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

[Keith Thompson]

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

"Glenn Hutchings" <zo*****@google mail.comwrote in message
news:11******** *************@a 75g2000cwd.goog legroups.com...

>On 4 Feb, 18:49, rich...@cogsci. ed.ac.uk (Richard Tobin) wrote:

>>For fun, try 23.140693 ^ i.

For even more fun, try computing i ^ i. Betcha didn't see that
coming!

I stated in this thread that I wanted to take e to the i pi and get as close
as I could with it using pre-defined types. I think i^i better suits this
purpose. LS

That makes no sense. e^(i*pi) is -1; i^i is a completely different
real value, approximately 0.20788. I won't waste my time trying to
figure out what you really meant.

--
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.

[Dave Vandervies]

In article <ln************ @nuthaus.mib.or g>,
Keith Thompson <ks***@mib.orgw rote:

[Snip everything C-related]

>Apparently i^i (where i is the imaginary square root of -1 and "^"
denotes exponentiation) is a real number. I'm sure there's a simple
mathematical proof of this, but I'm too lazy to track it down or
reconstruct it.

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

So:
i = e^(ln i)
i^i = (e^(ln i))^i
= e^(ln i * i)

Completing the proof is left as an exercise for the reader.
dave

--
Dave Vandervies dj******@csclub .uwaterloo.ca
I wonder if it has something to do with being able to accurately handle
an excruciating amount of detail without checking yourself into a mental
institution. --Logan Shaw in the scary devil monastery

[Lane Straatman]

"CBFalconer " <cb********@yah oo.comwrote in message
news:45******** *******@yahoo.c om...

Lane Straatman wrote:

>>

... snip ...

>>
I think spending any money on a computer for this guy is bad money.
He has no real choice about distributing software, as he lacks the
ability to manipulate the dominant OS.

Dominant? The cockroach population of NYC exceeds the human
population. Does that make them dominant?

Sitting here as I do on Martin Luther King blvd., you need to know that the
only answer I have to that is *yes.* LS

[Lane Straatman]

"Dave Vandervies" <dj******@caffe ine.csclub.uwat erloo.cawrote in message
news:eq******** **@rumours.uwat erloo.ca...

In article <ln************ @nuthaus.mib.or g>,
Keith Thompson <ks***@mib.orgw rote:

[Snip everything C-related]

>>Apparently i^i (where i is the imaginary square root of -1 and "^"
denotes exponentiation) is a real number. I'm sure there's a simple
mathematica l proof of this, but I'm too lazy to track it down or
reconstruct it.

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

So:
i = e^(ln i)
i^i = (e^(ln i))^i
= e^(ln i * i)

Completing the proof is left as an exercise for the reader.

That's a nice touch. LS

[Richard Heathfield]

Keith Thompson said:

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

<snip>

>z1 = cpow(z1, z1);
printf("%lf %lf ", z1);
return 0;}

The "%lf" format specifier is invalid.

Tiny nit time. That is not true in C99, which the above program clearly
must be. See 7.19.6.1(7): "l (ell) [...] has no effect on a following
a, A, e, E, f, F, g, or G conversion specifier."

So "%lf" is equivalent to "%f". So that has nothing to do with his bug
(which, as you noted elsewhere, is probably more to do with the fact
that he has two format specifiers but only one datum, and that being of
the wrong type!).

<snip>

--
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]

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

"Malcolm McLean" <re*******@btin ternet.comwrite s:
[...]

>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.

Do they *really* require you to agree to that? How can they enforce a
restriction like that? If you distribute your own source code, how
can the fact that you've fed it to their compiler affect your rights?
I suspect the situation isn't that simple.

You can always install Cygwin, which includes gcc (I'm not sure how
much support it has for Windows-specific programming, though).

I think actually the open source restriction applies to anything based on
their samples.
Since the API is so intricate the only sensible way to to take a working
program and modify it, this could be very broad. They are obviously scared
of GNU.

I didn't read the agreement very carefully. I doubt the thing has any legal
force because you are logging into a computer, not signing a binding
contract.

The saga continues. I installed the free C compiler. It compiles "hello
world". But it won't do a Windows program. I didn't buy Vista for its
command shell capabilities, so I download the SDK. Fair enough. The compiler
won't recognise it. After about two hour rooting about on the web, I find a
Microsoft page telling me how to edit various configuration files to get it
to work - two of them, plus paths. Meanwhile the OS merrily throws threats
at every edit. Of course I make a typing mistake. At 1.00 in the morning, I
finally get a "Hello World" - in C++, I haven't figiured out how to get C
mode yet.

I am a programmer not a hacker. Some people might see this as an interesting
challenge. Personally I just see it as a total nuisance which takes times
from what I should be doing, like adding subroutines to BASICdraw. Two days
wasted. This sort of thing is often rationalised as "teething troubles". In
fact it is a constant situation. At work I am struggling with two new
programming environments - R and a new Lisp compiler. In computing, you are
very frequently using software for the first time.

[Keith Thompson]

Richard Heathfield <rj*@see.sig.in validwrites:

Keith Thompson said:

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

<snip>

>>z1 = cpow(z1, z1);
printf("%lf %lf ", z1);
return 0;}

The "%lf" format specifier is invalid.

Tiny nit time. That is not true in C99, which the above program clearly
must be. See 7.19.6.1(7): "l (ell) [...] has no effect on a following
a, A, e, E, f, F, g, or G conversion specifier."

You're right, I missed that.

--
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.

[Malcolm McLean]

"Dave Vandervies" <dj******@caffe ine.csclub.uwat erloo.cawrote in message

>
So:
i = e^(ln i)
i^i = (e^(ln i))^i
= e^(ln i * i)

Wow.That's clever.

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though. Do you think he's trying to pull a fast one on us?

[Richard Tobin]

In article <ln************ @nuthaus.mib.or g>,
Keith Thompson <ks***@mib.orgw rote:

>Apparently i^i (where i is the imaginary square root of -1 and "^"
denotes exponentiation) is a real number. I'm sure there's a simple
mathematical proof of this, but I'm too lazy to track it down or
reconstruct it.

It's exp(-pi/2). Someone else showed the proof, but you can just type
"i^i" into Google to see the numerical value.

-- Richard

--
"Considerat ion shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[Richard Tobin]

In article <eL************ *************** ***@bt.com>,
Malcolm McLean <re*******@btin ternet.comwrote :

>Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

-- Richard
--
"Considerat ion shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.