Are "pi" and "e" related?

[DrJones]

Maybe this is not the best place to post it, but I didn't know where to do it.

I was playing with my calculator, when I discovered that:
PI * Cos 30 +-= e

Do you know any better relation between "Pi" and "e"?
Is this an important thing, or just a curiosity?

GoodBye!

[ralphmerridew]

The best relation between pi and e is almost certainly

e^(i pi) == -1

[groza528]

hehe, that's what I said in OT. It may get confusing to have to check two forums to read our responses though.

[ZutAlors!]

And, incidentally, PI*cos(3) != e

PI*cos(30) = 2.720699046
e = 2.718281828

Close (within 99.9%), but with enough judicious playing with available constants, you can approximate nearly anything.

[dave10000]

1000*[sin(Pi*1/49)*sin(Pi*2/47)]/Pi = e

(to .000104% accuracy!)

Find this and more at The Inverse Symbolic Calculator:

http://www.cecm.sfu.ca/projects/ISC/ISCmain.html

[Chuck]

12/5 + 1/pi = e

approximately.

[GH]

e/pi * pi/e = 1

*exactly*

[ralphmerridew]

I think that e^(pi sqrt(163)) is very close to an integer (the difference is < 10^-12)

[Griffin]

.0009 * pi^7 ~ e

[HyToFry]

Griffen lives?

[HappyMutant]

Explain why e^(i*pi) = 1.

(It's fun. I did it in calculus when no one was looking...)

While you're at it, solve for i^(i).

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Brunch - you'll love it. It's not quite breakfast, it's not quite lunch; but it comes with a slice of cantaloupe at the end.

[ralphmerridew]

That should be -1 in your post.

[tigg]

I think i^i is ambiguous.
for any integer k,
e^(-(4k+1)pi/2) = i^i.

[mole]

i ^ i = (e^(i.pi/2))^i
= e^i(i.pi/2)
= e^-pi/2

[groza528]

I proved last year that e^(pi*i)=-1. The easiest way to show it is to work backwards.

Upon learning about radians and polar notation I happened to noticed that cis(pi)=-1 (where cis(x)= cos(x)+i*sin(x)) and wondered if there was a connection. There was. I started with the proposition that perhaps e^x=cis(x/i) and my calculator confirmed this. So I found the algebra behind it.
e^x=cis(x/i)
original equation. x/i=-xi, combining that step with expansion yields
e^x=cos(-xi)+i*sin(-xi)
I used basic knowledge about sin and cos to get
e^x=cos(xi)-i*sin(xi)
After that I eliminated the i with hyperbolic trig.
e^x=cosh(x)+sinh(x)
Using the definitions of these hyperbolic trig functions
e^x=(e^x+e^(-x))/2+(e^x-e^(-x))/2
Then simplify
e^x=e^x
so e^(pi*i)=cis(pi)=-1

[This message has been edited by groza528 (edited 03-01-2002 06:51 AM).]

[ralphmerridew]

groza, your post uses circular reasoning, since the relations between cosh(x) and cos(ix) are proven using e^(ix) == cis(x).

The usual proofs are generally either:

1) Use various methods to show

code:

---

e^x = sum (n=0 to oo, x^n/n!)


sin(x) = sum(n=0 to oo, (-1)^n x^(2n+1) /(2n+1)!)


cos(x) = sum(n=0 to oo, (-1)^n x^(2n)/(2n)!)

---

Then compare the formulas for e^ix and cis(x)

2) Using differential equations:

y'' + y == 0, y(0)=1, y'(0)=i, y twice differentiable everywhere has a unique solution. But both y(x)=e^(ix) and y(x)=cis(x) satisfy it. Therefore both functions are equal everywhere.

(Edit to fix mistake noted by tigg)

[This message has been edited by ralphmerridew (edited 02-28-2002 03:24 PM).]

[tigg]

for 2), I think you want y'(0) = i

[groza528]

Thanks ralphmerridew, I didn't know that. I guess I did use circular reasoning. I just used my calculator as a reference that cos(i)=cosh(1) etc.

[Jen Aside]

Pi and e... together, they make a tasty treat!

[Ghost Post]

didnt einstein discover something like this.
it was apparantly his faviriote equation, because it relates two mysteries of maths, or something like that!

[Sparticus]

ralphmerridew: you're a C++ programmer, aren't you?

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"'Tis better to be thought an idiot than to open one's mouth and erase all doubt...No, my mouth was not open."

[ralphmerridew]

I am a programmer, but I prefer Java & C; I greatly dislike C++. Why do you ask?

[groza528]

I think it was the ==. I don't know much about programming, but I think == is used in C++

[CzarJ]

Well, yes, but surely in other languages as well.

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Unslumping yourself is not easily done.

[ctrlaltdel]

how does pi relate to i???

they are cousins, everybody knows that...

[Chuck]

193 × e + 45 × pi = 666.

Well, it's close.

[tigg]

Chuck- you're a strange man..

[Lucky Wizard]

I noticed today (while fooling with my calculator) that this is approximately true:

code:

---

  pi       2


(e  )-pi=(5 )-5

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johnny, for the record, the e^(i*pi)=-1 thing was discovered by Euler, not Einstein.