam computer.orgwro te in message
news:kY******* *********@newss vr23.news.prodi gy.net...

>>Lane Straatman wrote:
Since we have a bunch of
rationals in C, we have a bunch of transcendentals as the arctan's
of
these numbers. LS
In C, the atan() functions are a bunch of rationals. Ideally, in
certain ranges, with something resembling IEEE754 compliance, the
quality is measurable in ULPS difference between the actual result
and the
mathematicall y correct one. It looks like you jumped from C to
something else in mid-sentence.

You couldn't be more wrong.

Tim is right - atan() returns a double, and doubles have a finite number
of bits, so they literally *cannot* store irrational numbers. The best
they can do is store the closest rational approximation one can find
within the bits available.

The point is that if you have arctan of a rational then the *next* digit in
its expansion is not something that algebraic methods can get a hold of.
Where Dr. Kelly at the end of chp 24 of _Unleashed_ expands e to a page and
a half then has 99860 , do you have a bet what the next number would be? If
you expand one third for a page and a half, I could get that digit. LS

[Lane Straatman]

"Richard Bos" <rl*@hoekstra-uitgeverij.nlwr ote in message
news:45******** *********@news. xs4all.nl...

"Lane Straatman" <in*****@invali d.netwrote:

>Devcpp doesn't even have tgmath.h .

Of course it doesn't. It's a C89 implementation, not a C99 one.

"It" refers to what? ^^

>

> I copied it out of lcc

Bad idea, as you'd have known if you'd read the FAQ.

It's certainly not jacob's fault that I sniped his header and got errors by
the hundred.

>Is there something about tgmath.h that makes its inclusion here
ill-advised?

<http://c-faq.com/cpp/missinghdr.html >

I've never seen a stdbool.h that differed at all.
--
LS
"Hoekstra" is as common a name as "Jones" here in dutch Michigan. Pete
Hoekstra is the house representative from nearby Holland.

[Keith Thompson]

"Lane Straatman" <in*****@invali d.netwrites:

"Richard Bos" <rl*@hoekstra-uitgeverij.nlwr ote in message
news:45******** *********@news. xs4all.nl...

"Lane Straatman" <in*****@invali d.netwrote:

Devcpp doesn't even have tgmath.h .

Of course it doesn't. It's a C89 implementation, not a C99 one.

"It" refers to what? ^^

Um, devcpp? What else would it refer to?

I copied it out of lcc

Bad idea, as you'd have known if you'd read the FAQ.

It's certainly not jacob's fault that I sniped his header and got errors by
the hundred.

Nobody said it was jacob's fault.

Is there something about tgmath.h that makes its inclusion here
ill-advised?

<http://c-faq.com/cpp/missinghdr.html >

I've never seen a stdbool.h that differed at all.

That differed from what?

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Richard Heathfield]

Lane Straatman said:

"Richard Heathfield" <rj*@see.sig.in validwrote in message
news:jd******** *************@b t.com...

<snip>

>Tim is right - atan() returns a double, and doubles have a finite
number of bits, so they literally *cannot* store irrational numbers.
The best they can do is store the closest rational approximation one
can find within the bits available.

The point is that if you have arctan of a rational then the *next*
digit in its expansion is not something that algebraic methods can get
a hold of. Where Dr. Kelly

Actually, Ian is a very very very clever man, a fine fellow, an
excellent host, a talented musician, a superb programmer, and a quite
astounding linguist, but it has to be said, when all's said and done,
that he's not a doctor. I know the book says he is - at least three
times - but the book is wrong. That's probably my fault, because I
think I just kind of assumed he had a doctorate when I was setting up
the team, and it turns out he hasn't. If ever there were a failure of
the British education system, it is this: that it does not regard Ian
as a doctor. Having said that, I'm now going to stop making a fuss
about it. Instead, I would like to award Ian an honorary doctorate from
the University of Common Sense.

at the end of chp 24 of _Unleashed_ expands e to a page and
a half then has 99860 , do you have a bet what the next number would
be? If you expand one third for a page and a half, I could get that
digit.

Ian's expansion of e, however, was not done via calls to atan(), which
is what Tim Prince was talking about. atan() returns a double, and a
double cannot store an irrational number precisely. A circle can, but a
double can't.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Malcolm McLean]

"Richard Heathfield" <rj*@see.sig.in validwrote in message

Malcolm McLean said:

>>
"Richard Heathfield" <rj*@see.sig.in validwrote in message

>>>Microsoft have taken away my compiler. I bought a brand new Windows
Vista
machine, installed my copy of Visual Studio and - no executable.

You need Service Pack 7. Admittedly, Visual Studio won't work with that
either (possibly modulo emulators), but you get a perfectly capable C
compiler as part of the bundle. And of course it's completely free.

Seriously?

Er, no. Sorry, Malcolm, I thought you knew - "Service Pack 7" is the
traditional
name for Linux when being recommended as a fix for a Windows-specific
problem.

>I need a Windows library to compile my games and BASICdraw,

You have just discovered that Microsoft aren't all that interested in
supporting
"legacy software" (e.g. their own two-year-old compiler, if your account
is
anything to go by). Have you considered porting your stuff to a more
stable
platform?

They do offer a free compiler. It looks pretty horrid, but at least it
compiled "Hello World" - not without five minutes tweaking to get rid of a
file called stdafx.h it insisted on adding.

Anyway I have registered and agreed not to distribute any programs compiled
with it as open source, etc, etc, etc. I've no real choice. A computer isn't
a computer without a programming environment - it's just a glorified
typewriter.