Thread: Here's some fun

I'm in playfull mood, it being Monday night (huh?)

Anyway. I assert that there are five magical numbers. Here they are:

0, 1, e, i and pi

Test - how are they related?

Remember you may be asked to show your proof!

PS For goodness sake, somebody tell me how to HTML extensions to ASCII characters working!

Are you talking about stuff like this?

ƒ=(i²Π)

Originally Posted by (In)Sanity

Are you talking about stuff like this?

Æ’=(i²π)

Yes, exactly that. According to my HTML tables, your π should show as, well, what we expect "pi" to look like.

Sorry to drum it in, but it worked at *shhh* that other place!

Trying to find out why, but did get

Æ’=(i²Π)

Code:

Æ’=(i<sup>2</sup>Π)

Ohh and

♣ ♠ ♥ ♦

I don't see pi in the chart...
http://www.primeshop.com/html/jump3b.htm

But, assuming that it is and I'm just missing it, you have to use the number rather than the alternate designation.

For instance:
&# 181; (minus the space, of course.)
µ

µ
µ

I posted the question on the phpbb site to see if anyone has an answer why π doesn't work. It should IMO. If they don't have an answer I'll hack the code to make it work.

Originally Posted by invert_nexus

For instance:
&# 181; (minus the space, of course.)
µ

Thanks for that, but life's too short for that sort of sophistication.

Is it my brower's fault, perhaps (it's a standard MS issue - how I hate them!)

Originally Posted by Guitarist

Originally Posted by invert_nexus

For instance:
&# 181; (minus the space, of course.)
µ

Thanks for that, but life's too short for that sort of sophistication.

Is it my brower's fault, perhaps (it's a standard MS issue - how I hate them!)

I'll be sure π works. Just need a little time.

Originally Posted by (In)Sanity

I'll be sure π works. Just need a little time.

Sorry, boss. Wasn't paying attention. Yes, in your last pi showed as it should. But how? If I type π, what do you see? Is there some other button I need to press? I notice you have some " code" stuff at the bottom. What's that about?

Sorry to be dim about this, but the less I know about computing, the better I feel!

Originally Posted by Guitarist

0, 1, e, i and pi

Test - how are they related?

e^(i*pi)-1=0

Yes, in your last pi showed as it should. But how?

I suspect he used the character map.

Originally Posted by invert_nexus

Yes, in your last pi showed as it should. But how?

I suspect he used the character map.

Yah I did, #920 or something close to that. I have the phpbb guys looking at why π doesn't work. They can't get it working either, so I'm sure that will drive them nuts and I'll have my patch, and or other solution.

Originally Posted by quantumdude

Originally Posted by Guitarist

0, 1, e, i and pi

Test - how are they related?

e^(i*pi)-1=0

Almost. e^i*pi = -1. Remember?

Anyway, that equation really blows me away. Remember the proof? (via Euler?)

Originally Posted by Guitarist

Originally Posted by quantumdude

Originally Posted by Guitarist

0, 1, e, i and pi

Test - how are they related?

e^(i*pi)-1=0

Almost. e^i*pi = -1. Remember?

Duh, that should have been e^(i*pi)+1=0.

Anyway, that equation really blows me away. Remember the proof? (via Euler?)

I don't know exactly how Euler proved it, but I would do it with a power series for e^x, and then plug in ix for x.

Originally Posted by quantumdude

I don't know exactly how Euler proved it, but I would do it with a power series for e^x, and then plug in ix for x.

Hmm. I'm not sure Taylor would do it. Have to think about it.
I'm pretty sure your man used trig identities e.g. 2cos x = (e^ix + e^-ix).

Anyway, it was a bit of fun, to cheer up a Monday.


cheers.

Originally Posted by Guitarist

Hmm. I'm not sure Taylor would do it. Have to think about it.

It works alright, I make my Calculus II students do it.

I'm pretty sure your man used trig identities e.g. 2cos x = (e^ix + e^-ix).

The proof using power series is one way to derive that identity. If you do what I said and write down the expansion of e^x, and then put in ix in for x, then all the even powers will be real and all the odd powers will be imaginary. You will find that the sum of the real parts are precisely the power series for cos(x) and the sum of the imaginary parts are precisely the power series for sin(x).

Hence, e^(ix)=cos(x)+isin(x).

Originally Posted by quantumdude

It works alright, I make my Calculus II students do it.


Hence, e^(ix)=cos(x)+isin(x).

And don't they just love you for it?!

But yes, got there in the end.

Originally Posted by Guitarist

I'm in playfull mood, it being Monday night (huh?)

Anyway. I assert that there are five magical numbers. Here they are:

0, 1, e, i and pi

Test - how are they related?

Remember you may be asked to show your proof!

PS For goodness sake, somebody tell me how to HTML extensions to ASCII characters working!

did you get the "inspiration" for that question from a book? i've forgotten it but its abt this person who asked his student how are e, i ,pi, 0 and 1 related and the end of the poem is e^i pi + 1 = 0