As noted above, there's no such value. This is true for any nonzero base. We know from elsewhere that 0 = xy is only possible if x = 0 or y = 0. Well here you're saying 0 = x^k = x*x*..., essentially. So it follows that this is only possible if x=0. So e, and anything else not zero, doesn't work.

I just wanted to respond to you explaining my qualification below. x^k = x*x... (k-times) only when k is a positive integer. There isn't a natural expansion for other values, and exponentiation for all numbers is described by the exponential function. Since I'm no king whiz, I don't know how that function will behave using, say, a complex irrational.

Yes, I was just giving an example to bring out an intuition that should exist from previous experience in math.

There is no value k that will do so, real or complex. The limit of e^x as x approaches negative infinity is 0. That's the best you can do, since negative infinity isn't really a "value" per se, it only has meaning in the limit statement.

To refute the complex number possibility specifically:

Say complex w = xi + y for real x and y. e^w = e^(xi+y) = e^(xi) * e^y We know e^y cannot be 0, thus if e^w =0 it must be e^(xi) that is equal to 0.

But by Euler's formula: e^(xi) = i * sin(x) + cos(x) 0 = i * sin(x) + cos(x) => i * sin(x) = -cos(x) But neither sin(x) nor cos(x) may have an imaginary value for real x, thus one side of the equation is imaginary and the other is not. Therefore the equation has no solution.

Thus if e^x = 0 lacks a solution for real x, it also lacks one for complex x.

I'm not, it's just easier to relate to simple mathematical ideas about zero that way. I said "essentially" because that's not actually what exponentiation means, but getting into more wouldn't be particularly fruitful.

What exponent? Maybe I've spent too long in Babylon, but this is like asking "how many times must I multiply e against itself to equal 0", and the answer is "there's no answer", because the only number multiplied against itself any number of times, and yield zero, is zero.

Unless there's an imaginary number out there that suffices.

alsterellieMay 20 2009, 22:37:21 UTC 7 years ago

nekokazeMay 20 2009, 22:43:02 UTC 7 years ago

roadriverrailMay 20 2009, 22:49:16 UTC 7 years ago

nekokazeMay 20 2009, 23:04:09 UTC 7 years ago

There is no value k that will do so, real or complex. The limit of e^x as x approaches negative infinity is 0. That's the best you can do, since negative infinity isn't really a "value" per se, it only has meaning in the limit statement.

nekokazeMay 20 2009, 23:11:52 UTC 7 years ago

Say complex w = xi + y for real x and y.

e^w = e^(xi+y) = e^(xi) * e^y

We know e^y cannot be 0, thus if e^w =0 it must be e^(xi) that is equal to 0.

But by Euler's formula:

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

0 = i * sin(x) + cos(x) => i * sin(x) = -cos(x)

But neither sin(x) nor cos(x) may have an imaginary value for real x, thus one side of the equation is imaginary and the other is not. Therefore the equation has no solution.

Thus if e^x = 0 lacks a solution for real x, it also lacks one for complex x.

roadriverrailMay 20 2009, 23:19:52 UTC 7 years ago

This sort of thing doesn't come up often in, say, operating systems development, so I haven't touched these topics since college.

korean_guy_01May 20 2009, 22:53:06 UTC 7 years ago

nekokazeMay 20 2009, 23:05:24 UTC 7 years ago

roadriverrailMay 20 2009, 22:45:44 UTC 7 years ago

Unless there's an imaginary number out there that suffices.

makingthematrixMay 21 2009, 06:39:30 UTC 7 years ago

cr4kMay 21 2009, 18:38:04 UTC 7 years ago

triprcMay 21 2009, 19:55:54 UTC 7 years ago

cr4kMay 21 2009, 21:28:24 UTC 7 years ago

greedyalgorithmMay 21 2009, 20:52:36 UTC 7 years ago