e to the i pi

[Dave Vandervies]

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

I think you mean "power series expansion". It's only a polynomial if
it's finite.


dave
(knowing the power series expansions for sin and cos would also be helpful)

--
Dave Vandervies (e-mail address removed)

What would you give if somebody used a silly unit (hogsheads) and gave
a good justification for it? --Joe Zeff and Garrett Wollman

More thought. in the scary devil monastery

 

[Keith Thompson]

http://www.billfordx.net/screendumps/cstuff_7.htm

Clearly, this latest version is off. Heathfield says I have two specifiers
for one datum. Is it not a datum with a real and a complex part, both of
which need attention in the conversion specifiers?

[...]

This is in reference to your attempt to print a single complex value
using a printf format that expects two floating-point arguments. I
have tried at least twice to explain this to you. I will not waste my
time doing so again.

In fact, I will not waste my time responding to anything you post, or
offering you any sort of help after this. I just read your article in
comp.std.c,
<http://groups.google.com/group/comp.std.c/msg/d91bae8783a38bde>, in
which you replied to a polite, informative, and only mildly critical
article by (e-mail address removed) with:

| You have the manners of ferrets. F*** yourselves collectively and shove the
| standard up your ***. LS

[edited for content]

It's not inconceivable that that article was a forgery. If so, please
say so. (I don't believe it was.)

Until and unless your behavior improves dramatically, I judge you to
be a troll (deliberate or not, I don't care which), and I will ignore
you. I leave it to others to make their own judgements.

Yes, this is off-topic. I decided to post it anyway as an attempt to
improve this newsgroup's signal-to-noise ratio.

 

[Malcolm McLean]

I think you mean "power series expansion". It's only a polynomial if
it's finite.


dave
(knowing the power series expansions for sin and cos would also be
helpful)

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

 

[Malcolm McLean]

I've had a very good experience this last week with Devcpp. Up and
running immediately. I'll post the above link in an ng with MS in the
name and see what comes. LS

I bought Visual Studio when I was still a games programmer. It is handy to
have the same compiler you use at work.
Now I've started a PhD in computer modelling of peptides, and I use a Linux
system for scientific work. However I kept the Visual Studio for doing
little hobby programs, like BASICdraw. I do want people to run my stuff, but
the virus problem means that it is hard to distribute executables. I might
be better off on Linux - sometimes it is better to be a little fish in a
little pond than a little fish in a big one.

So I did actually have a choice, for the first time in many years. I just
assumed that everything would work the same on the new Windows. Now I am not
sure that redm,ond won't remotely delete a dual boot.

http://www.personal.leeds.ac.uk/~bgy1mm

 

[CBFalconer]

.... snip ...

It looks slick as all get out. Looks don't count for much though.
I can't find a run button or pull-down and don't want to have to
step outside the IDE to run the executable:

On Windoze, ALT-tab can bring you to the terminal window (dosbox),
on which you compile and run the application just edited. Easier
and quicker than fooling with a brain-dead IDE. You can apply the
same technique, unaltered, to Linux etc.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>

"A man who is right every time is not likely to do very much."
-- Francis Crick, co-discover of DNA
"There is nothing more amazing than stupidity in action."
-- Thomas Matthews

 

[Keith Thompson]

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

 

[robertwessel2]

Right, what is the minimum maximum? LS

The required minimums for long doubles are exactly the same as for
doubles. The minimums for both DBL_DIG and LDBL_DIG are both 10.

In practice, it's often considerably larger. For example, IEEE
doubles will get you 15 decimal digits, 80-bit Intel/x86 style
extended reals (assuming your compiler supports them) will get you 18
digits. Other implementations of extended reals exists - the double-
double format mentioned by Walter Robinson is fairly common in
software implementations, and gets you about 31 decimal digits, the
128 bit extended real on zSeries has a 112 bit mantissa and nets about
33 decimal digits. There is not really a standard for IEEE extended
doubles, but the standard does specify some minimums, which is
approximately what the 80-bit x86 format implements.

 

[Malcolm McLean]

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

Say we've got three points in three dimensions. Put a line between a, b, and
c. It is just a normal angle. Three points always lie in a plane.
So if our vertices are complex, that makes 4 dimensions, x, xi, y, yi for
each point. But we still have three points.
Do they still lie in a plane, or is the angle between them complex?

 

[Keith Thompson]

Say we've got three points in three dimensions. Put a line between a, b, and
c. It is just a normal angle. Three points always lie in a plane.
So if our vertices are complex, that makes 4 dimensions, x, xi, y, yi for
each point. But we still have three points.
Do they still lie in a plane, or is the angle between them complex?

I think that when you're dealing with trig functions applied to
complex numbers, you're no longer dealing with triangles. A triangle
in any number of dimensions lies in a plane. I'm not sure there's a
geometric interpretation for a triangle with complex sides and/or
angles.

Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically. There are infinite series for each of the
exp(), sin(), and cos() functions, derived using their properties for
real numbers. Applying the same series to complex numbers gives you
mathematically useful results -- including cool whacko stuff like
e^(i*pi) + 1 = 0.

I imagine that quaternions would yield some interesting results -- but
standard C doesn't support them, so that's off-topic.

 

[Dik T. Winter]

>
> I think that when you're dealing with trig functions applied to
> complex numbers, you're no longer dealing with triangles.

Indeed. cos(i.ln(10)) = 5.2.

> Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
> e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
> make sense algebraically.

Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

> I imagine that quaternions would yield some interesting results

Not really.

 

[Richard Tobin]

Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

It certainly *corresponds* to something geometrical - cos x and sin x
are the coordinates of exp(ix), but to *make sense* geometrically
you'd need an intuition as to why exp(ix) should be a circle for real x.

-- Richard

 

[Yevgen Muntyan]

[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)

Try this one:

e^(i*5*pi/2) = i =>
i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

Yevgen

 

[Dik T. Winter]

> > > Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
> > > e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
> > > make sense algebraically.

>
> >Also geometrically. Take the complex plane with real and imaginary
> >axis. Draw a circle with radius 1 around the origin. exp(i * x)
> >circles around this circle when x increases.

>
> It certainly *corresponds* to something geometrical - cos x and sin x
> are the coordinates of exp(ix), but to *make sense* geometrically
> you'd need an intuition as to why exp(ix) should be a circle for real x.[/QUOTE]

In mathematics it is easy to be lead astray by intuition.

 

[Dik T. Winter]

> Dave Vandervies wrote: ....

This is the main branch of ln.

> e^(i*5*pi/2) = i =>
> i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

In the complex numbers it is not necessarily true that (a^b)^c = a^(b*c):
-i = i^3 = ((-1)^(1/2))^3 = (-1)^(3/2) = ((-1)^3)^(1/2) = (-1)^(1/2) = i ?

 

[Yevgen Muntyan]

This is the main branch of ln.

Yep. Note how I didn't use ln() in my "proof".

And note how I did *exactly* same thing as in the original proof.
While "ln" (with small "l") is well-defined, something^something
isn't. You got to tell what it means (like man page does).
Anyway, point really was: "see how off-topic stuff everybody likes can
be so easily wrong?"
(I hope nobody will get a bright idea to cross-post to nonsense.math
or something).

Yevgen

 

[Guest]

This is the main branch of ln.


In the complex numbers it is not necessarily true that (a^b)^c = a^(b*c):
-i = i^3 = ((-1)^(1/2))^3 = (-1)^(3/2) = ((-1)^3)^(1/2) = (-1)^(1/2) = i ?

That's not specific to the complex numbers, is it? ((-1)^2)^(1/2) !=
(-1)^(2*(1/2)).

 

[Tak-Shing Chan]

It certainly *corresponds* to something geometrical - cos x and sin x
are the coordinates of exp(ix), but to *make sense* geometrically
you'd need an intuition as to why exp(ix) should be a circle for real x.

-- Richard

Do matrices look geometric enough for you?

[ 1 0] [ 0 1]
Define I = [ ] and J = [ ]. It can be proven that
[ 0 1] [-1 0]

aI + bJ is isomorphic to the complex number a + bi. It can
further be proven that

[ 0 x] [ cos x sin x]
exp [ ] = [ ]
[-x 0] [-sin x cos x]

which corresponds to exp (ix) = cos x + i sin x.

Tak-Shing

 

[¬a\\/b]

don't know if it is right:
i=0+i*1=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) it seems ok

i^i=(e^(i*pi/2))^i=e^(-pi/2)=0,20787957635076190854695561
i have seen that number in this thread...

|#include <stdio.h>
|#include <complex.h>
|int main(void)
|{
| const long double complex z1 = cpowl(I, I);
| printf("%Lf %Lf\n", creall(z1), cimagl(z1));
| return 0;
|}

|and its output:

|0.207880 0.000000


are you sure of that?
if the answer is yes i^i has more than 1 result

 

[¬a\\/b]

my book says if a, beR
a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eRx[0,2*pi)

if i have a+i*b than find
r=sqrt(a^2+b^2)
if r=0 than a+i*b=0*e^0
else resolve
cos(t)=a/r sin(t)=b/r te[0, 2*pi)
(note 1=a^2/r^2+b^2/r^2=cos(t)^2+sin(t)^2)
and write a+i*b=r*e^(i*t)

if i have r*e^(i*t)=r(cos(t)+i*sin(t))=
r*cos(t)+i*r*sin(t)=a+i*b

don't know if it is right:
i=0+i*1=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) it seems ok
i^i=(e^(i*pi/2))^i=e^(-pi/2)=0,20787957635076190854695561

it seems right
"e^(i*5*pi/2) = i" seems to me wrong because
not(5*pi/2e[0,2pi))

 

[¬a\\/b]

my book says if a, beR

my book don't say really this but seems right in this way

a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eRx[0,2*pi)

a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eR^(+)x(-pi,+pi]

if i have a+i*b than find
r=sqrt(a^2+b^2)
if r=0 than a+i*b=0*e^0
else resolve
cos(t)=a/r sin(t)=b/r te[0, 2*pi)

cos(t)=a/r sin(t)=b/r for te(-pi,+pi]

(note 1=a^2/r^2+b^2/r^2=cos(t)^2+sin(t)^2)
and write a+i*b=r*e^(i*t)

if i have r*e^(i*t)=r(cos(t)+i*sin(t))=
r*cos(t)+i*r*sin(t)=a+i*b

it seems right
"e^(i*5*pi/2) = i" seems to me wrong because
not(5*pi/2e[0,2pi))

not(5*pi/2e(-pi,pi])