The most interesting math problem I've seen in a long time

[stan394]

I have already given up looking it at the math way. I am looking at this as a trick question...

 

[slpaulson]

Originally posted by: JohnCU
I got bored in electromagnetics today so I came up with this.

Let z = x + jy, where j = sqrt(-1), a complex number

e^z = 0
e^(x+jy) = 0
(e^x)(e^jy) = 0
e^x(cosy + jsiny) = 0

Let's assume that, for a real value, e^x is never 0, so the only way this equation could be 0 is if cosy + jsiny = 0

=> cosy + jsiny = 0
cosy = -jsiny
1/(-j) = siny/cosy
j = tan(y)

arctan(j) = y

So, if z = j*arctan(j), then you get 0.

arctan(j) is infinity*j according to my calculator, so if you go by that the answer turns out to be -infinity, which the op doesn't want. I don't know if that's a convention or what. Of course I would think arctan(j) = 1/0...

 

[MrDudeMan]

Originally posted by: RaynorWolfcastle
The real question is:
does e^(0.99999...) = e?

ohhhh teh noeessss!!

 

[TheRyuu]

I bet the OP just wanted a 100 post thread.

I have yet to get one either

 

[Leper Messiah]

my limited math skillz say e^z=0 can be simplified to: zth root of e= zth root of 0? 0 divided by anything=0, so thats the answer? Hell if I know though.

 

[IHateMyJob2004]

Originally posted by: OdiN
42

ROFL ... VERY GOOD SIR!

 

[IHateMyJob2004]

WQEll, looks like a dead thread, but wouldn't this involve taking the dirivite of e^X till it's a real number? That is, till X basically fizzles out of the equation, should it ever actually happen?

 

[Soccer55]

Originally posted by: IHateMyJob2004
WQEll, looks like a dead thread, but wouldn't this involve taking the dirivite of e^X till it's a real number? That is, till X basically fizzles out of the equation, should it ever actually happen?

d/dx[e^x] = e^x

You'll never be able to get rid of it

The thread only seems dead because Eeezee hasn't posted his alleged solution

-Tom

 

[MrDudeMan]

Originally posted by: IHateMyJob2004
WQEll, looks like a dead thread, but wouldn't this involve taking the dirivite of e^X till it's a real number? That is, till X basically fizzles out of the equation, should it ever actually happen?

you can never change the power of e with derivatives to my knowledge.

 

[Eeezee]

Originally posted by: wizboy11

Originally posted by: mariok2006

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: DVK916
There is no solution. His plan worked. He wanted to get people to spend time trying to solve a problem that has no solution.

I vote for a ban

seconded

thirded

quadforthed

what ever the hell is for fifth (fifthed?)

Edit: Now show us the dam solution if you have one! :|

You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

 

[JohnCU]

Originally posted by: Eeezee

Originally posted by: wizboy11

Originally posted by: mariok2006

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: DVK916
There is no solution. His plan worked. He wanted to get people to spend time trying to solve a problem that has no solution.

I vote for a ban

seconded

thirded

quadforthed

what ever the hell is for fifth (fifthed?)

Edit: Now show us the dam solution if you have one! :|

You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

It's not interesting because it's very easy to see that there is no solution.

 

[Soccer55]

Originally posted by: Eeezee

Originally posted by: wizboy11

Originally posted by: mariok2006

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: stan394

Originally posted by: DVK916
There is no solution. His plan worked. He wanted to get people to spend time trying to solve a problem that has no solution.

I vote for a ban

seconded

thirded

quadforthed

what ever the hell is for fifth (fifthed?)

Edit: Now show us the dam solution if you have one! :|

You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

It's been 7 pages of talking about how there is no solution despite your claim that there is one. Prove us wrong!

-Tom

 

[TheRyuu]

7 fvck'n pages is long enough.

Just give us the dam solution :|

 

[MrDudeMan]

Originally posted by: Eeezee

You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

seriously, everyone is losing interest. just give us your 'solution' please.

 

[LordMorpheus]

e raised to a complex number can be written:

Euler's gives e^(j*Theta) = cos(theta) + j*sin(theta)

You can't find a theta to make that equal to zero.

you can say that e^j*theta + e^-j*theta = 0 when theta = pi/2, 3*pi/2, etc.

you can get the real or imaginary parts of that to be zero, but the abs(e^j*theta) will always be greater than zero. sorry.

 

[tikwanleap]

Originally posted by: LordMorpheus
e raised to a complex number can be written:

Euler's gives e^(j*Theta) = cos(theta) + j*sin(theta)

You can't find a theta to make that equal to zero.

you can say that e^j*theta + e^-j*theta = 0 when theta = pi/2, 3*pi/2, etc.

you can get the real or imaginary parts of that to be zero, but the abs(e^j*theta) will always be greater than zero. sorry.

You are correct

e^z is never ever equal to zero...

 

[stan394]

Originally posted by: Eeezee


You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

what the hell? you showed up, post some junk and still just leave us with the spider?!

 

[MrDudeMan]

Originally posted by: stan394

Originally posted by: Eeezee


You want to ban me because I got you thinking about an interesting math problem?

I said it was an interesting problem, and apparently 7 pages of posts is enough to prove that!

what the hell? you showed up, post some junk and still just leave us with the spider?!

seriously wtf.

this isnt even interesting anymore. now it is just annoying.

 

[dornick]

I think it's pretty obvious the OP just wanted to cause a ruckus, knowing full well there would be plenty of people here willing to dedicate plenty of time to the problem. Let the thread die. There is no solution.

 

[effowe]

-2304 is the answer.

 

[piasabird]

i^4=1
i^8=1
i^12=1

Any number divisible by 4 is a sloution if e=i, or does e in this case stand for something else?

i= square root of -1 and thus is an unreal number.

 

[ReliableData]

um.....e stands for e.
And looking at the graph of e^x, no real number solution exists, and i'm not going to try to do complex number stuff, but just looking at it, i would wager that no true solution exists in the complex realm.

Another vote for ban.

 

[eits]

Originally posted by: So

Originally posted by: ShadoWing
um.....z=1?

e^1 = e.

haha someone failed math

 

[eits]

Originally posted by: coomar
never fails to amuse

it's not rearranged into a triangle. it's rearranged into a heptagon.

 

[eits]

the answer is "lim z -> 0"