e^(i*pi)

**sleeplessdragn** · 03-28-2005, 09:14 PM

In my math class a discussion arose concerning this interesting problem. The solution of e^(i*pi) is equal to -1, yet not even my math teacher could explain exactly why this was. All that he could say was that in order to prove this, a derivative of e and the fractional value of pi was involved. I know that it involves a certain trig identity but not much more than that.

After searching through google I found not much more information, although I admit that i did not really google too deeply. I found, though, that a correlation between e and pi has to exist.

Does anyone know of any correlation between e and pi? Are there any practical uses for this equation? Discuss.

GuidoHunter · 03-28-2005, 10:00 PM

It's been said that that equation is the most magical equation in all of math because it incorporates several different types of math and relates them all. That is, the transcendental, the algebraic, and the imaginary.

I don't know much about it myself, and if you already googled it, I can't really furnish much more.

--Guido

http://andy.mikee385.com

Kefit · 03-28-2005, 10:18 PM

This equation is incredibly important to physics and differential equations. Lemme see if I can remember the proof:

First off, this proof uses infinite series. Put basically, any differentiable function can be represented by the sum of an infinite series. No, I don't expect you to know what that means. Just accept that:

cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + . . .

sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + . . .

e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + . . .

It follows that:

e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

i*sin(x) = (i*x) - (i*x^3)/(3!) + (i*x^5)/(5!) - (i*x^7)/(7!) + . . .

Now something interesting happens when you add cos(x) and i*sin(x) together. Lemme see if I can illustrate this clearly:

.....cos(x) = 1..........- (x^2)/(2!) ......................+ (x^4)/(4!) .................- (x^6)/(6!) + . . .
+ i*sin(x) = ..(i*x)..................... - (i*x^3)/(3!) .................+ (i*x^5)/(5!) - . . .
---------------------------------------------------------------------------------------------------------------
....e^(i*x) = 1 + (i*x) - (x^2)/(2!) - (i*x^3)/(3!) + (x^4)/(4!) + (i*x^5)/(5!) - (x^6)/(6!) + . . .

Don't be discouraged if the idea of infinite series is unfamiliar to you - just take the series I gave for sin(x), cos(x), and e^x for granted, and everything else I show above follows from simple arithmetic on the series.

Anyway, this creates the general equation:

e^(i*x) = cos(x) + i*sin(x)

if x = PI, then:

e^(i*PI) = cos(PI) + i*sin(PI) = -1 + i*0 = -1

Oh, and shame on your math teacher for not knowing this proof

**Cenright** · 03-28-2005, 10:57 PM

(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.)
I know I am looking at this from outside the equation, but still...

GuidoHunter · 03-28-2005, 11:36 PM

Uh, what? Where does zero come into this and when is it illegal to multiply by zero?

--Guido

http://andy.mikee385.com

Kefit · 03-28-2005, 11:41 PM

Well, some people use zero to set up this identity as:

e^(i*PI) + 1 = 0

And then they sometimes use that to try and prove the existence of God due to the existence of such a streamlined relation between what are more or less the five most important constants in all of math.

Of course, that is not what Cenright was talking about; rather this is the real way that zero gets involved with this equation.

Anticrombie0909 · 03-29-2005, 11:13 AM

*head explodes*

Omeganitros · 03-29-2005, 11:20 AM

Actually, I think I got a grasp on that.

However, I can see that me and Infinity will be having a frustrating relationship in the future.

**sleeplessdragn** · 03-29-2005, 07:47 PM

I heard about the mathematical proof of the existence of God, but i had NO clue that this equation was the root of that proof. That simply astounds me. I actually did a one eye brow raise when i read that.

**Kilgamayan** · 03-29-2005, 08:25 PM

Jesus, I couldn't have produced that and I'm a math major.

I'll have to remember that.

**Tasselfoot** · 03-29-2005, 08:34 PM

All I remember about that proof is that my fat ass math god buddy used to geek around talking about e^(i*Pi) instead of -1 any time -1 came up, or some such multiplier of that.

Needless to say, yes... he's a math god. and yes... he's a virgin. and yes... he weighs 300lbs.

And, with that, I go back to hating math math and back to loving finance math.

blahblah18 · 03-29-2005, 09:18 PM

OK ITS REAL SIMPLE...
kefit has it right...
they're called taylor series... anyone? anyone? Bueller? calculus?

GuidoHunter · 03-29-2005, 10:41 PM

Pfft, they're nothing but MacLaruin serieseses centered about zero.

--Guido

http://andy.mikee385.com

**igotrhythm** · 04-4-2005, 09:25 AM

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

banditcom · 04-4-2005, 12:50 PM

Quote:

Originally Posted by igotrhythm

Here's the proof I saw once for proving the existence of God. At least, I think this is how it goes.

0=0+0+0+0... and this stretches on to infinity. This can be rearranged to

0=(1-1)+(1-1)+(1-1)...

0=1-1+1-1+1-1...

=1+0+0+0+0...=1

Thus, something has been created out of nothing. However, this is not legitimate because the infinite series is divergent. That is, the series of partial sums (1, 0, 1, 0...) do not approach any specific number. If it did, this limit is defined as the sum of the infinite series.

lol... That's for stupid people who fail to realize that the -1 is still in there to counter the 1.

Anyways, nice proof Kefit. I totally forgot about the different series, even though they're really cool. Currently, I'm so sick of proofs. Most of them I feel like "this doesn't seem to prove anything" but they do.

Braydz · 04-7-2005, 07:34 AM

I actually understood all of that...

Is all of this uni/college level maths? I'm still in high school... Also, curious... Do those equations of sine and cosine work with both radians and degrees, or only one? Never seen sine and cosine shown that way before.

GuidoHunter · 04-8-2005, 01:46 AM

All calculus is done in radians, and yes, this is Cal II stuff.

--Guido

http://andy.mikee385.com

hatakikakashi · 04-8-2005, 03:19 AM

Quote:

Originally Posted by Cenright

(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.)
I know I am looking at this from outside the equation, but still...

I always thought that the rule was that you can't divide by zero. 50 multpied by zero equals (zero) 50 divided by zero equals (your not allowed to do that)

I never understood why you can't do that. Also all this math is beyond me so I'll shut up now.

GuidoHunter · 04-8-2005, 03:27 AM

Quote:

Originally Posted by hatakikakashi

I always thought that the rule was that you can't divide by zero. 50 multpied by zero equals (zero) 50 divided by zero equals (your not allowed to do that)

I never understood why you can't do that. Also all this math is beyond me so I'll shut up now.

See my post after that.

If you want to understand why you can't divide by zero, just look at limits. Since you can always make a number smaller and smaller, you'll never end, which is why there is no answer to dividing by zero.

--Guido

http://andy.mikee385.com

Braydz · 04-8-2005, 04:10 AM

When I think of dividing by zero, I just think of the unit circle. tan(pi/2) is undefined as tan(x) = sin(x)/cos(x), which would result in tan(pi/2) = 1/0. Now, as far as I was taught, the tan function is basically the length of the hypotenuse from the origin to a tangent running vertically from the point (1,0). If the hypotenuse is going up vertically from (0,0), they will never meet, or somehow meet at infinity. That is why a graph of tan(x) has asymptotets at pi/2 adding or subtracting any normal number of pi. The graph flies upwards, approaching infinite. And when you think of it, there is no number of 0's that can go into any number. Is infinite a real number?

Sorry if I didn't make any sense, I am trying to remember something my specialist maths teacher told me at the end of last year. The fact that I am also a "young" maths student doesn't help, along with the fact that I have trouble explaining myself. If anyone can please correct my foggy memory here, please do.

03-28-2005, 09:14 PM	#1
sleeplessdragn ~Bang that beat Harder~ Join Date: Jan 2004 Posts: 2,321	*e^(ipi)** In my math class a discussion arose concerning this interesting problem. The solution of e^(i*pi) is equal to -1, yet not even my math teacher could explain exactly why this was. All that he could say was that in order to prove this, a derivative of e and the fractional value of pi was involved. I know that it involves a certain trig identity but not much more than that. After searching through google I found not much more information, although I admit that i did not really google too deeply. I found, though, that a correlation between e and pi has to exist. Does anyone know of any correlation between e and pi? Are there any practical uses for this equation? Discuss. __________________

03-28-2005, 10:57 PM	#4
Cenright You thought I was a GUY?! Join Date: Sep 2003 Location: Beaumont (A town with 25 times fewer people than this site) Age: 38 Posts: 3,139	(It seems as you are just finding a way to rid yourself of the 'i', thus attempting to find a way to legally multiply by zero.) I know I am looking at this from outside the equation, but still... __________________ http://www.flashflashrevolution.com/...Cube_in_55.mpg

03-29-2005, 07:47 PM	#9
sleeplessdragn ~Bang that beat Harder~ Join Date: Jan 2004 Posts: 2,321	I heard about the mathematical proof of the existence of God, but i had NO clue that this equation was the root of that proof. That simply astounds me. I actually did a one eye brow raise when i read that. __________________

03-29-2005, 08:25 PM	#10
Kilgamayan Super Scooter Happy Join Date: Feb 2003 Location: Location, Location. Age: 41 Posts: 6,583	Jesus, I couldn't have produced that and I'm a math major. I'll have to remember that. __________________ I watched clouds awobbly from the floor o' that kayak. Souls cross ages like clouds cross skies, an' tho' a cloud's shape nor hue nor size don't stay the same, it's still a cloud an' so is a soul. Who can say where the cloud's blowed from or who the soul'll be 'morrow? Only Sonmi the east an' the west an' the compass an' the atlas, yay, only the atlas o' clouds.

03-29-2005, 08:34 PM	#11
Tasselfoot Retired BOSS Join Date: Jul 2003 Location: Widget Heaven Age: 42 Posts: 25,185	All I remember about that proof is that my fat ass math god buddy used to geek around talking about e^(i*Pi) instead of -1 any time -1 came up, or some such multiplier of that. Needless to say, yes... he's a math god. and yes... he's a virgin. and yes... he weighs 300lbs. And, with that, I go back to hating math math and back to loving finance math. __________________ RIP

03-28-2005, 10:18 PM	#3
Kefit FFR Player Join Date: Apr 2003 Posts: 1,517	This equation is incredibly important to physics and differential equations. Lemme see if I can remember the proof: First off, this proof uses infinite series. Put basically, any differentiable function can be represented by the sum of an infinite series. No, I don't expect you to know what that means. Just accept that: cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + . . . sin(x) = x - (x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + . . . e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + . . . It follows that: e^(ix) = 1 + (ix) - (x^2)/(2!) - (ix^3)/(3!) + (x^4)/(4!) + (ix^5)/(5!) - (x^6)/(6!) + . . . isin(x) = (ix) - (ix^3)/(3!) + (ix^5)/(5!) - (ix^7)/(7!) + . . . Now something interesting happens when you add cos(x) and isin(x) together. Lemme see if I can illustrate this clearly: .....cos(x) = 1..........- (x^2)/(2!) ......................+ (x^4)/(4!) .................- (x^6)/(6!) + . . . + isin(x) = ..(ix)..................... - (ix^3)/(3!) .................+ (ix^5)/(5!) - . . . --------------------------------------------------------------------------------------------------------------- ....e^(ix) = 1 + (ix) - (x^2)/(2!) - (ix^3)/(3!) + (x^4)/(4!) + (ix^5)/(5!) - (x^6)/(6!) + . . . Don't be discouraged if the idea of infinite series is unfamiliar to you - just take the series I gave for sin(x), cos(x), and e^x for granted, and everything else I show above follows from simple arithmetic on the series. Anyway, this creates the general equation: e^(ix) = cos(x) + isin(x) if x = PI, then: e^(iPI) = cos(PI) + isin(PI) = -1 + i*0 = -1 Oh, and shame on your math teacher for not knowing this proof

03-28-2005, 11:41 PM	#6
Kefit FFR Player Join Date: Apr 2003 Posts: 1,517	Well, some people use zero to set up this identity as: e^(i*PI) + 1 = 0 And then they sometimes use that to try and prove the existence of God due to the existence of such a streamlined relation between what are more or less the five most important constants in all of math. Of course, that is not what Cenright was talking about; rather this is the real way that zero gets involved with this equation.

03-29-2005, 11:13 AM	#7
Anticrombie0909 FFR Player Join Date: Jul 2003 Posts: 4,683	head explodes

03-29-2005, 11:20 AM	#8
Omeganitros auauauau Join Date: Jun 2003 Location: Hee-Haw! Age: 37 Posts: 8,897	Actually, I think I got a grasp on that. However, I can see that me and Infinity will be having a frustrating relationship in the future.

03-29-2005, 09:18 PM	#12
blahblah18 FFR Player Join Date: Aug 2004 Location: NOW BLAHTOPIA Posts: 1,662	OK ITS REAL SIMPLE... kefit has it right... they're called taylor series... anyone? anyone? Bueller? calculus? __________________ but for now... postCount++

04-7-2005, 07:34 AM	#16
Braydz FFR Player Join Date: Nov 2004 Location: Australia Posts: 92	I actually understood all of that... Is all of this uni/college level maths? I'm still in high school... Also, curious... Do those equations of sine and cosine work with both radians and degrees, or only one? Never seen sine and cosine shown that way before.

04-8-2005, 04:10 AM	#20
Braydz FFR Player Join Date: Nov 2004 Location: Australia Posts: 92	When I think of dividing by zero, I just think of the unit circle. tan(pi/2) is undefined as tan(x) = sin(x)/cos(x), which would result in tan(pi/2) = 1/0. Now, as far as I was taught, the tan function is basically the length of the hypotenuse from the origin to a tangent running vertically from the point (1,0). If the hypotenuse is going up vertically from (0,0), they will never meet, or somehow meet at infinity. That is why a graph of tan(x) has asymptotets at pi/2 adding or subtracting any normal number of pi. The graph flies upwards, approaching infinite. And when you think of it, there is no number of 0's that can go into any number. Is infinite a real number? Sorry if I didn't make any sense, I am trying to remember something my specialist maths teacher told me at the end of last year. The fact that I am also a "young" maths student doesn't help, along with the fact that I have trouble explaining myself. If anyone can please correct my foggy memory here, please do.

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