Survey posted 2003-06-06 23:32 by Henrik.

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after looking at the 'cool graphs' section on technoplaza...

are you sure? i couldn't find the EXACT graph anywhere at technoplaza.net

If there are some of you who don't know what they're talking about, click the link (if the filter didn't kill it), or click here, just paste the sections of the link together.
http://www.technoplaza.net/
graphs/index.cgi?
displaygraphs=1&howmany=15
&type=5&ppequ=1&ss=1

There is one that looks like a variation of the one at the ticalc.org logo, so that could be played around with...

I was talking about the "A Raindrop in Water". It turned out to be a little different, which I didn't realize until now, but I would guess it can be modified, although I haven't tried it myself.

It does look very similar... maybe it uses the same... idea to get the ripple effect, only... different.

Earlier, I was thinking of (x^2+y^2)^.5 and sin/cos, so I had the right idea, I guess...

no. have a nice day.

Yeah.

It said how to somewhere on here at some time...

I'm working on updating QuadForm in a way that could blow some peoples minds...

The graph looks like it is symetric around the z-axis: The further you go out from the z-axis the smaller it gets, but it is still wavy, so I think it is an exponentially decreasing sinusoid (or cosinusoid)

To see this in 2d graph y(x) = A*e^(-B*x) * cos(C*x)
Pick reasonable values for the Constants.

but instead of x we have distance from the z axis, so replace the xs in the above 2d formula with the distance formula to get:

z(x,y) = A*e^(-B*sqrt(x^2+y^2))* cos(C*sqrt(x^2+y^2))

And if you think about it some more the square root operators would not be needed to get radial symetry.


Any other theories?

this may sound stupid but whats the e?

2.78.... or something vaguely similar.
It is a value which engineering uses a fair bit.

e = the sum of 1/n! where n starts at 0 and continues on to infinity in incriments of 1. Also, you can solve for e from this equation:

e^(imaginary pi) = -1

I'd solve for e, but I'm too lazy :P

Very interesting, learn something new every day.

Something I am trying to figure out is that a couple weeks ago I was walking through some local 'park land' to get to uni, and someone had randomly written
e^i(pi)=-1 (obviously with the right symbols)
on the path in chalk.

Why? Was it his idea of revision?

Heh, that's funny, today outside of church, on the bench, I wrote "3.1415926535..." I would have done all that I knew (it's now over 150), but it takes a long time to make precise little marks on wood. It will probably become unoticable in a few days anyway.

So, why does e^(i*PI)=-1?
Hm...
ln(e^(i*PI))=ln(-1)
i*PI=ln(-1)
i*PI=ln(1)+i*PI
i*PI=i*PI

Oh.... It all makes sense now! I never got that before... :) Wow...

Um I think your proof may be circular?

Your simplification of ln(-1) into ln(1)+i*pi is probably based on the origional identiy you are trying to prove.

The proof I was tought for eulers identity is based on taylor expansions.

Nah, I've just always seen on my calc (in rectangular mode) that ln(x)=ln(-x)+pi*i (if x is less than zero)... I could be wrong, but that was my observation. Then, I remembered that, and it came up in the formula... I don't see WHY you'd add pi*i, but oh well :)

Here is the precise formula:

Ln(complex_number) = Ln(abs(complex_number)) + i*angle(complex_number)

The first part: Ln(abs(complex_number)) is real valued.
The second part: i*angle(complex_number) is imaginary


I would not consider using this advanced formula as proof of the euler formula to be a valid proof.

Well, then I had the right idea... I just tested numbers like

ln(-2) ln(-8.9) ln(-9) ...

I think I might try to remember that...

Wait never mind. I remembered the proof for ln(complex) = ln(abs(complex)) + i*angle(complex)

It is simple, convert the imaginary number to polar form: r*e^(i*angle) I hope you know what the polar form is.

Apply the ln function to this to get

ln(r*e^(i*angle))

Use a ln() identity:

ln(r) + ln(e^(i*angle))

And simplify the second ln()

ln(r) + i*angle

where r is abs(imaginary number) which is sqrt(real^2+imag^2)

DARN, you beat me to saying it! LOL

You could also put in

(1+1/x)^x in for Y1 and at the homescreen type

Y1(999999999)

and you'll get an approximation of e... of course, you could also just hit the e (or e^) key on your calculator ;-)

To the best of my knowledge, it goes like 2.71828182846... (is that right?)

Yup, that's what my TI-89 told me.

That's all I knew :) LOL

Its the natural log.

No it isn't, it's the base of the natural log.

Also known as 2.718281828459...

Even more techinically, its Leonhard Euler's number, hence e.

VP

Or the natural log of 1.

log |1| = 0
regardless of base.

it just so happens to be log |15.154....|
(but that is just log |e^e| so it is quite meaningless).

yep, that sounded stupid. ^_^