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

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coomar

Banned
Apr 4, 2005
2,431
0
0
the limit as z goes to zero is defined as 1
i think you were thinking about limit as z goes to negative infinity
 

MrDudeMan

Lifer
Jan 15, 2001
15,069
92
91
Originally posted by: musicman87
anyone got a ti-92 platinum on them? (solve e^z=1,z)

lol ti-92...

and dont you think at least 1 of the several EEs in this thread have tried methods like that?
 

Soccer55

Golden Member
Jul 9, 2000
1,660
4
81
Originally posted by: MBrown
has someone firgured it out yet?

Many of us agree that there is no solution to this problem. So yes, we've figured it out, but we would like to see Eeezee's solution since he claims that a solution exists.

-Tom
 

JASANITY

Senior member
Dec 10, 2000
504
0
0
Originally posted by: dullard
Originally posted by: hypn0tik
Heh. Nice. The slopes of the two triangular pieces aren't the same. The slope of the red one is 3/8 while the slope of the green one is 2/5.
Right, the big combination is NOT a triangle. Overlap the two pictures and you'll instantly see where the "hole" material went.

that is sweet. a slight difference in the slope = square hole.
 

letdown427

Golden Member
Jan 3, 2006
1,594
1
0
The only way it could work is if e doesn't equal the function we're all inclined to think of, but that he has just decided that e is a variable f his own choosing.

people have already explained why e^z != 0 and they're correct.

for e^z = 0 , e = 0 while z !=0

why z != 0? because, in another interesting maths thing,0, and everything else ^0, = 1. yes, 0^0 = 1
 

coomar

Banned
Apr 4, 2005
2,431
0
0
Originally posted by: JustAnAverageGuy
Proof

NO SOLUTION, it's an asymptote

thats on the real axis, the solution people were proposing as possible would have been on the imaginary axis

there should be a way to prove its not possible by contradiction
 
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